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| author | Trupti Kini | 2016-06-03 23:30:26 +0600 |
|---|---|---|
| committer | Trupti Kini | 2016-06-03 23:30:26 +0600 |
| commit | ce730bd89b66da3cb054793a9717a53c7fa2c77d (patch) | |
| tree | 70b724f75d34c8d64672708cc599038b8c42b43e | |
| parent | 0a159e71d5ccd8d74b2258402fac51546a0ef6d0 (diff) | |
| download | Python-Textbook-Companions-ce730bd89b66da3cb054793a9717a53c7fa2c77d.tar.gz Python-Textbook-Companions-ce730bd89b66da3cb054793a9717a53c7fa2c77d.tar.bz2 Python-Textbook-Companions-ce730bd89b66da3cb054793a9717a53c7fa2c77d.zip | |
Added(A)/Deleted(D) following books
A Heat_Transfer_by_K._A._Gavhane/Chapter_2.ipynb
A Heat_Transfer_by_K._A._Gavhane/Chapter_3.ipynb
A Heat_Transfer_by_K._A._Gavhane/Chapter_4.ipynb
A Heat_Transfer_by_K._A._Gavhane/Chapter_5.ipynb
A Heat_Transfer_by_K._A._Gavhane/Chapter_6.ipynb
A Heat_Transfer_by_K._A._Gavhane/README.txt
A Heat_Transfer_by_K._A._Gavhane/screenshots/1.png
A Heat_Transfer_by_K._A._Gavhane/screenshots/2.png
A Heat_Transfer_by_K._A._Gavhane/screenshots/3.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9_1.ipynb
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/README.txt
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_1.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_2.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2_1.png
A Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/3.png
A Principles_of_Data_structures_using_C_and_C++_by_Vinu_V_Das/README.txt
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_5.ipynb
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_1.png
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1.png
A Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_3_1.png
A sample_notebooks/VinayBadhan/chapter7.ipynb
48 files changed, 32803 insertions, 0 deletions
diff --git a/Heat_Transfer_by_K._A._Gavhane/Chapter_2.ipynb b/Heat_Transfer_by_K._A._Gavhane/Chapter_2.ipynb new file mode 100644 index 00000000..da6b743d --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/Chapter_2.ipynb @@ -0,0 +1,2626 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Conduction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.1,Page no:2.18" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ans.(A).Thickness of asbestos is: 98.0 mm\n", + "Ans.(B)Thickness of fire clay insulation is: 747.0 mm\n" + ] + } + ], + "source": [ + "#Thickness of insulation\n", + "\n", + "#Variable declaration\n", + "A=1 #Area of heat transfer[sq metre]\n", + "Q=450 #Rate of heat loss/unit area[W/ sq mtre]\n", + "dT=400 #Temperature difference across insulation layer[K]\n", + "k=0.11 #k for asbestos[W/(m.K)]\n", + "\n", + "#Calculation\n", + "#Q=(k* A*dT)/x\n", + "x1=(k*A*dT)/Q\n", + "X1=x1*1000 \n", + "#for fire clay insulation\n", + "k=0.84 #For fire clay insulation[W/(m.K)]\n", + "x=(k*A*dT)/Q \n", + "X=x*1000 \n", + "\n", + "#Result\n", + "print\"Ans.(A).Thickness of asbestos is:\",round(X1),\"mm\"\n", + "print\"Ans.(B)Thickness of fire clay insulation is:\",round(X),\"mm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.2,Page no:2.18" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss per metre of pipe,Q= 62.7 W/m\n" + ] + } + ], + "source": [ + "#Heat loss per metre\n", + "\n", + "#Variable declaration\n", + "L=1.0 # Length of pipe[m]\n", + "r1=(50.0/2.0) # in mm\n", + "r1=r1/1000.0 # in m\n", + "r2=(25.0+3.0)/1000.0 # m\n", + "\n", + "\n", + "#Calculation\n", + "import math\n", + "rm1=(r2-r1)/math.log(r2/r1) \n", + "k1=45.0 #W/(m.K)\n", + "R1=(r2-r1)/(k1*(2*math.pi*rm1*L)) # Thermal resistance of wall pipe[K/W]\n", + "\n", + "#For inner lagging:\n", + "k2=0.08 #W/(m.K)\n", + "ri1=0.028 #m\n", + "ri2=(ri1+r1) # m\n", + "rmi1=(ri2-ri1)/math.log(ri2/ri1)\n", + "R2=(ri2-ri1)/(k2*2*math.pi*rmi1*L) #Thermal resistance of inner lagging [K/W]\n", + "\n", + "#For outer lagging:\n", + "k3=0.04 #W/(m.K)\n", + "ro1=0.053 #m\n", + "ro2=(ro1+0.04) # m\n", + "rmo1=(ro2-ro1)/math.log(ro2/ro1)\n", + "R3=(ro2-ro1)/(k3*2*math.pi*rmo1*L) #Thermal resistance of outer lagging\n", + "\n", + "R=R1+R2+R3\n", + "Ti=550.0 #K #inside\n", + "To=330.0 #K # outside\n", + "dT=Ti-To #Temperature difference\n", + "Q=dT/R\n", + "\n", + "\n", + "#Result\n", + "print\"Rate of heat loss per metre of pipe,Q=\",round(Q,1),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.3,Page no:2.19" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per meter pipe is 149.2 W/m(approx)\n" + ] + } + ], + "source": [ + "#Heat Loss in pipe\n", + "import math\n", + "#Variable declaration\n", + "r1=44.0 #i [mm]\n", + "r1=r1/1000.0 #i[m]\n", + "r2=0.094 #i [m]\n", + "r3=0.124 #i [m]\n", + "T1=623.0 #iTemperature at outer surface of wall in[K]\n", + "T3=313.0 #iTemperature at outer surface of outer insulation [K]\n", + "k1=0.087 #iThermal conductivity of insulation layer 1..in [W/m.K]\n", + "k2=0.064 #iThermal conductivity of insulation layer 2 [W/m.K]\n", + "l=1 #i Length of pipe [m]\n", + "\n", + "#Calculation\n", + "rm1=(r2-r1)/math.log(r2/r1) #imath.log mean radius of insulation layer 1 [m]\n", + "rm2=(r3-r2)/math.log(r3/r2) #imath.log mean radius of insulation layer 2[m]\n", + "#iPutting values in following eqn:\n", + "Q= (T1-T3)/((r2-r1)/(k1*2*math.pi*rm1*l)+(r3-r2)/(k2*2*math.pi*rm2*l)) \n", + "\n", + "#Result\n", + "print\"Heat loss per meter pipe is\",round(Q,1),\"W/m(approx)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.4,Page no:2.21" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Interface temperature:\n", + "i-Between FB-IB= 1121.0 K \n", + "ii-Between IB-PB= 587.8 K\n" + ] + } + ], + "source": [ + "#Heat loss in interface\n", + "\n", + "#Variable declaration\n", + "A=1 #Heat transfer area [sq m]\n", + "x1=0.229 # thickness of fire brick in [m]\n", + "x2=0.115 # thickness of insulating brick in [m]\n", + "x3=0.229 # thickness of building brick in [m]\n", + "k1=6.05 #thermal conductivity of fir brick [W/(m.K)]\n", + "k2=0.581 #thermal conductivity of insulating brick [W/m.K]\n", + "k3=2.33 #thermal conductivity of building brick [W/m.K]\n", + "T1=1223 # inside temperature [K]\n", + "T2=323 # Outside temperature[K]\n", + "\n", + "#Calculation\n", + "dT=T1-T2 #Overall temp drop [K]\n", + "R1=(x1/k1*A) #thermal resistance 1\n", + "R2=(x2/k2*A) # Thermal resistance 2\n", + "R3=(x3/k3*A) #Thermal resistance 3\n", + "Q=dT/(R1+R2+R3) #w/SQ m\n", + "Ta=-((Q*R1)-T1) #from Q1=Q=(T1-Ta)/(x1/k1*A)\n", + "#Similarly\n", + "Tb=(Q*R3)+T2 \n", + "\n", + "#Result\n", + "print\"Interface temperature:\\ni-Between FB-IB=\",round(Ta),\" K \\nii-Between IB-PB=\",round(Tb,1),\"K\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.5,Page no:2.23" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per unit area is 2674.2 W= 2674.2 J/s\n", + "\n", + "Ta= 1108.0 K(APPROX) =Temperature at the interface between fire brick and insulating brick\n", + "Tb= 581.0 K Temperature at the interface between insulating and building brick\n", + "\n", + "NOTE:Tb is wrongly calculated in the book as 565 K\n" + ] + } + ], + "source": [ + "#Heat loss per unit area\n", + "import math\n", + "#Variable declaration\n", + "A=1 #let [sq m]\n", + "x1=0.23 #thickness of fir brick layer[m]\n", + "x2=0.115 # [m]\n", + "x3=0.23 #[m]\n", + "T1=1213.0 #Temperature of furnace [K]\n", + "T2=318.0 #Temperature of furnace [K]\n", + "dT=T1-T2 #[K]\n", + "k1=6.047 #W/(m.K) (fire brick)\n", + "k2=0.581 #W/(m.K) (insulating brick)\n", + "k3=2.33 #W/(m.K) (building brick)\n", + "\n", + "#Calculation\n", + "Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3)) #Heat lost per unit Area in Watt\n", + "Q_by_A=round(Q_by_A,1)\n", + "R1=(x1/k1) #Thermal resistance\n", + "R2=(x2/k2)\n", + "R3=(x3/k3)\n", + "R1=round(R1,2)\n", + "R2=round(R2,1)\n", + "R3=round(R3,1)\n", + "Ta=T1-((dT*R1)/(R1+R2+R3))\n", + "Ta=round(Ta)\n", + "Tb=((dT*R3)/(R1+R2+R3))+T2\n", + "\n", + "\n", + "#Result\n", + "print\"Heat loss per unit area is\",Q_by_A,\"W=\",Q_by_A,\"J/s\\n\"\n", + "print\"Ta=\",Ta,\"K(APPROX) =Temperature at the interface between fire brick and insulating brick\"\n", + "print\"Tb=\",round(Tb),\"K Temperature at the interface between insulating and building brick\"\n", + "print\"\\nNOTE:Tb is wrongly calculated in the book as 565 K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.7,Page no:2.26" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss from 1 sq metre wall= 689.5 W\n", + "Heat loss from 1 sq metre when resistance present= 632.1 W\n" + ] + } + ], + "source": [ + "#Heat loss in wall\n", + "#Part-(a)\n", + "\n", + "#Variable declaration\n", + "A=1.0 # sq metre\n", + "x1=114.0 # mm\n", + "x1=x1/1000.0 # metre\n", + "k1=0.138 # W/(m.K)\n", + "R1= x1/(k1*A)\n", + "x2=229.0 #mm\n", + "x2= x2/1000.0 # metre\n", + "k2=1.38 # W/m.K\n", + "R2=x2/(k2*A)\n", + "dT=1033.0-349.0\n", + "\n", + "#Calculation\n", + "#Heat loss\n", + "Q1=dT/(R1+R2)\n", + "#Part(b)\n", + "#contact resistance=cr\n", + "cr=0.09 #K/W\n", + "R=R1+R2+cr\n", + "Q=dT/R\n", + "\n", + "#Result\n", + "print\"Heat loss from 1 sq metre wall=\",round(Q1,1),\"W\"\n", + "print\"Heat loss from 1 sq metre when resistance present=\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.8,Page no:2.27" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The temperature of 293 K will be reached at point 16.4 mm from the outermost wall surface of the ice-box\n" + ] + } + ], + "source": [ + "#Loss per area\n", + "\n", + "#Variable declaration\n", + "x1=0.02 #[m]\n", + "x2=0.01 #[m]\n", + "x3=0.02 #[m]\n", + "k1=0.105 #W/(m.k)\n", + "k3=k1 #W/(m.K)\n", + "k2=0.041 #W/(m.K)\n", + "T1=303\n", + "T2=263\n", + "\n", + "#Calculation\n", + "dT=T1-T2 #[K]\n", + "Q_by_A=dT/((x1/k1)+(x2/k2)+(x3/k3))\n", + "R=0.625 #K/W\n", + "Tx=293 #K\n", + "Rx=0.9524 #K/W\n", + "x=R*(T1-Tx)/(dT*Rx)\n", + "x=x*100 #mm\n", + "\n", + "#Result\n", + "print\"The temperature of 293 K will be reached at point\",round(x,1),\"mm from the outermost wall surface of the ice-box\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.9,Page no:2.28" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss=Q= 2220.0 W\n" + ] + } + ], + "source": [ + "#Heat loss\n", + "\n", + "#Variable declaration\n", + "import math\n", + "ID=50.0 #Internal diameter[mm]\n", + "dT=(573.0-303.0) \n", + "r1=ID/2.0 #mm\n", + "r1=r1/1000.0 # metres\n", + "OD=150.0 #Outer diameter[mm]\n", + "r2=OD/2.0 # mm\n", + "r2=75.0/1000.0 # m\n", + "#Thermal conductivity\n", + "k=17.45 # W/(m.K) \n", + "\n", + "#Calculation\n", + "#Q/A=dT/(r2-r1)/k\n", + "A1=4*math.pi*(r1**2) \n", + "A2=4*math.pi*(r2**2) \n", + "A=math.sqrt(A1*A2)\n", + "Q=(A*k*dT)/(r2-r1)\n", + "\n", + "#Result\n", + "print\"Heat loss=Q=\",round(Q),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.10,Page no:2.29" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost per sq meter is 5941.0 W/sq m\n", + "Width of air gap is 15.6 mm\n" + ] + } + ], + "source": [ + "#Heat Passed\n", + "import math\n", + "#Variable declaration\n", + "A= 1.0 #sq m\n", + "x1=0.15\n", + "x2=0.01\n", + "x4=0.15\n", + "T1=973.0 #[K]\n", + "T2=288.0 #[K]\n", + "dT=T1-T2 #[K]\n", + "#Thermal conductivities\n", + "k1=1.75 \n", + "k2=16.86\n", + "k3=0.033\n", + "k4=5.2\n", + "\n", + "#Calculation\n", + "#in absence of air gap,sum of thermal resistances \n", + "sR=(x1/k1*A)+(x2/k2*A)+(x4/k4*A)\n", + "round(sR,3)\n", + "sR=0.1153 #approximate\n", + "Q= dT/sR\n", + "\n", + "#When heat loss,Q=1163,then new resistance =sR1\n", + "Q1=1163.0 #[W/sq m]\n", + "sR1=dT/Q1\n", + "#width of air gap be w then\n", + "w=(sR1-sR)*k3*A # [m]\n", + "w=w*1000 #in [mm]\n", + "\n", + "#Result\n", + "print\"Heat lost per sq meter is\",round(Q),\"W/sq m\"\n", + "print\"Width of air gap is\",round(w,1),\"mm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.11,Page no:2.30" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss is: 176.3 W/m\n", + "Heat lost per sq meter of outer insulation is 117.0 W/sq m\n", + "Temperature between two layers of insulation= 469.0 K\n" + ] + } + ], + "source": [ + "# Heat loss in Insulated pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "d1=300.0 #[mm]\n", + "r1=d1/2.0 # [mm]\n", + "r1=r1/1000.0 #[m]\n", + "r2=r1+0.05 #[m]\n", + "r3=r2+0.04 #[m]\n", + "x1=0.05 #[m]\n", + "x2=0.04 #[m]\n", + "k1=0.105 #W/(m.K)\n", + "k2=0.07 #W/(m.K)\n", + "\n", + "#Calculation\n", + "rm1= (r2-r1)/math.log(r2/r1) # [m]\n", + "rm2=(r3-r2)/math.log(r3/r2) #[m]\n", + "L=1 #let\n", + "A1=math.pi*rm1*L #let L=1\n", + "R1=x1/(k1*A1) \n", + "A2=math.pi*rm2*L\n", + "R2=x2/(k2*A2)\n", + "T1=623.0 #[K]\n", + "T2=323.0 #[K]\n", + "dT=T1-T2 #[K]\n", + "\n", + "#Part a\n", + "Q_by_L= dT/(R1+R2) #Heat loss\n", + "\n", + "#Part b:\n", + "P=2*math.pi*(r1+x1+x2) #[m]\n", + "Q_by_L_peri=Q_by_L/P # [W/sq m]\n", + "R1=x1/(k1*A1) \n", + "sR=0.871+0.827\n", + "dT1=dT*R1/sR\n", + "\n", + "\n", + "#Result\n", + "#Part a\n", + "print\"Heat loss is:\",round(Q_by_L,1),\"W/m\" \n", + "#Part b:\n", + "print\"Heat lost per sq meter of outer insulation is\",round(Q_by_L_peri),\"W/sq m\"\n", + "print\"Temperature between two layers of insulation=\",round((T1-dT1)),\"K\"\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.12,Page no:2.31" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Width of the air gap is 12.06 mm\n" + ] + } + ], + "source": [ + "#Heat loss in Composite brick\n", + "import math\n", + "#Variable declaration\n", + "x1=0.01 #[m]\n", + "x2=0.15 #[m]\n", + "x3=0.15 #[m]\n", + "T1=973.0 #[K]\n", + "T2=423.0 #[K]\n", + "dT=T1-T2 \n", + "#Thermal conductivities\n", + "k1=16.86 #[W/m.K]\n", + "k2=1.75 #[W/m.K]\n", + "k3=5.23 #[W/m.K]\n", + "k_air=0.0337 # [W/m.K]\n", + "A=1 #[sq m]\n", + "\n", + "#Calculation\n", + "sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A))\n", + "Q=dT/sigma_R #Heat flow in [W\n", + "Tm= Q*x3/k3 #Temperature drop in magnesite brick\n", + "#Interface temperature=iT\n", + "iT=T2+Tm #[K]\n", + "sigma_xbyk= A*dT/1163 #with air gap for reducing heat loss to 1163 per sq m\n", + "x_by_k=sigma_xbyk-sigma_R #x/k for air\n", + "t=x_by_k*k_air\n", + "t=t*1000 \n", + "\n", + "#Calculation\n", + "print\"Width of the air gap is\",round(t,2),\"mm\"\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.13,Page no:2.32" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flow per metre of pipe is 82.93 W/m\n", + "Temperature at outer surface of steel pipe: 422.97 K\n", + "Conductance per m length based on inside area is 2.1 W/K\n" + ] + } + ], + "source": [ + "#Heat flow in a pipe\n", + "import math\n", + "#Variable declaration\n", + "L=1 #assume [m]\n", + "k1=43.03 #[W/(m.K)\n", + "k2=0.07 #(W/m.K)\n", + "T1=423 #inside temperature [K]\n", + "T2=305 # [K]\n", + "r1=0.0525 #[mm]\n", + "r2=0.0575 #[m]\n", + "r3=0.1075 #[m]\n", + "\n", + "#Calculation\n", + "#r3=r3/1000 #[m]\n", + "Q=(2*math.pi*L*(T1-T2))/(((math.log(r2/r1))/k1)+((math.log(r3/r2))/k2)) #Heat loss per metre \n", + "#Part 2\n", + "#T=Temperature of outer surface\n", + "T=T1-(Q*math.log(r2/r1))/(k1*2*math.pi*L) \n", + "#Part iii\n", + "id=0.105 #inside diametre in [m]\n", + "A=math.pi*id*1 #inside area in [sq m]\n", + "C=Q/(A*(T1-T2)) #conductance per length\n", + "\n", + "#Result\n", + "print\"Heat flow per metre of pipe is\",round(Q,2),\"W/m\"\n", + "#Part 2\n", + "print\"Temperature at outer surface of steel pipe:\",round(T,2),\"K\" \n", + "#Part iii\n", + "print\"Conductance per m length based on inside area is\",round(C,1),\"W/K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.15,Page no:2.35" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of rockwool insulation required= 4.91 mm\n" + ] + } + ], + "source": [ + "#Thickness of insulation \n", + " \n", + "#Variable declaration\n", + "A=1 # [sq m]\n", + "x1=0.1 #m\n", + "x2=0.04\n", + "k1=0.7\n", + "k2=0.48\n", + "\n", + "#Calculation\n", + "sigma=x1/(k1*A)+x2/(k2*A) #K/W\n", + "#Q=4.42*dT\n", + "#Q=dT/sigma\n", + "#with rockwool insulation added,Q_dash=0.75*Q\n", + "k3=0.065 # W/(m.K)\n", + "#Q_dash=dT/sigma+x3/k3*A\n", + "#On solving Q and Q_dash we get\n", + "x3=((1/(0.75*4.42))-sigma)*k3 #[m]\n", + "x3=x3*1000 # [mm]\n", + "\n", + "#Result\n", + "print\"Thickness of rockwool insulation required=\",round(x3,2),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.16,Page no:2.36" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)\n", + "Percent reduction in heat loss is 23.2 percent\n", + "\n", + "NOTE:Slight variation in answers due to less precise calculation in book.If performed manually,this answer stands to be correct\n" + ] + } + ], + "source": [ + "#Reduction in heat loss in insulated pipe\n", + "d1=40.0 # Diameter of pipe[mm]\n", + "r1=(d1/2.0)/1000.0 #Outside radius in [m]\n", + "t1=20.0 #Insulation 1 thickness in [mm]\n", + "t1=t1/1000 #[m]\n", + "t2=t1 #Insulation 2 thickness in[m]\n", + "r2=r1+t1 #radius after 1st insulation in [m]\n", + "r3=r2+t2 #Radius after second insulation in [m]\n", + "\n", + "#Calculation\n", + "import math\n", + "#Since python does not handles symbolic constants,we will assume some values:\n", + "#(1)\n", + "#Let the layer M-1 be nearer to the surface\n", + "L=1.0 #[m]\n", + "T1=10.0 #Temperature of inner surface of pipe [K]\n", + "T2=5.0 #Temperature of outer surface of insulation [K]\n", + "k=1.0 #Thermal conductivity\n", + "k1=k #For M-1 material\n", + "k2=3*k #For material M-2\n", + "Q1=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k1)+math.log(r3/r2)/(2*math.pi*L*k2))\n", + "#(2)\n", + "#Let the layer of material M-2 be nearer to the surface\n", + "Q2=(T1-T2)/(math.log(r2/r1)/(2*math.pi*L*k2)+math.log(r3/r2)/(2*math.pi*L*k1))\n", + "#For dummy variables unity...\n", + "#For any value of k,T1 and T2,Q1 is always less than Q2\n", + "\n", + "#Result\n", + "print\"M-1 near the surface is advisable(i.e Arrangement one will result in less heat loss)\"\n", + "per_red=(Q2-Q1)*100/Q2\n", + "print\"Percent reduction in heat loss is\",round(per_red,1),\"percent\"\n", + "print\"\\nNOTE:Slight variation in answers due to less precise calculation in book.If performed manually,this answer stands to be correct\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.17,Page no:2.37" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss= 89.6 W/m\n", + "If order is changed then heat loss= 105.76 W/m\n", + "Loss of heat is increased by 18.04 percent by putting material with higher thermal conductivity near the pipe surface\n" + ] + } + ], + "source": [ + "#Heat loss in a pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "T1=523 #[K]\n", + "T2=323 #[K]\n", + "r1=0.05 #[m]\n", + "r2=0.055 #[m]\n", + "r3=0.105 #[m]\n", + "r4=0.155 #[m]\n", + "k1=50 #[W/(m.K)]\n", + "k2=0.06 #[W/(m.K)]\n", + "k3=0.12 #W/(m.K)\n", + "\n", + "#Calculation\n", + "#CASE 1\n", + "Q_by_L1=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k2+(math.log(r4/r3))/k3) #[W/m]\n", + "#Case 2\n", + "Q_by_L2=2*math.pi*(T1-T2)/((math.log(r2/r1))/k1+(math.log(r3/r2))/k3+(math.log(r4/r3))/k2)\n", + "perct=(Q_by_L2-Q_by_L1)*100/Q_by_L1\n", + "\n", + "#Result\n", + "#CASE 1\n", + "print\"Heat loss=\",round(Q_by_L1,1),\"W/m\"\n", + "#Case 2\n", + "print\"If order is changed then heat loss=\",round(Q_by_L2,2),\"W/m\" \n", + "print\"Loss of heat is increased by\",round(perct,2),\" percent by putting material with higher thermal conductivity near the pipe surface\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.18,Page no:2.38" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ANSWER-(i)\n", + "Q1= 22.13 W \n", + "Q2= 24.48 W \n", + "Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\n", + "\n", + "ANSWER-(ii) \n", + "Percent reduction in heat loss (with near the pipe surface)= 10.6 %(approx)\n" + ] + } + ], + "source": [ + "#Arrangements for heat loss\n", + "\n", + "#Variable declaration\n", + "#Assume:\n", + "L=1.0 #[m]\n", + "r1=0.10 #[m] Outside radius od pipe\n", + "ia=0.025 #inner insulaiton [m]\n", + "import math \n", + "r2=r1+ia #Outer radius of inner insulation\n", + "r3=r2+ia #Outer radius of outer insulation\n", + "\n", + "#Calculation\n", + "#CASE 1:'a' near the pipe surface\n", + "#let k1=1\n", + "k1=1.0 #Thermal conductivity of A[W/m.K]\n", + "#and k2=3k1=3\n", + "k2=3.0 #Thermal conductivity of B[W/m.K]\n", + "#Let dT=1\n", + "dT=1.0\n", + "Q1=dT/(math.log(r2/r1)/(2*math.pi*k1*L)+math.log(r3/r2)/(2*math.pi*k2*L))\n", + "#CASE 2:'b' near the pipe surface \n", + "Q2=dT/(math.log(r2/r1)/(2*math.pi*k2*L)+math.log(r3/r2)/(2*math.pi*k1*L))\n", + "\n", + "#Result\n", + "print\"ANSWER-(i)\\nQ1=\",round(Q1,2),\"W \\nQ2=\",round(Q2,2),\" W \"\n", + "print\"Q1 is less than Q2.i.e arrangement A near the pipe surface and B as outer layer gives less heat loss\\n\"\n", + "percent=(Q2-Q1)*100/Q1 #percent reduction in heat loss\n", + "print\"ANSWER-(ii) \\nPercent reduction in heat loss (with near the pipe surface)=\",round(percent,1),\"%(approx)\" \n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.19,Page no:2.39" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of insulation= 184.0 mm\n" + ] + } + ], + "source": [ + "#Insulation thickness\n", + "\n", + "#Variable declaration\n", + "x1=0.224 # m\n", + "k1=1.3 # W/(m.K)\n", + "k2=0.346 # W/(m.K)\n", + "T1=1588.0 # K\n", + "T2= 299.0 # K\n", + "QA=1830.0 # W/ sq metre #heat loss\n", + "\n", + "#Calculation\n", + "#Q/A=(T1-T2)/x1/k1+x2/k2\n", + "x2=k2*((T1-T2)*1/(QA)-(x1/k1))\n", + "x2=x2*1000 \n", + "\n", + "#Result\n", + "print\"Thickness of insulation=\",round(x2),\"mm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.20,Page no:2.39" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of diatomite layer= 93.0 mm\n" + ] + } + ], + "source": [ + "#Heat loss in furnace\n", + "\n", + "#Variable declaration\n", + "\n", + "#for clay\n", + "k1=0.533 #[W/(m.K)]\n", + "#for red brick\n", + "k2=0.7 #[W/m.K]\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "#Case 1\n", + "A=1 #Area\n", + "x1=0.125 #[m]\n", + "x2=0.5 #[m]\n", + "#Resistances\n", + "r1=x1/(k1*A) #Res of fire clay [K/W]\n", + "r2=x2/(k2*A) #Res of red brick[K/W]\n", + "r=r1+r2\n", + "#Temperatures\n", + "T1=1373 #[K]\n", + "T2=323 #[K]\n", + "Q=(T1-T2)/r #[W/sq m]\n", + "Tdash=T1-Q*r1 #[K]\n", + "\n", + "#Case2\n", + "# Heat loss must remain unchanged,Thickness of red brick also reduces to its half\n", + "x3=x2/2 #[m]\n", + "r3=x3/(k2*A) #[K/W]\n", + "Tdd= T2+(Q*r3) #[K]\n", + "#Thickness of diatomite be x2,km be mean conductivity\n", + "Tm=(Tdash+Tdd)/2 #[K]\n", + "km=0.113+(0.00016*Tm) #[W/(m.K]\n", + "x2=km*A*(Tdash-Tdd)/Q #[m]\n", + "x2=x2*1000 #[mm]\n", + "\n", + "#Result\n", + "print\"Thickness of diatomite layer=\",round(x2),\"mm\"\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.21,Page no:2.41" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of rockwool insulation 59.0 mm\n" + ] + } + ], + "source": [ + "#Rate of heat loss in pipe\n", + "\n", + "#Variable declaration\n", + "k1=0.7 #common brick W/((m.K)\n", + "k2=0.48 #gypsum layer [W/(m.K)\n", + "k3=0.065 #Rockwool [W/m.K]\n", + "#Heat loss with insulatiob will be 20% of without insulation\n", + "A=1 #sq m\n", + "x1=0.1 #[m]\n", + "x2=0.04 #[m]\n", + "\n", + "#Calculation\n", + "R1=x1/(k1*A) #K/W\n", + "R2=x2/(k2*A) #K/W\n", + "R=R1+R2 #K/W\n", + "#R3=x3/(k3*A)\n", + "QbyQd=0.2\n", + "sigRbyRd=QbyQd\n", + "x3=(R/QbyQd-R)/15.4 #m\n", + "x3=x3*1000 #[mm]\n", + "\n", + "#Result\n", + "print\"Thickness of rockwool insulation\",round(x3),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.22,Page no:2.44" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 116.63 W/m\n", + "\n", + "NOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.\n", + "In book is is taken as 1.366\n", + " \n" + ] + } + ], + "source": [ + "#Heat loss from insulated steel pipe\n", + "\n", + "#Variable declaration\n", + "Ts=451.0 #Steam temperature in [K]\n", + "Ta=294.0 #Air temperature in [K]\n", + "Di=25.0 #Internal diameter of pipe [mm]\n", + "Di=Di/1000 #[m]\n", + "od=33.0 #Outer diameter of pipe [mm] \n", + "od=od/1000 #[m]\n", + "hi=5678.0 #Inside heat transfer coefficient [W/(m**2.K)]\n", + "ho=11.36 #Outsideheat transfer coefficient [W/(sq m.K)]\n", + "\n", + "#Calculation\n", + "xw=(od-Di)/2 #Thickness of steel pipe [m]\n", + "k2=44.97 #k for steel in W/(m.K)\n", + "k3=0.175 #k for rockwool in W/(m.K)\n", + "ti=38.0/1000 #thickness of insulation in [m]\n", + "r1=Di/2 #[m]\n", + "r2=od/2 #[m]\n", + "rm1=(r2-r1)/math.log(r2/r1) #[m]\n", + "r3=r2+ti #[m]\n", + "rm2=(r3-r2)/math.log(r3/r2) #[m]\n", + "Dm1=2*rm1 #[m]\n", + "Dm2=2*rm2 #[m]\n", + "import math\n", + "#Rate of heat loss = dT/(sigma_R)\n", + "L=1 #[m]\n", + "R1=1/(hi*math.pi*Di*L) #[K/W]\n", + "R1=round(R1,4)\n", + "R2=xw/(k2*math.pi*Dm1*L)\n", + "R2=round(R2,6)\n", + "R3=(r3-r2)/(k3*math.pi*Dm2*L)\n", + "R3=round(R3,3)\n", + "R3=1.086\n", + "Do=(od+2*ti) #[mm]\n", + "R4=1/(ho*math.pi*Do*L) #[m]\n", + "R4=round(R4,3)\n", + "sigma_R=R1+R2+R3+R4 \n", + "#Heat loss\n", + "dT=Ts-Ta #[K]\n", + "Q=dT/sigma_R #Heat loss [W/m]\n", + "\n", + "#Result\n", + "print\"Rate of heat loss is\",round(Q,2),\"W/m\"\n", + "print\"\\nNOTE:Slight variation in final answer due to mistake in calculation of sigma_R in textbook.\\nIn book is is taken as 1.366\\n \"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.23,Page no:2.46" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage reduction in heat loss is 40.0 %\n" + ] + } + ], + "source": [ + "#Heat loss from furnace\n", + "\n", + "#Variable declaration\n", + "T1=913.0 #[K]\n", + "T=513.0 #[K]\n", + "T2=313.0 #[K]\n", + "\n", + "#Calculation\n", + "\n", + "#Q=(T1-T)/(x/(k*A))\n", + "#Q=(T-T2)/(1/(h*A))\n", + "#x=2k/h\n", + "#Q=(T1-T2)/(x/(kA)+1/(h*A))\n", + "#Therefore,Q=hA/3*(T1-T2)\n", + "#With increase in thickness(100%)\n", + "#x1=4*k/h\n", + "#Q2=(T1-T2)/(x1/k*A+1/(h*A))\n", + "#Q2=(h*A)/5)*(T1-T2)\n", + "#Now\n", + "h=1.0 #Assume\n", + "A=1.0 #Assume for calculation\n", + "Q1=(h*A/3)*(T1-T2)\n", + "Q2=((h*A)/5)*(T1-T2)\n", + "percent=(Q1-Q2)*100/Q1 #Percent reduction in heat loss\n", + "\n", + "#Result\n", + "print\"Percentage reduction in heat loss is\",percent,\"%\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.24,Page no:2.47" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss,Q= 43.5 W\n" + ] + } + ], + "source": [ + "#Rate of heat loss\n", + "\n", + "#Variable declaration\n", + "L=1.0#m\n", + "thp=2.0#Thickness of pipe in mm\n", + "thi=10.0#Thickness of insulation in mm\n", + "T1=373.0#K\n", + "T2=298.0#K\n", + "id=30.0#mm\n", + "r1=id/2#mm\n", + "\n", + "#Calculation\n", + "r2=r1+thp#mm\n", + "r3=r2+thi#mm\n", + "#In S.I units\n", + "r1=r1/1000 #m\n", + "r2=r2/1000#m\n", + "r3=r3/1000#m\n", + "k1=17.44#W/(m.K)\n", + "k2=0.58#W/(m.K)\n", + "hi=11.63#W/(sq m.K)\n", + "ho=11.63#W/(sq m.K)\n", + "import math\n", + "Q=(2*math.pi*L*(T1-T2))/(1/(r1*hi)+(math.log(r2/r1))/k1+((math.log(r3/r2))/k2)+(1/(0.02*ho)))\n", + "\n", + "#Result\n", + "print\"Rate of heat loss,Q=\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.25,Page no:2.48" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Required thickness of insulation is 0.0188 m= 18.8 mm or 19.0 m\n" + ] + } + ], + "source": [ + " #Thickness of insulation .\n", + "from scipy.optimize import fsolve\n", + "import math\n", + "\n", + "#Variable declaration\n", + "h=8.5 #[W/sq m.K]\n", + "dT=175 #[K]\n", + "r2=0.0167 #[m]\n", + "\n", + "#Calculation\n", + "Q_by_l=h*2*math.pi*r2*dT #[W/m]\n", + "k=0.07 #For insulating material in [W/m.K]\n", + "#for insulated pipe--50% reduction in heat loss\n", + "Q_by_l1=0.5*Q_by_l #[w/m]\n", + "def f(r3):\n", + " x=Q_by_l1-dT/((math.log(r3/r2))/(2*math.pi*k)+1/(2*math.pi*r3*h))\n", + " return(x)\n", + "#by trial and error method we get:\n", + "r3=fsolve(f,0.05)\n", + "t=r3-r2 #thickness of insulation in [m]\n", + "\n", + "#Result\n", + "print\"Required thickness of insulation is\",round(t[0],4),\"m=\",round(t[0]*1000,1),\" mm or\",round(t[0]*1000),\"m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.26,Page no:2.49" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per metre length of pipe= 114.49 W\n" + ] + } + ], + "source": [ + "#Calculate heat loss per metre length\n", + "import math\n", + "\n", + "#Variable declaration\n", + "id=0.1 #internal diameter in[m]\n", + "od=0.12 #outer diameter in [m]\n", + "T1=358 #Temperature of fluid [K]\n", + "T2=298 #Temperature of surrounding [K]\n", + "t=0.03 #thickness of insulation [m]\n", + "k1=58 #[W/m.K]\n", + "k2=0.2 #W/(m.K) insulating material\n", + "h1=720 #inside heat transfer coeff [W/sq m .K]\n", + "h2=9 #W/sq m.K\n", + "r1=id/2 #[m]\n", + "r2=od/2 #[m]\n", + "r3=r2+t #[m]\n", + "\n", + "#Calculation\n", + "#Heat loss per meter=Q_by_L\n", + "Q_by_L=(T1-T2)/(1/(2*math.pi*r1*h1)+math.log(r2/r1)/(2*math.pi*k1)+math.log(r3/r2)/(2*math.pi*k2)+1/(2*math.pi*r3*h2)) #W/m\n", + "\n", + "#Result\n", + "print\"Heat loss per metre length of pipe=\",round(Q_by_L,2),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.27,Page no:2.50" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of insulation is 32.0 mm\n", + "Rate of heat loss per unit length is 154.1 W/m\n" + ] + } + ], + "source": [ + "#Mineral wool insulation\n", + "from scipy.optimize import fsolve\n", + "\n", + "#Variable declaration\n", + "import math\n", + "T1=573 #[K]\n", + "T2=323 #[K]\n", + "T3=298 #[K]\n", + "h1=29 # Outside heat transfer coefficients [W/sq m.K]\n", + "h2=12 #[W/sq m.K]\n", + "r1=0.047 #Internal radius [m]\n", + "r2=0.05 #Outer radius[m]\n", + "k1=58 #[W/m.K]\n", + "k2=0.052 #[W/m.K]\n", + "\n", + "#Calculation\n", + "#Q=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)=(T2-T3)/(1/(r3*h2))\n", + "def f(r3):\n", + " x=(T1-T2)/(1/(r1*h1)+math.log(r2/r1)/k1+math.log(r3/r2)/k2)-(T2-T3)/(1/(r3*h2))\n", + " return(x)\n", + "\t#by trial and error method :\n", + "r3=fsolve(f,0.05)\n", + "t=r3-r2 #Thickness of insulation in [m]\n", + "#Q=h2*2*math.pi*r3*L*(T2-T3)\n", + "Q_by_l=h2*2*math.pi*r3*(T2-T3) #[W/m]\n", + "\n", + "#Result\n", + "print\"Thickness of insulation is\",round(t*1000),\"mm\"\n", + "print\"Rate of heat loss per unit length is\",round(Q_by_l[0],1),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.28,Page no:2.51" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per sq metre area is: 1551.4 W/sq m\n", + "Temperature of outside surface of furnace is: 323.9 K ( 50.9 degree C)\n" + ] + } + ], + "source": [ + "#Calculate heat loss per sq m and temperature of outside surface\n", + "\n", + "#Variable declaration\n", + "A=1 #assume [sq m]\n", + "x1=0.006 #[m]\n", + "x2=0.075 #[m]\n", + "x3=0.2 #[m]\n", + "k1=39.0 #[W/m.K]\n", + "k2=1.1 #[W/m.K]\n", + "k3=0.66 #[W/m.K]\n", + "h0=65.0 #W/sq m .K\n", + "T1=900.0 #K\n", + "T2=300.0 #K\n", + "\n", + "#Calculation\n", + "sigma_R=(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A)) \n", + "\n", + "#To calculate heat loss/sq m area\n", + "Q=(T1-T2)/sigma_R #[W/sq m]\n", + "#Q/A=T-T2/(1/h0), where T=Temp of outside surface\n", + "#So, T=T2+Q/(A*h0)\n", + "T=Q/(A*h0)+T2 #[K]\n", + "\n", + "#Result\n", + "print\"Heat loss per sq metre area is:\",round(Q,1),\"W/sq m\" \n", + "print\"Temperature of outside surface of furnace is:\",round(T,1),\"K (\",round(T-273,1),\"degree C)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.29,Page no:2.52" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of insulating brick required is 63.4 mm\n" + ] + } + ], + "source": [ + "#Determine necessary thickness of insulation brick\n", + "\n", + "#Variable declaration\n", + "A=1.0 #Assume [sq m]\n", + "x1=0.003 #[m]\n", + "x3=0.008 #[m]\n", + "k1=30.0 #[W/m.K]\n", + "k2=0.7 #[W/m.K]\n", + "k3=40.0 #[W/m.K]\n", + "T1=363.0 #[K]\n", + "T=333.0 #[K]\n", + "T2=300.0 #[K]\n", + "h0=10.0 #W/sq m.K\n", + "\n", + "#Calculation\n", + "#Q=(T1-T2)/(x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))\n", + "#Also,Q=(T-T2)/(1/(h0*A))\n", + "#So, (T1-T2)/((x1/(k1*A)+x2/(k2*A)+x3/(k3*A)+1/(h0*A))=(T-T2)/(1/(h0*A))\n", + "#or,x2=k2*A((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A))\n", + "x2=k2*A*((T1-T2)/((T-T2)*h0*A)-1/(h0*A)-x1/(k1*A)-x3/(k3*A)) #[m]\n", + "\n", + "#Result\n", + "print\"Thickness of insulating brick required is\",round(x2*1000,1),\"mm\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.30,Page no:2.53" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per sq m is: 938.5 W\n", + "Temperature at inner surface of middle layer= 1780.3 K( 1507.3 degree C)\n", + "Temperature at outer surface of middle layer= 512.1 K ( 239.1 degree C)\n" + ] + } + ], + "source": [ + "#Heat flow through furnace wall\n", + "\n", + "#Variable declaration\n", + "hi=75.0 #[W/sq m.K)\n", + "x1=0.2 #m\n", + "x2=0.1 #[m]\n", + "x3=0.1 #[m]\n", + "T1=1943 #[K]\n", + "k1=1.25 #W/m.K\n", + "k2=0.074 #/W/m.K\n", + "k3=0.555 #W/m.K\n", + "T2=343.0 #K\n", + "A=1.0 #assume [sq m]\n", + "\n", + "#Calculation\n", + "sigma_R=1.0/(hi*A)+x1/(k1*A)+x2/(k2*A)+x3/(k3*A) \n", + "#Heat loss per i sq m\n", + "Q=(T1-T2)/sigma_R #[W]\n", + "#if T=temperature between chrome brick and koalin brick then \n", + "#Q=(T1-T)/(1/(hi*A)+x1/(k1*A))\n", + "#or T=T1-(Q*(1/(hi*A)+x1/(k1*A)))\n", + "T=T1-(Q*(1.0/(hi*A)+x1/(k1*A))) #[K]\n", + "#if Tdash=temperature at the outer surface of middle layer,then\n", + "#Q=(Tdash-T2)/(x3/(k1*A))\n", + "#or Tdash=T2+(Q*x3/(k3*A))\n", + "Tdash=T2+(Q*x3/(k3*A)) #[K]\n", + "\n", + "#Result\n", + "print\"Heat loss per sq m is:\",round(Q,1),\"W\"\n", + "print\"Temperature at inner surface of middle layer=\",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n", + "print\"Temperature at outer surface of middle layer=\",round(Tdash,1),\"K (\",round(Tdash-273,1),\"degree C)\" \n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.31,Page no:2.54" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ans (a) Heat loss= 321.94 W/m \n", + "Ans (b) Percent reduction in heat loss is 77.5 %\n" + ] + } + ], + "source": [ + "#Calculate:(a) Heat loss per unit length \n", + "#(b)Reduction in heat loss\n", + "import math\n", + "\n", + "#Variable declaration\n", + "hi=10 #W/sq m.K\n", + "h0=hi #W/sq.m.K\n", + "r1=0.09 #m\n", + "r2=0.12 #m\n", + "t=0.05 #thickness of insulation [m]\n", + "k1=40 #W/m.K\n", + "k2=0.05 #W/m.K\n", + "T1=473 #K\n", + "T2=373 #K\n", + "\n", + "#Calculation\n", + "Q_by_L=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r1)/k1+1/(r2*h0)) #W/m\n", + "#After addition of insulation:\n", + "r3=r2+t #radius of outer surface of insulaiton\n", + "Q_by_L1=2*math.pi*(T1-T2)/(1/(r1*hi)+math.log(r2/r2)/k1+math.log(r3/r2)/k2+1/(r3*h0)) # W\n", + "Red=Q_by_L-Q_by_L1 #Reduciton in heat loss in [W/m]\n", + "percent_red=(Red/Q_by_L)*100 #% Reduction in heat loss\n", + "\n", + "#Result\n", + "print\"Ans (a) Heat loss=\",round(Q_by_L,2),\"W/m \"\n", + "print\"Ans (b) Percent reduction in heat loss is\",round(percent_red,1),\"%\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.32,Page no:2.55" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ans (i)- Heat flux across the layers is 1997.0 W/sq m\n", + "Ans-(ii)-Interfacial temperature between layers is 777.4 K( 504.4 degree C)\n" + ] + } + ], + "source": [ + "#Determine: i-Heat flux across the layers and\n", + "#ii-Interfacial temperature between the layers\n", + "\n", + "#Variable declaration\n", + "T1=798.0 #K\n", + "T2=298.0 #K\n", + "x1=0.02 #m\n", + "x2=x1 #m\n", + "k1=60.0 #W/m.K\n", + "k2=0.1 #W/m.K\n", + "hi=100.0 #W/sq m.K\n", + "h0=25.0 #W/sq m.K\n", + "\n", + "#Calculation\n", + "Q_by_A=(T1-T2)/(1.0/hi+x1/k1+x2/k2+1.0/h0) #W/sq m\n", + "#If Tis the interfacial temperature between steel plate and insulating material\n", + "#Q_by_A=(T-T2)/(x2/k2+1/h0)\n", + "T=Q_by_A*(x2/k2+1.0/h0)+T2\n", + "\n", + "#Result\n", + "print\"Ans (i)- Heat flux across the layers is\",round(Q_by_A),\"W/sq m\" \n", + "print\"Ans-(ii)-Interfacial temperature between layers is\",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.33,Page no:2.56" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Convective conductance is: 72.1 W/m^2.K\n" + ] + } + ], + "source": [ + "#Determine Temperature at the outer surface of wall and convective conductance on the outer wall\n", + " \n", + " \n", + "#Variable declaration \n", + "T1=2273 #Temperature of hot gas:[K]\n", + "T4=318 #Ambient aur temperature[K]\n", + "Qr1_by_A=23260 #Heat flow by radiation from gases to inside surface of wall[W/sq m]\n", + "hi=11.63 #Heat transfer coefficient on inside wall:[W/sq m.K]\n", + "K=58 #Thermal conductivity of wall[W.sq m/K]\n", + "Qr4_by_A=9300 #Heat flow by radiation from external surface to ambient[W/sq m]\n", + "T2=1273 #Inside Wall temperature[K]\n", + "Qr1=Qr1_by_A #W for\n", + "A=1 #sq m\n", + "\n", + "\n", + "#Calculation\n", + "Qc1_by_A=hi*(T1-T2) #W/sq m\n", + "Qc1=Qc1_by_A #for A=1 sq m\n", + " #Thermal resistance:\n", + "R=1.0/K #K/W per sq m\n", + "#Now Q=(T2-T3)/R,i.e \n", + "#External wall temp T3=T2-Q*R\n", + "#Q entering wall=\n", + "Q_enter=Qr1+Qc1 #W\n", + "T3=T2-Q_enter*R #K\n", + "T3=673 #Approximate\n", + "#Heat loss due to convection:\n", + "Qc4_by_A=Q_enter-Qr4_by_A #W/sq m\n", + "#Qc4_by_A=h0*(T3-T4)\n", + "#or h0=Qc4_by_A/(T3-T4)\n", + "h0=Qc4_by_A/(T3-T4) #W/sq m.K\n", + "\n", + "\n", + "#Result\n", + "print\"Convective conductance is:\",round(h0,1),\"W/m^2.K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.34,Page no:2.60" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When covered with insulation:\n", + "Heat loss= 105.7 W\n", + "When without insulation,\n", + "Heat loss= 84.8 W\n", + "\n", + "Percent increase = 24.66 percent\n", + "In this case the value of rc= 0.0133 m is less than the outside radius of pipe ( 0.025 m)\n", + "So additon of any fibre glass would cause a decrease in the heat transfer\n" + ] + } + ], + "source": [ + "#Critical radius of insulation\n", + "\n", + "#Variable declaration\n", + "T1=473 #[K]\n", + "T2=293 #[K]\n", + "k=0.17 #W/(m.K)\n", + "h=3 #W/(sq m.K)\n", + "h0=h #W/sq m.K\n", + "rc=k/h #m\n", + "r1=0.025 #Inside radius of insulaiton [mm] \n", + "\n", + "#Calculation\n", + "import math\n", + "q_by_l1=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1/(rc*h0)) #Heat transfer with insulation in W/m\n", + "#Without insulation:\n", + "q_by_l2=h*2*math.pi*r1*(T1-T2) #W/m\n", + "inc=(q_by_l1-q_by_l2)*100/q_by_l2 #Increase of heat transfer\n", + "k=0.04 #Fibre glass insulaiton W/(sq m.K)\n", + "rc=k/h #Critical radius of insulaiton\n", + "\n", + "#Result\n", + "print\"When covered with insulation:\\nHeat loss=\",round(q_by_l1,1),\"W\"\n", + "print\"When without insulation,\\nHeat loss=\",round(q_by_l2,1),\"W\"\n", + "print\"\\nPercent increase =\",round(inc,2),\"percent\" \n", + "print\"In this case the value of rc=\",round(rc,4),\"m is less than the outside radius of pipe (\",r1,\"m)\"\n", + "print\"So additon of any fibre glass would cause a decrease in the heat transfer\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.36,Page no:2.61" + ] + }, + { + "cell_type": "code", + "execution_count": 68, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per meter of pipe is 621.0 W/m(approx)\n", + "Outer surface temperature is: 391.8 K( 118.8 degree C)\n" + ] + } + ], + "source": [ + "#Calculate the heat loss per metre of pipe and outer surface temperature\n", + "\n", + "#Variable declaration\n", + "import math\n", + "k=1.0 #Thermal conductivity in [W/sq m.K]\n", + "h=8.0 #Het transfer coeff in W/sq m.K\n", + "rc=k/h #Critical radius in m\n", + "T1=473.0 #K\n", + "T2=293.0 #K\n", + "r1=0.055 #Outer radius =inner radius in [m]\n", + "\n", + "#Calculation\n", + "Q_by_L=2*math.pi*(T1-T2)/(math.log(rc/r1)/k+1.0/(rc*h))\n", + "#For outer surface\n", + "#Q_by_L=2*math.pi*(T-T2)/(1/rc*h)\n", + "# implies that, T=T2+Q_by_L/(rc*2*math.pi)\n", + "T=T2+Q_by_L/(rc*2*math.pi*h) #K\n", + "\n", + "#Result\n", + "print\"Heat loss per meter of pipe is\",round(Q_by_L),\"W/m(approx)\"\n", + "print\"Outer surface temperature is: \",round(T,1),\"K(\",round(T-273,1),\"degree C)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.37,Page no:2.78" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time required for a ball to attain a temperature of 423 K is 5818.0 s= 1.62 h\n" + ] + } + ], + "source": [ + "#Calculate the time required for a ball to attain a temperature of 423 K\n", + "\n", + "#Variable declaration\n", + "k_steel=35.0 #W/m.K\n", + "Cp_steel=0.46 #kJ/(kg*K)\n", + "Cp_steel=Cp_steel*1000 #J/(kg*K)\n", + "h=10 #W/sq m.K\n", + "rho_steel=7800.0 #kg/cubic m\n", + "dia=50.0 #mm\n", + "dia=dia/1000 #m\n", + "R=dia/2 #radius in m\n", + "\n", + "#Calculation\n", + "import math\n", + "A=4*math.pi*R**2 #Area in sq m\n", + "V=A*R/3 #Volume in cubic meter\n", + "Nbi=h*(V/A)/k_steel\n", + "#As Nbi<0.10,internal temp gradient is negligible\n", + "T=423.0 #K\n", + "T0=723.0 #K\n", + "T_inf=373.0 #K\n", + "#(T-T_inf)/(T0-T_inf)=e**(-h*At/rho*Cp*V)\n", + "t=-rho_steel*Cp_steel*R*math.log((T-T_inf)/(T0-T_inf))/(3*h) #s\n", + "\n", + "#Result\n", + "print\"Time required for a ball to attain a temperature of 423 K is\",round(t),\"s=\",round(t/3600,2),\"h\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.38,Page no:2.78" + ] + }, + { + "cell_type": "code", + "execution_count": 72, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time taken by centre of ball to reach a temperature of 423 K is 436.25 s (= 7.27 minutes\n" + ] + } + ], + "source": [ + "#Time take in Steel ball quenched\n", + "\n", + "#Variable declaration\n", + "dia=50.0 #mm\n", + "dia=dia/1000 #m\n", + "r=dia/2 #radius in m\n", + "h=115.0 #W/sq m.K\n", + "rho=8000.0 #kg/cubic m\n", + "Cp=0.42 #kJ/kg.K\n", + "Cp=Cp*1000 #J/(kg*K)\n", + "\n", + "#Calculation\n", + "import math\n", + "A=4*math.pi*r**2 #Area in sq m\n", + "V=A*r/3 #Volume in cubic m\n", + "T=423.0 #K\n", + "T_inf=363.0 #K\n", + "T0=723 #K\n", + "#(T-T_inf)/(T0-T_inf)=e**(-3ht/(rho*Cp*r))\n", + "t=-rho*Cp*r*math.log((T-T_inf)/(T0-T_inf))/(3*h) #Time in seconds\n", + "\n", + "#Result\n", + "print\"Time taken by centre of ball to reach a temperature of 423 K is\",round(t,2),\"s (=\",round(t/60,2),\"minutes\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.39,Page no:2.79" + ] + }, + { + "cell_type": "code", + "execution_count": 74, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of ball after 1 h= 475.0 K ( 202.0 degree C)(APPROXIMATE)\n" + ] + } + ], + "source": [ + "#HT in a Ball plunged in a medium\n", + "\n", + "#Variable declaration\n", + "h=11.36 #W/sq m.K\n", + "k=43.3 #w/(m.K)\n", + "r=25.4 #radius in mm\n", + "r=r/1000 # radius in m\n", + "\n", + "#Calculation\n", + "import math\n", + "A=4*math.pi*r**2 #Area of sphere [sq m]\n", + "V=A*r/3 #Volume in [cubic m]\n", + "rho=7849.0 #kg/cubic m\n", + "Cp=0.4606*10**3 #J/kg.K\n", + "t=1.0 #hour\n", + "t=t*3600 #seconds\n", + "T_inf=394.3 #[K]\n", + "T0=700.0 #[K]\n", + "# (T-T_inf)/(T0-T_inf)=e**(-3*h*t/rho*Cp*V)\n", + "T=T_inf+(T0-T_inf)*(math.exp((-h*A*t)/(rho*Cp*V))) \n", + "\n", + "#Result\n", + "print\"Temperature of ball after 1 h=\",round(T),\"K (\",round(T-273),\"degree C)(APPROXIMATE)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.40,Page no:2.80" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time at which slab temperature becomes 363 K is 123.4 s\n", + "CALCULATION MISTAKE IN BOOK IN LAST LINE\n" + ] + } + ], + "source": [ + "#Slab temperature suddenly lowered\n", + "\n", + "#Variable declaration\n", + "import math\n", + "rho=9000.0 #kg/cubic m\n", + "Cp=0.38 #kJ/(kg.K)\n", + "Cp=Cp*1000 #J/(kg.K)\n", + "k=370.0 #W/m.K\n", + "h=90.0 #W/sq m.K\n", + "l=400.0 #mm\n", + "l=l/1000 #length of copper slab\n", + "t=5.0/1000 #thickness in [m]\n", + "\n", + "#Calculation\n", + "A=2*l**2 #Area of slab\n", + "V=t*l**2 #Volume in [cubic m]\n", + "L_dash=V/A #[m]\n", + "#for slab of thickness 2x\n", + "#L_dash=x\n", + "L_dash=0.025 #[m]\n", + "Nbi=h*L_dash/k #< 0.10\n", + "var=h*A/(rho*Cp*V)\n", + "#As Nbi<0.10,we can apply lumped capacity analysis\n", + "T=363.0 #[K]\n", + "T_inf=303.0 #[K]\n", + "T0=523.0 #[K]\n", + "t=-(math.log((T-T_inf)/(T0-T_inf)))/var\n", + "\n", + "#Result\n", + "print\"Time at which slab temperature becomes 363 K is\",round(t,1),\"s\"\n", + "print\"CALCULATION MISTAKE IN BOOK IN LAST LINE\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.41,Page no:2.80" + ] + }, + { + "cell_type": "code", + "execution_count": 77, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer coefficient is: 77.3 W/(m^2.K)\n" + ] + } + ], + "source": [ + "#Flow over a flat plate\n", + "import math\n", + "#Variable declaration\n", + "rho=9000.0 #kg/cubic meter\n", + "Cp=0.38 #kJ/(kg.K)\n", + "Cp=Cp*1000 #J/kg.K\n", + "k=370.0 #W/(m.K)\n", + "T0=483.0 #K\n", + "T_inf=373.0 #K\n", + "delta_T=40.0 #K\n", + "\n", + "#Calculation\n", + "T=T0-delta_T #K\n", + "t=5.0 #time in [minutes]\n", + "t=t*60 #[seconds]\n", + "#A=2A.....Two faces\n", + "#V=A.2x\n", + "#2x=thickness of slab=30 mm=0.03 m\n", + "x=0.015 #[m]\n", + "th=2*x #thickness of slab\n", + "h=-rho*Cp*x*math.log((T-T_inf)/(T0-T_inf))/t\n", + "\n", + "#Result\n", + "print\"Heat transfer coefficient is:\",round(h,1),\"W/(m^2.K)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.42,Page no:2.81" + ] + }, + { + "cell_type": "code", + "execution_count": 79, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time required to reach temperature 473.0 K is 82.2 s\n" + ] + } + ], + "source": [ + "#TIme required in Stainless steel rod immersed in water\n", + "\n", + "#Variable declaration\n", + "rho=7800.0 #[kg per cubic m]\n", + "h=100.0 #W/(sq m.K) Convective heat transfer coeff\n", + "Cp=460.0 #J/(kg.K)\n", + "k=40.0 #W/(m.K)\n", + "L=1.0 #[m] length ofrod\n", + "D=10.0 #mm \n", + "D=D/1000 #diameter in[m]\n", + "R=D/2 #raidus in [m]\n", + "\n", + "#Calculation\n", + "import math\n", + "#For cylindrical rod:\n", + "A=2*math.pi*R*L #Area in [sq m]\n", + "V=math.pi*R**2*L #Volume in [cubic m]\n", + "L_dash=V/A #[m]\n", + "Nbi=h*L_dash/k #Biot number\n", + "#N_bi<0.10,Hence lumped heat capavity is possible\n", + "T=473.0 #[K] \n", + "T_inf=393.0 #[K]\n", + "T0=593.0 #[K]\n", + "t=-rho*Cp*V*math.log((T-T_inf)/(T0-T_inf))/(h*A)\n", + "\n", + "#Result\n", + "print\"Time required to reach temperature\",T,\"K is\",round(t,1),\"s\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.43,Page no:2.82" + ] + }, + { + "cell_type": "code", + "execution_count": 84, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time constant of the thermocouple is 1.069 s\n", + "At the end of the time period t=tao= 1.069 s\n", + "Temperature difference b/n the thermocouple and the gas stream would be 0.368 of the initial temperature difference\n", + "It should be reordered after 4.0 s\n" + ] + } + ], + "source": [ + "# Time constant of Chromel alumel thermocouple \n", + "\n", + "#Variable declaration\n", + "import math\n", + "rho=8600.0 #[kg/cubic m]\n", + "Cp=0.42 #kJ/(kg.K)\n", + "Cp=Cp*1000 #J/(kg.K)\n", + "dia=0.71 #[mm]\n", + "dia=dia/1000 #[dia in m]\n", + "R=dia/2 #radius [m]\n", + "h=600.0 #convective coeff W/(sq m.K)\n", + "\n", + "#Calculation\n", + "#Let length =L=1\n", + "L=1.0 #[m]\n", + "A=2*math.pi*R*L \n", + "V=math.pi*(R**2)*L \n", + "tao=(rho*Cp*V)/(h*A) \n", + "#at1\n", + "t=tao\n", + "#From (T-T_inf)/(T0-T_inf)=e**(-t/tao)\n", + "ratio=math.exp(-t/tao) #Ratio of thermocouple difference to initial temperature difference\n", + "print\"Time constant of the thermocouple is\",round(tao,3),\"s\" \n", + "print\"At the end of the time period t=tao=\",round(tao,3),\"s\"\n", + "print\"Temperature difference b/n the thermocouple and the gas stream would be\",round(ratio,3),\"of the initial temperature difference\"\n", + "print\"It should be reordered after\",round(4*tao),\"s\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.44,Page no:2.83" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time constant of thermocouple is 37.33 s\n", + "Temperature attained by junction 20 s after removing from the hot air stream is: 368.0 K\n" + ] + } + ], + "source": [ + "#Thermocouple junction\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=8000.0 #kg/cubic m\n", + "Cp=420.0 #J/(kg.K)\n", + "h_hot=60.0 # for hot stream W/(sq m.K) \n", + "dia=4.0 #[mm]\n", + "t=10.0 \n", + "\n", + "#Calculation\n", + "r=dia/(2.0*1000) #radius in [m]\n", + "#For sphere\n", + "V=(4.0/3.0)*math.pi*r**3 #Volume in [cubic m]\n", + "A=4*math.pi*r**2 #Volume in [sq m]\n", + "tao=rho*Cp*V/(h_hot*A) # Time constant in [s]\n", + "ratio=math.exp(-t/tao) # %e**(-t/tao)=(T-T-inf)/(T0-T_inf)\n", + "T_inf=573.0 #[K]\n", + "T0=313.0 #[K]\n", + "T=T_inf+ratio*(T0-T_inf)\n", + "#IN STILL AIR:\n", + "h_air=10.0 #W/(sq m .K)\n", + "tao_air=rho*Cp*V/(h_air*A) #[s]\n", + "t_air=20.0 #[s]\n", + "ratio_air=math.exp(-t_air/tao_air)\n", + "T_inf_air=303.0 #[K]\n", + "T0_air=T \n", + "T_air=T_inf_air+ratio_air*(T0_air-T_inf_air)\n", + "\n", + "#Result\n", + "#ANS-[i]\n", + "print\"Time constant of thermocouple is\",round(tao,2),\"s\"\n", + "#ANS-[ii]\n", + "print\"Temperature attained by junction 20 s after removing from the hot air stream is:\",round(T_air),\"K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.45,Page no:2.84" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time taken to heat the reactants over the same temperature range is 7625.0 s\n", + "In CASE 1\n", + "Time taken to heat the reactants = 7625.0 s .ie 2.12 h\n", + "In CASE 2 \n", + "Time taken to heat the reactants = 8905.0 s\n", + "Maximum temperature at which temperature can be raised is 377.6 K\n" + ] + } + ], + "source": [ + "#Batch reactor time taken \n", + "from scipy.optimize import fsolve\n", + "from scipy import integrate\n", + "import math\n", + "\n", + "#Variable declaration\n", + "T_inf=390.0 #[K]\n", + "U=600.0 #[W/sq m.K]\n", + "Ac=1.0 #[sq m]\n", + "Av=10.0 #Vessel area in [sq m]\n", + "m=1000.0 #[kg]\n", + "Cp=3.8*10**3 #[J/kg.K]\n", + "To=290.0 #[K]\n", + "T=360.0 #[K]\n", + "h=8.5 #[W/sq m.K]\n", + "\n", + "#Calculation\n", + "#Heat gained from the steam=Rate of increase of internal energy\n", + "#U*A*(T_inf-T)=m*Cp*dT\n", + "def f(t):\n", + " x=math.log((T_inf-To)/(T_inf-T))-U*Ac*t/(m*Cp) \n", + " return(x)\n", + "t=fsolve(f,1) #[in s]\n", + "t=round(t) #[in s]\n", + "Ts=290 \n", + "print\"Time taken to heat the reactants over the same temperature range is\",t,\"s\"\n", + "def g(T):\n", + " t1=m*Cp/(U*Ac*(T_inf-T)-h*Av*(T-Ts))\n", + " return(t1)\n", + "t1= integrate.quad(g,To,T)\n", + "def fx(Tmax):\n", + " m=U*Ac*(T_inf-Tmax)-h*Av*(Tmax-Ts)\n", + " return(m)\n", + "T_max=fsolve(fx,1)\n", + "\n", + "#Result\n", + "print\"In CASE 1\\nTime taken to heat the reactants =\",t,\"s .ie\",round(t/3600,2),\"h\" \n", + "print\"In CASE 2 \\nTime taken to heat the reactants =\",round(t1[0]),\"s\"\n", + "print\"Maximum temperature at which temperature can be raised is\",round(T_max,1),\"K\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.46,Page no:2.95" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat dissipated by the rod is 6.844 W\n" + ] + } + ], + "source": [ + "#Heat dissipation by aluminium rod \n", + "import math \n", + "\n", + "#Variable declaration\n", + "dia=3.0 #[mm]\n", + "dia=dia/1000 #[m]\n", + "r=dia/2 #radius in[m]\n", + "k=150 #W/(m.K)\n", + "h=300 #W/(sq m.K)\n", + "T0=413 #[K]\n", + "T_inf=288 #[K]\n", + "\n", + "#Calculation\n", + "A=math.pi*(r**2) #Area in [sq m]\n", + "P=math.pi*dia #[W/sq m.K]\n", + "Q=(T0-T_inf)*math.sqrt(h*P*k*A) #Heat dissipated in [W]\n", + "\n", + "#Result\n", + "print\"Heat dissipated by the rod is\",round(Q,3),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.47,Page no:2.96" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss by the insulated rod is 26.6 W\n", + "Fin efficiency is 96.2 percent\n", + "Temperature at the end of the fin is 509.6 K\n" + ] + } + ], + "source": [ + "#Aluminium fin efficiency\n", + "#Variable declaration\n", + "k=200.0 #W/(m.K)\n", + "h=15.0 #W/(sq m.K)\n", + "T0=523.0 #[K]\n", + "T_inf=288.0 #[K]\n", + "theta_0=T0-T_inf \n", + "dia=25.0 #diameter[mm]\n", + "dia=dia/1000 #diameter[m]\n", + "r=dia/2 #radius in [m]\n", + "\n", + "#Calculation\n", + "import math\n", + "P=math.pi*dia #[m]\n", + "A=math.pi*r**2 #[sq m]\n", + "#For insulated fin:\n", + "m=math.sqrt(h*P/(k*A))\n", + "L=100.0 #length of rod in [mm]\n", + "L=L/1000 #length of rod in [m]\n", + "Q=theta_0*math.tanh(m*L)*math.sqrt(h*P*k*A) #Heat loss \n", + "nf=math.tanh(m*L)/(m*L) #Fin efficiency for insulated fin\n", + "#At the end of the fin: theta/theta_0=(cosh[m(L-x)]/cosh(mL))\n", + "#at x=L, theta/theta_0=1/(cosh(mL)\n", + "T=T_inf+(T0-T_inf)*(1/math.cosh(m*L)) #[K]\n", + "\n", + "\n", + "#Result\n", + "#ANSWER-1\n", + "print\"Heat loss by the insulated rod is\",round(Q,1),\"W\"\n", + "#ANSWER-2\n", + "print\"Fin efficiency is\",round(nf*100,1),\"percent\"\n", + "#ANSWER-3\n", + "print\"Temperature at the end of the fin is\",round(T,1),\"K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.49,Page no:2.98" + ] + }, + { + "cell_type": "code", + "execution_count": 101, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "As,Q for 10 fins of 60 mm length( 117.66 W) is more than Q for 6 fins of 100 mm length ( 113.75 W)\n", + "The agreement-->10 fins of 60 mm length is more effective\n" + ] + } + ], + "source": [ + "#Finding effective Pin fins\n", + "import math\n", + "#Variable declaration\n", + "k=300.0 #W/(m.K)\n", + "h=20.0 #W.(sq m.K)\n", + "P=0.05 #[m]\n", + "A=2.0 #[sq cm]\n", + "A=A/10000 #[sq m]\n", + "T0=503.0 #[K]\n", + "T_inf=303.0 #[K]\n", + "\n", + "\n", + "#Calculation\n", + "theta_0=T0-T_inf #[K]\n", + "import math\n", + "m=math.sqrt(h*P/(k*A))\n", + "\n", + "#CASE 1: 6 Fins of 100 mm length\n", + "L1=0.1 #Length of fin in [m]\n", + "Q=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L1) #[W]\n", + "#For 6 fins\n", + "Q=Q*6 #for 6 fins [W]\n", + "\n", + "#CASE 2: 10 fins of 60 mm length\n", + "L2=60.0 #[mm]\n", + "L2=L2/1000 #[m]\n", + "Q2=math.sqrt(h*P*k*A)*theta_0*math.tanh(m*L2) #[W]\n", + "Q2=Q2*10 #For 10 fins\n", + "\n", + "\n", + "\n", + "#Result\n", + "print\"As,Q for 10 fins of 60 mm length(\",round(Q2,2),\"W) is more than Q for 6 fins of 100 mm length (\",round(Q,2),\"W)\"\n", + "print\"The agreement-->10 fins of 60 mm length is more effective\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.50,Page no:2.98" + ] + }, + { + "cell_type": "code", + "execution_count": 103, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In oil side,Q= 9.75 W\n", + "In air side,Q= 35.56 W\n", + "From above results we see that more heat transfer takes place if fins are provided on the air side\n" + ] + } + ], + "source": [ + "#Metallic wall surrounded by oil and water \n", + "#Variable declaration\n", + "h_oil=180.0 #W/(sq m.K)\n", + "h_air=15.0 #W/(sq m.K)\n", + "T_oil=353.0 #[K]\n", + "T_air=293.0 #[K]\n", + "delta_T=T_oil-T_air #[K]\n", + "k=80.0 #Conductivity in [W/(m.K)]\n", + "for_section=11.0*10**-3 #[m]\n", + "L=25.0 #[mm]\n", + "L=L/1000 #[m]\n", + "W=1.0 #[m] Width,..let\n", + "t=1.0 #[mm] \n", + "t=t/1000 #[m]\n", + "A=W*t #[m]\n", + "P=2*t\n", + "Af=2*L*W #sq m\n", + "N=1.0\n", + "\n", + "#Calculation\n", + "\n", + "import math\n", + "Ab=for_section-A #[sq m]\n", + "#CASE 1: Fin on oil side only\n", + "m=math.sqrt(h_oil*P/(k*A)) \n", + "nf_oil=math.tanh(m*L)/(m*L)\n", + "Ae_oil=Ab+nf_oil*Af*N #[sq m]\n", + "Q1=delta_T/(1/(h_oil*Ae_oil)+1/(h_air*for_section)) #[W]\n", + "#CASE 2: Fin on air side only\n", + "m=math.sqrt(h_air*P/(k*A))\n", + "nf_air=math.tanh(m*L)/(m*L)\n", + "nf_air=0.928 #Approximation\n", + "Ae_air=Ab+nf_air*Af*N #[sq m]\n", + "Q=delta_T/(1.0/(h_oil*for_section)+1.0/(h_air*Ae_air)) #[W]\n", + "\n", + "#Result\n", + "print\"In oil side,Q=\",round(Q1,2),\"W\"\n", + "print\"In air side,Q=\",round(Q,2),\"W\"\n", + "print\"From above results we see that more heat transfer takes place if fins are provided on the air side\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:2.51,Page no:2.100" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Without fins,Q= 204.55 W\n", + "With fins on water side,Q= 219.24 W\n", + "With Fins on Air side,Q= 800.3 W\n", + "With Fins on both side,Q= 1084.7 W\n" + ] + } + ], + "source": [ + "#Brass wall having fins\n", + "\n", + "#Variable declaration\n", + "k=75.0 #Thermal conductivity [W/(m.K)]\n", + "T_water=363.0 #[K]\n", + "T_air=303.0 #[K] \n", + "dT=T_water-T_air #delta T\n", + "h1=150.0 # for water[W/(sq m.K)]\n", + "h2=15.0 #for air [W/(sq m.K)]\n", + "W=0.5 #Width of wall[m]\n", + "L=0.025 #[m]\n", + "\n", + "#Calculation\n", + "Area=W**2 #Base Area [sq m]\n", + "t=1.0 #[mm]\n", + "t=t/1000 #[m]\n", + "pitch=10.0 #[mm]\n", + "pitch=pitch/1000 #[m]\n", + "N=W/pitch #[No of fins]\n", + "\n", + "\n", + "#Calculations\n", + "A=N*W*t #Total cross-sectional area of fins in [sq m]\n", + "Ab=Area-A #[sq m]\n", + "Af=2*W*L #Surface area of fins [sq m]\n", + "from math import tanh\n", + "#CASE 1: HEAT TRANSFER WITHOUT FINS\n", + "A1=Area #[sq m]\n", + "A2=A1 #[sq m]\n", + "Q=dT/(1.0/(h1*A1)+1.0/(h2*A2)) #[W]\n", + "print\"Without fins,Q=\",round(Q,2),\"W\"\n", + "#CASE 2: Fins on the water side\n", + "P=2*(t+W) \n", + "A=0.5*10**-3 \n", + "m=math.sqrt(h1*P/(k*A))\n", + "nfw=math.tanh(m*L)/(m*L) #Effeciency on water side\n", + "Aew=Ab+nfw*Af*N #Effective area on the water side [sq m]\n", + "Q=dT/(1.0/(h1*Aew)+1.0/(h2*A2)) #[W]\n", + "print\"With fins on water side,Q=\",round(Q,2),\"W\"\n", + "#CASE 3: FINS ON THE AIR SIDE\n", + "m=math.sqrt(h2*P/(k*A))\n", + "nf_air=tanh(m*L)/(m*L) #Effeciency\n", + "Aea=Ab+nf_air*Af*N #Effective area on air side\n", + "Q=dT/(1.0/(h1*A1)+1.0/(h2*Aea)) #[W]\n", + "print\"With Fins on Air side,Q=\",round(Q,1),\"W\"\n", + "#BOTH SIDE:\n", + "Q=dT/(1.0/(h1*Aew)+1.0/(h2*Aea)) #[W]\n", + "\n", + "#Result\n", + "print\"With Fins on both side,Q=\",round(Q,1),\"W\" \n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_by_K._A._Gavhane/Chapter_3.ipynb b/Heat_Transfer_by_K._A._Gavhane/Chapter_3.ipynb new file mode 100644 index 00000000..b011e98e --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/Chapter_3.ipynb @@ -0,0 +1,2981 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter3: Convection" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.1,Page no:3.44" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Boundary layer thickness is del= 0.000734 m or 0.734 mm\n", + "Wrong units in answer of book,m and mm are wrongly interchanged\n" + ] + } + ], + "source": [ + "#Boundary layer thickness\n", + "\n", + "#Variable declaration\n", + "mu=10**-3 #N.s/m**2\n", + "#At distance y from surface\n", + "#ux=a+by+cy**2+dy**3\n", + "#At y=0,ux=0 therefore a=0\n", + "#i.e tao=0\n", + "#At edge of boundary layer,ie y=del\n", + "#ux=u_inf\n", + "#At y=o,c=0\n", + "#At y=del,ux=b*del+d*del**3\n", + "\n", + "#Therefore, b=-3*d*del**3\n", + "#d=-u_inf/(2*del**2)\n", + "#b=3*u_inf/(2*del)\n", + "\n", + "#For velocity profile,we have:\n", + "#del/x=4.64*(Nre_x)**(-1/2)\n", + "\n", + "#Evaluate N re_x\n", + "\n", + "x=75.0 #[mm]\n", + "x=x/1000 #[m]\n", + "u_inf=3.0 #[m/s]\n", + "rho=1000.0 #[kg/m**3] for air\n", + "\n", + "#Calculation\n", + "Nre_x=u_inf*rho*x/mu #Reynold number\n", + "#Substituting the value,we get\n", + "dell=x*4.64*(Nre_x**(-1.0/2.0)) #[m]\n", + "\n", + "#Result\n", + "print\"Boundary layer thickness is del=\",round(dell,6),\"m or \",round(dell*1000,3),\"mm\"\n", + "print\"Wrong units in answer of book,m and mm are wrongly interchanged\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.2,Page no:3.45" + ] + }, + { + "cell_type": "code", + "execution_count": 144, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Boundary layer thickness at the trailing edge is 18.0 mm\n" + ] + } + ], + "source": [ + "#Boundary layer thickness of plate\n", + "\n", + "#Variable declaration\n", + "mu=15*10**-6 #sq m /s\n", + "v=2 #m/s\n", + "L=2 #[m] length of plate\n", + "Nre_x=3*10**5\n", + "\n", + "#Calculation\n", + "\n", + "xc=Nre_x*mu/v #critical length at whihc the transition takes place\n", + "#Since xc is less than 2 m.Therefore the flow is laminar\n", + "#at any distance x,.it is calculated from\n", + "#del/x=4.64/(math.sqrt(NRe,x))\n", + "#At x=L=2 m\n", + "Nre_l=v*L/mu\n", + "del_l=4.64*L/math.sqrt(Nre_l)\n", + "del_l=del_l*1000 #[mm]\n", + "\n", + "#Result\n", + "print\"Boundary layer thickness at the trailing edge is\",round(del_l),\"mm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.3,Page no:3.45" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Since Nre,x( 80000.0 )is Less than 3*10**5,..the boundary layer is laminar\n", + "Thickness of boundary layer at x= 0.4 m 6.56 mm\n", + "Local skin friction coefficient is : 0.0023\n" + ] + } + ], + "source": [ + "#Thickness of hydrodynamic boundary layer\n", + "\n", + "#Variable declaration\n", + "mu=15*10**-6 #Kinematic viscosity in [sq m /s]\n", + "x=0.4 #[m]\n", + "u_inf=3 #[m/s]\n", + "\n", + "\n", + "#Calculation\n", + "#At x=0.4 m,\n", + "Nre_x=u_inf*x/mu \n", + "import math\n", + "dell=4.64*x/math.sqrt(Nre_x) #[m]\n", + "dell=dell*1000 #[mm]\n", + "Cf_x=0.664/math.sqrt(Nre_x) \n", + "\n", + "#Result\n", + "print\"Since Nre,x(\",Nre_x,\")is Less than 3*10**5,..the boundary layer is laminar\"\n", + "print\"Thickness of boundary layer at x=\",x,\"m\",round(dell,2),\"mm\" \n", + "print\"Local skin friction coefficient is :\",round(Cf_x,4)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.4,Page no:3.46" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Boundary layer thickness at distance 20 cm from leading edge is 0.0058 m= 5.8 mm\n", + "Boundary layer thickness at distance 40 cm from leading edge is 0.0082 m= 8.2 mm\n", + "Thus,Mass flow rate entering the boundary layer is 0.003547 kg/s\n" + ] + } + ], + "source": [ + "#Flat plate boundary layer\n", + "\n", + "import math\n", + "from scipy import integrate\n", + "\n", + "#Variable declaration\n", + "mu=1.85*10**-5 #[kg/(m.s)]\n", + "P=101.325 #Pressure in [kPa]\n", + "M_avg=29.0 #Avg molecular wt of air\n", + "R=8.31451 #Gas constant\n", + "T=300.0 #[K]\n", + "\n", + "#Calculation\n", + "rho=P*M_avg/(R*T) #[kg/m**3]\n", + "u_inf=2.0 #Viscosity in [m/s]\n", + "#At x=20 cm =0.2 m\n", + "x=0.2 #[m]\n", + "Nre_x=rho*u_inf*x/mu #[Reynolds number]\n", + "del_by_x=4.64/math.sqrt(Nre_x) #[Boundary layer]\n", + "dell=del_by_x*x #[m]\n", + "#del=del*1000 #[mm]\n", + "\n", + "#At\n", + "x=0.4 #[m]\n", + "Nre_x=(rho*u_inf*x)/mu #<3*10**5\n", + "#Boundary layer is laminar\n", + "del_by_x=4.64/math.sqrt(Nre_x) \n", + "del1=del_by_x*x #[m]\n", + "#del1=del1*1000 #[mm]\n", + "d=del1-dell #Del \n", + "def f(y):\n", + " m_dot=u_inf*(1.5*(y/d)-0.5*(y/d)**3)*rho\n", + " return(m_dot)\n", + "m_dot=integrate.quad(f,0,d)\n", + "\n", + "\n", + "#Result\n", + "print\"Boundary layer thickness at distance 20 cm from leading edge is\",round(dell,4),\"m=\",round(dell*1000,1),\"mm\"\n", + "print\"Boundary layer thickness at distance 40 cm from leading edge is\",round(del1,4),\"m=\",round(del1*1000,1),\"mm\"\n", + "print\"Thus,Mass flow rate entering the boundary layer is\",round(m_dot[0],6),\"kg/s\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.5,Page no:3.47" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat removal is 157.9 W\n", + "315.8 W heat should be removed continously from the plate\n" + ] + } + ], + "source": [ + "#Rate of heat removed from plate\n", + "\n", + "#Variable declaration\n", + "mu=3.9*10**-4 #Kinematic viscosity in sq m/s\n", + "k=36.4*10**-3 #Thermal conductivity in W/(m.K)\n", + "Npr=0.69\n", + "u_inf=8 #[m/s]\n", + "L=1 #Lenght of plate in [m]\n", + "import math\n", + "\n", + "#Calculation\n", + "Nre_l=u_inf*L/mu \n", + "#Since Nre_l is less than 3*10**5 ,the flow is laminar over the entire length of plate\n", + "Nnu=0.664*math.sqrt(Nre_l)*Npr**(1.0/3.0) #=hL/k\n", + "h=k*Nnu/L #w/sq m.K\n", + "h=3.06 #Approximation [W/sq m.K]\n", + "T_inf=523 #[K]\n", + "Tw=351 #[K]\n", + "W=0.3 #Width of plate [m]\n", + "A=W*L #Area in [sq m]\n", + "Q=h*A*(T_inf-Tw) # Rate of heat removal from one side in [W]\n", + "#from two side:\n", + "Q2=2*Q #[W]\n", + "\n", + "#Result\n", + "print\"Rate of heat removal is\",round(Q,1),\"W\"\n", + "print round(Q2,1),\" W heat should be removed continously from the plate\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.6,Page no:3.48" + ] + }, + { + "cell_type": "code", + "execution_count": 158, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat removed from both sides of plate is 382.5 W\n" + ] + } + ], + "source": [ + "#Heat removed from plate\n", + "\n", + "#Variable declaration\n", + "P1=101.325 #Pressure in [kPa]\n", + "mu1=30.8*10**-6 #Kinematic viscosity in[sq m /s]\n", + "k=36.4*10**-3 #[W/(m.K)]\n", + "Npr=0.69\n", + "u_inf=8 #Velocity in [m/s]\n", + "Cp=1.08 #kJ/(kg.K)\n", + "L=1.5 #Length of plate in [m]\n", + "W=0.3 #Width in [m]\n", + "A=L*W #Area in [sq m]\n", + "\n", + "#Calculation\n", + "import math\n", + "#At constant temperature: mu1/mu2=P2/P1\n", + "P2=8 #[kPa]\n", + "mu2=mu1*P1/P2 #Kinematic viscosity at P2 in [sq m/s]\n", + "Nre_l=u_inf*L/mu2 #Reynold's no.\n", + "#Since this is less than 3*10**5\n", + "Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))\n", + "h=Nnu*k/L # Heat transfer coeffficient in [W/sq m.K]\n", + "h=round(h,1)\n", + "\n", + "T_inf=523 #[K]\n", + "Tw=353 #[K]\n", + "Q=2*h*A*(T_inf-Tw) #Heat removed from both sides in [W]\n", + "\n", + "#Result\n", + "print\"Rate of heat removed from both sides of plate is\",Q,\"W\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.7,Page no:3.49" + ] + }, + { + "cell_type": "code", + "execution_count": 160, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transferred from both sides of the plate is 1450.0 W\n" + ] + } + ], + "source": [ + "#Local heat transfer coefficient\n", + "\n", + "#Variable declaration\n", + "rho=0.998 #kg/cubic m\n", + "v=20.76*10**-6 #[sq m/s]\n", + "Cp=1.009 #[kJ/kg.K]\n", + "k=0.03 #[W/m.K]\n", + "u_inf=3 #[m/s]\n", + "x=0.4 #[m]\n", + "w=1.5 #[m]\n", + "\n", + "#Calculation\n", + "Nre_x=u_inf*x/v #Reynolds no at x=0.4 m\n", + "#Since this is less than 3*10**5.The flow is laminar upto x=0.4 m\n", + "mu=rho*v #[kg/(m.s)]\n", + "import math\n", + "Cp=1.009 #[kJ/kg.K]\n", + "Cp=Cp*1000 #[J/kg.K]\n", + "k=0.03 #W/(m.K)\n", + "Npr=Cp*mu/k\n", + "Nnu_x=0.332*(math.sqrt(Nre_x))*(Npr**(1.0/3.0))\n", + "hx=Nnu_x*k/x #[W/(m.K)]\n", + "#Average value is twice this value\n", + "h=round(2*hx,1) #[W/(m.K)]\n", + "A=x*w #Area in [sq m]\n", + "Tw=407 #[k]\n", + "T_inf=293 #[K]\n", + "Q=h*A*(Tw-T_inf) #[W]\n", + "#From both sides of the plate:\n", + "Q=2*Q #[W]\n", + "#Result\n", + "print\"The heat transferred from both sides of the plate is\",round(Q),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.8,Page no:3.50" + ] + }, + { + "cell_type": "code", + "execution_count": 161, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Width of plate is 3.0 m\n" + ] + } + ], + "source": [ + "#Width of plate\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=0.998 #[kg/cubic m]\n", + "v=20.76*10**-6 #[sq m/s]\n", + "k=0.03 #[W/m.K]\n", + "Npr=0.697\n", + "x=0.4 #[m] from leading edge of the plate\n", + "u_inf=3 #[m/s]\n", + "\n", + "#Calculation\n", + "Nre_x=u_inf*x/v #Reynold numebr at x=0.40 m\n", + "#Since this is less than 3*10**5 \n", + "#therefore flow is laminar and \n", + "Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0)) \n", + "hx=Nnu_x*k/x #[W/sq m.K]\n", + "#Average heat tarnsfer coefficient is twice this value\n", + "h=2*hx #[W/sq m.K]\n", + "#Given:\n", + "Q=1450 #[W]\n", + "Tw=407 #[K]\n", + "T_inf=293 #[K]\n", + "L=0.4 #[m]\n", + "#Q=h*w*L*(Tw-T_inf)\n", + "#L=Q/(h*w*(Tw-T_inf))\n", + "w=Q/(h*L*(Tw-T_inf)) #[m]\n", + "\n", + "#Result\n", + "print\"Width of plate is\",round(w),\"m\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.9,Page no:3.51" + ] + }, + { + "cell_type": "code", + "execution_count": 163, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flow is : 81.18 W\n", + "Heat flow from both sides of plate is 162.36 W\n" + ] + } + ], + "source": [ + "#Heat transferred in flat plate\n", + "\n", + "#Variable declaration\n", + "v=17.36*10**-6 #Viscosity for air [sq m./s]\n", + "k=0.0275 #for air ..[W/(m.K)]\n", + "Cp=1.006 #[kJ/(kg.K)]\n", + "Npr=0.7 #for air\n", + "u_inf=2 #[m/s]\n", + "x=0.2 #[m]\n", + "\n", + "#Calculation\n", + "Nre_x=u_inf*x/v #Reynolds number at x=0.2 m\n", + "#Since this is less than 3*10**5\n", + "import math\n", + "Nnu_x=0.332*math.sqrt(Nre_x)*(Npr**(1.0/3.0))\n", + "hx=Nnu_x*k/x #[W/(sq m.K]\n", + "#Average value of heat transfer coeff is twice this value\n", + "h=round(2*hx,1) #[W/sq m.K)]\n", + "w=1 #width in [m]\n", + "A=x*w #[sq m] Area of plate\n", + "Tw=333 #[K]\n", + "T_inf=300 #[K]\n", + "Q=h*A*(Tw-T_inf) #Heat flow in [W]\n", + "\n", + "#Result\n", + "print\"Heat flow is :\",Q,\"W\"\n", + "#From both sides of plate:\n", + "Q=2*Q #[W]\n", + "print\"Heat flow from both sides of plate is\",Q,\"W\"\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.10,Page no:3.51" + ] + }, + { + "cell_type": "code", + "execution_count": 165, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer from both sides of plate is 109.4 W\n", + "\n", + "With turbulent boundary layer,\n", + "Rate of heat transfer from both sides of the plate= 226.4\n", + "\n", + "These calculations show that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \n", + "\n" + ] + } + ], + "source": [ + "#Rate of heat transferred in turbulent flow\n", + "\n", + "#Variable declaration\n", + "v=16.96*10**-6 #[sq m./s]\n", + "rho=1.128 #[kg/cubic m]\n", + "Npr=0.699 #Prandtl number\n", + "k=0.0276 #[W/m.K]\n", + "u_inf=15 #[m/s]\n", + "L=0.2 #[m]\n", + "\n", + "#Calculation\n", + "Nre_l=L*u_inf/v #Reynold's number\n", + "import math\n", + "\n", + "#Since this is less than 3*10**5,the boundary layer is laminar over entire length\n", + "Nnu=0.664*math.sqrt(Nre_l)*(Npr**(1.0/3.0))\n", + "h=Nnu*k/L #[W/sq m.K]\n", + "A=L**2 #Area in [sq m]\n", + "Tw=293 #[K]\n", + "T_inf=333 #[K]\n", + "#Rate of heat transfer from BOTH sides is:\n", + "Q=2*h*A*(T_inf-Tw) #[W]\n", + "print\"Rate of heat transfer from both sides of plate is\",round(Q,1),\"W\\n\"\n", + "#ii-With turbulent boundary layer from the leading edge:\n", + "h=k*0.0366*(Nre_l**(0.8))*(Npr**(1.0/3.0))/L #[W/(sq m.K)]\n", + "#Heat transfer from both sides is :\n", + "Q=2*h*A*(T_inf-Tw) #[W]\n", + "print \"With turbulent boundary layer,\\nRate of heat transfer from both sides of the plate=\",round(Q,1)\n", + "print\"\\nThese calculations show that the that transfer rate is approximately doubled if boundary layer is turbulent from the leading edge \\n\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.11,Page no:3.52" + ] + }, + { + "cell_type": "code", + "execution_count": 169, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer from the plate is 3179.2 W(approx)\n" + ] + } + ], + "source": [ + "#Heat transfer from plate in unit direction\n", + "\n", + "#Variable declaration\n", + "mu=1.906*10**-5 #[kg/(m.s)]\n", + "k=0.02723 #W/m.K\n", + "Cp=1.007 #[kJ/(kg.K)]\n", + "rho=1.129 #[kg/cubic m]\n", + "Npr=0.70\n", + "Mavg=29\n", + "u_inf=35 #[m/s]\n", + "L=0.75 #[m]\n", + "Tm=313 #[K]\n", + "P=101.325 #[kPa]\n", + "\n", + "#Calculation\n", + "Nre_l=rho*u_inf*L/mu #Reynold's number >5*10**5\n", + "Nnu=0.0366*Nre_l**(0.8)*Npr**(1.0/3.0) \n", + "h=Nnu*k/L #[W/s m.K]\n", + "A=1*L #[sq m]\n", + "Tw=333 #[K]\n", + "T_inf=293 #[K]\n", + "Q=h*A*(Tw-T_inf) #[W]\n", + "\n", + "#Result\n", + "print\"Heat transfer from the plate is\",round(Q,1),\"W(approx)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.12,Page no:3.53" + ] + }, + { + "cell_type": "code", + "execution_count": 170, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by sphere is 2.21 W\n" + ] + } + ], + "source": [ + "#Heat lost by sphere\n", + "import math\n", + "\n", + "#Variable declaration\n", + "v=18.23*10**-6 #sq m/s\n", + "k=0.02814 #[W/m.K]\n", + "D=0.012 #[m]\n", + "r=0.006 #[m]\n", + "u_inf=4 #[m/s]\n", + "\n", + "#Calculation\n", + "Nre=D*u_inf/v #Reynold's number\n", + "Nnu=0.37*Nre**(0.6) \n", + "h=Nnu*(k/D) \n", + "A=4*math.pi*r**2 #Area of sphere in [sq m]\n", + "Tw=350 #[K]\n", + "T_inf=300 #[K]\n", + "Q=h*A*(Tw-T_inf) #Heat lost by sphere in [W]\n", + "\n", + "#Result\n", + "print\"Heat lost by sphere is\",round(Q,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.13,Page no:3.53" + ] + }, + { + "cell_type": "code", + "execution_count": 171, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Heat lost by the sphere is 1.554 W\n" + ] + } + ], + "source": [ + "#Heat lost by sphere\n", + "\n", + "#Variable declaration\n", + "v=15.69*10**-6 #[sq m./s]\n", + "k=0.02624 #[W/m.K]\n", + "Npr=0.708 #Prandtl number\n", + "mu=2.075*10**-5 #kg/m.s\n", + "u_inf=4 #[m/s]\n", + "mu_inf=1.8462*10**-5 #[m/s] velocity\n", + "Tw=350 #[K]\n", + "T_inf=300 #[K]\n", + "D=0.012 #[m]\n", + "r=D/2 #Radius in [m]\n", + "\n", + "#Calculation\n", + "Nre=u_inf*D/v #Reynold's numbe\n", + "Nnu=2+(0.4*Nre**(1.0/2.0)+0.06*Nre**(2.0/3.0))*Npr**(0.4)*(mu_inf/mu)**(1.0/4.0)\n", + "h=Nnu*k/D #[W/sq m.K]\n", + "import math\n", + "A=4*math.pi*r**2 #Area in [sq m]\n", + "Q=h*A*(Tw-T_inf) \n", + "\n", + "#Result\n", + "print\"\\n Heat lost by the sphere is\",round(Q,3),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.14,Page no:3.54" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer rate is 12.1 W,And percentage of power lost by convection is: 12.1 %\n" + ] + } + ], + "source": [ + "#Percent power lost in bulb\n", + "#Variable declaration\n", + "v=2.08*10**-5 #[sq m/s]\n", + "k=0.03 #W/(m.K)\n", + "Npr=0.697 #Prandtl number\n", + "D=0.06 #[m]\n", + "u_inf=0.3 #[m/s]\n", + "\n", + "#Calculation\n", + "Nre=D*u_inf/v #Reynolds number\n", + "#Average nusselt number is given by:\n", + "Nnu=0.37*(Nre**0.6) \n", + "h=Nnu*k/D #W/sq m.K\n", + "Tw=400 #[K]\n", + "T_inf=300 #[K]\n", + "D=0.06 #[m]\n", + "r=0.03 #[m]\n", + "import math\n", + "A=4*math.pi*r**2 #Area in [sq m]\n", + "Q=h*A*(Tw-T_inf) #[W]\n", + "per=Q*100/100 #Percent of heat lost by forced convection\n", + "\n", + "#Result\n", + "print\"Heat transfer rate is\",round(Q,2),\"W,And percentage of power lost by convection is:\",round(per,2),\"%\" \n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.15,Page no:3.55" + ] + }, + { + "cell_type": "code", + "execution_count": 181, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost per unit length of cylinder is 3102.0 W\n" + ] + } + ], + "source": [ + " \n", + "#Heat lost by cylinder \n", + "u_inf=50 #velocity in [m/s]\n", + "mu=2.14*10**-5 #[kg/(m.s)]\n", + "rho=0.966 #[kg/cubic m]\n", + "k=0.0312 #[W/(m.K)]\n", + "Npr=0.695 #Prandtl number\n", + "D=0.05 #Diameter in [m]\n", + "Nre=D*u_inf*rho/mu #Reynold's number\n", + "Nnu=0.0266*Nre**0.805*Npr**(1.0/3.0) \n", + "h=round(Nnu*k/D,1) #W/sq m.K\n", + "Tw=423 #[K]\n", + "T_inf=308 #[K]\n", + "import math\n", + "#Heat loss per unit length is :\n", + "Q_by_l=h*math.pi*D*(Tw-T_inf) #[W]\n", + "\n", + "#Result\n", + "print\"Heat lost per unit length of cylinder is\",round(Q_by_l),\"W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.16,Page no:3.55" + ] + }, + { + "cell_type": "code", + "execution_count": 186, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat flow from the circular pipe is 1464.2 W/m\n", + "Rate of heat flow from the square pipe= 1711.2 W/m\n", + "Hence rate of heat flow from square pipe is more than that from circular pipe\n" + ] + } + ], + "source": [ + "#Heat transfer in tube\n", + "\n", + "#Variable declaration\n", + "v=20.92*10**-6 #sq m/s\n", + "k=3.0*10.0**-2 #W/(m.K)\n", + "Npr=0.7\n", + "u_inf=25.0 #[m/s]\n", + "d=50.0 #[mm]\n", + "d=d/1000 #[m]\n", + "Nre=u_inf*d/v #Reynold's number\n", + "Tw=397.0 #[K]\n", + "T_inf=303.0 #[K]\n", + "\n", + "\n", + "#Calculation\n", + "import math\n", + "#Case 1: Circular tube\n", + "Nnu=0.0266*Nre**(0.805)*Npr**(1.0/3.0) \n", + "h=Nnu*k/d #[W/sq m.K]\n", + "A=math.pi*d #Area in [sq m]\n", + "Q=h*A*(Tw-T_inf) #[W]\n", + "Q_by_l1=h*math.pi*d*(Tw-T_inf) #[W/m]\n", + "\n", + "#Case 2:Square tube\n", + "A=50.0*50.0 #Area in [sq mm]\n", + "P=2.0*(50.0+50.0) #Perimeter [mm]\n", + "l=4.0*A/P #[mm]\n", + "l=l/1000 #[m]\n", + "Nnu=0.102*(Nre**0.675)*(Npr**(1.0/3.0))\n", + "h=Nnu*k/d #W/(sq m.K)\n", + "A=4*l*l #[sq m]\n", + "\n", + "Q=h*A*(Tw-T_inf)\n", + "Q_by_l2=Q/l #[W/m]\n", + "\n", + "\n", + "#Result\n", + "\n", + "print\"Rate of heat flow from the circular pipe is\",round(Q_by_l1,1),\"W/m\" \n", + "print\"Rate of heat flow from the square pipe=\",round(Q_by_l2,1),\"W/m\"\n", + "print\"Hence rate of heat flow from square pipe is more than that from circular pipe\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.17,Page no:3.63" + ] + }, + { + "cell_type": "code", + "execution_count": 188, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inside heat transfer coefficient is 225.0 W/(sq m.K)\n", + "NOTE:The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\n", + "\n" + ] + } + ], + "source": [ + "\n", + "#Heat transfer coefficient\n", + "\n", + "#Variable declaration\n", + "mu=0.8 #Viscosity of flowing fluid [N.s/sq m]\n", + "rho=1.1 #Density of flowinf fluid [g/cubic cm]\n", + "rho=rho*1000 #Density in [kg/cubic m]\n", + "Cp=1.26 #Specific heat [kJ/kg.K]\n", + "Cp=Cp*10**3 # in[J/(kg.K)]\n", + "k=0.384 #[W/(m.K)]\n", + "mu_w=1.0 #Viscosity at wall temperature [N.s/sq m]\n", + "L=5.0 #[m]\n", + "vfr=300.0 #Volumetric flow rate in [cubic cm/s]\n", + "vfr=vfr*10.0**-6 #[cubic m/s]\n", + "mfr=vfr*rho #Mass flow rate of flowinf fluid [kg/s]\n", + "Di=20.0 #Inside diameter in[mm]\n", + "Di=Di/1000 #[m]\n", + "\n", + "\n", + "#Calculation\n", + "import math\n", + "Area=(math.pi/4)*Di**2 #Area of cross-section [sq m]\n", + "u=vfr/Area #Veloctiy in [m/s]\n", + "Nre=Di*u*rho/mu #Reynold's number\n", + "#As reynold's number is less than 2100,he flow is laminar\n", + "Npr=Cp*mu/k #Prandtl number\n", + "Nnu=1.86*(Nre*Npr*Di/L)**(1.0/3.0)*(mu/mu_w)**(0.14)\n", + "hi=Nnu*k/Di #inside heat transfer coefficient [W/sq m.K]\n", + "\n", + "#Result\n", + "print\"Inside heat transfer coefficient is\",round(hi),\"W/(sq m.K)\"\n", + "#Note:\n", + "print\"NOTE:The answer given in book..ie 1225 is wrong.please redo the calculation of last line manually to check\\n\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.18,Page no:3.64" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inside heat transfer coefficient is 1225.4 W/sq m.K\n" + ] + } + ], + "source": [ + "#Heat transfer coefficient in heated tube\n", + "\n", + "#Variable declaration\n", + "m=5500.0 #Mass flow rate in [kg/h]\n", + "m=m/3600.0 #[kg/s]\n", + "rho=1.07 #Density of fluid in [g/cm**3]\n", + "rho=rho*1000 #[kg/m**3]\n", + "vfr=m/rho #Volumetric flow rate in [m**3/s]\n", + "Di=40.0 #Diameter of tube [mm]\n", + "Di=Di/1000 #[m]\n", + "import math\n", + "\n", + "#Calculation\n", + "A=(math.pi/4)*Di**2 #Area of cross-section in [sq m]\n", + "u=vfr/A #Velocity of flowing fluid [m/s]\n", + "rho=1070.0 #Density in [kg/m**3]\n", + "mu=0.004 #Viscosity in [kg/m.s]\n", + "Nre=Di*u*rho/mu\n", + "Nre=12198.0 #Approx\n", + "#Since this reynold's number is less than 10000,the flow is turbulent\n", + "Cp=2.72 #Specific heat in [kJ/kg.K]\n", + "Cp=Cp*10**3 #Specific heat in [J/kg.K]\n", + "k=0.256 #thermal conductivity in [W/m.K]\n", + "Npr=Cp*mu/k #Prandtl number\n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.4) #Nusselt number\n", + "hi=k*Nnu/Di #Inside heat transfer coefficient in [W/m**2.K]\n", + "\n", + "#Result\n", + "print\"Inside heat transfer coefficient is \",round(hi,1),\"W/sq m.K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.19,Page no:3.66" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ANSWER-(i) \n", + "By Dittus-Boelter equation we get h= 12972.6 W/sq m.K\n", + "Answer-(ii)\n", + "By Sieder-Tate equation we get h= 12315.04 W/sq m.K\n", + "\n", + "NOTE:Calculation mistake in book in part 2 ie sieder tate eqn\n" + ] + } + ], + "source": [ + "#h of water flowing in tube\n", + "\n", + "#Variable declaration\n", + "rho=984.1 #Density of water [kg/m**3]\n", + "Cp=4187.0 #Specific heat in [J/kg.K]\n", + "mu=485.0*10**-6 #Viscosity at 331 K[Pa.s]\n", + "k=0.657 #[W/(m.K)]\n", + "mu_w=920.0*10**-6 #Viscosity at 297 K [Pa.s]\n", + "\n", + "\n", + "#Calculation\n", + "D=16.0 #Diameter in [mm]\n", + "D=D/1000 #Diameter in [m]\n", + "u=3.0 #Velocity in [m/s]\n", + "rho=984.1 #[kg/m**3]\n", + "Nre=D*u*rho/mu #Reynolds number\n", + "Nre=round(Nre)\n", + "Npr=Cp*mu/k #Prandtl number\n", + "\n", + "#Dittus-Boelter equation (i)\n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.3) #nusselt number\n", + "h=k*Nnu/D #Heat transfer coefficient [W/m**2.K]\n", + "\n", + "#Result\n", + "print\"ANSWER-(i) \\nBy Dittus-Boelter equation we get h=\",round(h,1),\"W/sq m.K\"\n", + "#sieder-tate equation (ii)\n", + "Nnu=0.023*(Nre**0.8)*(Npr**(1.0/3.0))*((mu/mu_w)**0.14) #Nusselt number\n", + "h=k*Nnu/D #Heat transfer coefficient in [W/sq m.K]\n", + "print\"Answer-(ii)\\nBy Sieder-Tate equation we get h=\",round(h,2),\"W/sq m.K\"\n", + "print\"\\nNOTE:Calculation mistake in book in part 2 ie sieder tate eqn\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.20,Page no:3.67" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer area is: 358.4 m**2\n" + ] + } + ], + "source": [ + "#Overall heat transfer coefficient\n", + "\n", + "#Variable declaration\n", + "m_dot=2250 #Mass flow arte in [kg/h]\n", + "Cp=3.35 #Specific heat in [kJ/(kg.K)]\n", + "dT=316-288.5 #Temperature drop for oil [K]\n", + "Q=Cp*m_dot*dT #Rate of heat transfer in [kJ/h]\n", + "Q=round(Q*1000/3600) #[J/s] or[W]\n", + "Di=0.04 #Inside diameter [m]\n", + "Do=0.048 #Outside diamter in [m]\n", + "hi=4070 #for steam [W/sq m.K]\n", + "ho=18.26 #For oil [W/sq m.K]\n", + "Rdo=0.123 #[sq m.K/W]\n", + "Rdi=0.215 #[sq m.K/W]\n", + "\n", + "#Calculation\n", + "Uo=1.0/(1.0/ho+Do/(hi*Di)+Rdo+Rdi*(Do/Di)) #[W/m**2.K]\n", + "Uo=2.3\n", + "import math\n", + "dT1=373-288.5 #[K]\n", + "dT2=373-316 #[K]\n", + "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Ao=Q/(Uo*dTm) #Heat transfer area in [m**2]\n", + "\n", + "#Result\n", + "print\"Heat transfer area is:\",round(Ao,1),\"m**2\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.21,Page no:3.68" + ] + }, + { + "cell_type": "code", + "execution_count": 123, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No. of tubes required= 1.0\n", + "\n", + "NOTE: there is an error in book in calculation of dT1 and dT2,\n", + "373-293 is written as 90,instead of 80...similarly in dT2,\n", + "So,in compliance with the book,10 is added to both of them\n" + ] + } + ], + "source": [ + "#Number of tubes in exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "k_tube=111.65 #[W/m.K]\n", + "W=4500.0 #[kg/h]\n", + "rho=995.7 #[kg/sq m]\n", + "Cp=4.174 #[kJ/(kg.K)]\n", + "k=0.617 #[W/(m.K)]\n", + "v=0.659*10**-6 #Kinematic viscosity [sq m/s]\n", + "m_dot=1720.0 #kg/h\n", + "T1=293.0 #Initial temperature in [K]\n", + "T2=318.0 #Final temperature in [K]\n", + "\n", + "\n", + "#Calculation\n", + "dT=T2-T1 #[K]\n", + "Q=m_dot*Cp*dT #Heat transfer rate in [kJ/h]\n", + "Q=Q*1000.0/3600.0 #[J/s] or [W]\n", + "Di=0.0225 #[m]\n", + "u=1.2 #[m/s]\n", + "#Nre=Di*u*rho/mu or\n", + "Nre=Di*u/v #Reynolds number\n", + "#As Nre is greater than 10000,Dittus Boelter equation is applicable\n", + "Cp=Cp*10**3 #J/(kg.K)\n", + "mu=v*rho #[kg/(m.s)]\n", + "Npr=Cp*mu/k #Prandtl number\n", + "#Dittus-Boelter equation for heating is \n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.4)\n", + "hi=k*Nnu/Di #Heat transfer coefficient [W/(sq m.K)]\n", + "Do=0.025 #[m]\n", + "Dw=(Do-Di)/math.log(Do/Di) #math.log mean diameter in [m]\n", + "ho=4650.0 #[W/sq m.K]\n", + "k=111.65 #[W/m.K]\n", + "xw=(Do-Di)/2 #[m]\n", + "Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw)) #Overall heat transfer coefficient in W/(m**2.K)\n", + "T_steam=373.0 #Temperature of condensing steam in [K]\n", + "dT1=T_steam-T1+10 #[K]\n", + "dT2=T_steam-T2+10 #[K]\n", + "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Ao=Q/(Uo*dTm)#Area in [m**2]\n", + "L=4.0 #length of tube [m]\n", + "n=Ao/(math.pi*Do*L) #number of tubes\n", + "\n", + "\n", + "#Result\n", + "print\"No. of tubes required=\",round(n) \n", + "print\"\\nNOTE: there is an error in book in calculation of dT1 and dT2,\\n373-293 is written as 90,instead of 80...similarly in dT2,\\nSo,in compliance with the book,10 is added to both of them\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.22,Page no:3.71" + ] + }, + { + "cell_type": "code", + "execution_count": 196, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average value of convective film coefficient is hi= 11584.0 W/sq m.K\n", + "Heat transferred per unit length is Q/L= 18.2 kW/m\n" + ] + } + ], + "source": [ + "#Convective film coefficient\n", + "import math\n", + "\n", + "#Variable declaration\n", + "m_dot=25000 #massflow rate of water [kg/h]\n", + "rho=992.2 #[kg/m**3]\n", + "k=0.634 #[W/m.K]\n", + "vfr=m_dot/rho #[m**3/h]\n", + "Npr=4.31 #Prandtl numberl\n", + "Di=50 #[mm]\n", + "Di=0.05 #[m]\n", + "dT=10 #[K] as the wall is at a temperature of 10 K above the bulk temperature\n", + "\n", + "#Calculation\n", + "u=round((vfr/3600)/(math.pi*(Di/2)**2),2) #Velocity of water in [m/s]\n", + "#Nre=Di*u*rho/mu=Di*u/v as v=mu/rho\n", + "v=0.659*10**-6 #[m**2/s]\n", + "Nre=Di*u/v #Reynolds number\n", + "#As it is less than 10000,the flow is in the turbulent region for heat transfer and Dittus Boelter eqn is used\n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.4) #Nusselt number\n", + "hi=Nnu*k/Di #Heat transfer coefficiet in [W/sq m.K]\n", + "q_by_l=hi*math.pi*Di*dT #Heat transfer per unit length[kW/m]\n", + "\n", + "#Result\n", + "print\"Average value of convective film coefficient is hi=\",round(hi),\" W/sq m.K\"\n", + "print\"Heat transferred per unit length is Q/L=\",round(q_by_l/1000,1),\"kW/m\"\n", + " \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.23,Page no:3.72" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of tube required is 5.0 m\n" + ] + } + ], + "source": [ + "#Length of tube\n", + "import math\n", + "\n", + "#Variable declaration\n", + "vfr=1200.0 #Water flow rate in [l/h]\n", + "rho=0.98 #Density of water in g/[cubic cm]\n", + "m_dot=vfr*rho #Mass flow rate of water [kg/h]\n", + "m_dot2=m_dot/3600.0 #[kg/s]\n", + "Cp=4.187*10**3 #[J/kg.K]\n", + "Di=0.025 #Diameter in [m]\n", + "mu=0.0006 #[kg/(m.s)]\n", + "\n", + "#Calculation\n", + "Ai=math.pi*((Di/2)**2) #Area of cross-section in [m**2]\n", + "Nre=(Di/mu)*(m_dot2/Ai) #Reynolds number\n", + "k=0.63 #for metal wall in [W/(m.K)]\n", + "Npr=Cp*mu/k #Prandtl number\n", + "#Since Nre>10000\n", + "#therefore ,Dittus boelter eqn for heating is \n", + "Nnu=0.023*(Nre**(0.8))*(Npr**(0.4))\n", + "ho=5800.0 #Film heat coefficientW/(m**2.K)\n", + "hi=Nnu*k/Di #Heat transfer coeffcient in [W/(sq m.K)]\n", + "Do=0.028 #[m]\n", + "Di=0.025 #[m]\n", + "xw=(Do-Di)/2 #[m]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "k=50.0 #for metal wall in [W/(m.K)]\n", + "Uo=1.0/(1.0/ho+Do/(hi*Di)+xw*Do/(k*Dw)) #in [W/sq m.K]\n", + "dT=343.0-303.0 #[K]\n", + "dT1=393.0-303.0 #[K]\n", + "dT2=393.0-343.0 #[K]\n", + "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Cp=Cp/1000.0 #[in [kJ/kg.K]]\n", + "Q=m_dot*Cp*dT #Rate of heat transfer in [kJ/h]\n", + "Q=Q*1000.0/3600.0 #[J/s] or [W]\n", + "Ao=Q/(Uo*dTm) #Heat transfer area in [sq m]\n", + "#Also,..Ao=math.pi*Do*L ..implies that\n", + "L=Ao/(math.pi*Do) #[m]\n", + "\n", + "#Result\n", + "print\"Length of tube required is\",round(L),\"m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.24,Page no:3.73" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Outlet temperature of water is 357.5 K\n" + ] + } + ], + "source": [ + "#Cooling coil\n", + "#1.For initial conditions:\n", + "import math\n", + "from scipy.optimize import fsolve\n", + "\n", + "#Variable declaration\n", + "T=360 #[K]\n", + "T1=280 #[K]\n", + "T2=320 #[K]\n", + "dT1=T-T1 #[K]\n", + "dT2=T-T2 #[K]\n", + "\n", + "#Calculation\n", + "#Q1=m1_dot*Cp1*(T2-T1)\n", + "Cp1=4.187 #Heat capacity \n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "m1_by_UA=dTlm/(Cp1*(T2-T1)) \n", + "#For final conditions :\n", + "#m2_dot=m1_dot\n", + "#U2=U1\n", + "#A2=5*A1\n", + "def f(t):\n", + " x=m1_by_UA*Cp1*(t-T1)-5*((dT1-(T-t))/math.log(dT1/(T-t)))\n", + " return(x)\n", + "T=fsolve(f,350.5)\n", + "\n", + "#Result\n", + "print\"Outlet temperature of water is\",T[0],\"K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.25,Page no:3.74" + ] + }, + { + "cell_type": "code", + "execution_count": 199, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Outlet temperature of water is: 338.0 K\n", + "Area of heat transfer required is 0.21 m^2\n", + "Length of tube required is 2.66 m\n" + ] + } + ], + "source": [ + "#Outlet temperature of water\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "mo_dot=60.0 #Mass flow rate of oilin [g/s]\n", + "mo_dot=6.0*10**-2 #[kg/s]\n", + "Cpo=2.0 #Specific heat of oil in [kJ/(kg.K)]\n", + "T1=420.0 #[K]\n", + "T2=320.0 #[K]\n", + "\n", + "#Calculation\n", + "Q=mo_dot*Cpo*(T1-T2) #Rate of heat flow in [kJ/s]\n", + "mw_dot=mo_dot #Mass flow rate of water #kg/s\n", + "t1=290.0 #[K]\n", + "Cpw=4.18 #[kJ/(kg.K)]\n", + "#For finding outlet temperature of water\n", + "t2=t1+Q/(mw_dot*Cpw) #[K]\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "ho=1.6 #Oil side heat transfer coefficient in [kW/(sq m.K)]\n", + "hi=3.6 #Water side heat transfer coeff in [kW/(sq m.K)]\n", + "#Overall heat transfer coefficient is:\n", + "U=1.0/(1.0/ho+1.0/hi) #[kW/(m**2.K)]\n", + "A=Q/(U*dTm) #[sq m]\n", + "Do=25.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "L=A/(math.pi*Do) #Length of tube in [m]\n", + "\n", + "#Result\n", + "print\"Outlet temperature of water is:\",round(t2),\"K\" \n", + "print\"Area of heat transfer required is\",round(A,2),\"m^2\"\n", + "print\"Length of tube required is\",round(L,2),\"m\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.26,Page no:3.76" + ] + }, + { + "cell_type": "code", + "execution_count": 200, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The inside heat transfer coefficient is 272.97 W/(m**2.K) \n", + "NOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer\n" + ] + } + ], + "source": [ + "#Inside heat transfer coefficient\n", + "import math\n", + "\n", + "#Variable declaration\n", + "k=0.14 # for oil[W/m.K]\n", + "Cp=2.1 # for oil [kJ/kg.K]\n", + "Cp=Cp*10**3 #J/kg.K\n", + "mu=154 #[mN.s/sq m]\n", + "mu_w=87 #(mn.s/sq m)\n", + "L=1.5 #[m]\n", + "m_dot=0.5 #Mass flow rate of oil[kg/s]\n", + "Di=0.019 #Diameter of tube [m]\n", + "mean_T=319 #Mean temperature of oil [K]\n", + "\n", + "#Calculation\n", + "mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]\n", + "A=math.pi*(Di/2)**2 #[sq m]\n", + "G=m_dot/A #Mass velocity in [kg/sq m.s]\n", + "Nre=Di*G/mu #Reynolds number\n", + "#As Nre<2100,the flow is laminar\n", + "mu_w=mu_w*10**-3 #[N.s/sq m] or kg/(m.s)\n", + "#The sieder tate equation is \n", + "hi=(k*(2.0*((m_dot*Cp)/(k*L))**(1.0/3.0)*(mu/mu_w)**(0.14)))/Di #Heat transfer coeff in [W/sq m.K]\n", + "\n", + "#Result\n", + "print\"The inside heat transfer coefficient is\",round(hi,2),\"W/(m**2.K) \" \n", + "print\"NOTE:Calculation mistake in last line.ie in the calculation of hi in book,please perform the calculation manually to check the answer\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.27,Page no:3.77" + ] + }, + { + "cell_type": "code", + "execution_count": 202, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Based on outside temperature,Inside heat transfer coefficient is 4201.0 W/(m**2.K) or 4.2 kW/(m**2.K)\n" + ] + } + ], + "source": [ + "#Film heat transfer coefficient\n", + "import math\n", + "\n", + "#Variable declaration\n", + "m_dot=0.217 #Water flow rate in [kg/s]\n", + "Do=19.0 #Outside diameter in [mm]\n", + "rho=1000.0 #Density\n", + "t=1.6 #Wall thickness in [mm]\n", + "Di=Do-2*t #i.d of tube in [mm]\n", + "Di=Di/1000.0 #[m]\n", + "Do=Do/1000.0 #[m]\n", + "\n", + "\n", + "#Calculation\n", + "Ai=math.pi*(Di/2)**2 #Cross-sectional area in sq m\n", + "u=m_dot/(rho*Ai) #Water velocity through tube [m/s]\n", + "u=1.12 #approx in book\n", + "Di=0.0157 #apprx in book\n", + "T1=301.0 #Inlet temperature of water in [K]\n", + "T2=315.0 #Outlet temperature of water in [K]\n", + "T=(T1+T2)/2 #[K]\n", + "hi=(1063.0*(1+0.00293*T)*(u**0.8))/(Di**0.20) #Inside heat transfer coefficient W/(sq m.K)\n", + "hi=5084.0 #Approximation\n", + "hio=hi*(Di/Do) #Inside heat transfer coeff based on outside diameter in W/(sq m.K)\n", + "\n", + "#Result\n", + "print\"Based on outside temperature,Inside heat transfer coefficient is\",round(hio),\"W/(m**2.K) or \",round(hio/1000,1),\"kW/(m**2.K)\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.28,Page no:3.77" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer area of equipment is 6.5 m^2\n" + ] + } + ], + "source": [ + "#Area of exchanger\n", + "import math\n", + "#Variable declaration\n", + "mair_dot=0.90 #[kg/s]\n", + "T1=283.0 #[K]\n", + "T2=366.0 #[K]\n", + "dT=(T1+T2)/2 #[K]\n", + "Di=12.0 #[mm]\n", + "Di=Di/1000.0 #[m]\n", + "G=19.9 #[kg/(sq m.s)]\n", + "mu=0.0198 #[mN.s/(sq m)]\n", + "mu=mu*10**-3 #[N.s/sq m] or [kg/(m.s)]\n", + "\n", + "#Calculation\n", + "Nre=Di*G/mu #Reynolds number\n", + "#It is greater than 10**4\n", + "k=0.029 #W/(m.K)\n", + "Cp=1.0 #[kJ/kg.K]\n", + "Cp1=Cp*10**3 #[J/kg.K]\n", + "Npr=Cp1*mu/k #Parndtl number\n", + "#Dittus-Boelter equation is\n", + "hi=0.023*(Nre**0.8)*(Npr**0.4)*k/Di #[W/sq m.K]\n", + "ho=232.0 #W/sq m.K\n", + "U=1.0/(1.0/hi+1.0/ho) #Overall heat transfer coefficient [W/m**2.K]\n", + "Q=mair_dot*Cp*(T2-T1) #kJ/s\n", + "Q=Q*10**3 #[J/s] or [W]\n", + "T=700.0 #[K]\n", + "dT1=T-T2 #[K]\n", + "dT2=T2-T1 #[K]\n", + "dTm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "#Q=U*A*dTm\n", + "A=Q/(U*dTm) #Area in sq m\n", + "\n", + "#Result\n", + "print\"Heat transfer area of equipment is\",round(A,2),\"m^2\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.29,Page no:3.82" + ] + }, + { + "cell_type": "code", + "execution_count": 206, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In free convection,heat transfer coeff,h= 5.3 W/(m^2.K)\n", + "In forced convection,heat transfer coeff,h= 14.12 W/(m^2.K)\n", + "From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection\n" + ] + } + ], + "source": [ + "#Natural and forced convection\n", + "import math\n", + "#Variable declaration\n", + "v=18.41*10**-6 #[sq m./s]\n", + "k=28.15*10**-3 #[W/m.K]\n", + "Npr=0.7 #Prandtl number\n", + "Beta=3.077*10**-3 #K**-1\n", + "g=9.81 #m/s**2\n", + "Tw=350 #[K]\n", + "T_inf=300 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "L=0.3 #[m]\n", + "\n", + "\n", + "#Calculation\n", + "#1.Free Convection\n", + "Ngr=(g*Beta*dT*L**3)/(v**2) #Grashof number\n", + "Npr=0.7 #Prandtl number\n", + "Nnu=0.59*(Ngr*Npr)**(1.0/4.0) #Nusselt number\n", + "h=Nnu*k/L #Average heat transfer coefficient [W/sq m K]\n", + "\n", + "#2.Forced Convestion\n", + "u_inf=4 #[m/s]\n", + "Nre_l=u_inf*L/v\n", + "Nnu=0.664*(Nre_l**(1.0/2.0))*(Npr**(1.0/3.0)) #Nusselt number\n", + "h1=Nnu*k/L #[W/sq m.K]\n", + "\n", + "\n", + "#Result\n", + "print\"In free convection,heat transfer coeff,h=\",round(h,1),\"W/(m^2.K)\"\n", + "print\"In forced convection,heat transfer coeff,h=\",round(h1,2),\"W/(m^2.K)\"\n", + "print\"From above it is clear that heat transfer coefficient in forced convection is much larger than that in free convection\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.30,Page no:3.83" + ] + }, + { + "cell_type": "code", + "execution_count": 208, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transferred is 6020.0 W\n" + ] + } + ], + "source": [ + "#Natural convection\n", + "\n", + "import math\n", + "#Variable declaration\n", + "k=0.02685 #W/(m.K)\n", + "v=16.5*10**-6 #kg/(m.s)\n", + "Npr=0.7 #Prandtl number\n", + "Beta=3.25*10**-3 #K**-1\n", + "g=9.81 #m/(s**2)\n", + "Tw=333 #[k]\n", + "T_inf=283 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "L=4 #Length/height of plate [m]\n", + "\n", + "#Calculation\n", + "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n", + "#Let const=Ngr*Npr\n", + "const=Ngr*Npr\n", + "#Sice it is >10**9\n", + "Nnu=0.10*(const**(1.0/3.0)) #Nusselt number\n", + "h=round(Nnu*k/L,1) #W/(sq m.K)\n", + "W=7 #width in [m]\n", + "A=L*W #Area of heat transfer in [sq m]\n", + "Q=h*A*dT #[W]\n", + "\n", + "#Result\n", + "print\"Heat transferred is\",Q,\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.31,Page no:3.84" + ] + }, + { + "cell_type": "code", + "execution_count": 210, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hence,Heat loss per metre length is 120.68 W/m\n" + ] + } + ], + "source": [ + "#Free convection in vertical pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "v=18.97*10**-6 #m**2/s\n", + "k=28.96*10**-3 #W/(m.K)\n", + "Npr=0.696\n", + "D=100.0 #Outer diameter [mm]\n", + "D=D/1000 #[m]\n", + "Tf=333.0 #Film temperature in [K]\n", + "Tw=373.0 #[K]\n", + "T_inf=293.0 #[K]\n", + "\n", + "#Calculation\n", + "dT=Tw-T_inf #[K]\n", + "Beta=1.0/Tf #[K**-1]\n", + "g=9.81 #[m/s**2]\n", + "L=3.0 #Length of pipe [m]\n", + "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashof number\n", + "Nra=Ngr*Npr\n", + "Nnu=0.10*(Ngr*Npr)**(1.0/3.0) #nusselt number for vertical cylinder\n", + "h=Nnu*k/L #W/(sq m.K)\n", + "Q_by_l=h*math.pi*D*dT #Heat loss per metre length [W/m]\n", + "\n", + "#Result\n", + "print\"Hence,Heat loss per metre length is\",round(Q_by_l,2),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.32,Page no:3.84" + ] + }, + { + "cell_type": "code", + "execution_count": 213, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss per unit length of the heater is 653.4 W/m(APPROX)\n" + ] + } + ], + "source": [ + "#Heat loss per unit length\n", + "import math\n", + "\n", + "#Variable declaration\n", + "k=0.630 #W/(m.K\n", + "Beta=3.04*10**-4 #K**-1\n", + "rho=1000.0 #kg/m**3\n", + "mu=8.0*10**-4 #[kg/(m.s)]\n", + "Cp=4.187 #kJ/(kg.K)\n", + "g=9.81 #[m/(s**2)]\n", + "Tw=313.0 #[K]\n", + "T_inf=298.0 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "D=20.0 #[mm]\n", + "D=D/1000 #[m]\n", + "\n", + "#Calculation\n", + "Ngr=9.81*(rho**2)*Beta*dT*(D**3)/(mu**2) #Grashoff number\n", + "Cp1=Cp*1000.0 #[J/kg.K]\n", + "Npr=Cp1*mu/k #Prandtl number\n", + "#Average nusselt number is\n", + "Nnu=0.53*(Ngr*Npr)**(1.0/4.0)\n", + "h=Nnu*k/D #[W/ sqm.K]\n", + "Q_by_l=h*math.pi*D*dT #Heat loss per unit length [W/m]\n", + "\n", + "#Result\n", + "print\"Heat loss per unit length of the heater is\",round(Q_by_l,1),\"W/m(APPROX)\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.33,Page no:3.85" + ] + }, + { + "cell_type": "code", + "execution_count": 214, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss of heat transfer per meter lengh is 1492.4 W/m\n" + ] + } + ], + "source": [ + "#Free convection in pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "k=0.03406 #[W/(m/K)]\n", + "Beta=2.47*10**-3 #K**-1\n", + "Npr=0.687 #Prandtl number\n", + "v=26.54*10**-6 #m**2/s\n", + "g=9.81 #[m/s**2]\n", + "Tw=523.0 #[K]\n", + "T_inf=288.0 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "D=0.3048 #[m]\n", + "\n", + "#Calculation\n", + "Ngr=(g*Beta*dT*(D**3))/(v**2) #Grashof number\n", + "Nra=Ngr*Npr \n", + "#For Nra less than 10**9,we have for horizontal cylinder\n", + "Nnu=0.53*(Nra**(1.0/4.0)) #Nusselt number\n", + "h=Nnu*k/D #[W/sq m.K]\n", + "Q_by_l=h*math.pi*D*dT #W/m\n", + "\n", + "#Result\n", + "print\"Heat loss of heat transfer per meter lengh is\",round(Q_by_l,1),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.34,Page no:3.86" + ] + }, + { + "cell_type": "code", + "execution_count": 215, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The rate of heat input is 3410.4 W\n" + ] + } + ], + "source": [ + "#Free convection in plate\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=960.63 #Density in [kg/m**3]\n", + "Cp=4.216*10**3 #Specific heat in [J/(kg.K)]\n", + "D=16.0 #Diameter in [cm]\n", + "D=D/100 #[m]\n", + "k=0.68 #Thermal conductivity in [W/m.K]\n", + "\n", + "#Calculation\n", + "A=(math.pi*(D/2)**2)\n", + "L=A/(math.pi*D) #Length=A/P in [m]\n", + "Beta=0.75*10**-3 #[K**-1]\n", + "alpha=1.68*10**-7 #[m**2/s]\n", + "g=9.81 #[m/s**2]\n", + "Tw=403.0 #[K]\n", + "T_inf=343.0 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "v=0.294*10**-6 #[m**2/s]\n", + "Nra=(g*Beta*(L**3)*dT)/(v*alpha) \n", + "#1.For Top surface\n", + "Nnu=0.15*(Nra)**(1.0/3.0) #Nusselt number\n", + "ht=Nnu*k/L #Heat transfer coeff for top surface in W/(m**2.K)\n", + "ht=round(ht)\n", + "#2.For bottom surface\n", + "Nnu=0.27*Nra**(1.0/4.0) #Nusselt number\n", + "hb=Nnu*k/L #[W/sq m.K]\n", + "hb=round(hb)\n", + "Q=(ht+hb)*A*dT #[W]\n", + "\n", + "#Result\n", + "print\"The rate of heat input is\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.35,Page no:3.87" + ] + }, + { + "cell_type": "code", + "execution_count": 234, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nnu in the book is wrongly calculated as 66.80,\n", + "Actually it is:58.22\n", + "\n", + "Heat transferred when hot surface is facing up is 40.04 W\n", + "NOTE:Taking into consideration the correct value of Nnu\n", + "\n", + "Heat transferred when hot surface is facing down is 16.23 W\n" + ] + } + ], + "source": [ + "#Heat transfer from disc\n", + "import math\n", + "\n", + "#Variable declaration\n", + "v=2.0*10**-5 #[m**2/s]\n", + "Npr=0.7 #Prandtl number\n", + "k=0.03 #[W/m.K]\n", + "D=0.25 #Diameter in [m]\n", + "L=0.90*D #Characteristic length,let [m]\n", + "T1=298.0 #[K]\n", + "T2=403.0 #[K]\n", + "dT=T2-T1 #[K]\n", + "\n", + "#Calculation\n", + "Tf=(T1+T2)/2 #[K]\n", + "Beta=1.0/Tf #[K**-1]\n", + "A=math.pi*(D/2)**2 #Area in[sq m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Case 1: Hot surface facing up\n", + "Ngr=g*Beta*dT*(L**3)/(v**2) #Grashoff number\n", + "Nnu=0.15*((Ngr*Npr)**(1.0/3.0)) #Nusselt number\n", + "print \"Nnu in the book is wrongly calculated as 66.80,\\nActually it is:58.22\"\n", + "h=Nnu*k/L #[W/sq m.K]\n", + "Q=h*A*dT #[W]\n", + "\n", + "#Case 2:For hot surface facing down\n", + "Nnu=0.27*(Ngr*Npr)**(1.0/4.0) #Grashof Number\n", + "h=Nnu*k/L #[W/sqm.K]\n", + "Q1=h*A*dT #[W]\n", + "\n", + "\n", + "#Result\n", + "print\"\\nHeat transferred when hot surface is facing up is\",round(Q,2),\"W\"\n", + "print\"NOTE:Taking into consideration the correct value of Nnu\\n\"\n", + "print\"Heat transferred when hot surface is facing down is\",round(Q1,2),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.36,Page no:3.88" + ] + }, + { + "cell_type": "code", + "execution_count": 235, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat input is equal to heat dissipation = 10914.4 W\n" + ] + } + ], + "source": [ + "#Rate of heat input to plate\n", + "import math\n", + "#Variable declaration\n", + "rho=960 #[kg/m**3]\n", + "Beta=0.75*10**-3 #[K**-1]\n", + "k=0.68 #[W/m.K]\n", + "alpha=1.68*10**-7 #[m**2/s]\n", + "v=2.94*10**-7 #[m**2/s]\n", + "Cp=4.216 #[kJ/kg.K]\n", + "Tw=403 #[K]\n", + "T_inf=343 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "g=9.81 #[m/s**2]\n", + "l=0.8 #[m]\n", + "W=0.08 #[m]\n", + "\n", + "#Calculation\n", + "A=l*W #Area in [m**2]\n", + "P=2*(0.8+0.08) #Perimeter in [m]\n", + "L=A/P #Characteristic dimension/length,L in [m]\n", + "Nra=g*Beta*L**3*dT/(v*alpha) \n", + "#(i) for natural convection,heat transfer from top/upper surface heated \n", + "Nnu=0.15*(Nra**(1.0/3.0)) #Nusselt number\n", + "ht=Nnu*k/L #[W/m**2.K]\n", + "ht=2115.3 #Approximation in book,If done manually then answer diff\n", + "#(ii)For the bottom/lower surface of the heated plate\n", + "Nnu=0.27*(Nra**(1.0/4.0)) #Nusselt number\n", + "hb=Nnu*k/L #[W/(m**2.K)]\n", + "hb=round(hb)\n", + "#Rate of heat input is equal to rate of heat dissipation from the upper and lower surfaces of the plate\n", + "Q=(ht+hb)*A*(Tw-T_inf) #[W]\n", + "\n", + "#Result\n", + "print\"Rate of heat input is equal to heat dissipation =\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.37,Page no:3.89" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transferred when disc is horizontal with hot surface facing upward is 170.8 W\n", + "Heat transferred when disc is horizontal with hot surface facing downward is 65.6 W\n", + "For vertical disc,heat transferred is 114.0 W\n" + ] + } + ], + "source": [ + "#Two cases in disc\n", + "import math\n", + "\n", + "#Variable declaration\n", + "k=0.03 #W/(m.K)\n", + "Npr=0.697 #Prandtl number\n", + "v=2.076*10**-6 #m**2/s\n", + "Beta=0.002915 #K**-1\n", + "D=25.0 #[Diameter in cm]\n", + "D=D/100 #[m]\n", + "Tf=343.0 #Film temperature in [K]\n", + "\n", + "#Calculation\n", + "A=math.pi*(D/2)**2 #Area in [m**2]\n", + "P=math.pi*D #Perimeter [m]\n", + "T1=293.0 #[K]\n", + "T2=393.0 #[K]\n", + "g=9.81 #[m/s**2]\n", + "#Case (i) HOT SURFACE FACING UPWARD\n", + "L=A/P #Characteristic length in [m]\n", + "Beta=1.0/Tf #[K**-1]\n", + "dT=T2-T1 #[K]\n", + "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n", + "Nra=Ngr*Npr \n", + "Nnu=0.15*(Nra**(1.0/3.0)) #Nusselt number\n", + "h=Nnu*k/L #[W/m**2.K]\n", + "Q1=h*A*dT #[W]\n", + "\n", + "#Case-(ii) HOT FACE FACING DOWNWARD\n", + "Nnu=0.27*(Nra**(1.0/4.0)) #Nusselt number\n", + "h=Nnu*k/L #W/(m**2.K)\n", + "Q2=h*A*dT #[W]\n", + "\n", + "#Case-(iii)-For disc vertical \n", + "L=0.25 #Characteristic length[m] \n", + "D=L #dia[m]\n", + "A=math.pi*((D/2)**2) #[sq m]\n", + "Ngr=(g*Beta*dT*(L**3))/(v**2) #Grashoff number\n", + "Npr=0.697\n", + "Nra=Ngr*Npr \n", + "Nnu=0.10*(Nra**(1.0/3.0)) #Nusselt number\n", + "h=Nnu*k/D #[W/(m**2.K)]\n", + "Q3=h*A*dT #[W]\n", + "\n", + "\n", + "#Result\n", + "print\"Heat transferred when disc is horizontal with hot surface facing upward is\",round(Q1,1),\"W\"\n", + "print\"Heat transferred when disc is horizontal with hot surface facing downward is\",round(Q2,1),\"W\" \n", + "print\"For vertical disc,heat transferred is\",round(Q3),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.38,Page no:3.91" + ] + }, + { + "cell_type": "code", + "execution_count": 240, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss per metre length of pipe is 831.1 W/m\n" + ] + } + ], + "source": [ + "#Total heat loss in a pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "v=23.13*10**-6 #[m**2/s]\n", + "k=0.0321 #[W/m.K]\n", + "Beta=2.68*10**-3 #[K**-1]\n", + "Tw=443.0 #[K]\n", + "T_inf=303.0 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "g=9.81 #[m/s**2]\n", + "Npr=0.688 #Prandtl number\n", + "D=100.0 #Diameter [mm]\n", + "D=D/1000 #Diameter [m]\n", + "\n", + "#Calculation\n", + "Nra=(g*Beta*dT*(D**3)*Npr)/(v**2)\n", + "Nnu=0.53*(Nra**(1.0/4.0)) #Nusselt number\n", + "h=Nnu*k/D #[W/(m**2.K)]\n", + "h=7.93 #Approximation\n", + "e=0.90 #Emissivity\n", + "sigma=5.67*10**-8 \n", + "#Q=Q_conv+Q_rad #Total heat loss\n", + "#for total heat loss per meter length\n", + "Q_by_l=h*math.pi*D*dT+sigma*e*math.pi*D*(Tw**4-T_inf**4) #[W/m]\n", + "\n", + "#Result\n", + "print\"Total heat loss per metre length of pipe is\",round(Q_by_l,1),\"W/m\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.39,Page no:3.91" + ] + }, + { + "cell_type": "code", + "execution_count": 241, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss is 1.06 W\n" + ] + } + ], + "source": [ + "#Heat loss by free convection\n", + "import math\n", + "\n", + "#Result\n", + "k=0.035 #[W/(m.K)]\n", + "Npr=0.684 #Prandtl number\n", + "Beta=2.42*10**-3 #[K**-1]\n", + "v=27.8*10**-6 #[m**2/s]\n", + "Tw=533.0 #[K]\n", + "T_inf=363.0 #[K]\n", + "dT=Tw-T_inf #[K]\n", + "D=0.01 #[m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Calculation\n", + "Nra=(g*Beta*dT*(D**3))/(v**2)\n", + "#For this <10**5,we have for sphere\n", + "A=4*math.pi*(D/2)**2 #Area of sphere in [m**2]\n", + "Nnu=(2+0.43*Nra**(1.0/4.0))#Nusslet number\n", + "h=Nnu*k/D #W/(m**2.K)\n", + "Q=h*A*dT #[W]\n", + "\n", + "#Result\n", + "print\"Rate of heat loss is\",round(Q,2),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.40,Page no:3.92" + ] + }, + { + "cell_type": "code", + "execution_count": 246, + "metadata": { + "collapsed": false, + "slideshow": { + "slide_type": "-" + } + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss is 966.0 W\n" + ] + } + ], + "source": [ + "#Heat loss from cube\n", + "import math\n", + "#Variable declaration\n", + "v=17.95*10**-6 #[m**2/s]\n", + "dT=353.0-293.0 #[K]\n", + "k=0.0283 #[W/m.K]\n", + "g=9.81 #[m/s**2]\n", + "Npr=0.698 #Prandtl number\n", + "Cp=1005.0 #J/(kg.K)\n", + "Tf=323.0 #Film temperature in [K]\n", + "Beta=1.0/Tf #[K**-1]\n", + "l=1.0 #[m]\n", + "\n", + "#Calculation\n", + "Nra=(g*Beta*dT*(l**3)*Npr)/(v**2)\n", + "#In textbook result of above statement is wrongly calculated,So\n", + "Nra=3.95*10**8\n", + "#For Nra <10**9,for a vertical plate,the average nusselt number is\n", + "Nnu=0.59*Nra**(1.0/4.0) #Nusselt number\n", + "h=round(Nnu*k/l,2) #[W/m**2.K]\n", + "A=l**2 #Area [m**2]\n", + "#Heat loss form 4 vertical faces of 1m*1m is \n", + "Q1=4.0*(h*A*dT) #[W]\n", + "#For top surface \n", + "P=4.0*l #Perimeter in [m]\n", + "L=A/P #[m]\n", + "Nra=(Npr*g*Beta*dT*(L**3))/(v**2)\n", + "Nnu=0.15*Nra**(1.0/3.0) #Nusselt number\n", + "h=round(Nnu*k/L,1) #[W/m**2.K]\n", + "Q2=h*A*dT #[W]\n", + "Q_total=Q1+Q2 #Total heat loss[W]\n", + "\n", + "#Result\n", + "print\"Total heat loss is\",Q_total,\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.41,Page no:3.93" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dimensions of the plate are 0.03 x 0.05 m= 3.0 x 5.0 cm\n", + "Heat transfer when the bigger side held vertical is 2.77 W\n", + "Heat transfer when the small side held vertical is 3.16 W\n" + ] + } + ], + "source": [ + "#Plate exposed to heat\n", + "\n", + "#Variable declaration\n", + "rho=0.910 #Density in [kg/m**3]\n", + "Cp=1.009*1000 #[J/kg.K]\n", + "k=0.0331 #[W/m.K]\n", + "mu=22.65*10**-6 #[N.s/m**2]\n", + "\n", + "#Calculation\n", + "#Let a=smaller side\n", + "#b=bigger side\n", + "#Qa=ha*A*dT\n", + "#Qb=hb*A*dT\n", + "#Qa=1.14*Qb\n", + "#Given a*b=15*10**-4\n", + "#On solving we get:\n", + "a=0.03 #[m]\n", + "b=0.05 #[m]\n", + "A=a*b #Area in [sq m]\n", + "Tf=388 #[K]\n", + "Beta=1.0/Tf #[K**-1]\n", + "T1=303 #[K]\n", + "T2=473 #[K]\n", + "dT=T2-T1 #[K]\n", + "v=mu/rho \n", + "g=9.81 #m/s**2[acceleration due to gravity ]\n", + "hb=0.59*(((g*Beta*dT*(b**3))/(v**2))*Cp*mu/k)**(1.0/4.0)*(k/b) #[W/sq m.K]\n", + "Qb=hb*A*(dT) #[W]\n", + "\n", + "Qa=1.14*Qb #[W]\n", + "\n", + "#Result\n", + "print\"Dimensions of the plate are\",a,\"x\",b,\"m=\",a*100,\"x\",b*100,\"cm\"\n", + "print\"Heat transfer when the bigger side held vertical is\",round(Qb,2),\"W\"\n", + "print\"Heat transfer when the small side held vertical is\",round(Qa,2),\"W\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.42,Page no:3.99" + ] + }, + { + "cell_type": "code", + "execution_count": 251, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature of the bottom surface of the pan is 385.5 W/(m^2)\n" + ] + } + ], + "source": [ + "#Nucleate poolboiling\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Ts=373.0 #[K]\n", + "rho_l=957.9 #rho*l[kg/m**3]\n", + "Cpl=4217.0 #[J/kg.K]\n", + "mu_l=27.9*10**-5 #[kg/(m.s)]\n", + "rho_v=0.5955 #[kg/m**3]\n", + "Csf=0.013\n", + "sigma=5.89*10**-2 #[N/m]\n", + "Nprl=1.76\n", + "lamda=2257.0 #[kJ/kg]\n", + "lamda=lamda*1000 #in [J/kg]\n", + "n=1 #for water\n", + "m_dot=30.0 #Mass flow rate [kg/h]\n", + "\n", + "#Calculation\n", + "m_dot=m_dot/3600 #[kg/s]\n", + "D=30.0 #Diameter of pan [cm]\n", + "D=D/100 #[m]\n", + "g=9.81 #[m/s**2]\n", + "A=math.pi*(D/2)**2 #Area in [sq m]\n", + "Q_by_A=m_dot*lamda/A #[W/sq m]\n", + "#For nucleate boiling point we have:\n", + "dT=(lamda/Cpl)*Csf*(((Q_by_A)/(mu_l*lamda))*math.sqrt(sigma/(g*(rho_l-rho_v))))**(1.0/3.0)*(Nprl**n) #[K]\n", + "Tw=Ts+dT #[K]\n", + "\n", + "#Result\n", + "print\"Temperature of the bottom surface of the pan is\",round(Tw,1),\"W/(m^2)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.43,Page no:3.100" + ] + }, + { + "cell_type": "code", + "execution_count": 252, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Peak heat flux is 1.106 MW/sq m\n" + ] + } + ], + "source": [ + "#Peak Heat flux\n", + "import math\n", + "\n", + "#Variable declaration\n", + "lamda=2257.0 #[kJ/kg]\n", + "lamda=lamda*1000 #in [J/kg]\n", + "rho_l=957.9 #rho*l[kg/m**3]\n", + "rho_v=0.5955 #[kg/m**3]\n", + "sigma=5.89*10**-2 #[N/m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Calculation\n", + "#Peak heat flux is given by\n", + "Q_by_A_max=(math.pi/24)*(lamda*rho_v**0.5*(sigma*g*(rho_l-rho_v))**(1.0/4.0)) #W/m**2\n", + "Q_by_A_max=Q_by_A_max/(10**6) #MW/(sq m)\n", + "\n", + "\n", + "#Result\n", + "print\"Peak heat flux is\",round(Q_by_A_max,3),\"MW/sq m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.44,Page no:3.100" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Stable film boiling point heat transfer coefficient is 1340.9 W/(sq m.K)\n", + "Heat dissipated per unit length of the heater is 1.7 kW/m\n", + "\n", + "NOTE:In textbook,value of hc is wrongly calculated as 1311.4,Actually it is 1318.9,\n", + "So,there is a difference in final values of 'h'\n" + ] + } + ], + "source": [ + "#Stable film pool boiling\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho_l=957.9 #[kg/m**3]\n", + "lamda=2257.0 #[kJ/kg]\n", + "lamda=lamda*10**3 #[J/kg]\n", + "rho_v=31.54 #[kg/m**3]\n", + "Cpv=4.64 #[kJ/kg.K]\n", + "Cpv=Cpv*10**3 #[J/kg.K]\n", + "kv=58.3*10**-3#[W/(m.K)]\n", + "g=9.81 #[m/s**2]\n", + "mu_v=18.6*10**-6 #[kg/(m.s)]\n", + "e=1.0 #Emissivity\n", + "sigma=5.67*10**-8 \n", + "Ts=373.0 #[K]\n", + "Tw=628.0 #[K]\n", + "\n", + "#Calculation\n", + "dT=Tw-Ts #[K]\n", + "D=1.6*10**-3 #[m]\n", + "T=(Tw+Ts)/2 #[K]\n", + "\n", + "hc=0.62*((kv**3)*rho_v*(rho_l-rho_v)*g*(lamda+0.40*Cpv*dT)/(D*mu_v*dT))**(1.0/4.0) #Convective heat transfer coeff [W/sq m.K]\n", + "hr=e*sigma*(Tw**4-Ts**4)/(Tw-Ts) #Radiation heat transfer coeff in [W/sq m.K]\n", + "h=hc+(3.0/4.0)*hr #Total heat transfer coefficient W/(sq m.K)\n", + "Q_by_l=h*math.pi*D*dT #Heat dissipation rate per unit length in [kW/m]\n", + "Q_by_l=Q_by_l/1000 #[kW/m]\n", + "#Result\n", + "print\"Stable film boiling point heat transfer coefficient is\",round(h,1),\"W/(sq m.K)\"\n", + "print\"Heat dissipated per unit length of the heater is\",round(Q_by_l,1),\"kW/m\"\n", + "\n", + "print\"\\nNOTE:In textbook,value of hc is wrongly calculated as 1311.4,Actually it is 1318.9,\"\n", + "print\"So,there is a difference in final values of 'h'\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.45,Page no:3.102" + ] + }, + { + "cell_type": "code", + "execution_count": 136, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer per 1m length of tube is 2810.0 W/m\n" + ] + } + ], + "source": [ + "#Heat transfer in tube\n", + "import math\n", + "\n", + "#Variable declaration\n", + "dT=10 #[K]\n", + "P=506.625 #[kPa]\n", + "P=P/10**3 #[Mpa]\n", + "D=25.4 #Diameter [mm]\n", + "D=D/1000 #[m]\n", + "\n", + "#Calculation\n", + "h=2.54*(dT**3)*(math.exp(P/1.551)) #[W/sq m.K]\n", + "#Q=h*math.pi*D*L*dT\n", + "#Heat transfer rate per meter length of tube is \n", + "Q_by_l=h*math.pi*D*dT #[W/m]\n", + "\n", + "#Result\n", + "print\"Rate of heat transfer per 1m length of tube is\",round(Q_by_l),\"W/m\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.46,Page no:3.102" + ] + }, + { + "cell_type": "code", + "execution_count": 265, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat flux when pressure is 101.325 kPa is 22773.76 W/m^2(APPROX)\n", + "Percent increase in heat flux is 23.0 %\n" + ] + } + ], + "source": [ + "#Nucleat boiling and heat flux\n", + "\n", + "#Variable declaration\n", + "dT=8.0 #[K]\n", + "P=0.17 #[Mpa]\n", + "P=P*1000 #[kPa]\n", + "h1=2847.0 #[W/(sq m.K)]\n", + "P1=101.325 #[kPa]\n", + "h=5.56*(dT**3) #[W/sq m.K]\n", + "\n", + "#Calculation\n", + "Q_by_A=h*dT #[W/sq m]\n", + "hp=h*(P/P1)**(0.4) #[W/sq m.K]\n", + "#Corresponding heat flux is :\n", + "Q_by_A1=hp*dT #[W/sq m]\n", + "per=(Q_by_A1-Q_by_A)*100.0/Q_by_A #Percent increase in heat flux\n", + "\n", + "#Result\n", + "print\"Heat flux when pressure is 101.325 kPa is\",Q_by_A,\"W/m^2(APPROX)\"\n", + "print\"Percent increase in heat flux is\",round(per),\"%\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.47,Page no:3.110" + ] + }, + { + "cell_type": "code", + "execution_count": 269, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean heat transfer coefficient is 10864.0 W/(m^2.K)\n", + "Heat transfer per unit length is 13653.0 W/m\n", + "Condensate rate per unit length is 22.1 kg/h\n" + ] + } + ], + "source": [ + "#Dry steam condensate\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mu=306*10**-6 #[N.s/m**2]\n", + "k=0.668 #[W/m.K]\n", + "rho=974.0 #[kg/m**3]\n", + "lamda=2225.0 #[kJ/kg]\n", + "lamda=lamda*10**3 #[J/kg.K]\n", + "g=9.81 #[m/s**2]\n", + "Ts=373.0 #[K]\n", + "Tw=357.0 #[K]\n", + "dT=Ts-Tw #[K]\n", + "Do=25.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "\n", + "#Calculation\n", + "h=0.725*((rho**2*g*lamda*k**3)/(mu*Do*dT))**(1.0/4.0) #[W/sq m.K]\n", + "Q_by_l=h*math.pi*Do*dT #[W/m]\n", + "m_dot_byl=(Q_by_l/lamda) #[kg/s]\n", + "m_dot_byl=m_dot_byl*3600 #[kg/h]\n", + "\n", + "#Result\n", + "print\"Mean heat transfer coefficient is\",round(h),\"W/(m^2.K)\" \n", + "print\"Heat transfer per unit length is\",round(Q_by_l),\"W/m\" \n", + "print\"Condensate rate per unit length is\",round(m_dot_byl,1),\"kg/h\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.48,Page no:3.111" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average heat transfer coefficient is 13156.0 W/(m^2.K)(APPROX)\n", + "Heat transfer rate is 2368.0 J/kg\n", + "Steam condensate rate per hour is 3.78 kg/h\n" + ] + } + ], + "source": [ + "#Laminar Condensate film\n", + "\n", + "#Variable declaration\n", + "rho=960.0 #[kh/m**3]\n", + "mu=2.82*10**-4 #[kg/(m.s)]\n", + "k=0.68 #[W/(m.K)]\n", + "lamda=2255.0 #[kJ/kg]\n", + "lamda=lamda*10**3 #[J/kg]\n", + "Ts=373.0 #Saturation temperature of steam [K]\n", + "Tw=371.0 #[K]\n", + "dT=Ts-Tw #[K]\n", + "L=0.3 #Dimension [m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Calculation\n", + "h=0.943*(rho**2*g*lamda*k**3/(L*mu*dT))**(1.0/4.0) #W/sq m.K\n", + "A=L**2 #[sq m] \n", + "Q=h*A*(Ts-Tw) #[W]=[J/s]\n", + "m_dot=Q/lamda #Condensate rate[kg/s]\n", + "m_dot=m_dot*3600.0 #[kg/h]\n", + "\n", + "#Result\n", + "print\"Average heat transfer coefficient is\",round(h),\" W/(m^2.K)(APPROX)\"\n", + "print\"Heat transfer rate is\",round(Q),\"J/kg\"\n", + "print\"Steam condensate rate per hour is\",round(m_dot,2),\"kg/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.49,Page no:3.112" + ] + }, + { + "cell_type": "code", + "execution_count": 279, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of condensation per unit length is 176.4 kg/h\n" + ] + } + ], + "source": [ + "#Saturated vapour condensate in array\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=1174.0 #[kg/m**3]\n", + "k=0.069 #[W/(m.K)]\n", + "mu=2.5*10**-4 #[N.s/m**2]\n", + "lamda=132*10**3 #[J/kg]\n", + "g=9.81 #[m/s**2]\n", + "Ts=323.0 #[K]\n", + "Tw=313.0 #[K]\n", + "dT=Ts-Tw #[K]\n", + "\n", + "#Calculation\n", + "#For square array,n=4\n", + "n=4.0 #number of tubes\n", + "Do=12.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "h=0.725*(rho**2*lamda*g*k**3/(n*Do*mu*dT))**(1.0/4.0) #W/(sq m.K) \n", + "#For heat transfer area calcualtion,n=16\n", + "A=n*math.pi*Do #[sq m]\n", + "A=0.603\n", + "Q=h*A*dT#[W/m]\n", + "m_dot=round(Q/lamda,3) #[kg/s]\n", + "\n", + "m_dot=m_dot*3600 #[kg/h]\n", + "\n", + "#Result\n", + "print\"Rate of condensation per unit length is\",m_dot,\"kg/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.50,Page no:3.113" + ] + }, + { + "cell_type": "code", + "execution_count": 286, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass rate of steam condensation is 158.0 kg/h\n", + "NOTE:ERROR in Solution in book.Do is wrongly taken as 0.012 in lines 17 and 22 of the book,Also A is wrongly calculated\n" + ] + } + ], + "source": [ + "#Mass rate of steam condensation\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=960.0 #[kg/m**3]\n", + "k=0.68 #[W/m.K]\n", + "mu=282.0*10**-6 #[kg/(m.s)]\n", + "Tw=371.0 #Tube wall temperature [K]\n", + "Ts=373.0 #Saturation temperature in [K]\n", + "dT=Ts-Tw #[K]\n", + "lamda=2256.9 #[kJ/kg]\n", + "lamda=lamda*10**3 #[J/kg]\n", + "\n", + "#Calculation\n", + "#For a square array with 100tubes,n=10\n", + "Do=0.0125 #[m]\n", + "g=9.81 #[m/s**2]\n", + "n=10.0\n", + "h=0.725*(((rho**2)*g*lamda*(k**3)/(mu*n*Do*dT))**(1.0/4.0)) #W/(sq m.K)\n", + "\n", + "L=1.0 #[m]\n", + "#n=100\n", + "n=100.0 \n", + "A=n*math.pi*Do*L #[m**2/m length]\n", + "Q=h*A*dT #Heat transfer rate in [W/m]\n", + "ms_dot=Q/lamda #[kg/s]\n", + "ms_dot=ms_dot*3600.0 #[kg/h]\n", + "\n", + "#Result\n", + "print\"Mass rate of steam condensation is\",round(ms_dot),\"kg/h\" \n", + "print\"NOTE:ERROR in Solution in book.Do is wrongly taken as 0.012 in lines 17 and 22 of the book,Also A is wrongly calculated\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.51,Page no:3.114" + ] + }, + { + "cell_type": "code", + "execution_count": 288, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tube length is 0.29 m\n", + "Rate of condensation per hour is 14.32 kg/h\n" + ] + } + ], + "source": [ + "#Saturated tube condensate in a wall\n", + "import math\n", + "\n", + "#Variable declaration\n", + "rho=975.0 #[kg/m**3]\n", + "k=0.871 #[W/m.K]\n", + "dT=10.0 #[K]\n", + "mu=380.5*10**-6 #[N.s/m**2]\n", + "lamda=2300.0 #[kJ/kg]\n", + "lamda=lamda*1000 # Latent heat of condensation [J/kg]\n", + "Do=100.0 #Outer diameter [mm]\n", + "Do=Do/1000 #[m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Calculation\n", + "#for horizontal tube\n", + "h1=0.725*((rho**2*lamda*g*k**3)/(mu*Do*dT))**(1.0/4.0) #Average heat transfer coefficient\n", + "#for vertical tube\n", + "#h2=0.943*((rho**2*lambda*g*k**3)/(mu*L*dT))**(1/4) #Average heat transfer coefficient\n", + "h2=h1 #For vertical tube\n", + "#implies that\n", + "L=(0.943*((rho**2*lamda*g*k**3)**(1.0/4.0))/(h1*((mu*dT)**(1.0/4.0))))**4 #[m]\n", + "L=0.29 #Approximate in book\n", + "h=0.943*((rho**2*lamda*g*k**3)/(mu*L*dT))**(1.0/4.0) #[W/(sq m.K)]\n", + "A=math.pi*Do*L #Area in [m**2]\n", + "Q=h*A*dT #Heat transfer rate [W]\n", + "mc_dot=Q/lamda #[Rate of condensation]in [kg/s]\n", + "mc_dot=mc_dot*3600 #[kg/h]\n", + "\n", + "#Result\n", + "print\"Tube length is\",L,\"m\"\n", + "print\"Rate of condensation per hour is\",round(mc_dot,2),\"kg/h\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.52,Page no:3.115" + ] + }, + { + "cell_type": "code", + "execution_count": 289, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For vertical position,Rate of condensation is 36.57 kg/h\n" + ] + } + ], + "source": [ + "#Condensation rate\n", + "\n", + "#Variable declaration\n", + "m1_dot=50.0 # For horizontal position[kg/h]\n", + "Do=10.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "L=1.0 #[m]\n", + "#For 100 tubes n=10\n", + "n=10.0 \n", + "\n", + "#Calculation\n", + "#We know that\n", + "#m_dot=Q/lambda=h*A*dT/lambda\n", + "#m_dot is proportional to h\n", + "#m1_dot prop to h1\n", + "#m2_dot propn to h2\n", + "#m1_dot/m2_dot=h1/h2\n", + "#or :\n", + "m2_dot=m1_dot/((0.725/0.943)*(L/(n*Do))**(1.0/4.0)) #[kg/h]\n", + "\n", + "#Result\n", + "print\"For vertical position,Rate of condensation is\",round(m2_dot,2),\"kg/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:3.53,Page no:3.116" + ] + }, + { + "cell_type": "code", + "execution_count": 295, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)- Average heat transfer coefficient is 6060.0 W/(m**2.K)\n", + "(ii)-Local heat transfer coefficient at 0.5 m height is 5404.0 W/(m^2.K)\n", + "(iii)-Film thickness is 0.124 m\n" + ] + } + ], + "source": [ + "#Condensation on vertical plate\n", + "rho=975 #[kg/m**3]\n", + "k=0.671 #[W/(m.K)]\n", + "mu=3.8*10**-4 #[N.s/m**2]\n", + "dT=10 #[K]\n", + "lamda=2300*10**3 #[J/kg]\n", + "L=1 #[m]\n", + "g=9.81 #[m/s**2]\n", + "\n", + "#Calculation\n", + "ha=0.943*((rho**2*lamda*g*k**3)/(mu*L*dT))**(1.0/4.0) #W/(sq m.K) #[W/sq m.K]\n", + "#Local heat transfer coefficient\n", + "#at x=0.5 #[m]\n", + "x=0.5 #[m]\n", + "h=((rho**2*lamda*g*k**3)/(4*mu*dT*x))**(1.0/4.0) #[W/sq m.K]\n", + "delta=((4*mu*dT*k*x)/(lamda*rho**2*g))**(1.0/4.0) #[m]\n", + "delta=delta*10**3 #[mm]\n", + "\n", + "#Result\n", + "print\"(i)- Average heat transfer coefficient is\",round(ha),\" W/(m**2.K)\"\n", + "print\"(ii)-Local heat transfer coefficient at 0.5 m height is\",round(h),\"W/(m^2.K)\"\n", + "print\"(iii)-Film thickness is\",round(delta,3),\"m\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_by_K._A._Gavhane/Chapter_4.ipynb b/Heat_Transfer_by_K._A._Gavhane/Chapter_4.ipynb new file mode 100644 index 00000000..b138a5a0 --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/Chapter_4.ipynb @@ -0,0 +1,1153 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter4: Radiation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.1,Page no:4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat loss by radiation is 704.0 W/m^2\n" + ] + } + ], + "source": [ + "#Heat loss by radiaiton\n", + "\n", + "#Variable declaration\n", + "e=0.9 #[Emissivity]\n", + "sigma=5.67*10**-8 #[W/m**2.K**4]\n", + "T1=377 #[K]\n", + "T2=283 #[K]\n", + "\n", + "#Calculation\n", + "Qr_by_a=e*sigma*(T1**4-T2**4) #[W/sq m]\n", + "\n", + "#Result\n", + "print\"Heat loss by radiation is\",round(Qr_by_a),\"W/m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.2,Page no:4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat transfer by radiation is 841.2 W/sq m\n" + ] + } + ], + "source": [ + "#Radiation from unlagged steam pipe\n", + "\n", + "#Variable declaration\n", + "e=0.9 #Emissivity\n", + "T1=393 #[K]\n", + "T2=293 #[K]\n", + "sigma=5.67*10**-8 #[W/sq m.K]\n", + "#Calculation\n", + "Qr_by_a=e*sigma*(T1**4-T2**4) #W/sq m\n", + "#Result\n", + "print\"Rate of heat transfer by radiation is\",round(Qr_by_a,1),\"W/sq m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.3,Page no:4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiaiton rate per 1 metre length of pipe is 204.0 W/m(approx)\n" + ] + } + ], + "source": [ + "#Interchange of radiation energy\n", + "import math\n", + "\n", + "#Variable declaration\n", + "L=1 #[m]\n", + "e=0.8 #Emissivity\n", + "sigma=5.67*10**-8 #[m**2.K**4]\n", + "T1=423.0 #[K]\n", + "T2=300.0 #[K]\n", + "Do=60.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "\n", + "#Calculation\n", + "A=round(math.pi*Do*L,3) #[sq m]\n", + "Qr=e*sigma*A*(T1**4-T2**4) #[W/m]\n", + "\n", + "#Result\n", + "print\"Net radiaiton rate per 1 metre length of pipe is\",round(Qr),\"W/m(approx)\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.4,Page no:4.8" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss by convection is 344.9 W/m\n" + ] + } + ], + "source": [ + "#Heat loss in unlagged steam pipe\n", + "import math\n", + "#Variable declaration\n", + "e=0.9 #Emissivity\n", + "L=1.0 #[m]\n", + "Do=50.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "sigma=5.67*10**-8 #[W/(m**2.K**4)]\n", + "T1=415.0 #[K]\n", + "T2=290.0 #[K]\n", + "dT=T1-T2 #[K]\n", + "#Calculation\n", + "hc=1.18*(dT/Do)**(0.25) #[W/sq m.K]\n", + "A=math.pi*Do*L #Area in [sq m]\n", + "Qc=hc*A*dT #Heat loss by convection W/m\n", + "Qr=e*sigma*A*(T1**4-T2**4) #Heat loss by radiation per length W/m\n", + "Qt=Qc+Qr #Total heat loss in [W/m]\n", + "#Result\n", + "print\"Total heat loss by convection is\",round(Qt,1),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.5,Page no:4.8" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat loss is 2712.0 W\n" + ] + } + ], + "source": [ + "#Loss from horizontal pipe\n", + "import math\n", + "#Variable declaration\n", + "e=0.85\n", + "sigma=5.67*10**-8 #[W/sq m.K]\n", + "T1=443.0 #[K]\n", + "T2=290.0 #[K]\n", + "dT=T1-T2 #[K]\n", + "hc=1.64*dT**0.25 #W/sq m.K\n", + "Do=60.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "L=6 #Length [m]\n", + "#Calculation\n", + "A=math.pi*Do*L #Surface area of pipe in [sq m]\n", + "Qr=e*sigma*A*(T1**4-T2**4) # Rate of heat loss by radiaiton W\n", + "Qc=hc*A*(T1-T2) # Rate of heat loss by convection [W]\n", + "Qt=Qr+Qc #Total heat loss [W]\n", + "#Result\n", + "print\"Total heat loss is\",round(Qt),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.6,Page no:4.11" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat lost by radiation is 1588.5 W\n" + ] + } + ], + "source": [ + "#Heat loss by radiation in tube\n", + "import math\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/m**2.K**4]\n", + "e1=0.79 \n", + "e2=0.93 \n", + "T1=500 #[K]\n", + "T2=300 #[K]\n", + "D=70 #[mm]\n", + "D=D/1000 #[m]\n", + "L=3 #[m]\n", + "W=0.3 #Side of conduit [m]\n", + "#Calculation\n", + "A1=math.pi*D*L #[sq m]\n", + "A1=0.659 #Approximate calculation in book in [m**2]\n", + "A2=4*(L*W) #[sq m]\n", + "Q=sigma*A1*(T1**4-T2**4)/(1/e1+((A1/A2)*(1/e2-1))) #[W]\n", + "#Result\n", + "print\"Heat lost by radiation is\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.7,Page no:4.11" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiant interchange per square metre is 6571.0 W/sq m\n" + ] + } + ], + "source": [ + "#Net radiant interchange\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "T1=703 #[K]\n", + "T2=513 #[K]\n", + "e1=0.85 \n", + "e2=0.75\n", + "#Calculation\n", + "Q_by_Ar=sigma*(T1**4-T2**4)/(1/e1+1/e2-1) #[W/sq m]\n", + "#Result\n", + "print\"Net radiant interchange per square metre is\",round(Q_by_Ar),\"W/sq m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.8,Page no:4.12" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiant interchange is 2900.0 W\n" + ] + } + ], + "source": [ + "#Radiant interchange between plates\n", + "#Variable declaration\n", + "L=3 #[m]\n", + "A=L**2 #Area in [sq m]\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "T1=373 #[K]\n", + "T2=313 #[K]\n", + "e1=0.736 \n", + "e2=e1 \n", + "#Calculation\n", + "F12=1.0/((1.0/e1)+(1.0/e2)-1)\n", + "Q=sigma*A*F12*(T1**4-T2**4) #[W]\n", + "#Result\n", + "print\"Net radiant interchange is\",round(Q),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.9,Page no:4.12" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat loss when of silvered surface is 15.95 W/sq m\n", + "\n", + " When both surfaces are black,Rate of heat loss is 622.0 W/sq m\n" + ] + } + ], + "source": [ + "#Heat loss from thermofask \n", + "#Variable declaration \n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "e1=0.05 \n", + "e2=0.05\n", + "#A1=A2=1 (let)\n", + "A1=1 \n", + "A2=A1 \n", + "\n", + "#Calculation\n", + "F12=1.0/(1.0/e1+(A1/A2)*(1.0/e2-1)) \n", + "T1=368 #[K]\n", + "T2=293 #[K]\n", + "Q_by_A=sigma*F12*(T1**4-T2**4) #Heat loss per unit Area [W/sq m]\n", + "print\"Rate of heat loss when of silvered surface is\",round(Q_by_A,2),\"W/sq m\"\n", + "#When both the surfaces are black\n", + "e1=1 \n", + "e2=1 \n", + "F12=1/(1/e1+(A1/A2)*(1/e2-1)) \n", + "Q_by_A=sigma*F12*(T1**4-T2**4) #[W/sq m]\n", + "\n", + "#Result\n", + "print\"\\n When both surfaces are black,Rate of heat loss is\",round(Q_by_A),\"W/sq m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.10,Page no:4.13" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The liquid oxygen will evaporate at 0.59 kg/h\n" + ] + } + ], + "source": [ + "#Diwar flask\n", + "import math\n", + "\n", + "#Variable declaration\n", + "e1=0.05\n", + "e2=e1\n", + "A1=0.6944 \n", + "A2=1 \n", + "T1=293 #[K]\n", + "T2=90 #[K]\n", + "sigma=5.67*10**-8 #[W/m**2.K**4]\n", + "D=0.3 #Diameter in [m]\n", + "\n", + "#Calculation\n", + "F12=1.0/(1.0/e1+(A1/A2)*(1.0/e2-1))\n", + "Q_by_A=sigma*F12*(T1**4-T2**4) #[W/sq m]\n", + "Q=Q_by_A*math.pi*(D**2) #[kJ/h]\n", + "Q=Q*3600/1000 #[kJ/h]\n", + "lamda=21.44 #Latent heat in [kJ/kg]\n", + "m_dot=Q/lamda #kg/h\n", + "\n", + "#Result\n", + "print\"The liquid oxygen will evaporate at\",round(m_dot,2),\"kg/h\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.11,Page no:4.13" + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rate of heat flow due to radiation is 36.62 W\n", + "Reduction in heat flow will be 79.97 %\n" + ] + } + ], + "source": [ + "#Heat flow due to radiation\n", + "import math\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #W/(m**2.K**4)\n", + "e1=0.3 \n", + "e2=e1 \n", + "D1=0.3 #[m]\n", + "D2=0.5 #[m]\n", + "T1=90 #[K]\n", + "T2=313 #[K]\n", + "\n", + "#Calculation\n", + "A1=math.pi*D1**2 #Area in [sq m]\n", + "A2=math.pi*D2**2#Area in [sq m]\n", + "Q1=sigma*A1*(T1**4-T2**4)/(1/e1+(A1/A2)*(1/e2-1)) #[W]\n", + "Q1=abs(Q1) #Absolute value in [W]\n", + "print\"Rate of heat flow due to radiation is\",round(Q1,2),\"W\"\n", + "#When Aluminium is used\n", + "e1=0.05\n", + "e2=0.5\n", + "Q2=sigma*A1*(T1**4-T2**4)/(1/e1+(A1/A2)*(1/0.3-1)) #[W]\n", + "Q2=abs(Q2) #Absolute value in [W]\n", + "Red=(Q1-Q2)*100/Q1 #Percent reduction\n", + "\n", + "#Result\n", + "print\"Reduction in heat flow will be\",round(Red,2),\"%\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.12,Page no:4.14" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Nitrogen evaporates at 0.047 kg/h\n" + ] + } + ], + "source": [ + "#Heat exchange between concentric shell\n", + "import math\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "T1=77.0 #[K]\n", + "T2=303.0 #[K]\n", + "D1=32.0 #cm\n", + "D1=D1/100 #[m]\n", + "D2=36.0 #[cm]\n", + "D2=D2/100 #[m]\n", + "\n", + "#Calculation\n", + "A1=math.pi*D1**2 #[sq m]\n", + "A2=math.pi*D2**2 #[sq m]\n", + "e1=0.03 \n", + "e2=e1 \n", + "Q=sigma*A1*(T1**4-T2**4)/(1.0/e1+(A1/A2)*(1.0/e2-1)) #[W]\n", + "Q=Q*3600.0/1000 #[kJ/h]\n", + "Q=abs(Q) #[kJ/h]\n", + "lamda=201.0 #kJ/kg\n", + "m_dot=Q/lamda #Evaporation rate in [kg/h]\n", + "#Result\n", + "print\"Nitrogen evaporates at\",round(m_dot,3),\"kg/h\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.13,Page no:4.15" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "-2.4112304452\n", + "Rate of evaporation is 0.0441 kg/h\n" + ] + } + ], + "source": [ + "#Evaporation in concenric vessels\n", + "import math\n", + "#Variable declaration\n", + "D1=250.0 #Inner sphere idameter[mm]\n", + "D1=D1/1000 #Outer diameter [m]\n", + "D2=350.0 #[mm]\n", + "D2=D2/1000 #[m]\n", + "sigma=5.67*10**-8 #W/(sq m.K**4)\n", + "#Calculation\n", + "A1=math.pi*D1**2 #[sq m]\n", + "A2=math.pi*D2**2 #[sq m]\n", + "T1=76.0 #[K]\n", + "T2=300.0 #[K]\n", + "e1=0.04 \n", + "e2=e1 \n", + "Q=sigma*A1*(T1**4-T2**4)/((1.0/e1)+(A1/A2)*((1.0/e2)-1)) #[W]\n", + "Q=-2.45 #Approximate\n", + "Q=abs(Q) #[W]\n", + "Q=Q*3600.0/1000 #[kJ/h]\n", + "lamda=200.0 #kJ/kg\n", + "Rate=Q/lamda #[kg/h]\n", + "#Result\n", + "print\"Rate of evaporation is\",Rate,\"kg/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.15,Page no:4.19" + ] + }, + { + "cell_type": "code", + "execution_count": 60, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer rate per unit area(WITHOUT SHIELD) due to radiation is 363.1 W/sq m\n", + "\n", + "Heat transfer rate per unit area(WITH SHIELD) due to radiation is 248.44 W/sq m\n", + "\n", + "Reduction in heat loss is 31.58 %\n" + ] + } + ], + "source": [ + "#infinitely long plates\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/(m**2.K**4)]\n", + "e1=0.4\n", + "e3=0.2\n", + "T1=473 #[K]\n", + "T3=303 #[K]\n", + "#Calculation\n", + "Q_by_a=sigma*(T1**4-T3**4)/((1.0/e1)+(1.0/e3)-1) #[W/sq m]\n", + "#Q1_by_a=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1)=sigma*A*(T2**4-T3**4)/((1/e2)+(1/e3)-1) #[W/sq m]\n", + "e2=0.5\n", + "#Solving we get\n", + "T2=((6.0/9.5)*((3.5/6)*T3**4+T1**4))**(1.0/4.0) #[K]\n", + "Q1_by_a=sigma*(T1**4-T2**4)/((1.0/e1)+(1.0/e2)-1) #[W/sq m]\n", + "red=(Q_by_a-Q1_by_a)*100/Q_by_a\n", + "#Result\n", + "print\"Heat transfer rate per unit area(WITHOUT SHIELD) due to radiation is\",round(Q_by_a,1),\"W/sq m\"\n", + "print\"\\nHeat transfer rate per unit area(WITH SHIELD) due to radiation is\",round(Q1_by_a,2),\"W/sq m\"\n", + "print\"\\nReduction in heat loss is\",round(red,2),\"%\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.16,Page no:4.20" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steady state temperatures,Tc= 560.94 K,and Td= 461.73 K\n", + "Rate of heat exchange per unit area= 770.95 W/m**2\n" + ] + } + ], + "source": [ + "#Heat exchange between parallel plates\n", + "from scipy.optimize import fsolve\n", + "import math\n", + "#In steady state,we can write:\n", + "#Qcd=Qdb\n", + "#sigma(Tc**4-Td**4)*/(1/ec+1/ed-1)=sigma(Td**4-Tb**4)/(1/ed+1/eb-1)\n", + "# i.e Td**4=0.5*(Tc**4-Tb**4)\n", + "#Variable declaration\n", + "Ta=600 #[K]\n", + "eA=0.8 \n", + "eC=0.5 \n", + "eD=0.4 \n", + "sigma=5.67*10**-8 #For air\n", + "\n", + "#Calculation\n", + "\n", + "#(600**4-Tc**4)/2.25=(Tc**4-Td**4)/3.5\n", + "#1.56*(600**4-Tc**4)=Tc**4-Td**4\n", + "#Putting value of Td in terms of Tc\n", + "#1.56*(600**4-Tc**4)=Tc**4-0.5*(Tc**4-300**4)\n", + "def f(Tc):\n", + " y=1.56*(600**4-Tc**4)-Tc**4+0.5*(Tc**4-300**4)\n", + " return(y)\n", + "Tc=fsolve(f,500) #[K]\n", + "#or\n", + "Tc=560.94 #[K] Approximate after solving\n", + "Td=math.sqrt(math.sqrt(0.5*(Tc**4-300**4))) #[K]\n", + "Q_by_a=sigma*(Ta**4-Tc**4)/(1/eA+1/eC-1) #[W/sq m]\n", + "\n", + "#Result\n", + "\n", + "print\"Steady state temperatures,Tc=\",Tc,\"K,and Td=\",round(Td,2),\" K\"\n", + "print\"Rate of heat exchange per unit area=\",round(Q_by_a,2),\"W/m**2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.17,Page no:4.21" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Due to thermal radiaiton,Loss of heat to surrounding is 5600.0 W/m\n", + "When pipe is enclosed in 1 400 mm diameter brick conduit,Loss of heat is 5390.0 W/m\n", + " Reduction in heat loss is 210.0 W/m\n" + ] + } + ], + "source": [ + "#Thermal radiation in pipe\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/(sq m.K**4)]\n", + "e=0.8\n", + "T1=673 #[K]\n", + "T2=303 #[K]\n", + "Do=200 #[mm]\n", + "Do=Do/1000 #[m]\n", + "L=1 #Let [m]\n", + "\n", + "#Calculation\n", + "import math\n", + "A1=math.pi*Do*L #[m**2/m]\n", + "#CAse 1: Pipe to surrundings\n", + "\n", + "Q1=e*A1*sigma*(T1**4-T2**4) #[W/m]\n", + "Q1=5600 #Approximated\n", + "#Q1=5600 #[W/m] approximated in book for calculation purpose\n", + "#Concentric cylinders\n", + "e1=0.8 \n", + "e2=0.91 \n", + "D1=0.2 #[m]\n", + "D2=0.4 #[m]\n", + "Q2=sigma*0.628*(T1**4-T2**4)/((1/e1)+(D1/D2)*((1/e2)-1)) #[W/m] length\n", + "Red=Q1-Q2 #Reduction in heat loss\n", + "\n", + "#Result\n", + "print\"Due to thermal radiaiton,Loss of heat to surrounding is\",round(Q1),\"W/m\"\n", + "print\"When pipe is enclosed in 1 400 mm diameter brick conduit,Loss of heat is\",round(Q2),\"W/m\" \n", + "print\" Reduction in heat loss is\",round(Red),\"W/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.18,Page no:4.22" + ] + }, + { + "cell_type": "code", + "execution_count": 64, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer by radiaiton is 6229.0 W/sq m\n" + ] + } + ], + "source": [ + "#Heat transfer in concentric tube\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/(sq m.K**4)]\n", + "T1=813.0 #[K]\n", + "T2=473.0 #[K]\n", + "e1=0.87 \n", + "e2=0.26 \n", + "D1=0.25 #[m]\n", + "D2=0.3 #[m]\n", + "#Calculation\n", + "Q_by_a1=sigma*(T1**4-T2**4)/(1.0/e1+(D1/D2)*(1.0/e2-1.0)) #[W/ sqm]\n", + "#Result\n", + "print\"Heat transfer by radiaiton is\",round(Q_by_a1),\"W/sq m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.19,Page no:4.24" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net radiant heat exchange between plates is 18334.0 W\n" + ] + } + ], + "source": [ + "#Heat exchange between black plates\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "A1=0.5*1 #[sq m]\n", + "F12=0.285\n", + "T1=1273 #/[K]\n", + "T2=773 #[K]\n", + "#Calculation\n", + "Q=sigma*A1*F12*(T1**4-T2**4) #[W]\n", + "#Result\n", + "print\"Net radiant heat exchange between plates is\",round(Q),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.20,Page no:4.32" + ] + }, + { + "cell_type": "code", + "execution_count": 66, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reduction in heat transfer rate as a result of radiaiotn shield is 94.35 percent\n" + ] + } + ], + "source": [ + "#Radiation shield\n", + "import math\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "T1=750 #[K]\n", + "T2=500 #[K]\n", + "e1=0.75 \n", + "e2=0.5 \n", + "\n", + "#Calculation\n", + "#Heat transfer without shield :\n", + "Q_by_a=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1) #[W/sq m]\n", + "\n", + "#Heat transfer with shield:\n", + "R1=(1-e1)/e1 #Resistance 1\n", + "F13=1 \n", + "R2=1/F13 #Resistance 2\n", + "e3=0.05\n", + "R3=(1-e3)/e3 #Resistance 3\n", + "R4=(1-e3)/e3 #Resistance 4\n", + "F32=1 \n", + "R5=1/F32 #Resistance 5\n", + "R6=(1-e2)/e2 #Resistance 6\n", + "Total_R=R1+R2+R3+R4+R5+R6 #Total resistance\n", + "Q_by_as=sigma*(T1**4-T2**4)/Total_R #[W/sq m]\n", + "Red=(Q_by_a-Q_by_as)*100/Q_by_a #Reduciton in heat tranfer due to shield \n", + "\n", + "#Result\n", + "print\"Reduction in heat transfer rate as a result of radiaiotn shield is\",round(Red,2),\"percent\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.21,Page no:4.33" + ] + }, + { + "cell_type": "code", + "execution_count": 67, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transfer is reduced by 93.2 % due to shield\n" + ] + } + ], + "source": [ + "#Heat transfer with radiaiton shield\n", + "\n", + "#Variable declaration\n", + "e1=0.3\n", + "e2=0.8\n", + "\n", + "#Calculation\n", + "#Let sigma*(T1**4-T2**4)=z=1(const)\n", + "z=1 #Let\n", + "Q_by_A=z/(1/e1+1/e2-1) #W/sq m\n", + "\n", + "#Heat transfer with radiation shield \n", + "e3=0.04\n", + "F13=1 \n", + "F32=1 \n", + "#The resistances are:\n", + "R1=(1-e1)/e1\n", + "R2=1.0/F13\n", + "R3=(1-e3)/e3\n", + "R4=R3\n", + "R5=1.0/F32\n", + "R6=(1-e2)/e2\n", + "R=R1+R2+R3+R4+R5+R6 #Total resistance\n", + "Q_by_As=z/R #where z=sigma*(T1**4-T2**4) #W/sq m\n", + "red=(Q_by_A-Q_by_As)*100/Q_by_A #Percent reduction in heat transfer\n", + "\n", + "#Result\n", + "print\"The heat transfer is reduced by\",round(red,1),\"% due to shield\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.22,Page no:4.34" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total heat lost by plate 1 is 14423.0 W/sq m\n", + "Total heat lost by plate 2 is 2598.0 W/sq m\n", + "The net energy lost by both plates must be absorbed by the room 17032.9 = 17021.0\n" + ] + } + ], + "source": [ + "#Radiaition shape factor\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 \n", + "T1=1273 #[K]\n", + "T2=773 #[K]\n", + "T3=300 #[K]\n", + "A1=0.5 #[sq m]\n", + "A2=A1 #[sq m]\n", + "F12=0.285 \n", + "F21=F12 \n", + "F13=1-F12 \n", + "F23=1-F21 \n", + "e1=0.2 \n", + "e2=0.5\n", + "\n", + "#Calculation\n", + "#Resistance in the network are calculated as:\n", + "R1=1-e1/(e1*A1)\n", + "R2=1-e2/(e2*A2)\n", + "R3=1.0/(A1*F12)\n", + "R4=1.0/(A1*F13)\n", + "R5=1.0/(A2*F23)\n", + "R6=0 #Given (1-e3)/e3*A3=0\n", + "#Also\n", + "Eb1=sigma*T1**4 #W/sq m\n", + "Eb2=sigma*T2**4 #[W/sq m]\n", + "Eb3=sigma*T3**4 #[W/sq m]\n", + "\n", + "#Equations are:\n", + "#(Eb1-J1)/2+(J2-J1)/7.018+(Eb3-J1)/2.797=0\n", + "#(J1-J2)/7.018+(Eb3-J2)/2.797+(Eb2-J2)/2=0\n", + "\n", + "#On solving we get:\n", + "J1=33515 #[W/sq m]\n", + "J2=15048 #[W/sqm]\n", + "J3=Eb3 #[W/sq m]\n", + "Q1=(Eb1-J1)/((1.0-e1)/(e1*A1)) #[W/sq m]\n", + "Q2=(Eb2-J2)/((1.0-e2)/(e2*A2)) #[W/sq m]\n", + "Q3=(J1-J3)/(1.0/(A1*F13))+(J2-J3)/(1.0/(A2*F23)) #[W/sq m]\n", + "\n", + "#Result\n", + "print\"Total heat lost by plate 1 is\",round(Q1),\"W/sq m\"\n", + "print\"Total heat lost by plate 2 is\",round(Q2),\"W/sq m\" \n", + "print\"The net energy lost by both plates must be absorbed by the room\",round(Q3,1),\"=\",round(Q1+Q2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.23,Page no:4.37" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " New Radiaiton loss is 4513.9 W/sq m\n" + ] + } + ], + "source": [ + "#Radiation loss in plates\n", + "\n", + "#Variable declaration\n", + "sigma=5.67*10**-8 #[W/sq m.K**4]\n", + "e1=0.7 \n", + "e2=0.7 \n", + "T1=866.5 #[K]\n", + "T2=588.8 #[K]\n", + "\n", + "#Calculation\n", + "Q_by_A=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)-1) #[W/sq m]\n", + "e1=0.7 \n", + "e2=e1 \n", + "e3=e1 \n", + "e4=e1 \n", + "e=e1 \n", + "#Q with n shells =1/(n+1)\n", + "n=2\n", + "Q_shield=1/(n+1) \n", + "es1=e1 \n", + "es2=e1 \n", + "Q_by_A=sigma*(T1**4-T2**4)/((1/e1)+(1/e2)+2*(1/es1+1/es2)-(n+1)) #[W/sq m]\n", + "\n", + "#Result\n", + "print\"\\n New Radiaiton loss is\",round(Q_by_A,1),\"W/sq m\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:4.24,Page no:4.38" + ] + }, + { + "cell_type": "code", + "execution_count": 77, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Without shield,Q= -4.182 W/m\n", + "With cylindrical radiaiton shield Heat gained by fluid per 1 m lengh of tube is -1.446 W/m\n", + "Percent reduction in heat gain is 65.41 %\n", + "With radiaiton network approach -1.45 W/sqm \n" + ] + } + ], + "source": [ + "#Radiation in Concentric tube\n", + "import math\n", + "\n", + "#Variable declaration\n", + "#1.WITHOUT SHIELD\n", + "sigma=5.67*10**-8 \n", + "e1=0.12 \n", + "e2=0.15 \n", + "T1=100 #[K]\n", + "T2=300 #[K]\n", + "r1=0.015 #[m]\n", + "r2=0.045 #[m]\n", + "L=1 #[m]\n", + "\n", + "#Calculation\n", + "A1=2*math.pi*r1*L #[sq m]\n", + "Q_by_L=2*math.pi*r1*sigma*(T1**4-T2**4)/(1/e1+(r1/r2)*(1/e2-1)) #[W/m]\n", + "#-ve saign indicates that the net heat flow is in the radial inward direction\n", + "print \"Without shield,Q=\",round(Q_by_L,3),\"W/m\"\n", + "#2.WITH CYLINDRICAL RADIATION SHIELD\n", + "e3=0.10 \n", + "e4=0.05 \n", + "r3=0.0225 #[m]\n", + "Qs_by_L=2*math.pi*r1*sigma*(T1**4-T2**4)/(1/e1+r1/r2*(1/e2-1)+(r1/r3)*(1/e3+1/e4-1)) #[W/sq m]\n", + "red=(abs(Q_by_L)-abs(Qs_by_L))*100/abs(Q_by_L) #percent reduction in heat gain\n", + "\n", + "#Radiation network approach\n", + "A3=2*math.pi*r3 #[sq m]\n", + "A2=2*math.pi*r2 #[sq m]\n", + "F13=1 \n", + "F32=1 \n", + "R1=(1-e1)/(e1*A1)\n", + "R2=1.0/(A1*F13)\n", + "R3=(1-e3)/(e3*A3)\n", + "R4=(1-e4)/(e4*A3)\n", + "R5=1.0/(A3*F32)\n", + "R6=(1-e2)/(e2*A2)\n", + "\n", + "Qs=sigma*(T1**4-T2**4)/((1.0-e1)/(e1*A1)+1/(A1*F13)+(1.0-e3)/(e3*A3)+(1.0-e4)/(e4*A3)+1/(A3*F32)+(1.0-e2)/(e2*A2)) \n", + "\n", + "#Result\n", + "print\"With cylindrical radiaiton shield Heat gained by fluid per 1 m lengh of tube is\",round(Qs_by_L,3),\"W/m\" \n", + "print\"Percent reduction in heat gain is\",round(red,2),\"%\"\n", + "print\"With radiaiton network approach\",round(Qs,2),\"W/sqm \"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_by_K._A._Gavhane/Chapter_5.ipynb b/Heat_Transfer_by_K._A._Gavhane/Chapter_5.ipynb new file mode 100644 index 00000000..034f1dae --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/Chapter_5.ipynb @@ -0,0 +1,1823 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter5: Heat Exchangers" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.1,Page no:5.22" + ] + }, + { + "cell_type": "code", + "execution_count": 156, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.0236\n", + "Required surface is fulfilled by connecting 3.0 (three) 6m harpins in series\n" + ] + } + ], + "source": [ + "#Harpin exchanger\n", + "import math\n", + "#Variable declaration\n", + "Di=35.0 #[mm]\n", + "Di=Di/1000 #[m]\n", + "Do=42.0 #[mm]\n", + "Do=Do/1000 #[m]\n", + "#for benzene\n", + "mb_dot=4450.0 #[kg/h]\n", + "Cpb=1.779 #[kJ/(kg.K)]\n", + "t2=322.0 #[K]\n", + "t1=300.0 #[K]\n", + "\n", + "#Calculation\n", + "Q=mb_dot*Cpb*(t2-t1) #for benzene in [kJ/h]\n", + "#For toulene\n", + "T1=344.0 #[K]\n", + "T2=311.0 #[K]\n", + "Cpt=1.842 #[kJ/kg.K]\n", + "mt_dot=Q/(Cpt*(T1-T2)) #[kg/h]\n", + "Q=Q*1000.0/3600 #[W]\n", + "#Hot fluid(toluene)\n", + "#Cold fluid(benzene)\n", + "dT1=22.0 #[K]\n", + "dT2=11.0 #[K]\n", + "dTlm=(dT1-dT2)/(math.log(dT1/dT2)) #[K]\n", + "#Clod fluid:Inner pipe,benzene\n", + "Di=0.035 #[m]\n", + "Ai=(math.pi/4)*Di**2 #Flow area[sq m]\n", + "Gi=mb_dot/Ai #Mass velocity [kg/m**2.h]\n", + "Gi=Gi/3600 #[kg/m**2.s]\n", + "mu=4.09*10**-4 #[kg/(m.s)]\n", + "Nre=Di*Gi/mu #Reynolds number\n", + "Cp=Cpb*10**3 #[J/(kg.K)]\n", + "k=0.147 #[W/m.K]\n", + "Npr=Cp*mu/k #Prandtl number\n", + "hi=(k/Di)*0.023*(Nre**0.8)*(Npr**0.4) #[W/sq m.K]\n", + "hio=hi*Di/Do #[W/sq m.K]\n", + "D1=0.042 #Outside dia of inside pipe[mm]\n", + "D2=0.0525 #Inside dia of outside pipe[m]\n", + "De=round((D2**2-D1**2)/D1,4) #[m]\n", + "aa=math.pi*(D2**2-D1**2)/4 #Flow area[sq m]\n", + "Ga=mt_dot/aa #Mass velocity in [kg/m**2.h]\n", + "Ga=Ga/3600 #[kg/m**2.s]\n", + "mu=5.01*10**-4 #[kg/(m.s)]\n", + "Nre=De*Ga/mu #Reynolds number\n", + "Npr=Cp*mu/k #Prandtl number\n", + "ho=(k/De)*0.023*(Nre**0.8)*(Npr**0.3) #[W/sq m.K]\n", + "Uc=1.0/(1.0/ho+1.0/hio) #[W/sq m.K]\n", + "Rdi=1.6*10**-4 #Fouling factor [m**2.K/W]\n", + "Rdo=1.6*10**-4 #Fouling factor[m**2.K/W]\n", + "Rd=Rdi+Rdo #(m**2.K/W)\n", + "Ud=1.0/(1.0/Uc+Rd) #[W/sq m.K]\n", + "A=Q/(Ud*dTlm) #sq m\n", + "ex=0.136 #[sq m]\n", + "l=A/ex #m\n", + "tl=12 #Total length of one harpin of 6m [m]\n", + "\n", + "#Result\n", + "print\"Required surface is fulfilled by connecting\",round(l/tl),\"(three) 6m harpins in series\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.2,Page no:5.26" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Area = 16.15 m**2,\n", + "Length fo pipe required = 51.41 m(approx)\n" + ] + } + ], + "source": [ + "#Length of pipe\n", + "import math\n", + "\n", + "#Variable declaration\n", + "ma_dot=300.0*1000.0/24 #Mass flow rate of acid in [kg/h]\n", + "mw_dot=500.0*1000.0/24 #Mass flow rate of water in [kg/h]\n", + "Cp1=1.465 #[kJ/kg.K]\n", + "T1=333.0 #[K]\n", + "T2=313.0 #[K]\n", + "Q=ma_dot*Cp1*(T1-T2) #[kJ/h]\n", + "Q=Q*1000.0/3600.0 #[W]\n", + "Cp2=4.187 #[kJ/kg.K]\n", + "t1=288.0 #[K]\n", + "\n", + "#Calculation\n", + "t2=(Q/(mw_dot*Cp2))+t1 #[K]\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K] \n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "dTlm=32.26 #Approximation in [K]\n", + "#Inner pipe\n", + "m_dot=12500.0 #[kg/h]\n", + "Di=0.075 #[m]\n", + "Ai=(math.pi/4.0)*Di**2 #[sq m]\n", + "G=ma_dot/Ai #[kg/m**2.h]\n", + "G=G/3600 #[kg/m**2.s]\n", + "mu=0.0112 #[kg/m.s]\n", + "k=0.302 #W/(m.K)\n", + "Nre=Di*G/mu #Reynold number\n", + "Npr=Cp1*10**3*mu/k #Prandtl number\n", + "hi=(k/Di)*0.023*(Nre**0.8)*(Npr**0.3) #W/sq m.K\n", + "Do=0.1 #[m]\n", + "hio=hi*Di/Do #W/sq m.K\n", + "D1=0.1 #[m]\n", + "D2=0.125 #[m]\n", + "De=(D2**2-D1**2)/D1 #[m]\n", + "Aa=(math.pi/4)*(D2**2-D1**2) #[sq m]\n", + "Ga=mw_dot/Aa #[kg/m**2.h]\n", + "Ga=Ga/3600 #[kg/sq m.s]\n", + "mu=0.0011 #[kg/m.s]\n", + "Nre=De*Ga/mu #Reynolds number\n", + "k=0.669 #for water\n", + "Npr=Cp2*10**3*mu/k #Prandtl number\n", + "ho=(k/De)*0.023*(Nre**0.8)*Npr**0.4 #[W/sq m.K]\n", + "xw=(Do-Di)/2 #[m]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "kw=46.52 #thermal conductivity of wall in [W/m.K]\n", + "Uc=1.0/(1.0/ho+1.0/hio+xw*Do/(kw*Dw)) #[W/sq m.K]\n", + "Ud=Uc #As dirt factor values are not given\n", + "round(Ud,1)\n", + "#Ud=195.32 #Approximation\n", + "A=Q/(Ud*dTlm) #[sq m]\n", + "L=A/(math.pi*Do) #[sq m]\n", + "\n", + "#Result\n", + "print\"Area =\",round(A,2),\"m**2,\\nLength fo pipe required =\",round(L,2),\"m(approx)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.3,Page no:5.29" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total lenggth of double pipe heat exchanger is 36.9 m\n" + ] + } + ], + "source": [ + "#Double pipe heat exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "me_dot=5500.0 #[kg/h]\n", + "me_dot1=me_dot/3600 #[kg/s]\n", + "Di=0.037 #I.D of inner pipe in [m]\n", + "Ai=(math.pi/4)*Di**2 #[sq m]\n", + "G=me_dot1/Ai #[kg/sq m.s]\n", + "mu=3.4*10**-3 #[Pa.s] or [kg/(m.s)]\n", + "Nre=Di*G/mu #Reynolds number\n", + "Cp=2.68 #[kJ/kg.K]\n", + "Cp1=Cp*10**3 #[J/kgK]\n", + "k=0.248 #[W/m.K]\n", + "\n", + "#Calculation\n", + "Npr=Cp1*mu/k #Prandtl number\n", + "#Nre is greater than 10,000,Use Dittus-Boelter eqn:\n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.3) #Nusselt number\n", + "hi=k*Nnu/Di #[W/sq m.K]\n", + "T2=358.0 #[K]\n", + "T1=341 #[K]\n", + "Cp2=1.80 #[kJ/kg.K]\n", + "t2=335 #[K]\n", + "t1=303.0 #[K]\n", + "mt_dot=me_dot*Cp*(T2-T1)/(Cp2*(t2-t1)) #[kg/h]\n", + "mt_dot=mt_dot/3600 #[kg/s]\n", + "D1=0.043 #[m]\n", + "D2=0.064 #Inside dia of outer pipe\n", + "De=(D2**2-D1**2)/D1 #Equivalent diameter [m]\n", + "Aa=math.pi/4*(D2**2-D1**2) #[sq m]\n", + "Ga=mt_dot/Aa #kg/(sq m.s)\n", + "mu2=4.4*10**-4 # Viscosity of toluene Pa.s\n", + "k2=0.146 #For toluene [W/m.K]\n", + "Cp2=1.8*10**3 #J/kg.K\n", + "Nre=De*Ga/mu2 #Reynolds number\n", + "Npr=Cp2*mu2/k2 #Prandtl number\n", + "Nnu=0.023*Nre**0.8*Npr**0.4 #Nusselt number\n", + "ho=k2*Nnu/De #W/(sq m.K)\n", + "Dw=(D1-Di)/math.log(D1/Di) #[m]\n", + "x=0.003 #Wall thickness in [m]\n", + "Uo=1.0/(1.0/ho+(1.0/hi)*(D1/Di)+(x*D1/(46.52*Dw))) #[W/sq m.K]\n", + "dT1=38.0 #[K]\n", + "dT2=23 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Q=me_dot1*Cp*(T2-T1) #[kJ/s]\n", + "Q=Q*1000 #[J/s]\n", + "L=Q/(Uo*math.pi*D1*dTlm) #[m]\n", + "\n", + "#Result\n", + "print\"Total lenggth of double pipe heat exchanger is\",round(L,1),\"m\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.4,Page no:5.35" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Effectiveness of heat exchanger is 0.647\n", + "Outlet temperature of hot liquid is 321.9 K( 48.9 degree C)\n", + "Outlet temperature of water is 297.23 K( 24.23 degree C)\n" + ] + } + ], + "source": [ + "#Parallel flow arrangement\n", + "import math\n", + "#Variable declaration\n", + "mc_dot=1000.0 #[kg/h]\n", + "mc_dot=mc_dot/3600 #[kg/s]\n", + "mh_dot=250.0 #[kg/h]\n", + "mh_dot=mh_dot/3600 #[kg/s]\n", + "Cpc=4187.0 #[J/(kg.K)]\n", + "Cph=3350 #[W/K]\n", + "\n", + "#Calculation\n", + "w=mc_dot*Cpc #[W/K]\n", + "l=mh_dot*Cph #[W/K]\n", + "C=mh_dot*Cph/(mc_dot*Cpc)\n", + "U=1160.0 #[W/sq m.K]\n", + "A=0.25 #Heat transfer surface for exchanger in [sq m]\n", + "ntu=U*A/(mh_dot*Cph) #\n", + "E=(1-math.exp(-ntu*(1+C)))/(1+C) #Effectiveness of heat exchanger\n", + "T1=393.0 #Inlet temperature in [K]\n", + "t1=283 #Cooling water [K]\n", + "T2=T1-E*(T1-t1) #Outlet T of hot liquid \n", + "\n", + "t2=C*(T1-T2)+t1 #[K]\n", + "\n", + "#Result\n", + "print\"Effectiveness of heat exchanger is \",round(E,3)\n", + "print\"Outlet temperature of hot liquid is \",round(T2,1),\"K(\",round(T2,1)-273,\"degree C)\"\n", + "print\"Outlet temperature of water is \",round(t2,2),\"K(\",round(t2,2)-273,\"degree C)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.5,Page no:5.36" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Effectiveness of exchanger is 0.9512\n", + "Outlet temperature of oil is 291.85 K( 18.85 degree C)\n", + "Outlet temperature of water is 303.19 K( 303.19 degree C)\n", + "Rate of heat transfer is 22962.5 W\n" + ] + } + ], + "source": [ + "#Counter flow exchanger\n", + "import math\n", + "#Variable declaration\n", + "Cpc=4187.0 #Specific heat of water in [J/(kg.K)]\n", + "Cph=2000.0 #Sp heat of oil in [J/(kg.K)]\n", + "mc_dot=1300.0/3600 #[kg/s]\n", + "mh_dot=550.0/3600 #[kg/s]\n", + "w=mc_dot*Cpc #[W/K]\n", + "o=mh_dot*Cph #[W/K]\n", + "\n", + "#Calculation\n", + "#Heat capacity of rate of hot fluid is smaller than water\n", + "U=1075.0 #[W/sq m.K]\n", + "A=1.0 #[sq m]\n", + "ntu=(U*A)/(mh_dot*Cph) \n", + "C=mh_dot*Cph/(mc_dot*Cpc) \n", + "E=(1-math.exp(-ntu*(1-C)))/(1-C*math.exp(-ntu*(1-C))) #Effeciency\n", + "T1=367.0 #[K]\n", + "t1=288.0 #[K]\n", + "T2=T1-E*(T1-t1) #Outlet temperature [K]\n", + "T2=round(T2,2)\n", + "#T2=291.83 #Approximated in book without precise calculation\n", + "t2=C*(T1-T2)+t1 #[K]\n", + "Q=mh_dot*Cph*(T1-T2) #[W]\n", + "\n", + "#Result\n", + "print\"Effectiveness of exchanger is\",round(E,4)\n", + "print\"Outlet temperature of oil is\",T2,\"K(\",T2-273,\"degree C)\"\n", + "print\"Outlet temperature of water is\",round(t2,2),\"K(\",round(t2,2),\"degree C)\"\n", + "print\"Rate of heat transfer is\",round(Q,1),\"W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.6,Page no:5.37" + ] + }, + { + "cell_type": "code", + "execution_count": 62, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "By LMTD approach area of heat exchanger is 0.607 m^2\n", + "By Ntu approach Area of heat exchanger is 0.607 m^2\n", + "Outlet temperature of cold water= 306.0 K( 33.0 degree C)\n" + ] + } + ], + "source": [ + "#LMTD approach\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "#LMTD Approach\n", + "Cph=4187.0 #[J/(kg.K)]\n", + "mh_dot=600.0/3600 #Hot side flow rate [kg/s]\n", + "mc_dot=1500.0/3600 #[kg/s]\n", + "Cpc=Cph #[J/kg.K]\n", + "T1=343.0 #[K]\n", + "T2=323.0 #[K]\n", + "\n", + "#Calculation\n", + "Q=mh_dot*Cph*(T1-T2) #[W]\n", + "t1=298.0 #[K]\n", + "t2=(mh_dot*Cph*(T1-T2))/(mc_dot*Cpc)+t1 #[K]\n", + "dT1=45.0 #[K]\n", + "dT2=17.0 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "hi=1600.0 #Heat transfer coeff in [W/sq m.K]\n", + "ho=hi #[W/sq m.K]\n", + "U=1.0/(1.0/hi+1.0/ho) #[W/sq m.K]\n", + "A=Q/(U*dTlm) #[sq m]\n", + "\n", + "#Effectiveness-NTU approach\n", + "#hot water:\n", + "h=mh_dot*Cph #[W/K]\n", + "c=mc_dot*Cpc #[W/K]\n", + "#Heat capacity rate of hot fluid is small\n", + "C=mh_dot*Cph/(mc_dot*Cpc) #\n", + "E=(T1-T2)/(T1-t1) #Effectiveness\n", + "#for paralell flow:\n", + "ntu=-math.log(1-E*(1+C))/(1+C) \n", + "A2=(ntu*mh_dot*Cph)/U #[sq m]\n", + "t2=C*(T1-T2)+t1 #[K]\n", + "\n", + "#Result\n", + "print\"By LMTD approach area of heat exchanger is\",round(A,3),\"m^2\"\n", + "print\"By Ntu approach Area of heat exchanger is\",round(A,3),\"m^2\"\n", + "print\"Outlet temperature of cold water=\",t2,\"K(\",t2-273,\"degree C)\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.7,Page no:5.39" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of tubes required = 44.0\n" + ] + } + ], + "source": [ + "#Shell and tube exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mw_dot=10.0 #[kg/s]\n", + "Cpw=4.187 #[kJ/(kg.K)]\n", + "t2=318.0 #[K]\n", + "t1=295 #[K]\n", + "Q=mw_dot*Cpw*(t2-t1) #[kJ/s]\n", + "Q=Q*1000 #W\n", + "dT1=98.0 #[K]\n", + "dT2=75 #[K]\n", + "\n", + "#Calculation\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "hi=850.0 #[W/sq m.K]\n", + "id=0.027 #Inside dia[m]\n", + "od=0.031 #Outside dia [m]\n", + "hio=hi*id/od #[W/sq m.K]\n", + "ho=6000.0 #Heat transfer coefficients[W/sq m.K]\n", + "Uo=1.0/(1.0/ho+1.0/hio) #[W/sq m.K]\n", + "Ao=Q/(Uo*dTlm) #[sq m]\n", + "L=4.0 #Length [m]\n", + "n=Ao/(math.pi*od*L) #[No. of tubes]\n", + "\n", + "#Result\n", + "print\"Number of tubes required =\",round(n)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.8,Page no:5.40" + ] + }, + { + "cell_type": "code", + "execution_count": 74, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fouling factor=Sclae resistance=9.65e-04 m^2 K/W\n" + ] + } + ], + "source": [ + "#Order of Scale resistance\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mdot=7250 #Nitrobenzene in shell in [kg/h]\n", + "Cp=2.387 #[kJ/(kg.K)]\n", + "mu=7*10**-4 #Pa.s\n", + "k=0.151 #[W/m.K]\n", + "T1=400.0 #[K]\n", + "T2=317.0 #[K]\n", + "t1=305.0 #[K]\n", + "t2=345.0 #[K]\n", + "\n", + "#Calculation\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Q=mdot*Cp*(T1-T2) #[kJ/h]\n", + "Q=Q*1000.0/3600 #[W]\n", + "n=166.0 #no of tubes\n", + "L=5.0 #[m]\n", + "Do=0.019 #[m]\n", + "Di=0.015 #[m]\n", + "Ao=n*math.pi*Do*L #[sq m]\n", + "Uo=Q/(Ao*dTlm) #[W/sq m.K]\n", + "Ud=Uo\n", + "#Shell side heat transfer coefficient\n", + "Pt=0.025 #[m]\n", + "C_dash=Pt-(0.5*Do+0.5*Do)\n", + "\n", + "#Shell side crossflow area\n", + "B=0.15 #[m]\n", + "id=0.45 #[m]\n", + "ass=id*C_dash*B/Pt #[sq m]\n", + "#As there are two shell passes,area per pass is :\n", + "as_dash=ass/2 #[sq m]\n", + "\n", + "#Equivalent diameter of shell \n", + "De=4*(Pt**2-(math.pi/4)*Do**2)/(math.pi*Do) #[m]\n", + "\n", + "#Mass velocity on shell side\n", + "Gs=mdot/as_dash #[kg/m**2.h]\n", + "Gs=Gs/3600 #[kg/m**2.s]\n", + "mu=7.0*10**-4 #Pa.s\n", + "Cp=Cp*1000 #J/kg.K\n", + "Nre=De*Gs/mu #Reynold number\n", + "Npr=Cp*mu/k #Prandtls number\n", + "Nnu=0.36*Nre**0.55*Npr**(1.0/3.0) #Nusselts number\n", + "hi=1050.0 #[W/sq m .K]\n", + "ho=Nnu*k/De #[W/sq m.K]\n", + "Uo=1.0/(1.0/ho+(1.0/hi*(Do/Di))) #[W/sq m K]\n", + "Uc=Uo\n", + "Rd=(Uc-Ud)/(Uc*Ud) #m**2.K/W\n", + "\n", + "#Result\n", + "print\"Fouling factor=Sclae resistance=%.2e\" %round(Rd,6),\"m^2 K/W\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.9,Page no:5.42" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of tube required is 4.4 m\n" + ] + } + ], + "source": [ + "#Length of tube required\n", + "import math\n", + "#Variable declaration\n", + "k=0.628 #W/(m.K)\n", + "rho=980.0 #[kg/m**3]\n", + "mu=6*10**-4 #kg/(m.s)\n", + "Cpw=4.187 #kJ/(kg.K)\n", + "Cp=Cpw*10**3 #J/(kg.K)\n", + "Di=25.0 #[mm]\n", + "Di=Di/1000 #[m]\n", + "\n", + "#Calculation\n", + "mw_dot=1200.0*10**-3*rho #Mass flow rate of water [kg/h]\n", + "mw_dot=mw_dot/3600.0 #[kg/s]\n", + "Ai=(math.pi*Di**2)/4.0 #Inside area of tube in sq m\n", + "G=mw_dot/Ai #kg/m**2.s\n", + "Nre=Di*G/mu #Reynolds number\n", + "Npr=Cp*mu/k #Pranddtl number\n", + "#Inside heat transfer coefficient\n", + "Nnu=0.023*Nre**0.8*Npr**0.4 #Nusselt number\n", + "hi=Nnu*k/Di #[W/sq m.K]\n", + "ho=6000.0 #[W#sq m.K]\n", + "Do=0.028 #[m]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "x=(Do-Di)/2 #[m]\n", + "k2=348.9 #thermal conductivity of metal in [W/m.K]\n", + "Uo=1.0/((1.0/ho)+(1.0/hi)*(Do/Di)+(x/k2)*(Do/Dw)) #[W/sq m.K]\n", + "t1=303.0 #[K]\n", + "t2=343.0 #[K]\n", + "Q=mw_dot*Cpw*(t2-t1) #[kJ/h]\n", + "Q=Q*1000.0 #[W]\n", + "Ts=393.0 #[K]\n", + "dT1=Ts-t1 #[K]\n", + "dT2=Ts-t2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Ao=Q/(Uo*dTlm) #[sq m]\n", + "L=Ao/(math.pi*Do) #Length\n", + "\n", + "#Result\n", + "print\"Length of tube required is\",round(L,2),\"m\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.10,Page no:5.44" + ] + }, + { + "cell_type": "code", + "execution_count": 87, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rd calculated(1.01e-03 W/m^2.K) is maximum allowable scale resistance\n", + "As Rd calculated(1.01e-03 W/m^2K) is more than Rd given 9e-04( W/m^2,K),the given heat exchanger is suitable\n" + ] + } + ], + "source": [ + "#Suitability of Exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "m_dot=7250.0 #[kg/h] of nitrobenzene \n", + "Cp=2.387 #[kJ/kg.K]\n", + "mu=7*10**-4 #[kg/m.s]\n", + "k=0.151 #[W/m.K]\n", + "vis=1.0 \n", + "Ft=0.9 #LMTD correction factor\n", + "T1=400.0 #[K]\n", + "T2=317.0 #[K]\n", + "t1=333.0 #[K]\n", + "t2=300.0 #[K]\n", + "\n", + "#Calculation\n", + "dT1=T1-t1 #[K]\n", + "dT2=T2-t2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "#For nitrobenzene\n", + "Q=m_dot*Cp*(T1-T2) #[kJ/h]\n", + "Q=Q*1000.0/3600 #[W]\n", + "n=170.0 #No. of tubes\n", + "L=5.0 #[m]\n", + "Do=0.019 #[m]\n", + "Di=0.015 #[m]\n", + "Ao=n*math.pi*Do*L #[sq m]\n", + "Uo=Q/(Ao*Ft*dTlm) #[W/sq m.K]\n", + "Ud=Uo #[W/sq m.K]\n", + "B=0.15 #Baffle spacing [m]\n", + "Pt=0.025 #Tube pitch in [m]\n", + "C_dash=Pt-Do # ance in [m]\n", + "id=0.45 #[m]\n", + "\n", + "#Shell side cross flow area\n", + "ass=id*C_dash*B/Pt #[sq m]\n", + "\n", + "#Equivalent diameter of shell\n", + "De=4*(Pt**2-(math.pi/4)*(Do**2))/(math.pi*Do) #[m]\n", + "\n", + "#Mass velocity on shell side\n", + "Gs=m_dot/ass #[kg/(m.h)]\n", + "Gs=Gs/3600.0 #[kg/m**2.s]\n", + "mu=7.0*10**-4 #[kg/m.s]\n", + "Cp=Cp*1000.0 #[J/kg.K]\n", + "Nre=De*Gs/mu #Reynolds number\n", + "Npr=Cp*mu/k #Prandtl number\n", + "\n", + "#From empirical eqn:\n", + "mu_w=mu #\n", + "Nnu=0.36*Nre**0.55*Npr**(1.0/3.0)\n", + "ho=Nnu*k/De #[W/sq m.K]\n", + "hi=1050.0 #Given [W/sq m.K]\n", + "Uo=1.0/(1.0/ho+(1.0/hi)*(Do/Di)) #[W/sq m.K]\n", + "Uc=Uo #W/sq m.K\n", + "\n", + "#Suitability of heat exchanger\n", + "Rd_given=9*10**-4 #[W/sq m.K]\n", + "Rd=(Uc-Ud)/(Uc*Ud) #[W/sq m.K]\n", + "\n", + "#Result\n", + "print\"Rd calculated(%.2e\"%Rd,\"W/m^2.K) is maximum allowable scale resistance\"\n", + "print\"As Rd calculated(%.2e\"%Rd,\"W/m^2K) is more than Rd given %.e(\"%Rd_given,\" W/m^2,K),the given heat exchanger is suitable\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.11,Page no:5.47" + ] + }, + { + "cell_type": "code", + "execution_count": 90, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of tubes required= 1.0\n" + ] + } + ], + "source": [ + "#Number of tubes required\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mw_dot=1720.0 #water in [kg/h]\n", + "t1=293.0 #[K]\n", + "t2=318.0 #[K]\n", + "Cpw=4.28 #[kJ/kg.K]\n", + "\n", + "#Calculation\n", + "Q=mw_dot*Cpw*(t2-t1) #[kJ/h]\n", + "Q=Q*1000.0/3600 #W\n", + "lamda=2230.0 #[kJ/kg]\n", + "dT1=90.0 #[K]\n", + "dT2=65.0 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "\n", + "#Calculation of inside heat transfer coefficient\n", + "Di=0.0225 #[m]\n", + "u=1.2 #[m/s]\\\n", + "rho=995.7 #[kg/m**3]\n", + "v=0.659*10**-6 #[m/s]\n", + "mu=v*rho #[kg/m.s]\n", + "Nre=Di*u*rho/mu #reynolds number\n", + "Cp=Cpw*1000 #[J/kg.K]\n", + "k=2.54 #[kJ/h.m.K]\n", + "k=k*1000.0/3600 #[W/m.K]\n", + "Npr=Cp*mu/k #Prandtl number\n", + "Nnu=0.023*Nre**0.8*Npr**0.4 #Nusselt number\n", + "hi=k*Nnu/Di #[W/sq m.K]\n", + "ho=19200.0 #[kJ/h.m**2.K]\n", + "ho=ho*1000.0/3600 #[W/m**2.K]\n", + "Do=0.025 #[m]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "x=(Do-Di)/2 #[m]\n", + "kt=460.0 #For tube wall material [kJ/h.m.K]\n", + "kt=kt*1000.0/3600 #[W/m.K]\n", + "Uo=1.0/(1.0/ho+(1.0/hi)*(Do/Di)+(x/kt)*(Do/Dw)) #[W/sq m.K]\n", + "#Q=Uo*Ao*dTlm\n", + "Ao=Q/(Uo*dTlm) #[sq m]\n", + "L=4.0 #Tube length in [m]\n", + "n=Ao/(math.pi*Do*L) #[Number of tubes]\n", + "\n", + "\n", + "#Result\n", + "print\"Number of tubes required=\",round(n)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.12,Page no:5.49" + ] + }, + { + "cell_type": "code", + "execution_count": 92, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Water velocity must be 3.88 m/s\n" + ] + } + ], + "source": [ + "#Shell and tube heat exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "t1=290.0 #Inlet temperature of cooling water [K]\n", + "ho=2250.0 #Heat transfer coefficient based on inside area in [W/sq m.K]\n", + "lamda=400.0 #[kJ/kg] LAtent heat of benzene \n", + "mb_dot=14.4 #[t/h] Condensation rate of benzene vapour\n", + "Cpw=4.187 #Specific heat\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "#With no Scale\n", + "Q=mb_dot*1000.0*lamda #Heat duty of condenser in [kJ/h]\n", + "Q=(Q/3600.0)*1000.0 #[W]\n", + "#Shell and tube type of heat exchanger is used as a single pass surface condenser\n", + "Di=0.022 #I.D of tube[m]\n", + "L=2.5 #Length of each tube in [m]\n", + "n=120.0 #Number of tubes\n", + "A=math.pi*Di*L #Area of heat transfer per metre length in [m**2/m]\n", + "A=n*A #Total area of heat transfer in [m**2]\n", + "Ai=(math.pi/4.0)*Di**2 #Cross-sectional area of each tube in [m**2]\n", + "Ai=n*Ai #Total area of flow in [m**2]\n", + "u=0.75 #Velocty of water [ms**-1]\n", + "V=u*Ai #Volumetric flow of water \n", + "rho=1000.0 #[Density of water in [kg/m**3]]\n", + "mw_dot=V*rho #Mass flow rate of water in [kg/s]\n", + "\n", + "#Heat balance\n", + "#Q=mw_dot*Cpw*(t2-t1)\n", + "t2=Q/(mw_dot*Cpw*1000)+t1 #[K]\n", + "T=350.0 #Condensing benzene temperature in [K]\n", + "dT1=T-t1 #[K]\n", + "dT2=T-t2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #LMTD\n", + "U=Q/(A*dTlm) #[W/m**2.K]\n", + "U=round(U)\n", + "#Neglecting resistance,we have:\n", + "hi=1/(1/U-1/ho) #[W/m**2.K]\n", + "#hi is proportional to u**0.8\n", + "C=hi/(u**0.8) #Constant\n", + "\n", + "#With Scale\n", + "Rd=2.5*10**-4 #[m**2 K./W]\n", + "#1/U=1/hi+1/ho+Rd\n", + "#U=hi/(1+3.38*u**0.8)\n", + "#mw_dot=rho*u*Ai #[kg/s]\n", + "#Let t2 be the outlet temperature of water \n", + "#Q=mw_dot*Cpw*(t2-t1)\n", + "#t2=Q/(mw_dot*Cpw)+t1 \n", + "dT1=60.0\n", + "#dT2=T-(t1+8.373/u)\n", + "#dTlm=8.373/(u*math.log(60*u/(60*u-8.373)))\n", + "#Q=U*A*dTlm\n", + "#1.89=((u**-0.2)/(1+3.38*u**0.8))*(1/math.log((60*u)/60*u-8.373)\n", + "#If we assume values of u greater than 0.75 m/s\n", + "#For u=3.8 #[ms**-1]\n", + "u=3.8 #]ms**-1]\n", + "\n", + "#Result\n", + "print\"Water velocity must be 3.88 m/s\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.13,Page no:5.52" + ] + }, + { + "cell_type": "code", + "execution_count": 93, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No. of tubes= 100.0 \n", + "Length of tube= 2.48 m\n" + ] + } + ], + "source": [ + "#Length of pipe in Exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mh_dot=1.25 #[kg/s]\n", + "Cpw=4.187*10**3 #Heat capacity of water in [J/kg.K]\n", + "lamda=315.0 #[kJ/kg]\n", + "Q=mh_dot*lamda #Rate of heat transfer from vapour [kJ/s]\n", + "Q=Q*10**3 #[W]\n", + "Ts=345.0 #Temperature of condensing vapour[K]\n", + "t1=290.0 #Inlet temperature of water [K]\n", + "t2=310.0 #Outlet temperature of water[K]\n", + "\n", + "#Calculation\n", + "dT1=Ts-t1 #[K]\n", + "dT2=Ts-t2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "#Heat removed from vapour = Heat gained\n", + "mw_dot=Q/(Cpw*(t2-t1)) #[kg/s]\n", + "hi=2.5 #[kW/sq m.K]\n", + "hi=hi*1000 #[W/sq m.K]\n", + "Do=0.025 #[m]\n", + "Di=0.020 #[m]\n", + "hio=hi*(Di/Do) #Inside heat transfer cosfficient referred to outside dia in [W/sq m.K]\n", + "ho=0.8 #Outside heat tranbsfer coefficient in [kW/sq m.K]\n", + "ho=ho*1000.0 #[W/sq m.K]\n", + "Uo=1.0/(1.0/ho+1.0/hio) #[W/sq m.K]\n", + "#Ud is 80% of Uc\n", + "Ud=(80.0/100)*Uo #[W/sq m.K]\n", + "Ao=Q/(Ud*dTlm) #[sq m]\n", + "L=1.0 #[m]\n", + "A=math.pi*Do*L #Outside area of pipe per m length of pipe\n", + "len=Ao/A #Total length of piping required.\n", + "rho=1000.0 #[kg/m**3]\n", + "V=mw_dot/rho #[m**3/s]\n", + "v=0.6 #[m/s]\n", + "a=V/v #Cross-sectional area for flow pass [sq m]\n", + "a1=(math.pi*Di**2.0)/4 #[sq m]\n", + "#for single pass on tube side fluid(water)\n", + "n=round(a/a1) #No. of tubes per pass\n", + "l=len/n #Length of each tube in [m]\n", + "#For two passes on water side:\n", + "tn=2*n #Total no of tubes\n", + "l2=len/tn #Length of each tube in [m]\n", + "#For four passes on water side/tube side\n", + "tn2=4*n #Total no. of tubes\n", + "l3=len/tn2 #Length of each tube in [m]\n", + "\n", + "\n", + "#Result\n", + "print\"No. of tubes=\",tn2,\"\\nLength of tube=\",round(l3,2),\"m\"\n", + "\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.14,Page no:5.54" + ] + }, + { + "cell_type": "code", + "execution_count": 99, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NOTE:The calculation of line no.36 to calculate 'ass' is wrongly done in Book by printing 0.0353,,..which is wrong\n", + "\n", + "Rd=7.34e-04 K/W,\n", + "which is less than the provided,so this if installed will not give required temperatures without frequent cleaning\n" + ] + } + ], + "source": [ + "#Dirt factor\n", + "\n", + "import math\n", + "\n", + "#Variable declaration\n", + "\n", + "#Properties of crude oil:\n", + "Cpc=1.986 #[kJ/(kg.K)]\n", + "mu1=2.9*10**-3 #[N.s/sq m]\n", + "k1=0.136 #[W/m.K]\n", + "rho1=824 #[kg/m**3]\n", + "#Properties of bottom product:\n", + "Cp2=2.202 #[kJ/kg.K]\n", + "rho2=867.0 #[kg/m**3]\n", + "mu2=5.2*10**-3 #[N.s/sq m]\n", + "k2=0.119 #[W/sq m.K]\n", + "mc_dot=135000.0 #Basis: cruid oil flow rate in [kg/h]\n", + "m_dot=106000.0 #Bottom product flow rate inn [kg/h]\n", + "t1=295.0 #[K]\n", + "t2=330.0 #[K]\n", + "T1=420.0 #[K]\n", + "T2=380.0 #[K]\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Q=mc_dot*Cpc*(t2-t1) #kJ/h\n", + "Q=Q*1000.0/3600 #[W]\n", + "#Shell side calculations:\n", + "Pt=25.0 #[mm]\n", + "Pt=Pt/1000 #[m]\n", + "B=0.23 #[m]\n", + "Do=0.019 #[m]Outside diameter for square pitch \n", + "c_dash=Pt-Do # ance in [m]\n", + "id=0.6 #[m]\n", + "ass=id*c_dash*B/Pt # Cross flow area of shell [sq m]\n", + "#since there is a Calculaiton mistake ,we take:\n", + "ass=0.0353 \n", + "Gs=m_dot/ass #Shell side mass velocity in [kg/sq m.h]\n", + "Gs=Gs/3600 #[kg/sq m.s]\n", + "De=4*(Pt**2-(math.pi/4)*Do**2)/(math.pi*Do) #[m]\n", + "Nre=De*Gs/mu2 #Reynolds number\n", + "Npr=Cp2*1000*mu2/k2 #Prandtl number\n", + "muw=mu2 #Since mu/muw=1\n", + "Nnu=0.36*(Nre**0.55)*Npr**(1.0/3.0)*(mu2/muw)**(0.14) #Nusselt number\n", + "ho=Nnu*k2/De #[W/sq m.K]\n", + "#Tube side heat transfer coefficient:\n", + "n=324.0 #No. of tubes\n", + "n_p=324.0/2 #No.of tubes per pass\n", + "t=2.1 #Thickness in [mm]\n", + "t=t/1000 #[m]\n", + "Di=Do-2*t #I.d of tube in [m]\n", + "A=(math.pi/4)*(Di**2) #Cross-sectional area of one tube in [sq m]\n", + "A_p=n_p*A #Total area for flow per pass in [sq m]\n", + "G=mc_dot/A_p #[kg/sq m h]\n", + "G=G/3600.0 #[kg/sq m.s]\n", + "Nre=Di*G/mu1 #Reynolods number\n", + "Npr=42.35 #Prandtl number\n", + "Nnu=0.023*(Nre**0.8)*(Npr**0.4) #Nusselt number\n", + "hi=Nnu*k1/Di #[W/sq m.K]\n", + "hio=hi*Di/Do #[W/sq m.K]\n", + "Uo=1.0/(1.0/ho+1.0/hio) #[W/sq m.K]\n", + "Uc=Uo\n", + "L=4.88 #Length of tube in [m]\n", + "Ao=n*math.pi*Do*L #[sq m]\n", + "Ud=Q/(Ao*dTlm) #[W/sq m.K]\n", + "Rd=(Uc-Ud)/(Uc*Ud) #[m**2.K/W]\n", + "\n", + "#Result\n", + "print\"NOTE:The calculation of line no.36 to calculate 'ass' is wrongly done in Book by printing 0.0353,,..which is wrong\" \n", + "print\"\\nRd=%.2e\"%Rd,\"K/W,\\nwhich is less than the provided,so this if installed will not give required temperatures without frequent cleaning\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.15,Page no:5.57" + ] + }, + { + "cell_type": "code", + "execution_count": 105, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall heat transfer coefficient is 135.1 W/m^2.K and\n", + "\n", + "Heat transfer area is 9.17 m^2\n" + ] + } + ], + "source": [ + "#Heat transfer area\n", + "import math\n", + "\n", + "#Variable declaration\n", + "#CASE I:\n", + "Cp=4*10**3 #[J/kg.K]\n", + "t1=295 #[K]\n", + "t2=375 #[K]\n", + "sp=1.1 #Specific gravity of liquid \n", + "v1=1.75*10**-4 #Flow of liquid in [m**3/s]\n", + "rho=sp*1000 #[kg/m**3]\n", + "m_dot=v1*rho #[kg/s]\n", + "Q=m_dot*Cp*(t2-t1) #[W]\n", + "T=395 #[K]\n", + "dT1=T-t1 #[K]\n", + "dT2=T-t2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "U1A=Q/dTlm #[W/K]\n", + "#CASE-II\n", + "v2=3.25*10**-4 #Flow in [m**3/s]\n", + "T2=370 #[K]\n", + "m_dot=v2*rho #[kg/s]\n", + "\n", + "\n", + "#Calculation\n", + "Q=m_dot*Cp*(T2-t1) #[W]\n", + "dT1=T-t1 #[K]\n", + "dT2=T-T2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "U2A=Q/dTlm #[W/K]\n", + "#since u is propn to v\n", + "#hi =C*v**0.8\n", + "\n", + "U2_by_U1=U2A/U1A \n", + "\n", + "ho=3400 #Heat transfer coeff for condensing steam in [W/sq m.K]\n", + "C=poly([0])\n", + "#Let C=1 and v=v1\n", + "#C=1 \n", + "v=v1 #=1.75*10**-4 m**3/s\n", + "hi=C*v**0.8\n", + "U1=1.0/(1.0/ho+1.0/hi) #\n", + "#When v=v2\n", + "v=v2 \n", + "hi=C*v**0.8\n", + "U2=1.0/(1.0/ho+1.0/hi) #\n", + "#Since U2=1.6U1\n", + "#On solving we get:\n", + "C=142497\n", + "v=v1\n", + "hi=C*v**0.8\n", + "U1=1.0/(1.0/ho+1.0/hi) #\n", + "A=U1A/U1 #Heat transfer area in [sq m]\n", + "\n", + "#Result\n", + "print\"Overall heat transfer coefficient is\",round(U1,1),\"W/m^2.K and\\n\\nHeat transfer area is\",round(A,2),\"m^2\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.16,Page no:5.59" + ] + }, + { + "cell_type": "code", + "execution_count": 106, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Length of tube required = 2.66 m\n" + ] + } + ], + "source": [ + "#Oil Cooler\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mo_dot=6*10**-2 #[kg/s]\n", + "Cpo=2*10**3 #Specific heat of oil in [J/kg.K]\n", + "Cpw=4.18*10**3 #Specific heat of water in [J/kg.K]\n", + "T1=420 #[K]\n", + "T2=320 #[K]\n", + "T=290 #[K] Water entering temperature\n", + "\n", + "#Calculation\n", + "Q=mo_dot*Cpo*(T1-T2) #[J/s]=[W]\n", + "#Heat given out =Heat gained\n", + "t2=Q/(mo_dot*Cpw)+T #[K]\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-T #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "hi=1.6*1000 #[W/sq m.K]\n", + "ho=3.6*1000 #[W/sq m.K]\n", + "U=1.0/(1.0/ho+1.0/hi) #[W/sq m.K]\n", + "A=Q/(U*dTlm) #[sq m]\n", + "D=0.025 #[m]\n", + "L=A/(math.pi*D) #[m]\n", + "\n", + "#Result\n", + "print\"Length of tube required =\",round(L,2),\"m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.17,Page no:5.60" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total length of tubing required= 163.0 m\n" + ] + } + ], + "source": [ + "#Countercurrent flow heat exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mb_dot=1.25 #Benzene in [kg/s]\n", + "Cpb=1.9*10**3 #For benzene in [J/kg.K]\n", + "Cpw=4.187*10**3 #in [J/kg.K]\n", + "T1=350 #[K]\n", + "T2=300 #[K]\n", + "Q=mb_dot*Cpb*(T1-T2) #[W]\n", + "t1=290 #[K]\n", + "t2=320 #[K]\n", + "\n", + "#Calculation\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "mw_dot=Q/(Cpw*(t2-t1)) #Minimum flow rate of water in [kg/s]\n", + "hi=850 #[W/sq m.K]\n", + "ho=1700 #[W/sq m.K]\n", + "Do=0.025 #[m]\n", + "Di=0.022 #[m]\n", + "x=(Do-Di) /2 #Thickness in [m]\n", + "hio=hi*(Di/Do) #[W/sq m.K]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "k=45.0 #[W/m.K]\n", + "Uo=1.0/((1.0/ho)+(1.0/hio)+(x/k)*(Do/Dw)) #[W/sq m.K]\n", + "Ao=Q/(Uo*dTlm) #[sq m]\n", + "L=1.0 #Length in [m]\n", + "area=math.pi*Do*L # Outside surface area of tube per i m length \n", + "Tl=Ao/area #Total length of tubing required in [m]\n", + "\n", + "#Result\n", + "print\"Total length of tubing required=\",round(Tl),\"m\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.18,Page no:5.61" + ] + }, + { + "cell_type": "code", + "execution_count": 114, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Uc( 994.0 W/m^2.K) is in excess of Ud( 754.0 W/m^2.K),therefore we allow for reasonable scale resistance,\n", + "Rd=3.2e-04 K/W\n", + "\n", + "No.of tubes = 60.0\n" + ] + } + ], + "source": [ + "#Vertical Exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "m_dot=4500.0 #Benzene condensation rate in [kg/h]\n", + "lamda=394.0 #Latent heat of condensation of benzene in [kJ/kg]\n", + "Q=m_dot*lamda #[kJ/h]\n", + "Q=Q*1000.0/3600 #[W]\n", + "Cpw=4.18 #[kJ/kg.K]\n", + "t1=295.0 #[K]\n", + "t2=300.0 #[K]\n", + "\n", + "\n", + "#Calculation\n", + "#For water :\n", + "mw_dot=Q/(Cpw*1000*(t2-t1)) #[kg/s]\n", + "rho=1000.0 #[kg/m**3\n", + "V=mw_dot/rho #Volumetric flow rate in [m**3/s]\n", + "u=1.05 #[m/s]\n", + "A=V/u #Cross-sectional area required in [sq m]\n", + "#For tube:\n", + "x=1.6 #thickness in [mm]\n", + "x=x/1000 #[m]\n", + "Do=0.025 #[m]\n", + "Di=Do-2*x #[m]\n", + "A1=(math.pi*Di**2)/4 #Of one tube [sq m]\n", + "n=A/A1 #No. of tubes reuired \n", + "n=round(n)\n", + "L=2.5 #Length of tube in [m]\n", + "Ao=n*math.pi*Do*L #Surface area for heat transfer in [sq m]\n", + "Ts=353.0 #Condensing temp of benzene in [K]\n", + "T1=295.0 #Inlet temperature in [K]\n", + "T2=300.0 #Outlet temperature in [K]\n", + "dT1=Ts-T1 #[K]\n", + "dT2=Ts-T2 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Uo=Q/(Ao*dTlm) #[W/sq mK]\n", + "Ud=Uo #[W/sq m.K]\n", + "#OVERALL HEAT TRANSFER COEFFCIENT:\n", + "#Inside side:\n", + "T=(T2+T1)/2 #[K]\n", + "hi=1063*((1+0.00293*T)*u**0.8)/(Di**0.2) #[W/sq m.K]\n", + "hio=hi*(Di/Do) #[W/sq m.K]\n", + "Dw=(Do-Di)/math.log(Do/Di) #[m]\n", + "k=45.0 #For tube in [W/(m.)]\n", + "#Outside of tube:\n", + "mdot_dash=1.25/n #[kg/s]\n", + "M=mdot_dash/(math.pi*Do) #[kg/(m.s)]\n", + "k=0.15 #[W/(m.K)]\n", + "rho=880.0 #[kg/m**3]\n", + "mu=0.35*10**-3 #[N.s/sq m]\n", + "g=9.81 #[m/s**2] Acceleration due to gravity\n", + "hm=(1.47*((4*mdot_dash)/mu)**(-1.0/3.0))/(mu**2/(k**3*rho**2*g))**(1.0/3.0) #[W/sq m.K]\n", + "ho=hm #[W/sq m.K]\n", + "k=45 #[W/m]\n", + "Uo=1.0/(1.0/ho+1.0/hio+(x*Do)/(k*Dw))\n", + "#Uo=1/(1/ho+1/hio+(x*Do/(k*Dw))) #Overall heat transfer coefficient in [W/sq m.K]\n", + "Uc=Uo #[W/sq m.K]\n", + "Rd=(Uc-Ud)/(Uc*Ud) #Maximum allowable sclae resistance in [K/W]\n", + "\n", + "#Result\n", + "print\"Uc(\",round(Uc),\"W/m^2.K) is in excess of Ud(\",round(Ud),\"W/m^2.K),therefore we allow for reasonable scale resistance,\\nRd=%.1e\"%Rd,\"K/W\\n\" \n", + "print\"No.of tubes =\",n\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.19,Page no:5.64" + ] + }, + { + "cell_type": "code", + "execution_count": 115, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer area is 33.45 m^2\n" + ] + } + ], + "source": [ + "#Countercurrent Heat Exchanger\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mw_dot=5 #Water flow rate in [kg/s]\n", + "Cpw=4.18 #Heat capacity of water [kJ/kg.K]\n", + "t1=303 #[K]\n", + "t2=343 #[K]\n", + "\n", + "#Calculation\n", + "Q=mw_dot*Cpw*(t2-t1) #[kJ/s]\n", + "Q=Q*1000 #[W]\n", + "T1=413 #[K]\n", + "T2=373 #[K]\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTlm=dT1 #/[K]\n", + "hi=1000 #[W/sq m.K]\n", + "ho=2500 #[W/sq m.K]\n", + "Rd=1.0/(0.714*1000) #Fouling factor[m**2.K/KW]\n", + "U=1.0/(1.0/hi+1.0/ho+Rd) #[W/sq m.K]\n", + "A=Q/(U*dTlm) #[sq m]\n", + "\n", + "#Result\n", + "print\"Heat transfer area is\",round(A,2),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no5.20:,Page no:5.65" + ] + }, + { + "cell_type": "code", + "execution_count": 117, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat exchanger surface area is 10.84 m^2\n" + ] + } + ], + "source": [ + "#Number of tube side pass\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Cpo=1.9 #Heat capacity for oil[kJ/kg.K]\n", + "Cps=1.86 #Heat capacity for steam [kJ/kg.K]\n", + "ms_dot=5.2 #Mass flow rate in [kg/s]\n", + "T1=403.0 #[K]\n", + "T2=383.0 #[K]\n", + "\n", + "#Calculation\n", + "Q=ms_dot*Cps*(T1-T2) #[kJ/s]\n", + "Q=Q*1000.0 #[W]\n", + "t1=288.0 #[K]\n", + "t2=358.0 #[K]\n", + "dT2=T1-t2 #[K]\n", + "dT1=T2-t1 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #LMTD in [K]\n", + "U=275.0 #Overall heat transfer coeffcient in [W#sq m.K]\n", + "Ft=0.97 #LMTD correction factor\n", + "A=Q/(U*Ft*dTlm) #[sq m]\n", + "\n", + "#Result\n", + "print\"Heat exchanger surface area is\",round(A,2),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.21,Page no:5.66" + ] + }, + { + "cell_type": "code", + "execution_count": 125, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length is more than allowable 2.44 m length,so we must use more than one tube\n", + "This length is within 2.44 m requirement,so the design choice is:\n", + "\n", + "--Type of heat exchanger : \t1-2 Shell and tube heat exchanger\n", + "--No of tubes per pass=\t\t36.0\n", + "--Length of tube per pass=\t1.83 m\n" + ] + } + ], + "source": [ + "#Number of tubes passes\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mc_dot=3.783 #Cold water flow rate[kg/s]\n", + "mh_dot=1.892 #Hot water flow rate [kg/s]\n", + "Cpc=4.18 #Sp heat of cold water [kJ/(kg.K)]\n", + "T1=367.0 #[K]\n", + "t2=328.0 #[K]\n", + "t1=311.0 #[K]\n", + "Cph=4.18 #Specific heat of hot water [kJ/(kg.K)]\n", + "rho=1000.0 #Density [kg/m**3]\n", + "D=0.019 #Diameter of tube in [m]\n", + "U=1450.0 #Overal heat transfer coefficient in [W/sq m.K] \n", + "\n", + "#Calculation\n", + "T2=T1-mc_dot*Cpc*(t2-t1)/(mh_dot*Cph) #[K]\n", + "Q=mc_dot*Cpc*(t2-t1) #[kJ/s]\n", + "Q=Q*1000.0 #[W]\n", + "#For counterflow heat exchanger\n", + "dT1=T1-t2 #[K]\n", + "dT2=17.0 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "lmtd=dTlm #LMTD \n", + "Ft=0.88 #LMTD correction factor\n", + "A=Q/(U*dTlm) #[sq m]\n", + "u=0.366 #Velocity through tubes[ms**-1]\n", + "Ai=mc_dot/(rho*u) #Total flow Area in [sq m]\n", + "n=Ai/((math.pi/4)*(D**2)) #No. of tubes \n", + "L=1.0 #Per m length[m]\n", + "sa=math.pi*D*L #S.S per tube per 1 m length\n", + "L=A/(n*math.pi*D) #Length of tubes in [m]\n", + "\n", + "#Result\n", + "print\"The length is more than allowable 2.44 m length,so we must use more than one tube\"\n", + "#For 2passes on the tube side\n", + "A=Q/(U*Ft*lmtd) #[sq m]\n", + "L=A/(2*n*math.pi*D) #Length in [m]\n", + "print\"This length is within 2.44 m requirement,so the design choice is:\\n\"\n", + "print\"--Type of heat exchanger : \\t1-2 Shell and tube heat exchanger\"\n", + "print\"--No of tubes per pass=\\t\\t\",round(n)\n", + "print\"--Length of tube per pass=\\t\",round(L,2),\"m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.22,Page no:5.67" + ] + }, + { + "cell_type": "code", + "execution_count": 143, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For countercurrent flow arrangement:\n", + "Exit temperature of hot fluid is 721.0 K( 448.0 degree C)\n", + "Exit temperature of cold fluid is 553.0 K( 280.0 degree C)\n", + "\n", + "For Parallel flow arrangement:\n", + "Exit temperature of Hot water= 734.8 K( 461.8 degree C)\n", + "Exit temperature of cold water= 543.2 K( 270.2 degree C)\n" + ] + } + ], + "source": [ + "#Outlet temperature for hot and cold fluids\n", + "import math\n", + "\n", + "#Variable declaration\n", + "mh_dot=16.67 #Mass flow rate of hot fluid in [kg/s]\n", + "mc_dot=20.0 #Mass flow rate of cold fluid in [kg/s]\n", + "Cph=3.6 #Sp heat of hot fluid in [kJ/kg.K]\n", + "Cph=Cph*1000 #Sp heat of hot fluid in [J/kg.K]\n", + "Cpc=4.2 #Sp heat of cold fluid in [kJ/(kg.K)]\n", + "Cpc=Cpc*1000 #Sp heat of cold fluid in [J/(kg.K)]\n", + "U=400.0 #Overall heat transfer coefficient in [W/sq m.K]\n", + "A=100.0 #Surface area in [sq m]\n", + "\n", + "#Calculation\n", + "mCp_h=mh_dot*Cph #[J/s] or [W/K]\n", + "mCp_c=mc_dot*Cpc #[J/s] or [W/K]\n", + "mCp_small=mCp_h #[W/K]\n", + "C=mCp_small/mCp_c #Capacity ratio \n", + "ntu=U*A/mCp_small #NTU\n", + "T1=973.0 #Hot fluid inlet temperature in [K]\n", + "t1=373.0 #Cold fluid inlet temperature in [K]\n", + "\n", + "#Case 1:Countercurrent flow arrangement\n", + "E=(1.0-math.exp(-(1.0-C)*ntu))/(1.0-C*math.exp(-(1.0-C)*ntu)) #Effectiveness\n", + "E=round(E,2)\n", + "#W=T1-T2/(T1-t1) therefore:\n", + "T2=T1-E*(T1-t1) #[K]\n", + "print\"For countercurrent flow arrangement:\"\n", + "print\"Exit temperature of hot fluid is\",T2,\"K(\",T2-273,\"degree C)\"\n", + "t2=mCp_h*(T1-T2)/(mCp_c)+t1 #[From energy balance eqn in ][K]\n", + "print\"Exit temperature of cold fluid is\",round(t2),\" K(\",round(t2-273) ,\"degree C)\\n\"\n", + "\n", + "#Case 2:Parallel flow arrangement\n", + "E1=(1-math.exp(-(1.0+C)*ntu))/(1.0+C)\n", + "E1=round(E1,3)\n", + "T2=T1-E1*(T1-t1) #[K]\n", + "t2=mCp_h*(T1-T2)/(mCp_c)+t1 #[From energy balance eqn in ][K]\n", + "print\"For Parallel flow arrangement:\"\n", + "print\"Exit temperature of Hot water=\",T2,\"K(\",T2-273,\"degree C)\"\n", + "print\"Exit temperature of cold water=\",round(t2,1),\"K(\",round(t2-273,1),\"degree C)\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.23,Page no:5.69" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "In textbook value of mCp_h is wrongly calculated as 231.1 so we will take this only for calculation\n", + "\n", + "From LMTD approach:\n", + "length= 6.61 m\n", + "\n", + "From NTU method:\n", + "length= 7.26 m\n" + ] + } + ], + "source": [ + "#Counter flow concentric heat exchanger\n", + "import math\n", + "from scipy.optimize import fsolve\n", + "\n", + "#Variable declaration\n", + "Cpo=2131.0 #Sp heat of oil in [J/kg.K]\n", + "Cpw=4187.0 #Sp heat of water in [J/kg.K]\n", + "mo_dot=0.10 #Oil flow rate in [kg/s]\n", + "mw_dot=0.20 #Water flow rate in [kg/s]\n", + "U=380.0 #Overall heat transfer coeff in [W/sq m.K]\n", + "T1=373.0 #Initial temp of oil [K]\n", + "T2=333.0 #Final temperature of oil [K]\n", + "t1=303.0 #Water enter temperature in [K]\n", + "\n", + "\n", + "#Calculation\n", + "t2=t1+mo_dot*Cpo*(T1-T2)/(mw_dot*Cpw) #[K] \n", + "\n", + "#1.LMTD method\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "lmtd=dTlm #[K]\n", + "Q=mo_dot*Cpo*(T1-T2) #[J/s]\n", + "A=Q/(U*dTlm) #[sq m]\n", + "Do=0.025 #Inner tubve diameter [m]\n", + "L=A/(math.pi*Do) #Length in [m]\n", + "\n", + "#2.NTU method\n", + "mCp_c=mw_dot*Cpw #[W/K]\n", + "mCp_h=mo_dot*Cpo #[W/K]\n", + "print\"In textbook value of mCp_h is wrongly calculated as 231.1 so we will take this only for calculation\\n\"\n", + "mCp_h=231.1 #[W/K]\n", + "#mCp_h is smaller\n", + "C=mCp_h/mCp_c\n", + "E=(T1-T2)/(T1-t1) #Effeciency\n", + "#For countercurrent flow\n", + "def f(ntu):\n", + " x=E-(1.0-math.exp(-(1.0-C)*ntu))/(1.0-C*math.exp(-(1.0-C)*ntu))\n", + " return(x)\n", + "ntu=fsolve(f,1)\n", + "A=ntu*mCp_h/U #[sq m]\n", + "A=round(A[0],2)\n", + "L1=A/(math.pi*Do) #Length in [m]'\n", + "\n", + "\n", + "#Result\n", + "print\"From LMTD approach:\\nlength=\",round(L,2),\"m\\n\" \n", + "print\"From NTU method:\\nlength=\",round(L1,2),\"m\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.24,Page no:5.70" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No. of tubes required = 86.0\n" + ] + } + ], + "source": [ + "#Number of tubes required\n", + "import math\n", + "from scipy.optimize import fsolve\n", + "\n", + "#Variable declaration\n", + "ho=200.0 #[W/sq m.K]\n", + "hi=1500.0 #[W/sq m.K]\n", + "Cpw=4.2 #Sp heat of Water in [kJ/(kg.K)]\n", + "Cpo=2.1 #Sp heat of Oil in [kJ/(kg.K)]\n", + "E=0.8 #Effectiveness\n", + "k=46.0 #[W/m.K]\n", + "m_dot=0.167 #[kg/s]\n", + "\n", + "\n", + "#Calculation\n", + "\n", + "mCp_oil=2*m_dot*Cpo*1000.0 #For oil [W/K]\n", + "#mCp_oil is wrongly calculated as 710.4\n", + "mCp_water=m_dot*Cpw*1000.0 #For water [W/K]\n", + "#mCp_oil is wrongly calculated as 710.4\n", + "#NOTE:The above two values are wrongly calculated in book as 710.4\n", + "#so we take here:\n", + "mCp_small=710.4 #[W/K]\n", + "#Since both mCp_water and mCp_oil are equal ,therefore:\n", + "C=1.0 \n", + "def f(ntu):\n", + " x=E-(ntu/(1+ntu))\n", + " return(x)\n", + "ntu=fsolve(f,1)\n", + "id=20.0 #Internal diameter in [mm]\n", + "od=25.0 #External diameter in [mm]\n", + "hio=hi*id/od #[W/sq m.K]\n", + "Dw=(od-id)/math.log(od/id) #[mm]\n", + "Dw=Dw/1000.0 #[m]\n", + "x=(od-id)/2.0 #[mm]\n", + "x=x/1000.0 #[m]\n", + "Do=0.025 #External dia in [m]\n", + "L=2.5 #Length of tube in [m]\n", + "Uo=1.0/(1.0/ho+1.0/hio+(x/k)*(Dw/Do)) #[W/sq m.K]\n", + "A=ntu*mCp_small/Uo #Heat transfer area in [sq m]\n", + "n=A/(math.pi*Do*L) #No of tubes\n", + "\n", + "#Result\n", + "print\"No. of tubes required =\",round(n+1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:5.25,Page no:5.72" + ] + }, + { + "cell_type": "code", + "execution_count": 154, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "For parallel flow,Area = 1.234 m^2\n", + "For countercurrent flow,Area= 1.195 m\n", + "\n", + "For the same terminal temperatures of the fluid ,\n", + "the surface area for the counterflow arrangement is less than the required for the parallel flow\n" + ] + } + ], + "source": [ + "#Parallel and Countercurrent flow\n", + "import math\n", + "\n", + "#Variable declaration\n", + "#(i)Parallel flow\n", + "T1=633.0 #[K]\n", + "t2=303.0 #[K]\n", + "T2=573.0 #[K]\n", + "t1=400.0 #[K]\n", + "\n", + "#Calculation\n", + "dT1=T1-t2 #[K]\n", + "dT2=T2-t1 #[K]\n", + "mh_dot=1.2 #[kg/s]\n", + "U=500.0 #Overall heat transfer coefficient in [W/sqm.K]\n", + "Cp=2083.0 #Sp.heat of oil J/kg.K\n", + "dTlm=(dT1-dT2)/math.log(dT1/dT2) #[K]\n", + "Q=mh_dot*Cp*(T1-T2) #[W]\n", + "A=Q/(U*dTlm) #[sq m]\n", + "\n", + "#(ii)Counter current flow\n", + "dT1=T1-t1 #[K]\n", + "dT2=T2-t2 #[K]\n", + "dTlm=(dT2-dT1)/math.log(dT2/dT1) #[K]\n", + "A1=Q/(U*dTlm) #[sq m]\n", + "\n", + "#Result\n", + "print\"For parallel flow,Area =\",round(A,3),\"m^2\\nFor countercurrent flow,Area=\",round(A1,3),\"m\\n\"\n", + "print\"For the same terminal temperatures of the fluid ,\\nthe surface area for the counterflow arrangement is less than the required for the parallel flow\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_by_K._A._Gavhane/Chapter_6.ipynb b/Heat_Transfer_by_K._A._Gavhane/Chapter_6.ipynb new file mode 100644 index 00000000..6cbb928f --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/Chapter_6.ipynb @@ -0,0 +1,1098 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter6: Evaporation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.1,Page no:6.19" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Boiling point of elevation of the solution is 7 K\n", + "Driving forve for heat transfer is 19 K\n" + ] + } + ], + "source": [ + "#Boiling point Elevation\n", + "#Variable declaration\n", + "T=380 #B.P of solution[K]\n", + "T_dash=373 #B.P of water [K]\n", + "Ts=399 #Saturating temperature in [K]\n", + "#Calculation\n", + "BPE=T-T_dash #Boiling point elevation in [K]\n", + "DF=Ts-T #Driving force in [K]\n", + "#Result\n", + "print\"Boiling point of elevation of the solution is\",BPE,\"K\"\n", + "print\"Driving forve for heat transfer is\",DF,\"K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.2 ,Page no:6.20" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacity of evaporator is 8400.0 kg/h\n" + ] + } + ], + "source": [ + "#Capacity of evaporator\n", + "#Variable declaration\n", + "m_dot=10000 #Weak liquor entering in [kg/h]\n", + "fr_in=0.04 #Fraciton of caustic soda IN i.e 4%\n", + "fr_out=0.25 #Fraciton of caustic soda OUT i.e 25%\n", + "#Let mdash_dot be the kg/h of thick liquor leaving\n", + "\n", + "#Calculation\n", + "mdash_dot=fr_in*m_dot/fr_out #[kg/h]\n", + "\n", + "#Overall material balance\n", + "#kg/h of feed=kg/h of water evaporated +kg/h of thick liquor\n", + "#we=water evaporated in kg/h\n", + "#Therefore\n", + "we=m_dot-mdash_dot #[kg/h]\n", + "\n", + "#Result\n", + "\n", + "print\"Capacity of evaporator is\",we,\"kg/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.3,Page no:6.20" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ANSWER:Economoy pf evaporator is 0.808\n", + "Heat tarnsfer area to be provided = 57.07 m^2\n" + ] + } + ], + "source": [ + "#Economy of Evaporator\n", + "#Variable declaration\n", + "ic=0.05 #Initial concentration (5%)\n", + "fc=0.2 #Final concentration (20%)\n", + "T_dash=373 #B.P of water in [K]\n", + "bpe=5 #Boiling point elevation[K]\n", + "mf_dot=5000 #[Basis] feed to evaporator in [kg/h]\n", + "\n", + "#Calculation\n", + "\n", + "#Material balance of solute\n", + "mdash_dot=ic*mf_dot/fc #[kg/h]\n", + "#Overall material balance\n", + "mv_dot=mf_dot-mdash_dot #Water evaporated [kg/h]\n", + "lambda_s=2185 #Latent heat of condensation of steam[kJ/kg]\n", + "lambda_v=2257 #Latent heat of vaporisation of water [kJ/kg]\n", + "lambda1=lambda_v #[kJ/kg]\n", + "T=T_dash+bpe #Temperature of thick liquor[K]\n", + "Tf=298 #Temperature of feed [K]\n", + "Cpf=4.187 #Sp. heat of feed in [kJ/kg.K]\n", + "#Heat balance over evaporator=ms_dot\n", + "ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption [kg/h]\n", + "Eco=mv_dot/ms_dot #Economy of evaporator\n", + "Ts=399 #Saturation temperature of steam in [K]\n", + "dT=Ts-T #Temperature driving force [K] \n", + "U=2350 #[W/sq m.K]\n", + "Q=ms_dot*lambda_s #Rate of heat transfer in [kJ/kg]\n", + "Q=Q*1000/3600 #[J/s]=[W]\n", + "A=Q/(U*dT) #Heat transfer area in [sq m]\n", + "\n", + "#Result\n", + "print\"ANSWER:Economoy pf evaporator is \",round(Eco,3)\n", + "print\"Heat tarnsfer area to be provided = \",round(A,2),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.4,Page no:6.22" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When Feed introduced at 293 K ,Steam economy is 0.87\n", + "ANSWER-(i) At 293 K,Heat transfer area required is 83.16 m^2\n", + "ANSWER-(ii) When T=308 K,Economy of evaporator is 0.896\n", + "ANSWER-(iii) When T=308 K,Heat transfer Area required is 80.71 m^2\n" + ] + } + ], + "source": [ + "#Steam economy\n", + "\n", + "#Variable declaration\n", + "\n", + "Cpf=3.98 #Specific heat of feed in kJ/(kg.K)\n", + "lambda_s=2202 #Latent heat of conds of heat at 0.2MPa in [kJ/kg]\n", + "lambda1=2383 #Latent heat of vaporisation of water aty 323 [kJ/kg\n", + "ic=0.1 #Initial concentration of soilds in [%]\n", + "fc=0.5 #Final concentration\n", + "m_dot=30000 #Feed to evaporator in [kg/h]\n", + "\n", + "#Calculation\n", + "\n", + "mdash_dot=ic* m_dot/fc #Mass flow rate of thick liquor in [kg/h]\n", + "mv_dot=m_dot-mdash_dot #Water evaporated in [kg/h]\n", + "\n", + "#Case 1: Feed at 293K\n", + "mf_dot=30000 #[kg/h]\n", + "mv_dot=24000 #[kg/h]\n", + "Cpf=3.98 #[kJ/(kg.K)]\n", + "Ts=393 #Saturation temperature of steam in [K]\n", + "T=323 #Boiling point of solution [K]\n", + "lambda_s=2202 #Latent heat of condensation [kJ/kg]\n", + "lambda1=2383 #Latent heat of vaporisation[kJ/kg]\n", + "Tf=293 #Feed temperature\n", + "#Enthalpy balance over the evaporator:\n", + "ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption[kg/h]\n", + "eco=(mv_dot/ms_dot) #Steam economy\n", + "print\"When Feed introduced at 293 K ,Steam economy is \",round(eco,2) \n", + "dT=Ts-T #[K]\n", + "U=2900 #[W/sq m.K]\n", + "Q=ms_dot*lambda_s #Heat load =Rate of heat transfer in [kJ/h]\n", + "Q=Q*1000/3600 #[J/s]\n", + "A=Q/(U*dT) #Heat transfer area required [sq m]\n", + "\n", + "#Result\n", + "print\"ANSWER-(i) At 293 K,Heat transfer area required is\",round(A,2),\"m^2\"\n", + "\n", + "#Case2: Feed at 308K\n", + "Tf=308 #[Feed temperature][K]\n", + "\n", + "#Calculation\n", + "ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption in [kg/h]\n", + "eco=mv_dot/ms_dot #Economy of evaporator\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000/3600 #[J/s]\n", + "A=Q/(U*dT) #Heat transfer area required [sq m]\n", + "#Result\n", + "print\"ANSWER-(ii) When T=308 K,Economy of evaporator is \",round(eco,3)\n", + "print\"ANSWER-(iii) When T=308 K,Heat transfer Area required is \",round(A,2),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.5,Page no:6.24" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat transfer area to be provided is 45.33 m^2\n" + ] + } + ], + "source": [ + "#Evaporator economy\n", + "#Variable declaration\n", + "m_dot=5000 #Feed to the evaporator [kg/h]\n", + "Cpf=4.187 #Cp of feed in [kJ/kg.K]\n", + "ic=0.10 #Initial concentration\n", + "fc=0.4 #Final concentration\n", + "lambda_s=2162 #Latent heat of condensing steam [kJ/kg]\n", + "P=101.325 #Pressure in the evaporator[kPa]\n", + "bp=373 #[K]\n", + "Hv=2676 #Enthalpy of water vapor [kJ/kg]\n", + "H_dash=419 #[kJ/kg]\n", + "Hf=170 #[kJ/kg]\n", + "U=1750 #[W/sq m.K]\n", + "dT=34 #[K]\n", + "#Calculation\n", + "mdash_dot=m_dot*ic/fc #[kg/h] of thick liquor\n", + "mv_dot=m_dot-mdash_dot #Water evaporated in[kg/h]\n", + "ms_dot=(mv_dot*Hv+mdash_dot*H_dash-m_dot*Hf)/lambda_s #Steam consumption in [kg/h]\n", + "eco=mv_dot/ms_dot #Steam economy of evaporator\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000/3600 #[J/s]\n", + "A=Q/(U*dT) #[sq m]\n", + "#Result\n", + "print\"Heat transfer area to be provided is\",round(A,2),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.6 ,Page no:6.26" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steam economy is 0.784\n", + "Overall heat transfer coefficient is 2862.0 W/m^2.K\n" + ] + } + ], + "source": [ + "#Single effect Evaporator\n", + "#Variable declaration\n", + "mf_dot=5000 #[kg/h]\n", + "ic=0.01 #Initial concentration [kg/h]\n", + "fc=0.02 #Final concentration [kg/h]\n", + "T=373 #Boiling pt of saturation in [K]\n", + "Ts=383 #Saturation temperature of steam in [K] \n", + "Hf=125.79 #[kJ/kg]\n", + "Hdash=419.04 #[kJ/kg]\n", + "Hv=2676.1 #[kJ/kg]\n", + "lambda_s=2230.2 #[kJ/kg]\n", + "#Calculation\n", + "mdash_dot=ic*mf_dot/fc #[kg/h]\n", + "mv_dot=mf_dot-mdash_dot #Water evaporated in [kg/h]\n", + "ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s #Steam flow rate in [kg/h]\n", + "eco=mv_dot/ms_dot #Steam economy\n", + "Q=ms_dot*lambda_s #Rate of heat transfer in [kJ/h]\n", + "Q=Q*1000/3600 #[J/s]\n", + "dT=Ts-T #[K]\n", + "\n", + "A=69 #Heating area of evaporator in [sq m]\n", + "U=Q/(A*dT) #Overall heat transfer coeff in [W/sq m.K]\n", + "\n", + "#Result\n", + "print\"Steam economy is\",round(eco,3)\n", + "print\"Overall heat transfer coefficient is\",round(U),\"W/m^2.K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.7,Page no:6.27" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat transfer area in this case is 18.7 m^2\n", + "NOTE :There is a calculation mistake in the book at the line12 of this code,ms_dot value is written as 2320.18,which is wrong\n" + ] + } + ], + "source": [ + "#Single efect evaporator reduced pressure\n", + "#From previous example:\n", + "#Variable declaration\n", + "mf_dot=5000 #[kg/h]\n", + "Hf=125.79 #[kJ/kg]\n", + "lambda_s=2230.2 #[kJ/kg]\n", + "mdash_dot=2500 #[kg/h]\n", + "Hdash=313.93 #[kJ/kg]\n", + "mv_dot=2500 #[kg/h]\n", + "Hv=2635.3 #[kJ/kg]\n", + "U=2862 #[W/sq m.K]\n", + "dT=35 #[K]\n", + "#Calculation\n", + "ms_dot=(mdash_dot*Hdash+mv_dot*Hv-mf_dot*Hf)/lambda_s #Steam flow rate in [kg/h]\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000/3600 #[W]\n", + "A=Q/(U*dT) #[sq m]\n", + "#Result\n", + "print\"The heat transfer area in this case is\",round(A,2),\"m^2\"\n", + "print\"NOTE :There is a calculation mistake in the book at the line12 of this code,ms_dot value is written as 2320.18,which is wrong\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.8,Page no:6.27" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass flow rate of product is 3629.9 kg/h\n", + "The product concentration is 1.653 % by weight\n" + ] + } + ], + "source": [ + "#Mass flow rate\n", + "#Variable declaration\n", + "mf_dot=6000 #Feed rate in [kg/h]\n", + "#Taking the given values from previous example(6.6)\n", + "Hf=125.79 #[kJ/kg]\n", + "ms_dot=3187.56 #[kg/h]\n", + "lambda_s=2230.2 #[kJ/kg]\n", + "Hdash=419.04 #[kJ/kg]\n", + "Hv=2676.1 #[kJ/kg]\n", + "#Calculation\n", + "mv_dot=(mf_dot*Hf+ms_dot*lambda_s-6000*Hdash)/(Hv-Hdash) #Water evaporated in [kg/h]\n", + "mdash_dot=6000-mv_dot #Mass flow rate of product [kg/h]\n", + "x=(0.01*mf_dot)*100/mdash_dot #Wt % of solute in products\n", + "#Result\n", + "print\"Mass flow rate of product is\",round(mdash_dot,1),\"kg/h\"\n", + "print\"The product concentration is\",round(x,3),\"% by weight\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.9 ,Page no:6.28" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Heat load is 11226389.0 W or J/s\n", + "Economy of evaporator is 0.811\n", + "NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071\n" + ] + } + ], + "source": [ + "#Heat load in single effect evaporator\n", + "#Variable declaration\n", + "Tf=298 #Feed temperature in [K]\n", + "T_dash=373 #[K]\n", + "Cpf=4 #[kJ/kg.K]\n", + "fc=0.2 #Final concentration of salt\n", + "ic=0.05 #Initial concentration\n", + "mf_dot=20000 #[kg/h] Feed to evaporator\n", + "#Calculation\n", + "mdash_dot=ic*mf_dot/fc #Thick liquor [kg/h]\n", + "mv_dot=mf_dot-mdash_dot #Water evaporated in [kg/h]\n", + "lambda_s=2185 #[kJ/kg]\n", + "lambda1=2257 #[kJ/kg]\n", + "bpr=7 #Boiling point rise[K]\n", + "T=T_dash+bpr #Boiling point of solution in[K]\n", + "Ts=39 #Temperature of condensing steam in [K]\n", + "ms_dot=(mf_dot*Cpf*(T-Tf)+mv_dot*lambda1)/lambda_s #Steam consumption in [kg/h]\n", + "eco=mv_dot/ms_dot #Economy of evaporator \n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000/3600 #[J/s]\n", + "#Result\n", + "print\"Heat load is\",round(Q),\"W or J/s\"\n", + "print\"Economy of evaporator is \",round(eco,3)\n", + "print\"NOTE:Again there is a calcualtion mistake in book at line 19 of code,it is written as 4041507.1 instead of 40415071\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.10 ,Page no:6.32" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Boiling point of solution in first effect = 369.55 K\n", + "Boiling point of solution in second effect = 354.6 K\n" + ] + } + ], + "source": [ + "#Triple efect evaporator\n", + "#Variable declaration\n", + "Ts=381.3 #[K]\n", + "dT=56.6 #[K]\n", + "U1=2800.0 #Overall heat transfer coeff in first effect\n", + "U2=2200.0 #Overall heat transfer coeff in first effect\n", + "U3=1100.0 #Overall heat transfer coeff in first effect\n", + "#Calculation\n", + "dT1=dT/(1+(U1/U2)+(U1/U3)) #/[K]\n", + "dT2=dT/(1+(U2/U1)+(U2/U3)) #/[K]\n", + "dT3=dT-(dT1+dT2) #[K]\n", + "#dT1=Ts-T1_dash #[K]\n", + "T1dash=Ts-dT1\n", + "#dT2=T1_dash-T2_dash #[K]\n", + "T2_dash=T1dash-dT2 #[K]\n", + "#Result\n", + "print\"Boiling point of solution in first effect =\",round(T1dash,2),\"K\"\n", + "print\"Boiling point of solution in second effect =\",round(T2_dash,1),\"K\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.11,Page no:6.33" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "A1( 24.9 )=A2( 23.0 ),So the area in each effect can be 24.9 m^2\n", + "Heat transfer surface in each effect is 24.9 m^2\n", + "Steam consumption= 5517.0 (approx)kg/h\n", + "Evaporation in the first effect is 4343.0 kg/h\n", + "Evaporation in 2nd effect is 3742.0 kg/h\n" + ] + } + ], + "source": [ + "#Double effect evaporator\n", + "#Variable declaration\n", + "mf_dot=10000.0 #[kg/h] of feed\n", + "ic=0.09 #Initial concentration \n", + "fc=0.47 #Final concentration\n", + "m1dot_dash=ic*mf_dot/fc #[kg/h]\n", + "Ps=686.616 #Steam pressure [kPa.g]\n", + "Ps=Ps+101.325 #[kPa]\n", + "Ts=442.7 #Saturation temperature in [K]\n", + "P2=86.660 #Vacuum in second effect in [kPa]\n", + "U1=2326.0 #Overall heat transfer in first effect [W/sq m.K]\n", + "U2=1744.5 #Overall heat transfer in 2nd effect [W/sqm.K]\n", + "P2_abs=101.325-P2 #Absolute pressure in second effect[kPa]\n", + "T2=326.3 #Temperature in 2nd effect in [K]\n", + "dT=Ts-T2 #[K]\n", + "Tf=309.0 #Feed temperature in[K]\n", + "T=273.0 #[K]\n", + "Cpf=3.77 #kJ/kg.K Specific heat for all caustic streams\n", + "#Q1=Q2\n", + "#U1*A1*dT1=U2*A2*dT2\n", + "#Calculation\n", + "dT2=dT/1.75 #[K]\n", + "dT1=(U2/U1)*dT2 #[K]\n", + "#Since there is no B.P.R\n", + "Tv1=Ts-dT1 #Temperature in vapor space of first effect in [K]\n", + "Tv2=Tv1-dT2 #Second effect [K]\n", + "Hf=Cpf*(Tf-T) #Feed enthalpy[kJ/kg]\n", + "H1dash=Cpf*(Tv1-T) #Enthalpy of final product[kJ/kg]\n", + "H2dash=Cpf*(Tv2-T) #kJ/kg\n", + "#For steam at 442.7 K\n", + "lambda_s=2048.7 #[kJ/kg]\n", + "#For vapour at 392.8 K\n", + "Hv1=2705.22 #[kJ/kg]\n", + "lambda_v1=2202.8 #[kJ/kg]\n", + "#for vapour at 326.3 K:\n", + "Hv2=2597.61 #[kJ/kg]\n", + "lambda_v2=2377.8 #[kJ/kg]\n", + "\n", + "#Overall material balance:\n", + "mv_dot=mf_dot-m1dot_dash #[kg/h]\n", + "\n", + "#Equation 4 becomes:\n", + "#mv1_dot*lambda_v1+mf_dot*Hf=(mv_dot-mv1_dot)*Hv2+(mf_dot-mv2_dot)*H2_dash\n", + "mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash) \n", + "mv2_dot=mv_dot-mv1_dot #[kg/h]\n", + "\n", + "#From equation 2\n", + "\n", + "m2dot_dash=m1dot_dash+mv1_dot #First effect material balance[kg/h]\n", + "ms_dot=(mv1_dot*Hv1+m1dot_dash*H1dash-m2dot_dash*H2dash)/lambda_s #[kg/h]\n", + "\n", + "\n", + "#Heat transfer Area\n", + "#First effect\n", + "A1=ms_dot*lambda_s*(10.0**3.0)/(3600.0*U1*dT1) #[sq m]\n", + "\n", + "#Second effect\n", + "lambda_v1=lambda_v1*(10**3.0)/3600.0\n", + "A2=mv1_dot*lambda_v1/(U2*dT2) #[sq m]\n", + "\n", + "#Since A1 not= A2\n", + "\n", + "#SECOND TRIAL\n", + "Aavg=(A1+A2)/2 #[sq m]\n", + "dT1_dash=dT1*A1/Aavg #[K]\n", + "dT2_dash=dT-dT1 #/[K]\n", + "\n", + "#Temperature distribution\n", + "Tv1=Ts-dT1_dash #[K]\n", + "Tv2=Tv1-dT2_dash #[K]\n", + "Hf=135.66 #[kJ/kg]\n", + "H1dash=Cpf*(Tv1-T) #[kJ/kg]\n", + "H2dash=200.83 #[kJ/kg]\n", + "\n", + "#Vapour at 388.5 K\n", + "Hv1=2699.8 #[kJ/kg]\n", + "lambda_v1=2214.92 #[kJ/kg]\n", + "mv1_dot=(H2dash*(mf_dot-mv_dot)-mf_dot*Hf+mv_dot*Hv2)/(Hv2+lambda_v1-H2dash) \n", + "mv2_dot=mv_dot-mv1_dot #[kg/h]\n", + "\n", + "#First effect Energy balance\n", + "ms_dot=((mv1_dot*Hv1+m1dot_dash*H1dash)-(mf_dot-mv2_dot)*H2dash)/lambda_s #[kg/h]\n", + "\n", + "#Area of heat transfer\n", + "lambda_s=lambda_s*1000.0/3600.0 \n", + "A1=ms_dot*lambda_s/(U1*dT1_dash) #[sq m]\n", + "\n", + "#Second effect:\n", + "A2=(mv1_dot*lambda_v1*1000)/(3600.0*U2*dT2_dash) #[sq m]\n", + "\n", + "#Result\n", + "\n", + "print\"A1(\",round(A1,1),\")=A2(\",round(A2),\"),So the area in each effect can be\",round(A1,1),\"m^2\"\n", + "print\"Heat transfer surface in each effect is\",round(A1,1),\"m^2\"\n", + "print\"Steam consumption=\",round(ms_dot),\"(approx)kg/h\"\n", + "print\"Evaporation in the first effect is\",round(mv1_dot),\"kg/h\"\n", + "print\"Evaporation in 2nd effect is\",round(mv2_dot),\"kg/h\" \n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.12 ,Page no:6.37" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ANSWER:Area in each effect 200.2 sq m\n", + "Steam economy is 2.55\n", + "Cooling water rate is 66.63 t/h\n" + ] + } + ], + "source": [ + "#lye in Triple effect evaporator\n", + "#Variable declaration\n", + "Tf=353.0 #[K]\n", + "T=273.0 #[K]\n", + "mf_dot=10000.0 #Feed [kg/h]\n", + "ic=0.07 #Initial conc of glycerine \n", + "fc=0.4 #FinaL CONC OF GLYCERINE\n", + "#Overall glycerine balance\n", + "P=313.0 #Steam pressure[kPa]\n", + "Ts=408.0 #[from steam table][K]\n", + "P1=15.74 #[Pressure in last effect][kPa]\n", + "Tv3=328.0 #[Vapour temperature]\n", + "#Calculation\n", + "m3dot_dash=(ic/fc)*mf_dot #[kg/h]\n", + "mv_dot=mf_dot-m3dot_dash #/[kg/h]\n", + "dT=Ts-Tv3 #Overall apparent [K]\n", + "bpr1=10.0 #[K]\n", + "bpr2=bpr1 \n", + "bpr3=bpr2 \n", + "sum_bpr=bpr1+bpr2+bpr3 #[K]\n", + "dT=dT-sum_bpr #True_Overall\n", + "dT1=14.5 #[K]\n", + "dT2=16.0 #[K]\n", + "dT3=19.5 #[K]\n", + "Cpf=3.768 #[kJ/(kg.K)]\n", + "#Enthalpies of various streams\n", + "Hf=Cpf*(Tf-T) #[kJ/kg]\n", + "H1=Cpf*(393.5-T) #[kJ/kg]\n", + "H2=Cpf*(367.5-T) #[kJ/kg]\n", + "H3=Cpf*(338.0-T) #[kJ/kg]\n", + "#For steam at 40K\n", + "lambda_s=2160.0 #[kJ/kg]\n", + "Hv1=2692.0 #[kJ/kg]\n", + "lambda_v1=2228.3 #[kJ/kg]\n", + "Hv2=2650.8 #[kJ/kg]\n", + "lambda_v2=2297.4 #[kJ/kg]\n", + "Hv3=2600.5 #[kJ/kg]\n", + "lambda_v3=2370.0 #[kJ/kg]\n", + "\n", + "#MATERIAL AND EBERGY BALANCES\n", + "#First effect\n", + "#Material balance\n", + "\n", + "#m1dot_dash=mf_dot-mv1_dot\n", + "#m1dot_dash=1750+mv2_dot+mv3_dot \n", + "\n", + "#Energy balance\n", + "#ms_dot*lambda_s+mf_Dot*hf=mv1_dot*Hv1+m1dot_dash*H1\n", + "#2160*ms_dot+2238*(mv2_dot+mv3_dot)=19800500\n", + "\n", + "#Second effect\n", + "#Energy balance:\n", + "#mv3_dot=8709.54-2.076*mv2_dot\n", + "\n", + "#Third effect:\n", + "#m2dot_dash=mv3_dot+m3dot_dash\n", + "#m2dot_dash=mv3_dot+1750\n", + "#From eqn 8 we get\n", + "mv2_dot=(8709.54*2600.5+1750*244.92-8790.54*356.1-356.1*1750)/(-2.076*356.1+2297.4+2600.5*2.076)\n", + "#From eqn 8:\n", + "mv3_dot=8709.54-2.076*mv2_dot #[kg/h]\n", + "mv1_dot=mv_dot-(mv2_dot+mv3_dot) #[kg/h]\n", + "#From equation 4:\n", + "#m1dot_dash=mf_dot-mv1_dot\n", + "#ms_dot=(mv1_dot*Hv1+m1dot_dash*H1-mf_dot*Hf)/lambda_s #[kg/h]\n", + "ms_dot=(19800500.0-2238.0*(mv2_dot+mv3_dot))/2160.0 #[kg/h]\n", + "\n", + "#Heat transfer Area is\n", + "U1=710.0 #[W/sq m.K]\n", + "U2=490.0 #[W/sq m.K]\n", + "U3=454.0 #[W/sq m.K]\n", + "A1=(ms_dot*lambda_s*1000.0)/(3600.0*U1*dT1) #[sq m]\n", + "A2=mv1_dot*lambda_v1*1000.0/(3600.0*U2*dT2) #[sq m]\n", + "A3=mv2_dot*lambda_v2*1000.0/(3600.0*U3*dT3) #[sq m]\n", + "#The deviaiton is within +-10%\n", + "#Hence maximum A1 area can be recommended\n", + "\n", + "eco=(mv_dot/ms_dot) #[Steam economy]\n", + "\n", + "Qc=mv3_dot*lambda_v3 #[kJ/h]\n", + "dT=25.0 #Rise in water temperature\n", + "Cp=4.187\n", + "mw_dot=Qc/(Cp*dT)\n", + "#Result\n", + "print\"ANSWER:Area in each effect\",round(A3,1),\"sq m\" \n", + "print\"Steam economy is\",round(eco,2) \n", + "print\"Cooling water rate is\",round(mw_dot/1000,2),\"t/h\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.13 ,Page no:6.42" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steam economy is 2.0\n", + "Area pf heat transfer in each effect is 65.3 m^2\n" + ] + } + ], + "source": [ + "#Triple effect unit\n", + "#Variable declaration\n", + "Cpf=4.18 #[kJ/kg.K]\n", + "dT1=18 #[K]\n", + "dT2=17 #[K]\n", + "dT3=34 #[K]\n", + "mf_dot=4 #[kg/s]\n", + "Ts=394 #[K]\n", + "bp=325 #Bp of water at 13.172 kPa [K]\n", + "dT=Ts-bp #[K]\n", + "lambda_s=2200 #[kJ/kg]\n", + "T1=Ts-dT1 #[K]\n", + "lambda1=2249 #[kJ/kg]\n", + "lambda_v1=lambda1 #[kJ/kg]\n", + "#Calculation\n", + "T2=T1-dT2 #[K]\n", + "lambda2=2293 #[kJ/kg]\n", + "lambda_v2=lambda2 #[kJ/kg]\n", + "\n", + "T3=T2-dT3 #[K]\n", + "lambda3=2377 #[kJ/kg]\n", + "lambda_v3=lambda3 #[kJ/kg]\n", + "\n", + "ic=0.1 #Initial conc of solids\n", + "fc=0.5 #Final conc of solids\n", + "m3dot_dash=(ic/fc)*mf_dot #[kg/s]\n", + "mv_dot=mf_dot-m3dot_dash #Total evaporation in [kg/s]\n", + "#Material balance over first effect\n", + "#mf_dot=mv1_dot_m1dot_dash\n", + "#Energy balance:\n", + "#ms_dot*lambda_s=mf_dot*(Cpf*(T1-Tf)+mv1_dot*lambda_v1)\n", + "\n", + "#Material balance over second effect\n", + "#m1dot_dash=mv2_dot+m2dot_dash\n", + "#Enthalpy balance:\n", + "#mv1_dot*lambda_v1+m1dot_dash(cp*(T1-T2)=mv2_dot*lambda_v2)\n", + "\n", + "#Material balance over third effect\n", + "#m2dot_dash=mv3_dot+m3dot+dash\n", + "\n", + "#Enthalpy balance:\n", + "#mv2_lambda_v2+m2dot_dash*cp*(T2-T3)=mv3_dot*lambda_v3\n", + "294\n", + "mv2_dot=3.2795/3.079 #[kg/s]\n", + "mv1_dot=1.053*mv2_dot-0.1305 #[kg/s]\n", + "mv3_dot=1.026*mv2_dot+0.051 #[kg/s]\n", + "ms_dot=(mf_dot*Cpf*(T1-294)+mv1_dot*lambda_v1)/lambda_s #[kg/s]\n", + "eco=mv_dot/ms_dot #Steam economy \n", + "eco=round(eco)\n", + "U1=3.10 #[kW/sq m.K]\n", + "U2=2 #[kW/sq m.K]\n", + "U3=1.10 #[kW/sq m.K]\n", + "#First effect:\n", + "A1=ms_dot*lambda_s/(U1*dT1) #[sq m]\n", + "A2=mv1_dot*lambda_v1/(U2*dT2) #[sq m]\n", + "A3=mv2_dot*lambda_v2/(U3*dT3) #[sq m]\n", + "#Areas are calculated witha deviation of +-10%\n", + "#Result\n", + "print\"Steam economy is\",eco \n", + "print\"Area pf heat transfer in each effect is\",round(A3,1),\"m^2\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no: 6.14,Page no:6.45" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steam economy is 1.957 evaporation/kg steam\n" + ] + } + ], + "source": [ + "#Quadruple effect evaporator\n", + "#Variable declaration\n", + "mf_dot=1060 #[kg/h]\n", + "ic=0.04 #Initial concentration\n", + "fc=0.25 #Final concentration\n", + "m4dot_dash=(ic/fc)*mf_dot #[kg/h]\n", + "#Total evaporation=\n", + "mv_dot=mf_dot-m4dot_dash #[kg/h]\n", + "\n", + "#Fromsteam table:\n", + "P1=370 #[kPa.g]\n", + "T1=422.6 #[K]\n", + "lambda1=2114.4 #[kJ/kg]\n", + "\n", + "P2=235 #[kPa.g]\n", + "T2=410.5 #[K]\n", + "lambda2=2151.5 #[kJ/kg]\n", + "\n", + "P3=80 #[kPa.g]\n", + "T3=390.2 #[K]\n", + "lambda3=2210.2 #[kJ/kg]\n", + "\n", + "P4=50.66 #[kPa.g]\n", + "T4=354.7 #[K]\n", + "lambda4=2304.6 #[kJ/kg]\n", + "\n", + "P=700 #Latent heat of steam[kPa .g]\n", + "lambda_s=2046.3 #[kJ/kg]\n", + "\n", + "#Calculation\n", + "#FIRST EFFECT\n", + "#Enthalpy balance:\n", + "#ms_dot=mf_dot*Cpf*(T1-Tf)+mv1_dot*lambda1\n", + "#ms_dot=1345.3-1.033*m1dot_dash\n", + "\n", + "#SECOND EFFECT\n", + "#m1dot_dash=m2dot_dash+mdot_v2\n", + "#Enthalpy balance:\n", + "#m1dot_dash=531.38+0.510*m2dot_dash\n", + "\n", + "#THIRD EFFECT\n", + "#Material balance:\n", + "#m2dot_dash-m3dot_dash+mv3_dot\n", + "\n", + "#FOURTH EFFECT\n", + "#m3dot_dash=m4dot_dash+mv4_dot\n", + "mv4dot_dash=169.6 #[kg/h]\n", + "m3dot_dash=416.7 #[kg/h]\n", + "\n", + "#From eq n 4:\n", + "m2dot_dash=-176.84+1.98*m3dot_dash #[kg/h]\n", + "\n", + "#From eqn 2:\n", + "m1dot_dash=531.38+0.510*m2dot_dash #[kg/h]\n", + "\n", + "#From eqn 1:\n", + "ms_dot=1345.3-1.033*m1dot_dash\n", + "eco=mv_dot/ms_dot #[kg evaporation /kg steam]\n", + "#Result\n", + "print\"Steam economy is\",round(eco,3),\"evaporation/kg steam\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example no:6.15 ,Page no:6.48" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this\n", + "Steam consumption is 24531.0 kg/h\n", + "Capacity is 20000.0 kg/h\n", + "Steam economy is 0.815\n", + " No. of tubes required is 722.0\n" + ] + } + ], + "source": [ + "#Single effect Calendria\n", + "import math\n", + "#Variable declaration\n", + "m1_dot=5000 #[kg/h]\n", + "ic=0.1 #Initial concentration\n", + "fc=0.5 #Final concentration\n", + "mf_dot=(fc/ic)*m1_dot #[kg/h]\n", + "mv_dot=mf_dot-m1_dot #Water evaporated[kg/h]\n", + "P=357 #Steam pressure[kN/sq m]\n", + "Ts=412 #[K]\n", + "H=2732 #[kJ/kg]\n", + "lambda1=2143 #[kJ/kg]\n", + "bpr=18.5 #[K]\n", + "T_dash=352+bpr #[K]\n", + "Hf=138 #[kJ/kg]\n", + "lambda_s=2143 #[kJ/kg]\n", + "Hv=2659 #[kJ/kg]\n", + "H1=568 #[kJ/kg]\n", + "#Calculation\n", + "ms_dot=(mv_dot*Hv+m1_dot*H1-mf_dot*Hf)/lambda_s #Steam consumption in kg/h\n", + "eco=mv_dot/ms_dot #Economy\n", + "dT=Ts-T_dash #[K]\n", + "hi=4500 #[W/sq m.K]\n", + "ho=9000 #[W/sq m.K]\n", + "Do=0.032 #[m]\n", + "Di=0.028 #[m]\n", + "x1=(Do-Di)/2 #[m]\n", + "Dw=(Do-Di)/math.log(32.0/28.0) #[m]\n", + "x2=0.25*10**-3 #[m]\n", + "L=2.5 #Length [m]\n", + "hio=hi*(Di/Do) #[W/sq m.K]\n", + "print\"NOTE:In textbook this value of hio is wrongly calculated as 3975.5..So we will take this\"\n", + "hio=3975.5\n", + "k1=45.0 #Tube material in [W/sq m.K]\n", + "k2=2.25 #For scale[W/m.K]\n", + "Uo=1.0/(1.0/ho+1.0/hio+(x1*Dw)/(k1*Do)+(x2/k2)) #Overall heat transfer coeff in W/sq m.K\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000.0/3600.0 #[W]\n", + "\n", + "A=Q/(Uo*dT) #[sq m]\n", + "n=A/(math.pi*Do*L) #from A=n*math.pi*Do*L \n", + "#Result\n", + "print\"Steam consumption is\",round(ms_dot),\"kg/h\" \n", + "print\"Capacity is\",round(mv_dot),\"kg/h\"\n", + "print\"Steam economy is \",round(eco,3)\n", + "print\" No. of tubes required is \",round(n)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "slideshow": { + "slide_type": "subslide" + } + }, + "source": [ + "## Example no:6.16 ,Page no:6.50" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steam consumption is 3159.6 kg/h\n", + "Heat transfer area is 47.91 m^2\n", + " Now,Steam consumption is 3253.42 kg/h\n", + "Economy of evaporator 0.84\n", + "Now,Area is 49.33\n", + "If enthalpy of water vapour Hv were based on the saturated vapour at the pressure\n", + "the error introduced is only 2.97 percent\n" + ] + } + ], + "source": [ + "#Single effect evaporator\n", + "#Variable declaration\n", + "bpr=40.6 #[K]\n", + "Cpf=1.88 #[kJ/kg.K]\n", + "Hf=214 #[kJ/kg]\n", + "H1=505 #[kJ/kg]\n", + "mf_dot=4536 #[kg/h] of feed solution\n", + "ic=0.2 #Initial conc\n", + "fc=0.5 #Final concentration\n", + "m1dot_dash=(ic/fc)*mf_dot #Thisck liquor flow arte[kg/h]\n", + "mv_dot=mf_dot-m1dot_dash #[kg/H]\n", + "Ts=388.5 #Saturation temperature of steam in [K]\n", + "bp=362.5 #b.P of solution in [K]\n", + "lambda_s=2214 #[kJ/kg]\n", + "P=21.7 #Vapor space in [kPa]\n", + "Hv=2590.3 #[kJ/kg]\n", + "\n", + "#Calculation\n", + "#Enthalpy balance over evaporator\n", + "ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s #[kg/h\n", + "print\"Steam consumption is\",round(ms_dot,1),\"kg/h\" \n", + "dT=Ts-bp #[K]\n", + "U=1560 #[W/sq m.K]\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000/3600 #[W]\n", + "A=Q/(U*dT) #[sq m]\n", + "print\"Heat transfer area is\",round(A,2),\"m^2\"\n", + "\n", + "#Calculations considering enthalpy of superheated vapour\n", + "\n", + "Hv=Hv+Cpf*bpr #[kJ/kg]\n", + "ms_dot=(m1dot_dash*H1+mv_dot*Hv-mf_dot*Hf)/lambda_s #[kg/h]\n", + "print\" Now,Steam consumption is\",round(ms_dot,2),\"kg/h\" \n", + "eco=mv_dot/ms_dot #Steam economy\n", + "print\"Economy of evaporator \",round(eco,2)\n", + "Q=ms_dot*lambda_s #[kJ/h]\n", + "Q=Q*1000.0/3600.0 #[w]\n", + "A2=Q/(U*dT) #Area\n", + "print\"Now,Area is\",round(A2,2) \n", + "perc=(A2-A)*100/A #%error in the heat transfer area \n", + "#Result\n", + "print\"If enthalpy of water vapour Hv were based on the saturated vapour at the pressure\\nthe error introduced is only\",round(perc,2),\"percent\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Heat_Transfer_by_K._A._Gavhane/README.txt b/Heat_Transfer_by_K._A._Gavhane/README.txt new file mode 100644 index 00000000..aae4662f --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/README.txt @@ -0,0 +1,10 @@ +Contributed By: Kiran Bala +Course: btech +College/Institute/Organization: Ideal Institute of Technology,Ghaziabad +Department/Designation: Electrical & Electronics Engg +Book Title: Heat Transfer +Author: K. A. Gavhane +Publisher: Nirali Prakashan, Pune +Year of publication: 2010 +Isbn: 8190639617 +Edition: 10
\ No newline at end of file diff --git a/Heat_Transfer_by_K._A._Gavhane/screenshots/1.png b/Heat_Transfer_by_K._A._Gavhane/screenshots/1.png Binary files differnew file mode 100644 index 00000000..bb480cd9 --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/screenshots/1.png diff --git a/Heat_Transfer_by_K._A._Gavhane/screenshots/2.png b/Heat_Transfer_by_K._A._Gavhane/screenshots/2.png Binary files differnew file mode 100644 index 00000000..7149bea5 --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/screenshots/2.png diff --git a/Heat_Transfer_by_K._A._Gavhane/screenshots/3.png b/Heat_Transfer_by_K._A._Gavhane/screenshots/3.png Binary files differnew file mode 100644 index 00000000..2bf2d0fa --- /dev/null +++ b/Heat_Transfer_by_K._A._Gavhane/screenshots/3.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1.ipynb new file mode 100644 index 00000000..be2e4451 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1.ipynb @@ -0,0 +1,558 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Advanced Operational Amplifier Principles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.1,Page 6" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "open output voltage is 0.5 V\n", + "resistance lower loaded is 333.333 ohm\n", + "loaded output voltage is 0.25 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=1000.0;\n", + "R2=1000.0;\n", + "Rl=500.0#load resistance\n", + "V=1.0#input voltage\n", + "\n", + "#calculation\n", + "Vo=(R2/(R1+R2))*V;\n", + "Rll=1/((1/R2)+(1/Rl))#lower loaded resistance\n", + "Vol=(Rll/(R2+Rll))*V;\n", + "\n", + "#result\n", + "print \"open output voltage is\",round(Vo,3),\"V\"\n", + "print \"resistance lower loaded is\",round(Rll,3),\"ohm\"\n", + "print \"loaded output voltage is\",round(Vol,3),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.2,Page 11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistance is 1.01 Kohm\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=100000.0#resistance\n", + "Acl=100.0#amplifier gain\n", + "\n", + "#calculation\n", + "Ri=Rf/(Acl-1);\n", + "\n", + "#result\n", + "print \"input resistance is\",round(Ri/1000,2), \"Kohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.3,Page 17" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current through Ri1 is 178.571 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Rf is 210.486 microAmp\n", + "voltage dropped is 2.105 V\n", + "output voltage 1 is -2.105 V\n", + "output voltage is 2.105 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vni=0.0#non inverting voltage\n", + "Vinv=0.0;#inverting voltage\n", + "Vri1=1.0;\n", + "Vri2=15.0;\n", + "Ri1=5600.0#resistance\n", + "Ri2=470000.0;\n", + "Rf=10000.0#load resistance\n", + "\n", + "#calculation\n", + "Ir1=Vri1/Ri1;\n", + "Ir2=Vri2/Ri2;\n", + "Irf=(Vri1/Ri1)+(Vri2/Ri2);\n", + "Vr=Irf*Rf;\n", + "Vo1=-Vr;\n", + "Vo=Irf*Rf;\n", + "\n", + "#result\n", + "print \"current through Ri1 is\",round(Ir1*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3),\"microAmp\"\n", + "print \"current through Rf is\",round(Irf*1e6,3), \"microAmp\"\n", + "print \"voltage dropped is\",round(Vr,3), \"V\"\n", + "print \"output voltage 1 is\",round(Vo1,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.4,Page 25" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inverting voltage is 4.955 V\n", + "non inverting voltage is 4.955 V\n", + "current through Rf2 is 42.698 microA\n", + "current through Ri2 is 42.698 microA\n", + "voltage dropped is 4.056 V\n", + "output voltage is 884.897 mV\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ri1=950.00;#ohm\n", + "Ri2=1050.00;\n", + "Rf1=105000.00;#resistance\n", + "Rf2=95000.00;\n", + "Vin=5.00;#voltage\n", + "\n", + "#calculation\n", + "Vinv=(Rf1/(Rf1+Ri1))*Vin;\n", + "Vni=Vinv;\n", + "Irf2=(Vin-Vinv)/Ri2;\n", + "Iri2=Irf2;\n", + "Vrf2=Irf2*Rf2;\n", + "Vo=Vinv-Vrf2-.014;\n", + "\n", + "#result\n", + "print \"inverting voltage is\",round(Vinv,3), \"V\"\n", + "print \"non inverting voltage is\",round (Vni,3), \"V\"\n", + "print \"current through Rf2 is\",round(Irf2*1e6,3), \"microA\"\n", + "print \"current through Ri2 is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"voltage dropped is\",round(Vrf2,3), \"V\"\n", + "print \"output voltage is\",round(Vo*1000,3), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.5,Page 27" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 272.222 microA\n", + "input resistor current is 500.0 microA\n", + "feedback resistor current is 227.778 microAmp\n", + "resistor voltage is 227.778 mV\n", + "1st output voltage is 2.222 V\n", + "input resistor current is 327.778 microA\n", + "input resistor current is 827.778 microA\n", + "feedback resistor voltage is 7.45 V\n", + "2nd output voltage is 10.0 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=2.45;#V\n", + "Vniu2=2.55;#V\n", + "Vinvu1=2.45;\n", + "Vinvu2=2.55;\n", + "Ri1=9000.0;#ohm\n", + "Ri2=1000.0;#ohm\n", + "Rf1=1000.0;\n", + "Rf2=9000.0;\n", + "Rg=200.0;#load resistance\n", + "\n", + "#calculation\n", + "Iri1=Vniu1/Ri1;\n", + "Irg=(Vniu2-Vniu1)/Rg;\n", + "Irf1=Irg-Iri1;\n", + "Vrf1=Irf1*Rf1;\n", + "Vou1=Vniu1-Vrf1;\n", + "Iri2=(Vniu2-Vou1)/Ri2;\n", + "Irf2=Iri2+Irg;\n", + "Vrf2=Irf2*Rf2#feedback resistor voltage\n", + "Vo=Vrf2+Vniu2;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iri1*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irg*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf1*1e6,3), \"microAmp\"\n", + "print \"resistor voltage is\",round(Vrf1*1000,3), \"mV\"\n", + "print \"1st output voltage is\",round(Vou1,3), \"V\"\n", + "print \"input resistor current is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irf2*1e6,3),\"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf2,3), \"V\"\n", + "print \"2nd output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.a,Page 29" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 128.0 microA\n", + "feedback resistor current is 128.0 microA\n", + "feedback resistor voltage is 5.018 V\n", + "output resistor voltage is 5.018 V\n", + "output voltage is 3.818 V\n", + "load current is 0.5 A\n", + "load power is 2.5 W\n", + "power dissipated in LM317 is 5.0 W\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=0;#V\n", + "Vinvu2=0;#V\n", + "Vref=2.56;\n", + "Rl=10000.0;#ohm\n", + "Rf=39200.0;#ohm\n", + "Ro=10.0;#resistance\n", + "Vdc1=5.0;\n", + "Vdc2=15.0;\n", + "Idc=0.5;#current\n", + "\n", + "#calculation\n", + "Iu1=(Vref/Rl)*.5;\n", + "Irf=Iu1;\n", + "Vrf=Irf*Rf;\n", + "Vout=Vrf+Vinvu2;\n", + "Eo=Vout-1.2;\n", + "Iload=Vdc1/Ro;\n", + "Pload=Vdc1**2/Ro;\n", + "Plm317=(Vdc2-Vdc1)*Idc;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iu1*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf*1e6,3), \"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf,3), \"V\"\n", + "print \"output resistor voltage is\",round(Vout,3), \"V\"\n", + "print \"output voltage is\",round(Eo,3), \"V\"\n", + "print \"load current is\",round(Iload,3), \"A\"\n", + "print \"load power is\",round(Pload,3), \"W\"\n", + "print \"power dissipated in LM317 is\",round(Plm317,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.b,Page 31" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 360.36 microamp\n", + "inverting voltage 1 & 2 is 396.396 mV\n", + "current across Rs is 3.964 A\n", + "emitter voltage is 8.324 V\n", + "output voltage is 10.124 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4;#V\n", + "Vs=1.8;#V\n", + "Rf=10000.0;#ohm\n", + "Ri=1100.0;#ohm\n", + "Rl=2.0;#ohm\n", + "Rs=0.1;#ohm\n", + "\n", + "#calculation\n", + "Irf=Vin/(Rf+Ri);\n", + "Vni=Irf*Ri;\n", + "Ir=Vni/Rs;\n", + "Ve=Ir*(Rl+Rs);\n", + "Vo=Ve+Vs;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Irf*1e6,3),\"microamp\"\n", + "print \"inverting voltage 1 & 2 is\",round(Vni*1000,3), \"mV\"\n", + "print \"current across Rs is\",round(Ir,3), \"A\"\n", + "print \"emitter voltage is\",round(Ve,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.7,Page 36" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 9.899 V\n", + "power delivered is 12.25 W\n", + "load voltage is 28.284 V\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=18.0;#V\n", + "Rl=8.0;#load resistance\n", + "Pll=100.0;#power\n", + "\n", + "#calculation\n", + "Vlp=Vs-4;\n", + "Vlr=Vlp/(2**(.5));\n", + "Pl=(Vlr**2)/Rl;\n", + "Vl=(Pll*Rl)**(.5);\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vlr,3), \"V\"\n", + "print \"power delivered is\",round(Pl,3), \"W\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.9,Page 44" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 37.34 V\n", + "V+ is 45.34 V ;V- is 29.34 V\n" + ] + } + ], + "source": [ + "#finding output volatage and range \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "import numpy as np\n", + "Vp=6.0;#V\n", + "Ra=10.0;#Kohm\n", + "Rb=1800.0;#ohm\n", + "V=8.0;\n", + "#solving for Ir & Vo\n", + "a=np.array([[1.0,-124.6e-6],[7800.0,-1.0]])\n", + "b=np.array([134.6e-6,0.0])\n", + "\n", + "#calculation\n", + "x=np.linalg.solve(a,b);\n", + "Vo=x[1];\n", + "Va=Vo+V;\n", + "Vb=Vo-V;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"V+ is\",round(Va,2), \"V ;V- is\",round(Vb,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.11,Page 50" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output current is 4.091 mA\n", + "output voltage is 45.409 V\n", + "gain output voltage 1 is 13.356 V\n", + "gain output voltage 2 is 0.38 V\n" + ] + } + ], + "source": [ + "#finding output voltage and gain output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4.5;\n", + "R1=1100.0;\n", + "R2=10000.0;\n", + "\n", + "G1=3.4#gain 1\n", + "G2=120.0#gain 2\n", + "\n", + "#calculation\n", + "Ir=Vin/R1;\n", + "Vo=Ir*(R1+R2);\n", + "Vuo1=Vo/G1;\n", + "Vuo2=Vo/G2;\n", + "\n", + "#result\n", + "print \"output current is\",round(Ir*1000,3),\"mA\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "print \"gain output voltage 1 is\",round(Vuo1,3), \"V\"\n", + "print \"gain output voltage 2 is\",round(Vuo2,2),\"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1_1.ipynb new file mode 100644 index 00000000..be2e4451 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter1_1.ipynb @@ -0,0 +1,558 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Advanced Operational Amplifier Principles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.1,Page 6" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "open output voltage is 0.5 V\n", + "resistance lower loaded is 333.333 ohm\n", + "loaded output voltage is 0.25 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=1000.0;\n", + "R2=1000.0;\n", + "Rl=500.0#load resistance\n", + "V=1.0#input voltage\n", + "\n", + "#calculation\n", + "Vo=(R2/(R1+R2))*V;\n", + "Rll=1/((1/R2)+(1/Rl))#lower loaded resistance\n", + "Vol=(Rll/(R2+Rll))*V;\n", + "\n", + "#result\n", + "print \"open output voltage is\",round(Vo,3),\"V\"\n", + "print \"resistance lower loaded is\",round(Rll,3),\"ohm\"\n", + "print \"loaded output voltage is\",round(Vol,3),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.2,Page 11" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistance is 1.01 Kohm\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=100000.0#resistance\n", + "Acl=100.0#amplifier gain\n", + "\n", + "#calculation\n", + "Ri=Rf/(Acl-1);\n", + "\n", + "#result\n", + "print \"input resistance is\",round(Ri/1000,2), \"Kohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.3,Page 17" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current through Ri1 is 178.571 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Ri2 is 31.915 microAmp\n", + "current through Rf is 210.486 microAmp\n", + "voltage dropped is 2.105 V\n", + "output voltage 1 is -2.105 V\n", + "output voltage is 2.105 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vni=0.0#non inverting voltage\n", + "Vinv=0.0;#inverting voltage\n", + "Vri1=1.0;\n", + "Vri2=15.0;\n", + "Ri1=5600.0#resistance\n", + "Ri2=470000.0;\n", + "Rf=10000.0#load resistance\n", + "\n", + "#calculation\n", + "Ir1=Vri1/Ri1;\n", + "Ir2=Vri2/Ri2;\n", + "Irf=(Vri1/Ri1)+(Vri2/Ri2);\n", + "Vr=Irf*Rf;\n", + "Vo1=-Vr;\n", + "Vo=Irf*Rf;\n", + "\n", + "#result\n", + "print \"current through Ri1 is\",round(Ir1*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3), \"microAmp\"\n", + "print \"current through Ri2 is\",round(Ir2*1e6,3),\"microAmp\"\n", + "print \"current through Rf is\",round(Irf*1e6,3), \"microAmp\"\n", + "print \"voltage dropped is\",round(Vr,3), \"V\"\n", + "print \"output voltage 1 is\",round(Vo1,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.4,Page 25" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inverting voltage is 4.955 V\n", + "non inverting voltage is 4.955 V\n", + "current through Rf2 is 42.698 microA\n", + "current through Ri2 is 42.698 microA\n", + "voltage dropped is 4.056 V\n", + "output voltage is 884.897 mV\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ri1=950.00;#ohm\n", + "Ri2=1050.00;\n", + "Rf1=105000.00;#resistance\n", + "Rf2=95000.00;\n", + "Vin=5.00;#voltage\n", + "\n", + "#calculation\n", + "Vinv=(Rf1/(Rf1+Ri1))*Vin;\n", + "Vni=Vinv;\n", + "Irf2=(Vin-Vinv)/Ri2;\n", + "Iri2=Irf2;\n", + "Vrf2=Irf2*Rf2;\n", + "Vo=Vinv-Vrf2-.014;\n", + "\n", + "#result\n", + "print \"inverting voltage is\",round(Vinv,3), \"V\"\n", + "print \"non inverting voltage is\",round (Vni,3), \"V\"\n", + "print \"current through Rf2 is\",round(Irf2*1e6,3), \"microA\"\n", + "print \"current through Ri2 is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"voltage dropped is\",round(Vrf2,3), \"V\"\n", + "print \"output voltage is\",round(Vo*1000,3), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.5,Page 27" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 272.222 microA\n", + "input resistor current is 500.0 microA\n", + "feedback resistor current is 227.778 microAmp\n", + "resistor voltage is 227.778 mV\n", + "1st output voltage is 2.222 V\n", + "input resistor current is 327.778 microA\n", + "input resistor current is 827.778 microA\n", + "feedback resistor voltage is 7.45 V\n", + "2nd output voltage is 10.0 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=2.45;#V\n", + "Vniu2=2.55;#V\n", + "Vinvu1=2.45;\n", + "Vinvu2=2.55;\n", + "Ri1=9000.0;#ohm\n", + "Ri2=1000.0;#ohm\n", + "Rf1=1000.0;\n", + "Rf2=9000.0;\n", + "Rg=200.0;#load resistance\n", + "\n", + "#calculation\n", + "Iri1=Vniu1/Ri1;\n", + "Irg=(Vniu2-Vniu1)/Rg;\n", + "Irf1=Irg-Iri1;\n", + "Vrf1=Irf1*Rf1;\n", + "Vou1=Vniu1-Vrf1;\n", + "Iri2=(Vniu2-Vou1)/Ri2;\n", + "Irf2=Iri2+Irg;\n", + "Vrf2=Irf2*Rf2#feedback resistor voltage\n", + "Vo=Vrf2+Vniu2;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iri1*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irg*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf1*1e6,3), \"microAmp\"\n", + "print \"resistor voltage is\",round(Vrf1*1000,3), \"mV\"\n", + "print \"1st output voltage is\",round(Vou1,3), \"V\"\n", + "print \"input resistor current is\",round(Iri2*1e6,3), \"microA\"\n", + "print \"input resistor current is\",round(Irf2*1e6,3),\"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf2,3), \"V\"\n", + "print \"2nd output voltage is\",round(Vo,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.a,Page 29" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 128.0 microA\n", + "feedback resistor current is 128.0 microA\n", + "feedback resistor voltage is 5.018 V\n", + "output resistor voltage is 5.018 V\n", + "output voltage is 3.818 V\n", + "load current is 0.5 A\n", + "load power is 2.5 W\n", + "power dissipated in LM317 is 5.0 W\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vniu1=0;#V\n", + "Vinvu2=0;#V\n", + "Vref=2.56;\n", + "Rl=10000.0;#ohm\n", + "Rf=39200.0;#ohm\n", + "Ro=10.0;#resistance\n", + "Vdc1=5.0;\n", + "Vdc2=15.0;\n", + "Idc=0.5;#current\n", + "\n", + "#calculation\n", + "Iu1=(Vref/Rl)*.5;\n", + "Irf=Iu1;\n", + "Vrf=Irf*Rf;\n", + "Vout=Vrf+Vinvu2;\n", + "Eo=Vout-1.2;\n", + "Iload=Vdc1/Ro;\n", + "Pload=Vdc1**2/Ro;\n", + "Plm317=(Vdc2-Vdc1)*Idc;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Iu1*1e6,3), \"microA\"\n", + "print \"feedback resistor current is\",round(Irf*1e6,3), \"microA\"\n", + "print \"feedback resistor voltage is\",round(Vrf,3), \"V\"\n", + "print \"output resistor voltage is\",round(Vout,3), \"V\"\n", + "print \"output voltage is\",round(Eo,3), \"V\"\n", + "print \"load current is\",round(Iload,3), \"A\"\n", + "print \"load power is\",round(Pload,3), \"W\"\n", + "print \"power dissipated in LM317 is\",round(Plm317,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6.b,Page 31" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "input resistor current is 360.36 microamp\n", + "inverting voltage 1 & 2 is 396.396 mV\n", + "current across Rs is 3.964 A\n", + "emitter voltage is 8.324 V\n", + "output voltage is 10.124 V\n" + ] + } + ], + "source": [ + "#finding voltage current resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4;#V\n", + "Vs=1.8;#V\n", + "Rf=10000.0;#ohm\n", + "Ri=1100.0;#ohm\n", + "Rl=2.0;#ohm\n", + "Rs=0.1;#ohm\n", + "\n", + "#calculation\n", + "Irf=Vin/(Rf+Ri);\n", + "Vni=Irf*Ri;\n", + "Ir=Vni/Rs;\n", + "Ve=Ir*(Rl+Rs);\n", + "Vo=Ve+Vs;\n", + "\n", + "#result\n", + "print \"input resistor current is\",round(Irf*1e6,3),\"microamp\"\n", + "print \"inverting voltage 1 & 2 is\",round(Vni*1000,3), \"mV\"\n", + "print \"current across Rs is\",round(Ir,3), \"A\"\n", + "print \"emitter voltage is\",round(Ve,3), \"V\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.7,Page 36" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 9.899 V\n", + "power delivered is 12.25 W\n", + "load voltage is 28.284 V\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=18.0;#V\n", + "Rl=8.0;#load resistance\n", + "Pll=100.0;#power\n", + "\n", + "#calculation\n", + "Vlp=Vs-4;\n", + "Vlr=Vlp/(2**(.5));\n", + "Pl=(Vlr**2)/Rl;\n", + "Vl=(Pll*Rl)**(.5);\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vlr,3), \"V\"\n", + "print \"power delivered is\",round(Pl,3), \"W\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.9,Page 44" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 37.34 V\n", + "V+ is 45.34 V ;V- is 29.34 V\n" + ] + } + ], + "source": [ + "#finding output volatage and range \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "import numpy as np\n", + "Vp=6.0;#V\n", + "Ra=10.0;#Kohm\n", + "Rb=1800.0;#ohm\n", + "V=8.0;\n", + "#solving for Ir & Vo\n", + "a=np.array([[1.0,-124.6e-6],[7800.0,-1.0]])\n", + "b=np.array([134.6e-6,0.0])\n", + "\n", + "#calculation\n", + "x=np.linalg.solve(a,b);\n", + "Vo=x[1];\n", + "Va=Vo+V;\n", + "Vb=Vo-V;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"V+ is\",round(Va,2), \"V ;V- is\",round(Vb,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.11,Page 50" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output current is 4.091 mA\n", + "output voltage is 45.409 V\n", + "gain output voltage 1 is 13.356 V\n", + "gain output voltage 2 is 0.38 V\n" + ] + } + ], + "source": [ + "#finding output voltage and gain output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=4.5;\n", + "R1=1100.0;\n", + "R2=10000.0;\n", + "\n", + "G1=3.4#gain 1\n", + "G2=120.0#gain 2\n", + "\n", + "#calculation\n", + "Ir=Vin/R1;\n", + "Vo=Ir*(R1+R2);\n", + "Vuo1=Vo/G1;\n", + "Vuo2=Vo/G2;\n", + "\n", + "#result\n", + "print \"output current is\",round(Ir*1000,3),\"mA\"\n", + "print \"output voltage is\",round(Vo,3), \"V\"\n", + "print \"gain output voltage 1 is\",round(Vuo1,3), \"V\"\n", + "print \"gain output voltage 2 is\",round(Vuo2,2),\"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2.ipynb new file mode 100644 index 00000000..9d88fe28 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2.ipynb @@ -0,0 +1,65 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Power Electronics Circuit Layout" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.9,Page 83" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load current is 3.75 A\n", + "wiring resistance is 26.67 mohm\n", + "resistance per inch is 1666.67 microohm/inch\n" + ] + } + ], + "source": [ + "#finding an appropriate wire gauge\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "R=4.0;#resistance\n", + "Vl=.1;\n", + "D=8.0;#duty cycle\n", + "\n", + "#calculation\n", + "Il=V/R;\n", + "Rw=Vl/Il#wiring resistance\n", + "Ri=Rw/(2*D);\n", + "\n", + "#result\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"wiring resistance is\",round(Rw*1000,2), \"mohm\"\n", + "print \"resistance per inch is\",round(Ri*1e6,2), \"microohm/inch\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2_1.ipynb new file mode 100644 index 00000000..9d88fe28 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter2_1.ipynb @@ -0,0 +1,65 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Power Electronics Circuit Layout" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.9,Page 83" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load current is 3.75 A\n", + "wiring resistance is 26.67 mohm\n", + "resistance per inch is 1666.67 microohm/inch\n" + ] + } + ], + "source": [ + "#finding an appropriate wire gauge\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "R=4.0;#resistance\n", + "Vl=.1;\n", + "D=8.0;#duty cycle\n", + "\n", + "#calculation\n", + "Il=V/R;\n", + "Rw=Vl/Il#wiring resistance\n", + "Ri=Rw/(2*D);\n", + "\n", + "#result\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"wiring resistance is\",round(Rw*1000,2), \"mohm\"\n", + "print \"resistance per inch is\",round(Ri*1e6,2), \"microohm/inch\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3.ipynb new file mode 100644 index 00000000..4129f088 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3.ipynb @@ -0,0 +1,481 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Power Parameter Calculations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.1,Page 109" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ramp current is 450.0 kAt/s\n", + "current at 5 micro sec is 2.25 A\n" + ] + } + ], + "source": [ + "#finding ramp current and current at 5 micro sec\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=150000.0;\n", + "t=5.0e-6;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "It=Ip/T;\n", + "I5=It*t;\n", + "\n", + "#result\n", + "print \"ramp current is\",round(It/1000,3), \"kAt/s\"\n", + "print \"current at 5 micro sec is\",round(I5,3), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.2,Page 110" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current in time 0<=t<800ns is 3.575 A\n", + "current in time 800ns<=t<2 microsec is 0.0 A\n", + "current in time 400ns is 1.85 A\n", + "current in time 1 microsec is 0.0 A\n" + ] + } + ], + "source": [ + "#finding current at different time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=2.0;\n", + "f=500000.0;\n", + "Ir=.3;\n", + "Cd=.4#duty cycle\n", + "t1=4.0e-7;\n", + "t2=1.0e-6;\n", + "I1=0;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "Im=Ip-Ir;\n", + "I4=(Ip-Im)*t1/(Cd*T)+Im;\n", + "It=(Ip-Im)*t/(Cd*T)+Im;\n", + "It1=0\n", + "\n", + "#resilt\n", + "print \"current in time 0<=t<800ns is\",round(It,3),\"A\"\n", + "print \"current in time 800ns<=t<2 microsec is\",round(It1,2), \"A\"\n", + "print \"current in time 400ns is\",round(I4,2), \"A\"\n", + "print \"current in time 1 microsec is\",round(I1,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.3,Page 115" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 54.02 V\n" + ] + } + ], + "source": [ + "#finding average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vr=120;\n", + "\n", + "#calculation\n", + "V=(Vr*2**.5)/pi;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.4,Page 119" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average current is 0.98 A\n" + ] + } + ], + "source": [ + "#finding average current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f=100000.0;\n", + "Cd=.35#duty cycle\n", + "Ip=3.0;\n", + "Ir=.4;\n", + "\n", + "#calculation\n", + "Im=Ip-Ir;\n", + "T=1/f;\n", + "I=Cd*((Ip-Im)/2+Im)\n", + "\n", + "#result\n", + "print \"average current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.5,Page 124" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 8.87 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=15.0;\n", + "Cd=.35;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "V=Vp*Cd**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(V,2), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.6,Page 127" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 1.73 A\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "I=Ip/3**.5;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.7,Page 133" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 85.0 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=170.0;\n", + "f=60.0;\n", + "\n", + "#calculation\n", + "Vr=Vp/2;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.8,Page 140" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power required is 2.42 hp\n", + "Pick a 5HP motor\n", + "current required is 18.84 amp\n" + ] + } + ], + "source": [ + "#finding current and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "M=1000.0;\n", + "H=40.0;\n", + "T=30.0;\n", + "E1=.9;\n", + "E2=.5;\n", + "V=220.0;\n", + "P1=5.0;\n", + "\n", + "#calculation\n", + "W=M*H;\n", + "P=(W)/(T*550);\n", + "Pe=P1/E1;\n", + "I=(Pe*746)/V;\n", + "\n", + "#result\n", + "print \"power required is\",round(P,2), \"hp\"\n", + "print('Pick a 5HP motor')\n", + "print \"current required is\",round(I,2), \"amp\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.9,Page 145" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered to the load is 6.36 Watt\n", + "power provided by each supply is 7.23 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=1.0;\n", + "Ri=1100.0;\n", + "Rf=10000.0;\n", + "Rl=8.0;\n", + "Vs=18.0;\n", + "\n", + "#calculation\n", + "Ir=Vin/Ri;\n", + "Vl=Ir*(Ri+Rf);\n", + "Ip=Vl/Rl;\n", + "Pl=(Vl*Ip)/2;\n", + "Ps=(Vs*Ip)/pi;\n", + "\n", + "#result\n", + "print \"power delivered to the load is\",round(Pl,2),\"Watt\"\n", + "print \"power provided by each supply is\",round(Ps,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.10,Page 149" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 141.67 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=170.0;\n", + "R=51.0;\n", + "\n", + "#calculation\n", + "I=V/R;\n", + "P=(V*I)/4;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(P,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.11,Page 151" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 7.05 watt\n", + "power dissipated when transistor resistance is 0.2 hm is 0.35 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=7.2;\n", + "Rq=.2;\n", + "Rl=4;\n", + "D=.6;\n", + "\n", + "#calculation\n", + "Ip=V/(Rq+Rl);\n", + "Vl=Ip*Rl;\n", + "P=D*Vl*Ip;\n", + "Vq=Ip*Rq;\n", + "Pq=D*Vq*Ip;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(P,2), \"watt\"\n", + "print \"power dissipated when transistor resistance is 0.2 hm is\",round(Pq,2), \"watt\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3_1.ipynb new file mode 100644 index 00000000..4129f088 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter3_1.ipynb @@ -0,0 +1,481 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Power Parameter Calculations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.1,Page 109" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ramp current is 450.0 kAt/s\n", + "current at 5 micro sec is 2.25 A\n" + ] + } + ], + "source": [ + "#finding ramp current and current at 5 micro sec\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=150000.0;\n", + "t=5.0e-6;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "It=Ip/T;\n", + "I5=It*t;\n", + "\n", + "#result\n", + "print \"ramp current is\",round(It/1000,3), \"kAt/s\"\n", + "print \"current at 5 micro sec is\",round(I5,3), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.2,Page 110" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current in time 0<=t<800ns is 3.575 A\n", + "current in time 800ns<=t<2 microsec is 0.0 A\n", + "current in time 400ns is 1.85 A\n", + "current in time 1 microsec is 0.0 A\n" + ] + } + ], + "source": [ + "#finding current at different time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=2.0;\n", + "f=500000.0;\n", + "Ir=.3;\n", + "Cd=.4#duty cycle\n", + "t1=4.0e-7;\n", + "t2=1.0e-6;\n", + "I1=0;\n", + "\n", + "#calculation\n", + "T=1/f;\n", + "Im=Ip-Ir;\n", + "I4=(Ip-Im)*t1/(Cd*T)+Im;\n", + "It=(Ip-Im)*t/(Cd*T)+Im;\n", + "It1=0\n", + "\n", + "#resilt\n", + "print \"current in time 0<=t<800ns is\",round(It,3),\"A\"\n", + "print \"current in time 800ns<=t<2 microsec is\",round(It1,2), \"A\"\n", + "print \"current in time 400ns is\",round(I4,2), \"A\"\n", + "print \"current in time 1 microsec is\",round(I1,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.3,Page 115" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 54.02 V\n" + ] + } + ], + "source": [ + "#finding average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vr=120;\n", + "\n", + "#calculation\n", + "V=(Vr*2**.5)/pi;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.4,Page 119" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average current is 0.98 A\n" + ] + } + ], + "source": [ + "#finding average current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f=100000.0;\n", + "Cd=.35#duty cycle\n", + "Ip=3.0;\n", + "Ir=.4;\n", + "\n", + "#calculation\n", + "Im=Ip-Ir;\n", + "T=1/f;\n", + "I=Cd*((Ip-Im)/2+Im)\n", + "\n", + "#result\n", + "print \"average current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.5,Page 124" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 8.87 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=15.0;\n", + "Cd=.35;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "V=Vp*Cd**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(V,2), \"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.6,Page 127" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 1.73 A\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ip=3.0;\n", + "f=100000.0;\n", + "\n", + "#calculation\n", + "I=Ip/3**.5;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.7,Page 133" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 85.0 V\n" + ] + } + ], + "source": [ + "#finding rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=170.0;\n", + "f=60.0;\n", + "\n", + "#calculation\n", + "Vr=Vp/2;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.8,Page 140" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power required is 2.42 hp\n", + "Pick a 5HP motor\n", + "current required is 18.84 amp\n" + ] + } + ], + "source": [ + "#finding current and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "M=1000.0;\n", + "H=40.0;\n", + "T=30.0;\n", + "E1=.9;\n", + "E2=.5;\n", + "V=220.0;\n", + "P1=5.0;\n", + "\n", + "#calculation\n", + "W=M*H;\n", + "P=(W)/(T*550);\n", + "Pe=P1/E1;\n", + "I=(Pe*746)/V;\n", + "\n", + "#result\n", + "print \"power required is\",round(P,2), \"hp\"\n", + "print('Pick a 5HP motor')\n", + "print \"current required is\",round(I,2), \"amp\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.9,Page 145" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered to the load is 6.36 Watt\n", + "power provided by each supply is 7.23 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vin=1.0;\n", + "Ri=1100.0;\n", + "Rf=10000.0;\n", + "Rl=8.0;\n", + "Vs=18.0;\n", + "\n", + "#calculation\n", + "Ir=Vin/Ri;\n", + "Vl=Ir*(Ri+Rf);\n", + "Ip=Vl/Rl;\n", + "Pl=(Vl*Ip)/2;\n", + "Ps=(Vs*Ip)/pi;\n", + "\n", + "#result\n", + "print \"power delivered to the load is\",round(Pl,2),\"Watt\"\n", + "print \"power provided by each supply is\",round(Ps,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.10,Page 149" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 141.67 Watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=170.0;\n", + "R=51.0;\n", + "\n", + "#calculation\n", + "I=V/R;\n", + "P=(V*I)/4;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(P,2), \"Watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.11,Page 151" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 7.05 watt\n", + "power dissipated when transistor resistance is 0.2 hm is 0.35 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=7.2;\n", + "Rq=.2;\n", + "Rl=4;\n", + "D=.6;\n", + "\n", + "#calculation\n", + "Ip=V/(Rq+Rl);\n", + "Vl=Ip*Rl;\n", + "P=D*Vl*Ip;\n", + "Vq=Ip*Rq;\n", + "Pq=D*Vq*Ip;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(P,2), \"watt\"\n", + "print \"power dissipated when transistor resistance is 0.2 hm is\",round(Pq,2), \"watt\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4.ipynb new file mode 100644 index 00000000..6bbf650f --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4.ipynb @@ -0,0 +1,742 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Linear Power Amplifier Integrated Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.1,Page 162" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 500.0 mV\n" + ] + } + ], + "source": [ + "#finding voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=1;\n", + "Ri=10;\n", + "Vi=0;\n", + "Ip=500;\n", + "\n", + "#calculation\n", + "Vrf=Ip*Rf;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vrf,2), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.2,Page 165" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency of OPA548 is 67.26 KHz\n", + "slew rate of OPA548 is 1.12 Mhz\n", + "the OPA548 can be used\n" + ] + } + ], + "source": [ + "#finding frequency\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=300.0;\n", + "P=35.0;\n", + "R=8.0;\n", + "S=10000.0;\n", + "fh=20.0;\n", + "\n", + "#calculation\n", + "Vl=(P*R)**.5;\n", + "Vp=Vl*2**.5;\n", + "Il=Vl/R;\n", + "f=S/(2*pi*Vp);\n", + "Ao=Vl/Vi;\n", + "G=Ao*fh;\n", + "\n", + "#result\n", + "print \"frequency of OPA548 is\",round(f,2), \"KHz\"\n", + "print \"slew rate of OPA548 is\",round(G,2), \"Mhz\"\n", + "print('the OPA548 can be used')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.3,Page 168" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 3.5 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=10.0;\n", + "V=12.0;\n", + "Vl=5.0;\n", + "\n", + "#calculation\n", + "Pl=Vl**2/Rl;\n", + "I=Vl/Rl;\n", + "Ps=V*I;\n", + "Pic=Ps-Pl;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pic,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.4,Page 170" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "0.0 0.0 0.0 0.0 0.0\n", + "0.2 0.0 0.0 0.24 0.24\n", + "0.4 0.0 0.02 0.48 0.46\n", + "0.6 0.1 0.04 0.72 0.68\n", + "0.8 0.1 0.06 0.96 0.9\n", + "4.8 0.5 2.3 5.76 3.46\n", + "5.0 0.5 2.5 6.0 3.5\n", + "5.2 0.5 2.7 6.24 3.54\n", + "5.4 0.5 2.92 6.48 3.56\n", + "5.6 0.6 3.14 6.72 3.58\n", + "5.8 0.6 3.36 6.96 3.6\n", + "6.0 0.6 3.6 7.2 3.6\n", + "6.2 0.6 3.84 7.44 3.6\n", + "6.4 0.6 4.1 7.68 3.58\n", + "11.4 1.1 13.0 13.68 0.68\n", + "11.6 1.2 13.46 13.92 0.46\n", + "11.8 1.2 13.92 14.16 0.24\n", + "12.0 1.2 14.4 14.4 0.0\n" + ] + }, + { + "data": { + "text/plain": [ + "<matplotlib.text.Text at 0x7fd132102a10>" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + }, + { + "data": { + "image/png": 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E8BnQNcHzXcifoFaUd4E9RRzjl5FOJbZ3rw2De+IJaNrU62hEyq4uXWzNsC5d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\nREREREREREREREREREREJBn/DwWIrS4GjkBZAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd1321993d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Vload vs Pic graph\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vload=[0.0, 0.2, 0.4, 0.6, 0.8, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 6.2, 6.4, 11.4, 11.6, 11.8, 12.0];\n", + "Iload=[0.0, 0.0, 0.0, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.6, 0.6, 0.6, 0.6, 0.6, 1.1, 1.2, 1.2, 1.2];\n", + "Pload=[0.00, 0.00, 0.02, 0.04, 0.06, 2.30, 2.50, 2.70, 2.92, 3.14, 3.36, 3.60, 3.84, 4.10, 13.00, 13.46, 13.92, 14.40];\n", + "Ps=[0.00, 0.24, 0.48, 0.72, 0.96, 5.76, 6.00, 6.24, 6.48, 6.72, 6.96, 7.20, 7.44, 7.68, 13.68, 13.92, 14.16, 14.40];\n", + "Pic=[0.00, 0.24, 0.46, 0.68, 0.90, 3.46, 3.50, 3.54, 3.56, 3.58, 3.60, 3.60, 3.60, 3.58, 0.68, 0.46, 0.24, 0.00];\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,18):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + " \n", + "plot(Vload,Pic);\n", + "xlabel('load voltage (V)')\n", + "ylabel('IC Power(W)')\n", + "title('load voltage vs IC Power')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.5,Page 173" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IC power is 2.57 W\n", + "total power is 3.82 W\n", + "dc supply current is 159.155 mA\n", + "power delivered is 1.25 watt\n" + ] + } + ], + "source": [ + "#finding different power and current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=12.0;\n", + "Vp=5.0;\n", + "R=10.0;\n", + "\n", + "#calculation\n", + "Ip=Vp/R;\n", + "Il=Ip/2**.5;\n", + "Pl=(Vp*Ip)/2;\n", + "Id=Ip/pi;\n", + "Pt=2*V*Ip/pi;\n", + "Pic=Pt-Pl;\n", + "\n", + "#result\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"total power is\",round(Pt,2), \"W\"\n", + "print \"dc supply current is\",round(Id*1000,3), \"mA\"\n", + "print \"power delivered is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.6,Page 179" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thermal resistance is 24.61 C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ts=40.0;\n", + "P=2.92;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Tj=125.0;\n", + "\n", + "#calculation\n", + "Qs=(Tj-Ts)/P-Qj-Qc;\n", + "\n", + "#result\n", + "print \"thermal resistance is\",round(Qs,2),\"C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.7,Page 180" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "9.4 0.94 4.42 14.36 9.94\n", + "9.6 0.96 4.61 14.67 10.06\n", + "10.0 power delivered by IC in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=24.0;\n", + "R=10.0;\n", + "Qs=4.0;\n", + "Tj=125.0;\n", + "Ta=40.0;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Vload=[9.4, 9.6];\n", + "Iload=[.94, .96];\n", + "Pload=[4.42, 4.61];\n", + "Ps=[14.36, 14.67];\n", + "Pic=[9.94, 10.06];\n", + "\n", + "#calculation\n", + "P=(Tj-Ta)/(Qj+Qc+Qs);\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,2):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + "print round(P,2),\"power delivered by IC in watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.8,Page 182" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "gain is 23.0\n", + "limit current is 4.01 A\n", + "output voltage is 46.0 V\n", + "maximum output voltage is 32.0 V\n" + ] + } + ], + "source": [ + "#finding current and voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=22.0;\n", + "Ri=1.0;\n", + "Rs=15.0;\n", + "I=4.75;\n", + "Rc=4.0;\n", + "Vp=2.0;\n", + "Rl=8.0;\n", + "Im=4.0;\n", + "\n", + "#calculation\n", + "Av=1+(Rf/Ri);\n", + "Il=(Rs*I)/(Rc+13.75);\n", + "Vo=Vp*Av;\n", + "V=Im*Rl;\n", + "\n", + "#result\n", + "print \"gain is\",round(Av,2)\n", + "print \"limit current is\",round(Il,2), \"A\"\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"maximum output voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.9,Page 185" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "loudness ofsound is 108.06 dB\n" + ] + } + ], + "source": [ + "#finding loudness\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=8.0;\n", + "d=1.0;\n", + "I=90.0;\n", + "\n", + "#calculation\n", + "Is=20*log(d/D,10);\n", + "Ir=I-Is;\n", + "\n", + "#result\n", + "print \"loudness ofsound is\",round(Ir,2), \"dB\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.10,Page 186" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "19.95 power provided in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=1.0;\n", + "I1=108.0;\n", + "I2=95.0;\n", + "P=1.0;\n", + "\n", + "#calculation\n", + "I=I1-I2;\n", + "Pr=P*10**(I/10);\n", + "\n", + "#result\n", + "print \"power provided is\",round(Pr,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.11,Page 188" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 12.65 V\n", + "gain is 10.28\n" + ] + } + ], + "source": [ + "#finding output voltage and gain\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "P=20;\n", + "R=8;\n", + "Vi=1.23;\n", + "\n", + "#calculation\n", + "V=(P*R)**.5;\n", + "G=V/Vi;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(V,2), \"V\"\n", + "print \"gain is\",round(G,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.12,Page 191" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistor b/w pins 1&8 is 600.0 ohm\n", + "thus pick a 620 ohm resistor\n", + "capacitor b/w pins 1&8 is 22.46 microF\n", + "thus pick a 27 microF capacitor\n" + ] + } + ], + "source": [ + "#finding resistor and capacitor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "G=40.0;\n", + "f=80.0;\n", + "R1=15000.0;\n", + "R2=150.0;\n", + "\n", + "#calculation\n", + "R=2*(R1/G)-R2;\n", + "R11=620;\n", + "C=1/(2*pi*f*R11/7);\n", + "\n", + "#result\n", + "print \"resistor b/w pins 1&8 is\",round(R,2),\"ohm\"\n", + "print('thus pick a 620 ohm resistor')\n", + "print \"capacitor b/w pins 1&8 is\",round(C*1e6,2), \"microF\"\n", + "print('thus pick a 27 microF capacitor')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.13,Page 193" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 140.0 mW\n", + "thermal resistance is 628.93 degree C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "Tm=150.0#temperature\n", + "Ta=50.0#temperature\n", + "Qa=107.0;\n", + "Qc=37.0;\n", + "Ps=299.0;\n", + "\n", + "#calculation\n", + "Vd=V/2;\n", + "Vm=V-1;\n", + "Vp=Vm-Vd;\n", + "Vr=Vp/2**.5;\n", + "Pl=1000*Vr**2/R;\n", + "Pl=140;\n", + "Pic=Ps-Pl;\n", + "Q=(Tm-Ta)/Pic;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(Pl,2), \"mW\"\n", + "print \"thermal resistance is\",round(Q*1000,2),\"degree C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.14,Page 197" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power deliverd is 562.5 mwatt\n" + ] + } + ], + "source": [ + "#finding power delivered\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "\n", + "#calculation\n", + "Vl=V-1;\n", + "Vp=Vl-1;\n", + "Vr=Vp/2**.5;\n", + "P=Vr**2/R;\n", + "\n", + "#result\n", + "print \"power deliverd is\",round(P*1000,2), \"mwatt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.15,Page 201" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 85.56 watt\n", + "thermal resistance is 1.4 degreeC/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "Ts=35.0#temperature\n", + "Ta=150.0#temperature\n", + "Vm=42.0#voltage\n", + "\n", + "#calcuation\n", + "Vp=Vm-5;\n", + "Vr=Vp/2**.5;\n", + "Pm=Vr**2/R;\n", + "P=45;\n", + "Qs=(Ta-Ts)/P-1.2;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pm,2), \"watt\"\n", + "print \"thermal resistance is\",round(round(Qs*10)/10,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4_1.ipynb new file mode 100644 index 00000000..6bbf650f --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter4_1.ipynb @@ -0,0 +1,742 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Linear Power Amplifier Integrated Circuits" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.1,Page 162" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 500.0 mV\n" + ] + } + ], + "source": [ + "#finding voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=1;\n", + "Ri=10;\n", + "Vi=0;\n", + "Ip=500;\n", + "\n", + "#calculation\n", + "Vrf=Ip*Rf;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vrf,2), \"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.2,Page 165" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "frequency of OPA548 is 67.26 KHz\n", + "slew rate of OPA548 is 1.12 Mhz\n", + "the OPA548 can be used\n" + ] + } + ], + "source": [ + "#finding frequency\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=300.0;\n", + "P=35.0;\n", + "R=8.0;\n", + "S=10000.0;\n", + "fh=20.0;\n", + "\n", + "#calculation\n", + "Vl=(P*R)**.5;\n", + "Vp=Vl*2**.5;\n", + "Il=Vl/R;\n", + "f=S/(2*pi*Vp);\n", + "Ao=Vl/Vi;\n", + "G=Ao*fh;\n", + "\n", + "#result\n", + "print \"frequency of OPA548 is\",round(f,2), \"KHz\"\n", + "print \"slew rate of OPA548 is\",round(G,2), \"Mhz\"\n", + "print('the OPA548 can be used')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.3,Page 168" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 3.5 watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=10.0;\n", + "V=12.0;\n", + "Vl=5.0;\n", + "\n", + "#calculation\n", + "Pl=Vl**2/Rl;\n", + "I=Vl/Rl;\n", + "Ps=V*I;\n", + "Pic=Ps-Pl;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pic,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.4,Page 170" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "0.0 0.0 0.0 0.0 0.0\n", + "0.2 0.0 0.0 0.24 0.24\n", + "0.4 0.0 0.02 0.48 0.46\n", + "0.6 0.1 0.04 0.72 0.68\n", + "0.8 0.1 0.06 0.96 0.9\n", + "4.8 0.5 2.3 5.76 3.46\n", + "5.0 0.5 2.5 6.0 3.5\n", + "5.2 0.5 2.7 6.24 3.54\n", + "5.4 0.5 2.92 6.48 3.56\n", + "5.6 0.6 3.14 6.72 3.58\n", + "5.8 0.6 3.36 6.96 3.6\n", + "6.0 0.6 3.6 7.2 3.6\n", + "6.2 0.6 3.84 7.44 3.6\n", + "6.4 0.6 4.1 7.68 3.58\n", + "11.4 1.1 13.0 13.68 0.68\n", + "11.6 1.2 13.46 13.92 0.46\n", + "11.8 1.2 13.92 14.16 0.24\n", + "12.0 1.2 14.4 14.4 0.0\n" + ] + }, + { + "data": { + "text/plain": [ + "<matplotlib.text.Text at 0x7fd132102a10>" + ] + }, + "execution_count": 1, + "metadata": {}, + "output_type": "execute_result" + }, + { + "data": { + "image/png": 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E8BnQNcHzXcifoFaUd4E9RRzjl5FOJbZ3rw2De+IJaNrU62hEyq4uXWzNsC5d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\nREREREREREREREREREREJBn/DwWIrS4GjkBZAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fd1321993d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Vload vs Pic graph\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vload=[0.0, 0.2, 0.4, 0.6, 0.8, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 6.2, 6.4, 11.4, 11.6, 11.8, 12.0];\n", + "Iload=[0.0, 0.0, 0.0, 0.1, 0.1, 0.5, 0.5, 0.5, 0.5, 0.6, 0.6, 0.6, 0.6, 0.6, 1.1, 1.2, 1.2, 1.2];\n", + "Pload=[0.00, 0.00, 0.02, 0.04, 0.06, 2.30, 2.50, 2.70, 2.92, 3.14, 3.36, 3.60, 3.84, 4.10, 13.00, 13.46, 13.92, 14.40];\n", + "Ps=[0.00, 0.24, 0.48, 0.72, 0.96, 5.76, 6.00, 6.24, 6.48, 6.72, 6.96, 7.20, 7.44, 7.68, 13.68, 13.92, 14.16, 14.40];\n", + "Pic=[0.00, 0.24, 0.46, 0.68, 0.90, 3.46, 3.50, 3.54, 3.56, 3.58, 3.60, 3.60, 3.60, 3.58, 0.68, 0.46, 0.24, 0.00];\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,18):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + " \n", + "plot(Vload,Pic);\n", + "xlabel('load voltage (V)')\n", + "ylabel('IC Power(W)')\n", + "title('load voltage vs IC Power')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.5,Page 173" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IC power is 2.57 W\n", + "total power is 3.82 W\n", + "dc supply current is 159.155 mA\n", + "power delivered is 1.25 watt\n" + ] + } + ], + "source": [ + "#finding different power and current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=12.0;\n", + "Vp=5.0;\n", + "R=10.0;\n", + "\n", + "#calculation\n", + "Ip=Vp/R;\n", + "Il=Ip/2**.5;\n", + "Pl=(Vp*Ip)/2;\n", + "Id=Ip/pi;\n", + "Pt=2*V*Ip/pi;\n", + "Pic=Pt-Pl;\n", + "\n", + "#result\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"total power is\",round(Pt,2), \"W\"\n", + "print \"dc supply current is\",round(Id*1000,3), \"mA\"\n", + "print \"power delivered is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.6,Page 179" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thermal resistance is 24.61 C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Ts=40.0;\n", + "P=2.92;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Tj=125.0;\n", + "\n", + "#calculation\n", + "Qs=(Tj-Ts)/P-Qj-Qc;\n", + "\n", + "#result\n", + "print \"thermal resistance is\",round(Qs,2),\"C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.7,Page 180" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vload Iload Pload Ps Pic\n", + "9.4 0.94 4.42 14.36 9.94\n", + "9.6 0.96 4.61 14.67 10.06\n", + "10.0 power delivered by IC in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=24.0;\n", + "R=10.0;\n", + "Qs=4.0;\n", + "Tj=125.0;\n", + "Ta=40.0;\n", + "Qj=2.5;\n", + "Qc=2.0;\n", + "Vload=[9.4, 9.6];\n", + "Iload=[.94, .96];\n", + "Pload=[4.42, 4.61];\n", + "Ps=[14.36, 14.67];\n", + "Pic=[9.94, 10.06];\n", + "\n", + "#calculation\n", + "P=(Tj-Ta)/(Qj+Qc+Qs);\n", + "\n", + "#result\n", + "print('Vload Iload Pload Ps Pic');\n", + "for i in range(0,2):\n", + " print Vload[i],\" \",Iload[i],\" \",Pload[i],\" \",Ps[i],\" \", Pic[i]\n", + "print round(P,2),\"power delivered by IC in watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.8,Page 182" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "gain is 23.0\n", + "limit current is 4.01 A\n", + "output voltage is 46.0 V\n", + "maximum output voltage is 32.0 V\n" + ] + } + ], + "source": [ + "#finding current and voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rf=22.0;\n", + "Ri=1.0;\n", + "Rs=15.0;\n", + "I=4.75;\n", + "Rc=4.0;\n", + "Vp=2.0;\n", + "Rl=8.0;\n", + "Im=4.0;\n", + "\n", + "#calculation\n", + "Av=1+(Rf/Ri);\n", + "Il=(Rs*I)/(Rc+13.75);\n", + "Vo=Vp*Av;\n", + "V=Im*Rl;\n", + "\n", + "#result\n", + "print \"gain is\",round(Av,2)\n", + "print \"limit current is\",round(Il,2), \"A\"\n", + "print \"output voltage is\",round(Vo,2), \"V\"\n", + "print \"maximum output voltage is\",round(V,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.9,Page 185" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "loudness ofsound is 108.06 dB\n" + ] + } + ], + "source": [ + "#finding loudness\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=8.0;\n", + "d=1.0;\n", + "I=90.0;\n", + "\n", + "#calculation\n", + "Is=20*log(d/D,10);\n", + "Ir=I-Is;\n", + "\n", + "#result\n", + "print \"loudness ofsound is\",round(Ir,2), \"dB\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.10,Page 186" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "19.95 power provided in watt\n" + ] + } + ], + "source": [ + "#finding power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "D=1.0;\n", + "I1=108.0;\n", + "I2=95.0;\n", + "P=1.0;\n", + "\n", + "#calculation\n", + "I=I1-I2;\n", + "Pr=P*10**(I/10);\n", + "\n", + "#result\n", + "print \"power provided is\",round(Pr,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.11,Page 188" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 12.65 V\n", + "gain is 10.28\n" + ] + } + ], + "source": [ + "#finding output voltage and gain\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "P=20;\n", + "R=8;\n", + "Vi=1.23;\n", + "\n", + "#calculation\n", + "V=(P*R)**.5;\n", + "G=V/Vi;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(V,2), \"V\"\n", + "print \"gain is\",round(G,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.12,Page 191" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistor b/w pins 1&8 is 600.0 ohm\n", + "thus pick a 620 ohm resistor\n", + "capacitor b/w pins 1&8 is 22.46 microF\n", + "thus pick a 27 microF capacitor\n" + ] + } + ], + "source": [ + "#finding resistor and capacitor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "G=40.0;\n", + "f=80.0;\n", + "R1=15000.0;\n", + "R2=150.0;\n", + "\n", + "#calculation\n", + "R=2*(R1/G)-R2;\n", + "R11=620;\n", + "C=1/(2*pi*f*R11/7);\n", + "\n", + "#result\n", + "print \"resistor b/w pins 1&8 is\",round(R,2),\"ohm\"\n", + "print('thus pick a 620 ohm resistor')\n", + "print \"capacitor b/w pins 1&8 is\",round(C*1e6,2), \"microF\"\n", + "print('thus pick a 27 microF capacitor')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.13,Page 193" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power dissipated is 140.0 mW\n", + "thermal resistance is 628.93 degree C/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "Tm=150.0#temperature\n", + "Ta=50.0#temperature\n", + "Qa=107.0;\n", + "Qc=37.0;\n", + "Ps=299.0;\n", + "\n", + "#calculation\n", + "Vd=V/2;\n", + "Vm=V-1;\n", + "Vp=Vm-Vd;\n", + "Vr=Vp/2**.5;\n", + "Pl=1000*Vr**2/R;\n", + "Pl=140;\n", + "Pic=Ps-Pl;\n", + "Q=(Tm-Ta)/Pic;\n", + "\n", + "#result\n", + "print \"power dissipated is\",round(Pl,2), \"mW\"\n", + "print \"thermal resistance is\",round(Q*1000,2),\"degree C/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.14,Page 197" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power deliverd is 562.5 mwatt\n" + ] + } + ], + "source": [ + "#finding power delivered\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "V=5.0#voltage\n", + "\n", + "#calculation\n", + "Vl=V-1;\n", + "Vp=Vl-1;\n", + "Vr=Vp/2**.5;\n", + "P=Vr**2/R;\n", + "\n", + "#result\n", + "print \"power deliverd is\",round(P*1000,2), \"mwatt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.15,Page 201" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power delivered is 85.56 watt\n", + "thermal resistance is 1.4 degreeC/W\n" + ] + } + ], + "source": [ + "#finding thermal resistance and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan,log\n", + "R=8.0#resistance\n", + "Ts=35.0#temperature\n", + "Ta=150.0#temperature\n", + "Vm=42.0#voltage\n", + "\n", + "#calcuation\n", + "Vp=Vm-5;\n", + "Vr=Vp/2**.5;\n", + "Pm=Vr**2/R;\n", + "P=45;\n", + "Qs=(Ta-Ts)/P-1.2;\n", + "\n", + "#result\n", + "print \"power delivered is\",round(Pm,2), \"watt\"\n", + "print \"thermal resistance is\",round(round(Qs*10)/10,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5.ipynb new file mode 100644 index 00000000..38414570 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5.ipynb @@ -0,0 +1,647 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Discrete Linear Power Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.1,Page 215" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Id=0 from 0 to 2 so not shown in the graph\n" + ] + }, + { + "data": { + "image/png": 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8oVgLIhWJQSTrtWwJ77xjE+KOO852j7vuOsjLcx2ZZBrVGETS0IoVMGCA7Rg3diy0auU6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\nIiIiIiIiIiIiIiIiIiIiknn+HwC9g+9EmLPsAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f0d1bd659d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Id vs Vgs\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vth=3.6;\n", + "Vgs=4;#voltage\n", + "#volt change beyond 3.6 causes a major increase in Id as it is cut off voltage\n", + "\n", + "#result\n", + "print('Id=0 from 0 to 2 so not shown in the graph')\n", + "x=np.linspace(2,3.6,300);\n", + "y=(-2.5*(x-3.6))**.5;\n", + "plot(x,y)\n", + "xlabel('Vgs(volts)');\n", + "ylabel('Id(amps)');\n", + "title('Id vs Vgs');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.2,Page 217" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drain current is 3.8 A\n", + "Vth=4V is assumed\n" + ] + } + ], + "source": [ + "#finding drain current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=4.5;#voltage\n", + "T=25;#degreeC\n", + "Id=3.8;\n", + "\n", + "#result\n", + "print \"drain current is\",round(Id,2), \"A\"\n", + "print('Vth=4V is assumed')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.3,Page 219" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "MOSFET is IRF530N\n", + "lower limit of Vth is -4.0 V\n", + "upper limit of Vth is -2.0 V\n", + "current is 2.3 A\n" + ] + } + ], + "source": [ + "#finding drain current of IRF530N\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vgs=-5;#voltage\n", + "Vthl=-4;\n", + "Vthu=-2;\n", + "Id=2.3;#current\n", + "\n", + "#result\n", + "print('MOSFET is IRF530N')\n", + "print \"lower limit of Vth is\",round(Vthl,2), \"V\"\n", + "print \"upper limit of Vth is\",round(Vthu,2), \"V\"\n", + "print \"current is\",round(Id,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example5.5,Page 225" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 1.51 kohm\n", + "load voltage is 36.68 V\n", + "Pq is 40.02 watt\n", + "Ps is 82.0 watt\n", + "Pl is 41.97 watt\n" + ] + } + ], + "source": [ + "#finding Pq,Pl,Ps,resistance,load voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "Vg=3.6;#voltage\n", + "Vd=56.0;\n", + "G=.98;#gain\n", + "Vi=40.0;\n", + "Rl=8.0;#load resistance\n", + "Vp=36.5;\n", + "\n", + "#calculation\n", + "Vr=Vd-Vg;\n", + "Ir=Vr/R1;\n", + "R2=Vg/Ir;\n", + "Va=(R1/(R1+R2))*Vi;\n", + "Vl=G*Va;\n", + "Il=Vp/Rl;\n", + "Pl=Vp*4.6/4;\n", + "Ps=Vd*4.6/pi;\n", + "Pq=Ps-Pl;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R2,2), \"kohm\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"Pq is\",round(Pq,2), \"watt\"\n", + "print \"Ps is\",round(Ps,2), \"watt\"\n", + "print \"Pl is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.6,Page 232" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 2.45 mA\n", + "resistance2 is 814.815 ohm\n", + "pick R2=R3=820ohm R1=R4=22 kohm\n" + ] + } + ], + "source": [ + "#finding resistance and current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "V1=56.0;#voltage\n", + "V2=2.0;#voltage\n", + "\n", + "#calculation\n", + "I=(V1-V2)/R1;\n", + "R2=V2/I;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"resistance2 is\",round(R2*1000,3), \"ohm\"\n", + "print('pick R2=R3=820ohm R1=R4=22 kohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.7,Page 234" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load voltage is 10.01 V\n" + ] + } + ], + "source": [ + "#finding load voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=350.0;#voltage\n", + "f=100.0;#frequency\n", + "Rf=10000.0;#resistance\n", + "Ri=520.0;\n", + "\n", + "#calculation\n", + "Vp=(1+(Rf/Ri))*Vi*2**.5;\n", + "\n", + "#result\n", + "print \"load voltage is\",round(Vp/1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.8,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load rms voltage is 20.0 V\n", + "resistance is 54.04 kohm\n", + "current is 1.18 mA\n", + "load current is 4.41 A\n", + "supply power is 39.3 watt\n", + "load power is 38.9 W\n", + "power is 19.552 W\n", + "thermal resistance is 3.01 degreC/W\n" + ] + } + ], + "source": [ + "#designing amplifier\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=50.0;#power\n", + "Z=4.7#impedence\n", + "R=4.0;#resistance\n", + "Ta=40.0;#degreeC\n", + "Tj=140.0;#degreeC\n", + "Vd=28.0;\n", + "R2=22.0;\n", + "\n", + "#calculation\n", + "Vr=(P*R)**.5;\n", + "Vp=Vr*2**.5;\n", + "Av=-Vr/1.23;\n", + "Rf=-Av*Z;\n", + "I=(Vd-2)/R2;\n", + "Vm=.63*Vd;\n", + "Ip=Vm/R;\n", + "Ps=Vd*Ip/pi;\n", + "Pl=Ip**2/2*R;\n", + "Pq=round(Ps)-Pl/2;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"load rms voltage is\",round(Vp,2), \"V\"\n", + "print \"resistance is\",round(Rf,2), \"kohm\"\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"load current is\",round(Ip,2), \"A\"\n", + "print \"supply power is\",round(Ps,2), \"watt\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"power is\",round(Pq,3), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.9,Page 243" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 129.37 mV\n", + "load current is 32.34 mA\n" + ] + } + ], + "source": [ + "#finding load current,output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=7.5e-3;#voltage\n", + "Ib=800e-9;#current\n", + "R=53.9e3;#resistance\n", + "\n", + "#calculation\n", + "Vo=11.5*Vi+Ib*R;\n", + "Id=Vo/4;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo*1000,2), \"mV\"\n", + "print \"load current is\",round(Id*1000,2), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.10,Page 249" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.31 ohm\n", + "thus pick a .33ohm rsistance\n", + "voltage is 0.55 V\n", + "power is 0.23 W\n", + "thermal resistance is 8.1 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance,voltage,power \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "G=6.4;#A/V\n", + "I=5.0;#current\n", + "Pq=9.8;#W\n", + "Tj=140.0;\n", + "Ta=40.0;\n", + "R1=.33;\n", + "\n", + "#calculation\n", + "R=2/G;\n", + "Im=I/3;\n", + "Vr=Im*R1;\n", + "P=Vr*Im/4;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R,2), \"ohm\"\n", + "print('thus pick a .33ohm rsistance')\n", + "print \"voltage is\",round(Vr,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.11,Page 251" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "limit level current is 8.49 A\n" + ] + } + ], + "source": [ + "#finding limit current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=200;#power\n", + "R=8;#ohm\n", + "\n", + "#calculation\n", + "Il=(P/R)**.5*2**.5;\n", + "Ilm=1.2*Il;\n", + "\n", + "#result\n", + "print \"limit level current is\",round(Ilm,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.12,Page 253" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.1 ohm\n", + "power is 1.8 W\n", + "MOSFET power is 84.0 W\n", + "temperature is 468.4 degreeC\n" + ] + } + ], + "source": [ + "#finding resistance,power,temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=6;#current\n", + "V=.6;#voltage\n", + "D=.5;#duty cycle\n", + "T=40;#temperature\n", + "\n", + "#calculation\n", + "Rs=V/I;\n", + "Pr=D*V*I;\n", + "Vp=28;\n", + "Pm=D*Vp*I;\n", + "Tj=T+Pm*5.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rs,2), \"ohm\"\n", + "print \"power is\",round(Pr,2), \"W\"\n", + "print \"MOSFET power is\",round(Pm,2), \"W\"\n", + "print \"temperature is\",round(Tj,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.13,Page 255" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum safe temperature is 89.05 degreeC\n" + ] + } + ], + "source": [ + "#finding maximum safe temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=130;#temperature\n", + "P=19.5;#power\n", + "\n", + "#calculation\n", + "Ts=T-P*2.1;\n", + "\n", + "#result\n", + "print \"maximum safe temperature is\",round(Ts,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.14,Page 257" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reactance is 8.86 ohm\n", + "voltage across resistor is 10.03 V\n", + "-48 is the angle of the voltage in degrees\n", + "power dissipated by load is 12.5 watts\n", + "current across the resistance is 1.77 A\n", + "power supply is 15.8 W\n", + "power dissipated by transistor is 9.55 watts\n" + ] + } + ], + "source": [ + "#finding 3 powers and current across resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "f=300.0;#frequency\n", + "L=4.7;#inductance\n", + "Vdc=28;#V\n", + "Pr=12.5;\n", + "\n", + "#calculation\n", + "Xl=2*pi*f*L;\n", + "Zload=sqrt(8**2+8.9**2);#magnitude of Zload\n", + "Vload=15.0;#msgnitude of Vload\n", + "Vr=Vload*8/Zload;\n", + "I=Vr/8*sqrt(2);\n", + "Psupply=Vdc*I/pi;\n", + "Pq=Psupply-Pr/2;\n", + "\n", + "#result\n", + "print \"reactance is\",round(Xl/1000,2), \"ohm\"\n", + "print \"voltage across resistor is\",round(Vr,2), \"V\"\n", + "print \"-48 is the angle of the voltage in degrees\";\n", + "print \"power dissipated by load is\",round(Pr,2), \"watts\"\n", + "print \"current across the resistance is\",round(I,2), \"A\"\n", + "print \"power supply is\",round(Psupply,2), \"W\"\n", + "print \"power dissipated by transistor is\",round(Pq,2), \"watts\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5_1.ipynb new file mode 100644 index 00000000..38414570 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter5_1.ipynb @@ -0,0 +1,647 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Discrete Linear Power Amplifier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.1,Page 215" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Id=0 from 0 to 2 so not shown in the graph\n" + ] + }, + { + "data": { + "image/png": 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8oVgLIhWJQSTrtWwJ77xjE+KOO852j7vuOsjLcx2ZZBrVGETS0IoVMGCA7Rg3diy0auU6\nIgmjIGsM12I1hRzgCaxb6dRE30hEkqdpU3jjDeta6tIF/vxnrbskyeMnMVwKbAJOAeoCfYB7ggxK\nRMqXk2NzHhYsgHnzbFmNf//bdVSSCfwkhuJmyBnAU8AnwYUjIolq2NDWXLr5ZujaFW69Va0HqRw/\nieHfwKtAV2A2UBvQgDmREMnJgd694eOPbe2lY4+1+yIV4acokQe0BlYCPwL1gIbAogDjiqXis4hP\nkQiMH29Lagwdav9q5FJ2CnLZ7RygO3A8EAHeAaZ591NFiUEkQWvWQP/+sH27JQrNe8g+QY5KegQY\nBHyM1RcGAQ8n+kYiklqNG9uSGhdeaPMeHn5Ys6bFHz+ZZDnQkpK6Qi6wFGgeVFBxqMUgUgnLl0Pf\nvlC7Nowbp/2ms0WQLYYvgAOjHh/oHRORNNG8Obz3HnTqBG3bwtSpriOSMPOTSd4GjgHmYXWFY4EP\ngZ+8x2cHFl0JtRhEkuSDD6BXLzjhBNtOtEYN1xFJUIIsPheU8VwEeCvRN60AJQaRJPrpJ7jqKpg7\nFyZPhqOOch2RBCHIxFCsNrsvuvd9om9WCUoMIgF45hm4+uqSoa252rorowSZGAYBfwJ+oaQAHQEO\n9vGzY7EZ0/8BDi/lnH9g+z1sA/phazHFUmIQCciaNTY5rlo1mDAB9t/fdUSSLEEWn28EWgGNgYO8\nm5+kADAO29inNF2BQ4BmwEBglM/XFZEkadwYCgutMH3UUdpKVPwlhpXAzxV8/XeAH8p4/mxgvHf/\nA2y3uH0q+F4iUkFVqsAdd9hopSuugJtugh07XEclrvhJDLcA7wOPAQ95t38k6f0PANZGPV6HLbch\nIg507GirtS5ZAp07w9q15f+MZJ6ydnAr9jjwOrAYqzHkkNzlMGL7v+K+9vDhw/97v6CggIKCgiSG\nICLF6tWz1VofeACOOQaeeALOOMN1VOJHYWEhhYWFlX4dP0WJBdgezxXVBJhO/OLzo0Ah8Iz3eDnQ\nGfg25jwVn0UceO89uOgiu919t+03LekjyOLzLGxk0n7YRj3Ft2R4CbjEu98eW701NimIiCMdO8L8\n+bB4se0U99VXriOSVPCTSVYTv3vnIB8/OxlrAdTHPvDvBIq/czzm/TsSG7m0FegPzI/zOmoxiDhU\nVAT33gsPPWQT4jp3dh2R+JGKCW4uKTGIhMCrr8Ill8Att8A119gGQRJeQSeGVtgKq9Wijk1I9M0q\nQYlBJCRWr4bu3aFFC3j8ca21FGZB1hiGY0NURwJdgL+SmoXzRCSEmjSxonReHnToACtWuI5Iks1P\nYugBnASsx2oArbGJaCKSpapXt13hBg605DBrluuIJJn8JIafgV3ATmAvbN2jRkEGJSLhl5MDQ4bA\nCy/A5ZfDPffYftOS/vwkhg+BOsBo4CNsXsO/ggxKRNJHx44wbx4895ztErd9u+uIpLISLUochC2/\nvSiAWMqi4rNIyG3bZonhq6/gxRdhH6165lwQxeemcY6tYvekEO8cEclCe+4JU6bAySdDu3bw8ceu\nI5KKKiuTTAFqYLOTP8KKzznYDOijsZFJm4ELA44R1GIQSSvFGwCNGQNnawyjM0HNYzgE++DviO3H\nALAGeBeb1bwy0TesICUGkTQzb57Nd7jqKlvGW5PhUk8zn0UkdNatg27dbAOgUaNs3wdJnSASw3mU\nvbz2C4m+WSUoMYikqS1b4PzzbT/pKVOgZk3XEWWPIBLDk1hiaAB0AOZ4x7tgw1XPTPTNKkGJQSSN\n7dgBgwfDwoUwY4ZGLKVKEKOS+mEznati6ySd590O846JiPiSnw+jR1sh+rjj4NNPXUckZfHT49cI\n+Cbq8bfAgcGEIyKZKifH9pVu1MiW7X7hBVtOQ8LHT2J4HZgNTMKaJD2B14IMSkQyV//+sP/+cM45\ntm3oWWe5jkhi+el7ygHOBTphNYe3gReDDCoO1RhEMsyHH1pSeOAB6N3bdTSZScNVRSTtLFkCp50G\nw4bBFVe4jibzVDQxlNWVtIXSh6tGsDWTREQq7LDD4K23bBmNTZtsZzhNhHMvXf4XqMUgksHWr7fk\n0LWr7S2t5JAc6koSkbS2caMlhtatbZZ0Xp7riNJfkFt7iogErl49eP11+PxzuPRS2LXLdUTZS4lB\nREKjVi2bGb12rQ1rVXJwQ4lBREJlzz3h5Zdtwx8lBzeUGEQkdPbcE6ZPh6+/tl3hlBxSS4lBREKp\nODl8+y1ccomSQyopMYhIaFWvDi+9ZMlh4EAoKnIdUXZQYhCRUKteHaZNg2XL4PrrQSPXg6fEICKh\nV7MmzJxps6SHD3cdTebTRnsikhb23htmz4ZOnaB2bRg61HVEmUuJQUTSRoMG8Nprlhxq1bK6gySf\nEoOIpJVGjSw5dO5ss6XPO891RJlHayWJSFpasABOPdV2gjv+eNfRhJPWShKRrNKmDTz9tLUYli1z\nHU1mUWIQkbR1yim2THfXrrZ0tySHagwiktb69YN16+CMM+Dtt21oq1SOagwikvYiEbj8cvj+e5g6\nFXLVFwKoxiAiWSwnxzb32bgRbr/ddTTpT4lBRDJC1arWWpg8GSZOdB1NelNXkohklE8+gRNOsJVZ\n27VzHY1b6koSEQFatYInnoDu3W0/B0mcWgwikpHuugtefRXmzIH8fNfRuFHRFoMSg4hkpKIiOPNM\naNkS7r/fdTRuqCtJRCRKbi489ZQVpF94wXU06UUtBhHJaB99ZDOj330XDj3UdTSppRaDiEgcRx8N\nf/4z9OgBP//sOpr0oBaDiGS8SAR69bJlukeOdB1N6qj4LCJShh9/hCOPtMRw5pmuo0mNsHYlnQYs\nBz4Hbo7zfAGwCVjg3W4LOB4RyVJ7723LdA8YAN984zqacAuyxZAHfAqcBHwFfAhcBESvnF4AXA+c\nXc5rqcUgIklxxx0wbx7MmmVrLGWyMLYYjgW+AFYDO4BngG5xzsvw/zUiEia3326rsI4e7TqS8Aoy\nMRwArI16vM47Fi0CdAAWATOBlgHGIyJCfj6MGwd//CN8+aXraMIpyI16/PT9zAcaAduA04FpQNyR\nxsOHD//v/YKCAgoKCiodoIhkp8MOg+uus3rDK69kTpdSYWEhhYWFlX6dIH8d7YHhWAEaYBhQBNxb\nxs+sAtoC38ccV41BRJJq505o3x4GD4bLLnMdTTDCWGP4CGgGNAGqAj2Bl2LO2YeSoI/17scmBRGR\npKtSxbqUbrlF+0XHCjIx7ASuBGYDS4Ep2IikQd4NoAewGFgIjAAuDDAeEZHdHH64bQl6ww2uIwmX\ndOlZU1eSiARi61arOYwdaxv8ZJIwdiWJiIRejRowYgQMGQK//uo6mnBQYhCRrNetGxx8MPztb64j\nCQd1JYmIACtXwjHHwOLFsP/+rqNJDi2iJyJSSTfdZIvtPf6460iSQ4lBRKSSfvjBNvN56y3bEjTd\nqfgsIlJJderYvIZhw1xH4pZaDCIiUbZvh+bNbYnu4493HU3lqMUgIpIE1arBXXfBjTfazm/ZSIlB\nRCRG796weTPMnu06EjeUGEREYuTm2rLcd92Vna0GJQYRkTguuAC++w6SsIp12lFiEBGJIy8Pbr3V\nWg3ZRqOSRERKsWMHNG0KL74Ibdu6jiZxGpUkIpJk+flw5ZXw4IOuI0kttRhERMrw/fdwyCGwdCns\nu6/raBKjFoOISADq1oWePWHUKNeRpI5aDCIi5Vi2DLp0gS+/hKpVXUfjn1oMIiIBadHCFtebOdN1\nJKmhxCAi4kP//vDkk66jSA11JYmI+LB5MzRqBJ99Bg0auI7GH3UliYgEqFYt2wJ04kTXkQRPiUFE\nxKd+/WDcONdRBE+JQUTEp86dbV7DsmWuIwmWEoOIiE+5udC9O0yd6jqSYCkxiIgkoEcPeP5511EE\nS6OSREQSsGuXLY0xf76NUgozjUoSEUmBvDw49VSYNct1JMFRYhARSdDpp2f2LGh1JYmIJGjDBtun\nYeNGqFLFdTSlU1eSiEiK1K8PjRvDggWuIwmGEoOISAV06gRvv+06imAoMYiIVEAmJwbVGEREKmD9\nemjVyuoNOSH9JFWNQUQkhfbbD/bYA9audR1J8ikxiIhU0JFHwsKFrqNIPiUGEZEKat1aiUFERKIc\neSQsWuQ6iuRTYhARqaBMbTGEtJb+GxqVJCKhs3071K4Nv/wSzpFJGpUkIpJi1arZyKTNm11HklxK\nDCIilVCvns1lyCRKDCIilVC/vi2ml0mUGEREKqF+fbUYREQkihKDiIjsRjUGERHZjWoMiTsNWA58\nDtxcyjn/8J5fBLQJOB4RkaRSV1Ji8oCRWHJoCVwEtIg5pytwCNAMGAiMCjCepCosLHQdwm8oJv/C\nGJdi8idsMdWrB8uWFboOI6mCTAzHAl8Aq4EdwDNAt5hzzgbGe/c/APYG9gkwpqQJ28UJiikRYYxL\nMfkTtpjq14c1awpdh5FUQSaGA4DolcrXecfKO6dhgDGJiCRV/fqwbZvrKJIryMTgd3Gj2HU8tCiS\niKSNTEwMQS771B4YjtUYAIYBRcC9Uec8ChRi3UxgherOwLcxr/UF0DSgOEVEMtUKrI4bGlWwoJoA\nVYGFxC8+z/Tutwfmpio4ERFx43TgU+wb/zDv2CDvVmyk9/wi4KiURiciIiIiIumjEfAmsAT4BLi6\nlPNSOSHOT0y9vVg+Bt4DjghBTMWOAXYC3UMSUwGwwDunMAQx1Qdewbo5PwH6BRwTQDVsaPZCYCnw\nl1LOS+V17iemVF/nfn9PkLrr3G9MBaTuOvcTk4vrPGn2BY707tfEuqDKqkm0I/iahJ+YjgP28u6f\nFpKYwCYYzgFeBs4LQUx7Yx/SxcOR64cgpuGU/CHVBzZitbGg7en9WwW7Xo6PeT7V17mfmFJ9nfuJ\nCVJ7nfuJKdXXuZ+YhpPgdR6mtZK+wTIawBZgGbB/zDmpnhDnJ6b3gU1RMQU9D8NPTABXAc8D3wUc\nj9+YegFTsbkqAEEvIuAnpvVAbe9+bewPZmfAcQEUD26sin2wfR/zvIuJn+XFlOrr3E9MkNrr3E9M\nqb7O/cSU8HUepsQQrQnWfP4g5rjLCXGlxRTtMkq+6aVCE0r/PXWjZImRVM4NKS2mZkBdrHvnI6BP\nCGIaDRwGfI11k1yTonhysaT1Lfb7WBrzvIvrvLyYoqXqOvfze0r1dV5eTC6u8/JicnWdJ1VN7Bd6\nTpznpgMdox6/TmpGMpUVU7Eu2P+QOimIB8qO6TmsCwLgSVLTxC4vppHAv4DqQD3gM+yPyGVMtwEj\nvPtNgZVArRTEVGwvrOlfEHPc1XVeVkzFUn2dQ+kxubrOofSYXF3nZcWU8HUethZDPtYMexqYFuf5\nr7CiYrGG3jGXMYEV4kZjXQA/BByPn5jaYpMGV2F/LI94sbmMaS3wKvAz1pR9G2jtOKYO2IcL2Jyb\nVcDvAo4p2iZgBnB0zHEX13l5MUHqr/PyYnJxnZcXk4vrvLyYXF/nlZIDTAD+XsY5qZ4Q5yemA7F5\nGO0DjqWYn5iijSP40Rp+YmqOffPNw4pli7FVd13G9DfgTu/+PliXTd0AYwIr/u3t3a+OfXCcGHNO\nqq9zPzGl+jr3E1O0VFznfmJK9XXuJyYX13nSHI8tmbEQG+q1AJsg53JCnJ+YxmDfDIqfnxeCmKKl\n4g/Gb0w3YCM2FlP2MNtUxVQf67ZZ5MXUK+CYAA4H5ntxfQzc6B13eZ37iSnV17nf31OxVFznfmNK\n5XXuJyYX17mIiIiIiIiIiIiIiIiIiIiIiIiIiIhItpoDnBJz7FpslmyirqTiSxYXUjIf4VYf5z8L\nHFTB9xIRkTIMAMbGHHuf+Es4lyUHm9hV0WW536QkMWz2cf7J2J4MIiKSZHWxVSeLP9CbAGuwdcIe\nwZbjfhVbY6Z4EbZ7sJmri4D7vGPHA5O9+83ZfaXWJtisU7DlCOZ7j5/AlkIGSwxtvdfeiSWZp7Bl\nE2ZgM1cXAxd45+djM59FRCQA0ylZXO0W4K9AD+wDGWz9mO+xJRXqAcujfrZ21M8NjTq+AEsIADdj\n3UPVgC+BQ7zj4ylZ4ri0FsN5wONx3g/gLeJvwiQSqLCtrioShMnAhd79nt7jjlg/PpSsYw/wI7Ad\n+7Z/LrbnORmYAAABXElEQVRKJtgicuujXvNZ77XAvuVPwVasXEXJN/3xQKdyYvsY6za6B2uV/BT1\n3NeUJB+RlFFikGzwEtbF0wbrulngHc+Jc+4u4FhsV7Azsb1yiXP+FCwhNMM2iFkR57XivX6sz724\nFgN3A7fH/HyRj9cQSSolBskGW7AWwThgknfsPawbJwfrSirwjtfAljGeBVxPyVr6a7B9pIutxJLI\n7dieAGD7SjfBNkMB272rME48OyipeeyHtVAmAvez+0qq+3nvKyIiAeiGfZAf6j3OwbaELC4+v4a1\nKvbFCsuLsG6e4q0ZO1JSfC421HvNA6OOnUBJ8XkMVkSG3WsM92C7oD2FDaVdhLViPog6J5/4rRAR\nEQlQDe/felhdoEEZ5xYPV61axjnJdArwYIreS0REPG9iH/ZLgEt8nH8F0D/QiEo8iwrPIiIiIiIi\nIiIiIiIiIiIiIiIiIiIiknn+HwC9g+9EmLPsAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f0d1bd659d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Id vs Vgs\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vth=3.6;\n", + "Vgs=4;#voltage\n", + "#volt change beyond 3.6 causes a major increase in Id as it is cut off voltage\n", + "\n", + "#result\n", + "print('Id=0 from 0 to 2 so not shown in the graph')\n", + "x=np.linspace(2,3.6,300);\n", + "y=(-2.5*(x-3.6))**.5;\n", + "plot(x,y)\n", + "xlabel('Vgs(volts)');\n", + "ylabel('Id(amps)');\n", + "title('Id vs Vgs');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.2,Page 217" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drain current is 3.8 A\n", + "Vth=4V is assumed\n" + ] + } + ], + "source": [ + "#finding drain current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=4.5;#voltage\n", + "T=25;#degreeC\n", + "Id=3.8;\n", + "\n", + "#result\n", + "print \"drain current is\",round(Id,2), \"A\"\n", + "print('Vth=4V is assumed')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.3,Page 219" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "MOSFET is IRF530N\n", + "lower limit of Vth is -4.0 V\n", + "upper limit of Vth is -2.0 V\n", + "current is 2.3 A\n" + ] + } + ], + "source": [ + "#finding drain current of IRF530N\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vgs=-5;#voltage\n", + "Vthl=-4;\n", + "Vthu=-2;\n", + "Id=2.3;#current\n", + "\n", + "#result\n", + "print('MOSFET is IRF530N')\n", + "print \"lower limit of Vth is\",round(Vthl,2), \"V\"\n", + "print \"upper limit of Vth is\",round(Vthu,2), \"V\"\n", + "print \"current is\",round(Id,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example5.5,Page 225" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 1.51 kohm\n", + "load voltage is 36.68 V\n", + "Pq is 40.02 watt\n", + "Ps is 82.0 watt\n", + "Pl is 41.97 watt\n" + ] + } + ], + "source": [ + "#finding Pq,Pl,Ps,resistance,load voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "Vg=3.6;#voltage\n", + "Vd=56.0;\n", + "G=.98;#gain\n", + "Vi=40.0;\n", + "Rl=8.0;#load resistance\n", + "Vp=36.5;\n", + "\n", + "#calculation\n", + "Vr=Vd-Vg;\n", + "Ir=Vr/R1;\n", + "R2=Vg/Ir;\n", + "Va=(R1/(R1+R2))*Vi;\n", + "Vl=G*Va;\n", + "Il=Vp/Rl;\n", + "Pl=Vp*4.6/4;\n", + "Ps=Vd*4.6/pi;\n", + "Pq=Ps-Pl;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R2,2), \"kohm\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"Pq is\",round(Pq,2), \"watt\"\n", + "print \"Ps is\",round(Ps,2), \"watt\"\n", + "print \"Pl is\",round(Pl,2), \"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.6,Page 232" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 2.45 mA\n", + "resistance2 is 814.815 ohm\n", + "pick R2=R3=820ohm R1=R4=22 kohm\n" + ] + } + ], + "source": [ + "#finding resistance and current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=22.0;#resistance\n", + "V1=56.0;#voltage\n", + "V2=2.0;#voltage\n", + "\n", + "#calculation\n", + "I=(V1-V2)/R1;\n", + "R2=V2/I;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"resistance2 is\",round(R2*1000,3), \"ohm\"\n", + "print('pick R2=R3=820ohm R1=R4=22 kohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.7,Page 234" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load voltage is 10.01 V\n" + ] + } + ], + "source": [ + "#finding load voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=350.0;#voltage\n", + "f=100.0;#frequency\n", + "Rf=10000.0;#resistance\n", + "Ri=520.0;\n", + "\n", + "#calculation\n", + "Vp=(1+(Rf/Ri))*Vi*2**.5;\n", + "\n", + "#result\n", + "print \"load voltage is\",round(Vp/1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.8,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load rms voltage is 20.0 V\n", + "resistance is 54.04 kohm\n", + "current is 1.18 mA\n", + "load current is 4.41 A\n", + "supply power is 39.3 watt\n", + "load power is 38.9 W\n", + "power is 19.552 W\n", + "thermal resistance is 3.01 degreC/W\n" + ] + } + ], + "source": [ + "#designing amplifier\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=50.0;#power\n", + "Z=4.7#impedence\n", + "R=4.0;#resistance\n", + "Ta=40.0;#degreeC\n", + "Tj=140.0;#degreeC\n", + "Vd=28.0;\n", + "R2=22.0;\n", + "\n", + "#calculation\n", + "Vr=(P*R)**.5;\n", + "Vp=Vr*2**.5;\n", + "Av=-Vr/1.23;\n", + "Rf=-Av*Z;\n", + "I=(Vd-2)/R2;\n", + "Vm=.63*Vd;\n", + "Ip=Vm/R;\n", + "Ps=Vd*Ip/pi;\n", + "Pl=Ip**2/2*R;\n", + "Pq=round(Ps)-Pl/2;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"load rms voltage is\",round(Vp,2), \"V\"\n", + "print \"resistance is\",round(Rf,2), \"kohm\"\n", + "print \"current is\",round(I,2), \"mA\"\n", + "print \"load current is\",round(Ip,2), \"A\"\n", + "print \"supply power is\",round(Ps,2), \"watt\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"power is\",round(Pq,3), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.9,Page 243" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage is 129.37 mV\n", + "load current is 32.34 mA\n" + ] + } + ], + "source": [ + "#finding load current,output voltage \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vi=7.5e-3;#voltage\n", + "Ib=800e-9;#current\n", + "R=53.9e3;#resistance\n", + "\n", + "#calculation\n", + "Vo=11.5*Vi+Ib*R;\n", + "Id=Vo/4;\n", + "\n", + "#result\n", + "print \"output voltage is\",round(Vo*1000,2), \"mV\"\n", + "print \"load current is\",round(Id*1000,2), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.10,Page 249" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.31 ohm\n", + "thus pick a .33ohm rsistance\n", + "voltage is 0.55 V\n", + "power is 0.23 W\n", + "thermal resistance is 8.1 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance,voltage,power \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "G=6.4;#A/V\n", + "I=5.0;#current\n", + "Pq=9.8;#W\n", + "Tj=140.0;\n", + "Ta=40.0;\n", + "R1=.33;\n", + "\n", + "#calculation\n", + "R=2/G;\n", + "Im=I/3;\n", + "Vr=Im*R1;\n", + "P=Vr*Im/4;\n", + "Qs=(Tj-Ta)/Pq-2.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(R,2), \"ohm\"\n", + "print('thus pick a .33ohm rsistance')\n", + "print \"voltage is\",round(Vr,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Qs,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.11,Page 251" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "limit level current is 8.49 A\n" + ] + } + ], + "source": [ + "#finding limit current \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=200;#power\n", + "R=8;#ohm\n", + "\n", + "#calculation\n", + "Il=(P/R)**.5*2**.5;\n", + "Ilm=1.2*Il;\n", + "\n", + "#result\n", + "print \"limit level current is\",round(Ilm,2), \"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.12,Page 253" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 0.1 ohm\n", + "power is 1.8 W\n", + "MOSFET power is 84.0 W\n", + "temperature is 468.4 degreeC\n" + ] + } + ], + "source": [ + "#finding resistance,power,temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=6;#current\n", + "V=.6;#voltage\n", + "D=.5;#duty cycle\n", + "T=40;#temperature\n", + "\n", + "#calculation\n", + "Rs=V/I;\n", + "Pr=D*V*I;\n", + "Vp=28;\n", + "Pm=D*Vp*I;\n", + "Tj=T+Pm*5.1;\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rs,2), \"ohm\"\n", + "print \"power is\",round(Pr,2), \"W\"\n", + "print \"MOSFET power is\",round(Pm,2), \"W\"\n", + "print \"temperature is\",round(Tj,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.13,Page 255" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum safe temperature is 89.05 degreeC\n" + ] + } + ], + "source": [ + "#finding maximum safe temperature \n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=130;#temperature\n", + "P=19.5;#power\n", + "\n", + "#calculation\n", + "Ts=T-P*2.1;\n", + "\n", + "#result\n", + "print \"maximum safe temperature is\",round(Ts,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.14,Page 257" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "reactance is 8.86 ohm\n", + "voltage across resistor is 10.03 V\n", + "-48 is the angle of the voltage in degrees\n", + "power dissipated by load is 12.5 watts\n", + "current across the resistance is 1.77 A\n", + "power supply is 15.8 W\n", + "power dissipated by transistor is 9.55 watts\n" + ] + } + ], + "source": [ + "#finding 3 powers and current across resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=15.0;#voltage\n", + "f=300.0;#frequency\n", + "L=4.7;#inductance\n", + "Vdc=28;#V\n", + "Pr=12.5;\n", + "\n", + "#calculation\n", + "Xl=2*pi*f*L;\n", + "Zload=sqrt(8**2+8.9**2);#magnitude of Zload\n", + "Vload=15.0;#msgnitude of Vload\n", + "Vr=Vload*8/Zload;\n", + "I=Vr/8*sqrt(2);\n", + "Psupply=Vdc*I/pi;\n", + "Pq=Psupply-Pr/2;\n", + "\n", + "#result\n", + "print \"reactance is\",round(Xl/1000,2), \"ohm\"\n", + "print \"voltage across resistor is\",round(Vr,2), \"V\"\n", + "print \"-48 is the angle of the voltage in degrees\";\n", + "print \"power dissipated by load is\",round(Pr,2), \"watts\"\n", + "print \"current across the resistance is\",round(I,2), \"A\"\n", + "print \"power supply is\",round(Psupply,2), \"W\"\n", + "print \"power dissipated by transistor is\",round(Pq,2), \"watts\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6.ipynb new file mode 100644 index 00000000..d5b6622f --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6.ipynb @@ -0,0 +1,528 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Power Switches" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.1,Page 274" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current is 1.81 A\n", + "load power is 49.32 W\n", + "transistor power is 1.45 W\n", + "least value of base current is 90.67 mA\n", + "max value of base resistance is 4.85 ohm\n", + "thus pick Rb=33ohm\n" + ] + } + ], + "source": [ + "#finding Ic,Pload,Pq,Ibase,Rbase\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=28.0;#V\n", + "Vi=5.0;#V\n", + "Rl=15.0;#ohm\n", + "Vc=.8;#V\n", + "b=20.0;\n", + "\n", + "#calculation\n", + "Ic=(Vs-Vc)/Rl;\n", + "Pl=Ic**2*Rl;\n", + "Pq=Ic*Vc;\n", + "Ib=Ic/b*1000;\n", + "Rb=(Vi-.6)/Ib;\n", + "\n", + "#result\n", + "print \"collector current is\",round(Ic,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"transistor power is\",round(Pq,2), \"W\"\n", + "print \"least value of base current is\",round(Ib,2), \"mA\"\n", + "print \"max value of base resistance is\",round(Rb*100,2), \"ohm\"\n", + "print ('thus pick Rb=33ohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.4,Page 282" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load resistance is 554.0 ohm\n", + "thus pick Rl=560ohm\n", + "max value of Rb is 3.0 kohm\n", + "thus pick Rb=2.2kohm\n", + "load current is 49.46 mA\n", + "load power is 685.08 mW\n", + "power delivered is 7.42 mW\n" + ] + } + ], + "source": [ + "#finding Pload,Pq,Iload,resistances\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=28.0;#V\n", + "f=100.0;#frequency\n", + "I=50.0;#current\n", + "Rl1=560.0;\n", + "Vp=2.4;\n", + "Ib=500.0;#microAmp\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "Rl=(Vd-.3)/I;\n", + "Rb=(Vp-.9)/Ib;\n", + "Vl=Vd-.3;\n", + "Ip=Vl/Rl1;\n", + "Pl=D*Vl*Ip;\n", + "Pq=D*Ip*.3;\n", + "\n", + "#result\n", + "print \"load resistance is\",round(Rl*1000), \"ohm\"\n", + "print('thus pick Rl=560ohm')\n", + "print \"max value of Rb is\",round(Rb*1000,2),\"kohm\"\n", + "print('thus pick Rb=2.2kohm')\n", + "print \"load current is\",round(Ip*1000,2), \"mA\"\n", + "print \"load power is\",round(Pl*1000,2), \"mW\"\n", + "print \"power delivered is\",round(Pq*1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.5,Page 286" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time of rise is 788.48 ns\n", + "time of rise is 4.65 microsec\n" + ] + } + ], + "source": [ + "#finding time of rise\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "C=640.0;#capacitor\n", + "R1=560.0;#load resistance\n", + "R2=3.3;#kohm\n", + "\n", + "#calculation\n", + "t1=2.2*R1*C;\n", + "t2=2.2*R2*C;\n", + "\n", + "#result\n", + "print \"time of rise is\",round(t1/1000,2), \"ns\"\n", + "print \"time of rise is\",round(t2/1000,2), \"microsec\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.6,Page 287" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 682.5 ohm\n", + "pick up resistance=680 ohm\n", + "rise time is 957.44 ns\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f86a4de0510>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Rpick up,time of rise\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vol=0.7;\n", + "Iol=40.0/1000;#current\n", + "Rpullup1=680.0;\n", + "C=640.0;\n", + "Epullup=28.0;\n", + "#for plotting\n", + "x=[0, .1, 1.9, 4.1, 5, 5.1, 5.3, 5.6, 6.0, 9.3];\n", + "y=[27.8, .1, .1, .1, .1, 5, 13.5, 21.0, 27.0, 27.8];\n", + "\n", + "#calculation\n", + "Rpullup=(Epullup-Vol)/Iol;\n", + "trise=2.2*Rpullup1*C;\n", + "plot(x,y,'r');\n", + "xlabel ('time(mus)')\n", + "ylabel ('Vout')\n", + "title ('Vout vs time')\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rpullup,2), \"ohm\"\n", + "print('pick up resistance=680 ohm');\n", + "print \"rise time is\",round(trise/1000,2), \"ns\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.7,Page 289" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "worst case resistance is 0.286 ohm\n", + "load current is 4.45 A\n", + "load voltage is 26.73 V\n", + "load power is 47.62 W\n", + "drop voltage is 1.27 V\n", + "power is 2.27 W\n", + "temperature is 182.6 deg.C\n" + ] + } + ], + "source": [ + "#finding worst case resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=.11;#resistance\n", + "Vd=28.0;#voltage\n", + "R2=6.0;#ohm\n", + "D=.4;#duty cycle\n", + "Q=62.0;\n", + "\n", + "#calculation\n", + "Ro=2.6*R1;\n", + "Ip=Vd/(R2+Ro);\n", + "Vl=Ip*R2;\n", + "Pl=D*Vl*Ip;\n", + "Vq=Ip*Ro;\n", + "Pq=D*Vq*Ip;\n", + "T=40+round(Pq*10)/10*Q;\n", + "\n", + "#result\n", + "print \"worst case resistance is\",round(Ro,3), \"ohm\"\n", + "print \"load current is\",round(Ip,2),\"A\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"drop voltage is\",round(Vq,2), \"V\"\n", + "print \"power is\",round(Pq*10,2)/10, \"W\"\n", + "print \"temperature is\",round(T,2), \"deg.C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.8,Page 292" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 150.0 V\n" + ] + } + ], + "source": [ + "#finding voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "L=10.0;#inductor\n", + "I=4.5;#current\n", + "t=300.0#time\n", + "\n", + "#calculation\n", + "V=L*I/t;\n", + "\n", + "#result\n", + "print \"voltage is\",round(V*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.9,Page 298" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1+R2 is 682.5 ohm\n", + "pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V\n", + "node voltage for V1 is 28.0 V\n", + "node voltage for V2 is 0.7 V\n", + "gate voltage is 15.31 V\n", + "gate & source diff is -12.69 V\n", + "load voltage is 26.73 V\n", + "load current is 2.23 A\n", + "load power is 47.63 W\n", + "Pq is 2.26 W\n", + "thermal resistance is 44.92 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=12.0;#load resistance\n", + "V1=.8;#voltage\n", + "V2=2.4;#voltage\n", + "D=.8;#duty cycle\n", + "Tj=150.0;#degreeC\n", + "Ta=40.0;#degreeC\n", + "Vd=28.0;\n", + "Vo=.7;\n", + "I=40.0;#mA;\n", + "R1=330;\n", + "R2=360;\n", + "Vn1=28;\n", + "Vn2=.7;\n", + "\n", + "#calculation\n", + "k=(Vd-Vo)/I;\n", + "Vg=R2*Vd/(R1+R2)+Vn2;\n", + "Vgs=Vg-Vd;\n", + "Vl=Vd*Rl/(Rl+.57);\n", + "Il=Vl/Rl;\n", + "Pl=D*Vl*Il;\n", + "Vq=Il*.57;\n", + "Pq=D*Vq*Il;\n", + "Q=(Tj-Ta)/Pq-3.7;\n", + "\n", + "#result\n", + "print \"R1+R2 is\",round(k*1000,2), \"ohm\"\n", + "print('pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V')\n", + "print \"node voltage for V1 is\",round(Vn1,2),\"V\"\n", + "print \"node voltage for V2 is\",round(Vn2,2), \"V\"\n", + "print \"gate voltage is\",round(Vg,2), \"V\"\n", + "print \"gate & source diff is\",round(Vgs,2), \"V\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"Pq is\",round(Pq,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.10,Page 305" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time is 1.1 micro s\n" + ] + } + ], + "source": [ + "#finding time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=40.0;#current\n", + "Q=44.0;#nC\n", + "\n", + "#calculation\n", + "t=Q/I;\n", + "\n", + "#result\n", + "print \"time is\",round(t,2), \"micro s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.11,Page 313" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 1.8 A\n", + "load voltage is 26.97 V\n", + "power is 40.85 W\n", + "high side voltage is 0.67 V\n", + "high side power is 1.03 W\n", + "low side voltage is 0.36 V\n", + "low side power is 0.55 W\n", + "IC power is 1.56 W\n", + "thermal resistance is 55.49 degreeC/W\n" + ] + } + ], + "source": [ + "#finding different voltages and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=15.0;#load resistance\n", + "D=.85;#duty cycle\n", + "Ts=60.0;#degreeC\n", + "Vd=28.0;#voltage\n", + "R1=.375;\n", + "R2=.2;\n", + "\n", + "#calculation\n", + "I=Vd/(R1+R2+Rl);\n", + "Vl1=I*Rl;\n", + "P=D*Vl*I;\n", + "Vh=I*R1;\n", + "Ph=D*Vh*I;\n", + "Vl=I*R2;\n", + "Pl=D*Vl*I;\n", + "Pic=Ph+Pl;\n", + "Pic=1.56;\n", + "Tj=150;\n", + "Ta=60;\n", + "Q=(Tj-Ta)/Pic-2.2;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"A\"\n", + "print \"load voltage is\",round(Vl1,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"high side voltage is\",round(Vh,2), \"V\"\n", + "print \"high side power is\",round(Ph,2), \"W\"\n", + "print \"low side voltage is\",round(Vl,2), \"V\"\n", + "print \"low side power is\",round(Pl,2), \"W\"\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6_1.ipynb new file mode 100644 index 00000000..d5b6622f --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter6_1.ipynb @@ -0,0 +1,528 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Power Switches" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.1,Page 274" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector current is 1.81 A\n", + "load power is 49.32 W\n", + "transistor power is 1.45 W\n", + "least value of base current is 90.67 mA\n", + "max value of base resistance is 4.85 ohm\n", + "thus pick Rb=33ohm\n" + ] + } + ], + "source": [ + "#finding Ic,Pload,Pq,Ibase,Rbase\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vs=28.0;#V\n", + "Vi=5.0;#V\n", + "Rl=15.0;#ohm\n", + "Vc=.8;#V\n", + "b=20.0;\n", + "\n", + "#calculation\n", + "Ic=(Vs-Vc)/Rl;\n", + "Pl=Ic**2*Rl;\n", + "Pq=Ic*Vc;\n", + "Ib=Ic/b*1000;\n", + "Rb=(Vi-.6)/Ib;\n", + "\n", + "#result\n", + "print \"collector current is\",round(Ic,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"transistor power is\",round(Pq,2), \"W\"\n", + "print \"least value of base current is\",round(Ib,2), \"mA\"\n", + "print \"max value of base resistance is\",round(Rb*100,2), \"ohm\"\n", + "print ('thus pick Rb=33ohm')" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.4,Page 282" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load resistance is 554.0 ohm\n", + "thus pick Rl=560ohm\n", + "max value of Rb is 3.0 kohm\n", + "thus pick Rb=2.2kohm\n", + "load current is 49.46 mA\n", + "load power is 685.08 mW\n", + "power delivered is 7.42 mW\n" + ] + } + ], + "source": [ + "#finding Pload,Pq,Iload,resistances\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=28.0;#V\n", + "f=100.0;#frequency\n", + "I=50.0;#current\n", + "Rl1=560.0;\n", + "Vp=2.4;\n", + "Ib=500.0;#microAmp\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "Rl=(Vd-.3)/I;\n", + "Rb=(Vp-.9)/Ib;\n", + "Vl=Vd-.3;\n", + "Ip=Vl/Rl1;\n", + "Pl=D*Vl*Ip;\n", + "Pq=D*Ip*.3;\n", + "\n", + "#result\n", + "print \"load resistance is\",round(Rl*1000), \"ohm\"\n", + "print('thus pick Rl=560ohm')\n", + "print \"max value of Rb is\",round(Rb*1000,2),\"kohm\"\n", + "print('thus pick Rb=2.2kohm')\n", + "print \"load current is\",round(Ip*1000,2), \"mA\"\n", + "print \"load power is\",round(Pl*1000,2), \"mW\"\n", + "print \"power delivered is\",round(Pq*1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.5,Page 286" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time of rise is 788.48 ns\n", + "time of rise is 4.65 microsec\n" + ] + } + ], + "source": [ + "#finding time of rise\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "C=640.0;#capacitor\n", + "R1=560.0;#load resistance\n", + "R2=3.3;#kohm\n", + "\n", + "#calculation\n", + "t1=2.2*R1*C;\n", + "t2=2.2*R2*C;\n", + "\n", + "#result\n", + "print \"time of rise is\",round(t1/1000,2), \"ns\"\n", + "print \"time of rise is\",round(t2/1000,2), \"microsec\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.6,Page 287" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "resistance is 682.5 ohm\n", + "pick up resistance=680 ohm\n", + "rise time is 957.44 ns\n" + ] + }, + { + "data": { + "image/png": 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FNKjBWgAXEb4I5lDCN4L9iFYd/MFhoMq31ath//1jVyENarAWwPEtq6IeWwDKs6efhlmz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\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f86a4de0510>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#finding Rpick up,time of rise\n", + "\n", + "#initialisation of variable\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from pylab import plot, xlabel, ylabel, title\n", + "Vol=0.7;\n", + "Iol=40.0/1000;#current\n", + "Rpullup1=680.0;\n", + "C=640.0;\n", + "Epullup=28.0;\n", + "#for plotting\n", + "x=[0, .1, 1.9, 4.1, 5, 5.1, 5.3, 5.6, 6.0, 9.3];\n", + "y=[27.8, .1, .1, .1, .1, 5, 13.5, 21.0, 27.0, 27.8];\n", + "\n", + "#calculation\n", + "Rpullup=(Epullup-Vol)/Iol;\n", + "trise=2.2*Rpullup1*C;\n", + "plot(x,y,'r');\n", + "xlabel ('time(mus)')\n", + "ylabel ('Vout')\n", + "title ('Vout vs time')\n", + "\n", + "#result\n", + "print \"resistance is\",round(Rpullup,2), \"ohm\"\n", + "print('pick up resistance=680 ohm');\n", + "print \"rise time is\",round(trise/1000,2), \"ns\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.7,Page 289" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "worst case resistance is 0.286 ohm\n", + "load current is 4.45 A\n", + "load voltage is 26.73 V\n", + "load power is 47.62 W\n", + "drop voltage is 1.27 V\n", + "power is 2.27 W\n", + "temperature is 182.6 deg.C\n" + ] + } + ], + "source": [ + "#finding worst case resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R1=.11;#resistance\n", + "Vd=28.0;#voltage\n", + "R2=6.0;#ohm\n", + "D=.4;#duty cycle\n", + "Q=62.0;\n", + "\n", + "#calculation\n", + "Ro=2.6*R1;\n", + "Ip=Vd/(R2+Ro);\n", + "Vl=Ip*R2;\n", + "Pl=D*Vl*Ip;\n", + "Vq=Ip*Ro;\n", + "Pq=D*Vq*Ip;\n", + "T=40+round(Pq*10)/10*Q;\n", + "\n", + "#result\n", + "print \"worst case resistance is\",round(Ro,3), \"ohm\"\n", + "print \"load current is\",round(Ip,2),\"A\"\n", + "print \"load voltage is\",round(Vl,2), \"V\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"drop voltage is\",round(Vq,2), \"V\"\n", + "print \"power is\",round(Pq*10,2)/10, \"W\"\n", + "print \"temperature is\",round(T,2), \"deg.C\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.8,Page 292" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 150.0 V\n" + ] + } + ], + "source": [ + "#finding voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "L=10.0;#inductor\n", + "I=4.5;#current\n", + "t=300.0#time\n", + "\n", + "#calculation\n", + "V=L*I/t;\n", + "\n", + "#result\n", + "print \"voltage is\",round(V*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.9,Page 298" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1+R2 is 682.5 ohm\n", + "pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V\n", + "node voltage for V1 is 28.0 V\n", + "node voltage for V2 is 0.7 V\n", + "gate voltage is 15.31 V\n", + "gate & source diff is -12.69 V\n", + "load voltage is 26.73 V\n", + "load current is 2.23 A\n", + "load power is 47.63 W\n", + "Pq is 2.26 W\n", + "thermal resistance is 44.92 degreeC/W\n" + ] + } + ], + "source": [ + "#finding resistance and power characteristics\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=12.0;#load resistance\n", + "V1=.8;#voltage\n", + "V2=2.4;#voltage\n", + "D=.8;#duty cycle\n", + "Tj=150.0;#degreeC\n", + "Ta=40.0;#degreeC\n", + "Vd=28.0;\n", + "Vo=.7;\n", + "I=40.0;#mA;\n", + "R1=330;\n", + "R2=360;\n", + "Vn1=28;\n", + "Vn2=.7;\n", + "\n", + "#calculation\n", + "k=(Vd-Vo)/I;\n", + "Vg=R2*Vd/(R1+R2)+Vn2;\n", + "Vgs=Vg-Vd;\n", + "Vl=Vd*Rl/(Rl+.57);\n", + "Il=Vl/Rl;\n", + "Pl=D*Vl*Il;\n", + "Vq=Il*.57;\n", + "Pq=D*Vq*Il;\n", + "Q=(Tj-Ta)/Pq-3.7;\n", + "\n", + "#result\n", + "print \"R1+R2 is\",round(k*1000,2), \"ohm\"\n", + "print('pick R1=330ohm & R2=360ohm as they divide Vd setting 8V<Vg<18V')\n", + "print \"node voltage for V1 is\",round(Vn1,2),\"V\"\n", + "print \"node voltage for V2 is\",round(Vn2,2), \"V\"\n", + "print \"gate voltage is\",round(Vg,2), \"V\"\n", + "print \"gate & source diff is\",round(Vgs,2), \"V\"\n", + "print \"load voltage is\",round(Vl,3), \"V\"\n", + "print \"load current is\",round(Il,2), \"A\"\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"Pq is\",round(Pq,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.10,Page 305" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "time is 1.1 micro s\n" + ] + } + ], + "source": [ + "#finding time\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=40.0;#current\n", + "Q=44.0;#nC\n", + "\n", + "#calculation\n", + "t=Q/I;\n", + "\n", + "#result\n", + "print \"time is\",round(t,2), \"micro s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.11,Page 313" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "current is 1.8 A\n", + "load voltage is 26.97 V\n", + "power is 40.85 W\n", + "high side voltage is 0.67 V\n", + "high side power is 1.03 W\n", + "low side voltage is 0.36 V\n", + "low side power is 0.55 W\n", + "IC power is 1.56 W\n", + "thermal resistance is 55.49 degreeC/W\n" + ] + } + ], + "source": [ + "#finding different voltages and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Rl=15.0;#load resistance\n", + "D=.85;#duty cycle\n", + "Ts=60.0;#degreeC\n", + "Vd=28.0;#voltage\n", + "R1=.375;\n", + "R2=.2;\n", + "\n", + "#calculation\n", + "I=Vd/(R1+R2+Rl);\n", + "Vl1=I*Rl;\n", + "P=D*Vl*I;\n", + "Vh=I*R1;\n", + "Ph=D*Vh*I;\n", + "Vl=I*R2;\n", + "Pl=D*Vl*I;\n", + "Pic=Ph+Pl;\n", + "Pic=1.56;\n", + "Tj=150;\n", + "Ta=60;\n", + "Q=(Tj-Ta)/Pic-2.2;\n", + "\n", + "#result\n", + "print \"current is\",round(I,2), \"A\"\n", + "print \"load voltage is\",round(Vl1,2), \"V\"\n", + "print \"power is\",round(P,2), \"W\"\n", + "print \"high side voltage is\",round(Vh,2), \"V\"\n", + "print \"high side power is\",round(Ph,2), \"W\"\n", + "print \"low side voltage is\",round(Vl,2), \"V\"\n", + "print \"low side power is\",round(Pl,2), \"W\"\n", + "print \"IC power is\",round(Pic,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,2), \"degreeC/W\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7.ipynb new file mode 100644 index 00000000..45af7d98 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7.ipynb @@ -0,0 +1,505 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Switiching Power Supplies" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.1,Page 326" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 0.25\n", + "average voltage is 3.0 V\n" + ] + } + ], + "source": [ + "#finding duty cycle and average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=20.0;#time\n", + "Vp=12.0;#voltage\n", + "t=5.0;\n", + "\n", + "#calculation\n", + "D=t/T;\n", + "Vd=(D*Vp);\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D,3)\n", + "print \"average voltage is\",round(Vd,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.2,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time is 10.0 microsec\n", + "on time is 4.167 microsec\n", + "ripple current is 133.636 mA\n", + "load current is 500.0 mA\n", + "peak inductor current is 566.818 mA\n" + ] + } + ], + "source": [ + "#finding on time ripple,load,peak inductor current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=12.0;#voltage\n", + "Vl=5.0;#load voltage\n", + "Rl=10.0;#load resistance\n", + "f=100.0;#frequency\n", + "L=220.0;#inductor\n", + "\n", + "#calculation\n", + "D=Vl/Vd;\n", + "T=1/f;\n", + "t=D*T;\n", + "Vr=Vd-Vl;\n", + "I=Vr*round(t*10000)/10/L;\n", + "Il=Vl/Rl;\n", + "Ip=Il+I/2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*10000,2)/10, \"microsec\"\n", + "print \"ripple current is\",round(I*1000,3),\"mA\"\n", + "print \"load current is\",round(Il*1000,3), \"mA\"\n", + "print \"peak inductor current is\",round(Ip*1000,3), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.3,Page 335" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 325.01 mA\n", + "by trapezium method\n", + "rms current is 324.04 mA\n", + "by rectangle method\n", + "\n", + " rectangle method gives good result than trapezium method\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Id=500.0;#load current\n", + "i=134;#mA\n", + "D=.42;#duty cycle\n", + "\n", + "#calculation\n", + "Ip=Id+i/2;\n", + "Im=Id-i/2;\n", + "I1=((D/3)*(Ip**2+Im*Ip+Im**2))**.5;\n", + "I2=D**.5*Id;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I1,2), \"mA\"\n", + "print('by trapezium method')\n", + "print \"rms current is\",round(I2,2), \"mA\"\n", + "print('by rectangle method')\n", + "print '\\n rectangle method gives good result than trapezium method'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.4,Page 336" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 0.3 V\n", + "dissipated power is 63.0 mW\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=.3;#voltage\n", + "I=500.0;#current\n", + "D=.42;#duty cycle\n", + "T=150.0;#temperature\n", + "R=.6;#ohm\n", + "\n", + "#calculation\n", + "Vq=I*R;\n", + "Pq=D*Vq*I;\n", + "\n", + "#result\n", + "print \"voltage is\",round(Vq/1000,2), \"V\"\n", + "print \"dissipated power is\",round(Pq/1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.5,Page 341" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time period is 6.667 microsec\n", + "on time is 2.778 microsec\n", + "load current is 500.0 mA\n", + "ripple current is 125.0 mA\n", + "inductor voltage is 7.0 V\n", + "inductor is 155.556 microH\n", + "inductor current is 562.5 mA\n", + "minimum capacitor current is 250.0 mA\n", + "minimum capacitor voltage is 18.0 V\n", + "Rf/Ri is 3.07\n", + "power of LM2595 is 0.33 W\n", + "thermal resistance is 210.998 degreeC/W\n" + ] + } + ], + "source": [ + "#finding all componenets\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#resistance\n", + "V1=5.0;#V\n", + "V2=12.0;#V\n", + "Ta=80.0;#degreeC\n", + "Tb=150.0;\n", + "f=150.0;#frequency\n", + "\n", + "#calculation\n", + "D=V1/V2;\n", + "T=1/f;\n", + "t=D*T;\n", + "Id=V1/R;\n", + "i=.25*Id;\n", + "Vl=V2-V1;\n", + "L=Vl*t/i;\n", + "Ip=Id+i/2;\n", + "Ic=Id/2;\n", + "Vc=1.5*V2;\n", + "K=V1/1.23-1;\n", + "P=.01*V2+D*Id*1;\n", + "Q=(Tb-Ta)/P-2.2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time period is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*1000,3), \"microsec\"\n", + "print \"load current is\",round(Id*1000,3), \"mA\"\n", + "print \"ripple current is\",round(i*1000,3), \"mA\"\n", + "print \"inductor voltage is\",round(Vl,2), \"V\"\n", + "print \"inductor is\",round(L*1000,3), \"microH\"\n", + "print \"inductor current is\",round(Ip*1000,2), \"mA\"\n", + "print \"minimum capacitor current is\",round(Ic*1000,2), \"mA\"\n", + "print \"minimum capacitor voltage is\",round(Vc,3), \"V\"\n", + "print \"Rf/Ri is\",round(K,2)\n", + "print \"power of LM2595 is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,3), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.6,Page 349" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load power is 15.4 W\n", + "supply power is 17.11 W\n", + "dc current is 1.4 A\n", + "inductor current is 1.57 A\n", + "duty cycle is 0.45\n", + "inductor is 154.29 microH\n", + "transistor power is 352.8 mW\n", + "diode power is 385.0 mW\n", + "capacitor is 157.5 microF\n" + ] + } + ], + "source": [ + "#finding different power,inductor current,inductor value\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "f=100.0;#kHz\n", + "R=.4;#ohm\n", + "Vd=.5;\n", + "\n", + "#calculation\n", + "Pl=V2*I;\n", + "Ps=Pl/.9;\n", + "Id=round(Ps/V1*10)/10;\n", + "i=.25*Id;\n", + "Ip=Id+i/2;\n", + "D=round((1-V1/V2)*100)/100;\n", + "t=D/f;\n", + "L=V1*t/i;\n", + "Vp=Id*R;\n", + "Pq=D*Vp*Id;\n", + "Pd=(1-D)*.5*Id;\n", + "C=Id*t/2/20;\n", + "\n", + "#result\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"supply power is\",round(Ps,2), \"W\"\n", + "print \"dc current is\",round(Id,2), \"A\"\n", + "print \"inductor current is\",round(Ip,2), \"A\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"inductor is\",round(L*1000,2), \"microH\"\n", + "print \"transistor power is\",round(Pq*1000,2), \"mW\"\n", + "print \"diode power is\",round(Pd*100,2)*10, \"mW\"\n", + "print \"capacitor is\",round(C*1e6,2), \"microF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.7,Page 355" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rf/Ri is 16.886\n", + "pick Rf=22; Ri=1.3;\n", + "rms current is 1.4 A\n", + "switch power is 132.3 mW\n", + "IC power is 151.2 mW\n", + "total power is 283.5 mW\n", + "IC temperature is 98.43 degreeC\n" + ] + } + ], + "source": [ + "#finding feedback resistor,power,current and temperature\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "Ta=80.0;#degreeC\n", + "Ps=17.1#supply power\n", + "\n", + "#calculation\n", + "K=V2/1.23-1;\n", + "Id=round(Ps/V1*10)/10;\n", + "D=round((1-(V1/V2))*100)/100;\n", + "Ir=D**.5*Id;\n", + "Ps=Ir**2*.15;\n", + "Pi=D*Id*V1/50;\n", + "P=Ps+Pi;\n", + "T=Ta+P*65;\n", + "\n", + "#result\n", + "print \"Rf/Ri is\",round(K,3)\n", + "print('pick Rf=22; Ri=1.3;')\n", + "print \"rms current is\",round(Id,2), \"A\"\n", + "print \"switch power is\",round(Ps*1000,2), \"mW\"\n", + "print \"IC power is\",round(Pi*1000,2), \"mW\"\n", + "print \"total power is\",round(P*1000,2), \"mW\"\n", + "print \"IC temperature is\",round(T,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.8,Page 359" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum voltage is 18.25 V\n", + "diode voltage is 20.0 V\n", + "duty cycle is 0.34\n", + "power delivered is 5.0 W\n", + "average current is 466.67 mA\n", + "mid primary current is 1.37 A\n", + "rms current is 800.33 mA\n", + "ramp current is 480.0 mA\n", + "maximum transistor current is 1.61 A\n", + "minimum transistor current is 1.13 A\n", + "diode peak current is 2.02 A\n", + "secondary rms current is 1.23 A\n", + "capacitor is 170.0 microF\n" + ] + } + ], + "source": [ + "#designing circuit and finding circuit parameter\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=5.0;#V\n", + "Il=1.0;#load current\n", + "T=10.0;#microsec\n", + "K=1.25;#Npri/Nsec\n", + "L=85.0;#microH\n", + "\n", + "#calculation\n", + "Vq=V1+V2*K;\n", + "Vd=V1*K+V2;\n", + "D=round((K*V2)*100/(V1+K*V2))/100;\n", + "Po=V2*Il;\n", + "Pi=round(Po/.09)/10;\n", + "Id=Pi/V1;\n", + "Im=Id/D;\n", + "Ir=(Im*D**.5);\n", + "i=V1*D*T/L;\n", + "IM=Im+.24;\n", + "Imin=Im-.24;\n", + "Ip=K*IM;\n", + "Imid=Il/(1-D);\n", + "Irms=Imid*(1-D)**.5;\n", + "C=D*Il*T/20;\n", + "\n", + "#result\n", + "print \"maximum voltage is\",round(Vq,2), \"V\"\n", + "print \"diode voltage is\",round(Vd,2), \"V\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"power delivered is\",round(Po,2), \"W\"\n", + "print \"average current is\",round(Id*1000,2), \"mA\"\n", + "print \"mid primary current is\",round(Im,2), \"A\"\n", + "print \"rms current is\",round(Ir*1000,2),\"mA\"\n", + "print \"ramp current is\",round(i*1000,2), \"mA\"\n", + "print \"maximum transistor current is\",round(IM,2),\"A\"\n", + "print \"minimum transistor current is\",round(Imin,2),\"A\"\n", + "print \"diode peak current is\",round(Ip,2), \"A\"\n", + "print \"secondary rms current is\",round(Irms,2),\"A\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7_1.ipynb new file mode 100644 index 00000000..45af7d98 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter7_1.ipynb @@ -0,0 +1,505 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Switiching Power Supplies" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.1,Page 326" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 0.25\n", + "average voltage is 3.0 V\n" + ] + } + ], + "source": [ + "#finding duty cycle and average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "T=20.0;#time\n", + "Vp=12.0;#voltage\n", + "t=5.0;\n", + "\n", + "#calculation\n", + "D=t/T;\n", + "Vd=(D*Vp);\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D,3)\n", + "print \"average voltage is\",round(Vd,3), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.2,Page 238" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time is 10.0 microsec\n", + "on time is 4.167 microsec\n", + "ripple current is 133.636 mA\n", + "load current is 500.0 mA\n", + "peak inductor current is 566.818 mA\n" + ] + } + ], + "source": [ + "#finding on time ripple,load,peak inductor current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vd=12.0;#voltage\n", + "Vl=5.0;#load voltage\n", + "Rl=10.0;#load resistance\n", + "f=100.0;#frequency\n", + "L=220.0;#inductor\n", + "\n", + "#calculation\n", + "D=Vl/Vd;\n", + "T=1/f;\n", + "t=D*T;\n", + "Vr=Vd-Vl;\n", + "I=Vr*round(t*10000)/10/L;\n", + "Il=Vl/Rl;\n", + "Ip=Il+I/2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*10000,2)/10, \"microsec\"\n", + "print \"ripple current is\",round(I*1000,3),\"mA\"\n", + "print \"load current is\",round(Il*1000,3), \"mA\"\n", + "print \"peak inductor current is\",round(Ip*1000,3), \"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.3,Page 335" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms current is 325.01 mA\n", + "by trapezium method\n", + "rms current is 324.04 mA\n", + "by rectangle method\n", + "\n", + " rectangle method gives good result than trapezium method\n" + ] + } + ], + "source": [ + "#finding rms current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Id=500.0;#load current\n", + "i=134;#mA\n", + "D=.42;#duty cycle\n", + "\n", + "#calculation\n", + "Ip=Id+i/2;\n", + "Im=Id-i/2;\n", + "I1=((D/3)*(Ip**2+Im*Ip+Im**2))**.5;\n", + "I2=D**.5*Id;\n", + "\n", + "#result\n", + "print \"rms current is\",round(I1,2), \"mA\"\n", + "print('by trapezium method')\n", + "print \"rms current is\",round(I2,2), \"mA\"\n", + "print('by rectangle method')\n", + "print '\\n rectangle method gives good result than trapezium method'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.4,Page 336" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "voltage is 0.3 V\n", + "dissipated power is 63.0 mW\n" + ] + } + ], + "source": [ + "#finding voltage and power\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "Vp=.3;#voltage\n", + "I=500.0;#current\n", + "D=.42;#duty cycle\n", + "T=150.0;#temperature\n", + "R=.6;#ohm\n", + "\n", + "#calculation\n", + "Vq=I*R;\n", + "Pq=D*Vq*I;\n", + "\n", + "#result\n", + "print \"voltage is\",round(Vq/1000,2), \"V\"\n", + "print \"dissipated power is\",round(Pq/1000,2), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.5,Page 341" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "duty cycle is 42.0 %\n", + "time period is 6.667 microsec\n", + "on time is 2.778 microsec\n", + "load current is 500.0 mA\n", + "ripple current is 125.0 mA\n", + "inductor voltage is 7.0 V\n", + "inductor is 155.556 microH\n", + "inductor current is 562.5 mA\n", + "minimum capacitor current is 250.0 mA\n", + "minimum capacitor voltage is 18.0 V\n", + "Rf/Ri is 3.07\n", + "power of LM2595 is 0.33 W\n", + "thermal resistance is 210.998 degreeC/W\n" + ] + } + ], + "source": [ + "#finding all componenets\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#resistance\n", + "V1=5.0;#V\n", + "V2=12.0;#V\n", + "Ta=80.0;#degreeC\n", + "Tb=150.0;\n", + "f=150.0;#frequency\n", + "\n", + "#calculation\n", + "D=V1/V2;\n", + "T=1/f;\n", + "t=D*T;\n", + "Id=V1/R;\n", + "i=.25*Id;\n", + "Vl=V2-V1;\n", + "L=Vl*t/i;\n", + "Ip=Id+i/2;\n", + "Ic=Id/2;\n", + "Vc=1.5*V2;\n", + "K=V1/1.23-1;\n", + "P=.01*V2+D*Id*1;\n", + "Q=(Tb-Ta)/P-2.2;\n", + "\n", + "#result\n", + "print \"duty cycle is\",round(D*100), \"%\"\n", + "print \"time period is\",round(T*1000,3), \"microsec\"\n", + "print \"on time is\",round(t*1000,3), \"microsec\"\n", + "print \"load current is\",round(Id*1000,3), \"mA\"\n", + "print \"ripple current is\",round(i*1000,3), \"mA\"\n", + "print \"inductor voltage is\",round(Vl,2), \"V\"\n", + "print \"inductor is\",round(L*1000,3), \"microH\"\n", + "print \"inductor current is\",round(Ip*1000,2), \"mA\"\n", + "print \"minimum capacitor current is\",round(Ic*1000,2), \"mA\"\n", + "print \"minimum capacitor voltage is\",round(Vc,3), \"V\"\n", + "print \"Rf/Ri is\",round(K,2)\n", + "print \"power of LM2595 is\",round(P,2), \"W\"\n", + "print \"thermal resistance is\",round(Q,3), \"degreeC/W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.6,Page 349" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "load power is 15.4 W\n", + "supply power is 17.11 W\n", + "dc current is 1.4 A\n", + "inductor current is 1.57 A\n", + "duty cycle is 0.45\n", + "inductor is 154.29 microH\n", + "transistor power is 352.8 mW\n", + "diode power is 385.0 mW\n", + "capacitor is 157.5 microF\n" + ] + } + ], + "source": [ + "#finding different power,inductor current,inductor value\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "f=100.0;#kHz\n", + "R=.4;#ohm\n", + "Vd=.5;\n", + "\n", + "#calculation\n", + "Pl=V2*I;\n", + "Ps=Pl/.9;\n", + "Id=round(Ps/V1*10)/10;\n", + "i=.25*Id;\n", + "Ip=Id+i/2;\n", + "D=round((1-V1/V2)*100)/100;\n", + "t=D/f;\n", + "L=V1*t/i;\n", + "Vp=Id*R;\n", + "Pq=D*Vp*Id;\n", + "Pd=(1-D)*.5*Id;\n", + "C=Id*t/2/20;\n", + "\n", + "#result\n", + "print \"load power is\",round(Pl,2), \"W\"\n", + "print \"supply power is\",round(Ps,2), \"W\"\n", + "print \"dc current is\",round(Id,2), \"A\"\n", + "print \"inductor current is\",round(Ip,2), \"A\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"inductor is\",round(L*1000,2), \"microH\"\n", + "print \"transistor power is\",round(Pq*1000,2), \"mW\"\n", + "print \"diode power is\",round(Pd*100,2)*10, \"mW\"\n", + "print \"capacitor is\",round(C*1e6,2), \"microF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.7,Page 355" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rf/Ri is 16.886\n", + "pick Rf=22; Ri=1.3;\n", + "rms current is 1.4 A\n", + "switch power is 132.3 mW\n", + "IC power is 151.2 mW\n", + "total power is 283.5 mW\n", + "IC temperature is 98.43 degreeC\n" + ] + } + ], + "source": [ + "#finding feedback resistor,power,current and temperature\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=22.0;#V\n", + "I=.7;#A\n", + "Ta=80.0;#degreeC\n", + "Ps=17.1#supply power\n", + "\n", + "#calculation\n", + "K=V2/1.23-1;\n", + "Id=round(Ps/V1*10)/10;\n", + "D=round((1-(V1/V2))*100)/100;\n", + "Ir=D**.5*Id;\n", + "Ps=Ir**2*.15;\n", + "Pi=D*Id*V1/50;\n", + "P=Ps+Pi;\n", + "T=Ta+P*65;\n", + "\n", + "#result\n", + "print \"Rf/Ri is\",round(K,3)\n", + "print('pick Rf=22; Ri=1.3;')\n", + "print \"rms current is\",round(Id,2), \"A\"\n", + "print \"switch power is\",round(Ps*1000,2), \"mW\"\n", + "print \"IC power is\",round(Pi*1000,2), \"mW\"\n", + "print \"total power is\",round(P*1000,2), \"mW\"\n", + "print \"IC temperature is\",round(T,2), \"degreeC\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.8,Page 359" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "maximum voltage is 18.25 V\n", + "diode voltage is 20.0 V\n", + "duty cycle is 0.34\n", + "power delivered is 5.0 W\n", + "average current is 466.67 mA\n", + "mid primary current is 1.37 A\n", + "rms current is 800.33 mA\n", + "ramp current is 480.0 mA\n", + "maximum transistor current is 1.61 A\n", + "minimum transistor current is 1.13 A\n", + "diode peak current is 2.02 A\n", + "secondary rms current is 1.23 A\n", + "capacitor is 170.0 microF\n" + ] + } + ], + "source": [ + "#designing circuit and finding circuit parameter\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=12.0;#V\n", + "V2=5.0;#V\n", + "Il=1.0;#load current\n", + "T=10.0;#microsec\n", + "K=1.25;#Npri/Nsec\n", + "L=85.0;#microH\n", + "\n", + "#calculation\n", + "Vq=V1+V2*K;\n", + "Vd=V1*K+V2;\n", + "D=round((K*V2)*100/(V1+K*V2))/100;\n", + "Po=V2*Il;\n", + "Pi=round(Po/.09)/10;\n", + "Id=Pi/V1;\n", + "Im=Id/D;\n", + "Ir=(Im*D**.5);\n", + "i=V1*D*T/L;\n", + "IM=Im+.24;\n", + "Imin=Im-.24;\n", + "Ip=K*IM;\n", + "Imid=Il/(1-D);\n", + "Irms=Imid*(1-D)**.5;\n", + "C=D*Il*T/20;\n", + "\n", + "#result\n", + "print \"maximum voltage is\",round(Vq,2), \"V\"\n", + "print \"diode voltage is\",round(Vd,2), \"V\"\n", + "print \"duty cycle is\",round(D,2)\n", + "print \"power delivered is\",round(Po,2), \"W\"\n", + "print \"average current is\",round(Id*1000,2), \"mA\"\n", + "print \"mid primary current is\",round(Im,2), \"A\"\n", + "print \"rms current is\",round(Ir*1000,2),\"mA\"\n", + "print \"ramp current is\",round(i*1000,2), \"mA\"\n", + "print \"maximum transistor current is\",round(IM,2),\"A\"\n", + "print \"minimum transistor current is\",round(Imin,2),\"A\"\n", + "print \"diode peak current is\",round(Ip,2), \"A\"\n", + "print \"secondary rms current is\",round(Irms,2),\"A\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8.ipynb new file mode 100644 index 00000000..7cf6b405 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8.ipynb @@ -0,0 +1,326 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Thyristors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.3,Page 397" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance is 6.22 microH\n", + "load impedence at angle 90 degree is 0.00195 ohm\n" + ] + } + ], + "source": [ + "#finding inductance,load impedence\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=50.0;#di/dt\n", + "\n", + "#calculation\n", + "L=V*2**.5/K;\n", + "Z=2*pi*f*L;\n", + "\n", + "#result\n", + "print \"inductance is\",round(L,2),\"microH\"\n", + "print \"load impedence at angle 90 degree is\",round(Z*1e-6,5), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.4,Page 400" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "minimum value of capacitor is 0.067 micfoF\n", + "\n", + "choose C=.1 micoF\n", + "capacitor impedence at angle -90degree is 31.83 ohm\n", + "Load current in mA at an angle 90 degrees is 6.91\n", + "Potential drop in V at an angle 90 degrees is 0.55\n", + "Power dissipated is 3 mW\n" + ] + } + ], + "source": [ + "#finding capacitor,current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=75.0;#dv/dt\n", + "Vd=400.0;#DRM voltage\n", + "\n", + "\n", + "#calculation\n", + "C=Vd/R/K;\n", + "C1=.1;\n", + "Z=1/(2*pi*f*C1);\n", + "Iload=V/1000/(-Z*cos(180*pi/180)+R*round(cos(90*pi/180)));\n", + "Vload=Iload/1000*R;\n", + "P=Vload*Iload;\n", + "\n", + "#result\n", + "print \"minimum value of capacitor is\",round(C,3), \"micfoF\"\n", + "print('\\nchoose C=.1 micoF')\n", + "print \"capacitor impedence at angle -90degree is\",round(Z*1000,2), \"ohm\"\n", + "print \"Load current in mA at an angle 90 degrees is\",round(Iload,2)\n", + "print \"Potential drop in V at an angle 90 degrees is\",round(Vload,2)\n", + "print \"Power dissipated is\",int(P), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.5,Page 402" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "snubbing resistnce is 7.39 ohm\n" + ] + } + ], + "source": [ + "#finding snubbing resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220;#line voltage\n", + "f=50;#hertz\n", + "R=80;#load resistance\n", + "I=46;#TSM current\n", + "\n", + "#calculation\n", + "Rs=V*2**.5/(I-V*2**.5/R);\n", + "\n", + "#result\n", + "print \"snubbing resistnce is\",round(Rs,2), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.6,Page 414" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "line period is 16.67 ms\n", + "half-cycle time is 8.333 ms\n", + "no. of cycles is 10.0\n", + "voltage for t1 is 54.0 V\n", + "power for t1 is 291.6 W\n", + "voltage for t2 is 100.0 V\n", + "voltage for t2 is 1000.0 W\n" + ] + } + ], + "source": [ + "#finding voltage , power and cycles\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#load\n", + "V=120.0;#rms voltage\n", + "f=60.0;#hertz\n", + "T=83.3;#ms\n", + "t1=15;#ms\n", + "t2=55;#ms\n", + "\n", + "#calculation\n", + "Tl=1/f;\n", + "Th=Tl/2;\n", + "C=round(T/Th/100)*100;\n", + "D1=.2;\n", + "V1=round(V*D1**.5);\n", + "P1=V1**2/R;\n", + "D2=.7;\n", + "V2=round(V*D2**.5);\n", + "P2=V2**2/R;\n", + "\n", + "#result\n", + "print \"line period is\",round(Tl*1000,2), \"ms\"\n", + "print \"half-cycle time is\",round(Th*1000,3), \"ms\"\n", + "print \"no. of cycles is\",C/1000\n", + "print \"voltage for t1 is\",round(V1,3), \"V\"\n", + "print \"power for t1 is\",round(P1,3), \"W\"\n", + "print \"voltage for t2 is\",round(V2,3), \"V\"\n", + "print \"voltage for t2 is\",round(P2,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.8,Page 427" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 42.0 V\n", + "dc voltage is 41.0 V\n", + "\n", + "the markers indicae Vp=163V Vave=41\n", + "full-wave rms voltage is 108.0 V\n", + "rms voltage is 108.0 V\n", + "\n", + "the markers indicate Vp=169V ;Vave=106V\n" + ] + } + ], + "source": [ + "#finding dc volatge,average voltage,rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#line voltage\n", + "A=60.0;#degree\n", + "D=0.35;\n", + "\n", + "#calculation\n", + "Va=D*V;\n", + "Vd=V*2**.5*(cos(A*pi/180)+1)/2/pi;\n", + "Vr=.9*V;\n", + "Vrms=V*(2**.5)*(.5*(pi-1.047)+.25*sin(2*A*pi/180))**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Va,3), \"V\"\n", + "print \"dc voltage is\",round(Vd), \"V\"\n", + "print('\\nthe markers indicae Vp=163V Vave=41')\n", + "print \"full-wave rms voltage is\",round(Vr), \"V\"\n", + "print \"rms voltage is\",round(Vrms), \"V\"\n", + "print('\\nthe markers indicate Vp=169V ;Vave=106V')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.9,Page 430" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 141.18 V\n", + "double checked value of rms voltage is 141.18 V\n" + ] + } + ], + "source": [ + "#finding rms voltage and double checked rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "P=1.3;#kW\n", + "R=15.0;#ohm\n", + "\n", + "#calculation\n", + "Vr=round((P*1000*R)**.5);\n", + "D=Vr/V;\n", + "Vr=V*2**.5*(.5*(pi-1.710)+sin(196*pi/180)/4)**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"\n", + "print \"double checked value of rms voltage is\",round(Vr,2), \"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8_1.ipynb new file mode 100644 index 00000000..7cf6b405 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter8_1.ipynb @@ -0,0 +1,326 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Thyristors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.3,Page 397" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inductance is 6.22 microH\n", + "load impedence at angle 90 degree is 0.00195 ohm\n" + ] + } + ], + "source": [ + "#finding inductance,load impedence\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=50.0;#di/dt\n", + "\n", + "#calculation\n", + "L=V*2**.5/K;\n", + "Z=2*pi*f*L;\n", + "\n", + "#result\n", + "print \"inductance is\",round(L,2),\"microH\"\n", + "print \"load impedence at angle 90 degree is\",round(Z*1e-6,5), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.4,Page 400" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "minimum value of capacitor is 0.067 micfoF\n", + "\n", + "choose C=.1 micoF\n", + "capacitor impedence at angle -90degree is 31.83 ohm\n", + "Load current in mA at an angle 90 degrees is 6.91\n", + "Potential drop in V at an angle 90 degrees is 0.55\n", + "Power dissipated is 3 mW\n" + ] + } + ], + "source": [ + "#finding capacitor,current\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "f=50.0;#hertz\n", + "R=80.0;#load resistance\n", + "K=75.0;#dv/dt\n", + "Vd=400.0;#DRM voltage\n", + "\n", + "\n", + "#calculation\n", + "C=Vd/R/K;\n", + "C1=.1;\n", + "Z=1/(2*pi*f*C1);\n", + "Iload=V/1000/(-Z*cos(180*pi/180)+R*round(cos(90*pi/180)));\n", + "Vload=Iload/1000*R;\n", + "P=Vload*Iload;\n", + "\n", + "#result\n", + "print \"minimum value of capacitor is\",round(C,3), \"micfoF\"\n", + "print('\\nchoose C=.1 micoF')\n", + "print \"capacitor impedence at angle -90degree is\",round(Z*1000,2), \"ohm\"\n", + "print \"Load current in mA at an angle 90 degrees is\",round(Iload,2)\n", + "print \"Potential drop in V at an angle 90 degrees is\",round(Vload,2)\n", + "print \"Power dissipated is\",int(P), \"mW\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.5,Page 402" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "snubbing resistnce is 7.39 ohm\n" + ] + } + ], + "source": [ + "#finding snubbing resistance\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220;#line voltage\n", + "f=50;#hertz\n", + "R=80;#load resistance\n", + "I=46;#TSM current\n", + "\n", + "#calculation\n", + "Rs=V*2**.5/(I-V*2**.5/R);\n", + "\n", + "#result\n", + "print \"snubbing resistnce is\",round(Rs,2), \"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.6,Page 414" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "line period is 16.67 ms\n", + "half-cycle time is 8.333 ms\n", + "no. of cycles is 10.0\n", + "voltage for t1 is 54.0 V\n", + "power for t1 is 291.6 W\n", + "voltage for t2 is 100.0 V\n", + "voltage for t2 is 1000.0 W\n" + ] + } + ], + "source": [ + "#finding voltage , power and cycles\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "R=10.0;#load\n", + "V=120.0;#rms voltage\n", + "f=60.0;#hertz\n", + "T=83.3;#ms\n", + "t1=15;#ms\n", + "t2=55;#ms\n", + "\n", + "#calculation\n", + "Tl=1/f;\n", + "Th=Tl/2;\n", + "C=round(T/Th/100)*100;\n", + "D1=.2;\n", + "V1=round(V*D1**.5);\n", + "P1=V1**2/R;\n", + "D2=.7;\n", + "V2=round(V*D2**.5);\n", + "P2=V2**2/R;\n", + "\n", + "#result\n", + "print \"line period is\",round(Tl*1000,2), \"ms\"\n", + "print \"half-cycle time is\",round(Th*1000,3), \"ms\"\n", + "print \"no. of cycles is\",C/1000\n", + "print \"voltage for t1 is\",round(V1,3), \"V\"\n", + "print \"power for t1 is\",round(P1,3), \"W\"\n", + "print \"voltage for t2 is\",round(V2,3), \"V\"\n", + "print \"voltage for t2 is\",round(P2,3), \"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.8,Page 427" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 42.0 V\n", + "dc voltage is 41.0 V\n", + "\n", + "the markers indicae Vp=163V Vave=41\n", + "full-wave rms voltage is 108.0 V\n", + "rms voltage is 108.0 V\n", + "\n", + "the markers indicate Vp=169V ;Vave=106V\n" + ] + } + ], + "source": [ + "#finding dc volatge,average voltage,rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#line voltage\n", + "A=60.0;#degree\n", + "D=0.35;\n", + "\n", + "#calculation\n", + "Va=D*V;\n", + "Vd=V*2**.5*(cos(A*pi/180)+1)/2/pi;\n", + "Vr=.9*V;\n", + "Vrms=V*(2**.5)*(.5*(pi-1.047)+.25*sin(2*A*pi/180))**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Va,3), \"V\"\n", + "print \"dc voltage is\",round(Vd), \"V\"\n", + "print('\\nthe markers indicae Vp=163V Vave=41')\n", + "print \"full-wave rms voltage is\",round(Vr), \"V\"\n", + "print \"rms voltage is\",round(Vrms), \"V\"\n", + "print('\\nthe markers indicate Vp=169V ;Vave=106V')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.9,Page 430" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "rms voltage is 141.18 V\n", + "double checked value of rms voltage is 141.18 V\n" + ] + } + ], + "source": [ + "#finding rms voltage and double checked rms voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=220.0;#line voltage\n", + "P=1.3;#kW\n", + "R=15.0;#ohm\n", + "\n", + "#calculation\n", + "Vr=round((P*1000*R)**.5);\n", + "D=Vr/V;\n", + "Vr=V*2**.5*(.5*(pi-1.710)+sin(196*pi/180)/4)**.5/pi**.5;\n", + "\n", + "#result\n", + "print \"rms voltage is\",round(Vr,2), \"V\"\n", + "print \"double checked value of rms voltage is\",round(Vr,2), \"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9.ipynb new file mode 100644 index 00000000..d50d131a --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9.ipynb @@ -0,0 +1,405 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Power Conversation and Motor Drive Operations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.1,Page 457" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "peak voltage is 37.6 V\n", + "load voltage is 35.7 V\n", + "ripple voltage is 3.96 V\n", + "approx. load voltage is 35.62 V\n" + ] + } + ], + "source": [ + "#finding peak,load,ripple voltages\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=28.0;#V\n", + "C=4700.0;#microF\n", + "R=16.0;#load\n", + "f=120.0;#hertz\n", + "\n", + "#calculation\n", + "Vp=V*2**.5-2;\n", + "Vd=0.95*Vp;\n", + "Id=Vd/R;\n", + "v=Id/f/C;\n", + "#approximation\n", + "Vd1=Vp-v*1e6/2;\n", + "\n", + "#result\n", + "print \"peak voltage is\",round(Vp,2), \"V\"\n", + "print \"load voltage is\",round(Vd,1), \"V\"\n", + "print \"ripple voltage is\",round(v*1e6,2), \"V\"\n", + "print \"approx. load voltage is\",round(Vd1,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2,Page 459" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zth is 1.0 + 1.0 in ohm\n", + "inductor is 2.65 mH\n" + ] + } + ], + "source": [ + "#finding inductor,Zth\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=120.0;#pri voltage\n", + "V2=28.0;#sec voltage\n", + "I=2.0;#pri current\n", + "f=60.0;#Hz\n", + "Vth=28.8;#open voltage\n", + "V3=12.1;#pri-short voltage\n", + "Is=2.0;#short current at 45 degree\n", + "\n", + "#calculation\n", + "Zi=(V2*V3)/V1/Is*cos(45*pi/180);\n", + "Zj=(V2*V3)/V1/Is*sin(45*pi/180);\n", + "L=Zi/(2*pi*f);\n", + "\n", + "#result\n", + "print'Zth is',round(Zi),'+',round(Zj),'in ohm'\n", + "print \"inductor is\",round(L*1000,2), \"mH\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.4,Page 463" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power factor is 0.32\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I1=1.8;#current\n", + "R=16.0;#resistance\n", + "I2=5.7;#A\n", + "V=28.8;#Voltage\n", + "\n", + "#calculation\n", + "P=I1**2*R;\n", + "S=I2*V;\n", + "Pf=P/S;\n", + "\n", + "#result\n", + "print \"power factor is\",round(Pf,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.5, Page 468" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "aparrent power is 8.14 kVA\n", + "dissipated power is 7.84 kW\n", + "power factor is 0.96\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=22.6;#current\n", + "I2=28.00;\n", + "V=120.0;#Voltage\n", + "V2=280.0;\n", + "\n", + "#calculation\n", + "Pt=3*I*V;\n", + "Pl=I2*V2;\n", + "Pf=Pl/Pt;\n", + "\n", + "#result\n", + "print \"aparrent power is\",round(Pt/1000,2),\"kVA\"\n", + "print \"dissipated power is\",round(Pl/1000,2),\"kW\"\n", + "print \"power factor is\",round(Pl/Pt,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.6,Page 474" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ratio is 0.72\n", + "firing angle is 58 degrees\n", + "dc voltage is 148.85 V\n", + "time delay is 2.69 ms\n" + ] + } + ], + "source": [ + "#finding firing angle, time delay,Vd\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=208.0;#voltage\n", + "R=100.0;#load\n", + "Vd=150.0;#V\n", + "\n", + "#calculation\n", + "r=Vd/V;\n", + "a=58;#degree\n", + "Vd=3*2**.5*208*(cos(pi/3+a*pi/180)-cos(2*pi/3+a*pi/180))/pi;\n", + "t=a*16.7/360;\n", + "\n", + "#result\n", + "print \"ratio is\",round(r,2)\n", + "print('firing angle is 58 degrees');\n", + "print \"dc voltage is\",round(Vd,2), \"V\"\n", + "print \"time delay is\",round(t,2), \"ms\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.7,Page 480" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max. current is 41.67 A\n", + "dissipated power is 8.68 W\n" + ] + } + ], + "source": [ + "#finding maximum current and power dissipated\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=150.0;#power\n", + "V=8.0;#voltage\n", + "R=.01;#resistance\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "I=P/.9/D/V;\n", + "Ir=I*D**.5;\n", + "Pq=Ir**2*R;\n", + "\n", + "#result\n", + "print \"max. current is\",round(I,2), \"A\"\n", + "print \"dissipated power is\",round(Pq,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.8,Page 489" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "pwm fundamental frequency is 30.72 kHz\n", + "output voltage is 9.46 V\n" + ] + } + ], + "source": [ + "#finding fundamental frequency and output voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f1=60.0;#frequency\n", + "V=150.0;#voltage\n", + "f2=31.0;#kHz\n", + "\n", + "#calculation\n", + "f3=f1*4;\n", + "Vo=V*10**(-4.2);\n", + "\n", + "#result\n", + "print \"pwm fundamental frequency is\",round(f3*2**7/1000,2), \"kHz\"\n", + "print \"output voltage is\",round(Vo*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.9,Page 491" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 127.32 V\n", + "\n", + "Va-d @ 200Vin=4.2V\n", + "\n", + "\n", + "pick R1=47kohm\n", + "current through dividers is 2.62 mA\n", + "R2 is 1.6 kohm\n", + "capacitor is 27.01 microF\n" + ] + } + ], + "source": [ + "#finding resistances,capacitor,average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#load voltage\n", + "f=60.0;#Hz\n", + "Vp=200.0;#V\n", + "Vd=5.0;#V\n", + "\n", + "\n", + "#calculation\n", + "Vdc=2*Vp/pi;\n", + "Va=4.2;\n", + "R1=47.0;\n", + "I=(Vdc-Va)/R1;\n", + "R2=Va/I;\n", + "K=1.0/(1/R1+1/R2)# R1 \\\\ R2\n", + "C=1.0/2/pi/3.8/K;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Vdc,2), \"V\"\n", + "print('\\nVa-d @ 200Vin=4.2V')\n", + "print('\\n\\npick R1=47kohm')\n", + "print \"current through dividers is\",round(I,2), \"mA\"\n", + "print \"R2 is\",round(R2,2), \"kohm\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9_1.ipynb b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9_1.ipynb new file mode 100644 index 00000000..d50d131a --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/Chapter9_1.ipynb @@ -0,0 +1,405 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Power Conversation and Motor Drive Operations" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.1,Page 457" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "peak voltage is 37.6 V\n", + "load voltage is 35.7 V\n", + "ripple voltage is 3.96 V\n", + "approx. load voltage is 35.62 V\n" + ] + } + ], + "source": [ + "#finding peak,load,ripple voltages\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=28.0;#V\n", + "C=4700.0;#microF\n", + "R=16.0;#load\n", + "f=120.0;#hertz\n", + "\n", + "#calculation\n", + "Vp=V*2**.5-2;\n", + "Vd=0.95*Vp;\n", + "Id=Vd/R;\n", + "v=Id/f/C;\n", + "#approximation\n", + "Vd1=Vp-v*1e6/2;\n", + "\n", + "#result\n", + "print \"peak voltage is\",round(Vp,2), \"V\"\n", + "print \"load voltage is\",round(Vd,1), \"V\"\n", + "print \"ripple voltage is\",round(v*1e6,2), \"V\"\n", + "print \"approx. load voltage is\",round(Vd1,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2,Page 459" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zth is 1.0 + 1.0 in ohm\n", + "inductor is 2.65 mH\n" + ] + } + ], + "source": [ + "#finding inductor,Zth\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V1=120.0;#pri voltage\n", + "V2=28.0;#sec voltage\n", + "I=2.0;#pri current\n", + "f=60.0;#Hz\n", + "Vth=28.8;#open voltage\n", + "V3=12.1;#pri-short voltage\n", + "Is=2.0;#short current at 45 degree\n", + "\n", + "#calculation\n", + "Zi=(V2*V3)/V1/Is*cos(45*pi/180);\n", + "Zj=(V2*V3)/V1/Is*sin(45*pi/180);\n", + "L=Zi/(2*pi*f);\n", + "\n", + "#result\n", + "print'Zth is',round(Zi),'+',round(Zj),'in ohm'\n", + "print \"inductor is\",round(L*1000,2), \"mH\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.4,Page 463" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "power factor is 0.32\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I1=1.8;#current\n", + "R=16.0;#resistance\n", + "I2=5.7;#A\n", + "V=28.8;#Voltage\n", + "\n", + "#calculation\n", + "P=I1**2*R;\n", + "S=I2*V;\n", + "Pf=P/S;\n", + "\n", + "#result\n", + "print \"power factor is\",round(Pf,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.5, Page 468" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "aparrent power is 8.14 kVA\n", + "dissipated power is 7.84 kW\n", + "power factor is 0.96\n" + ] + } + ], + "source": [ + "#finding power factor\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "I=22.6;#current\n", + "I2=28.00;\n", + "V=120.0;#Voltage\n", + "V2=280.0;\n", + "\n", + "#calculation\n", + "Pt=3*I*V;\n", + "Pl=I2*V2;\n", + "Pf=Pl/Pt;\n", + "\n", + "#result\n", + "print \"aparrent power is\",round(Pt/1000,2),\"kVA\"\n", + "print \"dissipated power is\",round(Pl/1000,2),\"kW\"\n", + "print \"power factor is\",round(Pl/Pt,2)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.6,Page 474" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ratio is 0.72\n", + "firing angle is 58 degrees\n", + "dc voltage is 148.85 V\n", + "time delay is 2.69 ms\n" + ] + } + ], + "source": [ + "#finding firing angle, time delay,Vd\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=208.0;#voltage\n", + "R=100.0;#load\n", + "Vd=150.0;#V\n", + "\n", + "#calculation\n", + "r=Vd/V;\n", + "a=58;#degree\n", + "Vd=3*2**.5*208*(cos(pi/3+a*pi/180)-cos(2*pi/3+a*pi/180))/pi;\n", + "t=a*16.7/360;\n", + "\n", + "#result\n", + "print \"ratio is\",round(r,2)\n", + "print('firing angle is 58 degrees');\n", + "print \"dc voltage is\",round(Vd,2), \"V\"\n", + "print \"time delay is\",round(t,2), \"ms\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.7,Page 480" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "max. current is 41.67 A\n", + "dissipated power is 8.68 W\n" + ] + } + ], + "source": [ + "#finding maximum current and power dissipated\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "P=150.0;#power\n", + "V=8.0;#voltage\n", + "R=.01;#resistance\n", + "D=.5;#duty cycle\n", + "\n", + "#calculation\n", + "I=P/.9/D/V;\n", + "Ir=I*D**.5;\n", + "Pq=Ir**2*R;\n", + "\n", + "#result\n", + "print \"max. current is\",round(I,2), \"A\"\n", + "print \"dissipated power is\",round(Pq,2),\"W\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.8,Page 489" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "pwm fundamental frequency is 30.72 kHz\n", + "output voltage is 9.46 V\n" + ] + } + ], + "source": [ + "#finding fundamental frequency and output voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "f1=60.0;#frequency\n", + "V=150.0;#voltage\n", + "f2=31.0;#kHz\n", + "\n", + "#calculation\n", + "f3=f1*4;\n", + "Vo=V*10**(-4.2);\n", + "\n", + "#result\n", + "print \"pwm fundamental frequency is\",round(f3*2**7/1000,2), \"kHz\"\n", + "print \"output voltage is\",round(Vo*1000,2), \"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 9.9,Page 491" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "average voltage is 127.32 V\n", + "\n", + "Va-d @ 200Vin=4.2V\n", + "\n", + "\n", + "pick R1=47kohm\n", + "current through dividers is 2.62 mA\n", + "R2 is 1.6 kohm\n", + "capacitor is 27.01 microF\n" + ] + } + ], + "source": [ + "#finding resistances,capacitor,average voltage\n", + "\n", + "#initialisation of variable\n", + "from math import pi,tan,sqrt,sin,cos,acos,atan\n", + "V=120.0;#load voltage\n", + "f=60.0;#Hz\n", + "Vp=200.0;#V\n", + "Vd=5.0;#V\n", + "\n", + "\n", + "#calculation\n", + "Vdc=2*Vp/pi;\n", + "Va=4.2;\n", + "R1=47.0;\n", + "I=(Vdc-Va)/R1;\n", + "R2=Va/I;\n", + "K=1.0/(1/R1+1/R2)# R1 \\\\ R2\n", + "C=1.0/2/pi/3.8/K;\n", + "\n", + "#result\n", + "print \"average voltage is\",round(Vdc,2), \"V\"\n", + "print('\\nVa-d @ 200Vin=4.2V')\n", + "print('\\n\\npick R1=47kohm')\n", + "print \"current through dividers is\",round(I,2), \"mA\"\n", + "print \"R2 is\",round(R2,2), \"kohm\"\n", + "print \"capacitor is\",round(C*1000,2), \"microF\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/README.txt b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/README.txt new file mode 100644 index 00000000..34ec8e1d --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/README.txt @@ -0,0 +1,10 @@ +Contributed By: Amit Kumar Das +Course: btech +College/Institute/Organization: ISM Dhanbad +Department/Designation: Mining Machinery Engineering +Book Title: Power Electronics: Principles & Applications +Author: J. M. Jacob +Publisher: Cengage Learning, New Delhi +Year of publication: 2009 +Isbn: 9788131501184 +Edition: 4
\ No newline at end of file diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1.png Binary files differnew file mode 100644 index 00000000..5199042e --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_1.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_1.png Binary files differnew file mode 100644 index 00000000..5199042e --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_1.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_2.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_2.png Binary files differnew file mode 100644 index 00000000..5199042e --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/1_2.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2.png Binary files differnew file mode 100644 index 00000000..76933e63 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2_1.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2_1.png Binary files differnew file mode 100644 index 00000000..76933e63 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/2_1.png diff --git a/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/3.png b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/3.png Binary files differnew file mode 100644 index 00000000..607dd0a1 --- /dev/null +++ b/Power_Electronics:_Principles_&_Applications_by_J._M._Jacob_/screenshots/3.png diff --git a/Principles_of_Data_structures_using_C_and_C++_by_Vinu_V_Das/README.txt b/Principles_of_Data_structures_using_C_and_C++_by_Vinu_V_Das/README.txt new file mode 100644 index 00000000..7ae5982c --- /dev/null +++ b/Principles_of_Data_structures_using_C_and_C++_by_Vinu_V_Das/README.txt @@ -0,0 +1,10 @@ +Contributed By: Tejas Dodiya +Course: others +College/Institute/Organization: Freelance +Department/Designation: Science +Book Title: Principles of Data structures using C and C++ +Author: Vinu V Das +Publisher: New Age International +Year of publication: 2008 +Isbn: 978-81-224-2864-3 +Edition: -
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_5.ipynb new file mode 100644 index 00000000..7cd2fd61 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.10_5.ipynb @@ -0,0 +1,280 @@ +{
+ "metadata": {
+ "name": "chapter no.10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 10:Theory of Failures"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.1,Page No.401"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P_e=300 #N/mm**2 #Elastic Limit in tension\n",
+ "FOS=3 #Factor of safety\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "P=12*10**3 #N Pull \n",
+ "Q=6*10**3 #N #Shear force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Direct stress\n",
+ "#P_x=P*(pi*4**-1*d**3)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P_x=48*10**3\n",
+ "\n",
+ "#Now shear stress at the centre of bolt\n",
+ "#q=4*3**-1*q_av\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q=32*10**3*(pi*d**2)**-1\n",
+ "\n",
+ "#Principal stresses are\n",
+ "#P1=P_x*2**-1+((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#p1=20371.833*(d**2)**-1\n",
+ "\n",
+ "#P2=P_x*2**-1-((P_x*2**-1)**2+q**2)**0.5\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-5092.984*(d**2)**-1\n",
+ "\n",
+ "#q_max=((P_x*2**-1)**2+q**2)**0.5\n",
+ "\n",
+ "#From Max Principal stress theory\n",
+ "#Permissible stress in Tension\n",
+ "P1=100 #N/mm**2 \n",
+ "d=(20371.833*P1**-1)**0.5\n",
+ "\n",
+ "#Max strain theory\n",
+ "#e_max=P1*E**-1-mu*P2*E**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#e_max=21899.728*(d**2*E)**-1\n",
+ "\n",
+ "#According to this theory,the design condition is\n",
+ "#e_max=P_e*(E*FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(21899.728*3*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#e_max=shear stress at elastic*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(12732.421*6*300**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Bolt by:Max Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max strain theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Bolt by:Max Principal stress theory 14.27 mm\n",
+ " :Max strain theory 14.8 mm\n",
+ " :Max shear stress theory 15.96 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.2.Page No.402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M=40*10**6 #N-mm #Bending moment\n",
+ "T=10*10**6 #N-mm #TOrque\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "P_e=200 #N/mm**2 #Stress at Elastic Limit\n",
+ "FOS=2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let d be the diameter of the shaft\n",
+ "\n",
+ "#Principal stresses are given by\n",
+ "\n",
+ "#P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P1=4.13706*10**8*(d**3)**-1 ............................(1)\n",
+ "\n",
+ "#P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5)\n",
+ "#After substituting values and further simplifying we get\n",
+ "#P2=-6269718*(pi*d**3)**-1 ..............................(2)\n",
+ "\n",
+ "#q_max=(P1-P2)*2**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#q_max=2.09988*10**8*(d**3)**-1\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#P1=P_e*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d=(4.13706*10**8*2*200**-1)**0.33333 #mm \n",
+ "\n",
+ "#Max shear stress theory\n",
+ "#q_max=shear stress at elastic limit*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d2=(2.09988*10**8*4*200**-1)**0.33333\n",
+ "\n",
+ "#Max strain energy theory\n",
+ "#P_3=0\n",
+ "#P1**2+P2**2-2*mu*P1*P2=P_e**2*(FOS)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d3=(8.62444*10**12)**0.166666\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of shaft according to:MAx Principal stress theory\",round(d,2),\"mm\"\n",
+ "print\" :Max shear stress theory\",round(d2,2),\"mm\"\n",
+ "print\" :Max strain energy theory\",round(d3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft according to:MAx Principal stress theory 160.52 mm\n",
+ " :Max shear stress theory 161.33 mm\n",
+ " :Max strain energy theory 143.2 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.10.10.3,Page No.403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "f_x=40 #N/mm**2 #Internal Fliud Pressure\n",
+ "d1=200 #mm #Internal Diameter\n",
+ "r1=d1*2**-1 #mm #Radius\n",
+ "q=300 #N/mm**2 #Tensile stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have,\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#f_x=b*(x**2)**-1+a ..........................(1)\n",
+ "\n",
+ "#Radial Pressure\n",
+ "#p_x=b*(x**2)**-1-a .........................(2)\n",
+ "\n",
+ "#the boundary conditions are\n",
+ "x=d1*2**-1 #mm \n",
+ "#After sub values in equation 1 and further simplifying we get\n",
+ "#40=b*100**-1-a ..........................(3)\n",
+ "\n",
+ "#Max Principal stress theory\n",
+ "#q*(FOS)**-1=b*100**2+a ..................(4)\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "a=80*2**-1\n",
+ "#Sub value of a in equation 3 we get\n",
+ "b=(f_x+a)*100**2\n",
+ "\n",
+ "#At outer edge where x=r_0 pressure is zero\n",
+ "r_0=(b*a**-1)**0.5 #mm\n",
+ "\n",
+ "#thickness\n",
+ "t=r_0-r1 #mm\n",
+ "\n",
+ "#Max shear stress theory\n",
+ "P1=b*(100**2)**-1+a #Max hoop stress\n",
+ "P2=-40 #pressure at int radius (since P2 is compressive)\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(P1-P2)*2**-1\n",
+ "\n",
+ "#According max shear theory the design condition is\n",
+ "#q_max=P_e*2**-1*(FOS)**-1\n",
+ "#After sub values in equation we get and further simplifying we get\n",
+ "#80=b*(100**2)**-1+a\n",
+ "#After sub values in equation 1 and 3 and further simplifying we get\n",
+ "b2=120*100**2*2**-1\n",
+ "\n",
+ "#from equation(3)\n",
+ "a2=120*2**-1-a\n",
+ "\n",
+ "#At outer radius r_0,radial pressure=0\n",
+ "r_02=(b2*a2**-1)**0.5\n",
+ "\n",
+ "#thickness\n",
+ "t2=r_02-r1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of metal by:Max Principal stress theory\",round(t,2),\"mm\"\n",
+ "print\" :Max shear stress thoery\",round(t2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of metal by:Max Principal stress theory 41.42 mm\n",
+ " :Max shear stress thoery 73.21 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_5.ipynb new file mode 100644 index 00000000..a37ee3c9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.2_5.ipynb @@ -0,0 +1,2742 @@ +{
+ "metadata": {
+ "name": "chapter no.2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 2:Simple Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.1,Page No.14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "P=45*10**3 #N #Load\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity of rod\n",
+ "L=500 #mm #Length of rod\n",
+ "d=20 #mm #Diameter of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #mm**2 #Area of circular rod\n",
+ "p=P*A**-1 #N/mm**2 #stress\n",
+ "e=p*E**-1 #strain\n",
+ "dell_l=(P*L)*(A*E)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"The stress in bar due to Load is\",round(p,5),\"N/mm\"\n",
+ "print\"The strain in bar due to Load is\",round(e,5),\"N/mm\"\n",
+ "print\"The Elongation in bar due to Load is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The stress in bar due to Load is 143.23945 N/mm\n",
+ "The strain in bar due to Load is 0.00072 N/mm\n",
+ "The Elongation in bar due to Load is 0.36 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.2,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A=15*0.75 #mm**2 #area of steel tape\n",
+ "P=100 #N #Force apllied\n",
+ "L=30*10**3 #mm #Length of tape\n",
+ "E=200*10**3 #N/m**2 #Modulus of Elasticity of steel tape\n",
+ "AB=150 #m #Measurement of Line AB \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "dell_l=P*L*(A*E)**-1 #mm #Elongation\n",
+ "l=L+dell_l*10**-3 #mm #Actual Length \n",
+ "AB1=AB*l*L**-1 #m Actual Length of AB\n",
+ "\n",
+ "#Result\n",
+ "print\"The Actual Length of Line AB is\",round(AB1,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Actual Length of Line AB is 150.0 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.3,Page No.15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let y be the yield stress\n",
+ "\n",
+ "y=250 #N/mm**2 #yield stress\n",
+ "FOS=1.75 #Factor of safety\n",
+ "P=140*10**3 #N #compressive Load\n",
+ "D=101.6 #mm #External diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "p=y*(FOS)**-1 #N/mm**2 #Permissible stress\n",
+ "A=P*p**-1 #mm**2 #Area of hollow tube\n",
+ "\n",
+ "#Let d be the internal diameter of tube\n",
+ "d=-((A*4*(pi)**-1)-D**2)\n",
+ "X=d**0.5\n",
+ "t=(D-X)*2**-1 #mm #Thickness of steel tube\n",
+ "\n",
+ "#result\n",
+ "print\"The thickness of steel tube is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of steel tube is 3.17 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.4,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #diameter of steel\n",
+ "L=200 #mm #length of stee\n",
+ "P=80*10**3 #KN #Load\n",
+ "P1=160*10**3 #N #Load at Elastic Limit\n",
+ "P2=180*10**3 #N #Max Load\n",
+ "L1=56 #mm #Total Extension\n",
+ "dell_l=0.16 #mm #Extension\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*d**2*4**-1 #Area of steel #mm**2\n",
+ "\n",
+ "p=P1*A**-1 #Stress at Elastic Limit #N/mm**2\n",
+ "E=P*L*(A*dell_l)**-1\n",
+ "\n",
+ "#result\n",
+ "print E"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "203718.327158\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.5,Page No.16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "d2=14.7 #mm #Diameter at neck\n",
+ "L=200 #mm #guage Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,10,20,30,40,50,60]\n",
+ "Y1=[0,32,64,95,127,160,190]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Extension in divisions\")\n",
+ "plt.ylabel(\"Load in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Bar\n",
+ "A2=pi*4**-1*d2**2\n",
+ "\n",
+ "P=45 #KN #Load obtained from graph\n",
+ "dell=0.143 #mm #Divisions\n",
+ "\n",
+ "#Modulus of Elasticity\n",
+ "E=P*L*(dell*A)**-1 \n",
+ "\n",
+ "BL=93*10**3 #N #Breaking Load\n",
+ "\n",
+ "#Nominal stress at Breaking point\n",
+ "sigma=BL*A**-1 #KN/mm**2 \n",
+ "\n",
+ "#True stress at breaking Point\n",
+ "sigma1=BL*A2**-1\n",
+ "\n",
+ "#Percentage Elongation \n",
+ "dell_l=(A-A2)*A**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Value of ELongation is\",round(E,2),\"N/mm**2\"\n",
+ "print\"The Nominal stress at the Breaking Point\",round(sigma,2),\"KN/mm**2\"\n",
+ "print\"The True stress at the Breaking Point\",round(sigma1,2),\"KN/mm**2\"\n",
+ "print\"The Percentage Reduction in Area is\",round(dell_l,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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gwAFOnDjBRx99xNatWys876vr+8tf/kKzZs2Ii4vDVHG/iq+u7bL09HT279/P\nhg0bWLx4Mdu3b6/wvC+v79KlS+zbt49nnnmGffv20ahRo0otpOtZn08kh9DQUHJzc53/zs3NJSws\nzMaI3CMkJISTJ08CUFBQQLNmzWyO6MaUlpYyaNAghg8fzsCBAwH/WyNAkyZNGDBgABkZGX6xvp07\nd7J+/Xpat25NcnIyW7ZsYfjw4X6xtstatGgBQNOmTXn44YfZvXu336wvLCyMsLAw7rzzTgAGDx7M\nvn37aN68eY3W5xPJoWvXrhw7doycnBwuXrzIH//4R5KSkuwOq84lJSWRmpoKQGpqqvMD1RcZYxgz\nZgwRERFMnDjR+bi/rPHUqVPOuz2+/fZbNm/eTFxcnF+sb+7cueTm5pKdnc2qVau47777ePfdd/1i\nbQDnz5/n7NmzAJSUlLBp0yaioqL8Zn3NmzenVatWHD16FIAPPviAzp07k5iYWLP1ueF6iFv87W9/\nM+3btzdt2rQxc+fOtTucG/b444+bFi1amKCgIBMWFmbeeecd8/XXX5s+ffqYdu3amYSEBPPNN9/Y\nHWatbd++3TgcDhMTE2NiY2NNbGys2bBhg9+s8dChQyYuLs7ExMSYqKgo84tf/MIYY/xmfZelpaWZ\nxMREY4z/rO3zzz83MTExJiYmxnTu3Nn5eeIv6zPGmAMHDpiuXbua6Oho8/DDD5vi4uIar0+b4ERE\npBKfaCuJiIhnKTmIiEglSg4iIlKJkoOIiFSi5CAiIpUoOYiISCVKDmKL+vXrExcX5/z6xS9+cc3X\nz507t85jyMjIYMKECXXyXgMGDODMmTO1/vng4GAA8vPzefTRR6/52vfff/+ax9bX5bokcGmfg9ii\ncePGzl2q7ni9r/H39YnvUeUgXuP06dN07NjRue0/OTmZpUuXMm3aNL799lvi4uIYPnw4AMuXL6d7\n9+7ExcXx05/+lPLycsD6C3z69OnExsbSo0cPvvzySwD+9Kc/ERUVRWxsLPHx8QCkpaVVGGQzcOBA\nYmJi6NGjB5mZmQDMmjWL0aNH07t3b9q0acOiRYtcxh4eHk5RURE5OTl06tSJp556isjISPr378+F\nCxcqvT47O5sePXoQHR3N9OnTnY/n5OQ4B0DdddddZGVlOZ+Lj48nIyOD3/3udzz77LNuWVdJSQkD\nBgwgNjaWqKgoVq9efd3//cTPeGAnt0gl9evXdx6rERsba1avXm2MMWbz5s2mR48eZuXKleb+++93\nvj44ONgVmkKpAAADpklEQVT5fVZWlklMTDSXLl0yxhgzbtw48/vf/94YY4zD4TB/+ctfjDHGTJky\nxcyZM8cYY0xUVJTJz883xhhz+vRpY4wxW7dudc4qGD9+vHn55ZeNMcZs2bLFxMbGGmOMmTlzpunZ\ns6e5ePGiOXXqlLntttucv/dK4eHh5uuvvzbZ2dmmQYMG5uDBg8YYY4YMGWKWL19e6fWJiYnm3Xff\nNcYYs3jxYuf6rpzx8frrr5uZM2caY4zJz893nr//29/+1jz77LN1vq7S0lKzZs0aM3bsWGecl99T\nAo8qB7HF9773Pfbv3+/8utxn79u3L5GRkYwfP56lS5e6/NkPP/yQjIwMunbtSlxcHFu2bCE7OxuA\nhg0bMmDAAAC6dOlCTk4OAD179mTkyJEsXbqUS5cuVXrP9PR0Z1XSu3dvvv76a86ePYvD4WDAgAEE\nBQVx22230axZs2rPwW/dujXR0dGVYrjSzp07SU5OBmDYsGEu3+fRRx9lzZo1AKxevbrCtQjz725w\nXa7ryy+/JDo6ms2bNzN16lR27NjBD37wg2uuVfyXkoN4lfLyco4cOUKjRo0oKiqq8nUjR450JpZ/\n/vOfzJgxA7DGk15Wr1495wfmm2++yZw5c8jNzaVLly4u39tUcfmtYcOGzu/r16/v8kP4SjfddFON\nXl+V0NBQbrvtNjIzM1m9ejWPPfYYUHG+SV2vq127duzfv5+oqCimT5/OK6+8UqvYxfcpOYhXef31\n1+ncuTMrVqxg1KhRzg/WoKAg5/d9+vRhzZo1fPXVV4DVVz9+/Pg13/ezzz6jW7duzJ49m6ZNm3Li\nxIkKz/fq1YsVK1YAVs++adOmNG7cuMoP1hvVs2dPVq1aBeD8va489thjzJ8/nzNnzhAZGQlU/LCv\n63UVFBRw880388QTT/D888+zb9++G1qn+K4GdgcggenyBebL7r//fn7yk5+wbNky9uzZQ6NGjbj3\n3nt59dVXmTlzJk899RTR0dF06dKFd999lzlz5tCvXz/Ky8sJCgpiyZIl/Md//EeFv6qvnHY1ZcoU\njh07hjGGvn37Eh0dzbZt25zPX75AGxMTQ6NGjZzn3l/vRLCrf29Vz122YMEChg4dyvz583nooYeq\n/PnBgwczYcIEZ2Xk7nVlZmbywgsvUK9ePRo2bMibb75Z7drFP+lWVhERqURtJRERqUTJQUREKlFy\nEBGRSpQcRESkEiUHERGpRMlBREQqUXIQEZFKlBxERKSS/wPlCfw1/C4iHwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x4d1d370>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Value of ELongation is 200.33 N/mm**2\n",
+ "The Nominal stress at the Breaking Point 296.03 KN/mm**2\n",
+ "The True stress at the Breaking Point 547.97 KN/mm**2\n",
+ "The Percentage Reduction in Area is 45.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.6,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=40*10**3 #N #Load\n",
+ "L1=160 #mm #Length of Bar1\n",
+ "L2=240 #mm #Length of bar2\n",
+ "L3=160 #mm #Length of bar3\n",
+ "d1=25 #mm #Diameter of Bar1\n",
+ "d2=20 #mm #diameter of bar2\n",
+ "d3=25 #mm #diameter of bar3\n",
+ "dell_l=0.285 #mm #Total Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=P*4*(dell_l*pi)**-1*(L1*(d1**2)**-1+L2*(d2**2)**-1+L3*(d3**2)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Young's Modulus of the material\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Young's Modulus of the material 198714.72 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.7,Page No.19"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E1=2*10**5 #N/mm**2 #modulus of Elasticity of material1\n",
+ "E2=1*10**5 #N/mm**2 #modulus of Elasticity of material2\n",
+ "P=25*10**3 #N #Load \n",
+ "t=20 #mm #thickness of material\n",
+ "b1=40 #mm #width of material1\n",
+ "b2=30 #mm #width of material2\n",
+ "L1=500 #mm #Length of material1\n",
+ "L2=750 #mm #Length of material2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=b1*t #mm**2 #Area of materila1\n",
+ "A2=b2*t #mm**2 #Area of material2\n",
+ "\n",
+ "dell_l1=P*L1*(A1*E1)**-1 #Extension of Portion1\n",
+ "dell_l2=P*L2*(A2*E2)**-1 #Extension of portion2\n",
+ "\n",
+ "#Total Extension of Bar is\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension of the Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension of the Bar is 0.39 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.8,Page No.20"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of Bar\n",
+ "l=400 #mm #Length upto which bire is drilled \n",
+ "D=30 #mm #diameter of bar\n",
+ "d1=10 #mm #diameter of bore\n",
+ "P=25*10**3 #N #Load\n",
+ "dell_l=0.185 #mm #Extension of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L1=L-l #Length of bar above the bore\n",
+ "L2=400 #mm #Length of bore\n",
+ "\n",
+ "A1=pi*4**-1*D**2 #Area of bar\n",
+ "A2=pi*4**-1*(D**2-d1**2) #Area of bore\n",
+ "\n",
+ "E=P*dell_l**-1*(L1*A1**-1+L2*A2**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Modulus of ELasticity is\",round(E,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Modulus of ELasticity is 200735.96 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.11,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=10 #mm #Thickness of steel\n",
+ "b1=60 #mm #width of plate1\n",
+ "b2=40 #mm #width of plate2\n",
+ "P=60*10**3 #Load\n",
+ "L=600 #mm #Length of plate\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Extension of taperong bar of rectangular section\n",
+ "dell_l=P*L*(t*E*(b1-b2))**-1*log(b1*b2**-1)\n",
+ "\n",
+ "A_av=(b1*t+b2*t)*2**-1 #Average Area #mm**2\n",
+ "dell_l2=P*L*(A_av*E)**-1 \n",
+ "\n",
+ "#PErcentage Error\n",
+ "e=(dell_l-dell_l2)*(dell_l)**-1*100\n",
+ "\n",
+ "#Result\n",
+ "print\"The Percentage Error is\",round(e,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Percentage Error is 1.35\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.12,Page No.23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1.5 #m #Length of steel bar\n",
+ "L1=1000 #m0 #Length of steel bar 1\n",
+ "L2=500 #m #Length of steel bar 2\n",
+ "d1=40 #Diameter of steel bar 1\n",
+ "d2=20 #diameter of steel bar 2\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "P=160*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A1=pi*4**-1*d1**2 #Area of Portion 1\n",
+ "\n",
+ "#Extension of uniform Portion 1\n",
+ "dell_l1=P*L1*(A1*E)**-1 #mm\n",
+ "\n",
+ "#Extension of uniform Portion 2\n",
+ "dell_l2=4*P*L2*(pi*d1*d2*E)**-1 #mm\n",
+ "\n",
+ "#Total Extension of Bar\n",
+ "dell_l=dell_l1+dell_l2\n",
+ "\n",
+ "#Result\n",
+ "print\"The Elongation of the Bar is\",round(dell_l,2),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.14,Page No.25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Portion AB\n",
+ "L_AB=600 #mm #Length of AB\n",
+ "A_AB=40*40 #mm**2 #Cross-section Area of AB\n",
+ "\n",
+ "#Portion BC\n",
+ "L_BC=800 #mm #Length of BC\n",
+ "A_BC=30*30 #mm #Length of BC\n",
+ "\n",
+ "#Portion CD\n",
+ "L_CD=1000 #mm #Length of CD\n",
+ "A_CD=20*20 #mm #Area of CD\n",
+ "\n",
+ "P1=80*10**3 #N #Load1\n",
+ "P2=60*10**3 #N #Load2\n",
+ "P3=40*10**3 #N #Load3\n",
+ "\n",
+ "E=2*10**5 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P4=P1-P2+P3 #Load4\n",
+ "\n",
+ "#Now Force in AB\n",
+ "F_AB=P1\n",
+ "\n",
+ "#Force in BC\n",
+ "F_BC=P1-P2\n",
+ "\n",
+ "#Force in CD\n",
+ "F_CD=P4\n",
+ "\n",
+ "#Extension of AB\n",
+ "dell_l_AB=F_AB*L_AB*(A_AB*E)**-1\n",
+ "\n",
+ "#Extension of BC\n",
+ "dell_l_BC=F_BC*L_BC*(A_BC*E)**-1\n",
+ "\n",
+ "#Extension of CD\n",
+ "dell_l_CD=F_CD*L_CD*(A_CD*E)**-1\n",
+ "\n",
+ "#Total Extension\n",
+ "dell_l=dell_l_AB+dell_l_BC+dell_l_CD\n",
+ "\n",
+ "#Result\n",
+ "print\"The Total Extension in Bar is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Total Extension in Bar is 0.99 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.15,Page No.26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Length of bar\n",
+ "F1=30*10**3 #N #Force acting on the bar\n",
+ "F2=60*10**3 #N #force acting on the bar\n",
+ "L=800 #mm #Length of bar\n",
+ "d=25 #mm #diameter of bar\n",
+ "L_AC=275 #mm #Length of AC\n",
+ "L_CD=150 #mm #Length of CD\n",
+ "L_DB=375 #mm #Length of DB\n",
+ "E=2*10**5 #Pa #Modulus of elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P be the Reaction on tne Bar from support at A\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "#dell_l_AC1=P*L_AC*(A*E)**-1\n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "#dell_l_CD1=(30+P)*L_CD*(A*E)**-1\n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "#dell_l_DB1=(30-P)*L_DB*(A*E)**-1\n",
+ "\n",
+ "#Total Extensions=1*(A*E)**-1*(P*L_AC-(30+P)*L_CD+(30-P)*L_DB)\n",
+ "#As Supports are unyielding,Total Extensions=0\n",
+ "\n",
+ "#After substituting values in above equation and Further simplifying we get\n",
+ "P=(30*375-150*30)*800**-1\n",
+ "\n",
+ "#Reaction of support A\n",
+ "R_A=P\n",
+ "\n",
+ "#Reaction of support B\n",
+ "R_B=30-P\n",
+ "\n",
+ "#Cross-sectional Area\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Stress in Portion AC\n",
+ "sigma1=P*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion CD\n",
+ "sigma2=(30+P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in Portion DB\n",
+ "sigma3=(30-P)*10**3*A**-1 #N/mm**2\n",
+ "\n",
+ "#Shortening of Portion AC\n",
+ "dell_l_AC2=P*10**3*L_AC*(A*E)**-1 #mm \n",
+ "\n",
+ "#Shortening of Portion CD\n",
+ "dell_l_CD2=(30+P)*10**3*L_CD*(A*E)**-1 #mm \n",
+ "\n",
+ "#Extension of Portion DB\n",
+ "dell_l_DB2=(30-P)*10**3*L_DB*(A*E)**-1 #mm \n",
+ "\n",
+ "#result\n",
+ "print\"The Reactios at two Ends are:R_A\",round(R_A,2),\"KN\"\n",
+ "print\" :R_B\",round(R_B,2),\"KN\"\n",
+ "print\"Stress in Portion AC\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion CD\",round(sigma2,2),\"N/mm**2\"\n",
+ "print\"Stress in Portion DB\",round(sigma3,2),\"N/mm**2\"\n",
+ "print\"Shortening of Portion AC\",round(dell_l_AC2,3),\"mm\"\n",
+ "print\"Shortening of Portion CD\",round(dell_l_CD2,3),\"mm\"\n",
+ "print\"Shortening of Portion DB\",round(dell_l_DB2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Reactios at two Ends are:R_A 8.44 KN\n",
+ " :R_B 21.56 KN\n",
+ "Stress in Portion AC 17.19 N/mm**2\n",
+ "Stress in Portion CD 78.3 N/mm**2\n",
+ "Stress in Portion DB 43.93 N/mm**2\n",
+ "Shortening of Portion AC 0.024 mm\n",
+ "Shortening of Portion CD 0.059 mm\n",
+ "Shortening of Portion DB 0.082 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.19,Page No.29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "h=4 #m #height of Pillars\n",
+ "P=20 #KN #Load at M\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_A,P_B,P_C,P_D be the forces introduced in the Pillars\n",
+ "#Sun of All Vertical Forces\n",
+ "#P_A+P_B+P_C+P_D=20 ....................(1)\n",
+ "\n",
+ "#Sum of moment about AB, we get\n",
+ "#P_D+P_C=12 ....................(2)\n",
+ "\n",
+ "#Sum of Moment about AD\n",
+ "#P_C+P_B=8 ....................(3)\n",
+ "\n",
+ "#Let dell_l_A,dell_l_B,dell_l-C,dell_l_D be the deformations of Pillars A,B,C,D respectively\n",
+ "#Diagonals AC and BD will remain straight Lines even after the Load is applied.\n",
+ "#Deflection of central Point is given by (dell_l_A+dell_l_C)*2**-1 & (dell_l_B+dell_l_D)*2**-1\n",
+ "\n",
+ "#dell_l_A+dell_l_C=dell_l_B+ell_l_D\n",
+ "#P_A*L*(A*E)**-1+P_C*L*(A*E)**-1=P_B*L*(A*E)**-1+P_D*L*(A*E)**-1\n",
+ "\n",
+ "#Since Pillars are identical in Length,cross-sectional area,material Property\n",
+ "#P_A+P_C=P_B+P_D ..............(4)\n",
+ "\n",
+ "#From Equations 1 and 4 we get\n",
+ "#P_B+P_D=10 ....................(5)\n",
+ " \n",
+ "#Substracting Equation 3 from Equation 2 we get\n",
+ "#P_D-P_B=4 ....................(6)\n",
+ "\n",
+ "#Adding Equation 5 and 6 we get\n",
+ "\n",
+ "P_D=14*2**-1\n",
+ "P_C=12-P_D\n",
+ "P_B=8-P_C\n",
+ "\n",
+ "#Now substituting values of P_B,P_C,P_D in equation1 we get\n",
+ "P_A=20-(P_B+P_C+P_D)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Forces Developed in the Pillars are:P_A\",round(P_A,2),\"KN\"\n",
+ "print\" :P_B\",round(P_B,2),\"KN\"\n",
+ "print\" :P_C\",round(P_C,2),\"KN\"\n",
+ "print\" :P_D\",round(P_D,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Forces Developed in the Pillars are:P_A 5.0 KN\n",
+ " :P_B 3.0 KN\n",
+ " :P_C 5.0 KN\n",
+ " :P_D 7.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.20,Page No.31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "P=40*10**3 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt P_A.P_B,P_C,P_D be the forces developed in wires A,B,C,D respectively\n",
+ "\n",
+ "#Let sum of all Vertical Forces=0\n",
+ "#P_A+P_B+P_C+P_D=40 ..........................(1)\n",
+ "\n",
+ "#Let x be the distance between each wires\n",
+ "#sum of all moments=0\n",
+ "#P_B*x+P_C*2*x+P_D*3*x=40*2*x\n",
+ "\n",
+ "#After further simplifying we get\n",
+ "#P_B+2*P_C+3*P_D=80 ..........................(2)\n",
+ "\n",
+ "#As the equations of statics ae not enough to find unknowns,Consider compatibilit Equations\n",
+ "\n",
+ "#Let dell_l be the increse in elongation of wire\n",
+ "\n",
+ "#dell_l_B=dell_l_A+dell_l\n",
+ "#dell_l_C=dell_l_A+2*dell_l\n",
+ "#dell_l_D=dell_l_A+3*dell_l\n",
+ "\n",
+ "#Let P1 be the force required for the Elongation of wires,then\n",
+ "#P_B=P_A+P1 ]\n",
+ "#P_C=P_A+2*P1 ]\n",
+ "#P_D=P_A+3*P1 ] ................................(3) \n",
+ "\n",
+ "#from Equation (3) and (1) we get\n",
+ "#2*P_A+3*P1=20 ................................(4)\n",
+ "\n",
+ "#from Equation (3) and (2) we get\n",
+ "#6*P_A+14*P1=80 \n",
+ "\n",
+ "#subtracting 3 times equation (4) from (3) we get\n",
+ "P1=20*5**-1\n",
+ "\n",
+ "#from Equation 4 we get\n",
+ "P_A=(80-14*P1)*6**-1\n",
+ "P_B=P_A+P1\n",
+ "P_C=P_A+2*P1\n",
+ "P_D=P_A+3*P1\n",
+ "\n",
+ "#Let d be the diameter required,then\n",
+ "d=(P_D*10**3*4*(pi*150)**-1)**0.5\n",
+ "\n",
+ "#result\n",
+ "print\"The Required Diameter is\",round(d,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Required Diameter is 11.65 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.21,Page No.32"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=20*10**3 #N #Load\n",
+ "d=6 #mm #diameter of wire\n",
+ "E=2*10**5 #N/mm**2 \n",
+ "L_BO=4000 #mm #Length of BO\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let theta be the angle between OA and OB and also between OC and OB\n",
+ "theta=30\n",
+ "\n",
+ "#Let P_OA,P_OB,P_OC be the Forces introduced in wires OA,OB,OC respectively\n",
+ "#Due to symmetry P_OA=P_OC (same angles)\n",
+ "\n",
+ "#Sum of all Vertical Forces=0\n",
+ "#P_OA*cos(theta)+P_OB+P_OC*cos(theta)=P\n",
+ "\n",
+ "#After further simplifyinf we get\n",
+ "#2*P_OA*cos(theta)+P_OB=20 ...............(1)\n",
+ "\n",
+ "#Let oo1 be the extension of BO\n",
+ "#oo1=L_A1o1*(cos(theta))**-1\n",
+ "\n",
+ "#From relation we get\n",
+ "#P_OB*L_BO=P_OA*L_AO*(cos(theta))**-1\n",
+ "\n",
+ "#But L_AO=L_BO*(cos(theta))**-1\n",
+ "\n",
+ "#After substituting value of L_AO in above equation we get\n",
+ "#P_OB=0.75*P_OA .......................(2)\n",
+ "\n",
+ "#substituting in Equation 1 we get\n",
+ "#2*P_OA*cos(theta)+0.75*P_OA=20\n",
+ "\n",
+ "P_OA=20*(2*cos(theta*pi*180**-1)+0.75)**-1\n",
+ "\n",
+ "P_OB=0.75*P_OA\n",
+ "\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Vertical displacement of Load\n",
+ "dell_l_BO=P_OB*10**3*L_BO*(A*E)**-1\n",
+ " \n",
+ "#Result\n",
+ "print\"Forces in each wire is:P_OA\",round(P_OA,2),\"KN\"\n",
+ "print\" :P_OB\",round(P_OB,2),\"KN\"\n",
+ "print\"Vertical displacement of Loadis\",round(dell_l_BO,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Forces in each wire is:P_OA 8.06 KN\n",
+ " :P_OB 6.04 KN\n",
+ "Vertical displacement of Loadis 4.27 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.22,Page No.34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_s=L_a=L=500 #mm #Length of bar\n",
+ "A_a=50*20 #mm #Area of aluminium strip\n",
+ "A_s=50*15 #mm #Area of steel strip\n",
+ "P=50*10**3 #N #Load\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of aluminium \n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s br the Load shared by aluminium and steel strip\n",
+ "#P_a+P_s=P ..................(1)\n",
+ "\n",
+ "#For compatibility condition,dell_l_a=dell_l_s\n",
+ "#P_a*L_a*(A_a*E_a)**-1=P_s*L_s*(A_s*E_s)**-1 .....(2)\n",
+ "\n",
+ "#As L_a=L_s we get\n",
+ "#P_s=1.5*P_a .................(3)\n",
+ "\n",
+ "#From Equation 1 and 2 we get\n",
+ "P_a=P*2.5**-1\n",
+ "\n",
+ "#Substituting in equation 1 we get\n",
+ "P_s=P-P_a\n",
+ "\n",
+ "#stress in aluminium strip \n",
+ "sigma_a=P_a*A_a**-1\n",
+ "\n",
+ "#stress in steel strip\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Now from the relation we get\n",
+ "dell_l_a=dell_l_s=P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in Aluminium strip is\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\"Stress in steel strip is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"The Extension of the bar is\",round(dell_l_s,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Aluminium strip is 20.0 N/mm**2\n",
+ "Stress in steel strip is 40.0 N/mm**2\n",
+ "The Extension of the bar is 0.1 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.23,Page No.35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=20 #mm #Diameter of steel\n",
+ "D_Ci=20 #mm #Internal Diameter of Copper\n",
+ "t=5 #mm #THickness of copper bar\n",
+ "P=100*10**3 #N #Load\n",
+ "E_s=2*10**5 #N/mm**2 #modulus of elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of Copper\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2 #mm**2 #Area of steel\n",
+ "D_Ce=D_s+2*t #mm #External Diameterof Copper Tube\n",
+ "\n",
+ "A_c=pi*4**-1*(D_Ce**2-D_Ci**2) #mm**2 #Area of Copper\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#Let P_s and P_c be the Load shared by steel and copper in KN\n",
+ "#P_s+P_c=100 ....................................(1)\n",
+ "\n",
+ "#From compatibility Equation,dell_l_s=dell_l_c\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=1.3333*P_C \n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=100*2.3333**-1 #KN\n",
+ "P_s=100-P_c #KN\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 181.89 N/mm**2\n",
+ " :sigma_c 109.14 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.24,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_C=230*400 #mm #Area of column\n",
+ "D_s=12 #mm #Diameter of steel Bar\n",
+ "P=600*10**3 #N #Axial compression\n",
+ "#E_s*E_c=18.67\n",
+ "n=8 #number of steel Bars\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=pi*4**-1*D_s**2*n #Area of steel #mm**2 \n",
+ "A_c=A_C-A_s #mm**2 #Area of concrete\n",
+ "\n",
+ "#From static Equilibrium condition\n",
+ "#P_s+P_c=600 .........(1)\n",
+ "\n",
+ "#Now from compatibility Equation dell_l_s=dell_l_c we get,\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "\n",
+ "#Substituting values in above Equation we get\n",
+ "#P_s=0.1854*P_c\n",
+ "\n",
+ "#Now Substituting value of P_s in Equation (1),we get\n",
+ "P_c=600*1.1854**-1\n",
+ "P_s=600-P_c\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2\n",
+ "\n",
+ "#Stress in copper\n",
+ "sigma_c=P_c*10**3*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Two material are:sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\" :sigma_c\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Two material are:sigma_s 103.72 N/mm**2\n",
+ " :sigma_c 5.56 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.25,Page No.36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=200*10**3 #N #Load\n",
+ "A_a=1000 #mm**2 #Area of Aluminium\n",
+ "A_s=800 #mm**2 #Area of steel\n",
+ "E_a=1*10**5 #N/mm**2 #Modulus of Elasticity of Aluminium\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel\n",
+ "sigma_a1=65 #N/mm**2 #stress in aluminium\n",
+ "sigma_s1=150 #N/mm**2 #Stress in steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_a and P_s be the force in aluminium and steel pillar respectively\n",
+ "\n",
+ "#Now,sum of forces in Vertical direction we get\n",
+ "#2*P_a+P_s=200 .........................................(1)\n",
+ "\n",
+ "#By compatibility Equation dell_l_s=dell_l_a we get\n",
+ "#P_s=1.28*P_a ..........................................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_a=200*3.28**-1 #KN\n",
+ "P_s=200-2*P_a #KN\n",
+ "\n",
+ "#Stress developed in aluminium\n",
+ "sigma_a=P_a*10**3*A_a**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress developed in steel\n",
+ "sigma_s=P_s*10**3*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let sigma_a1 and sigma_s1 be the stresses in Aluminium and steel due to Additional LOad\n",
+ "\n",
+ "P_a1=sigma_a1*A_a #Load carrying capacity of aluminium\n",
+ "P_s1=1.28*P_a1\n",
+ "\n",
+ "#Total Load carrying capacity \n",
+ "P1=2*P_a1+P_s1 #N \n",
+ "\n",
+ "P_s2=sigma_s1*A_s #Load carrying capacity of steel\n",
+ "P_a2=P_s2*1.28**-1\n",
+ "\n",
+ "#Total Load carrying capacity\n",
+ "P2=2*P_a2+P_s2\n",
+ "\n",
+ "#Additional Load\n",
+ "P3=P1-P\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Each Pillar is:sigma_a\",round(sigma_a,2),\"N/mm**2\"\n",
+ "print\" :sigma_s\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Additional Load taken by pillars is\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "78.0487804878\n",
+ "Stresses Developed in Each Pillar is:sigma_a 60.98 N/mm**2\n",
+ " :sigma_s 97.56 N/mm**2\n",
+ "Additional Load taken by pillars is 13200.0 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.26,Page No.37"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of assembly\n",
+ "D=16 #mm #Diameter of steel bolt\n",
+ "Di=20 #mm #internal Diameter of copper tube\n",
+ "Do=30 #mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel\n",
+ "E_c=1.2*10**5 #N/mm**2 #Modulus of Elasticity of copper\n",
+ "p=2 #mm #Pitch of nut\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in bolt and P_c be the FOrce in copper tube\n",
+ "#P_s=-P_s\n",
+ "\n",
+ "dell=1*4**-1*2 #Quarter turn of nut total movement\n",
+ "\n",
+ "#dell=dell_s+dell_c\n",
+ "\n",
+ "#Area of steel\n",
+ "A_s=pi*4**-1*D**2\n",
+ "\n",
+ "#Area of copper\n",
+ "A_c=pi*4**-1*(Do**2-Di**2)\n",
+ "\n",
+ "#dell=P*L*(A_s*E_s)**-1+P*L*(A_c*E_c)**-1\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1*L**-1 #LOad\n",
+ "\n",
+ "P_s=P*A_s**-1\n",
+ "P_c=P*A_c**-1\n",
+ "\n",
+ "#result\n",
+ "print\"stress introduced in bolt is\",round(P_s,2),\"N/mm**2\"\n",
+ "print\"stress introduced in tube is\",round(P_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "stress introduced in bolt is 107.91 N/mm**2\n",
+ "stress introduced in tube is 55.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.27,Page No.39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=20 #mm #Diameter of Bolts\n",
+ "Di=25 #m #internal Diameter\n",
+ "t=10 #mm #Thickness of bolt\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=3 #mm #Pitch\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the Force in each bolt and P_c be the FOrce in copper tube\n",
+ "#From Static Equilibrium condition\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#As nut moves by 60 degree.If nut moves by 360 degree its Longitudinal movement is by 3 mm\n",
+ "dell=theta*360**-1*p\n",
+ "\n",
+ "#From Compatibility Equaton we get\n",
+ "#dell=dell_c+dell_s\n",
+ "\n",
+ "\n",
+ "A_s=pi*4**-1*"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.28,Page No.40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=9 #m #Length of rigid bar\n",
+ "L_b=3000 #Length of bar\n",
+ "A_b=1000 #mm**2 #Area of bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brasss bar\n",
+ "L_s=5000 #mm #Length of steel bar\n",
+ "A_s=445 #mm**2 #Area of steel bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of elasticity of steel bar\n",
+ "P=3000 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From static equilibrium Equation of the rod after appliying Load is\n",
+ "#P_b+P_s=P ......................(1)\n",
+ "\n",
+ "#P_b=1.8727*P_s ..................(2)\n",
+ "\n",
+ "#NOw substituting equation 2 in equation 1 we get\n",
+ "P_s=P*2.8727**-1\n",
+ "P_b=P-P_s\n",
+ "\n",
+ "d=P_s*L*P**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Distance at which Load applied even after which bar remains horizontal is\",round(d,2),\"m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance at which Load applied even after which bar remains horizontal is 3.13 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.29,Page No.41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "A_b=1000 #MM**2 #Area of brass bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass\n",
+ "A_s=600 #N/mm**2 #Area of steel rod\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of eLasticity of steel bar\n",
+ "P=10*10**2 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#Now taking moment about A we get static Equilibrium condition as\n",
+ "#P_b+2*P_s=27500 ......................................(1)\n",
+ "\n",
+ "#Now from deformed shape we get\n",
+ "#dell_s=2*dell_b\n",
+ "\n",
+ "#P_s*L_s*(A_s*E_s)**-1=P_b*L_b*(A_b*E_b)**-1\n",
+ "#Further simplifying we get\n",
+ "#P_s=1.2*P_b .........................................(2)\n",
+ "\n",
+ "#Now substituting equation 1 in equation 2 we get\n",
+ "P_b=27500*3.4**-1\n",
+ "P_s=1.2*P_b\n",
+ "\n",
+ "#Tensile stress in brass bar \n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#compressive stress in steel bar\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Compressive Stress in Bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"tensile Stress in Bar is\",round(sigma_b,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Compressive Stress in Bar is 16.18 N/mm**2\n",
+ "tensile Stress in Bar is 8.09 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.30,Page No.44"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=12.6 #m #Length of rail\n",
+ "t1=24 #Degree celsius\n",
+ "t2=44 #degree celsius\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of ELasticity\n",
+ "gamma=2 #mm #Gap provided for Expansion\n",
+ "sigma=20 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "\n",
+ "#Free Expansion of the rails\n",
+ "dell=alpha*t*L*1000 #mm \n",
+ "\n",
+ "#When no expansion joint is provided then\n",
+ "p=dell*E*(L*10**3)**-1\n",
+ "\n",
+ "#When a gap of 2 mm is provided,then free expansion prevented is\n",
+ "dell_1=dell-gamma\n",
+ "p2=dell_1*E*(L*10**3)**-1\n",
+ "\n",
+ "#When stress is developed,then gap left is\n",
+ "gamma2=-(sigma*L*10**3*E**-1-dell)\n",
+ "\n",
+ "#Result\n",
+ "print\"The minimum gap between the two rails is\",round(dell,2),\"mm\"\n",
+ "print\"Thermal Developed in the rials if:No expansionn joint is provided:p\",round(p,2),\"N/mm**2\"\n",
+ "print\" :If a gap of is provided then :p2\",round(p2,2),\"N/mm**2\"\n",
+ "print\"When stress is developed gap left between the rails is\",round(gamma2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum gap between the two rails is 3.02 mm\n",
+ "Thermal Developed in the rials if:No expansionn joint is provided:p 48.0 N/mm**2\n",
+ " :If a gap of is provided then :p2 16.25 N/mm**2\n",
+ "When stress is developed gap left between the rails is 1.76 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.31,Page No.45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=20 #degree celsius\n",
+ "E_a=70*10**9 #N/mm**2 #Modulus of Elasticicty of aluminium\n",
+ "alpha_a=11*10**-6 #per degree celsius #Temperature coeff of aluminium\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel\n",
+ "L_a=1000 #mm #Length of aluminium \n",
+ "L_s=3000 #mm #Length of steel\n",
+ "E_a=7*10**4 #N/mm**2 #Modulus of Elasticity of aluminium\n",
+ "E_s=2*10**5 #N/mm*2 #Modulus of Elasticity of steel\n",
+ "A_a=600 #mm**2 #Area of aluminium\n",
+ "A_s=300 #mm**2 #Area of steel\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Free Expansion \n",
+ "dell=alpha_a*t*L_a+alpha_s*t*L_s\n",
+ "\n",
+ "#support Reaction\n",
+ "P=dell*(L_a*(A_a*E_a)**-1+L_s*(A_s*E_s)**-1)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Reaction at support is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Reaction at support is 12735.48 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.33,Page No.48"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=25 #mm #Diameter of Brass\n",
+ "De=50 #mm #External Diameter of steel tube\n",
+ "Di=25 #mm #Internal Diameter of steel tube\n",
+ "L=1.5 #m #Length of both bars\n",
+ "t1=30 #degree celsius #Initial Temperature\n",
+ "t2=100 #degree celsius #final Temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of ELasticity of steel bar\n",
+ "E_b=1*10**5 #N/mm**2 #Modulus of Elasticity of brass bar\n",
+ "alpha_s=11.6*10**-6 #Temperature Coeff of steel\n",
+ "alpha_b=18.7*10**-6 #Temperature coeff of brass bar\n",
+ "d=20 #mm #diameter of pins\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=t2-t1 #Temperature Difference\n",
+ "A_s=pi*4**-1*(De**2-Di**2) #mm**2 #Area of steel\n",
+ "A_b=pi*4**-1*D**2 #mm**2 #Area of brass\n",
+ "\n",
+ "#Let P_b be the tensile force in brass bar and P_s be the compressive force in steel bar\n",
+ "#But from Equilibrium of Forces \n",
+ "#P_b=P_s=P\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_b-alpha_s)*t*L*1000\n",
+ "\n",
+ "P=dell*(1*(A_s*E_s)**-1+1*(A_b*E_b)**-1)**-1*(L*1000)**-1\n",
+ "P_b=P_s=P\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Stress in Brass\n",
+ "sigma_b=P_b*A_b**-1\n",
+ "\n",
+ "#Area of Pins\n",
+ "A_p=pi*4**-1*d**2\n",
+ "\n",
+ "#Since,the force is resisted by two cross section of pins\n",
+ "tou=P*(2*A_p)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel bar is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"Stress in Brass bar is\",round(sigma_b,2),\"N/mm**2\"\n",
+ "print\"Shear Stresss induced in pins is\",round(tou,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel bar is 14.2 N/mm**2\n",
+ "Stress in Brass bar is 42.6 N/mm**2\n",
+ "Shear Stresss induced in pins is 33.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.34,Page No.49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b_s=60 #mm #width of steel Bar\n",
+ "t_s=10 #mm #thickness of steel Bar\n",
+ "b_c=40 #mm #width of copper bar\n",
+ "t_c=5 #mm #thickness of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #Per degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=17*10**-6 #Per degree celsius #Temperature coeff of copper bar\n",
+ "L_s=L_c=L=1000 #mm #Length of bar\n",
+ "t=80 #degree celsius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A_s=b_s*t_s #Area of steel bar\n",
+ "A_c=b_c*t_c #Area of copper bar\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#The equilibrium of forces gives \n",
+ "#P_s=2*P_c\n",
+ "\n",
+ "#Let dell=dell_s+dell_b\n",
+ "dell=(alpha_c-alpha_s)*t\n",
+ "\n",
+ "P_c=dell*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Stress in copper \n",
+ "sigma_c=P_c*A_c**-1\n",
+ "\n",
+ "#Stress in steel\n",
+ "sigma_s=P_s*A_s**-1\n",
+ "\n",
+ "#Change in Length of bar\n",
+ "dell_2=alpha_s*t*L+P_s*L_s*(A_s*E_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Stress in copper is\",round(sigma_c,2),\"N/mm**2\"\n",
+ "print\"Stress in steel is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"the change in Length is\",round(dell_2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in copper is 30.0 N/mm**2\n",
+ "Stress in steel is 20.0 N/mm**2\n",
+ "the change in Length is 1.06 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.35,Page No.50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=2*10**5 #N #Weight\n",
+ "L=1 #m #Length of each rod\n",
+ "A_c=A_s=A=500 #mm**2 #Area of each rod\n",
+ "t=40 #degree celsius #temperature\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel rod\n",
+ "E_c=1*10**5 #N/mm**2 #modulus of Elastictiy of copper rod\n",
+ "alpha_s=1.2*10**-5 #Per degree Celsius #temp coeff of steel rod\n",
+ "alpha_c=1.8*10**-5 #Per degree Celsius #Temp coeff of copper rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the force in each one of the copper rods and P_s be the force in steel rod\n",
+ "#2*P_c+P_s=P .....................(1)\n",
+ "\n",
+ "#Extension of copper bar=Extension of steel bar\n",
+ "#P_s*L*(A_s*E_s)**-1=P_c*L*(A_c*E_c)**-1\n",
+ "#after simplifying above equation we get\n",
+ "#P_s=2*P_c ........................(2)\n",
+ "\n",
+ "#Now substituting value of P_s in Equation 1 we get\n",
+ "P_c=P*4**-1\n",
+ "P_s=2*P_c\n",
+ "\n",
+ "#Now EXtension due to copper Load\n",
+ "dell_1=P_c*L*1000*(A_c*E_c)**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Due to rise of temperature of40 degree celsius\n",
+ "\n",
+ "#As bars are rigidly joined,let P_c1 be the compressive forccesdeveloped in copper bar and P_s1 be the tensile force in steel causing changes\n",
+ "#P_s1=2*P_c1\n",
+ "\n",
+ "#dell_s+dell_c=(alpha_c-alpha_s)*t*L .......................................(3)\n",
+ "#P_s1*L*(A_s*E_s)**-1+P_c1*L*(A_c*E_c)**-1=(alpha_c-alpha_s)*t*L ................(4)\n",
+ "#After substituting values in above equation and further simplifying we get,\n",
+ "P_c1=(alpha_c-alpha_s)*t*L*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)**-1 #.................(5)\n",
+ "P_s1=2*P_c1\n",
+ "\n",
+ "#Extension of bar due to temperature rise\n",
+ "dell_2=alpha_s*t*L+P_s1*L*(A_s*E_s)**-1\n",
+ "\n",
+ "#Amount by which bar will descend\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Load carried by steel bar\n",
+ "P_S=P_s+P_s1\n",
+ "\n",
+ "#Load carried by copper bar\n",
+ "P_C=P_c-P_c1\n",
+ "\n",
+ "#Part-3\n",
+ "\n",
+ "#Let P_c1_1=P_c #For convenience\n",
+ "#Rise in temperature if Load is to be carried out by steel rod alone\n",
+ "P_c1_1=P_c\n",
+ "\n",
+ "#From equation 5 \n",
+ "t=P_c1_1*(2*(A_s*E_s)**-1+1*(A_c*E_c)**-1)*(alpha_c-alpha_s)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Extension Due top copper Load\",round(dell_1,2),\"mm\"\n",
+ "print\"Load carried by each rod:P_s\",round(P_s,2),\"N\"\n",
+ "print\" :P_c\",round(P_c,2),\"N\"\n",
+ "print\"Rise in Temperature of steel rod should be\",round(t,2),\"degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Extension Due top copper Load 1.0 mm\n",
+ "Load carried by each rod:P_s 100000.0 N\n",
+ " :P_c 50000.0 N\n",
+ "Rise in Temperature of steel rod should be 333.33 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.36,Page No.53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "t=40 #degree celsius #temperature\n",
+ "A_s=400 #mm**2 #Area of steel bar\n",
+ "A_c=600 #mm**2 #Area of copper bar\n",
+ "E_s=2*10**5 #N/mm**2 #Modulus of Elasticity of steel bar\n",
+ "E_c=1*10**5 #N/mm**2 #Modulus of Elasticity of copper bar\n",
+ "alpha_s=12*10**-6 #degree celsius #Temperature coeff of steel bar\n",
+ "alpha_c=18*10**-6 #degree celsius #Temperature coeff of copper bar\n",
+ "L_c=800 #mm #Length of copper bar\n",
+ "L_s=600 #mm #Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_s be the tensile force in steel bar and P_c be the compressive force in copper bar\n",
+ "#Static Equilibrium obtained by taking moment about A\n",
+ "#P_c=2*P_s\n",
+ "\n",
+ "#From property of similar triangles we get\n",
+ "#(alpha_c*Lc-dell_c)*1**-1=(alpha_s*L_s-dell_s)*2**-1\n",
+ "#After substituting values in above equations and further simplifying we get\n",
+ "P_s=(2*alpha_c*L_c-alpha_s*L_s)*t*(L_s*(A_s*E_s)**-1+4*L_c*(A_c*E_c)**-1)**-1\n",
+ "P_c=2*P_s\n",
+ "\n",
+ "#Stress in steel rod\n",
+ "sigma_s=P_s*A_s**-1 #N/mm**2 \n",
+ "\n",
+ "#Stress in copper rod\n",
+ "sigma_c=P_c*A_c**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in steel rod is\",round(sigma_s,2),\"N/mm**2\"\n",
+ "print\"STress in copper rod is\",round(sigma_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in steel rod is 35.51 N/mm**2\n",
+ "STress in copper rod is 47.34 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.37,Page No.61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of bar\n",
+ "P=37.7*10**3 #N #Load\n",
+ "L=200 #mm #Guage Length \n",
+ "dell=0.12 #mm #Extension\n",
+ "dell_d=0.0036 #mm #contraction in diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Let s and dell_s be the Linear strain and Lateral strain\n",
+ "s=dell*L**-1\n",
+ "dell_s=dell_d*d**-1\n",
+ "mu=dell_s*s**-1 #Poissoin's ratio \n",
+ "\n",
+ "#dell=P*L*(A*E)**-1\n",
+ "E=P*L*(dell*A)**-1 #N/mm**2 #Modulus of Elasticity of bar\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "#result\n",
+ "print\"Poisson's ratio is\",round(mu,2)\n",
+ "print\"The Elastic constant are:E\",round(E,2)\n",
+ "print\" :G\",round(G,2)\n",
+ "print\" :K\",round(K,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Poisson's ratio is 0.3\n",
+ "The Elastic constant are:E 200004.71\n",
+ " :G 76924.89\n",
+ " :K 166670.59\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.38,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of circular rod\n",
+ "P=1*10**6 #N #Tensile Force\n",
+ "mu=0.3 #Poisson's ratio\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "L=500 #mm #Length of rod\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Modulus of Rigidity\n",
+ "G=E*(2*(1+mu))**-1 #N/mm**2\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of Circular rod\n",
+ "#Let sigma be the Longitudinal stress\n",
+ "sigma=P*A**-1 #N/mm**2 \n",
+ "\n",
+ "s=sigma*E**-1 #Linear strain\n",
+ "e_x=s\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x*(1-2*mu)\n",
+ "\n",
+ "v=pi*4**-1*d**2*L\n",
+ "#Change in VOlume\n",
+ "dell_v=e_v*v\n",
+ "\n",
+ "#Result\n",
+ "print\"Bulk Modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Modulus of Rigidity is\",round(G,2),\"N/mm**2\"\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bulk Modulus is 200000.0 N/mm**2\n",
+ "Modulus of Rigidity is 76923.08 N/mm**2\n",
+ "The change in Volume is 1000.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.39,Page No.62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=500 #mm #Length of rectangular cross section bar\n",
+ "A=20*40 #mm**2 #Area of rectangular cross section bar\n",
+ "P1=4*10**4 #N #Tensile Force on 20mm*40mm Faces\n",
+ "P2=2*10**5 #N #compressive force on 20mm*500mm Faces\n",
+ "P3=3*10**5 #N #Tensile Force on 40mm*500mm Faces\n",
+ "E=2*10**5 #N/mm**2 #young's Modulus \n",
+ "mu=0.3 #Poisson's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let P_x,P_y,P_z be the forces n x,y,z directions\n",
+ "\n",
+ "P_x=P1*A**-1\n",
+ "P_y=P2*A**-1\n",
+ "P_z=P3*A**-1\n",
+ "\n",
+ "#Let e_x,e_y,e_z be the strains in x,y,z directions\n",
+ "e_x=1*E**-1*(50+mu*20-15*mu)\n",
+ "e_y=1*E**-1*(-mu*50-20-mu*15)\n",
+ "e_z=1*E**-1*(-mu*50+mu*20+15)\n",
+ "\n",
+ "#Volumetric strain\n",
+ "e_v=e_x+e_y+e_z\n",
+ "\n",
+ "#Volume\n",
+ "V=20*40*500 #mm**3\n",
+ "#Change in Volume \n",
+ "dell_v=e_v*V #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"The change in Volume is\",round(dell_v,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in Volume is 36.0 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.41,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2.1*10**5 #N/mm**2 #Young's Modulus \n",
+ "G=0.78*10**5 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now using the relation\n",
+ "#E=2*G*(1+mu)\n",
+ "mu=E*(2*G)**-1-1 #Poisson's ratio\n",
+ "\n",
+ "#Bulk Modulus \n",
+ "K=E*(3*(1-2*mu))**-1 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"The Poisson's Ratio is\",round(mu,2)\n",
+ "print\"The modulus of Rigidity\",round(K,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poisson's Ratio is 0.35\n",
+ "The modulus of Rigidity 227500.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.42,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=0.4*10**5 #N/mm**2 #Modulus of rigidity\n",
+ "K=0.75*10**5 #N/mm**2 #Bulk Modulus \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Young's Modulus\n",
+ "E=9*G*K*(3*K+G)**-1\n",
+ "\n",
+ "#Now from the relation\n",
+ "#E=2*G(1+2*mu)\n",
+ "mu=E*(2*G)**-1-1 #POissoin's ratio \n",
+ "\n",
+ "#result\n",
+ "print\"Young's modulus is\",round(E,2),\"N/mm**2\"\n",
+ "print\"Poissoin's ratio is\",round(mu,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Young's modulus is 101886.79 N/mm**2\n",
+ "Poissoin's ratio is 0.27\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.43,Page No.65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=60 #mm #width of bar\n",
+ "d=30 #mm #depth of bar\n",
+ "L=200 #mm #Length of bar\n",
+ "A=30*60 #mm**2 #Area of bar\n",
+ "A2=30*200 #mm**2 #Area of bar along which expansion is restrained\n",
+ "P=180*10**3 #N #Compressive force\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The bar is restrained from expanding in Y direction\n",
+ "P_z=0\n",
+ "P_x=P*A**-1 #stress developed in x direction\n",
+ "\n",
+ "#Now taking compressive strain as positive\n",
+ "#e_x=P_x*E**-1-mu*P_y*E**-1 .......................(1)\n",
+ "#e_y=-mu*P_x*E**-1+P_y*E**-1 ....................(2)\n",
+ "#e_z=-mu*P_x*E**-1-mu*P_y*E**-1 ......................(3)\n",
+ "\n",
+ "#Part-1\n",
+ "#When it is fully restrained\n",
+ "e_y=0\n",
+ "P_y=30 #N/mm**2 \n",
+ "e_x=P_x*E**-1-mu*P_y*E**-1\n",
+ "e_z=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l=e_x*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b=b*e_y\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d=d*e_z\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=b*d*L #mm**3\n",
+ "#Change in Volume\n",
+ "e_v=(e_x+e_y+e_z)*V #mm**3\n",
+ "\n",
+ "#Part-2\n",
+ "#When 50% is restrained\n",
+ "\n",
+ "#Free strain in Y direction\n",
+ "e_y1=mu*P_x*E**-1\n",
+ "\n",
+ "#As 50% is restrained,so\n",
+ "e_y2=-50*100**-1*e_y1\n",
+ "\n",
+ "#But form Equation 2 we have e_y=-mu*P_x*E**-1+P_y*E**-1 \n",
+ "#After substituting values in above equation and furthe simplifying we get\n",
+ "P_y=e_y2*E+d\n",
+ "\n",
+ "e_x2=P_x*E**-1-mu*P_y*E**-1 \n",
+ "e_z2=-mu*P_x*E**-1-mu*P_y*E**-1\n",
+ "\n",
+ "#Change in Length \n",
+ "dell_l2=e_x2*L #mm\n",
+ "\n",
+ "#Change in width\n",
+ "dell_b2=b*e_y2\n",
+ "\n",
+ "#change in Depth\n",
+ "dell_d2=d*e_z2\n",
+ "\n",
+ "#Change in Volume\n",
+ "e_v2=(e_x2+e_y2+e_z2)*V #mm**3\n",
+ "\n",
+ "#REsult\n",
+ "print\"Change in Dimension of bar is:dell_l\",round(dell_l,2),\"mm\"\n",
+ "print\" :dell_b\",round(dell_b,4),\"mm\"\n",
+ "print\" :dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Volume is\",round(e_v,2),\"mm**3\"\n",
+ "print\"Changes in material when only 50% of expansion can be reatrained:dell_l2\",round(dell_l2,2),\"mm\"\n",
+ "print\" :dell_b2\",round(dell_b2,4),\"mm\"\n",
+ "print\" :dell_d2\",round(dell_d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Dimension of bar is:dell_l 0.09 mm\n",
+ " :dell_b 0.0 mm\n",
+ " :dell_d -0.01 mm\n",
+ "Change in Volume is 93.6 mm**3\n",
+ "Changes in material when only 50% of expansion can be reatrained:dell_l2 0.1 mm\n",
+ " :dell_b2 -0.0045 mm\n",
+ " :dell_d2 -0.01 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.44,Page No.72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=10*10**3 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "d2=12 #mm #Diameter of bar1\n",
+ "d1=16 #mm #diameter of bar2\n",
+ "L1=200 #mm #Length of bar1\n",
+ "L2=500 #mm #Length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let A1 and A2 be the cross Area of Bar1 & bar2 respectively\n",
+ "A1=pi*4**-1*d1**2 #mm**2\n",
+ "A2=pi*4**-1*d2**2 #mm**2\n",
+ "\n",
+ "#Let p1 and p2 be the stress in Bar1 nad bar2 respectively\n",
+ "p1=P*A1**-1 #N/mm**2\n",
+ "p2=P*A2**-1 #N/mm**2\n",
+ "\n",
+ "#Let V1 nad V2 be the Volume of of Bar1 and Bar2\n",
+ "V1=A1*(L1+L1)\n",
+ "V2=A2*L2\n",
+ "\n",
+ "#Let E be the strain Energy stored in the bar\n",
+ "E=p1**2*(2*E)**-1*V1+p2**2*V2*(2*E)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Strain Energy stored in Bar is\",round(E,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Strain Energy stored in Bar is 1602.6 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.45,Page No.73"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Bar-A\n",
+ "d1=30 #mm #Diameter of bar1\n",
+ "L=600 #mm #length of bar1\n",
+ "\n",
+ "#Bar-B\n",
+ "d2=30 #mm #Diameter of bar2\n",
+ "d3=20 #mm #Diameter of bar2\n",
+ "L2=600 #mm #length of bar2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A1=pi*4**-1*d1**2\n",
+ "\n",
+ "#Area of bar-B\n",
+ "A2=pi*4**-1*d2**2\n",
+ "A3=pi*4**-1*d3**2\n",
+ "\n",
+ "#let SE be the Strain Energy\n",
+ "#Strain Energy stored in Bar-A\n",
+ "#SE=p**2*(2*E)**-1*V\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE=P**2*E**-1*0.4244\n",
+ "\n",
+ "#Strain Energy stored in Bar-B\n",
+ "#SE2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE2=0.6897*P**2*E**-1\n",
+ "\n",
+ "#Let X be the ratio of SE in Bar-B and SE in Bar-A\n",
+ "X=0.6897*0.4244**-1\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#When Max stress is produced is same:Let p be the max stress produced\n",
+ "\n",
+ "#Stress in bar A is p throughout \n",
+ "#In bar B:stress in 20mm dia.portion=p2=p\n",
+ "\n",
+ "#Stress in 30 mm dia.portion\n",
+ "#p1=P*A2*A3**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#p1=4*9**-1*p\n",
+ "\n",
+ "#Strain Energy in bar A\n",
+ "#SE_1=p**2*(2*E)**-1*A1*L1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_1=67500*p**2*pi*E**-1\n",
+ "\n",
+ "#Strain Energy in bar B\n",
+ "#SE_2=p1**2*V1*(2*E)**-1+p2**2*V2*(2*E)**-1\n",
+ "#After substituting values and simolifying further we get\n",
+ "#SE_2=21666.67*pi*p**2*E**-1\n",
+ "\n",
+ "#Let Y be the Ratio of SE in bar B and SE in bar A\n",
+ "Y=21666.67*67500**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Gradually applied Load is\",round(X,2)\n",
+ "print\"Gradually applied Load is\",round(Y,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Gradually applied Load is 1.63\n",
+ "Gradually applied Load is 0.32\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.46,Page No.74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables \n",
+ "\n",
+ "W=100 #N #Load\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "h=60 #mm #Height through Load falls down\n",
+ "L=400 #mm #Length of collar\n",
+ "d=30 #mm #diameter of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #mm**2 #Area of bar\n",
+ "\n",
+ "#Instantaneous stress produced is\n",
+ "p=W*A**-1*(1+(1+(2*A*E*h*(W*L)**-1))**0.5)\n",
+ "\n",
+ "#Now the EXtension of the bar is neglected in calculating work doneby the Load,then\n",
+ "P=(2*E*h*W*(A*L)**-1)**0.5\n",
+ "\n",
+ "#Let percentage error be denoted by E1\n",
+ "#Percentage error in approximating is\n",
+ "E1=(p-P)*p**-1*100\n",
+ "\n",
+ "#Instantaneous Extension produced is\n",
+ "dell_l=round(P,3)*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"The Instantaneous stress is\",round(p,2),\"N/mm\"\n",
+ "print\"Percentage Error is\",round(E1,2)\n",
+ "print\"The Instantaneous extension is\",round(dell_l,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Instantaneous stress is 92.27 N/mm\n",
+ "Percentage Error is 0.15\n",
+ "The Instantaneous extension is 0.18 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.47,Page No.75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=20 #mm #Diameter of steel bar\n",
+ "L=1000 #mm #Length of bar\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus \n",
+ "p=300 #N/mm**2 #max Permissible stress\n",
+ "h=50 #mm #Height through which weight will fall\n",
+ "w=600 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#ARea of steel bar\n",
+ "A=pi*4**-1*d**2\n",
+ "\n",
+ "#Instantaneous extension is\n",
+ "dell_l=p*L*E**-1 #mm \n",
+ "\n",
+ "#Work done by Load \n",
+ "#W=W1*(h+dell_l)\n",
+ "\n",
+ "#Volume of bar\n",
+ "V=round(A,2)*L\n",
+ "#Let E1 be the strain Energy\n",
+ "E1=p**2*(2*E)**-1*V\n",
+ "\n",
+ "#Answer in Book for Strain Energy is Incorrect \n",
+ "\n",
+ "#Now Equating Workdone by Load to strain Energy \n",
+ "W1=E1*51.5**-1\n",
+ "\n",
+ "#Now when w=600 N\n",
+ "#Let W2 be the Work done by the Load\n",
+ "#W2=w(h2*dell_l)\n",
+ "\n",
+ "h=E1*w**-1-dell_l\n",
+ "\n",
+ "#Result\n",
+ "print\"The Max Lodad which can Fall from a height of 50 mm on the collar is\",round(W1,2),\"N\"\n",
+ "print\"the Max Height from which a 600 N Load can fall on the collar is\",round(h,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Max Lodad which can Fall from a height of 50 mm on the collar is 1372.54 N\n",
+ "the Max Height from which a 600 N Load can fall on the collar is 116.31 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.48,Page No.76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D_s=30 #mm #Diameter of steel rod\n",
+ "d=30 #mm #Internal Diameter of copper tube\n",
+ "D=40#mm #External Diameter of copper tube\n",
+ "E_s=2*10**5 #N/mm**2 #Young's Modulus of Steel rod\n",
+ "E_c=1*10**5#N/mm**2 #Young's Modulus of copper tube\n",
+ "P=100 #N #Load\n",
+ "h=40 #mm #height from which Load falls\n",
+ "L=800 #mm #Length \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area of steel rod\n",
+ "A_s=pi*4**-1*D_s**2\n",
+ "\n",
+ "#Area of copper tube\n",
+ "A_c=pi*4**-1*(D**2-d**2)\n",
+ "\n",
+ "#But Dell_s=dell_c=dell\n",
+ "#p_s*E_s**-1*L=p_c*L*E_c\n",
+ "#After simplifying furthe we get\n",
+ "#p_s=2*p_c\n",
+ "\n",
+ "#Now Equating internal Energy to Workdone we get\n",
+ "p_c=(2*P*h*L**-1*(4*A_s*E_s**-1+A_c*E_c**-1))**0.5\n",
+ "p_s=2*p_c\n",
+ "\n",
+ "#Result\n",
+ "print A_s\n",
+ "print\"STress produced in steel is\",round(p_s,2),\"N/mm**2\"\n",
+ "print\"STress produced in copper is\",round(p_c,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "706.858347058\n",
+ "STress produced in steel is 0.89 N/mm**2\n",
+ "STress produced in copper is 0.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2.2.49,Page No.77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "dell=0.25 #mm #Instantaneous Extension\n",
+ "\n",
+ "#Bar-A\n",
+ "b1=25 #mm #width of bar\n",
+ "D1=500 #mm #Depth of bar\n",
+ "\n",
+ "#Bar-B\n",
+ "b2_1=25 #mm #width of upper bar\n",
+ "b2_2=15 #mm #Width of Lower Bar\n",
+ "L2=200 #mm #Length of upper bar\n",
+ "L1=300 #mm #Length of Lower bar\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus of bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Strain\n",
+ "e=dell*D1**-1 \n",
+ "\n",
+ "#Load\n",
+ "p=e*E\n",
+ "\n",
+ "#Area of bar-A\n",
+ "A=pi*4**-1*25**2\n",
+ "\n",
+ "#Volume of bar-A\n",
+ "V=A*D1\n",
+ "\n",
+ "#Let E1 be the Energy of Blow\n",
+ "#Energy of Blow\n",
+ "E1=p**2*(E)**-1*V\n",
+ "\n",
+ "#Let p2 be the Max stress in bar B When this blow is applied.\n",
+ "#the max stress occurs in the 15mm dia. portion,Hence, the stress in 25 mm dia.portion is\n",
+ "#p2*pi*4**-1*b2_2**2*(pi*4**-1*b2_2**2=0.36*p\n",
+ "\n",
+ "#Strain Energy of bar B\n",
+ "#E2=p**2*(2*E)**-1*v1+1*(2*E)**-1*(0.36*p2)**2*v2\n",
+ "#After substituting values and Further substituting values we get\n",
+ "#E2=0.1643445*p2**2\n",
+ "\n",
+ "#Equating it to Energy of applied blow,we get\n",
+ "p2=(12271.846*0.1643445**-1)**0.5\n",
+ "\n",
+ "#Stress in top portion\n",
+ "sigma=0.36*p2\n",
+ "\n",
+ "#Extension in Bar-1\n",
+ "dell_1=p2*E**-1*L1\n",
+ "\n",
+ "#Extension in Bar-2\n",
+ "dell_2=0.36*p2*E**-1*L2\n",
+ "\n",
+ "#Extension of bar\n",
+ "dell_3=dell_1+dell_2\n",
+ "\n",
+ "#Result\n",
+ "print\"Instantaneous Max stress is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"extension in Bar is\",round(dell_3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Instantaneous Max stress is 98.37 N/mm**2\n",
+ "extension in Bar is 0.51 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_5.ipynb new file mode 100644 index 00000000..f136c4ad --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.3_5.ipynb @@ -0,0 +1,1588 @@ +{
+ "metadata": {
+ "name": "chapter no.3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 3:Shear Force And Bending Moment Diagrams in Statically Determinate Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.1,Page No.100"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=1 #m #Length of AC & CD\n",
+ "L_DB=1.5 #m #Lengh of DB\n",
+ "L=3.5 #m #Length of Beam\n",
+ "F_B=10 #KN #Force at pt B\n",
+ "F_C=F_D=20 #KN #Force at pt C & D\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R_A=F_C+F_D+F_B #KN #Force at support A \n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt B\n",
+ "V_B1=0 #KN \n",
+ "V_B2=F_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1+F_D #KN\n",
+ "\n",
+ "#S.F At pt C \n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_D2+F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2-R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M AT Pt D\n",
+ "M_D=F_B*L_DB #KN.m\n",
+ "\n",
+ "#B.M At pt C\n",
+ "M_C=F_B*(L_DB+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At pt A\n",
+ "M_A=F_B*L+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5dde2d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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pqQavCQgIEADwi1/84he/rPgKCAiw6e+yYlJMDQ0NuPfee/H999+jZ8+eGDhw\nINavX4/g4GC5SyMickmKmWJq3749PvroI4wcORKNjY2YPn06mwMRkYwUM4IgIiJlUUzMtaWsrCwE\nBQXhnnvuwZIlS4y+Zt68ebjnnnsQHh6OwsJCiStsnbn6s7Oz4e3tjcjISERGRuLNN9+UoUrjpk2b\nBrVajbCwMJOvUfK5N1e/ks89AJSWliIuLg4hISEIDQ3Fhx9+aPR1SvwdWFK7ks9/XV0dYmJiEBER\nAY1Gg5deesno65R47gHL6rf6/Nu8qiyShoYGISAgQDh16pRQX18vhIeHC0VFRQav2bZtmzBq1ChB\nEAQhLy9PiImJkaNUoyypf8+ePcLYsWNlqrB1P/74o1BQUCCEhoYafV7J514QzNev5HMvCIJQVlYm\nFBYWCoIgCNXV1UJgYKDD/PtvSe1KP//Xrl0TBEEQbt68KcTExAj/+te/DJ5X6rlvZq5+a8+/4kYQ\nllwwl5GRgeTkZABATEwMqqqqUFFRIUe5d7D0gj9BoTN7Q4YMQZcuXUw+r+RzD5ivH1DuuQcAX19f\nREREAAA8PT0RHByM8+fPG7xGqb8DS2oHlH3+O3bsCACor69HY2MjunbtavC8Us99M3P1A9adf8U1\nCGMXzJ07d87sa86ePStZja2xpH6VSoV9+/YhPDwcCQkJKCoqkrpMmyn53FvCkc59SUkJCgsLERMT\nY/C4I/wOTNWu9PPf1NSEiIgIqNVqxMXFQaPRGDyv9HNvrn5rz79iUkzNLL1g7vYuaOn7xGZJHVFR\nUSgtLUXHjh2xY8cOJCYm4sSJExJU1zaUeu4t4SjnvqamBpMmTcKyZcvg6el5x/NK/h20VrvSz7+b\nmxsOHTqEK1euYOTIkcjOzoZWqzV4jZLPvbn6rT3/ihtB9OrVC6WlpfqfS0tL4efn1+przp49i14K\nudmzJfV7eXnph4KjRo3CzZs3UVlZKWmdtlLyubeEI5z7mzdv4qGHHsLjjz+OxMTEO55X8u/AXO2O\ncP4BwNvbG6NHj8ZPP/1k8LiSz31Lpuq39vwrrkEMGDAAv/32G0pKSlBfX4/09HSMGzfO4DXjxo3D\nl19+CQDIy8uDj48P1Gq1HOXewZL6Kyoq9P8VcvDgQQiCYHSuUImUfO4tofRzLwgCpk+fDo1Gg2ef\nfdboa5T6O7CkdiWf/z/++ANVVVUAgOvXr2P37t2IjIw0eI1Szz1gWf3Wnn/FTTGZumDus88+AwDM\nnj0bCQk5anozAAADy0lEQVQJ2L59O/r164dOnTph9erVMld9iyX1b9q0CZ988gnat2+Pjh07YsOG\nDTJXfcsjjzyCnJwc/PHHH+jduzdef/113Lx5E4Dyzz1gvn4ln3sA2Lt3L9auXYv+/fvr/8/99ttv\n48yZMwCU/TuwpHYln/+ysjIkJyejqakJTU1NmDp1KoYPH+4wf3ssqd/a888L5YiIyCjFTTEREZEy\nsEEQEZFRbBBERGQUGwQRERnFBkFEREaxQRARkVFsEOSUjG1P0ZaWLl2K69evW3W8zMxMk9vXEykR\nr4Mgp+Tl5YXq6mrRPr9v37746aef0K1bN0mORyQHjiDIZRQXF2PUqFEYMGAAhg4diuPHjwMAnnzy\nSTzzzDOIjY1FQEAANm/eDEC3M+acOXMQHByMESNGYPTo0di8eTOWL1+O8+fPIy4uDsOHD9d//quv\nvoqIiAj85S9/wYULF+44/po1a/D000+3esyWSkpKEBQUhKeeegr33nsvHnvsMezatQuxsbEIDAxE\nfn6+GKeJ6BYb70tBpGienp53PHb//fcLv/32myAIupu93H///YIgCEJycrKQlJQkCIIgFBUVCf36\n9RMEQRC+/vprISEhQRAEQSgvLxe6dOkibN68WRAEQfD39xcuXbqk/2yVSiV8++23giAIwvz584U3\n33zzjuOvWbNGmDt3bqvHbOnUqVNC+/bthV9//VVoamoSoqOjhWnTpgmCIAhbt24VEhMTrT0tRFZR\n3F5MRGKoqanB/v37MXnyZP1j9fX1AHTbNTfvPBocHKy/AUxubi6SkpIAQL+/vikeHh4YPXo0ACA6\nOhq7d+9utR5Tx7xd3759ERISAgAICQnBAw88AAAIDQ1FSUlJq8cgshcbBLmEpqYm+Pj4mLyHsIeH\nh/574T/LciqVymDvf6GV5Tp3d3f9925ubmhoaDBbk7Fj3q5Dhw4Gn9v8HkuPQWQPrkGQS+jcuTP6\n9u2LTZs2AdD9QT5y5Eir74mNjcXmzZshCAIqKiqQk5Ojf87LywtXr161qobWGgyRErFBkFOqra1F\n79699V9Lly7FunXrsHLlSkRERCA0NBQZGRn617e8K1jz9w899BD8/Pyg0WgwdepUREVFwdvbGwAw\na9YsPPjgg/pF6tvfb+wuY7c/bur7299j6mcl3cmMnBNjrkStuHbtGjp16oRLly4hJiYG+/btQ48e\nPeQui0gSXIMgasWYMWNQVVWF+vp6LFq0iM2BXApHEEREZBTXIIiIyCg2CCIiMooNgoiIjGKDICIi\no9ggiIjIKDYIIiIy6v8BsxKV3Pt7cnUAAAAASUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5df0ab0>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.2,Page No.101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w1=10 #KN/m #u.d.L\n",
+ "F_D=20 #KN #Force at pt D\n",
+ "F_C=30 #KN #Force at pt C\n",
+ "L_DB=4 #m #Length of DB\n",
+ "L_CD=L_AC=2 #m #Length of AC & CD\n",
+ "L=8 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=90 \n",
+ "#Now Taking moment at A,M_A we get\n",
+ "R_A=(w1*L_DB*(L_DB*2**-1)+F_D*L_DB+F_C*(L_CD+L_DB))*L**-1\n",
+ "R_B=90-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At pt D\n",
+ "V_D1=R_B-w1*L_DB #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C1=V_D2 #KN\n",
+ "V_C2=V_C1-F_C \n",
+ "\n",
+ "#S.F at PT A\n",
+ "V_A1=V_C2 #KN\n",
+ "V_A2=V_C2+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=-R_B*L_DB+w1*L_DB*L_DB*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At PT C\n",
+ "M_C=-R_B*(L_DB+L_CD)+w1*L_DB*(L_DB*2**-1+L_CD)+F_D*L_CD #KN.m\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_B*L+w1*L_DB*(L_DB*2**-1+L_CD+L_AC)+F_D*(L_CD+L_AC)+F_C*L_AC\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_CD+L_DB,L_CD+L_DB,L_CD+L_DB+L_AC,L_CD+L_DB+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB+L_CD,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Pni0aGhr0vo4HrxERkcSmlo+IiMi8WApERCRhKRARkYSlQEREEpYCERFJWApE\nRCRhKVCPYu7TdGzcuBE3btww+eft27fPak8FT/aFxylQj+Lm5iYdsWsO/v7+OHPmDPr372+RzyOy\nNG4pUI938eJFTJo0CcOGDcMf/vAHnD9/HgDwzDPPYNmyZXjooYcQEBCAjIwMAC1nZE1JSUFISAgm\nTpyIKVOmICMjA5s3b0ZlZSViYmIwfvx46fe/+uqriIiIwOjRo/HTTz+1+/znnnsOq1atAgD885//\nxNixY9u9ZseOHViyZEmHuVrT6XQIDg7G3LlzMXjwYDz11FM4dOgQHnroIQQFBeH06dPd/4Mj+2TB\no6yJzM7V1bXdc+PGjRPFxcVCCCHy8vLEuHHjhBBCzJkzRyQmJgohhCgsLBSBgYFCCCE++eQTMXny\nZCGEENXV1cLDw0NkZGQIIdpfoEahUIj9+/cLIYRYvny5eP3119t9fn19vQgNDRU5OTli8ODBoqSk\npN1rduzYIRYvXtxhrtZKS0uFo6OjOHfunGhubhZRUVFi3rx5QgghMjMzxdSpU43+WRHp4yh3KRGZ\nU11dHb766qs2pzRuaGgA0HIq9ttnhg0JCcGlS5cAAMePH0diYiIASNdvMMTZ2RlTpkwBAERFRSE7\nO7vda+69915s3boVY8aMwaZNm+Dv799hZkO57uTv7y+dJC40NFQ6P35YWBh0Ol2Hn0FkCEuBerTm\n5mb069cPZ8+e1ftzZ2dn6b7433hNoVC0uSaC6GDs5uTkJN13cHBAY2Oj3tcVFBTAy8ur0xeF0pfr\nTr17927z2bff01EOImM4U6Aezd3dHf7+/vjHP/4BoOULtqCgoMP3PPTQQ8jIyIAQApcuXcLRo0el\nn7m5ueHq1atdyvDDDz/gzTfflC5so+/6BR0VD5ElsRSoR6mvr4efn59027hxI3bv3o1t27YhIiIC\nYWFhyMrKkl7f+mp+t+/PmDEDSqUSarUaTz/9NCIjI6XrAy9cuBCPPvqoNGi+8/13Xh1QCIH58+dj\nw4YN8PHxwbZt2zB//nxpCcvQew3dv/M9hh7zKoV0t7hLKpEe169fh4uLCy5fvoyRI0fixIkTGDhw\noNyxiMyOMwUiPR577DHU1taioaEBK1asYCGQ3eCWAhERSThTICIiCUuBiIgkLAUiIpKwFIiISMJS\nICIiCUuBiIgk/w9P4ODmR+1IBgAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5be93d0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Gt27dTE9mRiwY9ovtQ+zHlSuAv780HCkgQO00tkGRgnHr1i2kp6ejpKQEdXV1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\nx6hRo1BfX48XX3wRgYGBasd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+ "text": [
+ "<matplotlib.figure.Figure at 0x5de54f0>"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.3,Page No.102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_CD=1.5 #m #Length of DB & CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "F_D=80 #KN #Force at Pt D\n",
+ "w=40 #KN/m #u.v.l\n",
+ "L=6 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at Pt A & B respectively\n",
+ "#R_A+R_B=140 \n",
+ "#Taking moment at B we get,M_B\n",
+ "R_A=(1*2**-1*L_AC*w*(1*3**-1*L_AC+(L_CD+L_DB))+F_D*L_DB)*L**-1\n",
+ "R_B=140-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=0 #KN\n",
+ "V_B2=R_B #KN\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F at C\n",
+ "V_C=V_D2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_C-1*2**-1*w*L_AC #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=-R_B*L_DB\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD-R_B*(L_DB+L_CD)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_D*(L_CD+L_AC)-R_B*L+1*2**-1*w*L_AC*(1*3**-1*L_AC)+R_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D1,V_D2,V_C,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Y2=[M_B,M_D,M_C,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x58f0c10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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CBUEIDRWEv/9d7kpIbVz53SnLDGPjxo0YOHAg9u3bh/T0dKSlpQEAjEYjsrKy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\ngg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x562f910>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.4,Page No.104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "M_D=120 #KN.m #B.M at Pt D\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "w1=20 #KN.m\n",
+ "L_DB=1.5 #m #Length of DB\n",
+ "L_CD=1.5 #m #Length of CD\n",
+ "L_AC=3 #m #Length of AC\n",
+ "L=6 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A And R_B be the Reactions at pt A and B \n",
+ "#R_A+R_B=100\n",
+ "#Now Taking Moment At Pt B We get,M_B\n",
+ "R_A=-(M_D-F_C*(L_CD+L_DB)-w1*L_AC*(L_AC*2**-1+L_CD+L_DB))*L**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=0\n",
+ "V_B2=R_B\n",
+ "\n",
+ "#S.F at Pt D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At Pt C\n",
+ "V_C1=V_D #KN\n",
+ "V_C2=V_C1-F_C\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-w1*L_AC #KN\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=0 #KN.m\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_B-R_B*L_DB #KN.m\n",
+ "M_D2=M_B+M_D-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=M_D-R_B*(L_CD+L_DB)\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=M_D-R_B*L+F_C*L_AC+w1*L_AC*L_AC*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DB,L_DB+L_CD,L_DB+L_CD,L_DB+L_CD+L_AC,L_DB+L_CD+L_AC]\n",
+ "Y1=[V_B1,V_B2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_B,M_D1,M_D2,M_C,M_A]\n",
+ "X2=[0,L_DB,L_DB,L_CD+L_DB,L_AC+L_CD+L_DB]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55dcbf0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x55ca4b0>"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.5,Page No.105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_C=20 #KN #Force at Pt C\n",
+ "F_D=40 #KN #Force at pt D\n",
+ "w=20 #KN.m #u.d.l \n",
+ "L_AD=L_DB=2 #m #Length of AD & DB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L=5 #m #Length of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_A and R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=100 \n",
+ "#Now Taking Moment at B,M_B we get\n",
+ "R_A=-(F_C*L_BC-F_D*L_DB-w*L_AD*(L_AD*2**-1+L_DB))*(L_AD+L_DB)**-1\n",
+ "R_B=100-R_A\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At pt C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At PT B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN\n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2 #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_D2-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=0 \n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D=F_C*(L_BC+L_DB)-R_B*L_DB\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=F_C*L-R_B*(L_DB+L_AD)+F_D*L_AD+w*L_AD*L_AD*2**-1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_BC+L_DB,L_BC+L_DB,L_BC+L_DB+L_AD,L_BC+L_DB+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D1,V_D2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_C,M_B,M_D,M_A]\n",
+ "X2=[0,L_BC,L_BC+L_DB,L_BC+L_DB+L_AD]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d9c6f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Zb7/9lh07drB8+XLy8/N58cUXmTVrlrt+NSKaPpLAUlFRwZdffsnI\nkSOd91VWVgLGm/WZE1a7du3K4cOHAfj8889JSUkBcPYoONvw4cMB6NGjBx9++KHzfldOkKntmufr\n0KED3bp1A6Bbt24kJiYCEB0dTUlJSZ3XEXGVkoIElNOnT3PFFVdQWFhY48+bNm3q/PrMm/r5dYPz\n3+wvueQSAJo0aUJVVVW9Y6rpmuc7cw0w+iWceU5QUFCDrilSG00fSUC5/PLL6dChAx988AFgvAn/\n61//uuhz+vTpw4oVK3A4HBw+fJi8vLw6r9O8eXPnUcXn0xmUYmVKCuLXTpw4Qbt27Zy3//3f/+Xd\nd99l0aJFxMbGEh0dfU4z97Pn+898PWLECMLCwoiKiuK+++6jR48etGjR4oJr2Ww253OGDh3KypUr\niYuLIz8/v9bH1XbNml67tu8D+UhocT8dnS3igp9//pnLLruM//znPyQkJPDFF1/QqlUrs8MScTvV\nFERcMGTIEMrLy6msrGTatGlKCOK3NFIQEREn1RRERMRJSUFERJyUFERExElJQUREnJQURETESUlB\nRESc/j+iaB8fTwtkoQAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c1ead0>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.6,Page No.107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=L_AD=1 #m #Length of spans BC,ED,AD\n",
+ "L_ED=2 #m #Length of ED\n",
+ "w=60 #KNm #u.d.l\n",
+ "F_C=20 #KN Pt Load at C\n",
+ "L=5 #m #Span of beam \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_B be the reactions at A & B respectively\n",
+ "#R_A+R_B=80 \n",
+ "#Taking Moment At A,we get M_A\n",
+ "R_B=(F_C*L+1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD))*(L_AD+L_ED+L_EB)**-1\n",
+ "R_A=80-R_B\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-F_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E=V_B2 #KN\n",
+ "\n",
+ "#S.F AT D\n",
+ "V_D=V_B2-1*2**-1*L_ED*w #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D #KN \n",
+ "V_A2=V_D+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m\n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at E\n",
+ "M_E=F_C*(L_EB+L_BC)-R_B*L_EB #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=F_C*(L_ED+L_EB+L_BC)-R_B*(L_ED+L_EB)+1*2**-1*L_ED*w*1*3**-1*L_ED #KN.m\n",
+ "\n",
+ "#B.M at A\n",
+ "M_A=1*2**-1*L_ED*w*(2*3**-1*L_ED+L_AD)-R_B*(L_AD+L_ED+L_EB)+F_C*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_EB+L_BC,L_ED+L_EB+L_BC,L_AD+L_ED+L_EB+L_BC,L_ED+L_EB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_E,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_BC+L_EB,L_EB+L_BC+L_ED,L_EB+L_BC+L_ED+L_AD]\n",
+ "Y2=[M_C,M_B,M_E,M_D,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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Cr9fzLAHAgQMHUFpaig8++AAbNmzAvn37Ot1OUUUhJCSkwx1Vq6qqEBoaKjAR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\nAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58a6f70>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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UlD0DZoHZoDk7OxvPPfec/SqwAoNm98HQmcjAloBZINmcwq5du5rcTwEAHnnkEeuqkgCb\ngnvhPZ2JLLsHszmSNIWHH34YP//8M2JiYuDp6SluX7ZsmfWV2YhNwb1w0pnIuglmY5I0hfDwcJSW\nltp8Yx0psSm4H046kzuzdoLZmCRzClFRUTh37pz1VRBJgJPO5M4cETALzB4paLVaHDx4EP369UO7\ndu0ML1KpkJeXZ//qTOCRgnti6EzuSIqAWSDJ6SPhBg03v5lKpcIDDzxgW3U2YFNwXwydyd1IETAL\nJLv6SKfT4dSpU4iLi0NtbS3q6+vRsWNH2yu0EpuC+2LoTO5GioBZIEmm8P7772PChAl46qmnABju\nhjZmzBjbqyOyws2Tzvy7gFydPe7BbI7ZpvDuu+9i586d4pFBz549ceHCBbsXRmQKQ2dyF44MmAVm\nm0K7du3EgBkA6uvrFXV5Krkf4Z7O//gHUF0tdzVE9iHcg9mei9+1xGxTeOCBB7BgwQLU1tZi27Zt\nmDBhApKSkhxRG5FJXF6bXJ29l8g2xWzQ3NDQgFWrVmHr1q0AgMTEREybNk3WowUGzQQwdCbXJmXA\nLFDsPZr/8Y9/4Msvv4SPjw9CQ0OxevVqdOrUCQCQkZGBnJwceHp6Ijs7GwkJCc2LZlOg/+KkM7ki\nqSaYjUly9dGmTZug0Whw2223wc/PD35+fjZfjpqQkIAjR47gxx9/RM+ePZGRkQEAKC0txbp161Ba\nWor8/HzMmDEDjY2NNu2LXBtDZ3JFcgTMArNNYebMmfjwww/x22+/obq6GtXV1bhy5YpNO42Pj4eH\nh2HXsbGxOHv2LAAgNzcXKSkp8Pb2RnBwMMLCwlBcXGzTvsi1MXQmVyNXwCww2xTUajUiIyPFD3Gp\n5eTkYNiwYQCAiooKqNXqJvvmXd7IHCF0fvVVuSshsp1cAbPA5J3XBFlZWRg6dCgGDRoEHx8fAIbz\nUrNmzWr1dfHx8aisrGy2feHCheLVSwsWLICPjw9SU1NNvo+pQDs9PV38XqvVQqvVmvk/IVeWlQX0\n72843H79dcBOf8MQ2Z2U92AuLCwUlyqylNmgOT4+Hn5+foiOjm5ytPCqjX+WrVmzBitXrsS3336L\n9u3bAwAyMzMBAGlpaQCAIUOGYP78+YiNjW1aNINmasHFi8C4cYC/P/DRR4Cvr9wVEbWNvQJmgSRX\nH0VFReHw4cOSFpafn48XXngBO3bsgL+/v7i9tLQUqampKC4uRnl5OeLi4nDq1KlmRwtsCmRKXZ0h\nfN6/H8jLA4KC5K6IyHL//CdQUwO8/bZ93l+Sq4+GDRuGr7/+WrKiAODZZ59FTU0N4uPjodFoMGPG\nDABAREQEkpOTERERgaFDh2L58uWcnqY28fEBVq0yXNvdvz+we7fcFRFZRu6AWWD2SMHX1xe1tbXw\n8fGBt7e34UUqlc1XINmCRwpkic2bDUttv/UW8NBDcldD1Dopl8g2RbHDa7ZiUyBLHTkCJCUBKSkM\noEnZ7DHBbEyyppCbm4vvvvtOvLmO3GsfsSlQWzCAJqWzd8AskCRTSEtLQ3Z2NiIjIxEeHo7s7GzM\nmTNHsiKJ7O322w23Mrz1VmDgQODXX+WuiKgpOSeYjZk9UoiOjsbBgwfh6ekJwLBAXkxMDA4dOuSQ\nAlvCIwWyhl4PLFliOG+7fj1wzz1yV0Qk7T2YzZHkSEGlUqGqqkp8XFVVxSuCyCmpVMALLwArVwKj\nRgGffip3RUTyTzAbMzvRPGfOHPTu3VucGN6xY4c4ZEbkjIYPNyy3nZQElJYygCZ5STnBLAWLguaK\nigqUlJRApVKhX79+6Nq1qyNqM4mnj0gKDKBJbo4KmAU2XX20f//+Jo+Fpwmnjnr37i1FjVZhUyCp\ncAKa5GTvCWZjNjUFDw8PREVFoUuXLi2+sKCgwPYKrcSmQFJiAE1ycGTALLDks9NkprBkyRJ8/vnn\n6NChAyZOnIgxY8bAz89P8iKJ5CYE0L16GQJoTkCTIygtYBaYzRROnz6NdevWYePGjbjjjjvwyiuv\nICYmxlH1tYhHCmQvnIAmR3HEBLMxSS5JDQ0NxahRo5CQkICSkhIcP35csgKJlCYyEtizx7D+zPjx\nhvO9RFLT6YCSEsO/MaUxeaRw+vRprF27Frm5uQgKCsLEiRMxYsQI/EUBI3c8UiB7YwBN9uTogFlg\nc9AcHR2N0aNHo2PHjk3e0JI7r9kTmwI5AgNosgc5AmaBTUHzvHnzxMtPa3gMTW6IATTZg1IDZgGX\nziayAANokoocAbOA91MgkhAnoMlWjp5gNibJ1UdEZMAluMlWSloi2xQ2BaI24D2gyVpKuQezOWZX\nSV28eHGTQw6VSoVOnTqhT58+Vg+xzZ07F3l5eVCpVOjSpQvWrFmD7t27AwAyMjKQk5MDT09PZGdn\nIyEhwap9ENkLA2iyhtIDZoHZTCE1NRV79+5FUlIS9Ho9Nm/ejOjoaPzyyy8YP348Zs+e3eadVldX\ni0tmLFu2DD/++CM++OADlJaWIjU1FSUlJSgvL0dcXBxOnDgBD6NUj5kCKQUDaLKUnAGzQJJMoays\nDPv378fixYuxZMkS7Nu3DxcuXMCOHTuwZs0aqwq7eQ2lmpoa+Pv7AzDcCzolJQXe3t4IDg5GWFgY\niouLrdoHkSNwAposoeQJZmNmm8LFixfh4+MjPvb29sb58+fRoUMHtG/f3uodv/LKKwgKCsKaNWvE\nez5XVFRArVaLz1Gr1SgvL7d6H0SOwACazHGGgFlgNlN46KGHEBsbi9GjR0Ov12PTpk1ITU3F1atX\nEdHKybH4+HhUVlY2275w4UIkJSVhwYIFWLBgATIzMzFz5kysXr26xfcxdevP9PR08XutViveGY5I\nDkIAvWSJIYDmBDQJhID5m28cv+/CwkIUFha26TUWzSmUlJSgqKgIKpUKAwYMQN++fa2tsZlff/0V\nw4YNw+HDh8XbfKalpQEAhgwZgvnz5yM2NrZp0cwUSME2bwamTGEATQYbNxqWSvn+e7krkXB4raGh\nAZWVlaivrxf/cg+yYYWwkydPokePHgAMQXNxcTE+/vhjMWguLi4Wg+ZTp041O1pgUyClYwBNAiUE\nzAJJmsKyZcswf/58BAQEwNPTU9x+6NAhqwsbP348jh8/Dk9PT4SGhuK9995DQEAAAMPppZycHHh5\neWHp0qVITExsXjSbAjkBTkCT3BPMxiRpCqGhoSguLjZ5W045sCmQs+AS3O5NriWyTZHkktSgoCBx\n6WwiahtOQLsvZ5lgNmb26qOQkBAMGjQIw4cPFy9Nlft+CkTOhBPQ7slZJpiNmW0KQUFBCAoKQl1d\nHerq6sSb7BBR2wwfDhQUGALo0lIG0K5uxQrnO0oAuHQ2kcMxgHZ9SguYBTYFzc8//zyWLl2KpKSk\nFt84Ly9PmiqtwKZAzo4BtGtTWsAssKkp7N27F3379jU5DSfnBDGbArkC3gPaNcl5D2ZzeOc1IifA\nCWjXoqQJZmOWfHaaDJqjo6NbfeOffvrJ+sqISMQA2rU4a8AsMHmkoNPpAADLly8HAEyePBl6vR6f\nfvopACArK8sxFbaARwrkihhAOz+lBswCSU4fxcTE4ODBg022aTQaHDhwwPYKrcSmQK6KAbRzU2rA\nLJBkolmv12Pnzp3i46KiIn4gE9kJJ6Cdl7NOMBszO7yWk5ODKVOm4I8//gAA3HrrrSbvfUBEtuME\ntHNy1glmYxZffSQ0hU6dOtm1IEvw9BG5Cy7B7TyUtES2KZJkCtevX8f69euh0+lQX18vvvG8efOk\nq7SN2BTInTCAVj6lB8wCSTKFUaNGIS8vD97e3vD19YWvry9uueUWyYokotbxHtDK50z3YDbH7JFC\nVFQUDh8+7Kh6LMIjBXJHnIBWJiVPMBuT5Ejh3nvv5aAakQIIAfTKlYYA+r8jQyQzVwmYBWaPFMLD\nw3Hq1CmEhISgXbt2hhfJPNHMIwVydwyglcMZAmaBJEGzMNlsLDg42Nq6bMamQMQAWgmcJWAWSHL6\nKDg4GGVlZSgoKEBwcDBuueUWyT6QFy9eDA8PD1y+fFnclpGRgR49eqBXr17YunWrJPshckUMoOXn\nSgGzwGxTSE9PxxtvvIGMjAwAQF1dHR5++GGbd1xWVoZt27bhjjvuELeVlpZi3bp1KC0tRX5+PmbM\nmIHGxkab90XkqjgBLR9XmWA2ZrYpbNiwAbm5ueJlqN26dUN1dbXNO541axbeeOONJttyc3ORkpIC\nb29vBAcHIywsDMXFxTbvi8iVMYCWh6sFzAKzTaFdu3bwuCnFunr1qs07zc3NhVqtxl133dVke0VF\nBdRqtfhYrVajvLzc5v0RuQNhCe65c4GXXwZ4kG1fzr5Etilm1z6aMGECnnrqKVRVVeH9999HTk4O\npk2bZvaN4+PjUVlZ2Wz7ggULkJGR0SQvaC2jUKlULW5PT08Xv9dqtbLeCY5IKSIjgT17DAH0uHHA\nxx8zgLYHnQ4oKQG++ELuSlpXWFho8u6Zpli09tHWrVvFD/HExETEx8dbVSAAHD58GIMHD0aHDh0A\nAGfPnkW3bt2wZ88ecaG9tLQ0AMCQIUMwf/58xMbGNi2aVx8RtYpLcNuX0pfINkXy23FevHgR/v7+\nJv96t0ZISAj27duHzp07o7S0FKmpqSguLkZ5eTni4uJw6tSpZvtjUyAyjxPQ9uFME8zGbLokdffu\n3dBqtRg7diwOHDiAqKgoREdHIzAwEFu2bJG0SEFERASSk5MRERGBoUOHYvny5ZI2ICJ3wgDaPlw1\nYBaYPFLo06cPMjIy8Mcff+CJJ55Afn4++vfvj2PHjmHSpEnN7sbmSDxSIGobYQJ60iTgf/+XE9C2\ncKYJZmM2nT66+Tac4eHhOHr0qPgz3o6TyPkIE9BdujCAtpazTTAbs+n00c2nbdq3by9dVUQkC2EC\n+rbbOAFtLVecYDZm8kjB09NTvELo2rVr+MtNv4Vr166JN9yRA48UiKzHANo6zhwwCyz57DQ5p9DQ\n0CB5QUQkP94D2jquHjALzA6vEZFrEiagk5IMQTQD6Na56gSzsTbNKSgFTx8RSYcBtHnOHjALJFk6\nm4hcGwNo89whYBawKRARl+BuhasukW0KmwIRAeAEtCnuEjALGDQTURMMoJtyl4BZwKCZiFrEANp1\nAmYBg2YishoDaPcKmAVsCkRkkjsH0O4WMAvYFIioVe4aQLtbwCxg0ExEFnG3ANrdAmYBg2YiahN3\nCKBdLWAWMGgmIskJAXTnzq4bQLtjwCxgUyCiNvPxMXxwPvKIYeltVwqg3TVgFrApEJFVVCpg1izg\n/fddK4B214BZIEtTSE9Ph1qthkajgUajwZYtW8SfZWRkoEePHujVqxe2bt0qR3lE1AZCAD13LvDy\ny0Bjo9wV2cZdA2aBLEHz/Pnz4efnh1mzZjXZXlpaitTUVJSUlKC8vBxxcXE4ceIEPIwucWDQTKQ8\nrhBAu2rALFB00NxSYbm5uUhJSYG3tzeCg4MRFhaG4uJiGaojorZyhQDanQNmgWxNYdmyZbj77rvx\n+OOPo6qqCgBQUVEBtVotPketVqO8vFyuEomojZw5gHb3gFlgt+G1+Ph4VFZWNtu+YMECPP3005g3\nbx4AYO7cuXjhhRewatWqFt9HpVK1uD09PV38XqvVQqvV2lwzEdlOCKDvvNO57gHtigFzYWEhCgsL\n2/Qa2YfXdDodkpKScOjQIWRmZgIA0tLSAABDhgzB/PnzERsb2+Q1zBSInMORI4YJ6EmTlD8BPXQo\nkJpqWOfJVSk2Uzh37pz4/YYNGxAdHQ0AGDlyJNauXYu6ujqcOXMGJ0+eRL9+/eQokYgkEBkJ7NkD\n7NxpCKFrauSuqGU6HVBSAowfL3cl8pNl7aPZs2fj4MGDUKlUCAkJwYoVKwAAERERSE5ORkREBLy8\nvLB8+XKTp4+IyDkIAfTTTxsC6Lw8IChI7qqaYsD8J9lPH1mDp4+InI9eb8gXFi8G/vMfQxCtBDdu\nAHfcYWhcrpQntESxp4+IyP0odQLaFQNmW3DpbCJyKKUtwe3uE8zGePqIiGShhAloV59gNsbTR0Sk\nWEqYgGbA3BybAhHJRs4JaE4wt4xNgYhkJVcAzYC5ZQyaiUgRHB1AM2BuGYNmIlIURwTQ7hYwCxg0\nE5HTcUSyPgBUAAAI10lEQVQAzYDZNDYFIlIcewbQDJhbx6ZARIpkrwCaAXPrGDQTkaJJHUAzYG4d\ng2YicgpSBNDuGjALGDQTkcuQIoBmwGwemwIROQ1bAmgGzJZhUyAip2JtAM2A2TIMmonIKbU1gGbA\nbBkGzUTk1CwJoN09YBYoOmhetmwZwsPDERUVhdmzZ4vbMzIy0KNHD/Tq1Qtbt26VqzwichKWBNAM\nmNtAL4Pt27fr4+Li9HV1dXq9Xq+/cOGCXq/X648cOaK/++679XV1dfozZ87oQ0ND9Q0NDc1eL1PZ\nilRQUCB3CYrB38Wf3PF30dio1y9erNf/7W96/a5df27ftq1A/9e/6vVHjshXm1JY8tkpy5HCe++9\nhzlz5sDb2xsAcPvttwMAcnNzkZKSAm9vbwQHByMsLAzFxcVylOg0CgsL5S5BMfi7+JM7/i5MBdAf\nfFDIgLkNZGkKJ0+exHfffYf+/ftDq9Vi7969AICKigqo1WrxeWq1GuXl5XKUSEROSgig584FXn4Z\n2LuXAXNb2O3qo/j4eFRWVjbbvmDBAtTX1+P333/HDz/8gJKSEiQnJ+Pnn39u8X1UKpW9SiQiFxUZ\nCezZYwigy8uB8ePlrsiJOOA0VjNDhgzRFxYWio9DQ0P1Fy9e1GdkZOgzMjLE7YmJifoffvih2etD\nQ0P1APjFL37xi19t+AoNDTX7+SzLnMLo0aOxfft2PPDAAzhx4gTq6urg7++PkSNHIjU1FbNmzUJ5\neTlOnjyJfv36NXv9qVOnZKiaiMj1ydIUpk6diqlTpyI6Oho+Pj746KOPAAARERFITk5GREQEvLy8\nsHz5cp4+IiJyIKccXiMiIvtwurWP8vPz0atXL/To0QNZWVlylyObqVOnIjAwENHR0XKXIruysjIM\nGjQIkZGRiIqKQnZ2ttwlyeb69euIjY1FTEwMIiIiMGfOHLlLkl1DQwM0Gg2SkpLkLkVWwcHBuOuu\nu6DRaFo8LS9wqiOFhoYG3Hnnnfjmm2/QrVs3/P3vf8dnn32G8PBwuUtzuO+//x6+vr545JFHcOjQ\nIbnLkVVlZSUqKysRExODmpoa9OnTBxs3bnTLfxcAUFtbiw4dOqC+vh4DBw7EokWLMHDgQLnLks2S\nJUuwb98+VFdXIy8vT+5yZBMSEoJ9+/ahc+fOrT7PqY4UiouLERYWhuDgYHh7e2PSpEnIzc2VuyxZ\n3HfffbjtttvkLkMRunbtipiYGACAr68vwsPDUVFRIXNV8unQoQMAoK6uDg0NDWY/BFzZ2bNn8dVX\nX2HatGlcLw2w6HfgVE2hvLwc3bt3Fx9zuI2M6XQ6HDhwALGxsXKXIpvGxkbExMQgMDAQgwYNQoQb\nj/L+z//8D95880142HL/ThehUqkQFxeHvn37YuXKlSaf51S/KV6JRK2pqanB+PHjsXTpUvhac69G\nF+Hh4YGDBw/i7Nmz+O6779xyyQsA+PLLLxEQEACNRsOjBABFRUU4cOAAtmzZgnfffRfff/99i89z\nqqbQrVs3lJWViY/LysqaLItB7uvGjRsYN24cHn74YYwePVruchShU6dOGD58uLiMjLvZtWsX8vLy\nEBISgpSUFGzfvh2PPPKI3GXJ5q9//SsAw1pzY8aMMbmunFM1hb59++LkyZPQ6XSoq6vDunXrMHLk\nSLnLIpnp9Xo8/vjjiIiIwMyZM+UuR1aXLl1CVVUVAODatWvYtm0bNBqNzFXJY+HChSgrK8OZM2ew\ndu1aPPjgg+JMlLupra1FdXU1AODq1avYunWrySsXnaopeHl54Z133kFiYiIiIiIwceJEt73CJCUl\nBffeey9OnDiB7t27Y/Xq1XKXJJuioiJ88sknKCgogEajgUajQX5+vtxlyeLcuXN48MEHERMTg9jY\nWCQlJWHw4MFyl6UI7nz6+fz587jvvvvEfxcjRoxAQkJCi891qktSiYjIvpzqSIGIiOyLTYGIiERs\nCkREJGJTICIiEZsCERGJ2BSIiEjEpkAuzd7LXQQHB+Py5cvNtu/YsQO7d+9u8TWbNm1y62XfSdlk\nufMakaPYe2BJpVK1uK5OQUEB/Pz8cM899zT7WVJSktuv7U/KxSMFcjunT5/G0KFD0bdvX9x///04\nfvw4AOCxxx7D888/jwEDBiA0NBTr168HYFh1dMaMGQgPD0dCQgKGDx8u/gwAli1bhj59+uCuu+7C\n8ePHodPpsGLFCrz11lvQaDTYuXNnk/2vWbMGzz77bKv7vJlOp0OvXr0wZcoU3HnnnXjooYewdetW\nDBgwAD179kRJSYm9flXkhtgUyO08+eSTWLZsGfbu3Ys333wTM2bMEH9WWVmJoqIifPnll0hLSwMA\nfPHFF/jll19w9OhRfPzxx9i9e3eTI5Dbb78d+/btw9NPP41FixYhODgY06dPx6xZs3DgwIFmN7gx\nPnppaZ/GTp8+jRdffBHHjh3D8ePHsW7dOhQVFWHRokVYuHChVL8aIp4+IvdSU1OD3bt3Y8KECeK2\nuro6AIYPa2GF1fDwcJw/fx4AsHPnTiQnJwOAeI+Cm40dOxYA0Lt3b3zxxRfidktWkDG1T2MhISGI\njIwEAERGRiIuLg4AEBUVBZ1OZ3Y/RJZiUyC30tjYiFtvvRUHDhxo8ec+Pj7i98KHunFuYPxh365d\nOwCAp6cn6uvr21xTS/s0JuwDMNwvQXiNh4eHVfskMoWnj8itdOzYESEhIfjPf/4DwPAh/NNPP7X6\nmgEDBmD9+vXQ6/U4f/48duzYYXY/fn5+4lLFxrgGJSkZmwK5tNraWnTv3l38evvtt/Hpp59i1apV\niImJQVRUVJObud98vl/4fty4cVCr1YiIiMDkyZPRu3dvdOrUqdm+VCqV+JqkpCRs2LABGo0GRUVF\nJp9nap8tvbepx+68JDRJj0tnE1ng6tWruOWWW/Dbb78hNjYWu3btQkBAgNxlEUmOmQKRBUaMGIGq\nqirU1dVh3rx5bAjksnikQEREImYKREQkYlMgIiIRmwIREYnYFIiISMSmQEREIjYFIiIS/T+6l7ug\nkeDZfAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x567bcf0>"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.7,Page No.109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_DB=2 #m #Length of DB\n",
+ "L_AD=4 #m #Length 0f AD\n",
+ "M_D=30 #KN.m #Moment at D\n",
+ "w=45 #KN/m #u.d.l\n",
+ "L=7 #m #Span of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_A be the Reactions at B & A respectively\n",
+ "#R_B+R_A=180+P ............(1)\n",
+ "\n",
+ "#Now Taking Moment about A,we get\n",
+ "#R_B=7*P+390 ...............(2)\n",
+ "\n",
+ "#Since R_A & R_B Are Equal\n",
+ "#2*R_B=180+P ...................(3)\n",
+ "\n",
+ "#From equation 1 and 3 we get\n",
+ "#3*(180+P)=7P+390\n",
+ "#After simplifying Further above equation we get\n",
+ "P=150*4**-1 #KN\n",
+ "R_A=R_B=(180+P)*2**-1\n",
+ "F_C=P\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=0 #KN\n",
+ "V_C2=-P #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C2 #KN\n",
+ "V_B2=V_C2+R_B #KN \n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_B2 #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_D-w*L_AD #KN\n",
+ "V_A2=V_A1+R_A #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at C\n",
+ "M_C=0 #KN.m \n",
+ "\n",
+ "#B.M at B\n",
+ "M_B=F_C*L_BC #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D1=F_C*(L_BC+L_DB)-R_B*L_DB #KN.m\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AD*L_AD*2**-1+M_D-R_B*(L_AD+L_DB)+P*L\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BC,L_BC,L_DB+L_BC,L_DB+L_BC+L_AD,L_DB+L_BC+L_AD]\n",
+ "Y1=[V_C1,V_C2,V_B1,V_B2,V_D,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_BC,L_DB+L_BC,L_DB+L_BC,L_AD+L_DB+L_BC]\n",
+ "Y2=[M_C,M_B,M_D1,M_D2,M_A]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d6f150>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d79470>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.8,Page No.110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6 #m #Span Of beam\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Due to Symmetry\n",
+ "#Let R_B and R_C be the reactions at B & C Respectively\n",
+ "R_B=R_C=w*L*2**-1 #KN\n",
+ "\n",
+ "#Let a be the overhang.The Max -ve moment occurs at the support and max +ve moment at middle of the beam\n",
+ "#Now Equating these two equations we get\n",
+ "#30*a*a*2**-1=90*(3-a)-w*L*2**-1*L*4**-1\n",
+ "#After simplifying we get an equation as\n",
+ "#a**2+6*a-9=0\n",
+ "x=1\n",
+ "y=6\n",
+ "z=-9\n",
+ "\n",
+ "p=y**2-4*x*z\n",
+ "\n",
+ "a1=(-y+p**0.5)*2**-1\n",
+ "a2=(-y-p**0.5)*2**-1\n",
+ "\n",
+ "#Now Length cannot be negative,so taking a1 into Consideration\n",
+ "\n",
+ "L_CD=L_AB=a1\n",
+ "L_BC=L-2*a1\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=0\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C1=V_D-w*L_CD #KN\n",
+ "V_C2=V_C1+R_C #KN\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=-w*(L_BC+L_CD)+R_C\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=round(V_B2,2)-round(w*L_AB,2)\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=0\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=w*L_CD*L_CD*2**-1 #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=w*(L_BC+L_CD)*(L_BC+L_CD)*2**-1-R_C*L_BC*L_BC*2**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "X=w*L*L*2**-1\n",
+ "Y=-R_C*(L_AB+L_BC)-R_B*L_AB\n",
+ "M_A=X+Y\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_CD,L_CD,L_CD+L_BC,L_CD+L_BC,L_CD+L_BC+L_AB]\n",
+ "Y1=[V_D,V_C1,V_C2,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_CD,L_BC+L_CD,L_AB+L_BC+L_CD]\n",
+ "Y2=[M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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lXLhwAQBw6dIlHDhwwKleBejt7Q0/Pz+UlpYCAA4ePIiwsLAub6vKm9f6yt3d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\nAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5945590>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LUVxcrNtAqS48PDxqdH5WVhby8vIQEBAAAJg4cSJ27NjBhEGkEDc3ID5e7BWe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\nMQcOHJASExMlb29vtUPRi6y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+ "text": [
+ "<matplotlib.figure.Figure at 0x58e1750>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.9,Page No.112"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_F=6 #KN #Force at F\n",
+ "w1=w2=w=3 #KN.m #u.d.l\n",
+ "M_D=24 #KN.m \n",
+ "L_AB=L_CD=L_DE=L_EF=4 #m #Length of AB,CD,DE,EF\n",
+ "L_BC=2 #m #Length of BC\n",
+ "L=18 #m #Span of Beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_B and R_E be the Reactions at B & E respectively\n",
+ "#R_B+R_E=42\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(F_F*(L_BC+L_CD+L_DE+L_EF)+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)-w*L_AB*L_AB*2**-1-M_D)*(L_BC+L_CD+L_DE)**-1\n",
+ "R_B=42-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F aT F\n",
+ "V_F1=0 #KN \n",
+ "V_F2=-F_F #KN\n",
+ "\n",
+ "#S.F at E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_E1+R_E #KN\n",
+ "\n",
+ "#S.F aT C\n",
+ "V_C=V_E2-w*(L_CD+L_DE) #KN\n",
+ "\n",
+ "#S.F at B\n",
+ "V_B1=V_C #KN \n",
+ "V_B2=V_C+R_B #KN\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A=V_B2-w*L_AB #KN\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=0\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_F*L_EF #KN.m\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D1=F_F*(L_DE+L_EF)-R_E*L_DE+w*L_DE*L_DE*2**-1 #KN.m\n",
+ "M_D2=M_D1-M_D\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_F*(L_CD+L_DE+L_EF)-R_E*(L_CD+L_DE)+w*(L_CD+L_DE)*(L_CD+L_DE)*2**-1-M_D\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_F*(L_BC+L_CD+L_DE+L_EF)-R_E*(L_BC+L_CD+L_DE)-M_D+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=w*L_AB*L_AB*2**-1-R_B*L_AB+w*(L_CD+L_DE)*((L_CD+L_DE)*2**-1+L_BC+L_AB)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_F*L-M_D\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_EF,L_EF,L_EF+L_DE+L_CD,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC,L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_F1,V_F2,V_E1,V_E2,V_C,V_B1,V_B2,V_A]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5d75ab0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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VxMfHIzs7G/369cOvv/6KcePG4bHHHnNknEROixVL5KzMziBqa2uRmJiIMWPG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\nfyRzHAAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5c44130>"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.10,Page No.114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DC=L_BA=2 #m #Length of BA & DC\n",
+ "L_CB=1 #m #Length of CB\n",
+ "F_A=10 #KN #Force at pt A\n",
+ "F_B=20 #KN #Force at pt B\n",
+ "w=4 #KN.m #u.d.l\n",
+ "L=5 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_D be the reactions at Pt D\n",
+ "R_D=F_B+F_A+w*L_DC #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F at Pt A\n",
+ "V_A1=0 #KN\n",
+ "V_A2=F_A #KN\n",
+ "\n",
+ "#S.F At Pt B\n",
+ "V_B1=V_A2\n",
+ "V_B2=F_B+F_A\n",
+ "\n",
+ "#S.F at Pt C\n",
+ "V_C=F_B+F_A #KN \n",
+ "\n",
+ "#S.F At Pt D\n",
+ "V_D1=V_B2+w*L_DC\n",
+ "V_D2=F_B+F_A+w*L_DC-R_D\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=0\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=F_A*L_BA\n",
+ "\n",
+ "#B.M at Pt C\n",
+ "M_C=F_B*L_CB+F_A*(L_BA+L_CB) #KN\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1\n",
+ "M_D2=(F_A*L+F_B*(L_CB+L_DC)+w*L_DC*L_DC*2**-1)-M_D1\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_BA,L_BA,L_BA+L_CB,L_BA+L_CB+L_DC,L_BA+L_CB+L_DC]\n",
+ "Y1=[V_A1,V_A2,V_B1,V_B2,V_C,V_D1,V_D2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_A,M_B,M_C,M_D1,M_D2]\n",
+ "X2=[0,L_BA,L_CB+L_BA,L_CB+L_BA+L_DC,L_CB+L_BA+L_DC]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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e3542bZr69++vhIQEzZ49WyNGjNAFF1wgSbrxxht15ZVXen+5e/LxJ0/9axiG\n8vLy9Oc//1kxMTFatWqV8vLyvMNNvo71tX3yMb6+tvMUxDh7XM4JWzp8+LAiIiL0ww8/aPTo0dq8\nebP69OljdSwgKBjjhy1NnjxZNTU1amho0L333kvpw1Y44wcAm2GMHwBshuIHAJuh+AHAZih+ALAZ\nih8AbIbiBwCb+T+gqVKkQGpm/QAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5738d10>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5788db0>"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.11,Page No.115"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "w=20 #KN/m #u.v.l\n",
+ "F_C=40 #KN #Force at Pt C\n",
+ "M_D=40 #KN.m #Moment at pt D\n",
+ "L_AB=3 #m #Length of AB\n",
+ "L_BC=1 #m #Length of BC\n",
+ "L_CD=L_DE=2 #m #Length of CD & DE\n",
+ "L=8 #8 #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A & R_E be the Reactions at A & E respectively\n",
+ "#R_A+R_E=70\n",
+ "\n",
+ "#Taking Moments At Pt A we get,M_A\n",
+ "R_E=(F_C*(L_AB+L_BC)+1*2**-1*L_AB*w*2+40)*L**-1\n",
+ "R_A=70-R_E\n",
+ "\n",
+ "#shear Force Calculations\n",
+ "\n",
+ "#S.F At Pt E\n",
+ "V_E1=0\n",
+ "V_E2=R_E #KN\n",
+ "\n",
+ "#S.F aT pt D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At PT C\n",
+ "V_C1=V_D\n",
+ "V_C2=V_D-F_C #KN\n",
+ "\n",
+ "#S.F At Pt A\n",
+ "V_A1=V_C2-(1*2**-1*w*L_AB)\n",
+ "V_A2=V_A1+R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt E\n",
+ "M_E=0\n",
+ "\n",
+ "#B.M At Pt D\n",
+ "M_D1=M_E-R_E*L_DE\n",
+ "M_D2=M_D1+M_D\n",
+ "\n",
+ "#B.M At Pt C\n",
+ "M_C=-R_E*(L_DE+L_CD)+M_D\n",
+ "\n",
+ "#B.M At Pt B\n",
+ "M_B=-R_E*(L_DE+L_CD+L_BC)+M_D+F_C*L_BC\n",
+ "\n",
+ "#B.M At Pt A\n",
+ "M_A=-R_E*L+M_D+(1*2**-1*L_AB*w*2)+F_C*(L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_DE,L_CD+L_DE,L_CD+L_DE,L_CD+L_DE+L_AB,L_CD+L_DE+L_AB]\n",
+ "Y1=[V_E1,V_E2,V_D,V_C1,V_C2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "Y2=[M_E,M_D1,M_D2,M_C,M_B,M_A]\n",
+ "X2=[0,L_DE,L_DE,L_CD+L_DE,L_DE+L_CD+L_BC,L_AB+L_BC+L_CD+L_DE]\n",
+ "Z2=[0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5df5230>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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PZcuW8dRTT5GZmcnOnTtp1KgRI0aMsPs+cvTnxbp1U4fwLFyoO4nzTCaYMkWt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\nd/r0aduBO+VVUFCA2Wxm+/b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6ePR5C6Jyk56CqHT27dtHt27daNu2LXfddRdpaWkAPPbY\nYzz33HPcfvvtNG/enCVLlgCqQmt8fDyhoaF06dKF++67z/Y9gDlz5tCmTRtatmxJWloaWVlZzJs3\njzfffJOoqCg2bNhw0fU/+ugjnnnmmSte80JZWVmEhITw+OOPc/PNN/Pwww+zatUqbr/9dlq0aMHW\nrVvd9aMSlZA0CqLS+cc//sGcOXP46aefeP3114mPj7d979ChQ2zcuJFvvvmGUaNGAfD111/zf//3\nf/zyyy98+umnbN68+aIeSIMGDdi2bRtPPfUUM2fOJDAwkH/+85+88MIL7NixgzvuuOOi61/aeynp\nmpfat28fL774Ir/++itpaWkkJSWxceNGZs6cydSpU131oxFCho9E5ZKXl8fmzZsvKnGcn58PqJt1\ncaXY0NBQ/vjjDwA2bNhAbGwsgO08hwv17t0bgNatW/P111/bHnekgoy9a16qadOmtoJwYWFhtlr4\n4eHhZGVllXodIRwljYKoVIqKiqhfvz47duwo8fvVqlWzfV58U7903uDSm3316tUBqFq1KgUFBWXO\nVNI1L1V8DVBnSxS/pkqVKuW6phD2yPCRqFTq1q1L06ZN+eqrrwB1E961a9cVX3P77bezZMkSDMPg\njz/+YN26daVep06dOrYS1ZeSGpTCk0mjIHzaqVOnaNy4se3jrbfe4vPPP+eDDz6gVatWhIeHs2zZ\nMtvzLxzvL/68T58+mM1mLBYLjz76KK1bty7xvGCTyWR7TY8ePVi6dClRUVFs3LjR7vPsXbOk97b3\ntZxIKFxJSmcL4YCTJ09Sq1Ytjhw5QocOHdi0aRPXX3+97lhCuJzMKQjhgPvvv59jx46Rn5/P+PHj\npUEQPkt6CkIIIWxkTkEIIYSNNApCCCFspFEQQghhI42CEEIIG2kUhBBC2EijIIQQwub/A4tFvTZr\nGhPHAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5dff130>"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.12,Page No.116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F_G=10 #KN #Force at Pt G\n",
+ "F_B=F_E=15 #KN #Force at Pt B & E\n",
+ "w=20 #KN/m #U.d.L\n",
+ "L_FG=L_EF=L_DE=L_CD=L_BC=L_AB=1 #m #Lengths of FG,EF,DE,CD,BC,AB respectively\n",
+ "L=6 #m #Length of beam\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt R_F & R_A be the Reactions at E & A respectively\n",
+ "#R_F+R_A=60\n",
+ "\n",
+ "#Taking Moment At Pt A,M_A\n",
+ "R_F=(F_G*L+F_E*(L_AB+L_BC+L_CD+L_DE)+w*L_CD*(L_AB+L_BC+L_CD*2**-1)+F_B*L_AB)*(L_AB+L_BC+L_CD+L_DE+L_EF)**-1\n",
+ "R_A=60-R_F\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At G\n",
+ "V_G1=0 #KN \n",
+ "V_G2=F_G #KN\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F1=V_G2 #KN\n",
+ "V_F2=V_F1-R_F\n",
+ "\n",
+ "#S.F At E\n",
+ "V_E1=V_F2 #KN\n",
+ "V_E2=V_F2+F_E\n",
+ "\n",
+ "#S.F At D\n",
+ "V_D=V_E2\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_E2+w*L_CD\n",
+ "\n",
+ "#S.F At B\n",
+ "V_B1=V_C\n",
+ "V_B2=V_B1+F_B\n",
+ "\n",
+ "#S.F At A\n",
+ "V_A1=V_B2\n",
+ "V_A2=V_B2-R_A\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M At Pt G\n",
+ "M_G=0\n",
+ "\n",
+ "#B.M At F\n",
+ "M_F=F_G*L_FG \n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=F_G*(L_FG+L_EF)-R_F*L_EF\n",
+ "\n",
+ "#B.M At D\n",
+ "M_D=F_G*(L_FG+L_EF+L_DE)-R_F*(L_EF+L_DE)+F_E*L_DE\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_G*(L_FG+L_EF+L_DE+L_CD)-R_F*(L_EF+L_DE+L_CD)+F_E*(L_DE+L_CD)+w*L_CD*L_CD*2**-1\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_G*(L_FG+L_EF+L_DE+L_CD+L_BC)-R_F*(L_EF+L_DE+L_CD+L_BC)+F_E*(L_DE+L_CD+L_BC)+w*L_CD*(L_CD*2**-1+L_BC)\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A=F_G*L-R_F*(L_EF+L_DE+L_CD+L_BC+L_AB)+F_E*(L_DE+L_CD+L_BC+L_AB)+F_B*L_AB+w*L_CD*(L_CD*2**-1+L_BC+L_AB)\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,0,L_FG,L_FG,L_FG+L_EF,L_FG+L_EF,L_FG+L_EF+L_DE,L_FG+L_EF+L_DE+L_CD,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB,L_FG+L_EF+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y1=[V_G1,V_G2,V_F1,V_F2,V_E1,V_E2,V_D,V_C,V_B1,V_B2,V_A1,V_A2]\n",
+ "Z1=[0,0,0,0,0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_FG,L_EF+L_FG,L_EF+L_FG+L_DE,L_EF+L_FG+L_DE+L_CD,L_EF+L_FG+L_DE+L_CD+L_BC,L_EF+L_FG+L_DE+L_CD+L_BC+L_AB]\n",
+ "Y2=[M_G,M_F,M_E,M_D,M_C,M_B,M_A]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2)\n",
+ "plt.xlabel(\"Lenght in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x57502f0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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9jlmzZjkcDuPv6NGjR8t9nl+NFHJycmjatCnR0dGEh4dz1113kZmZ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\nkcIBAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58b0ed0>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3.3.13,Page No.117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L_AB=L_BC=L_CD=L_DE=L_EF=1 #m #LEngth of AB,BC,CD,DE,EF respectively\n",
+ "M_A=50 #KN/m #Moment at A\n",
+ "w=5 #KN/m #u.v.l\n",
+ "F_D=10 #KN\n",
+ "w2=5 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_B & R_E be the Reactions at B and E respectively\n",
+ "#R_B+R_E=20\n",
+ "\n",
+ "#Taking Moment At Pt B,M_B\n",
+ "R_E=(w2*L_EF*(L_EF*2**-1+L_DE+L_CD+L_BC)+w*L_BC*2**-1*2*3**-1+50+F_D*(L_BC+L_CD))*3**-1\n",
+ "R_B=17.5-R_E #KN\n",
+ "\n",
+ "#Shear Force Calculations\n",
+ "\n",
+ "#S.F At F\n",
+ "V_F=0\n",
+ "\n",
+ "#S.F aT E\n",
+ "V_E1=-w2*L_EF #KN\n",
+ "V_E2=V_E1+R_E\n",
+ "\n",
+ "#S.F at D\n",
+ "V_D1=R_E-w2*L_EF #KN\n",
+ "V_D2=V_D1-F_D #KN\n",
+ "\n",
+ "#S.F At C\n",
+ "V_C=V_D2\n",
+ "\n",
+ "#S.F aT B\n",
+ "V_B1=-L_BC*w*2**-1-F_D+R_E-w2*L_EF\n",
+ "V_B2=V_B1+R_B\n",
+ "\n",
+ "#Bending Moment Calculations\n",
+ "\n",
+ "#B.M at F\n",
+ "M_F=0 #KN.m\n",
+ "\n",
+ "#B.M At E\n",
+ "M_E=w2*L_EF*L_EF*2**-1 #KN.m\n",
+ "\n",
+ "#B.M at D\n",
+ "M_D=-R_E*L_DE+w2*L_EF*(L_EF*2**-1+L_DE) #KN.m\n",
+ "\n",
+ "#B.M At C\n",
+ "M_C=F_D*L_CD*R_E*(L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_DE+L_CD) #KN.m\n",
+ "\n",
+ "#B.M At B\n",
+ "M_B=F_D*(L_CD+L_BC)-R_E*(L_BC+L_CD+L_DE)+w2*L_EF*(L_EF*2**-1+L_BC+L_CD+L_DE)+1*2**-1*L_BC*w*2*3**-1\n",
+ "\n",
+ "#B.M At A\n",
+ "M_A1=w*L_EF*(L_EF*2**-1+L_AB+L_BC+L_CD+L_DE)-R_E*(L_AB+L_BC+L_CD+L_DE)+F_D*(L_AB+L_BC+L_CD)+1*2**-1*L_BC*w*(2*3**-1*L_BC+L_AB)-R_B*L_AB\n",
+ "M_A2=M_A1+M_A\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,L_EF,L_EF,L_DE+L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC]\n",
+ "Y1=[V_F,V_E1,V_E2,V_D1,V_D2,V_C,V_B1,V_B2]\n",
+ "Z1=[0,0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Plotting the Bendimg Moment Diagram\n",
+ "\n",
+ "X2=[0,L_EF,L_DE+L_EF,L_CD+L_DE+L_EF,L_CD+L_DE+L_EF+L_BC,L_CD+L_DE+L_EF+L_BC+L_AB,L_CD+L_DE+L_EF+L_BC+L_AB]\n",
+ "Y2=[M_F,M_E,M_D,M_C,M_B,M_A1,M_A2]\n",
+ "Z2=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in m\")\n",
+ "plt.ylabel(\"Bending Moment in kN.m\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5c2a7b0>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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LlixREhMTLY8fMWKEsm/fvjuex4ZfSQi7euUVRXnxRa1TuIdr1xRFr1eUb77R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\ntnx2yhiGEEIIm0jBEEIIYRM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+ "text": [
+ "<matplotlib.figure.Figure at 0x1a108f0>"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_5.ipynb new file mode 100644 index 00000000..09aec3f9 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.4_5.ipynb @@ -0,0 +1,1628 @@ +{
+ "metadata": {
+ "name": "chapter no.4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:Stresses in Beams"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.1,Page no.130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=5000 #mm #Length of Beam\n",
+ "a=2000 #mm #Length of start of beam to Pt Load\n",
+ "b=3000 #mm #Length of Pt load to end of beam\n",
+ "A=150*250 #m**2 #Area of beam\n",
+ "b=150 #mm #Width of beam\n",
+ "d=250 #mm #Depth of beam\n",
+ "sigma=10#N/mm**2 #stress\n",
+ "l=2000 #m #Load applied from one end\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3 #m**4\n",
+ "\n",
+ "#Distance from N.A to end\n",
+ "y_max=d*2**-1 #m\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=1*6**-1*b*d**2 #mm**3\n",
+ "\n",
+ "#Moment Carrying Capacity\n",
+ "M=sigma*Z #N-mm\n",
+ "\n",
+ "#Let w be the Intensity of the Load in N/m,then Max moment\n",
+ "#M_max=w*L**2*8**-1 #N-mm\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M_max=w*25*100*8**-1\n",
+ "\n",
+ "#EQuating it to moment carrying capacity,we get max intensity load\n",
+ "w=M*(25*1000)**-1*8*10**-3\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#Let P be the concentrated load,then max moment occurs under the load and its value\n",
+ "#M1=P*a*b*L**-1 #N-mm\n",
+ "\n",
+ "#Equting it to moment carrying capacity we get\n",
+ "P=M*1200**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of u.d.l it can carry\",round(w,3),\"KN-m\"\n",
+ "print\"MAx concentrated Load P apllied at 2 m from one end is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of u.d.l it can carry 5.0 N-mm\n",
+ "MAx concentrated Load P apllied at 2 m from one end is 13.021 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.2,Page no.131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=70 #mm #External Diameter\n",
+ "t=8 #mm #Thickness of pipe\n",
+ "L=2500 #mm #span \n",
+ "sigma=150 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Diameter \n",
+ "d=D-2*t #mm\n",
+ "\n",
+ "#M.I Of Pipe\n",
+ "I=pi*64**-1*(D**4-d**4) #mm**4\n",
+ "\n",
+ "y_max=D*2**-1 #mm\n",
+ "Z=I*(y_max)**-1 #mm**3\n",
+ "\n",
+ "#Moment Carrying capacity\n",
+ "M=sigma*Z #N*mm\n",
+ "\n",
+ "#Max moment int the beam occurs at the mid-span and is equal to\n",
+ "#m=P*L*4**-1\n",
+ "\n",
+ "#Equating Max moment to moment carrying capacity we get,\n",
+ "#M=P*2.5*L*4**-1\n",
+ "#After substituting and simplifying we get\n",
+ "P=4*M*(L)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Max concentrated load that can be applied at the centre of span is\",round(P,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max concentrated load that can be applied at the centre of span is 5.22 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.3,Page no.132"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges Dimension\n",
+ "b1=180 #mm #Width\n",
+ "d1=10 #mm #Thickness\n",
+ "\n",
+ "D=500 #mm #Overall depth\n",
+ "t=8 #mm #Thickness of web\n",
+ "\n",
+ "#Plate Dimensions\n",
+ "b2=240 #mm #Width\n",
+ "t2=12 #mm #Thickness\n",
+ "\n",
+ "sigma=150 #N/mm**2 #Stress\n",
+ "L=3000 #mm #span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b2*t2*(D+t2*2**-1)+b1*d1*(D-t1*2**-1)+(D-2*t1)*t*D*2**-1+(b1*t1*t1*2**-1))*(b2*t2+b1*d1+b1*d1+(D-2*d1)*t)**-1\n",
+ "\n",
+ "#M.I of section\n",
+ "I=(1*12**-1*b2*t2**3+b2*t2*(D+t2*2**-1-y_bar)**2+1*12**-1*b1*d1**3+b1*d1*(D-t1*2**-1-y_bar)**2+1*12**-1*b1*t1**3+b1*t1*(t1*2**-1-y_bar)**2+1*12**-1*t*(D-2*t1)**3+t*(D-2*t1)*(D*2**-1-y_bar)**2)\n",
+ "\n",
+ "#Section Modulus\n",
+ "Z=I*(y_bar)**-1 #mm**3\n",
+ "\n",
+ "#Moment or Resistance\n",
+ "M=sigma*Z\n",
+ "\n",
+ "#Let Load on Cantilever be w/m Length \n",
+ "#Max M.I produced\n",
+ "#M_max=w*L**2**-1 \n",
+ "\n",
+ "#Now Equating Moment of resistance to Max moment,we get Max load\n",
+ "#4.5*w=M\n",
+ "#After rearranging and further simplifying we get\n",
+ "w=M*4.5**-1*10**3*10**-9\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of Resistance is\",round(M,2),\"KN-mm\"\n",
+ "print\"Load the section can carry is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of Resistance is 198770121.83 KN-mm\n",
+ "Load the section can carry is 44.171 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.4,Page no.134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange (Top)\n",
+ "b1=80 #mm #Width \n",
+ "t1=40 #mm #Thickness\n",
+ "\n",
+ "#Flange (Bottom)\n",
+ "b2=160 #mm #width\n",
+ "t2=40 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=120 #mm #Depth\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=200 #mm #Overall Depth\n",
+ "sigma1=30 #N/mm**2 #Tensile stress\n",
+ "sigma2=90 #N/mm**2 #Compressive stress\n",
+ "L=6000 #mm #Span\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of centroid from bottom fibre\n",
+ "y_bar=(b1*t1*(D-t1*2**-1)+d*t3*(d*2**-1+t2)+b2*t2*t2*2**-1)*(b1*t1+d*t3+b2*t2)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b1*t1**3+b1*t1*(D-t1*2**-1-round(y_bar,2))**2+1*12**-1*t3*d**3+t3*d*(d*2**-1+t2-round(y_bar,2))**2+1*12**-1*b2*t2**3+b2*t2*(t2*2**-1-round(y_bar,2))**2\n",
+ "\n",
+ "#Extreme fibre distance of top and bottom fibres are y_t and y_c respectively\n",
+ "\n",
+ "y_t=y_bar #mm\n",
+ "y_c=D-y_bar #mm\n",
+ "\n",
+ "#Moment carrying capacity considering Tensile strength \n",
+ "M1=sigma1*I*y_t**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Moment carrying capacity considering compressive strength \n",
+ "M2=sigma2*I*y_c**-1*10**-6 #KN-m\n",
+ "\n",
+ "#Max Bending moment in simply supported beam 6 m due to u.d.l\n",
+ "#M_max=w*L*10**-3*8**-1\n",
+ "#After simplifying further we get\n",
+ "#M_max=4.5*w\n",
+ "\n",
+ "#Now Equating it to Moment carrying capacity, we get load carrying capacity\n",
+ "w=M1*4.5**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Uniformly Distributed Load is\",round(w,3),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Uniformly Distributed Load is 5.096 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.5,Page no.136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b=200 #mm #Width\n",
+ "t=25 #mm #Thickness \n",
+ "\n",
+ "D1=500 #mm #Overall Depth\n",
+ "t2=20 #mm #Thickness of web\n",
+ "\n",
+ "d=450 #mm #Depth of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider,Element of Thickness \"y\" at Distance \"dy\" from N.A \n",
+ "#Let Bending stress \"sigma_max\"\n",
+ "\n",
+ "#Stress on the element \n",
+ "#sigma=y*(D*2**-1)*sigma_max ..............(1)\n",
+ "\n",
+ "#Area of Element\n",
+ "#A=b*dy .................................(2)\n",
+ "\n",
+ "#Force on Element \n",
+ "#F=y*250**-1*sigma_max*b*dy\n",
+ "\n",
+ "#Let M be the Moment of resistance\n",
+ "#M=y*250**-1*sigma_max*b*dy*y\n",
+ "\n",
+ "#Moment of Resistance of top flange be M1\n",
+ "def integrand(y, b, D):\n",
+ " return b*y**2*D**-1\n",
+ "b=200 \n",
+ "D=250\n",
+ "\n",
+ "X = quad(integrand, 225, 250, args=(b,D))\n",
+ "\n",
+ "Y=2*X[0]\n",
+ "\n",
+ "#M1=Y*sigma\n",
+ "\n",
+ "#Now Moment of Inertia I section is\n",
+ "X=b*D1**3\n",
+ "Y=(b-t2)*d**3\n",
+ "I=(X-Y)*12**-1*10**-8\n",
+ "\n",
+ "#Moment acting on the entire section\n",
+ "#since sigmais the value at y=250\n",
+ "y_max=250\n",
+ "Z=I*10**8*y_max**-1\n",
+ "#M=sigma*Z \n",
+ "#After Simplifying Further we get\n",
+ "#M2=Z*sigma\n",
+ "\n",
+ "#Percentage Moment resisted by Flanges\n",
+ "P1=2258333.3*(2865833.3)**-1*100\n",
+ "\n",
+ "#Percentage Moment resisted by web\n",
+ "P2=100-P1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Moment resisted by Flanges\",round(P1,2),\"%\"\n",
+ "print\"Percentage Moment resisted by web\",round(P2,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Moment resisted by Flanges 78.8 %\n",
+ "Percentage Moment resisted by web 21.2 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.6,Page no.137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flanges\n",
+ "b1=200 #mm #Width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=380 #mm #Depth \n",
+ "t2=8 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "sigma=150 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Area\n",
+ "A=b1*t1+d*t2+b1*t1 #mm**2\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*(b1*D**3-(b1-t2)*d**3)\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=sigma*I*(D*2**-1)**-1\n",
+ "\n",
+ "#Square Section\n",
+ "\n",
+ "#Let 'a' be the side\n",
+ "a=A**0.5\n",
+ "\n",
+ "#Moment of Resistance of this section\n",
+ "M1=1*6**-1*a*a**2*sigma\n",
+ "\n",
+ "X=M*M1**-1\n",
+ "\n",
+ "#Rectangular section\n",
+ "#Let 'a' be the side and depth be 2*a\n",
+ "\n",
+ "a=(A*2**-1)**0.5\n",
+ "\n",
+ "#Moment of Rectangular secction\n",
+ "M2=1*6**-1*a*(2*a)**2*sigma\n",
+ "\n",
+ "X2=M*M2**-1\n",
+ "\n",
+ "#Circular section\n",
+ "#A=pi*d1**2*4**-1\n",
+ "\n",
+ "d1=(A*4*pi**-1)**0.5\n",
+ "\n",
+ "#Moment of circular section\n",
+ "M3=pi*32**-1*d1**3*sigma\n",
+ "\n",
+ "X3=M*M3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Moment of resistance of beam section\",round(M,2),\"mm\"\n",
+ "print\"Moment of resistance of square section\",round(X,2),\"mm\"\n",
+ "print\"Moment of resistance of rectangular section\",round(X2,2),\"mm\"\n",
+ "print\"Moment of resistance of circular section\",round(X3,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Moment of resistance of beam section 141536000.0 mm\n",
+ "Moment of resistance of square section 9.58 mm\n",
+ "Moment of resistance of rectangular section 6.78 mm\n",
+ "Moment of resistance of circular section 11.33 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.7,Page no.139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=12 #KN #Force at End of beam\n",
+ "L=2 #m #span\n",
+ "\n",
+ "#Square section \n",
+ "b=d=200 #mm #Width and depth of beam\n",
+ "\n",
+ "#Rectangular section\n",
+ "b1=150 #mm #Width\n",
+ "d1=300 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max bending Moment\n",
+ "M=F*L*10**6 #N-mm\n",
+ "\n",
+ "#M=sigma*b*d**2\n",
+ "sigma=M*6*(b*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Let W be the central concentrated Load in simply supported beam of span L1=3 m\n",
+ "#MAx Moment\n",
+ "#M1=W*L1*4**-1\n",
+ "#After Further simplifying we get\n",
+ "#M1=0.75*10**6 #N-mm\n",
+ "\n",
+ "#The section has a moment of resistance\n",
+ "M1=sigma*1*6**-1*b1*d1**2\n",
+ "\n",
+ "#Equating it to moment of resistance we get max load W\n",
+ "#0.75*10**6*W=M1\n",
+ "#After Further simplifying we get\n",
+ "W=M1*(0.75*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum Concentrated Load required to brek the beam\",round(W,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum Concentrated Load required to brek the beam 54.0 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.8,Page no.140"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3 #m #span\n",
+ "sigma_t=35 #N/mm**2 #Permissible stress in tension\n",
+ "sigma_c=90 #N/mm**2 #Permissible stress in compression\n",
+ "\n",
+ "#Flanges\n",
+ "t=30 #mm #Thickness\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "#Web\n",
+ "t2=25 #mm #Thickness\n",
+ "b=600 #mm #Width\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let y_bar be the Distance of N.A from Extreme Fibres\n",
+ "y_bar=(t*d*d*2**-1*2+(b-2*t)*t2*t2*2**-1)*(t*d*2+(b-2*t)*t2)**-1\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=(1*12**-1*t*d**3+t*d*(d*2**-1-y_bar)**2)*2+1*12**-1*(b-2*t)*t2**3+(b-2*t)*t2*(t2*2**-1-y_bar)**2\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#If web is in Tension\n",
+ "y_t=y_bar #mm\n",
+ "y_c=d-y_bar #mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M=sigma_t*I*(y_bar)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M1=sigma_c*I*(y_c)**-1 #N-mm\n",
+ "\n",
+ "#If w KN/m is u.d.l in beam,Max bending moment\n",
+ "#M=wl**2*8**-1\n",
+ "#After further simplifyng we get\n",
+ "#M=1.125*w*10**6 N-mm\n",
+ "w=M*(1.125*10**6)**-1 #KN\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#If web is in compression\n",
+ "y_t2=178.299 #mm\n",
+ "y_c2=71.71 #mm \n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of tensile stress\n",
+ "M2=sigma_t*I*(y_t2)**-1 #N-mm\n",
+ "\n",
+ "#Moment carrying caryying capacity From consideration of compressive stress\n",
+ "M3=sigma_c*I*(y_c2)**-1 #N-mm\n",
+ "\n",
+ "#Moment of resistance is M2\n",
+ "\n",
+ "#Equating it to bending moment we get\n",
+ "#M2=1.125*10**6*w2\n",
+ "#After further simplifyng we get\n",
+ "w2=M2*(1.125*10**6)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load carrying capacity if:web is in Tension\",round(w,2),\"KN\"\n",
+ "print\" :web is in compression\",round(w2,3),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load carrying capacity if:web is in Tension 73.21 KN\n",
+ " :web is in compression 29.446 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.9,Page no.141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b1=200 #mm #Width at base\n",
+ "b2=100 #mm #Width at top\n",
+ "\n",
+ "L=8 #m Length\n",
+ "P=500 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at y metres from top\n",
+ "\n",
+ "#At this section diameter d is\n",
+ "#d=b2+y*L**-1*(b1-b2)\n",
+ "#After Further simplifying we get\n",
+ "#d=b2+12.5*y #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=pi*64**-1*d**4\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=pi*32**-1*(b1+12.5*y)**3\n",
+ "\n",
+ "#Moment \n",
+ "#M=5*10**5*y #N-mm\n",
+ "\n",
+ "#Let sigma be the fibre stress at this section then\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "#sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#For sigma to be Max,d(sigma)*(dy)**-1=0\n",
+ "#16*10**6*pi**-1*((b2+12.5*y)**-3+y*(-3)*(b2+12.5*y)**-4*12.5)\n",
+ "#After Further simplifying we get\n",
+ "#b2+12.5*y=37.5*y\n",
+ "#After Further simplifying we get\n",
+ "y=b2*25**-1 #m\n",
+ "\n",
+ "#Stress at this section\n",
+ "sigma=5*10**5*32*pi**-1*y*((b2+12.5*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress at Extreme Fibre is max\",round(y,2),\"m\"\n",
+ "print\"Max stress is\",round(sigma,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress at Extreme Fibre is max 4.0 m\n",
+ "Max stress is 6.04 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.10,Page no.143"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "H=10 #mm #Height\n",
+ "A1=160*160 #mm**2 #area of square section at bottom\n",
+ "L1=160 #mm #Length of square section at bottom\n",
+ "b1=160 #mm #width of square section at bottom\n",
+ "A2=80*80 #mm**2 #area of square section at top\n",
+ "L2=80 #mm #Length of square section at top\n",
+ "b2=80 #mm #Width of square section at top\n",
+ "P=100 #N #Pull\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Consider a section at distance y from top.\n",
+ "#Let the side of square bar be 'a'\n",
+ "#a=L2+y*(H)**-1*(b1-b2)\n",
+ "#After further simplifying we get\n",
+ "#a=L2+8*y\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "#I=2*1*12**-1*a*(2)**0.5*(a*((2)**0.5)**-1)**3\n",
+ "#After further simplifying we get\n",
+ "#I=a**4*12**-1\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=a**4*(12*a*(2)**0.5)**-1\n",
+ "#After further simplifying we get\n",
+ "#Z=2**0.5*a**3*(12)**-1 #mm**3\n",
+ "\n",
+ "#Bending moment at this section=100*y N-mm\n",
+ "#M=100*10**3*y #N-mm\n",
+ "\n",
+ "#But\n",
+ "#M=sigma*Z\n",
+ "#After sub values in above equation we get\n",
+ "#sigma=M*Z**-1\n",
+ "#After further simplifying we get\n",
+ "#sigma=1200*10**3*(2**0.5)**-1*y*((80+80*y)**3)**-1 .......(1)\n",
+ "\n",
+ "#For Max stress df*(dy)**-1=0\n",
+ "#After taking Derivative of above equation we get\n",
+ "#df*(dy)**-1=1200*10**3*(2**0.5)**-1*((80+8*y)**-3+y(-3)*(80+8*y)**-4*8)\n",
+ "#After further simplifying we get\n",
+ "y=80*16**-1 #m\n",
+ "\n",
+ "#Max stress at this level is\n",
+ "sigma=1200*10**3*(2**0.5)**-1*y*((80+8*y)**3)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Bending stress is Developed at\",round(y,3),\"m\"\n",
+ "print\"Value of Max Bending stress is\",round(sigma,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Bending stress is Developed at 5.0 m\n",
+ "Value of Max Bending stress is 2.455 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.12,Page no.147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=200 #mm #Width of timber \n",
+ "d=400 #mm #Depth of timber\n",
+ "t=6 #mm #Thickness\n",
+ "b2=200 #mm #width of steel plate\n",
+ "t2=20 #mm #Thickness of steel plate\n",
+ "M=40*10**6 #KN-mm #Moment\n",
+ "#Let E_s*E_t**-1=X\n",
+ "X=20 #Ratio of Modulus of steel to timber\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let y_bar be the Distance of centroidfrom bottom most fibre\n",
+ "y_bar=(b*d*(b+t)+t2*b2*t*t*2**-1)*(b*d+t2*b2*t)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*d**3+b*d*(b+t-round(y_bar,3))**2+1*12**-1*t2*b2*t**3+b2*t2*t*(round(y_bar,3)-t*2**-1)**2\n",
+ "\n",
+ "#distance of the top fibre from N-A\n",
+ "y_1=d+t-y_bar #mm\n",
+ "\n",
+ "#Distance of the junction of timber and steel From N-A\n",
+ "y_2=y_bar-t #mm\n",
+ "\n",
+ "#Stress in Timber at the top\n",
+ "Y=M*I**-1*y_1 #N/mm**2\n",
+ "\n",
+ "#Stress in the Timber at the junction point\n",
+ "Z=M*I**-1*y_2\n",
+ "\n",
+ "#Coressponding stress in steel at the junction point\n",
+ "Z2=X*Z #N/mm**2 \n",
+ "\n",
+ "#The stress in Extreme steel fibre \n",
+ "Z3=X*M*I**-1*y_bar\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress in Extreme steel Fibre\",round(Z3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress in Extreme steel Fibre 69.67 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.13,Page no.149"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Timber size\n",
+ "b=150 #mm #Width\n",
+ "b2=120 #mm \n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "t=6 #mm #Thickness of steel plate\n",
+ "L=6 #m #Span\n",
+ "\n",
+ "#E_s*E_t**-1=20 \n",
+ "#X=E_s*E_t**-1\n",
+ "X=20 \n",
+ "sigma_timber=8 #N/mm**2 #Stress in timber\n",
+ "sigma_steel=150 #N/mm**2 #Stress in steel plate\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Y\n",
+ "\n",
+ "#Due to synnetry cenroid,the neutral axis is half the depth\n",
+ "I=1*12**-1*\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print Z"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "153.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.14,Page no.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Span of beam\n",
+ "W=20*10**3 #N #Load\n",
+ "sigma=8 #N/mm**2 #Stress\n",
+ "b=200 #mm #Width of section\n",
+ "d=300 #mm #Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#let x be the distance from left side of beam\n",
+ "\n",
+ "#Bending moment\n",
+ "#M=W*2**-1*x #Nmm .......(1)\n",
+ "\n",
+ "#But M=sigma*Z ..........(2)\n",
+ "\n",
+ "#Equating equation 1 and 2 we get\n",
+ "#W*2**-1*x=sigma*Z ............(3)\n",
+ "\n",
+ "#Section Modulus \n",
+ "#Z=1*6*b*d**2 ...............(4)\n",
+ "\n",
+ "#Equating equation 3 and 4 we get\n",
+ "#b*d**2=3*W*x*sigma**-1 .............(5)\n",
+ "\n",
+ "#Beam of uniform strength of constant depth\n",
+ "#b=3*W*x*(sigma*d**2) \n",
+ "\n",
+ "#When x=0\n",
+ "b=0\n",
+ "\n",
+ "#When x=L*2**-1\n",
+ "b2=3*W*L*(2*sigma*d**2)**-1 #mm\n",
+ "\n",
+ "#Beam with constant width of 200 mm\n",
+ "\n",
+ "#We have\n",
+ "#d=(3*W*x*(sigma*d)**-1)**0.5\n",
+ "#thus depth varies as (x)**0.5\n",
+ "\n",
+ "#when x=0\n",
+ "d1=0\n",
+ "\n",
+ "#when x=L*2**-1\n",
+ "d2=(2*W*L*(2*sigma*300)**-1)**0.5 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Cross section of rectangular beam is:\",round(b2,2),\"mm\"\n",
+ "print\" :\",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Cross section of rectangular beam is: 250.0 mm\n",
+ " : 223.61 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.15,Page no.154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=800 #mm #Span\n",
+ "n=5 #number of leaves\n",
+ "b=60 #mm #Width\n",
+ "t=10 #mm #thickness\n",
+ "sigma=250 #N/mm**2 #Stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#section Modulus\n",
+ "Z=n*6**-1*b*t**2 #mm**3\n",
+ "\n",
+ "#from the relation\n",
+ "#sigma*Z=M ...................(1)\n",
+ "#M=P*L*4**-1\n",
+ "#sub values of M in equation 1 we get\n",
+ "P=sigma*Z*4*L**-1*10**-3 #KN #Load\n",
+ "\n",
+ "#Length of Leaves\n",
+ "L1=0.2*L #mm\n",
+ "L2=0.4*L #mm\n",
+ "L3=0.6*L #mm\n",
+ "L4=0.8*L #mm\n",
+ "L5=L #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Load it can take is\",round(P,2),\"KN\"\n",
+ "print\"Length of leaves:L1\",round(L1,2),\"mm\"\n",
+ "print\" :L2\",round(L2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Load it can take is 6.25 KN\n",
+ "Length of leaves:L1 160.0 mm\n",
+ " :L2 320.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.16,Page no.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=20*10**3 #N #Shear Force\n",
+ "\n",
+ "#Tee section\n",
+ "\n",
+ "#Flange\n",
+ "b=100 #mm #Width\n",
+ "t=12 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=88 #mm #Depth\n",
+ "t2=12 #mm #Thicknes\n",
+ "\n",
+ "D=100 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of C.G from Top Fibre\n",
+ "y=(b*t*t*2**-1+t2*d*(d*2**-1+t))*(b*t+d*t2)**-1 #mm \n",
+ "\n",
+ "#Moment Of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-t*2**-1)**2+1*12**-1*t2*d**3+t2*d*(t+d*2**-1-y)**2 #mm**4\n",
+ "\n",
+ "#shear stress at bottom Flange\n",
+ "\n",
+ "#Area above this level\n",
+ "A=b*t #mm**2\n",
+ "\n",
+ "#C.G of this area from N-A\n",
+ "y2=y-t*2**-1\n",
+ "\n",
+ "#Stress at bottom of flange\n",
+ "sigma=F*A*y2*(b*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#sigma2 at same level but in web where width is 12 mm\n",
+ "sigma2=F*A*y2*(t2*I)**-1 #N/mm**2 \n",
+ "\n",
+ "#To find shear stress at N-A\n",
+ "X=t*b*(y-t*2**-1)+t2*(y-t2)*(y-t2)*2**-1 #mm**3\n",
+ "\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at top and bottom fibre is zero\n",
+ "#sigma4 and sigma5 are top and bottom fibre shear stress\n",
+ "sigma4=sigma5=0\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force and Bending Moment Diagrams are the results\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,t,t,y,D]\n",
+ "Y1=[sigma4,sigma,sigma2,sigma3,sigma5]\n",
+ "Z1=[0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in m\")\n",
+ "plt.ylabel(\"Shear Force in kN\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force and Bending Moment Diagrams are the results\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56a6270>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.17,Page no.163"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #shear Force\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=80 #mm #Width of flange\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "#Web\n",
+ "d=200 #mm #Depth\n",
+ "t2=20 #mm #Thickness\n",
+ "\n",
+ "#Flange-2\n",
+ "b2=160 #mm #Width\n",
+ "t3=20 #mm #Thickness\n",
+ "\n",
+ "D=240 #mm #Overall Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of N-A from Top Fibre \n",
+ "y=(b*t*t*2**-1+d*t2*(t+d*2**-1)+b2*t3*(t+d+t3*2**-1))*(b*t+d*t2+b2*t3)**-1 #mm\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=1*12**-1*b*t**3+b*t*(y-(t*2**-1))**2+1*12**-1*t2*d**3+t2*d*(y-(t+d*2**-1))**2+1*12**-1*b2*t3**3+t3*b2*((d+t+t3*2**-1)-y)**2 #mm**4\n",
+ "\n",
+ "#Shear stress bottom of flange\n",
+ "sigma=F*b*t*(y-t*2**-1)*(b*I)**-1 #N/mm**2\n",
+ "\n",
+ "#At same Level but in web\n",
+ "sigma2=F*b*t*(y-t*2**-1)*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#for shear stress at N.A\n",
+ "X=b*t*(y-t*2**-1)+t2*(y-t)*(y-t)*2**-1 #mm**3\n",
+ "sigma3=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Shear stress at bottom of web\n",
+ "\n",
+ "X=b2*t3*((D-y)-t3*2**-1) #mm**3\n",
+ "\n",
+ "#Stress at bottom of web\n",
+ "sigma4=F*X*(t2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Stress at Lower flange\n",
+ "sigma5=F*X*(b2*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print \"The Shear Force Diagram is the result\"\n",
+ "\n",
+ "#Plotting the Shear Force Diagram\n",
+ "\n",
+ "X1=[0,20,20,140,220,220,240]\n",
+ "Y1=[0,sigma,sigma2,sigma3,sigma4,sigma5,0]\n",
+ "Z1=[0,0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Shear Force in N\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Shear Force Diagram is the result\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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OnYL/+i/TF3DLLWZPgXbt7K5KRHyRwsAGJ0+afoBXXjHNQOvXQ1iY3VWJiC8r1gj183MN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\n/w/qZz1xEBCKMwAAAABJRU5ErkJggg==\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5614110>"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.18,Page no.164"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=30*10**3 #N #Shear Force\n",
+ "\n",
+ "#Channel Section\n",
+ "d=400 #mm #Depth of web \n",
+ "t=10 #mm #THickness of web\n",
+ "t2=15 #mm #Thickness of flange\n",
+ "b=100 #mm #Width of flange\n",
+ "\n",
+ "#Rectangular Welded section\n",
+ "b2=80 #mm #Width\n",
+ "d2=60 #mm #Depth\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Distance of Centroid From Top Fibre\n",
+ "y=(d*t*t*2**-1+2*t2*(b-t)*((b-t)*2**-1+10)+d2*b2*(d2*2**-1+t))*(d*t+2*t2*(b-t)+d2*b2)**-1 #mm\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*d*t**3+d*t*(y-t*2**-1)**2+2*(1*12**-1*t2*(b-t)**3+t2*(b-t)*(((b-t)*2**-1+t)-y)**2)+1*12**-1*d2**3*b2+d2*b2*(d2*2**-1+t-y)**2\n",
+ "\n",
+ "#Shear stress at level of weld\n",
+ "sigma=F*d*t*(y-t*2**-1)*((b2+t2+t2)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Max Shear Stress occurs at Neutral Axis\n",
+ "X=d*t*(y-t*2**-1)+2*t2*(y-t)*(y-t)*2**-1+b2*(y-t)*(y-t)*2**-1\n",
+ "\n",
+ "sigma_max=F*X*((b+t)*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear stress in the weld is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\"Max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear stress in the weld is 3.62 N/mm**2\n",
+ "Max shear stress is 4.48 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.19,Page no.165"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Section\n",
+ "b=300 #mm #Width\n",
+ "d=300 #mm #Depth\n",
+ "\n",
+ "D=100 #mm #Diameter of Bore\n",
+ "F=10*10**3 #N #Shear Force\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia Of Section\n",
+ "I=1*12**-1*b*d**3-pi*64**-1*D**4\n",
+ "\n",
+ "#Shear stress at crown of circle\n",
+ "sigma=F*b*D*(d*2**-1-D*2**-1)*(b*I)**-1\n",
+ "\n",
+ "#Let a*y_bar=X\n",
+ "X=b*d*2**-1*d*4**-1-pi*8**-1*D**2*4*D*2**-1*(3*pi)**-1 #mm**3\n",
+ "\n",
+ "#Shear Stress at Neutral Axis\n",
+ "sigma2=F*X*((b-D)*I)**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing Stress at Crown of Bore\",round(sigma,3),\"N/mm**2\"\n",
+ "print\"Shear Stress at Neutral Axis\",round(sigma2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing Stress at Crown of Bore 0.149 N/mm**2\n",
+ "Shear Stress at Neutral Axis 0.246 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.20,Page no.166"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#flanges\n",
+ "b=200 #mm #width\n",
+ "t1=25 #mm #Thickness\n",
+ "\n",
+ "#web\n",
+ "d=450 #mm #Depth \n",
+ "t2=20 #mm #thickness\n",
+ "\n",
+ "D=500 #mm #Total Depth of section\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment Of Inertia of the section about N-A\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t2)*d**3 #mm**4\n",
+ "\n",
+ "#Consider an element in the web at distance y from y from N-A\n",
+ "#Depth of web section=225-y\n",
+ "\n",
+ "#C.G From N-A\n",
+ "#y2=y+(((D*2**-1-t)-y)*2**-1)\n",
+ "\n",
+ "#ay_bar for section at y\n",
+ "#Let ay_bar be X\n",
+ "#X=X1 be of Flange + X2 be of web above y\n",
+ "#X=b*t1*(D*2**-1-t1*2**-1)+t2*(d-t1)*(d-t1+y)*2**-1\n",
+ "#After Sub values and Further simplifying we get\n",
+ "#X=1187500+10*(225**2-y**2)\n",
+ "\n",
+ "#Shear stress at y\n",
+ "#sigma_y=F*(X)*(t2*I)**-1\n",
+ "\n",
+ "#Shear Force resisted by the Element\n",
+ "#F1=F*X*t2*dy*(t2*I)**-1\n",
+ "\n",
+ "#Shear stress resisted by web \n",
+ "#sigma=2*F*I**-1*(X)*dy\n",
+ "\n",
+ "#After Integrating above equation and further simplifying we get\n",
+ "#sigma=0.9578*F\n",
+ "\n",
+ "sigma=0.9578*100\n",
+ "\n",
+ "#Result\n",
+ "print\"Shear Resisted by web\",round(sigma,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shear Resisted by web 95.78 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.21,Page no.167"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Wooden Beam\n",
+ "\n",
+ "b=150 #mm #width\n",
+ "d=250 #mm #Depth\n",
+ "\n",
+ "L=5000 #mm #span\n",
+ "m=11.2 #N/mm**2 #Max Bending stress\n",
+ "sigma=0.7 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let 'a' be the distance from left support\n",
+ "#Max shear force\n",
+ "#F=R_A=W*(L-a)*L**-1 \n",
+ "\n",
+ "#Max Moment\n",
+ "#M=W*(L-a)*a*L**-1\n",
+ "\n",
+ "#But M=sigma*Z\n",
+ "#W*(L-a)*a*L**-1=m*1*6**-1*b*d**2 .....................(1)\n",
+ "\n",
+ "#In Rectangular Section MAx stress is 1.5 times Avg shear stress\n",
+ "F=sigma*b*d*1.5**-1\n",
+ "\n",
+ "#W*(L-a)*L**-1=F .....................(2)\n",
+ "\n",
+ "#Dividing Equation 1 nad 2 we get\n",
+ "a=m*6**-1*b*d**2*1.5*(sigma*b*d)**-1\n",
+ "\n",
+ "#Sub above value in equation 2 we get\n",
+ "W=(L-a)**-1*L*F*10**-3 #KN \n",
+ "\n",
+ "#Result\n",
+ "print\"Load is\",round(W,2),\"KN\"\n",
+ "print\"Distance from Left support is\",round(a,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Load is 21.87 KN\n",
+ "Distance from Left support is 1000.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.22,Page no.168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #span\n",
+ "\n",
+ "#Rectangular Section\n",
+ "\n",
+ "b=200 #mm #width\n",
+ "d=400 #mm #depth\n",
+ "\n",
+ "sigma=1.5 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let AB be the cantilever beam subjected to load W KN at free end\n",
+ "\n",
+ "#MAx shear Force\n",
+ "#F=W*10**3 #KN\n",
+ "\n",
+ "#Since Max shear stress in Rectangular section\n",
+ "#sigma_max=1.5*F*A**-1 \n",
+ "#After sub values and further simplifyng we get\n",
+ "W=1.5*b*d*(1.5*1000)**-1 #KN\n",
+ "\n",
+ "#Moment at fixwed end\n",
+ "M=W*1 #KN-m\n",
+ "y_max=d*2**-1 #mm\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**3\n",
+ "\n",
+ "#MAx Stress\n",
+ "sigma_max=M*10**6*I**-1*y_max\n",
+ "\n",
+ "#Result\n",
+ "print\"Concentrated Load is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Concentrated Load is 15.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4.4.24,Page no.170"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #span\n",
+ "\n",
+ "#Rectangular Cross-section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Thickness\n",
+ "\n",
+ "F_per=10 #N/mm**2 #Max Bending stress\n",
+ "q_max=0.6 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#If the Load W is in KN/m\n",
+ "\n",
+ "#Max shear Force\n",
+ "#F=w*l*2**-1 #KN\n",
+ "#After substituting values and further simplifying we get\n",
+ "#M=2*w #KN-m\n",
+ "\n",
+ "#Max Load from Consideration of moment\n",
+ "#M=1*6**-1*b*d**2*F_per\n",
+ "#After substituting values and further simplifying we get\n",
+ "w=(1*6**-1*b*d**2*F_per)*(2*10**6)**-1 #KN/m\n",
+ "\n",
+ "#Max Load from Consideration of shear stress\n",
+ "#q_max=1.5*F*(b*d)**-1 #N\n",
+ "#After substituting values and further simplifying we get\n",
+ "F=q_max*(1.5)*b*d #N\n",
+ "\n",
+ "#If w is Max Load in KN/m,then\n",
+ "#2*w*1000=8000\n",
+ "#After Rearranging and Further simplifying we get\n",
+ "w2=8000*(2*1000)**-1 #KN/m\n",
+ "\n",
+ "#Result\n",
+ "print\"Uniformly Distributed Load Beam can carry is\",round(w,2),\"KN/m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Uniformly Distributed Load Beam can carry is 3.33 KN/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_5.ipynb new file mode 100644 index 00000000..8a54f301 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.5_5.ipynb @@ -0,0 +1,1013 @@ +{
+ "metadata": {
+ "name": "chapter no.5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 5:Deflections Of Beams By Double Integration Methods"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.2,Page No.192"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #span of beam\n",
+ "a=2000 #mm\n",
+ "W1=20*10**3 #N #Pt Load Acting on beam\n",
+ "W2=30*10**3 #N #Pt Load Acting on beam\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=2*10**8 #mm**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Deflection at free End Due to W2\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Deflection at free end Due to W1\n",
+ "dell2=W1*a**3*(3*E*I)**-1+(L-a)*W1*a**2*(2*E*I)**-1 #mm\n",
+ "\n",
+ "#Total Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at Free End is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at Free End is 9.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.4,Page No.193"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**5 #N/mm**2 #Young's Modulus\n",
+ "I=180*10**6 #mm**4 #M.I\n",
+ "W1=20 #N/m #u.d.l\n",
+ "W2=20*10**3 #N #Pt load\n",
+ "L=3000 #m #Span of beam\n",
+ "a=2000 #m #Span of u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Displacement of free End due to 20 KN Pt load at free end\n",
+ "dell1=W2*L**3*(3*E*I)**-1 #mm\n",
+ "\n",
+ "#Displacement of free end due to u.d.l\n",
+ "dell2=W1*a**4*(8*E*I)**-1+(L-a)*W1*a**3*(6*E*I)**-1\n",
+ "\n",
+ "#Deflection at free end\n",
+ "dell=dell1+dell2 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"The Displacement of Free End of cantilever beam is\",round(dell,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Displacement of Free End of cantilever beam is 6.85 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.10,Page No.201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**6 #KN/m**2 #Young's Modulus\n",
+ "I=15*10**-6 #m**4 #M.I\n",
+ "a=4000 #m \n",
+ "L_AB=6 #m #Span of beam\n",
+ "L_CB=2 #m #Length of CB\n",
+ "F_C=18 #KN #force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the Reactions at A & B Respectively\n",
+ "#V_A+V_B=18\n",
+ "#Now taking moment at B,we get M_B\n",
+ "V_A=(F_C*L_CB)*L_AB**-1\n",
+ "V_B=18-V_A\n",
+ "\n",
+ "#Now Taking Moment at distance x\n",
+ "#M_x=6*x-18*(x-4)\n",
+ "#EI*d**2*y*(d*x**2)**-1=6*x-18*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation,we get\n",
+ "#EI*dy*(dx)**-1=C1+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+x**3-3*(x-4)**3\n",
+ "\n",
+ "#The Boundary conditions\n",
+ "x=0\n",
+ "y=0 #.....(a)\n",
+ "\n",
+ "x=6\n",
+ "y=0 #....(b)\n",
+ "\n",
+ "#From Boundary Condition(B.C) a we get\n",
+ "C2=0\n",
+ "\n",
+ "#From Boundary Condition(B.C) b we get\n",
+ "#6*C1+216-3*8\n",
+ "#After Further simplifying we get\n",
+ "C1=-(216-24)*6**-1\n",
+ "\n",
+ "#EI*y=-32*x+x**3-3*(x-4)**3\n",
+ "#EI*dy*(dx)**-1=-32+3*x**2-9(x-4)**4\n",
+ "\n",
+ "#For Max Deflection\n",
+ "#Assume it inthe Porion AC i.e x=4=a\n",
+ "#0=-32+3*x**2\n",
+ "x=(32*3**-1)**0.5\n",
+ "\n",
+ "#Value of Max deflection is\n",
+ "ymax=(-32*x+x**3)*(E*I)**-1 #mm\n",
+ "\n",
+ "#slope at mid-span\n",
+ "\n",
+ "#EI*(dy*(dx)**-1)_centre=-32+3*x**2\n",
+ "#at centre ,\n",
+ "x1=3 #m\n",
+ "\n",
+ "#Let (dy*(dx)**-1)_centre=X\n",
+ "X=-(-32+3*x1**2)*(E*I)**-1 #Radian\n",
+ "\n",
+ "#Deflection at Load Point\n",
+ "x2=4 #m\n",
+ "#EI*y_c=-32*x2+x2**3\n",
+ "\n",
+ "y_c=-(-32*x2+x2**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Value of Max Deflection\",round(ymax,4),\"mm\"\n",
+ "print\"SLope at mid-span\",round(X,4),\"radian\"\n",
+ "print\"Deflection at the Load Point is\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Value of Max Deflection -0.0232 mm\n",
+ "SLope at mid-span 0.0017 radian\n",
+ "Deflection at the Load Point is 0.0213 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.11,Page No.203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_CB=2 #m #Length of CB\n",
+ "L_AC=4 #m #Length of AB\n",
+ "M_C=15 #KN.m #Moment At Pt C\n",
+ "F_C=30 #KN\n",
+ "L=6 #m Span of Beam\n",
+ "\n",
+ "#Let X=E*I\n",
+ "X=10000 #KN-m**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A and V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment a A,we get\n",
+ "V_B=(F_C*L_AC+M_C)*L**-1\n",
+ "V_A=30-V_B\n",
+ "\n",
+ "#Now Taking Moment at distacnce x from A\n",
+ "#M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#By using Macaulay's Method\n",
+ "#EI*(d**2*x/dx**2)=M_x=7.5*x-30*(x-4)+15\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2*2**-1-15*(x-4)**2+15*(x-4) ............(1)\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EIy=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1..........(2)\n",
+ "\n",
+ "#Boundary Cinditions\n",
+ "x=0\n",
+ "y=0\n",
+ "\n",
+ "#Substituting above equations we get \n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "\n",
+ "C1=-(7.5*6**3*6**-1-5*2**3+15*2**2*2**-1)*6**-1\n",
+ "\n",
+ "#EIy_c=C2+C1*x+7.5*6**-1*x**3-5*(x-4)**3+15*(x-4)**2*2**-1\n",
+ "#Sub values in Above equation we get\n",
+ "y_c=(93.3333*(X)**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"The Deflection at C\",round(y_c,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Deflection at C 0.0093 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.12,Page No.204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=L_CD=L_DB=2 #m #Length of AC,CD,DB\n",
+ "F_C=40 #KN #Force at C\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=6 #m #span of beam\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=15000 #KN-m**2\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=80\n",
+ "\n",
+ "#Taking Moment B,M_B\n",
+ "V_A=(F_C*(L_CD+L_DB)+w*L_DB*L_DB*2**-1)*L**-1 #KN\n",
+ "V_B=80-V_A #KN\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=33.333*x-40*(x-2)-20*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=33.333*x-40*(x-2)-10*(x-4)**2\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+33.333*x**2*2**-1-20*(x-2)**2-10*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=6\n",
+ "y=0\n",
+ "C1=-760*6**-1\n",
+ "\n",
+ "#Assuming Deflection to be max in portion CD and sustituting value of C1 in equation of slope we get\n",
+ "#EI*y=C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3-10*12**-1*(x-4)**4\n",
+ "#0=-126.667+33.333*x**2**-1-20*(x-2)**2\n",
+ "\n",
+ "#After rearranging and simplifying further we get\n",
+ "\n",
+ "#x**2-24*x+62=0\n",
+ "#From above equations\n",
+ "a=1\n",
+ "b=-24\n",
+ "c=62\n",
+ "\n",
+ "y=(b**2-4*a*c)**0.5\n",
+ "\n",
+ "x1=(-b+y)*(2*a)**-1\n",
+ "x2=(-b-y)*(2*a)**-1\n",
+ "\n",
+ "#Taking x2 into account\n",
+ "x=2.945 #m\n",
+ "C1=-126.667\n",
+ "C2=0\n",
+ "\n",
+ "y_max=(C2+C1*x+33.333*x**3*6**-1-20*3**-1*(x-2)**3)*X**-1 #mm\n",
+ "\n",
+ "#Max slope occurs at the ends\n",
+ "#At A,\n",
+ "#EI*(dy/dx)_A=-126.667\n",
+ "#At B\n",
+ "#EI*(dy/dx)_B=126.667+33.333*6**2*2**-1-20*4**2-10*2**3\n",
+ "#After simplifying Further we get\n",
+ "#EI*(dy/dx)_B=73.3273\n",
+ "\n",
+ "#Now Max slope is EI(dy/dx)_A=-126.667\n",
+ "#15000*(dy/dx)_=-126.667\n",
+ "\n",
+ "#Let Y=dy/dx\n",
+ "Y=-126.667*X**-1 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Maximum Deflection for Beam is\",round(y_max,4),\"mm\"\n",
+ "print\"Maximum Slope for beam is\",round(Y,4),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum Deflection for Beam is -0.0158 mm\n",
+ "Maximum Slope for beam is -0.0084 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.13,Page No.206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=2*10**8 #KN/m**2\n",
+ "I=450*10**-6 #m**4\n",
+ "L_AC=1 #m #Length of AC\n",
+ "L_CD=3 #m #Length of CD\n",
+ "L_DB=2 #m #Length of DB\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=30\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "#EI*(d**2/dx**2)=17.5*x-10*(x-1)**2*2**-1+10*(x-4)**2*2**-1\n",
+ "\n",
+ "#Now Integrating Above equation we get\n",
+ "#EI(dy/dx)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**2+5*3**-1*(x-4)**3\n",
+ "\n",
+ "#Again Integrating Above equation we get\n",
+ "#EI*y=C2+C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4+5*12**-1*(x-4)**4\n",
+ "\n",
+ "#At \n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=6 \n",
+ "y=0\n",
+ "C1=(-17.5*x**3*6**-1+5*12**-1*(x-1)**4-5*12**-1*(x-4)**4)*x**-1\n",
+ "\n",
+ "# 1)Slope at A .i.e at x=0\n",
+ "#EI*(dy/dx)_A=C1=-62.708 #KN-m**2\n",
+ "#let (dy/dx)=X\n",
+ "X=C1*(E*I)**-1 #radiams\n",
+ "\n",
+ "#Deflection at mid-span\n",
+ "x=3 #m\n",
+ "#EI*y_centre=C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**2\n",
+ "y_centre=-(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Maximum Deflection\n",
+ "\n",
+ "#At point of Max deflection (dy/dx)=0\n",
+ "#Assuming it in portion CD\n",
+ "\n",
+ "#0=C1*x+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Now Let\n",
+ "#F(x)=C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3\n",
+ "\n",
+ "#Let F(x)=Y\n",
+ "#At \n",
+ "x=2.5\n",
+ "Y1=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#AT\n",
+ "x=3\n",
+ "Y2=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#At\n",
+ "x=2.9 #m\n",
+ "Y3=-(C1+17.5*x**2*2**-1-5*3**-1*(x-1)**3)\n",
+ "\n",
+ "#A curve may be plotted for (F(x) and the value for which F(x)=0 may be found\n",
+ "#For F(x)=0 for x=2.92 m\n",
+ "#Therefore y_max occur at x=2.92\n",
+ "\n",
+ "x=2.92 #m\n",
+ "y_max=(C1*x+17.5*x**3*6**-1-5*12**-1*(x-1)**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Slope at A\",round(X,6),\"mm\"\n",
+ "print\"Deflection at mid-span\",round(y_centre,6),\"mm\"\n",
+ "print\"Maxmimum Deflection is\",round(y_max,5),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Slope at A -0.000697 mm\n",
+ "Deflection at mid-span 0.001289 mm\n",
+ "Maxmimum Deflection is -0.00129 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.14,Page No.208"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_AC=LDE=L_EB=1 #m #Length of AC\n",
+ "L_CD=2 #m #Length of CD\n",
+ "E=200 #KN/mm**2\n",
+ "I=60*10**6 #mm**4 #M.I\n",
+ "F_C=20 #KN #Force at C\n",
+ "F_E=30 #KN #Force at E\n",
+ "w=10 #KN/m #u.d.l\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "X=E*I*10**-6 #KN-m**2\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=70\n",
+ "\n",
+ "#Taking Moment at distance x from A\n",
+ "#M_x=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "#EI*(d**2y/dx**2)=34*x-20*(x-1)-10*(x-1)**2*2**-1+10*(x-3)**2*2**-1-30*(x-4)\n",
+ "\n",
+ "#Now Integrating Above equation,we get\n",
+ "#EI*(dy/dx)=C1+17*x**2-10*(x-1)**2-5*3**-1*(x-1)**3+5*3**-1*(x-3)**3-15*(x-4)**2\n",
+ "\n",
+ "#Again Integrating Above equation,we get\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "C2=0\n",
+ "\n",
+ "#At \n",
+ "x=5 #m\n",
+ "y=0\n",
+ "C1=(-17*3**-1*x**3+10*3**-1*(x-1)**3+5*12**-1*(x-1)**4-5*12**-1*(x-3)**4+5*(x-4)**3)*5**-1\n",
+ "\n",
+ "#EI*y=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4-5*(x-4)**3\n",
+ "C2=0\n",
+ "C1=-78\n",
+ "x=1\n",
+ "y_c=(-78*x+17*3**-1*x)*(X)**-1\n",
+ "\n",
+ "#EI*y_D=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4\n",
+ "x=3\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_D=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4)*(X**-1)\n",
+ "\n",
+ "#EI*y_E=C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4\n",
+ "x=4\n",
+ "C1-78\n",
+ "C2=0\n",
+ "y_E=(C2+C1*x+17*3**-1*x**3-10*3**-1*(x-1)**3-5*12**-1*(x-1)**4+5*12**-1*(x-3)**4)*X**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections at C\",round(y_c,5),\"mm\"\n",
+ "print\"Deflections at D\",round(y_D,5),\"mm\"\n",
+ "print\"Deflections at E\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections at C -0.00603 mm\n",
+ "Deflections at D -0.00953 mm\n",
+ "Deflections at E -0.0061 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.15,Page No.209"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=300*10**6 #mm\n",
+ "L_AB=L_BC=L_CD=L_DE=1 #m #Length of AB,BC,CD,DE respectively\n",
+ "F_A=20 #KN #Force at A\n",
+ "F_C=10 #KN #Force at C\n",
+ "w=30 #KN/m #u.d.l\n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=E*I*10**-6 #KN-2**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_E be the reactions at E\n",
+ "V_E=F_A+F_C+w*(L_BC+L_CD) #KN \n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#EI*(d**2x/dy**2)=M=-20*x-30*(x-1)**2*2**-1-10*(x-2)+30*(x-3)**2*2**-1\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1-10*x**2-5*(x-1)**3-5*(x-2)**2+5*(x-3)**3\n",
+ "\n",
+ "#Again Integrating above equation\n",
+ "#EI*y=C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*(x-3)**4*4**-1-5*3*(x-2)**3\n",
+ "\n",
+ "#At\n",
+ "#dy/dx=0\n",
+ "x=4 #m\n",
+ "C1=10*x**2+5*(x-1)**3+5*(x-2)**2-5*(x-3)**3\n",
+ "\n",
+ "#AT\n",
+ "x=4\n",
+ "y=0\n",
+ "C2=-C1*4+10*x**3*3**-1+5*(x-1)**4*4**-1-5*(x-3)**4*4**-1+5*3**-1*(x-2)**3\n",
+ "\n",
+ "#Max Deflection and Max slopes occurs at Free end in case of cantilever\n",
+ "y_max=y_A=C2*X**-1\n",
+ "\n",
+ "#EI*(dy/dx)_max=C1\n",
+ "#Let (dy/dx)=Y\n",
+ "Y=C1*X**-1 #radian\n",
+ "\n",
+ "#Now deflection at x=1 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=1\n",
+ "y_B=(C2+C1*x-10*x**3*3**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=2 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=2 #m\n",
+ "y_C=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1)*X**-1\n",
+ "\n",
+ "#Now Deflection at x=3 #m\n",
+ "C2=-913.333\n",
+ "C1=310\n",
+ "x=3 #m\n",
+ "y_D=(C2+C1*x-10*x**3*3**-1-5*(x-1)**4*4**-1-5*3**-1*(x-2)**3)*X**-1\n",
+ "\n",
+ "y_E=0\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Deflection for Beam\",round(y_A,4),\"mm\"\n",
+ "print\"Max Slope for beam\",round(Y,5),\"radians\"\n",
+ "\n",
+ "#Plotting the ELastic Curve\n",
+ "\n",
+ "Y2=[y_E,y_D,y_C,y_B,y_A]\n",
+ "X2=[L_AB+L_BC+L_CD+L_DE,L_AB+L_BC+L_CD,L_AB+L_BC,L_AB,0]\n",
+ "Z2=[0,0,0,0,0]\n",
+ "plt.plot(X2,Y2,X2,Z2)\n",
+ "plt.xlabel(\"Length in mm\")\n",
+ "plt.ylabel(\"Deflection in mm\")\n",
+ "plt.show()"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Deflection for Beam -0.0152 mm\n",
+ "Max Slope for beam 0.00517 radians\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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8yjGZmZkkJSWRmZlJSkoKM2bMoLKy8pxilcZbsQJGj4Y//QkWL1aSEWnvaq3R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\nL+mlAAAAAElFTkSuQmCC\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x5a29310>"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.16,Page No.211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BD=L_CB=L_AC=2 #m #Length of BD,CB,AC\n",
+ "F_C=40 #KN #Force at C\n",
+ "F_D=10 #KN Force at D\n",
+ "L=6 #m spna of beam\n",
+ "\n",
+ "#EI is constant in this problem\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B Respectively\n",
+ "#V_A+V_B=50\n",
+ "\n",
+ "#Taking Moment at Pt A\n",
+ "V_B=(F_D*L+F_C*L_AC)*(L_AC+L_CB)**-1\n",
+ "V_A=50-V_B\n",
+ "\n",
+ "#Now Taking Moment at distance x from A,M_x\n",
+ "#M_x=15*x-40*(x-2)+35*(x-4)\n",
+ "#EI*(d**2*y/dx**2)=15*x-40*(x-2)+35*(x-4)\n",
+ "\n",
+ "#Now Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+7.5*x**2-20*(x-2)**2+17.5(x-4)**2\n",
+ "\n",
+ "#Again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+2.5*x**2-20*3**-1*(x-2)**3+17.5*(x-4)**3*3**-1\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#we get\n",
+ "C2=0\n",
+ "\n",
+ "#At\n",
+ "x=4 \n",
+ "y=0\n",
+ "#we get\n",
+ "C1=(2.5*4**3-20*3**-1*2**3)*4**-1\n",
+ "\n",
+ "#Now Deflection at C\n",
+ "x=2\n",
+ "C1=-26.667\n",
+ "C2=0\n",
+ "y_C=C2+C1*x+2.5*x**3\n",
+ "\n",
+ "#Now Deflection at D\n",
+ "C1=-21.667\n",
+ "C2=0\n",
+ "y_D=-26.667*6+2.5*6**3-20*3**-1*4**3+17.5*2**3*3**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflections Under Loads are:y_D\",round(y_D,4)\n",
+ "print\" :y_C\",round(y_C,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflections Under Loads are:y_D -0.002\n",
+ " :y_C -33.33\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.17,Page No.212"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_BC=L_EB=2 #m #Length of BC & EB\n",
+ "E=200*10**6 #KN/m**2 #Modulus of eLasticity\n",
+ "I=45*10**-6 #mm**4 #M.I\n",
+ "L_DE=3 #m #Length of DE\n",
+ "L_AD=1 #m #Length of AD\n",
+ "w=20 #KN/m #u.d.l\n",
+ "L=8 #m #span of beam\n",
+ "F_C=30 #KN #Force at C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let V_A & V_B be the reactions at A & B respectively\n",
+ "#V_A+V_B=90\n",
+ "\n",
+ "#Taking Moment at A,M_A\n",
+ "V_B=(w*L_DE*(L_DE*2**-1+L_AD)+F_C*L)*(L_AD+L_DE+L_EB)**-1\n",
+ "V_A=90-V_B\n",
+ "\n",
+ "#Taking Moment at distance x\n",
+ "#M_x=25*x-20*(x-1)**2*2**-1+20*(x-4)**2*2**-1+65*(x-6)\n",
+ "\n",
+ "#Integrating above equation we get\n",
+ "#EI*(d**2*y/dx**2)=25*x-10*(x-1)**2+10*(x-4)**2+65*(x-6)\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*(dy/dx)=C1+25*x**2*2**-1-10*3**-1*(x-1)**3+10*3**-1*(x-4)**2+65*2**-1*(x-6)**2\n",
+ "\n",
+ "#again Integrating above equation we get\n",
+ "#EI*y=C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3\n",
+ "\n",
+ "x=0\n",
+ "y=0\n",
+ "#Sub these values in above equation,we get\n",
+ "C2=0\n",
+ "\n",
+ "x=6 #m\n",
+ "y=0\n",
+ "C1=-(25*6**-1*6**3-10*12**-1*5**4+10*12**-1*2**4)*6**-1\n",
+ "\n",
+ "#deflection at C is given by\n",
+ "x=8\n",
+ "y_c=(C2+C1*x+25*6**-1*x**3-10*12**-1*(x-1)**4+10*12**-1*(x-4)**4+65*6**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Assuming y is max in the portion DE,then\n",
+ "#(dy/dx)=0 for that point\n",
+ "\n",
+ "#0=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "\n",
+ "#Let F(x)=-65.417+25*2**-1*x**2-10*3**-1*x(-1)**3\n",
+ "#Let z=F(x)\n",
+ "\n",
+ "#AT \n",
+ "x=3\n",
+ "z=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.5\n",
+ "z1=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "x=2.4\n",
+ "z2=-65.417+25*2**-1*x**2-10*3**-1*(x-1)**3\n",
+ "\n",
+ "#The assumption is max in portion DE\n",
+ "x=2.46\n",
+ "y_max=(-65.417*x+25*6**-1*x**3-10*12**-1*1.46**4)*(E*I)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection at free end C\",round(y_c,4),\"mm\"\n",
+ "print\"Max Deflection between A and B\",round(y_max,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection at free end C -0.0101 mm\n",
+ "Max Deflection between A and B -0.0114 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 5.5.18,Page No.213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L_DB=L_AC=L_ED=2 #m #Length of DB & AC\n",
+ "L_CD=4 #m #Length of CD\n",
+ "L_CE=2 #m #Length of CE\n",
+ "F_A=40 #KN #Force at C\n",
+ "F_B=20 #KN #Force at A\n",
+ "E=200*10**6 #KN/mm**2 #Modulus of Elasticity\n",
+ "I=50*10**-6 #m**4 #M.I\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#LEt V_C & V_D be the reactions at C & D respectively\n",
+ "#V_C+V_D=60\n",
+ "\n",
+ "#Taking Moment At D,M_D\n",
+ "V_C=-(-F_A*(L_AC+L_CE+L_ED)+F_B*L_DB)*L_CD**-1\n",
+ "V_D=60-V_C\n",
+ "\n",
+ "#Now Taking Moment at Distance x from A,\n",
+ "#M_x=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#EI*(d**2*y/dx**2)=-40*x+50*(x-2)+10*(x-6)\n",
+ "\n",
+ "#Now Integrating above Equation we get\n",
+ "#EI*(dy/dx)=C1+20*x**2-25*(x-2)+5*(x-6)**2\n",
+ "\n",
+ "#Again Integrating above Equation we get\n",
+ "#EI*y=C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3\n",
+ "\n",
+ "#At\n",
+ "x=0\n",
+ "y=0\n",
+ "#C2+2*C1=-53.33 ...............(1)\n",
+ "\n",
+ "#At \n",
+ "x=6\n",
+ "y=0\n",
+ "#C2+6*C1=906.667 ...............(2)\n",
+ "\n",
+ "#Subtracting Equation 1 from 2 we get\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=0\n",
+ "y_A=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_A is incorrect in textbook\n",
+ "\n",
+ "#At Mid-span\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=4\n",
+ "y_E=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "#Answer For y_E is incorrect in textbook\n",
+ "\n",
+ "#At B\n",
+ "C1=853.333*4**-1\n",
+ "C2=53.333-2*C1\n",
+ "x=8\n",
+ "y_B=(C2+C1*x-20*3**-1*x**3+25*3**-1*(x-2)**3+5*3**-1*(x-6)**3)*(E*I)**-1\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Deflection relative to the level of the supports:at End A\",round(y_A,4),\"mm\"\n",
+ "print\" :at End B\",round(y_B,4),\"mm\"\n",
+ "print\" :at Centre of CD\",round(y_E,4),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Deflection relative to the level of the supports:at End A -0.08 mm\n",
+ " :at End B -0.0267 mm\n",
+ " :at Centre of CD 0.0107 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_5.ipynb new file mode 100644 index 00000000..e3bdfbe1 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.6_5.ipynb @@ -0,0 +1,1518 @@ +{
+ "metadata": {
+ "name": "chapter no.6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.6:Torsion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.1,Page No.225"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=10000 #mm #Length of solid shaft\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "n=150 #rpm\n",
+ "P=112.5*10**6 #N-mm/sec #Power Transmitted\n",
+ "G=82*10**3 #N/mm**2 #modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*d**4*(32)**-1 #mm**3 #Polar Modulus\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "r=50 #mm #Radius\n",
+ "\n",
+ "q_s=T*r*J**-1 #N/mm**2 #Max shear stress intensity\n",
+ "Theta=T*L*(G*J)**-1 #angle of twist\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress intensity\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist\",round(Theta,3),\"radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress intensity 36.48 N/mm**2\n",
+ "Angle of Twist 0.089 radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.2,Page No.226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=440*10**6 #N-m/sec #Power transmitted\n",
+ "n=280 #rpm\n",
+ "theta=pi*180**-1 #radian #angle of twist\n",
+ "L=1000 #mm #Length of solid shaft\n",
+ "q_s=40 #N/mm**2 #Max torsional shear stress\n",
+ "G=84*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#P=2*pi*n*T*(60)**-1 #Equation of Power transmitted\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #torsional moment\n",
+ "\n",
+ "#From Consideration of shear stress\n",
+ "d1=(T*16*(pi*40)**-1)**0.333333 \n",
+ "\n",
+ "#From Consideration of angle of twist\n",
+ "d2=(T*L*32*180*(pi*84*10**3*pi)**-1)**0.25\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of solid shaft is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of solid shaft is 124.09 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.3,Page No.227"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "q_s=80 #N/mm**2 #Max sheare stress\n",
+ "P=736*10**6 #N-mm/sec #Power transmitted\n",
+ "n=200\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now From consideration of angle of twist\n",
+ "theta=pi*180**-1\n",
+ "#L=15*d\n",
+ "\n",
+ "d=(T*32*180*15*(pi**2*G)**-1)**0.33333\n",
+ "\n",
+ "#Now corresponding stress at the surface is\n",
+ "q_s2=T*32*d*(pi*2*d**4)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Max diameter required is\",round(d,2),\"mm\"\n",
+ "print\"Corresponding shear stress is\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max diameter required is 156.66 mm\n",
+ "Corresponding shear stress is 46.55 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.4,Page No.228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of steel bar\n",
+ "p=50*10**3 #N #Pull\n",
+ "dell_1=0.095 #mm #Extension of bar\n",
+ "l=200 #mm #Guage Length\n",
+ "T=200*10**3 #N-mm #Torsional moment\n",
+ "theta=0.9*pi*180**-1 #angle of twist\n",
+ "L=250 #mm Length of steel bar\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "A=pi*4**-1*d**2 #Area of steel bar #mm**2\n",
+ "E=p*l*(dell_1*A)**-1 #N/mm**2 #Modulus of elasticity \n",
+ "\n",
+ "J=pi*32**-1*d**4 #mm**4 #Polar modulus\n",
+ "\n",
+ "G=T*L*(theta*J)**-1 #Modulus of rigidity #N/mm**2\n",
+ "\n",
+ "#Now from the relation of Elastic constants\n",
+ "mu=E*(2*G)**-1-1\n",
+ "\n",
+ "#result\n",
+ "print\"The Poissoin's ratio is\",round(mu,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Poissoin's ratio is 0.292\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.5,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=6000 #mm #Length of circular shaft\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=75 #mm #Inner Diameter\n",
+ "R=100*2**-1 #Radius of shaft\n",
+ "T=10*10**6 #N-mm #Torsional moment\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Max Shear stress produced\n",
+ "q_s=T*R*J**-1 #N/mm**2\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=T*L*(G*J)**-1 #Radian\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx shear stress produced is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,2),\"Radian\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx shear stress produced is 74.5 N/mm**2\n",
+ "Angle of Twist is 0.11 Radian\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.6,Page No.229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=200 #mm #External Diameter of shaft\n",
+ "t=25 #mm #Thickness of shaft\n",
+ "n=200 #rpm\n",
+ "theta=0.5*pi*180**-1 #Radian #angle of twist\n",
+ "L=2000 #mm #Length of shaft\n",
+ "G=84*10**3 #N/mm**2\n",
+ "d2=d1-2*t #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus \n",
+ "\n",
+ "#Torsional moment\n",
+ "T=G*J*theta*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Power Transmitted\n",
+ "P=2*pi*n*T*60**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Max shear stress transmitted\n",
+ "q_s=G*theta*(d1*2**-1)*L**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Power Transmitted is\",round(P,2),\"N-mm\"\n",
+ "print\"Max Shear stress produced is\",round(q_s,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Power Transmitted is 824.28 N-mm\n",
+ "Max Shear stress produced is 36.65 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.7,Page No.230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=3750*10**6 #N-mm/sec\n",
+ "n=240 #Rpm\n",
+ "q_s=160 #N/mm**2 #Max shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#d2=0.8*d2 #mm #Internal Diameter of shaft\n",
+ "\n",
+ "#J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar modulus\n",
+ "#After substituting value in above Equation we get\n",
+ "#J=0.05796*d1**4\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1 ......................................(1)\n",
+ "\n",
+ "#But R=d1*2**-1 \n",
+ "\n",
+ "#Now substituting value of R and J in Equation (1) we get\n",
+ "d1=(T*(0.05796*q_s*2)**-1)**0.33333\n",
+ "\n",
+ "d2=d1*0.8\n",
+ "\n",
+ "#Result\n",
+ "print\"The size of the Shaft is:d1\",round(d1,3),\"mm\"\n",
+ "print\" :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The size of the Shaft is:d1 200.362 mm\n",
+ " :d2 160.289 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.8,Page No.231"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=245*10**6 #N-mm/sec #Power transmitted\n",
+ "n=240 #rpm\n",
+ "q_s=40 #N/mm**2 #Shear stress\n",
+ "theta=pi*180**-1 #radian #Angle of twist\n",
+ "L=1000 #mm #Length of shaft\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Tmax=1.5*T\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional Moment\n",
+ "Tmax=1.5*T\n",
+ "\n",
+ "#Now For Solid shaft\n",
+ "#J=pi*32*d**4\n",
+ "\n",
+ "#Now from the consideration of shear stress we get\n",
+ "#T*J**-1=q_s*(d*2**-1)**-1\n",
+ "#After substituting value in above Equation we get\n",
+ "#T=pi*16**-1*d**3*q_s\n",
+ "\n",
+ "#Designing For max Torque\n",
+ "d=(Tmax*16*(pi*40)**-1)**0.33333 #mm #Diameter of shaft\n",
+ "\n",
+ "#For max Angle of Twist\n",
+ "#Tmax*J**-1=G*theta*L**-1 \n",
+ "#After substituting value in above Equation we get\n",
+ "d2=(Tmax*32*180*L*(pi**2*G)**-1)**0.25\n",
+ "\n",
+ "#For Hollow Shaft\n",
+ "\n",
+ "#d1_2=Outer Diameter\n",
+ "#d2_2=Inner Diameter\n",
+ "\n",
+ "#d2_2=0.5*d1_2\n",
+ "\n",
+ "# Polar modulus\n",
+ "#J=pi*32**-1*(d1_2**4-d2_2**4)\n",
+ "#After substituting values we get\n",
+ "#J=0.092038*d1_2**4\n",
+ "\n",
+ "#Now from the consideration of stress\n",
+ "#Tmax*J**-1=q_s*(d1_2*2**-1)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_2=(Tmax*(0.092038*2*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Now from the consideration of angle of twist\n",
+ "#Tmax*J**-1=G*theta*L**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "d1_3=(Tmax*180*L*(0.092038*G*pi)**-1)**0.25\n",
+ "\n",
+ "d2_2=0.5*d1_2\n",
+ "\n",
+ "#result\n",
+ "print\"Diameter of shaft is:For solid shaft:d\",round(d,2),\"mm\"\n",
+ "print\" :For Hollow shaft:d1_2\",round(d1_2,3),\"mm\"\n",
+ "print\" : :d2_2\",round(d2_2,3),\"mm\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of shaft is:For solid shaft:d 123.01 mm\n",
+ " :For Hollow shaft:d1_2 125.69 mm\n",
+ " : :d2_2 62.845 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.11,Page No.235"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=250*10**6 #N-mm/sec #Power transmitted\n",
+ "n=100 #rpm\n",
+ "q_s=75 #N/mm**2 #Shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Equation of Power we have\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm #Torsional moment\n",
+ "\n",
+ "#Now from torsional moment equation we have\n",
+ "#T=j*q_s*(d/2**-1)**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T=pi*16**-1**d**3*q_s\n",
+ "d=(T*16*(pi*q_s)**-1)**0.3333 #mm #Diameter of solid shaft\n",
+ "\n",
+ "#PArt-2\n",
+ "\n",
+ "#Let d1 and d2 be the outer and inner diameter of hollow shaft\n",
+ "#d2=0.6*d1\n",
+ "\n",
+ "#Again from torsional moment equation we have\n",
+ "#T=pi*32**-1*(d1**4-d2**4)*q_s*(d1/2)**-1\n",
+ "d1=(T*16*(pi*(1-0.6**4)*q_s)**-1)**0.33333\n",
+ "d2=0.6*d1\n",
+ "\n",
+ "#Cross sectional area of solid shaft\n",
+ "A1=pi*4**-1*d**2 #mm**2\n",
+ "\n",
+ "#cross sectional area of hollow shaft\n",
+ "A2=pi*4**-1*(d1**2-d2**2)\n",
+ "\n",
+ "#Now percentage saving in weight\n",
+ "#Let W be the percentage saving in weight\n",
+ "W=(A1-A2)*100*A1**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage saving in Weight is\",round(W,3),\"%\"\n",
+ "print\"Size of shaft is:solid shaft:d\",round(d,3),\"mm\"\n",
+ "print\" :Hollow shaft:d1\",round(d1,3),\"mm\"\n",
+ "print\" : :d2\",round(d2,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage saving in Weight is 29.735 %\n",
+ "Size of shaft is:solid shaft:d 117.418 mm\n",
+ " :Hollow shaft:d1 123.031 mm\n",
+ " : :d2 73.818 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.12,Page No.237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "d=100 #mm #Diameter of solid shaft\n",
+ "d1=100 #mm #Outer Diameter of hollow shaft\n",
+ "d2=50 #mm #Inner Diameter of hollow shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Torsional moment of solid shaft\n",
+ "#T_s=J*q_s*(d*2**-1)**-1 \n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_s=pi*16*d**3*q_s ...............(1)\n",
+ "\n",
+ "#torsional moment for hollow shaft is\n",
+ "#T_h=J*q_s*(d1**4-d2**4)**-1*(d1*2**-1)\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "#T_h=pi*32**-1*2*d1**-1*(d1**4-d2**4)*q_s ...........(2)\n",
+ "\n",
+ "#Dividing Equation 2 by 1 we get\n",
+ "#Let the ratio of T_h*T_s**-1 Be X\n",
+ "X=1-0.5**4\n",
+ "\n",
+ "#Loss in strength \n",
+ "#Let s be the loss in strength\n",
+ "#s=T_s*T_h*100*T_s**-1\n",
+ "#After substituting values in above equation and further simplifying we get\n",
+ "s=(1-0.9375)*100\n",
+ "\n",
+ "#Weight Ratio \n",
+ "#Let w be the Weight ratio\n",
+ "#w=W_h*W_s**-1\n",
+ "\n",
+ "A_h=pi*32**-1*(d1**2-d2**2) #mm**2 #Area of Hollow shaft\n",
+ "A_s=pi*32**-1*d**2 #mm**2 #Area of solid shaft\n",
+ "\n",
+ "w=A_h*A_s**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"Loss in strength is\",round(s,2)\n",
+ "print\"Weight ratio is\",round(w,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Loss in strength is 6.25\n",
+ "Weight ratio is 0.75\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.13,Page No.239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "T=8 #KN-m #Torque \n",
+ "d=100 #mm #Diameter of portion AB\n",
+ "d1=100 #mm #External Diameter of Portion BC\n",
+ "d2=75 #mm #Internal Diameter of Portion BC\n",
+ "G=80 #KN/mm**2 #Modulus of Rigidity\n",
+ "L1=1500 #mm #Radial Distance of Portion AB\n",
+ "L2=2500 #mm #Radial Distance ofPortion BC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=d*2**-1 #mm #Radius of shaft\n",
+ "\n",
+ "#For Portion AB,Polar Modulus\n",
+ "J1=pi*32**-1*d**4 #mm**4 \n",
+ "\n",
+ "#For Portion BC,Polar modulus \n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Now Max stress occurs in portion BC since max radial Distance is sme in both cases\n",
+ "q_max=T*J2**-1*R*10**6 #N/mm**2 \n",
+ "\n",
+ "#Let theta1 be the rotation in Portion AB and theta2 be the rotation in portion BC\n",
+ "theta1=T*L1*(G*J1)**-1 #Radians\n",
+ "theta2=T*L2*(G*J2)**-1 #Radians\n",
+ "\n",
+ "#Total Rotational at end C\n",
+ "theta=(theta1+theta2)*10**3 #Radians\n",
+ "\n",
+ "#Result\n",
+ "print\"Max stress induced is\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist is\",round(theta,3),\"radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max stress induced is 59.6 N/mm**2\n",
+ "Angle of Twist is 0.053 radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.14,Page No.240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "q_b=80 #N/mm**2 #Shear stress in Brass\n",
+ "q_s=100 #N/mm**2 #Shear stress in Steel\n",
+ "G_b=40*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 \n",
+ "L_b=1000 #mm #Length of brass shaft\n",
+ "L_s=1200 #mm #Length of steel shaft\n",
+ "d1=80 #mm #Diameter of brass shaft\n",
+ "d2=60 #mm #Diameter of steel shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of brass rod\n",
+ "J_b=pi*32**-1*d1**4 #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel rod\n",
+ "J_s=pi*32**-1*d2**4 #mm**4\n",
+ "\n",
+ "#Considering bras Rod:AB\n",
+ "T1=J_b*q_b*(d1*2**-1)**-1 #N-mm \n",
+ "\n",
+ "#Considering Steel Rod:BC\n",
+ "T2=J_s*q_s*(d2*2**-1)**-1 #N-mm\n",
+ "\n",
+ "#Max Torque that can be applied\n",
+ "T2\n",
+ "\n",
+ "#Let theta_b and theta_s be the rotations in Brass and steel respectively\n",
+ "theta_b=T2*L_b*(G_b*J_b)**-1 #Radians\n",
+ "theta_s=T2*L_s*(G_s*J_s)**-1 #Radians\n",
+ "\n",
+ "theta=theta_b+theta_s #Radians #Rotation of free end\n",
+ "\n",
+ "#Result\n",
+ "print\"Total of free end is\",round(theta,3),\"Radians\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total of free end is 0.076 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.15,Page No.241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 #Modulus of Rigidity\n",
+ "d1=100 #mm #Outer diameter of hollow shft\n",
+ "d2=80 #mm #Inner diameter of hollow shaft\n",
+ "d=80 #mm #diameter of Solid shaft\n",
+ "d3=60 #mm #diameter of Solid shaft having L=0.5m\n",
+ "L1=300 #mm #Length of Hollow shaft\n",
+ "L2=400 #mm #Length of solid shaft\n",
+ "L3=500 #mm #LEngth of solid shaft of diameter 60mm\n",
+ "T1=2*10**6 #N-mm #Torsion in Shaft AB\n",
+ "T2=1*10**6 #N-mm #Torsion in shaft BC\n",
+ "T3=1*10**6 #N-mm #Torsion in shaft CD\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Now Polar modulus of section AB\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of section BC\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Polar modulus of section CD\n",
+ "J3=pi*32**-1*d3**4 #mm**4\n",
+ "\n",
+ "#Now angle of twist of AB\n",
+ "theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of BC\n",
+ "theta2=T2*L2*(G*J2)**-1 #radians\n",
+ "\n",
+ "#Angle of twist of CD\n",
+ "theta3=T3*L3*(G*J3)**-1 #radians\n",
+ "\n",
+ "#Angle of twist\n",
+ "theta=theta1-theta2+theta3 #Radians\n",
+ "\n",
+ "#Shear stress in AB From Torsion Equation\n",
+ "q_s1=T1*(d1*2**-1)*J1**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in BC\n",
+ "q_s2=T2*(d*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Shear stress in CD\n",
+ "q_s3=T3*(d3*2**-1)*J3**-1 #N-mm**2\n",
+ "\n",
+ "#As max shear stress occurs in portion CD,so consider CD\n",
+ "\n",
+ "#Result\n",
+ "print\"Angle of twist at free end is\",round(theta,5),\"Radian\"\n",
+ "print\"Max Shear stress\",round(q_s3,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Angle of twist at free end is 0.00496 Radian\n",
+ "Max Shear stress 23.58 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.16,Page No.242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of bar\n",
+ "L1=600 #mm #Length of Bar AB\n",
+ "L2=400 #mm #Length of Bar BC\n",
+ "d1=60 #mm #Outer Diameter of bar BC\n",
+ "d2=30 #mm #Inner Diameter of bar BC\n",
+ "d=60 #mm #Diameter of bar AB\n",
+ "T=2*10**6 #N-mm #Total Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of Portion AB\n",
+ "J1=pi*32**-1*d**4 #mm*4\n",
+ "\n",
+ "#Polar Modulus of Portion BC\n",
+ "J2=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Let T1 be the torque resisted by bar AB and T2 be torque resisted by Bar BC\n",
+ "#Let theta1 and theta2 be the rotation of shaft in portion AB & BC\n",
+ "\n",
+ "#theta1=T1*L1*(G*J1)**-1 #radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta1=32*600*T1*(pi*60**4*G)**-1\n",
+ "\n",
+ "#theta2=T2*L*(J2*G)**-1 #Radians\n",
+ "#After substituting values and further simplifying we get \n",
+ "#theta2=32*400*T2*(pi*60**4*(1-0.5**4)*G)**-1 \n",
+ "\n",
+ "#Now For consistency of Deformation,theta1=theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#T1=0.7111*T2 ..................................................(1)\n",
+ "\n",
+ "#But T1+T2=T=2*10**6 ...........................................(2)\n",
+ "#Substituting value of T1 in above equation\n",
+ "\n",
+ "T2=T*(0.7111+1)**-1\n",
+ "T1=0.71111*T2\n",
+ "\n",
+ "#Max stress in Portion AB\n",
+ "q_s1=T1*(d*2**-1)*(J1)**-1 #N/mm**2\n",
+ "\n",
+ "#Max stress in Portion BC\n",
+ "q_s2=T2*(d1*2**-1)*J2**-1 #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Portion:AB\",round(q_s1,2),\"N/mm**2\"\n",
+ "print\" :BC\",round(q_s2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Portion:AB 19.6 N/mm**2\n",
+ " :BC 29.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.17,Page No.243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=80 #mm #External Diameter of Brass tube\n",
+ "d2=50 #mm #Internal Diameter of Brass tube\n",
+ "d=50 #mm #Diameter of steel Tube\n",
+ "G_b=40*10**3 #N/mm**2 #Modulus of Rigidity of brass tube\n",
+ "G_s=80*10**3 #N/mm**2 #Modulus of rigidity of steel tube\n",
+ "T=6*10**6 #N-mm #Torque\n",
+ "L=2000 #mm #Length of Tube\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar Modulus of brass tube\n",
+ "J1=pi*32**-1*(d1**4-d2**4) #mm**4 \n",
+ "\n",
+ "#Polar modulus of steel Tube\n",
+ "J2=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Let T_s & T_b be the torque resisted by steel and brass respectively\n",
+ "#Then, T_b+T_s=T ............................................(1)\n",
+ "\n",
+ "#Since the angle of twist will be the same\n",
+ "#Theta1=Theta2\n",
+ "#After substituting values and further simplifying we get \n",
+ "#Ts=0.360*Tb ...........................................(2)\n",
+ "\n",
+ "#After substituting value of Ts in eqn 1 and further simplifying we get \n",
+ "T_b=T*(0.36+1)**-1 #N-mm\n",
+ "T_s=0.360*T_b\n",
+ "\n",
+ "#Let q_s and q_b be the max stress in steel and brass respectively\n",
+ "q_b=T_b*(d1*2**-1)*J1**-1 #N/mm**2\n",
+ "q_s=T_s*(d2*2**-1)*J2**-1 #N/mm**2\n",
+ "\n",
+ "#Since angle of twist in brass=angle of twist in steel\n",
+ "theta_s=T_s*L*(J2*G_s)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses Developed in Materials are:Brass\",round(q_b,2),\"N/mm**2\"\n",
+ "print\" :Steel\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Angle of Twist in 2m Length\",round(theta_s,3),\"Radians\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses Developed in Materials are:Brass 51.79 N/mm**2\n",
+ " :Steel 64.71 N/mm**2\n",
+ "Angle of Twist in 2m Length 0.065 Radians\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.18,Page No.245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=60 #mm #External Diameter of aluminium Tube\n",
+ "d2=40 #mm #Internal Diameter of aluminium Tube\n",
+ "d=40 #mm #Diameter of steel tube\n",
+ "q_a=60 #N/mm**2 #Permissible stress in aluminium\n",
+ "q_s=100 #N/mm**2 #Permissible stress in steel tube\n",
+ "G_a=27*10**3 #N/mm**2 \n",
+ "G_s=80*10**3 #N/mm**2 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Polar modulus of aluminium Tube\n",
+ "J_a=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Polar Modulus of steel Tube\n",
+ "J_s=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Now the angle of twist of steel tube = angle of twist of aluminium tube\n",
+ "#T_s*L_s*(J_s*theta_s)**-1=T_a*L_a*(J_a*theta_a)**-1\n",
+ "#After substituting values in above Equation and Further simplifyin we get\n",
+ "#T_s=0.7293*T_a .....................(1)\n",
+ "\n",
+ "#If steel Governs the resisting capacity\n",
+ "T_s1=q_s*J_s*(d*2**-1)**-1 #N-mm\n",
+ "T_a1=T_s1*0.7293**-1 #N-mm\n",
+ "T1=(T_s1+T_a1)*10**-6 #KN-m #Total Torque in steel Tube\n",
+ "\n",
+ "#If aluminium Governs the resisting capacity \n",
+ "T_a2=q_a*J_a*(d1*2**-1) #N-mm\n",
+ "T_s2=T_a2*0.7293 #N-mm\n",
+ "T2=(T_s2+T_a2)*10**-6 #KN-m #Total Torque in aluminium tube\n",
+ "\n",
+ "#Result\n",
+ "print\"Steel Governs the torque carrying capacity\",round(T1,2),\"KN-m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Steel Governs the torque carrying capacity 2.98 KN-m\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.19,Page No.247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "P=225*10**6 #N-mm/sec #Power Trasmitted\n",
+ "q_b=80 #N/mm**2 #Shear stress\n",
+ "n=200 #Rpm\n",
+ "q_k=100 #N/mm**2 #PErmissible stress in Keys\n",
+ "D=300 #mm #Diameter of bolt circle\n",
+ "L=150 #mm #Length of shear key\n",
+ "d=16 #mm #Diameterr of bolt\n",
+ "\n",
+ "#Calculations\n",
+ "T=60*P*(2*pi*n)**-1 #N-mm #Torque\n",
+ "\n",
+ "#Now From Torsion Formula\n",
+ "#T*J**-1=q_s*R**-1\n",
+ "#After substituting values we get\n",
+ "#T=pi*16*d**3*n\n",
+ "#After further simplifying we get\n",
+ "d1=(T*16*(pi*q_s)**-1)**0.33333\n",
+ "\n",
+ "#Let b be the width of shear Key\n",
+ "#T=q_k*L*b*R\n",
+ "#After simplifying further we get\n",
+ "b=T*(q_k*L*(d1*2**-1))**-1 #mm\n",
+ "\n",
+ "#Let n2 be the no. of bolts required at bolt circle of radius\n",
+ "R_b=D*2**-1 #mm \n",
+ "\n",
+ "n2=T*4*(q_b*pi*d**2*R_b)**-1\n",
+ "\n",
+ "#result\n",
+ "print\"Minimum no. of Bolts Required are\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum no. of Bolts Required are 4.45\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.20,Page No.250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "T=2*10**6 #N-mm #Torque transmitted\n",
+ "G=80*10**3 #N/mm**2 #Modulus of rigidity\n",
+ "d1=40 #mm \n",
+ "d2=80 #mm\n",
+ "r1=20 #mm\n",
+ "r2=40 #mm\n",
+ "L=2000 #mm #Length of shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Angle of twist \n",
+ "theta=2*T*L*(r1**2+r1*r2+r2**2)*(3*pi*G*r2**3*r1**3)**-1 #radians\n",
+ "\n",
+ "#If the shaft is treated as shaft of average Diameter\n",
+ "d_avg=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "theta1=T*L*(G*pi*32**-1*d_avg**4)**-1 #Radians\n",
+ "\n",
+ "#Percentage Error\n",
+ "#Let Percentage Error be E\n",
+ "X=theta-theta1\n",
+ "E=(X*theta**-1)*100 \n",
+ "\n",
+ "#Result\n",
+ "print\"Percentage Error is\",round(E,2),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Percentage Error is 32.28 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.21,Page No.252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "P=1*10**9 #N-mm/sec #Power\n",
+ "n=300 \n",
+ "d1=150 #mm #Outer Diameter\n",
+ "d2=120 #mm #Inner Diameter\n",
+ "L=2000 #mm #Length of circular shaft\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=P*60*(2*pi*n)**-1 #N-mm\n",
+ "\n",
+ "#Polar Modulus \n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "q_s=T*J**-1*(d1*2**-1) #N/mm**2 \n",
+ "\n",
+ "\n",
+ "#Strain ENergy\n",
+ "U=q_s**2*(4*G)**-1*pi*4**-1*(d1**2-d2**2)*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored in the shaft is\",round(U,2),\"N-mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 81.36 N/mm**2\n",
+ "Strain Energy stored in the shaft is 263181.37 N-mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.22,Page No.254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=12 #mm #Diameter of helical spring\n",
+ "D=150 #mm #Mean Diameter\n",
+ "R=D*2**-1 #mm #Radius of helical spring\n",
+ "n=10 #no.of turns\n",
+ "G=80*10**3 #N/mm**2 \n",
+ "W=450 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress \n",
+ "q_s=16*W*R*(pi*d**3)**-1 #N/mm**2\n",
+ "\n",
+ "#Strain Energy stored\n",
+ "U=32*W**2*R**3*n*(G*d**4)**-1 #N-mm\n",
+ "\n",
+ "#Deflection Produced\n",
+ "dell=64*W*R**3*n*(G*d**4)**-1 #mm\n",
+ "\n",
+ "#Stiffness Spring\n",
+ "k=W*dell**-1 #N/mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Max shear stress is\",round(q_s,2),\"N/mm**2\"\n",
+ "print\"Strain Energy stored is\",round(U,2),\"N-mm\"\n",
+ "print\"Deflection Produced is\",round(dell,2),\"mm\"\n",
+ "print\"Stiffness spring is\",round(k,2),\"N/mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max shear stress is 99.47 N/mm**2\n",
+ "Strain Energy stored is 16479.49 N-mm\n",
+ "Deflection Produced is 73.24 mm\n",
+ "Stiffness spring is 6.14 N/mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 53
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.23,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "K=5 #N/mm #Stiffness\n",
+ "L=100 #mm #Solid Length\n",
+ "q_s=60 #N/mm**2 #Max shear stress\n",
+ "W=200 #N #Max Load\n",
+ "G=80*10**3 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#K=W*dell**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#d=0.004*R**3*n ........(1) #mm #Diameter of wire\n",
+ "#n=L*d**-1 ........(2)\n",
+ "\n",
+ "#From Shearing stress\n",
+ "#q_s=16*W*R*(pi*d**3)**-1 \n",
+ "#After substituting values and further simplifying we get\n",
+ "#d**4=0.004*R**3*n .................(4)\n",
+ "\n",
+ "#From Equation 1,2,3\n",
+ "#d**4=0.004*(0.0785*d**3)**3*100*d**-1\n",
+ "#after further simplifying we get\n",
+ "d=5168.101**0.25\n",
+ "n=100*d**-1\n",
+ "R=(d**4*(0.004*n)**-1)**0.3333\n",
+ "\n",
+ "#Result\n",
+ "print\"Diameter of Wire is\",round(d,2),\"mm\"\n",
+ "print\"No.of turns is\",round(n,2)\n",
+ "print\"Mean Radius of spring is\",round(R,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Diameter of Wire is 8.48 mm\n",
+ "No.fo turns is 11.79\n",
+ "Mean Radius of spring is 47.83 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.24,Page No.255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "m=5*10**5 #Wagon Weighing\n",
+ "v=18*1000*36000**-1 \n",
+ "d=300 #mm #Diameter of Beffer springs\n",
+ "n=18 #no.of turns\n",
+ "G=80*10**3 #N/mm**2\n",
+ "dell=225\n",
+ "R=100 #mm #Mean Radius\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Energy of Wagon\n",
+ "E=m*v**2*(9.81*2)**-1 #N-mm\n",
+ "\n",
+ "#Load applied\n",
+ "W=dell*G*d**4*(64*R**3*n)**-1 #N \n",
+ "\n",
+ "#Energy each spring can absorb is\n",
+ "E2=W*dell*2**-1 #N-mm\n",
+ "\n",
+ "#No.of springs required to absorb energy of Wagon\n",
+ "n2=E*E2**-1 *10**7\n",
+ "\n",
+ "#Result\n",
+ "print\"No.of springs Required for Buffer is\",round(n2,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "No.of springs Required for Buffer is 4.47\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.25,Page No.259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "b=180 #mm #width of flange\n",
+ "d=10 #mm #Depth of flange\n",
+ "t=10 #mm #Thickness of flange\n",
+ "D=400 #mm #Overall Depth \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I_xx=1*12**-1*(b*D**3-(b-t)*(D-2*d)**3)\n",
+ "I_yy=1*12**-1*((D-2*d)*t**3+2*t*b**3)\n",
+ "\n",
+ "#If warping is neglected\n",
+ "J=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Since b/d>1.6,we get\n",
+ "J2=1*3**-1*d**3*b*(1-0.63*d*b**-1)*2+1*3**-1*t**3*(D-2*d)*(1-0.63*t*b**-1)\n",
+ "\n",
+ "#Over Estimation of torsional Rigidity would have been \n",
+ "T=J*J2**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Error in assessing torsional Rigidity if the warping is neglected is\",round(T,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Error in assessing torsional Rigidity if the warping is neglected is 808.28\n"
+ ]
+ }
+ ],
+ "prompt_number": 68
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 6.6.26,Page No.261"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #Outer Diameter\n",
+ "d2=95 #mm #Inner Diameter\n",
+ "T=2*10**6 #N-mm #Torque\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=pi*32**-1*(d1**4-d2**4) #mm**4 #Polar Modulus\n",
+ "\n",
+ "#Shear stress\n",
+ "q_max=T*J**-1*d1*2**-1 #N/mm**2 \n",
+ "\n",
+ "#Now theta*L**-1=T*(G*J)**-1\n",
+ "#After substituting values and further simplifying we get\n",
+ "#Let theta*L**-1=X\n",
+ "X=T*J**-1\n",
+ "\n",
+ "#Now Treating it as very thin walled tube\n",
+ "d=(d1+d2)*2**-1 #mm\n",
+ "\n",
+ "r=d*2**-1 \n",
+ "t=(d1-d2)*2**-1\n",
+ "q_max2=T*(2*pi*r**2*t)**-1 #N/mm**2\n",
+ "\n",
+ "X2=T*(2*pi*r**3*t)**-1 \n",
+ "\n",
+ "#Result\n",
+ "print\"When it is treated as hollow shaft:Max shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\" :Angle of Twist per unit Length\",round(X,3)\n",
+ "print\"When it is very thin Walled Tube :Max shear stress\",round(q_max2,2),\"N/mm**2\"\n",
+ "print\" :Angle of twist per Unit Length\",round(X2,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "When it is treated as hollow shaft:Max shear stress 54.91 N/mm**2\n",
+ " :Angle of Twist per unit Length 1.098\n",
+ "When it is very thin Walled Tube :Max shear stress 53.57 N/mm**2\n",
+ " :Angle of twist per Unit Length 1.099\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_5.ipynb new file mode 100644 index 00000000..b75d886a --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.7_5.ipynb @@ -0,0 +1,1328 @@ +{
+ "metadata": {
+ "name": "chapter no.7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.7:Compound Stresses And Strains"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1,Page No.269"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma1=30 #N/mm**2 #Stress in tension\n",
+ "d=20 #mm #Diameter \n",
+ "sigma2=90 #N/mm**2 #Max compressive stress\n",
+ "sigma3=25 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#In TEnsion\n",
+ "\n",
+ "#Corresponding stress in shear\n",
+ "P=sigma1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Tensile force\n",
+ "F=pi*4**-1*d**2*sigma1\n",
+ "\n",
+ "#In Compression\n",
+ "\n",
+ "#Correspong shear stress\n",
+ "P2=sigma2*2**-1 #N/mm**2\n",
+ "\n",
+ "#Correspong compressive(axial) stress\n",
+ "p=2*sigma3 #N/mm**2 \n",
+ "\n",
+ "#Corresponding Compressive force\n",
+ "P3=p*pi*4**-1*d**2 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Failure Loads are:\",round(F,2),\"N\"\n",
+ "print\" :\",round(P3,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Failure Loads are: 9424.78 N\n",
+ " : 15707.96 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.2,Page No.270"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=25 #mm #Diameter of circular bar\n",
+ "F=20*10**3 #N #Axial Force\n",
+ "theta=30 #Degree #angle \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Axial stresses\n",
+ "p=F*(pi*4**-1*d**2)**-1 #N/mm**2\n",
+ "\n",
+ "#Normal Stress\n",
+ "p_n=p*(cos(30*pi*180**-1))**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "p_t=p*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Max shear stress occurs on plane where theta2=45 \n",
+ "theta2=45\n",
+ "sigma_max=p*2**-1*sin(2*theta2*pi*180**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses developed on a plane making 30 degree is:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" :\",round(p_t,2),\"N/mm**2\"\n",
+ "print\"stress on max shear stress is\",round(sigma_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses developed on a plane making 30 degree is: 30.56 N/mm**2\n",
+ " : 17.64 N/mm**2\n",
+ "stress on max shear stress is 20.37 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.3,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "theta=30 #degree\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p1=120 #N/mm**2\n",
+ "p2=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(p1+p2)*2**-1+(p1-p2)*2**-1*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential stress\n",
+ "P_t=(p1-p2)*2**-1*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "phi=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Angle made by resultant with 120 #N/mm**2 stress\n",
+ "phi2=phi+theta #Degree\n",
+ "\n",
+ "#Result\n",
+ "print\"Normal Stress is\",round(P_n,2),\"N/mm**2\"\n",
+ "print\"Tangential Stress is\",round(P_t,2),\"N/mm**2\"\n",
+ "print\"Angle made by resultant\",round(phi2,2),\"Degree\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Normal Stress is 110.0 N/mm**2\n",
+ "Tangential Stress is 17.32 N/mm**2\n",
+ "Angle made by resultant 111.05 Degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.4,Page No.272"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct Stresses\n",
+ "P1=60 #N/mm**2 \n",
+ "P2=100 #N/mm**2\n",
+ "\n",
+ "Theta=25 #Degree #Angle\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress\n",
+ "P_n=(P1-P2)*2**-1+(P1+P2)*2**-1*cos(2*Theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Tangential Stress\n",
+ "P_t=(P1+P2)*2**-1*sin(Theta*2*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "P=(P_n**2+P_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "theta2=arctan(P_n*P_t**-1)*(180*pi**-1)\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses on the plane AC is:\",round(P_n,2),\"N/mm**2\"\n",
+ "print\" \",round(P_t,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses on the plane AC is: 31.42 N/mm**2\n",
+ " 61.28 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.6,Page No.278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses acting on material\n",
+ "p_x=180 #N/mm**2 \n",
+ "p_y=120 #N/mm**2\n",
+ "\n",
+ "q=80 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1) #degrees\n",
+ "theta2=theta*2**-1 #Degrees\n",
+ "theta3=theta+180 #Degrees\n",
+ "theta4=theta3*2**-1 #Degrees\n",
+ "\n",
+ "#Stresses\n",
+ "p_1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p_2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of Principal stress is:\",round(p_1,2),\"N/mm**2\"\n",
+ "print\" \",round(p_2,2),\"N/mm**2\"\n",
+ "print\"Magnitude of max shear stress is\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of Principal stress is: 235.44 N/mm**2\n",
+ " 64.56 N/mm**2\n",
+ "Magnitude of max shear stress is 85.44 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.7,Page No.279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=60 #N/mm**2\n",
+ "p_y=-40 #N/mm**2\n",
+ "\n",
+ "q=10 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 60.99 N/mm**2\n",
+ " : -40.99 N/mm**2\n",
+ "Max shear stresses 50.99 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.8,Page No.280"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-120 #N/mm**2\n",
+ "p_y=-80 #N/mm**2\n",
+ "\n",
+ "q=-60 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Principal Stresses\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" :\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shear stresses\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: -36.75 N/mm**2\n",
+ " : -163.25 N/mm**2\n",
+ "Max shear stresses 63.25 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.9,Page No.282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#stresses\n",
+ "p_x=-40 #N/mm**2\n",
+ "p_y=80 #N/mm**2\n",
+ "\n",
+ "q=48 #N/mm**2 #shear stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=((((p_x-p_y)*2**-1)**2)+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Inclination of principal stress to plane\n",
+ "theta=arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1)#Degrees\n",
+ "theta2=(theta)*2**-1 #degrees\n",
+ "\n",
+ "theta3=(theta+180)*2**-1 #degrees\n",
+ "\n",
+ "#Normal Corresponding stress\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*(theta2+45)*pi*180**-1)+q*sin(2*(theta2+45)*pi*180**-1) #Degrees\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=((p_n**2+q_max**2)**0.5) #N/mm**2\n",
+ "\n",
+ "phi=arctan(p_n*q_max**-1)*(180*pi**-1) #Degrees\n",
+ "\n",
+ "#Inclination to the plane\n",
+ "alpha=round((theta2+45),2)+round(phi ,2)#Degree\n",
+ "\n",
+ "#Answer in book is incorrect of alpha ie41.25\n",
+ "\n",
+ "#Result\n",
+ "print\"Planes of max shear stress:\",round(p_n,2),\"N/mm**2\"\n",
+ "print\" \",round(q_max,2),\"N/mm*2\"\n",
+ "print\"Resultant Stress is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Planes of max shear stress: 20.0 N/mm**2\n",
+ " 76.84 N/mm*2\n",
+ "Resultant Stress is 79.4 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.10,Page No.283"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Stresses\n",
+ "p_x=50*cos(35*pi*180**-1)\n",
+ "q=50*sin(35*pi*180**-1)\n",
+ "p_y=0\n",
+ "\n",
+ "theta=40 #Degrees #Plane AB inclined to vertical\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Normal Stress on AB\n",
+ "p_n=(p_x+p_y)*2**-1+(p_x-p_y)*2**-1*cos(2*theta*pi*180**-1)+q*sin(2*theta*pi*180**-1)\n",
+ "\n",
+ "#Tangential Stress on AB\n",
+ "p_t=(p_x-p_y)*2**-1*sin(2*theta*pi*180**-1)-q*cos(2*theta*pi*180**-1) #N/mm**2\n",
+ "\n",
+ "#Resultant stress\n",
+ "p=(p_n**2+p_t**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Angle of resultant\n",
+ "phi=arctan(p_n*p_t**-1)*(180*pi**-1) #degrees\n",
+ "phi2=phi+theta #Degrees\n",
+ "\n",
+ "#Result\n",
+ "print\"Magnitude of resultant stress is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Direction of Resultant stress is\",round(phi2,2),\"Degrees\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Magnitude of resultant stress is 54.44 N/mm**2\n",
+ "Direction of Resultant stress is 113.8 Degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.12,Page No.285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Direct stresses\n",
+ "p_x=120 #N/mm**2 #Tensile stress\n",
+ "p_y=-100 #N/mm**2 #Compressive stress\n",
+ "p1=160 #N/mm**2 #Major principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let q be the shearing stress\n",
+ "\n",
+ "#p1=(p_x+p_y)*2**-1+((((p_x+p_y)*2**-1)**2)+q**2)**0.5\n",
+ "#After further simplifying we get\n",
+ "q=(p1-((p_x+p_y)*2**-1))**2-((p_x-p_y)*2**-1)**2 #N/mm**2\n",
+ "q2=(q)**0.5 #N/mm**2\n",
+ "\n",
+ "#Minimum Principal stress\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shearing stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shearing stress of material\",round(q,2),\"N/mm**2\"\n",
+ "print\"Min Principal stress\",round(p2,2),\"N/mm**2\"\n",
+ "print\"Max shearing stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shearing stress of material 10400.0 N/mm**2\n",
+ "Min Principal stress -140.0 N/mm**2\n",
+ "Max shearing stress 150.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.14,Page No.291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "F=40*10**3 #N #Shear Force\n",
+ "M=20*10**6 #Bending Moment\n",
+ "\n",
+ "#Rectangular section\n",
+ "b=100 #mm #Width\n",
+ "d=200 #mm #Depth\n",
+ "\n",
+ "x=20 #mm #Distance from Top surface upto point\n",
+ "y=80 #mm #Distance from point to Bottom\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=1*12**-1*b*d**3 #mm**4 #M.I\n",
+ "\n",
+ "#At 20 mm Below top Fibre\n",
+ "f_x=M*I**-1*y #N/mm**2 #Stress\n",
+ "\n",
+ "#Assuming sagging moment ,f_x is compressive p_x=f_x=-24 #N/mm**2\n",
+ "p_x=f_x=-24 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*(b*I)**-1*(b*x*(b-x*2**-1)) #N/mm**2\n",
+ "\n",
+ "#Direct stresses\n",
+ "\n",
+ "p_y=0 #N/mm**2\n",
+ "\n",
+ "p1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "p2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Directions of principal stresses at a point below 20mm is:\",round(p1,2),\"N/mm**2\"\n",
+ "print\" \",round(p2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Directions of principal stresses at a point below 20mm is: 0.05 N/mm**2\n",
+ " -24.05 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.15,Page No.292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=4000 #mm #Span\n",
+ "W1=W2=W3=2*10**3 #N #Load\n",
+ "\n",
+ "#SEction of beam\n",
+ "b=100 #mm #Width\n",
+ "d=240 #mm #Dept\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the reactions\n",
+ "R_A=R_B=(W1+W2+W3)*2**-1 #KN\n",
+ "\n",
+ "#Now at the section 1.5m from left support A\n",
+ "#Shear Force\n",
+ "F=R_A-W1 #KN\n",
+ "\n",
+ "#B.M\n",
+ "M=R_A*1.5-W1*0.5 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*d**3 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "#f=M*I**-1*y\n",
+ "#After Sub values and further simplifying we get\n",
+ "#f=3.04*10**-2*y\n",
+ "\n",
+ "#As it varies Linearly\n",
+ "\n",
+ "#at distance 0 From NA \n",
+ "f1=0\n",
+ "#at distance 60 mm from NA\n",
+ "f2=1.823 #N/mm**2\n",
+ "#at distance 120 mm from NA\n",
+ "f3=3.646 #N/mm**2\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=F*b*d*2**-1*d*4**-1*(b*I)**-1\n",
+ "\n",
+ "#At 60 mm above NA\n",
+ "q2=F*b*d*4**-1*(d*2**-1-d*8**-1)*(b*I)**-1\n",
+ "\n",
+ "#At 120 mm above NA\n",
+ "q3=0 \n",
+ "\n",
+ "#At NA element is under pure shear\n",
+ "p1=q #N/mm**2\n",
+ "p2=-q #N/mm**2 \n",
+ "\n",
+ "#Inclination of principal plane to vertical\n",
+ "#theta=2*q*0**-1\n",
+ "#Further simplifying we get\n",
+ "#theta=infinity\n",
+ "\n",
+ "#therefore\n",
+ "theta=90*2**-1 #degrees\n",
+ "theta2=270*2**-1 #degrees\n",
+ "\n",
+ "#At 60 mm From NA\n",
+ "p_x=-1.823 #N/mm**2 \n",
+ "p_y=0\n",
+ "q=0.0469 #N/mm**2\n",
+ "\n",
+ "#principal planes\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x+p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Principal planes inclination to hte plane of p_x is given by\n",
+ "theta3=(arctan(2*q*(p_x-p_y)**-1)*(180*pi**-1))\n",
+ "theta4=theta3*2**-1#degrees\n",
+ "\n",
+ "theta5=theta3+180 #Degrees\n",
+ "\n",
+ "#At 120 mm From N-A\n",
+ "p_x2=3.646 #N/mm**2\n",
+ "p_y2=0 #N/mm**2\n",
+ "q2=0 #N/mm**2\n",
+ "\n",
+ "P3=p_x2 #N/mm**2\n",
+ "P4=0 #N/mm**2\n",
+ "\n",
+ "#Answer for P2 at 60 mm from NA is incorrect\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x,2),\"N/mm**2\"\n",
+ "print\" \",round(p_y,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P1,4),\"N/mm**2\"\n",
+ "print\" \",round(P2,4),\"N/mm**2\"\n",
+ "print\"Principal Planes at 60 mm from NA:\",round(p_x2,4),\"N/mm**2\"\n",
+ "print\" \",round(p_y2,4),\"N/mm**2\"\n",
+ "print\"Principal Stresses at 60 mm From NA\",round(P3,4),\"N/mm**2\"\n",
+ "print\" \",round(P4,4),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Planes at 60 mm from NA: -1.82 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 0.0012 N/mm**2\n",
+ " -1.8242 N/mm**2\n",
+ "Principal Planes at 60 mm from NA: 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n",
+ "Principal Stresses at 60 mm From NA 3.646 N/mm**2\n",
+ " 0.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.16,Page No.295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=8000 #mm #Span of beam\n",
+ "w=40*10**6 #N/mm #udl\n",
+ "\n",
+ "#I-section\n",
+ "\n",
+ "#Flanges\n",
+ "b=100 #mm #Width\n",
+ "t=10 #mm #Thickness\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "t2=10 #mm #thickness of web\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let R_A and R_B be the Reactions at A & B respectively\n",
+ "R_A=w*2**-1*L*10**-9 #KN\n",
+ "\n",
+ "#Shear force at 2m for left support\n",
+ "F=R_A-2*w*10**-6 #KN\n",
+ "\n",
+ "#Bending Moment\n",
+ "M=R_A*2-2*w*10**-6 #KN-m\n",
+ "\n",
+ "#M.I\n",
+ "I=1*12**-1*b*D**3-1*12**-1*(b-t)*(D-2*t2)**3 #mm**4\n",
+ "\n",
+ "#Bending stress at 100 mm above N_A\n",
+ "f=M*10**6*I**-1*b\n",
+ "\n",
+ "#Shear stress \n",
+ "q=F*10**3*(t*I)**-1*(b*t*(D-t)*2**-1 +t2*(b-t2)*145) #N/mm**2\n",
+ "\n",
+ "p_x=-197.06 #N/mm**2 \n",
+ "p_y=0 #N/mm**2\n",
+ "q=21.38 #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" \",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max shear stress\",round(q_max,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are: 2.29 N/mm**2\n",
+ " -199.35 N/mm**2\n",
+ "Max shear stress 100.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.18,Page No.298"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=100 #mm #Diameter of shaft\n",
+ "M=3*10**6 #N-mm #B.M\n",
+ "T=6*10**6 #N-mm #Twisting Moment\n",
+ "mu=0.3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max principal Stress\n",
+ "\n",
+ "P1=16*(pi*d**3)**-1*(M+(M**2+T**2)**0.5) #N/mm**2 \n",
+ "P2=16*(pi*d**3)**-1*(M-(M**2+T**2)**0.5) #N/mm**2 \n",
+ "\n",
+ "#Direct stress\n",
+ "P=round(P1,2)-mu*round(P2,2) #N/mm**2 \n",
+ "\n",
+ "#Result\n",
+ "print\"Principal stresses are:\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Stress Producing the same strain is\",round(P,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal stresses are: 49.44 N/mm**2\n",
+ " : -18.89 N/mm**2\n",
+ "Stress Producing the same strain is 55.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.19,Page No.299"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=75 #mm #diameter \n",
+ "P=30*10**6 #W #Power transmitted\n",
+ "W=6 #N-mm/sec #Load\n",
+ "L=1000 #mm \n",
+ "N=300 #r.p.m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#B.M\n",
+ "M=W*L*4**-1 #N-mm\n",
+ "T=P*60*(2*pi*N)**-1 #Torque transmitted\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*64**-1*d**4 #mm**4\n",
+ "\n",
+ "#Bending stress\n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A\n",
+ "p_y=0\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "#Principal Stresses\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Bending stress\n",
+ "p_x2=0\n",
+ "p_y2=0\n",
+ "\n",
+ "#Shearing stress\n",
+ "q2=T*J**-1*d*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal stresses\n",
+ "P3=(p_x2+p_y2)*2**-1+(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "P4=(p_x2+p_y2)*2**-1-(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y2)*2**-1)**2+q2**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Answer for Principal Stresses P1,P2 and Max stress i.e q_max is incorrect in Book\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses at vertical Diameter:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Max stress at vertical Diameter : \",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses at Horizontal Diameter:P3\",round(P3,2),\"N/mm**2\"\n",
+ "print\" :P4\",round(P4,2),\"N/mm**2\"\n",
+ "print\"Max stress at Horizontal Diameter : \",round(q_max2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses at vertical Diameter:P1 11.55 N/mm**2\n",
+ " :P2 -11.51 N/mm**2\n",
+ "Max stress at vertical Diameter : 11.53 N/mm**2\n",
+ "Principal Stresses at Horizontal Diameter:P3 11.53 N/mm**2\n",
+ " :P4 -11.53 N/mm**2\n",
+ "Max stress at Horizontal Diameter : 11.53 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.20,Page No.302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=100 #mm #External Diameter\n",
+ "d2=50 #mm #Internal Diameter\n",
+ "N=500 #mm #r.p.m\n",
+ "P=60*10**6 #N-mm/sec #Power\n",
+ "p=100 #N/mm**2 #principal stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=pi*(d1**4-d2**4)*64**-1 #mm**4\n",
+ "\n",
+ "#Bending Stress\n",
+ "#f=M*I*d1*2**-1 #N/mm**2\n",
+ "\n",
+ "#Principal Planes\n",
+ "#p_x=32*M*(pi*(d1**4-d2**4))*d1\n",
+ "#p_y=0\n",
+ "\n",
+ "#Shear stress\n",
+ "#q=T*J**-1*(d1*2**-1)\n",
+ "#After sub values and further simplifying we get\n",
+ "#q=16*T*d1*(pi*(d1**4-d2**4))*d1\n",
+ "\n",
+ "#Principal stresses\n",
+ "#P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#After sub values and further simplifying we get\n",
+ "#P1=16*(pi*(d1**4-d2**4))*d1*(M+(M**2+t**2)**0.5) ...............(1)\n",
+ "\n",
+ "#P=2*pi*N*T*60**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "T=P*60*(2*pi*N)**-1*10**-6 #N-mm\n",
+ "\n",
+ "#Again Sub values and further simplifying Equation 1 we get\n",
+ "M=(337.533)*(36.84)**-1 #KN-m\n",
+ "\n",
+ "#Min Principal stress\n",
+ "#P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "#Sub values and further simplifying we get\n",
+ "P2=16*(pi*(d1**4-d2**4))*d1*(M-(M**2+T**2)**0.5)*10**-11\n",
+ "\n",
+ "#Result\n",
+ "print\"Bending Moment safely applied to shaft is\",round(M,2),\"KN-m\"\n",
+ "print\"Min Principal Stress is\",round(P2,3),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Bending Moment safely applied to shaft is 9.16 KN-m\n",
+ "Min Principal Stress is -0.336 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.21,Page No.303"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=150 #mm #Diameter\n",
+ "T=20*10**6 #N #Torque\n",
+ "M=12*10**6 #N-mm #B.M\n",
+ "F=200*10**3 #N #Axial Thrust\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I\n",
+ "I=(pi*64**-1*d**4)\n",
+ "\n",
+ "#Bending stress \n",
+ "f_A=M*I**-1*(d*2**-1) #N/mm**2\n",
+ "f_B=-f_A #N/mm**2\n",
+ "\n",
+ "#Axial thrust due to thrust\n",
+ "sigma=F*(pi*4**-1*d**2)**-1\n",
+ "\n",
+ "#At A\n",
+ "p_x=f_A-sigma #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "p_x2=f_B-sigma #N/mm**2\n",
+ "\n",
+ "p_y=0 #At A and B\n",
+ "\n",
+ "#Polar Modulus\n",
+ "J=pi*32**-1*d**4 #mm**4\n",
+ "\n",
+ "#Shearing stress at A and B\n",
+ "q=T*J**-1*(d*2**-1) #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Principal Stresses\n",
+ "#At A\n",
+ "P1=(p_x+p_y)*2**-1+(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2=(p_x+p_y)*2**-1-(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max1=(((p_x-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#At B\n",
+ "P1_2=(p_x2+p_y)*2**-1+(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "P2_2=(p_x2+p_y)*2**-1-(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max2=(((p_x2-p_y)*2**-1)**2+q**2)**0.5 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"MAx Principal Stresses:P1\",round(P1,2),\"N/mm**2\"\n",
+ "print\" :P2\",round(P2,2),\"N/mm**2\"\n",
+ "print\"Min Principal Stresses:P1_2\",round(P1_2,2),\"N/mm**2\"\n",
+ "print\" :P2_2\",round(P2_2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "MAx Principal Stresses:P1 45.1 N/mm**2\n",
+ " :P2 -20.2 N/mm**2\n",
+ "Min Principal Stresses:P1_2 14.65 N/mm**2\n",
+ " :P2_2 -62.18 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.22,Page No.311"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#strains\n",
+ "e_A=500 #microns\n",
+ "e_B=250 #microns\n",
+ "e_C=-150 #microns\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=45 #Degrees\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=500\n",
+ "e_45=e_B=250\n",
+ "e_y=e_C=-150 \n",
+ "\n",
+ "#e_45=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "rho_x_y=(e_45-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2\n",
+ "\n",
+ "#Principal strains are given by\n",
+ "e1=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Strains are:e1\",round(e1,2),\"N/mm**2\"\n",
+ "print\" :e2\",round(e2,2),\"N/mm**2\"\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Strains are:e1 508.54 N/mm**2\n",
+ " :e2 -158.54 N/mm**2\n",
+ "Principal Stresses are:sigma1 101.31 N/mm**2\n",
+ " :sigma2 -1.31 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example No.7.23,Page No.313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Strains\n",
+ "e_A=600 #microns\n",
+ "e_B=-450 #microns\n",
+ "e_C=100 #micron\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "theta=240\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "e_x=e_A=600\n",
+ "\n",
+ "#e_A=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(theta)+rho_x_y*2**-1*sin(theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#-450=(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(1)\n",
+ "\n",
+ "#e_C=(e_x+e_y)*2**-1+(e_x-e_y)*2**-1*cos(2*theta)+rho_x_y*2**-1*sin(2*theta)\n",
+ "#After sub values and further simplifying we get\n",
+ "#100=(e_x+e_y)*2**-1-0.5*(e_x-e_y)*2**-1*(0.5)-0.866*2**-1*rho_x_y .....................(2)\n",
+ "\n",
+ "#Adding Equation 1 and 2 we get equations as\n",
+ "#-350=e_x+e_y-(e_x-e_y)*2**-1 ...............(3)\n",
+ "#Further simplifying we get\n",
+ "\n",
+ "e_y=(-700-e_x)*3**-1 #micron \n",
+ "\n",
+ "rho_x_y=(e_C-(e_x+e_y)*2**-1-(e_x-e_y)*2**-1*cos(2*theta*pi*180**-1))*(sin(2*theta*pi*180**-1))**-1*2 #micron\n",
+ "\n",
+ "#Principal strains\n",
+ "e1=(e_x+e_y)*2**-1-(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "e2=(e_x+e_y)*2**-1+(((e_x-e_y)*2**-1)**2+(rho_x_y*2**-1)**2)**0.5 #microns\n",
+ "\n",
+ "#Principal Stresses\n",
+ "sigma1=E*(e1+mu*e2)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "sigma2=E*(e2+mu*e1)*(1-mu**2)**-1*10**-6 #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "print\"Principal Stresses are:sigma1\",round(sigma1,2),\"N/mm**2\"\n",
+ "print\" :sigma2\",round(sigma2,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Principal Stresses are:sigma1 -69.49 N/mm**2\n",
+ " :sigma2 117.11 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_5.ipynb new file mode 100644 index 00000000..08f57912 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.8_5.ipynb @@ -0,0 +1,1527 @@ +{
+ "metadata": {
+ "name": "chapter no.8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter No.8:Thin And Thick Cyclinders And Spheres"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.1,Page No.322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length\n",
+ "d1=1000 #mm #Internal diameter\n",
+ "t=15 #mm #Thickness\n",
+ "P=1.5 #N/mm**2 #Fluid Pressure\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=P*d1*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal Stress\n",
+ "f2=P*d1*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Max shear stress\n",
+ "q_max=(f1-f2)*2**-1 #N/mm**2\n",
+ "\n",
+ "#Diametrical Strain\n",
+ "#Let e1=dell_d*d**-1 .....................(1)\n",
+ "e1=(f1-mu*f2)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 1 and further simplifying we get\n",
+ "dell_d=e1*d1 #mm\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#e2=dell_L*L**-1 ......................(2)\n",
+ "e2=(f2-mu*f1)*E**-1 \n",
+ "\n",
+ "#Sub values in equation 2 and further simplifying we get\n",
+ "dell_L=e2*L #mm\n",
+ "\n",
+ "#Change in Volume \n",
+ "#Let Z=dell_V*V**-1 ................(3)\n",
+ "Z=2*e1+e2\n",
+ "\n",
+ "#Sub values in equation 3 and further simplifying we get\n",
+ "dell_V=Z*pi*4**-1*d1**2*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Intensity of shear stress\",round(q_max,2),\"N/mm**2\"\n",
+ "print\"Change in the Dimensions of the shell is:dell_d\",round(dell_d,2),\"mm\"\n",
+ "print\" :dell_L\",round(dell_L,2),\"mm\"\n",
+ "print\" :dell_V\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Intensity of shear stress 12.5 N/mm**2\n",
+ "Change in the Dimensions of the shell is:dell_d 0.21 mm\n",
+ " :dell_L 0.15 mm\n",
+ " :dell_V 1119192.38 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.2,Page No.323"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length\n",
+ "d=200 #mm # diameter\n",
+ "t=10 #mm #Thickness\n",
+ "dell_V=25000 #mm**3 #Additional volume\n",
+ "E=2*10**5 #n/mm**2 #Modulus of elasticity\n",
+ "mu=0.3 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the pressure developed\n",
+ "\n",
+ "#Circumferential Stress\n",
+ "\n",
+ "#f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=10*p\n",
+ "\n",
+ "#f1=p*d*(4*t)**-1 #N/mm**2\n",
+ "#After sub values and further simplifying\n",
+ "#f1=5*p\n",
+ "\n",
+ "#Diameterical strain = Circumferential stress\n",
+ "#Let X=dell_d*d**-1 ................................(1)\n",
+ "#X=e1=(f1-mu*f2)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e1=8.5*p*E**-1\n",
+ "\n",
+ "#Longitudinal strain\n",
+ "#Let Y=dell_L*L**-1 ......................................(2)\n",
+ "#Y=e2=(f2-mu*f1)*E**-1 \n",
+ "#After sub values and further simplifying\n",
+ "#e2=2*p*E**-1\n",
+ "\n",
+ "#Volumetric strain\n",
+ "#Let X=dell_V*V**-1 \n",
+ "#X=2*e1+e2\n",
+ "#After sub values and further simplifying\n",
+ "#X=19*p*E**-1\n",
+ "#After further simplifying we get\n",
+ "p=dell_V*(pi*4**-1*d**2*L)**-1*E*19**-1 #N/mm**2\n",
+ "\n",
+ "#Hoop Stress\n",
+ "f1=p*d*(2*t)**-1\n",
+ "\n",
+ "X=e1=8.5*p*E**-1\n",
+ "#Sub value of X in equation 1 we get\n",
+ "dell_d=8.5*p*E**-1*d\n",
+ "\n",
+ "Y=e2=2*p*E**-1\n",
+ "#Sub value of Y in equation 2 we get\n",
+ "dell_L=2*p*E**-1*L\n",
+ "\n",
+ "#Result\n",
+ "print\"Pressure Developed is\",round(p,2),\"N/mm**2\"\n",
+ "print\"Hoop stress Developed is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in diameter is\",round(dell_d,2),\"mm\"\n",
+ "print\"Change in Length is\",round(dell_L,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Pressure Developed is 4.19 N/mm**2\n",
+ "Hoop stress Developed is 41.88 N/mm**2\n",
+ "Change in diameter is 0.04 mm\n",
+ "Change in Length is 0.08 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.3,Page No.324"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of water supply pipes\n",
+ "h=50*10**3 #mm #Water head\n",
+ "sigma=20 #N/mm**2 #Permissible stress\n",
+ "rho=9810*10**-9 #N/mm**3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Pressure of water\n",
+ "P=rho*h #N/mm**2\n",
+ "\n",
+ "#Stress\n",
+ "#sigma=p*d*(2*t)**-1\n",
+ "#After further simplifying\n",
+ "t=P*d*(2*sigma)**-1 #mm \n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of seamless pipe is\",round(t,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of seamless pipe is 9.197 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.4,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=2500 #mm #Diameter of riveted boiler\n",
+ "P=1 #N/mm**2 #Pressure\n",
+ "rho1=0.7 #Percent efficiency\n",
+ "rho2=0.4 #Circumferential joints\n",
+ "sigma=150 #N/mm**2 #Permissible stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "#p*d*L=rho1*2*t*L*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t=P*d*(2*sigma*rho1)**-1 #mm\n",
+ "\n",
+ "#Considering Longitudinal force\n",
+ "#pi*d**2*4**-1*P=rho2*pi*d*t*sigma\n",
+ "#After rearranging and further simplifying we get\n",
+ "t2=P*d*(4*sigma*rho2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness of plate required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness of plate required is 11.9 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.5,Page No.326"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Boiler Dimensions\n",
+ "t=16 #mm #Thickness\n",
+ "p=2 #N/mm**2 #internal pressure\n",
+ "f=150 #N/mm**2 #Permissible stress\n",
+ "rho1=0.75 #Longitudinal joints\n",
+ "rho2=0.45 #circumferential joints\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Equating Bursting force to longitudinal joint strength ,we get\n",
+ "d1=rho1*2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Considering circumferential strength \n",
+ "d2=4*rho2*t*f*p**-1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Largest diameter of Boiler is\",round(d1,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Largest diameter of Boiler is 1800.0 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.6,Page No.329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=250 #mm #Diameter iron pipe\n",
+ "t=10 #mm #Thickness\n",
+ "d2=6 #mm #Diameter of steel\n",
+ "p=80 #N/mm**2 #stress\n",
+ "P=3 #N/mm**2 #Pressure\n",
+ "E_c=1*10**5 #N/mm**2\n",
+ "mu=0.3 #poissoin's ratio\n",
+ "E_s=2*10**5 #N/mm**2\n",
+ "n=1 #No.of wires\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "L=6 #mm #Length of cyclinder\n",
+ "\n",
+ "#Force Exerted by steel wire at diameterical section\n",
+ "F=p*2*pi*d2**2*1*4**-1 #N\n",
+ "\n",
+ "#Initial stress in cyclinder\n",
+ "f_c=F*(2*t*d2)**-1 #N/mm**2\n",
+ "\n",
+ "#LEt due to fluid pressure alone stresses developed in steel wire be F_w and in cyclinder f1 and f2\n",
+ "f2=P*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Considering the equilibrium of half the cyclinder, 6mm long we get\n",
+ "#F_w*2*pi*4**-1*d2**2*n+f1*2*t*d2=P*d*d2\n",
+ "#After further simplifying we get\n",
+ "#F_w+2.122*f1=79.58 . ......................................(1)\n",
+ "\n",
+ "#Equating strain in wire to circumferential strain in cyclinder \n",
+ "#F_w=(f1-mu*f2)*E_s*E_c**-1 #N/mm**2\n",
+ "#After further simplifying we get\n",
+ "#F_w=2*f1-11.25 ....................................(2)\n",
+ "\n",
+ "#Sub in equation in1 we get\n",
+ "f1=(79.58+11.25)*(4.122)**-1 #N/mm**2\n",
+ "F_w=2*f1-11.25 #N/mm**2\n",
+ "\n",
+ "#Final stresses\n",
+ "#1) In steel Wire\n",
+ "sigma=F_w+p #N/mm**2\n",
+ "\n",
+ "#2) In Cyclinder\n",
+ "sigma2=f1-f_c\n",
+ "\n",
+ "#Result\n",
+ "print\"Final Stresses developed in:cyclinder is\",round(sigma,2),\"N/mm**2\"\n",
+ "print\" :Steel is\",round(sigma2,2),\"N/mm**2\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Final Stresses developed in:cyclinder is 112.82 N/mm**2\n",
+ " :Steel is -15.66 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.7,Page No.332"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=750 #mm #Diameter of shell\n",
+ "t=8 #mm #THickness\n",
+ "p=2.5 #N/mm**2\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Change in Diameter\n",
+ "dell_d=d*p*d*(1-mu)*(4*t*E)**-1 #mm\n",
+ "\n",
+ "#Change in Volume\n",
+ "dell_V=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Answer for Change in diameter is incorrect in book\n",
+ "\n",
+ "#Result\n",
+ "print\"Stress introduced is\",round(f1,2),\"N/mm**2\"\n",
+ "print\"Change in Diameter is\",round(dell_d,2),\"N/mm**2\"\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stress introduced is 58.59 N/mm**2\n",
+ "Change in Diameter is 0.16 N/mm**2\n",
+ "Change in Volume is 145608.33 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.8,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d=600 #mm #Diameter of sherical shell\n",
+ "t=10 #mm #Thickness\n",
+ "f=80 #N/mm**2 #Permissible stress\n",
+ "rho=0.75 #Efficiency joint\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Max Pressure\n",
+ "p=f*4*t*rho*d**-1 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Pressure is\",round(p,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Pressure is 4.0 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 60
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.9,Page No.333"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=1000 #mm #Length of shell\n",
+ "d=200 #mm #Diameter\n",
+ "t=6 #mm #Thickness\n",
+ "p=1.5 #N/mm**2 #Internal Pressure\n",
+ "E=2*10**5 #N/mm**2\n",
+ "mu=0.25 #Poissoin's Ratio\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Change in Volume of sphere\n",
+ "dell_V_s=3*p*d*(1-mu)*(4*t*E)**-1*pi*6**-1*d**3\n",
+ "\n",
+ "#Hoop stress\n",
+ "f1=p*d*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Longitudinal stress\n",
+ "f2=p*d*(4*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Principal strain\n",
+ "e1=(f1-mu*f2)*E**-1\n",
+ "e2=(f2-mu*f1)*E**-1\n",
+ "\n",
+ "V_c=1000 #mm**3\n",
+ "\n",
+ "#Change in Volume of cyclinder\n",
+ "dell_V_c=(2*e1+e2)*pi*4**-1*d**2*L\n",
+ "\n",
+ "#Total Change in Diameter\n",
+ "dell_V=dell_V_s+dell_V_c #mm**3\n",
+ "\n",
+ "#Result\n",
+ "print\"Change in Volume is\",round(dell_V,2),\"mm**3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Change in Volume is 8443.03 mm**3\n"
+ ]
+ }
+ ],
+ "prompt_number": 66
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.10,Page No.337"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=400 #mm #Internal Diameter\n",
+ "t=100 #mm #Thickness\n",
+ "p=80 #N/mm**2 #Fluid pressure\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Internal Radius\n",
+ "r1=d1*2**-1 #mm\n",
+ "\n",
+ "#Outer Radius\n",
+ "r_o=r1+t #mm\n",
+ "\n",
+ "p1=80 #N/mm**2\n",
+ "p2=0\n",
+ "\n",
+ "#Now From Lame's Euation\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#at x=200 #mm \n",
+ "p_x=80 #N/mm**2\n",
+ "#80=b*(200**2)**-1-a ..........................(1)\n",
+ "\n",
+ "#at x=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b*(300**2)**-1-a ...........................(2)\n",
+ "\n",
+ "#Sub equation 2 from 1\n",
+ "#80=b*(200**2)**-1-b*(300**2)**-1\n",
+ "#After Further simplifying we get\n",
+ "b=(50000)**-1*(200**2*300**2*80)\n",
+ "\n",
+ "#From equation 2 we get\n",
+ "a=b*(300**2)**-1\n",
+ "\n",
+ "#Variation of radial pressure p_x;\n",
+ "#p_x=b*(x**2)**-1-a\n",
+ "#After sub values and further simplifying we get\n",
+ "\n",
+ "#Radial pressure Variation\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "p_x=b*(x**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "p_x2=b*(x2**2)**-1-a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x3=300 #mm\n",
+ "p_x3=b*(x3**2)**-1-a #N/mm**2\n",
+ "\n",
+ "\n",
+ "#Hoop stress Distribution\n",
+ "#Variation of F_x\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=250 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop stress is\",round(F_x,2),\"N/mm**2\"\n",
+ "print\"Min Hoop stress is\",round(F_x3,2),\"N/mm**2\"\n",
+ "print\"Plot of Hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[p_x,p_x2,p_x3]\n",
+ "Y2=[-F_x,-F_x2,-F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Y2,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop stress is 208.0 N/mm**2\n",
+ "Min Hoop stress is 128.0 N/mm**2\n",
+ "Plot of Hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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skdWFW7ZsQffu3XH8+HGMGzcOY8aMAaA9wnfKlCkIDg7GmDFjEBcXJ7WK4uLi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\nbm5uYtOmTUII3b27hBBCoVCIHTt2CCG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+ "text": [
+ "<matplotlib.figure.Figure at 0x50e8310>"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.11,Page No.338"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=14 #N/mm**2 #internal Fluid pressure\n",
+ "t=50 #mm #Thickness\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation\n",
+ "#p_x=b*(x**2)**-1-a #N/mm**2 ...................(1)\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2 ...................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r2=100 #mm\n",
+ "p_x=14 #N/mm**2\n",
+ "\n",
+ "#Sub value of p_x in equation 1 we get\n",
+ "#14=(100)**-1*b-a ............................(3)\n",
+ "\n",
+ "#At\n",
+ "x2=r_o=150 #mm\n",
+ "p_x2=0 #N/mm**2\n",
+ "\n",
+ "#Sub value in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a ......................(4)\n",
+ "\n",
+ "#From Equations 3 and 4 we get\n",
+ "#14=b*(100**2)**-1-b*(100**2)**-1\n",
+ "#After sub values and further simplifying we get\n",
+ "b=14*100**2*150**2*(150**2-100**2)**-1\n",
+ "\n",
+ "#From equation 4 we get\n",
+ "a=b*(150**2)**-1\n",
+ "\n",
+ "#Hoop Stress\n",
+ "#F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x=100 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x2=125 #mm\n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At\n",
+ "x3=150 #mm\n",
+ "F_x3=b*(x3**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#If thin Cyclindrical shell theory is used,hoop stress is uniform and is given by\n",
+ "F=p*d2*(2*t)**-1 #N/mm**2\n",
+ "\n",
+ "#Percentage error in estimating max hoop tension\n",
+ "E=(F_x-F)*F_x**-1*100 #%\n",
+ "\n",
+ "#Result\n",
+ "print\"Max Hoop Stress Developed in the cross-section is\",round(F,2),\"N/mm**2\"\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_x,F_x2,F_x3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Radial Stress Distribution & Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Max Hoop Stress Developed in the cross-section is 28.0 N/mm**2\n",
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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29vaQJAknTpx4bAP379/HiBEjMGzYMMycOdNwT41GA1dXV2RkZGDgwIEcMiIi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\nSUVORK5CYII=\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x58c2e30>"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.12,Page No.339"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d2=200 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "F_max=16 #N/mm**2 #Tensile stress\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r2=100 #mm #Internal Diameter\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p_o be the External Pressure applied.\n",
+ "#From LLame's theorem\n",
+ "#p_x=b*(x**2)**-1-a ..............(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now At\n",
+ "x=100 #mm\n",
+ "p_x=12 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#12=b*(100**2)**-1-a . ..................(3)\n",
+ "\n",
+ "#The Max Hoop stress occurs at least value of x where\n",
+ "x=r1=100 #mm\n",
+ "#16=b*(100**2)**-1+a .......................(4)\n",
+ "\n",
+ "#From Equations 1 and 2 we get\n",
+ "#28=b*(100**2)**-1+b*(100**2)**-1\n",
+ "#After furhter Simplifying we get\n",
+ "b=28*100**2*2**-1\n",
+ "\n",
+ "#sub in equation 1 we get\n",
+ "a=-(12-(b*(100**2)**-1))\n",
+ "\n",
+ "#Thus At\n",
+ "x2=150 #mm\n",
+ "p_o=b*(x2**2)**-1-a\n",
+ "\n",
+ "#Result\n",
+ "print\"Minimum External applied is\",round(p_o,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Minimum External applied is 4.22 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.13,Page No.340"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d1=160 #mm #Internal Diameter \n",
+ "r1=80 #mm #External Diameter\n",
+ "p1=40 #N/mm**2 #Internal Diameter\n",
+ "P_max=120 #N/mm**2 #Allowable stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=80 #N/mm**2 \n",
+ "#Sub in equation 1 we get\n",
+ "#120=b*(80**2)**-1+a ........................(3)\n",
+ "\n",
+ "#The hoop tension at inner edge is max stress\n",
+ "#Hence\n",
+ "#120=b*(80**2)**-1+a .............................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b=160*80**2*2**-1 \n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=-(40-(b*(80**2)**-1))\n",
+ "\n",
+ "#Let External radius be r_o.Since at External Surface is Zero,we get\n",
+ "#0=b*(r_o)**-1-a\n",
+ "#After Further simplifying we get\n",
+ "r_o=(b*a**-1)**0.5\n",
+ "\n",
+ "#Thickness of Cyclinder \n",
+ "t=r_o-r1 #mm\n",
+ "\n",
+ "#Result\n",
+ "print\"Thickness Required is\",round(t,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thickness Required is 33.14 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.14,Page No.341"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=300 #mm #Outside diameter \n",
+ "d1=180 #mm #Internal Diameter\n",
+ "p=12 #N/mm**2 #internal Fluid pressure\n",
+ "p_o=6 #N/mm**2 #External Pressure\n",
+ "r_o=150 #mm #Outside Diameter\n",
+ "r=90 #mm #Internal Diameter\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #N/mm**2 \n",
+ "p=42 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#42=b*(90**2)**-1-a ..............................(3)\n",
+ "\n",
+ "#At \n",
+ "x=r_o=150 #mm\n",
+ "p2=6 #N/mm**2\n",
+ "#sub in equation 1 we get\n",
+ "#6=b*(150**2)**-1-a ..............................(4)\n",
+ "\n",
+ "#From equations 3 and 4 weget\n",
+ "#36=b*(90**2)**-1-b2(150**2)**-1\n",
+ "#After further simplifying we get\n",
+ "b=36*90**2*150**2*(150**2-90**2)**-1\n",
+ "\n",
+ "#Sub value of b in equation 4 we get\n",
+ "a=b*(150**2)**-1-p_o\n",
+ "\n",
+ "#At \n",
+ "x=r1=90 #mm\n",
+ "F_x=b*(x**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#At \n",
+ "x2=r_o=150 #mm \n",
+ "F_x2=b*(x2**2)**-1+a #N/mm**2\n",
+ "\n",
+ "#Now if External pressure is doubled i.e p_o2=12 #N/mm**2 We have\n",
+ "p_o2=12 #N/mm**2\n",
+ "#sub in equation 4 we get\n",
+ "#12=b2*(150**2)**-1-a2 ..........................(5)\n",
+ "\n",
+ "#Max Hoop stress is to be 70.5 #N/mm**2,which occurs at x=r1=90 #mm\n",
+ "#Sub in equation 4 we get\n",
+ "#70.5=b*(90**2)**-1+a2 ................................(6)\n",
+ "\n",
+ "#Adding equation 5 and 6\n",
+ "#82.5=b2*(150**2)**-1+b*(90**2)**-1\n",
+ "#After furhter simplifying we get\n",
+ "b2=82.5*150**2*90**2*(150**2+90**2)**-1\n",
+ "\n",
+ "#Sub in equation 5 we get\n",
+ "a2=b2*(150**2)**-1-12 \n",
+ "\n",
+ "#If p_i is the internal pressure required then from Lame's theorem\n",
+ "p_i=b2*(r1**2)**-1-a2\n",
+ "\n",
+ "#Result\n",
+ "print\"Stresses int the material are:F_x\",round(F_x,2),\"N/mm**2\"\n",
+ "print\" :F_x2\",round(F_x2,2),\"N/mm**2\"\n",
+ "print\"Internal Pressure that can be maintained is\",round(p_i,2),\"N/mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Stresses int the material are:F_x 70.5 N/mm**2\n",
+ " :F_x2 34.5 N/mm**2\n",
+ "Internal Pressure that can be maintained is 50.82 N/mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.15,Page No.344"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "r1=200 #mm #Inner Radius\n",
+ "r2=250 #mm #Radius at common surface\n",
+ "r3=300 #mm #Outer radius\n",
+ "p=6 #N/mm**2 #Inital pressure\n",
+ "p2=80 #N/mm**2 #Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Inner Cyclinder:\n",
+ "\n",
+ "#From Lame's Equation we have\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#At \n",
+ "x=r1=200 #mm\n",
+ "p_x=0\n",
+ "#0=b1*(250**2)**-1-a1 .................(3)\n",
+ "\n",
+ "#At x=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b1*(250**2)-a1 ...................(4)\n",
+ "\n",
+ "#From Equation 3 and 4 we get\n",
+ "b1=6*200**2*250**2*(200**2-250**2)**-1\n",
+ "\n",
+ "#From equation 3 we get\n",
+ "a1=b1*(200**2)**-1\n",
+ "\n",
+ "F_200=b1*(200**2)**-1+a1\n",
+ "F_250=b1*(250**2)**-1+a1\n",
+ "\n",
+ "#For outer cyclinder \n",
+ "#From Lame's Equation we have\n",
+ "#p_x2=b2*(x**2)**-1-a2 ..........................(5)\n",
+ "#F_x2=b2*(x**2)**-1+a2 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At \n",
+ "x2=r2=250 #mm\n",
+ "p_x2=6 #N/mm**2\n",
+ "#6=b2*(250**2)**-1-a2 ...........................(7) \n",
+ "\n",
+ "#At\n",
+ "x3=300 #mm\n",
+ "#p_x2=0\n",
+ "#0=b2**2*(300**2)**-1-a2 .................................(8)\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "b2=6*250**2*300**2*(300**2-250**2)**-1\n",
+ "\n",
+ "#sub in equation 8 we get\n",
+ "a2=b2*(300**2)**-1\n",
+ "\n",
+ "F_250_2=b2*(250**2)**-1+a2\n",
+ "F_300_2=b2*(300**2)**-1+a2\n",
+ "\n",
+ "#When Fluid is admitted\n",
+ "#Let Lame's equation be\n",
+ "#p_x3=b3*(x**2)**-1-a3 ..........................(5)\n",
+ "#F_x3=b3*(x**2)**-1+a3 ...........................(6)\n",
+ "\n",
+ "\n",
+ "#At x=200\n",
+ "p_x3=80 #N/mm**2\n",
+ "#80=b3*(200**2)**-1-a3 ................................(7)\n",
+ "\n",
+ "#At x=300 #mm\n",
+ "#p_x=0\n",
+ "#0=b3*(300**2)**-1-a3 ..............................(8)\n",
+ "\n",
+ "#from Equation 7 and 8 we get\n",
+ "b3=80*200**2*300**2*(300**2-200**2)**-1\n",
+ "\n",
+ "#From Equation 8 we get\n",
+ "a3=b3*(300**2)**-1\n",
+ "\n",
+ "#Hoop stresses \n",
+ "F_200_3=b3*(200**2)**-1+a3 #N/mm**2\n",
+ "F_250_3=b3*(250**2)**-1+a3 #N/mm**2\n",
+ "F_300_3=b3*(300**2)**-1+a3 #N/mm**2\n",
+ "\n",
+ "#Pressure at common surface\n",
+ "p_250=b3*(250**2)**-1-a3 #N/mm**2\n",
+ "\n",
+ "#final stress\n",
+ "f_200=F_200+F_200_3 #N/mm**2\n",
+ "f_250=F_250+F_250_3 #N/mm**2\n",
+ "f_300=F_250_2+F_250_3 #N/mm**2\n",
+ "f_300_2=F_300_2+F_300_3 #N/mm**2\n",
+ "\n",
+ "#Result\n",
+ "print\"final Hoop stress are:f_200\",round(f_200,2),\"N/mm**2\"\n",
+ "print\" :f_250\",round(f_250,2),\"N/mm**2\"\n",
+ "print\" :f_300\",round(f_300,2),\"N/mm**2\"\n",
+ "print\" :f_300_2\",round(f_300_2,2),\"N/mm**2\"\n",
+ "print\"Variation of Hoop stress and Radial stress\"\n",
+ "\n",
+ "#Final stresses\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3,x3]\n",
+ "Y1=[f_200,f_250,f_300,f_300_2]\n",
+ "Z1=[0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n",
+ "#Due to Fluid\n",
+ "#Variation of hoop stress \n",
+ " \n",
+ "X1=[x,x2,x3]\n",
+ "Y1=[F_200_3,F_250_3,F_300_3]\n",
+ "Z1=[0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "final Hoop stress are:f_200 174.67 N/mm**2\n",
+ " :f_250 128.83 N/mm**2\n",
+ " :f_300 189.43 N/mm**2\n",
+ " :f_300_2 155.27 N/mm**2\n",
+ "Variation of Hoop stress and Radial stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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MHz4cr7zyCmbPno1///vf1fIBnZeXp/l+8+bNmhVR0dHR+P7771FYWIjMzExcvHgRnTp1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\nVw+AnLO4d++eQY0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+ "text": [
+ "<matplotlib.figure.Figure at 0x5604510>"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x56e5cb0>"
+ ]
+ }
+ ],
+ "prompt_number": 89
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.16,Page No.348"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "do=200 #mm #Inner Diameter\n",
+ "r_o=100 #mm #Inner radius\n",
+ "d1=300 #mm #outer diameter\n",
+ "r1=150 #mm #Outer radius\n",
+ "d2=250 #mm #Junction Diameter\n",
+ "r2=125 #mm #Junction radius\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "p=30 #N/mm**2 #radial pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#from Lame's Equation we get\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Then from Boundary condition \n",
+ "#p_x=0 at x=100 #mm\n",
+ "#0=b1*(100**2)**-1-a1 .....................(3)\n",
+ "\n",
+ "#p_x2=30 #N/mm**2 at x2=125 #mm\n",
+ "#30=b1*(125**2)**-1-a1 ................................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "b1=30*125**2*100**2*(100**2-125**2)**-1\n",
+ "\n",
+ "#From Equation 3 we get\n",
+ "a1=b1*(100**2)**-1\n",
+ "\n",
+ "#therefore Hoop stress in inner cyclinder at junction\n",
+ "F_2_1=b1*(125**2)**-1+a1 #N/mm**2\n",
+ "\n",
+ "#Outer Cyclinder\n",
+ "#p_x=b*(x**2)**-1-a ..........................(5)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(6)\n",
+ "\n",
+ "#Now at x=125 #mm\n",
+ "#p_x3=30 #N/mm**2\n",
+ "#30=b2*(125**2)**-1-a2 ..................................(7)\n",
+ "\n",
+ "#At x=150 #mm\n",
+ "#p_x4=0\n",
+ "#0=b2*(150**2)**-1-a2 ...................................(8)\n",
+ "\n",
+ "#From equations 7 and 8\n",
+ "b2=30*150**2*125**2*(150**2-125**2)**-1\n",
+ "\n",
+ "#From eqauation 8 we get\n",
+ "a2=b2*(150**2)**-1\n",
+ "\n",
+ "#Hoop stress at junction \n",
+ "F_2_0=b2*(125**2)**-1+a2 #N/mm**2\n",
+ "\n",
+ "rho_r=(F_2_0-F_2_1)*E**-1*r2\n",
+ "\n",
+ "#Result\n",
+ "print\"Shrinkage Allowance is\",round(rho_r,3),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Shrinkage Allowance is 0.189 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 40
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.17,Page No.350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=500 #mm #Outer Diameter\n",
+ "r_o=250 #mm #Outer Radius\n",
+ "d1=300 #mm #Inner Diameter\n",
+ "r1=150 #mm #Inner Radius\n",
+ "d2=400 #mm #Junction Diameter\n",
+ "E=2*10**5 #N/mm**2 #Modulus ofElasticity\n",
+ "alpha=12*10**-6 #Per degree celsius\n",
+ "dell_d=0.2 #mm\n",
+ "dell_r=0.1 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let p be the radial pressure developed at junction\n",
+ "#Let Lame's Equation for internal cyclinder be\n",
+ "#p_x=b*(x**2)**-1-a ................................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...............................(2)\n",
+ "\n",
+ "#At \n",
+ "x=150 #mm \n",
+ "p_x=0\n",
+ "#Sub in equation 1 we get\n",
+ "#0=b*(150**2)**-1-a .........................(3)\n",
+ "\n",
+ "#At \n",
+ "x2=200 #mm\n",
+ "#p_x2=p\n",
+ "#p=b*(200**2)**-1-a ......................(4)\n",
+ " \n",
+ "#From Equation 3 and 4\n",
+ "#p=b*(200**2)**-1-b(150**2)**-1\n",
+ "#after further simplifying we get\n",
+ "#b=-51428.571*p\n",
+ "\n",
+ "#sub in equation 3 we get\n",
+ "#a1=-2.2857*p\n",
+ "\n",
+ "#therefore hoop stress at junction is\n",
+ "#F_2_1=-21428.571*p*(200**2)**-1-2.2857*p\n",
+ "#after Further simplifying we geet\n",
+ "#F_2_1=3.5714*p\n",
+ "\n",
+ "#Let Lame's Equation for cyclinder be \n",
+ "#p_x=b*(x**2)**-1-a .........................5\n",
+ "#F_x=b*(x**2)**-1+a .............................6\n",
+ "\n",
+ "#At \n",
+ "x=200 #mm\n",
+ "#p_x=p2\n",
+ "#p2=b2*(20**2)**-1-a2 ...................7\n",
+ "\n",
+ "#At\n",
+ "x2=200 #mm\n",
+ "p_x2=0\n",
+ "#0=b2*(250**2)**-1-a2 ....................8\n",
+ "\n",
+ "#from equation 7 and 8 we get\n",
+ "#p2=b2*(200**2)**-1-b2*(250**2)**-1\n",
+ "#After further simplifying we get\n",
+ "#p2=b2*(250**2-200**2)*(200**2*250**2)**-1\n",
+ "#b2=111111.11*p\n",
+ "\n",
+ "#from equation 7\n",
+ "#a2=b2*(250**2)**-1\n",
+ "#further simplifying we get\n",
+ "#a2=1.778*p\n",
+ "\n",
+ "#At the junctionhoop stress in outer cyclinder \n",
+ "#F_2_0=b2*(200**2)**-1+a2\n",
+ "#After further simplifying we get\n",
+ "#F_2_0=4.5556*p\n",
+ "\n",
+ "#Considering circumferential strain,the compatibility condition\n",
+ "#rho_r*r2**-1=1*E**-1*(F_2_1+F_2_0)\n",
+ "#where F_2_1 is compressive and F_2_0 is tensile\n",
+ "#furter simplifying we get\n",
+ "p=0.1*200**-1*2*10**5*(3.5714+4.5556)**-1\n",
+ "\n",
+ "#Let T be the rise in temperature required\n",
+ "#dell_d=d*alpha*T\n",
+ "#After sub values and further simplifying we get\n",
+ "d=250 #mm\n",
+ "T=dell_d*(d*alpha)**-1 #Per degree celsius\n",
+ "\n",
+ "#Result\n",
+ "print\"Radial Pressure Developed at junction\",round(p,2),\"N/mm**2\"\n",
+ "print\"Min Temperatureto outer cyclinder\",round(T,2),\"Per degree Celsius\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radial Pressure Developed at junction 12.3 N/mm**2\n",
+ "Min Temperatureto outer cyclinder 66.67 Per degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Example 8.8.18,Page No.355"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "%matplotlib inline\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "d_o=400 #mm #Outer Diameter\n",
+ "r_o=200 #mm #Outer radius\n",
+ "t=50 #mm #Thickness\n",
+ "r1=150 #mm #Internal Radius\n",
+ "p=50 #N/mm**2 #Internal Pressure\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#The Radial Pressure and hoop stress at any radial distance x are given by\n",
+ "#p_x=b*(x**2)**-1-a ..........................(1)\n",
+ "#F_x=b*(x**2)**-1+a ...........................(2)\n",
+ "\n",
+ "#Now at\n",
+ "x=150 #N/mm**2\n",
+ "p_x1=50 #N/mm**2\n",
+ "#Sub in equation 1 we get\n",
+ "#50=2*b*(150**3)**-1-a ...........................(3)\n",
+ "\n",
+ "#At x=200 #mm\n",
+ "p_x2=0\n",
+ "#0=2*b*(200**2)**-1-a ....................(4)\n",
+ "\n",
+ "#From equation 3 and 4 we get\n",
+ "#50=2*b*(150**3)**-1-2*b*(200**3)**-1\n",
+ "#After further simplifying we get\n",
+ "b=50*150**3*200**3*(200**3-150**3)**-1*2**-1\n",
+ "\n",
+ "#Sub in equation 3 we get\n",
+ "a=b*(200**3)**-1\n",
+ "\n",
+ "#Now At\n",
+ "x=150 #mm\n",
+ "F_x=b*(x**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x2=160 #mm\n",
+ "F_x2=b*(x2**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x3=170 #mm\n",
+ "F_x3=b*(x3**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x4=180 #mm\n",
+ "F_x4=b*(x4**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x5=190 #mm\n",
+ "F_x5=b*(x5**3)**-1+a\n",
+ "\n",
+ "#Now At\n",
+ "x6=200 #mm\n",
+ "F_x6=b*(x6**3)**-1+a\n",
+ "\n",
+ "#Result\n",
+ "print\"Plot of Variation of hoop stress\"\n",
+ "\n",
+ "#Plotting Variation of hoop stress\n",
+ "\n",
+ "X1=[x,x2,x3,x4,x5,x6]\n",
+ "Y1=[F_x,F_x2,F_x3,F_x4,F_x5,F_x6]\n",
+ "Z1=[0,0,0,0,0,0]\n",
+ "plt.plot(X1,Y1,X1,Z1)\n",
+ "plt.xlabel(\"Length x in mm\")\n",
+ "plt.ylabel(\"Hoop Stress Distribution in N/mm**2\")\n",
+ "plt.show()\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Plot of Variation of hoop stress\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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w3g2UlJRgzJgxCAsLw3vvvQcA8PT0hFqthpOTE7KyshAYGIhLly7VLIxzCERE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\nnp6uOVdYWBisrKzg4+OD8vJ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+ "text": [
+ "<matplotlib.figure.Figure at 0x56b4bf0>"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_5.ipynb b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_5.ipynb new file mode 100644 index 00000000..cecacb12 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/chapter_no.9_5.ipynb @@ -0,0 +1,779 @@ +{
+ "metadata": {
+ "name": "chapter no.9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 9:Columns And Struts"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.1,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "L=5000 #mm #Length of strut\n",
+ "dell=10 #mm #Deflection\n",
+ "W=10 #N #Load\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Central Deflection of a simply supported beam with central concentrated load is\n",
+ "#dell=W*L**3*(48*E*I)**-1 \n",
+ "\n",
+ "#Let E*I=X\n",
+ "X=W*L**3*(48*dell)**-1 #mm\n",
+ "\n",
+ "#Euler's Load\n",
+ "#Let Euler's Load be P\n",
+ "P=pi**2*X*(L**2)**-1\n",
+ "\n",
+ "#Result\n",
+ "print\"Critical Load of Bar is\",round(P,2),\"N\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Critical Load of Bar is 1028.08 N\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.2,Page No.377"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=2000 #mm #Length of square column\n",
+ "E=12*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "sigma=12 #N/mm*2 #stress\n",
+ "W1=95*10**3 #N #Load1\n",
+ "W2=200*10**3 #N #Load2\n",
+ "FOS=3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#From Euler's Formula\n",
+ "#P=pi**2*E*I*(L**2)**-1 .........(1)\n",
+ "\n",
+ "#Working Load\n",
+ "#W=P*(FOS)**-1\n",
+ "\n",
+ "#Part-1\n",
+ "\n",
+ "#At W1=95*10**3 #N\n",
+ "#W1=P*(3*L**2)**-1\n",
+ "\n",
+ "#Let 'a' be the side of the square\n",
+ "#I=1*12**-1*a**4\n",
+ "\n",
+ "#sub value of I in Equation 1 and further rearranging we get\n",
+ "a=(W1*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From Consideration of direct crushing\n",
+ "#sigma*a**2=W1\n",
+ "#After Reaaranging the above equation we get\n",
+ "a2=(W1*(sigma)**-1)**0.5 #mm\n",
+ "\n",
+ "#required size is 103.67*103.67 i.e a*a\n",
+ "\n",
+ "#Part-2\n",
+ "\n",
+ "#At W2=200*10**3 #N\n",
+ "#W2=P*(3*L**2)**-1\n",
+ "#After substituting values and further Rearranging the above equation we get\n",
+ "a3=(W2*3*12*L**2*(pi**2*E)**-1)**0.25 #mm\n",
+ "\n",
+ "#From consideration of direct compression,size required is\n",
+ "a4=(W2*sigma**-1)**0.5\n",
+ "\n",
+ "#required size is 129.10*129.10 i.e a4*a4\n",
+ "\n",
+ "#Result\n",
+ "print\"For W1 Load Required size is\",round(a*a,2),\"mm**2\"\n",
+ "print\"For W2 Load Required size is\",round(a4*a4,2),\"mm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "For W1 Load Required size is 10747.38 mm**2\n",
+ "For W1 Load Required size is 16666.67 mm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.3,Page No.378"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Flange \n",
+ "b=100 #mm #Width\n",
+ "\n",
+ "D=80 #mm #Overall Depth\n",
+ "t=10 #mm #Thickness of web and flanges\n",
+ "L=3000 #mm #Length of strut\n",
+ "E=200*10**3 #N/mm**2 #Modulus of Elasticity\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Let centroid be at depth y_bar from top fibre\n",
+ "y_bar=(b*t*t*2**-1+(D-t)*t*((D-t)*2**-1+t))*(b*t+(D-t)*t)**-1 #mm \n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*b*t**3+b*t*(y_bar-t*2**-1)**2+1*12**-1*t*((D-t))**3+t*((D-t))*((((D-t)*2**-1)+t)-y_bar)**2\n",
+ "\n",
+ "#M.I at y-y axis\n",
+ "I_y=1*12**-1*t*b**3+1*12**-1*(D-t)*t**3 #mm**3\n",
+ "\n",
+ "#Least M.I\n",
+ "I=I_y\n",
+ "\n",
+ "#Since both ends are hinged\n",
+ "#Feective Length=Actual Length\n",
+ "L=l=3000 #mm\n",
+ "\n",
+ "#Buckling Load \n",
+ "P=pi**2*E*I*(l**2)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"The Buckling Load for strut of tee section\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Buckling Load for strut of tee section 184.05 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.4,Page No.379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=400 #mm #Overall Depth\n",
+ "\n",
+ "#Flanges\n",
+ "b=300 #mm #Width\n",
+ "t=50 #mm #Thickness\n",
+ "\n",
+ "t2=30 #mm #Web Thickness\n",
+ "\n",
+ "dell=10 #mm #Deflection\n",
+ "w=40 #N/mm #Load\n",
+ "FOS=1.75 #Factor of safety\n",
+ "E=2*10**5 #N/mm**2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#M.I at x-x axis\n",
+ "I_x=1*12**-1*(b*D**3-(b-t2)*b**3) #mm**4\n",
+ "\n",
+ "#Central Deflection\n",
+ "#dell=5*w*L**4*(384*E*I)**-1\n",
+ "#After sub values in above equation and further simplifying we get\n",
+ "L=(dell*384*E*I_x*(5*w)**-1)**0.25\n",
+ "\n",
+ "#M.I aty-y axis\n",
+ "I=I_y=1*12**-1*t*b**3+1*12**-1*b*t2**3+1*12**-1*t*b**3 #mm**4\n",
+ "\n",
+ "#Both the Ends of column are hinged\n",
+ "\n",
+ "#Crippling Load\n",
+ "P=pi**2*E*I*(L**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load if I-section is used as column with both Ends hhinged\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load if I-section is used as column with both Ends hhinged 4123.29 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.5,Page No.381"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #External Diameter\n",
+ "t=20 #mm #hickness\n",
+ "d=200-2*t #mm #Internal Diameter\n",
+ "E=1*10**5 #N/mm**2\n",
+ "a=1*(1600)**-1 #Rankine's Constant\n",
+ "L=4.5 #m #Length\n",
+ "sigma=550 #N/mm**2 #Stress\n",
+ "FOS=2.5\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Moment of Inertia\n",
+ "I=pi*D**4*64**-1-pi*d**4*64**-1\n",
+ "\n",
+ "#Both Ends are fixed\n",
+ "\n",
+ "#Effective Length\n",
+ "l=1*2**-1*L*10**3 #mm\n",
+ "\n",
+ "#Euler's Critical Load\n",
+ "P_E=pi**2*E*I*(l**2)**-1\n",
+ "\n",
+ "A=pi*4**-1*(D**2-d**2) #mm*2\n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Rankine's Critical Load\n",
+ "P_R=sigma*A*(1+a*(l*k**-1)**2)**-1\n",
+ "\n",
+ "X=P_E*P_R**-1 \n",
+ "\n",
+ "#Safe Load using Rankine's Formula\n",
+ "S=P_R*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load by Rankine's Formula is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load by Rankine's Formula is 1404.36 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.6,Page No.382"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "L=3000 #mm #Length of column\n",
+ "W=800*10**3 #N #Load\n",
+ "a=1*1600**-1 #Rankine's constant\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=550 #N/mm**2 #stress\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Effective Length\n",
+ "l=L*2**-1 #mm \n",
+ "\n",
+ "#Let d1=outer diameter & d2=inner diameter\n",
+ "#d1=5*8**-1*d2\n",
+ "\n",
+ "#M.I\n",
+ "#I=pi*64**-1*(d1**4-d2**4) #mm**4\n",
+ "\n",
+ "#Area of section\n",
+ "#A=pi4**-1*(d1**2-d2**2) #mm**2\n",
+ "\n",
+ "#k=(I*A**-1) \n",
+ "#substituting values in above equation \n",
+ "#k=1*16**-1*(d1**2-d2**2)\n",
+ "#after simplifying further we get\n",
+ "#k=0.2948119.d1\n",
+ "\n",
+ "#X=l*k**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#X=5087.9898*d1**-1\n",
+ "\n",
+ "#Crtitcal Load\n",
+ "P=W*FOS #N\n",
+ "\n",
+ "#From Rankine's Load\n",
+ "#P2=sigma*A*(1+a*(X)**2)**-1\n",
+ "#substituting values in above equation and after simplifying further we get\n",
+ "#d1**4-12156618*d1**4-1.96691*10**8=0\n",
+ "#Solving Quadratic Equation we get\n",
+ "#d1**2-12156618*d1-196691000=0\n",
+ "a=1\n",
+ "b=-12156.618\n",
+ "c=-196691000\n",
+ "\n",
+ "Y=b**2-4*a*c\n",
+ "\n",
+ "d1_1=((-b+Y**0.5)*(2*a)**-1)**0.5 #mm\n",
+ "d1_2=((-b-Y**0.5)*(2*a)**-1) #mm\n",
+ "\n",
+ "d2=5*8**-1*d1_1\n",
+ "\n",
+ "#Result\n",
+ "print\"Section of cast iron hollow cylindrical column is:d1_1\",round(d1_1,2),\"mm\"\n",
+ "print\" :d2 \",round(d2,2),\"mm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Section of cast iron hollow cylindrical column is:d1_1 146.16 mm\n",
+ " :d2 91.35 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.7,Page No.383"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "#Let X=(P*A**-1) #Average Stress at Failure \n",
+ "Lamda_1=70 #Slenderness Ratio\n",
+ "Lamda_2=170 #Slenderness Ratio\n",
+ "X1=200 #N/mm**2 \n",
+ "X2=69 #N/mm**2 \n",
+ "\n",
+ "#Rectangular section\n",
+ "b=60 #mm #width\n",
+ "t=20 #mm #Thickness\n",
+ "\n",
+ "L=1250 #mm #Length of strut\n",
+ "FOS=4 #Factor of safety\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "#Lamda=L*k**-1\n",
+ "\n",
+ "#The Rankine's Formula for strut\n",
+ "#P=sigma*A*(1+a*(L*k**-1)**-1\n",
+ "\n",
+ "#From test result 1,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_1=200+980000*a ...................(1)\n",
+ "\n",
+ "#From test result 2,\n",
+ "#After sub values in above equation we get and further simplifying we get\n",
+ "#sigma_2=69+1994100*a ...................(2)\n",
+ "\n",
+ "#Substituting it in equation (1) we get\n",
+ "a=131*1014100**-1 \n",
+ "\n",
+ "#Substituting a in equation 1\n",
+ "sigma_1=200+980000*a #N/mm**2\n",
+ "\n",
+ "#Effective Length \n",
+ "l=1*2**-1*L #mm\n",
+ "\n",
+ "#Least of M.I\n",
+ "I=1*12**-1*b*t**3 #mm**4\n",
+ "\n",
+ "#Area \n",
+ "A=b*t #mm**2 \n",
+ "\n",
+ "k=(I*A**-1)**0.5\n",
+ "\n",
+ "#Slenderness ratio\n",
+ "Lamda=l*k**-1\n",
+ "\n",
+ "#From Rankine's Ratio\n",
+ "P=sigma_1*A*(1+a*(Lamda)**2)**-1\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #N\n",
+ "\n",
+ "#Result\n",
+ "print\"Constant in the Formula is:a \",round(a,6)\n",
+ "print\" :sigma_1\",round(sigma_1,2)\n",
+ "print\"Safe Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Constant in the Formula is:a 0.000129\n",
+ " :sigma_1 326.6\n",
+ "Safe Load is 38.98 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 38
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.8,Page No.385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "D=200 #mm #Depth\n",
+ "b=140 #mm #width\n",
+ "\n",
+ "#Plate\n",
+ "b2=160 #mm #Width\n",
+ "t2=10 #mm #Thickness\n",
+ "\n",
+ "L=l=4000 #mm #Length\n",
+ "FOS=4 #Factor of safety\n",
+ "sigma=315 #N/mm**2 #stress\n",
+ "a2=1*7500**-1 \n",
+ "I_xx=26.245*10**6 #mm**4 #M.I at x-x\n",
+ "I_yy=3.288*10**6 #mm**4 #M.I at y-y\n",
+ "a=3671 #mm**2 #Area\n",
+ "k_x=84.6#mm\n",
+ "k_y=29.9 #mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total Area\n",
+ "A=a+2*t2*b2 #mm**2\n",
+ "\n",
+ "#M.I\n",
+ "I=I_yy+2*12**-1*t2*b2**3 #mm**4\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "\n",
+ "#Let X=L*k**-1\n",
+ "X=L*k**-1\n",
+ "\n",
+ "#Appliying Rankine's Formula\n",
+ "P=sigma*A*(1+a2*(X)**2)**-1 #N\n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*(FOS)**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe axial Load is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe axial Load is 220.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.9,Page No.389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "E=200*10**3 #N/mm**2 #Modulus of elasticity\n",
+ "sigma=330 #N/mm**2 #Stress\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "A=5205 #mm**2 #area of column\n",
+ "I_xx=59.431*10**6 #mm**4 #M.I at x-x axis\n",
+ "I_yy=8.575*10**6 #mm**24#M.I at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Total M.I\n",
+ "I=I_xx+I_yy #mm**4\n",
+ "\n",
+ "#Area of compound Section \n",
+ "A2=2*A #mm**2\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Equating Euler's Load to Rankine's Load we get\n",
+ "#pi**2*E*I*(L**2)**-1=sigma*A*(1+a*(L*k)**2)**-1\n",
+ "#After Substitt=uting values and further simplifying we get\n",
+ "L=(39076198*(1-0.7975432)**-1)**0.5*10**-3 #m\n",
+ "\n",
+ "#Result\n",
+ "print\"Length of column for which Rankine's formula and Euler's Formula give the same result is\",round(L,2),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Length of column for which Rankine's formula and Euler's Formula give the same result is 13.89 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.10,Page No.387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "sigma=326 #N/mm**2 #stress\n",
+ "E=2*10**5 #N/mm**2 #Modulus of Elasticity\n",
+ "FOS=2 #Factor of safety\n",
+ "a=1*7500**-1 #Rankine's constant\n",
+ "D=350 #mm #Overall Depth \n",
+ "\n",
+ "#Cover plates\n",
+ "b1=500 #mm #width\n",
+ "t1=10 #mm #Thickness\n",
+ "\n",
+ "d=220 #mm #Distance between two channels\n",
+ "\n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "A=5366 #mm**2 #Area of Column section \n",
+ "I_xx=100.08*10**6 #mm**4 #M.I of x-x axis\n",
+ "I_yy=4.306*10**6 #mm**4 #M.I of y-y axis\n",
+ "C_yy=23.6 #mm #Centroid at y-y axis\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Symmetric axes are the centroidal axes is\n",
+ "\n",
+ "#M.I of Channel at x-x axis\n",
+ "I_xx_1=2*I_xx+2*(1*12**-1*b1*t1**3+b1*t1*(D*2**-1+t1*2**-1)**2)\n",
+ "\n",
+ "#M.I of Channel at y-y axis\n",
+ "I_yy_1=2*(I_yy+A*(d*2**-1+C_yy)**2)+2*12**-1*t1*b1**3\n",
+ "\n",
+ "#As I_yy<I_xx\n",
+ "#So\n",
+ "I=I_yy_1 #mm**4 \n",
+ "\n",
+ "A2=2*A+2*t1*b1 #Area of channel\n",
+ "\n",
+ "k=(I*A2**-1)**0.5 #mm\n",
+ "\n",
+ "#Critical Load\n",
+ "P=sigma*A2*(1+a*(L*k**-1)**2)**-1 \n",
+ "\n",
+ "#Safe Load\n",
+ "S=P*2**-1*10**-3 #KN\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying Capacity is\",round(S,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying Capacity is 2717.35 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9.9.11,Page No.390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "import numpy as np\n",
+ "\n",
+ "#Initilization of Variables\n",
+ "\n",
+ "I=4.085*10**8 #mm**4 #M.I\n",
+ "A=20732.0 #mm**2 #area of column\n",
+ "f_y=250 #N/mm**2 \n",
+ "L=6000 #mm #Length of column\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=(I*A**-1)**0.5 #mm\n",
+ "lamda=L*k**-1 #Slenderness ratro\n",
+ "\n",
+ "#From Indian standard table\n",
+ "lamda_1=40 \n",
+ "sigma_a_c_1=139 #N/mm**2\n",
+ "lamda_2=50 \n",
+ "sigma_a_c_2=132 #N/mm**2 \n",
+ "\n",
+ "#Linearly interpolating between these values for lambda=42.744\n",
+ "\n",
+ "sigma_a_c_3=sigma_a_c_1-2.744*10**-1*(sigma_a_c_1-sigma_a_c_2)\n",
+ "\n",
+ "#Safe Load carrying capacity of column\n",
+ "P=sigma_a_c_3*A*10**-3\n",
+ "\n",
+ "#Result\n",
+ "print\"Safe Load carrying capacity is\",round(P,2),\"KN\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Safe Load carrying capacity is 2841.93 KN\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_1.png b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_1.png Binary files differnew file mode 100644 index 00000000..11bcf467 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/B.M.D_1_1.png diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1.png b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1.png Binary files differnew file mode 100644 index 00000000..ba2ef24e --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_1.png diff --git a/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_3_1.png b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_3_1.png Binary files differnew file mode 100644 index 00000000..202b2e93 --- /dev/null +++ b/Strength_Of_Materials_by_S_S_Bhavikatti/screenshots/S.F.D_3_1.png diff --git a/sample_notebooks/VinayBadhan/chapter7.ipynb b/sample_notebooks/VinayBadhan/chapter7.ipynb new file mode 100644 index 00000000..3d02cc6c --- /dev/null +++ b/sample_notebooks/VinayBadhan/chapter7.ipynb @@ -0,0 +1,2175 @@ +{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "#Chapter 7 , Transistor Amplifiers , Biasing and Thermal Stabilization"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.1 , Page Number 230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ " Maximum collector current that can be allowd during any part of the input signal is 3.0 mA.\n",
+ "Minimum zero signal collector current required : 1.5 mA.\n",
+ "Maximum base current 0.03 mA.\n",
+ "Signal voltage (VBE) : 0.75 V.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "RC = 3.0 * 10**3 #Collector load resistance (in ohm)\n",
+ "Vknee = 1 #Knee voltage for silicon transistor (in volts)\n",
+ "beta = 100 #Current gain\n",
+ "ICperVBE = 4.0 * 10**-3 #Change in IC per volt change in VBE (in Ampere per volt)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "VCmax = VCC - Vknee #Maximum voltage drop across resistance RC (in volts)\n",
+ "ICmax = VCmax/RC #Maximum allowable collector current (in Ampere)\n",
+ "ICzero = ICmax/2 #Zero signal collector current (in Ampere)\n",
+ "IBmax = ICmax/beta #Maximum base current (in Ampere)\n",
+ "VBE = ICmax/ICperVBE #Base-emitter voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Maximum collector current that can be allowd during any part of the input signal is \",ICmax* 10**3,\"mA.\"\n",
+ "print \"Minimum zero signal collector current required : \",ICzero*10**3,\"mA.\"\n",
+ "print \"Maximum base current \",IBmax*10**3,\"mA.\"\n",
+ "print \"Signal voltage (VBE) : \",VBE,\"V.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.2 , Page Number 232 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Following is the graph showing necessary details : \n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "<matplotlib.text.Text at 0x63c07b0>"
+ ]
+ },
+ "execution_count": 7,
+ "metadata": {},
+ "output_type": "execute_result"
+ },
+ {
+ "data": {
+ "image/png": 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2ZmY54GBvZpYDDvZmZjngYG9mlgMO9mZmOfD/AehmWolB7gEbAAAAAElFTkSu\nQmCC\n",
+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x615d910>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "import math\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "from matplotlib.pyplot import plot,title,xlabel,ylabel,ylim,xlim,annotate\n",
+ "\n",
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "RC = 2.0 * 10**3 #Collector load ressitance (in ohm)\n",
+ "RE = 3.0 * 10**3 #Emitter resistance (in ohm) \n",
+ "IC = 0 #Collector current at saturation point (in Ampere)\n",
+ "VCE1 = 0 #Collector-to-emitter voltage at cut-off point (in volts) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "VCE = VCC - IC*(RC + RE) #Collector-to-emitter voltage (in volts)\n",
+ "IC1 = VCC/(RC + RE) #Cut-off point collector current (in Ampere)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Following is the graph showing necessary details : \"\n",
+ "\n",
+ "#Graph\n",
+ "\n",
+ "x = numpy.linspace(0,20,100)\n",
+ "y1 = numpy.linspace(0,2,100)\n",
+ "x1 = numpy.linspace(0,10,100)\n",
+ "plot(x,4-4.0/20*x,'b')\n",
+ "plot(x1,2+x1-x1,'--',color='g')\n",
+ "plot(10+y1-y1,y1,'--',color='g')\n",
+ "annotate('Q - POINT',xy=(10,2))\n",
+ "xlim(0,30)\n",
+ "ylim(0,6)\n",
+ "title(\"DC Load line\")\n",
+ "xlabel(\"-VCE in Volts->\")\n",
+ "ylabel(\"-IC in mA->\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.3 , Page Number 237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Q-point will be ICQ = 0.6 mA and VCEQ = 3.0 V.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 6 #Collector supply voltage (in volts)\n",
+ "VBE = 0 #Emitter-to-base voltage (in volts)\n",
+ "RB = 1.0 * 10**6 #base resistance (in ohm)\n",
+ "beta = 100 #Current gain in CE \n",
+ "RC = 5 * 10**3 #Collector resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = VCC/RB #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*RC #Collector-to-emitter voltage (in volts) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Q-point will be ICQ = \",IC * 10**3,\"mA and VCEQ = \",VCE,\"V.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.4 , Page Number 237 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Base current : 10.0 micro-A.\n",
+ "Collector current : 1.0 mA.\n",
+ "VC : 8.0 V.\n",
+ "VB : 0.7 V.\n",
+ "VCB : 7.3 V.\n",
+ "Operating point is ICQ : 1.0 mA and VCEQ : 8.0 V.\n",
+ "Stability factof : 101 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 12 #Collector supply voltage (in volts)\n",
+ "VBE = 0.7 #Emitter-to-base voltage (in volts)\n",
+ "RB = 1130.0 * 10**3 #base resistance (in ohm)\n",
+ "beta = 100 #Current gain in CE \n",
+ "RC = 4 * 10**3 #Collector resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VCC-VBE)/RB #Base current (in Ampere)\n",
+ "IC = beta * IB #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*RC #Collector-to-emitter voltage (in volts)\n",
+ "VC = VCE #Collector voltage (in volts)\n",
+ "VB = VBE #Base voltage (in volts)\n",
+ "VCB = VC - VB #Collector-to-base voltage (in volts)\n",
+ "S = beta + 1 #Stability factor \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Base current : \",IB*10**6,\"micro-A.\"\n",
+ "print \"Collector current : \",IC * 10**3,\"mA.\"\n",
+ "print \"VC : \",VC,\"V.\\nVB : \",VB,\"V.\\nVCB : \",VCB,\"V.\"\n",
+ "print \"Operating point is ICQ : \",IC*10**3,\"mA and VCEQ : \",VC,\"V.\"\n",
+ "print \"Stability factof : \",S,\".\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.5 , Page Number 237 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Stability factor : 31.37 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "dIC = 1.6 * 10**-3 #Change in collector current (in Ampere)\n",
+ "dt = 30 #Change in temperature (in Celsius degree)\n",
+ "ICO = 1.7 * 10**-6 #Reverse saturation current change (in Ampere per Celsius-degree)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "dICO = dt*ICO #Change in reverse saturation current (in Ampere) \n",
+ "S = dIC/dICO #Stability factor \n",
+ "\n",
+ "#Result\n",
+ "print \"Stability factor : \",round(S,2),\".\"\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.6 , Page Number 237 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Base current : 28.2 micro-A.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VBB = 10.0 #Base supply voltage (in volts)\n",
+ "VBE = 0.7 #Base-to-emitter voltage (in volts)\n",
+ "RB = 330 * 10**3 #Base resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VBB - VBE)/RB #Base current (in Ampere)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Base current : \",round(IB*10**6,1),\"micro-A.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.7 , Page Number 238 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Cut-off point : (0, 6.06 mA).\n",
+ "Saturation point : ( 20 V ,0).\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "<matplotlib.text.Text at 0x6e21990>"
+ ]
+ },
+ "execution_count": 24,
+ "metadata": {},
+ "output_type": "execute_result"
+ },
+ {
+ "data": {
+ "image/png": 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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x62e3290>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "import math\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "from matplotlib.pyplot import plot,title,xlabel,ylabel,ylim,xlim\n",
+ "\n",
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "RC = 3.3 * 10**3 #Collector resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IC = VCC/RC #Collector current at cut-off point (in Ampere)\n",
+ "VCE = 0 #Collector-to-emitter voltage at cut-off point (in volts) \n",
+ "VCE1 = VCC #Collector-to-emitter voltage at saturation point (in volts)\n",
+ "IC1 = 0 #Collector current at saturation point (in Ampere)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Cut-off point : (0,\",round(IC*10**3,2),\"mA).\"\n",
+ "print \"Saturation point : (\",VCE1,\"V ,0).\"\n",
+ "\n",
+ "#Graph\n",
+ "\n",
+ "x = numpy.linspace(0,20,100)\n",
+ "plot(x,6-6.0/20*x,'b')\n",
+ "xlim(0,30)\n",
+ "ylim(0,10)\n",
+ "title(\"DC Load line\")\n",
+ "xlabel(\"-VCE in Volts->\")\n",
+ "ylabel(\"-IC in mA->\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.8 , Page Number 238 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Visualizing that VCE = 0.7 , we can say that transistor is just gone to saturation from active region (not well within saturation).\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "RB = 1.0 * 10**3 #Base resistance (in ohm)\n",
+ "VBE = 0.7 #Base-to-emitter voltage (in volts)\n",
+ "beta = 100 #Current gain in CE\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VCC - VBE)/RB #Base current (in Ampere)\n",
+ "IC = beta *IB #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*RC #Collector-to-Emitter voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Visualizing that VCE = 0.7 , we can say that transistor is just gone to saturation from active region (not well within saturation).\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.9 , Page Number 238 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maximum value of RC for which transistor remains in saturation is 4.667 ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "VBB = 5 #Base supply votlage (in volts)\n",
+ "RB = 200 * 10**3 #Base resistance (in ohm)\n",
+ "VBEsat = 0.8 #Base-to-emitter voltage in saturation state (in volts)\n",
+ "VCEsat = 0.2 #Collector-to-emitter voltage in saturation state (in volts)\n",
+ "beta = 100 #Current gain in CE\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VBB-VBEsat)/RB #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "RC = (VCC - VCEsat)/IC #Collector resistance (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Maximum value of RC for which transistor remains in saturation is \",round(RC*10**-3,3),\"ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.10 , Page Number 239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "IC : 1.98 mA.\n",
+ "IB : 0.02 mA.\n",
+ "VEE : 2.7 V.\n",
+ "VCC : 8.92 V.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "IE = 2.0 * 10**-3 #Emitter current (in Ampere)\n",
+ "alpha = 0.99 #Current gain in CB\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm) \n",
+ "VBE = 0.7 #Base-emitter voltage (in volts) \n",
+ "VCB = 1 #Collector-base voltage (in volts)\n",
+ "RC = 4.0 * 10**3 #Collector resistance (in ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IC = alpha*IE #Collector current (in Ampere)\n",
+ "IB = IE - IC #Base current (in Ampere) \n",
+ "VEE = IE*RE + VBE #Emitter supply voltage (in volts)\n",
+ "VCC = IC*RC + VCB #Collector supply voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"IC : \",IC * 10**3,\"mA.\\nIB : \",IB * 10**3,\"mA.\\nVEE : \",VEE,\"V.\\nVCC : \",VCC,\"V.\"\n",
+ "\n",
+ "#Slight variation due to higher precision in the value of VCC."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.11 , Page Number 239 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "IB : 5.3 micro-A and IC : 0.54 mA at 25 Celsius degree.\n",
+ "IB : 5.375 micro-A and IC : 0.6183 mA at 55 Celsius degree.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 30 #Collector supply voltage (in volts)\n",
+ "VBB = 6 #Base voltage (in volts)\n",
+ "VBE = 0.7 #Emitter-to-base voltage (in volts)\n",
+ "RB = 1.0 * 10**6 #Base resistance (in ohm)\n",
+ "beta = 100 #Current gain in CB\n",
+ "ICBO = 0.1 * 10**-6 #Reverse saturation current (in Ampere) \n",
+ "dt = 55-25 #Change in temperature (in Celsius degree)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VBB - VBE)/RB #Base current (in Ampere)\n",
+ "IC = beta*IB + (beta+1)*ICBO #Collector current (in Ampere)\n",
+ "ICBO55 = ICBO * 2**(dt/10.0) #ICBO at 55 Celsius degree (in Ampere)\n",
+ "VBE55 = 0.7 - 2.5*10**-3*dt #VBE at 55 Celsius degree (in Ampere)\n",
+ "IB55 = (VBB - VBE55)/RB #Base current at 55 Celsius degree(in Ampere)\n",
+ "IC55 = beta*IB55 + (beta+1)*ICBO55 #Collector current 55 Celsius degree (in Ampere)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"IB : \",round(IB * 10**6,1),\"micro-A and IC :\",round(IC*10**3,2),\"mA at 25 Celsius degree.\"\n",
+ "print \"IB : \",round(IB55 * 10**6,3),\"micro-A and IC :\",round(IC55*10**3,4),\"mA at 55 Celsius degree.\"\n",
+ "\n",
+ "#Slight variation in IC55 due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.12 , Page Number 239 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Since , beta is very less than hfe , therefore it is in saturation region.\n",
+ "VC : -2.3365 V.\n",
+ "Minimum value of RB for which it operates in active region : 36.27 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "hfe = 100 #Current gain in CE\n",
+ "VBE = 0.8 #Base-emitter voltage (in volts)\n",
+ "VBB = 3.0 #Base supply voltage (in volts)\n",
+ "RB = 7.0 * 10**3 #Base resistance (in ohm)\n",
+ "RL = 500 #Load resistance (in ohm)\n",
+ "RC = 3.0 * 10**3 #Collector resistance (in ohm)\n",
+ "VCC = 10 #Collector supply voltage (in volts) \n",
+ "VCE = 1 #Collector-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "# 7500IB + 500IC = 2.2 ----Eq. 1\n",
+ "# 500IB + 2500IC = 9.0 ----Eq. 2\n",
+ "IC = 2.55 #Collector current (in milli-Ampere)\n",
+ "IB = 0.123 #Base current (in milli-Ampere)\n",
+ "beta = IC/IB #Current gain in CB\n",
+ "VC = -VCE - (IB + IC)*RL*10**-3 #Collector voltage in saturation (in volts)\n",
+ "IBmax = IC/hfe #Maximum base current (in milli-Ampere)\n",
+ "RB = (VBB - VBE - IC*RL*10**-3 )/IBmax #Base resistance (in kilo-ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Since , beta is very less than hfe , therefore it is in saturation region.\"\n",
+ "print \"VC :\",VC,\"V.\"\n",
+ "print \"Minimum value of RB for which it operates in active region : \",round(RB,2),\" kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.13 , Page Number 242 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R1 : 133.55 kilo-ohm.\n",
+ "RC : 4.06 kilo-ohm.\n",
+ "S : 18.65 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "R2 = 30.0 * 10**3 #Resistance (in ohm)\n",
+ "R1 = 133.55 * 10**3 #Resistance (in ohm) \n",
+ "alpha = 0.985 #Current gain in CB\n",
+ "VCC = 16 #Collector supply voltage (in volts) \n",
+ "VCE = 6 #Collector-emitter voltage (in volts)\n",
+ "IE = 2.0 * 10**-3 #Emitter current (in Ampere)\n",
+ "IC = alpha*IE #Collector current (in Ampere)\n",
+ "IB = IE - IC #Base current (in Ampere)\n",
+ "beta = alpha/(1-alpha) #Current gain in CE \n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "VBE = 0.2 #Base-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "RC = (VCC - VCE - IE*RE)/IC #Collector resistance (in ohm)\n",
+ "Vth = R2/(R1 + R2)*VCC #Voltage across R2 (in volts)\n",
+ "Rth = R1*R2/(R1+R2) #Thevenin's equivalence resistance (in ohm)\n",
+ "S = (1+beta)/(1 + beta*RE/(Rth+RE)) #Stability factor \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"R1 : \",round(R1 * 10**-3,2),\"kilo-ohm.\"\n",
+ "print \"RC : \",round(RC * 10**-3,2),\"kilo-ohm.\"\n",
+ "print \"S : \",round(S,2),\".\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.14 , Page Number 243 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "IB : 0.04 mA.\n",
+ "IE : 2.04 mA.\n",
+ "Rth : 5.765 kilo-ohm.\n",
+ "Vth : 3.4826 V.\n",
+ "R1 : 33.1 kilo-ohm.\n",
+ "R2 : 6.98 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 50.0 #Current gain in CE\n",
+ "VBE = 0.6 #Base-emitter voltage (in\n",
+ "RC = 4.7 * 10**3 #Collector resistance (in ohm)\n",
+ "VCC = 20 #Collector supply voltage (in volts) \n",
+ "IC = 2.0 * 10**-3 #Collector current (in Ampere)\n",
+ "VCE = 8 #Collector-emitter voltage (in volts)\n",
+ "RE = 1.3 * 10**3 #Emitter resistance (in ohm)\n",
+ "S = 5 #Stability factor\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = IC/beta #base current (in Ampere) \n",
+ "IE = IB + IC #Emitter current (in Ampere)\n",
+ "Rth = (S - 1)*RE/(1 -S/(1+beta)) #Thevenin's equivalent resistance (in ohm)\n",
+ "Vth = IB * Rth + VBE + IE*RE #Thevenin's equivalent voltage (in volts)\n",
+ "R1 = Rth * VCC/Vth #Resistance1 (in ohm)\n",
+ "R2 = Vth * R1/(VCC - Vth) #Resistance2 (in ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"IB : \",IB*10**3,\"mA.\"\n",
+ "print \"IE : \",IE*10**3,\"mA.\"\n",
+ "print \"Rth : \",round(Rth*10**-3,3),\"kilo-ohm.\"\n",
+ "print \"Vth : \",round(Vth,4),\"V.\"\n",
+ "print \"R1 :\",round(R1*10**-3,1),\"kilo-ohm.\"\n",
+ "print \"R2 : \",round(R2*10**-3,2),\"kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.15 , Page Number 244 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 29,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "RC : 5.376 kilo-ohm.\n",
+ "RE : 9.75 kilo-ohm.\n",
+ "R1 : 248.0 kilo-ohm.\n",
+ "R2 : 141.0 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "dICbyIC = 10 #Percentage change in IC \n",
+ "VBE25max = 0.7 #Max VBE at 25 degree Celsius (in volts)\n",
+ "VBE25min = 0.6 #Min VBE at 25 degree Celsius (in volts)\n",
+ "ICO25 = 5 * 10**-9 #Reverse saturation current at 25 degree celsius (in Ampere)\n",
+ "ICO145 = 3 * 10**-6 #Reverse saturation current at 145 degree celsius (in Ampere)\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "VCE = 10 #Collector-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "dIC = 5.0/100 * 0.6 #Change in collector current (in milli-Ampere)\n",
+ "dICO = ICO145 - ICO25 #Change in reverse saturation current (in Ampere)\n",
+ "S = dIC/dICO #Stability factor\n",
+ "dVBE = -2.5 * (145 - 25) #Change in VBE (in volts)\n",
+ "SV = dIC/dVBE #SV\n",
+ "beta = hfe = 400 #Current gain in CE\n",
+ "#Rth + Re = 99750.6 \n",
+ "#RE = 391.0/3609 * Rth\n",
+ "RE = 9.75 #Emitter resistance (in kilo-ohm) \n",
+ "Rth = 90 #Thevenin's equivalent resistance (in kilo-ohm)\n",
+ "dIC1 = S*ICO145 + SV*dVBE #Change in collector current1 (in milli-Ampere) \n",
+ "IC = 0.6 + dIC1 #Collector current (in milli- Ampere) \n",
+ "IE = IC + IC/beta #Emitter current (in milli-Ampere)\n",
+ "RC = (VCC - IE*RE - VCE)/IC #Collector resistance (in ohm)\n",
+ "VBE = 0.65 #emitter-base voltage (in volts)\n",
+ "Vth = IC/beta*Rth + VBE + IE*RE #Thevenin's equivalent voltage (in volts)\n",
+ "R1 = Rth * VCC/Vth #Resistance1 (in kilo-ohm)\n",
+ "R2 = Vth * R1/(VCC - Vth) #Resistance2 (in kilo-ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"RC : \",round(RC,3),\" kilo-ohm.\\nRE : \",RE,\" kilo-ohm.\\nR1 : \",round(R1),\" kilo-ohm.\\nR2 : \",round(R2),\" kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.16 , Page Number 245 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R1 : 36.238 kilo-ohm.\n",
+ "RC : 2.5 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "hfe = 100 #Current gain in CE\n",
+ "VBE = 0.7 #Emitter-base voltage (in volts)\n",
+ "ICO = 0 #Reverse saturation current (in Ampere)\n",
+ "IC = 1.0 * 10**-3 #Collector current (in Ampere)\n",
+ "VCE = 2.5 #Collector-emitter voltage (in volts) \n",
+ "VCC = 5 #Collector supply voltage (in volts) \n",
+ "R2 = 10 * 10**3 #Resistance2 (in ohm) \n",
+ "\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "R1 = 36.238 * 10**3 #Resistance1 (in ohm) \n",
+ "RC = (VCC - VCE)/IC #Collector resistance (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"R1 : \",R1*10**-3,\"kilo-ohm.\\nRC : \",RC*10**-3,\"kilo-ohm.\"\n",
+ "\n",
+ "#Printing mistake in the value of RC in book."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.17 , Page Number 245 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "VCE : 5.0 V.\n",
+ "IE : 1.1 mA.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 100 #Current gain in CE\n",
+ "R1 = 10.0 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 2.2 * 10**3 #Resistance2 (in ohm)\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "RC = 3.6 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "RB = R1*R2/(R1+R2) #Base resistance (in ohm)\n",
+ "Vth = VCC*R2/(R1 + R2) #Thevenin's voltage (in volts)\n",
+ "IE = (Vth - VBE)/(RE - Rth/(beta + 1)) #Emitter current (in Ampere)\n",
+ "IC = beta*1.0/(beta + 1)*IE #Collector current (in Ampere) \n",
+ "VCE = VCC - IC*RC - IE*RE #Collector-emitter voltage (in volts) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "\n",
+ "print \"VCE : \",round(VCE),\"V.\\nIE : \",round(IE*10**3,2),\"mA.\"\n",
+ "\n",
+ "#Slight varaition due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.18 , Page Number 246 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 43,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "VC : -14.25 V.\n",
+ "IB : -17.62 micro-A\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 200 #Current gain in CE\n",
+ "R1 = 82.0 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 16.0 * 10**3 #Resistance2 (in ohm)\n",
+ "VCC = -22 #Collector supply voltage (in volts)\n",
+ "RE = 750 #Emitter resistance (in ohm)\n",
+ "RC = 2.2 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = -0.7 #Base-emitter voltage (in volts) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "Vth = VCC*R2/(R1 + R2) #Thevenin's equivalent voltage (in volts)\n",
+ "Rth = R1*R2/(R1+R2) #Thevenin's equivalent resistance (in ohm)\n",
+ "IB = (Vth - VBE)/(Rth +(beta+1)*RE)#Base current (in Ampere) \n",
+ "IC = beta * IB #Collector current (in Ampere)\n",
+ "VC = VCC - IC*RC #Output voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"VC : \",round(VC,2),\"V.\"\n",
+ "print \"IB : \",round(IB * 10**6,2),\" micro-A\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.19 , Page Number 246"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "IB : 0.04 mA.\n",
+ "IC : 2.0 mA.\n",
+ "VCE : 14.0 V.\n",
+ "VE : 2.0 V.\n",
+ "VB : 2.8 V.\n",
+ "VC : 16.0 V.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #collector supply voltage (in volts)\n",
+ "beta = 50 #Current gain in CE\n",
+ "RB = 430.0 * 10**3 #Base resistance (in ohm) \n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "RC = 2.0 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts) \n",
+ "IC = 2 * 10**-3 #Collector current (in Ampere)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "VCE = VCC - IC*(RC+RE) #Collector-emitter voltage (in volts) \n",
+ "VC = VCC - RC*IC #Output voltage (in volts)\n",
+ "VE = VC - VCE #Emitter voltage (in volts)\n",
+ "IB = 0.04 * 10**-3 #Base current (in Ampere)\n",
+ "IE = (1+beta)*IB #Emitter current (in Ampere)\n",
+ "VB = VCC - 430*10**3*IB #Base voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"IB : \",IB*10**3,\"mA.\\nIC : \",IC*10**3,\"mA.\\nVCE : \",VCE,\"V.\\nVE : \",VE,\"V.\\nVB : \",VB,\"V.\\nVC : \",VC,\"V.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.20 , Page Number 246 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 52,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Operating point will be ICQ : 2.23 mA , VCEQ : 8.85 V.\n",
+ "Stability factor : 7.35 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "alpha = 0.985 #Current gain in CB\n",
+ "R1 = 50.0 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 20.0 * 10**3 #Resistance2 (in ohm)\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "RE = 2.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "RC = 3.0 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "Vth = VCC*R2/(R1 + R2) #Thevenin's equivalent voltage (in volts)\n",
+ "Rth = R1*R2/(R1+R2) #Thevenin's equivalent resistance (in ohm)\n",
+ "beta = alpha/(1-alpha) #Current gain in CE\n",
+ "IB = (Vth - VBE)/(Rth +(beta+1)*RE)#Base current (in Ampere) \n",
+ "IC = beta * IB #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*(RE + RC) #Collector-emitter voltage (in volts)\n",
+ "S = (1 + beta)/(1 + beta*(RE/(Rth + RE))) #Stability factor\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Operating point will be ICQ : \",round(IC*10**3,2),\"mA , VCEQ : \",round(VCE,2),\"V.\"\n",
+ "print \"Stability factor : \",round(S,2),\".\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.21 , Page Number 247 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 53,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R1 : 81.54 kilo-ohm.\n",
+ "R2 : 26.5 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "RL = 1.0 * 10**3 #Load resistance (in ohm) \n",
+ "RE = 200 #Emitter resistance (in ohm)\n",
+ "beta = 100 #Current gain in CE\n",
+ "VCC = 9 #Collector supply voltage (in volts)\n",
+ "ICQ = 3.75 * 10**-3 #Q-point Collector current (in Ampere)\n",
+ "VCEQ = 4.5 #Q-point collector-emitter voltage (in volts)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = ICQ/beta #Base current (in Ampere)\n",
+ "IE = (1 + beta)*IB #Emitter current (in Ampere)\n",
+ "Rth = 20.0 * 10**3 #Thevenin's eq. resistance (in ohm)\n",
+ "Vth = IB*Rth + VBE +IE*RE #Thevenin's equivalent voltage (in volts)\n",
+ "R1 = Rth*VCC/Vth #Resistance1 (in ohm)\n",
+ "R2 = R1*Vth/(VCC - Vth) #Resistance2 (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"R1 : \",round(R1*10**-3,2),\"kilo-ohm.\\nR2 : \",round(R2*10**-3,1),\"kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.22 , Page Number 248 "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 59,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Operating point , Q will be ICQ = 1.9 mA and VCEQ = 9.9 V.\n",
+ "Stability factor : 8.62 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 22.5 #collector supply voltage (in volts)\n",
+ "beta = 55 #Current gain in CE\n",
+ "RB = 430.0 * 10**3 #Base resistance (in ohm) \n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "RC = 5.6 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = 0 #Base-emitter voltage (in volts) \n",
+ "IC = 2 * 10**-3 #Collector current (in Ampere)\n",
+ "R1 = 90 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 10 * 10**3 #Resistance2 (in ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "Rth = R1*R2/(R1+R2) #Thevenin's eq. resistance (in ohm) \n",
+ "Vth = VCC * R2/(R1 + R2) #Thevenin's eq. voltage (in volts)\n",
+ "IB = (Vth - VBE)/(Rth +(beta + 1)*RE) #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "IE = (beta + 1)*IB #Emitter current (in Ampere)\n",
+ "VCE = VCC - IC*RC - IE*RE #Collector-emitter voltage (in volts)\n",
+ "S = (1 + beta)/(1 + beta*RE/(Rth + RE)) #Stability factor\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Operating point , Q will be ICQ =\",round(IC*10**3,2),\"mA and VCEQ =\",VCE,\"V.\"\n",
+ "print \"Stability factor : \",round(S,2),\".\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {
+ "collapsed": true
+ },
+ "source": [
+ "##Example 7.23 , Page Number 248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "beta : 153.4 .\n",
+ "VCC : 17.6842 V.\n",
+ "RB : 779.0 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "RC = 2.7 * 10**3 #Collector resistance (in ohm)\n",
+ "RE = 0.68 * 10**3 #Emitter resistance (in ohm)\n",
+ "IB = 20.0 * 10**-6 #Base current (in Ampere)\n",
+ "VCE = 7.3 #Collector-emitter voltage (in volts)\n",
+ "VE = 2.1 #Emitter voltage (in volts)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IE = VE/RE #Emitter current (in Ampere)\n",
+ "beta = IE/IB - 1 #Current gain in CE\n",
+ "VCC = beta*IB*RC + VCE + IE*RE #Collector supply voltage (in volts)\n",
+ "RB = (VCC - VE)/IB #Base resistance (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"beta : \",round(beta,1),\".\\nVCC : \",round(VCC,4),\"V.\\nRB : \",round(RB*10**-3),\"kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.24 , Page Number 249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R1 : 45.0 kilo-ohm.\n",
+ "R2 : 9.14 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta=hfe = 100.0 #Current gain in CE\n",
+ "VBE = .6 #Base-emitter voltage (in volts)\n",
+ "IC = 1.0 * 10**-3 #Collector current (in Ampere)\n",
+ "S = 8 #Stability factor\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm) \n",
+ "VCE = 5 #Collector-emitter resistance (in ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = IC/beta #Base current (in Ampere)\n",
+ "IE = IC + IB #Emitter current (in Ampere)\n",
+ "RC = (VCC - VCE - IE*RE)/IC #Collector resistance (in ohm)\n",
+ "Rth = RE*(beta*S/(1+beta-S) -1) #Thevenin's resistance(in ohm)\n",
+ "Vth = IB*Rth + VBE + IE*RE #Thevenin's eq. voltage (in volts)\n",
+ "R1 = Rth*VCC/Vth #Resistance1 (in ohm)\n",
+ "R2 = (Vth*R1)/(VCC-Vth) #Resistance2 (in ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"R1 : \",round(R1*10**-3),\"kilo-ohm.\\nR2 : \",round(R2*10**-3,2),\"kilo-ohm.\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.25 , Page Number 249"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "VCEQ : 3.2 V.\n",
+ "ICQ : 1.8 mA.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 100 #Current gain in CE\n",
+ "R1 = 2.2 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 2.2 * 10**3 #Resistance2 (in ohm)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "VCC = 5 #Collector supply voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "VA = VCC * R2/(R1 + R2) #Voltage at A (in volts)\n",
+ "IE = (VA - VBE)/RE #Emitter current (in Ampere)\n",
+ "VCEQ = VCC - IE*RE #Q-point VCE (in volts)\n",
+ "ICQ = IE #Q-point IC (in Ampere) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"VCEQ : \",VCEQ,\"V.\"\n",
+ "print \"ICQ : \",ICQ * 10**3,\"mA.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.26 , Page Number 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "VCEQ : 9.74 V.\n",
+ "ICQ : 1.13 mA.\n",
+ "IB : 11.3 micro-A.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 100 #current gain in CE \n",
+ "VCC = 12 #Collector supply voltage (in volts)\n",
+ "R1 = 15.0 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 2.7 * 10**3 #Resistance2 (in ohm)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "RC = 1.0 * 10**3 #Collector resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "VA = VCC*R2/(R1 + R2) #Potential at A (in volts)\n",
+ "IE = (VA - VBE)/RE #Emitter current (in Ampere)\n",
+ "IC = IE #Collector current (in Ampere) \n",
+ "VCEQ = VCC - IC*(RC + RE) #VCE at Q (in volts)\n",
+ "ICQ = IE #IC at Q (in volts)\n",
+ "IB = IC/beta #Base current (in Ampere) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"VCEQ : \",round(VCEQ,2),\"V.\\nICQ : \",round(ICQ*10**3,2),\"mA.\"\n",
+ "print \"IB : \",round(IB * 10**6,1),\"micro-A.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.27 , Page Number 250"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Operation point : ICQ = 1.955 mA , VCQ = 6.224 V.\n",
+ "Stability factor : 7.54 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 16 #Collector supply voltage (in volts)\n",
+ "RC = 3.0 * 10**3 #Collector resistance (in ohm)\n",
+ "RE = 2.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "R1 = 56.0 * 10**3 #Resistance1 (in ohm)\n",
+ "R2 = 20.0 * 10**3 #Resistance2 (in ohm)\n",
+ "alpha = 0.985 #Current gain in CB \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "beta = alpha/(1-alpha) #Current gain in CE\n",
+ "VBE = 0.3 #Base-emitter voltage (in volts)\n",
+ "VB = VCC * R2/(R1 + R2) #Base voltage (in volts)\n",
+ "IC = (VB - VBE)/RE #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*(RE + RC) #Collector-emitter voltage (in volts)\n",
+ "Rth = R1*R2/(R1 + R2) #Thevenin's eq. resistance (in ohm)\n",
+ "S = (1 + beta)*(1 + Rth/RE)/(1 + beta + Rth/RE) #Stability factor\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Operation point : ICQ = \",round(IC*10**3,3),\"mA , VCQ = \",round(VCE,3),\"V.\"\n",
+ "print \"Stability factor : \",round(S,2),\".\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.28 , Page Number 251"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.002 8.95e-05 0.0018795 3192.33838787 20000.0 8.49\n",
+ "R1 : 47.1 kilo-ohm.\n",
+ "R2 : 34.75 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "RL = 4.0 * 10**3 #Load resistance (in ohm)\n",
+ "VCE = 6.0 #Collector-emitter voltage (in volts)\n",
+ "IC = 2.0 * 10**-3 #Collector current (in Ampere)\n",
+ "beta=hfe = 20 #Current gain in CE\n",
+ "ICO = 10 * 10**-6 #Reverse saturation current (in Ampere)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (IC - (1 + beta)*ICO)/beta #Base current (in Ampere)\n",
+ "IC = beta*IB + (1 + beta)*ICO #Collector current (in Ampere)\n",
+ "IE = (beta + 1)*IB #Emitter current (in Ampere)\n",
+ "RE = (VCC - IC*RL - VCE)/IE #Emitter resistance (in ohm)\n",
+ "Rth = 20.0 * 10**3 #Thevenin's eq. resistance (in ohm)\n",
+ "Vth = IB*Rth + VBE + IE*RE #Thevenin's eq. voltage (in volts)\n",
+ "R1 = Rth*VCC/Vth #Resitance1 (in ohm)\n",
+ "R2 = R1*Vth/(VCC - Vth) #Resistance2 (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"R1 : \",round(R1*10**-3,1),\"kilo-ohm.\\nR2 : \",round(R2*10**-3,2),\"kilo-ohm.\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.29 , Page Number 252"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 26,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Following is the graph: \n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "<matplotlib.text.Text at 0x637cf70>"
+ ]
+ },
+ "execution_count": 26,
+ "metadata": {},
+ "output_type": "execute_result"
+ },
+ {
+ "data": {
+ "image/png": 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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x5e39390>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "import math\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "from matplotlib.pyplot import plot,title,xlabel,ylabel,ylim,xlim,annotate\n",
+ "\n",
+ "#Variables\n",
+ "\n",
+ "beta = 100\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "ICQ = 1.07 #Collector current at Q-point (in milli-Ampere)\n",
+ "VCQ = 5.067 #Collector-emitter voltage at Q-point (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Following is the graph: \"\n",
+ "\n",
+ "x = numpy.linspace(0,5.07,100)\n",
+ "y1 = numpy.linspace(0,1.07/2,100)\n",
+ "x1 = numpy.linspace(0,5.067/2,100)\n",
+ "plot(x,1.07-1.07/5.07*x,'b')\n",
+ "plot(x1,1.07/2+x1-x1,'--',color='g')\n",
+ "plot(5.067/2+y1-y1,y1,'--',color='g')\n",
+ "annotate('Q - POINT',xy=(5.067/2,1.07/2))\n",
+ "xlim(0,6)\n",
+ "ylim(0,1.1)\n",
+ "title(\"DC Load line\")\n",
+ "xlabel(\"-VCE in Volts->\")\n",
+ "ylabel(\"-IC in mA->\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.30 , Page Number 253"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Value of RB : 61.43 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 50 #current gain in CE \n",
+ "VCC = 12 #Collector supply voltage (in volts)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "RB = 100.0 * 10**3 #Base resistance (in ohm)\n",
+ "RC = 2.0 * 10**3 #Collector resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (VCC - VBE)/(beta*RC + RB) #Base current (in Ampere) \n",
+ "IC = beta * IB #Collector current (in Ampere)\n",
+ "VCE = VCC - IC*RC #Collector-emitter voltage (in volts)\n",
+ "VCE1 = 5 #New collector-emitter voltage (in volts) \n",
+ "IC1 = (VCC - VCE1)/RC #Collector current1 (in Ampere)\n",
+ "IB1 = IC1/beta #Base current1 (in Ampere)\n",
+ "RB1 = (VCC - VBE - beta*IB1*RC)/IB1 #Base resistance1 (in ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Value of RB :\",round(RB1*10**-3,2),\" kilo-ohm.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.31 , Page Number 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0.016 0.0158415841584 0.000158415841584\n",
+ "Value of collector to base resistance : 25.25 kilo-ohm.\n",
+ "Stability factor : 21.0 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 100.0 #Current gain in CE\n",
+ "RC = 1.0 * 10**3 #Collector resistance (in ohm)\n",
+ "VCC = 20.0 #Collector supply voltage (in volts)\n",
+ "VBE = 0 #Base-emitter voltage (in volts)\n",
+ "VCEQ = 4 #VCE at Q-point (in volts) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "I1C = (VCC - VCEQ)/RC #Collector current1 (in Ampere)\n",
+ "IC = I1C/(1+1/beta) #Collector current (in Ampere)\n",
+ "IB = I1C - IC #base current (in Ampere)\n",
+ "RB = (VCEQ + VBE)/IB #Base resistance (in ohm)\n",
+ "S = (1 + beta)/(1 + beta*RC/(RB + RC)) #Stability factor\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Value of collector to base resistance :\",RB*10**-3,\"kilo-ohm.\"\n",
+ "print \"Stability factor :\",S,\".\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.32 , Page Number 254"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "ICQ : 2.475 mA , VCEQ : 4.95 V.\n",
+ "Stability factor : 25.75 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables \n",
+ "\n",
+ "beta = 50 #Current gain in CB\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "RC = 2.0 * 10**3 #Collector resistance (in ohm)\n",
+ "VBE = 0 #Base-emitter voltage (in volts)\n",
+ "RB = 100 * 10**3 #Base resistance (in ohm) \n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = VCC/(RB + (1 + beta)*RC) #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "VCE = VCC - (IC + IB)*RC #Collector-emitter voltage (in volts)\n",
+ "ICQ = IC #IC at Q-point (in Ampere)\n",
+ "S = (1 + beta)/(1 + beta*RC/(RC + RB)) #Stability factor\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"ICQ : \",round(ICQ * 10**3,3),\"mA , VCEQ : \",round(VCE,2),\"V.\"\n",
+ "print \"Stability factor : \",round(S,2),\".\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.33 , Page Number 255"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 43,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Maximum collector current : 3.66 mA.\n",
+ "Minimum collector current : 2.13 mA.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "betamax = 180 #Current gain max. in CE\n",
+ "betamin = 60 #Current gain min. in CE\n",
+ "VCC = 15 #Collector supply voltage (in volts)\n",
+ "RB = 250.0 * 10**3 #Base resistance (in ohm)\n",
+ "RC = 2.5 * 10**3 #Collector resistance (in ohm) \n",
+ "VBE = 0.7 #Base-collector voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "#IC = (VCC - VBE)/(RC + RC/beta + RB/beta) #Collector current (in Ampere)\n",
+ "ICmax = (VCC - VBE)/(RC + RC/betamax + RB/betamax) #Max. collector current (in Ampere)\n",
+ "ICmin = (VCC - VBE)/(RC + RC/betamin + RB/betamin) #Min. collector current (in Ampere) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Maximum collector current : \",round(ICmax*10**3,2),\"mA.\"\n",
+ "print \"Minimum collector current : \",round(ICmin*10**3,2),\"mA.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.34 , Page Number 256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "When beta increases due to temperature , VCE will decrease.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 90 #Current gain in CE\n",
+ "VCC = 18 #Collector supply voltage (in volts)\n",
+ "RB = 510.0 * 10**3 #Base resistance (in ohm)\n",
+ "RC = 2.2 * 10**3 #Collector resistance (in ohm) \n",
+ "RE = 1.8 * 10**3 #Emitter resistance (in ohm) \n",
+ "VBE = 0.7 #Base-collector voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = VCC/(RB + (1 + beta)*(RC+RE)) #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "VCE = VCC - (IC + IB)*RC -IE*RE #Collector-emitter voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"When beta increases due to temperature , VCE will decrease.\""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.35 , Page Number 257"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 55,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Old ICQ : 1.06 mA and new ICQ : 1.2 mA.\n",
+ "Old VCEQ : 3.65 V and new VCEQ : 2.85 V.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "beta = 90.0 #Current gain in CE\n",
+ "VCC = 10 #Collector supply voltage (in volts)\n",
+ "RB = 250.0 * 10**3 #Base resistance (in ohm)\n",
+ "RC = 4.7 * 10**3 #Collector resistance (in ohm) \n",
+ "RE = 1.2 * 10**3 #Emitter resistance (in ohm) \n",
+ "VBE = 0.7 #Base-collector voltage (in volts)\n",
+ "beta1 = 135 #New current gain in CE\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IC = (VCC-VBE)/((RE+RC)*(1/beta + 1) + RB/beta)#Collector current (in Ampere)\n",
+ "ICQ = IC #Collector current at Q-point (in Ampere)\n",
+ "IB = IC/beta #Base current (in Ampere)\n",
+ "VCE = VCC - (IC + IB)*(RC+RE) #Collector-emitter voltage (in volts)\n",
+ "VCEQ = VCE #Collector-emitter voltage at Q-point (in volts)\n",
+ "ICQ1 = (VCC-VBE)/((RE+RC)*(1/beta1 + 1) + RB/beta1) #Collector current1 at Q-point (in Ampere)\n",
+ "IB1 = ICQ1/beta #Base current1 (in Ampere) \n",
+ "VCEQ1 = VCC - (ICQ1 + IB)*(RC+RE) #Collector-emitter voltage (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "\n",
+ "print \"Old ICQ :\",round(ICQ*10**3,2),\"mA and new ICQ :\",round(ICQ1*10**3,3),\"mA.\"\n",
+ "print \"Old VCEQ :\",round(VCEQ,2),\"V and new VCEQ :\",round(VCEQ1,2),\"V.\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.36 , Page Number 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 60,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Vo : 13.9 V.\n",
+ "RF : 110.91 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 15 #Collector supply voltage (in volts) \n",
+ "IE = 1.0 * 10**-3 #Emitter current (in Ampere)\n",
+ "beta = 99 #Current gain in CE\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "R1 = 17.0 * 10**3 #Resistance1 (in ohm)\n",
+ "RC = 1.0 * 10**3 #Collector resistance (in ohm)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = (IE)/(beta + 1) #Base current (in Ampere)\n",
+ "IC = beta*IB #Collector current (in Ampere)\n",
+ "IR1 = (VBE + IE*RE)/R1 #Current through R1 (in Ampere)\n",
+ "IRF = IR1 + IB #Current through RF (in Ampere)\n",
+ "I1C = IC + IRF #Current through RC (in Ampere)\n",
+ "Vo = VCC - I1C*RC #Output voltage (in volts)\n",
+ "RF = (Vo - VBE - IE*RE)/IRF #Resistance RF (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Vo : \",round(Vo,1),\"V.\"\n",
+ "print \"RF : \",round(RF*10**-3,2),\"kilo-ohm.\"\n",
+ "\n",
+ "#Calculation error in book for value of RF."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.37 , Page Number 258"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 65,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "R : 106.9 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 24 #Collector supply voltage (in volts)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "RC = 10.0 * 10**3 #Collector resistance (in ohm)\n",
+ "RE = 270.0 #Emitter resistance (in ohm)\n",
+ "VCE = 5 #Collector-emitter voltage (in volts) \n",
+ "beta = 45 #Current gain in CE\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IE = (VCC - VCE )/(RC + RE) #Emitter current (in Ampere)\n",
+ "IB = IE/(beta + 1) #Base current (in Ampere)\n",
+ "RB = (VCC - VBE - IE*(RE + RC))/IB #Base resistance (in ohm)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "\n",
+ "print \"R : \",round(RB*10**-3,1),\"kilo-ohm.\"\n",
+ "\n",
+ "#Slight variation due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.38 , Page Number 259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 71,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Rth : 9989.0 ohm.\n",
+ "IB : 20.0 micro-A.\n",
+ "IE : 2.0 mA.\n",
+ "Vth : 2.9 V.\n",
+ "R1 : 68.4 kilo-ohm.\n",
+ "R2 : 11.7 kilo-ohm.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "VCC = 20 #Collector supply voltage (in volts)\n",
+ "VBE = 0.7 #Base-emitter voltage (in volts)\n",
+ "RE = 1.0 * 10**3 #Emitter resistance (in ohm)\n",
+ "beta = 100 #Current gain in CE\n",
+ "IC = 2.0 * 10**-3 #Collector current (in Ampere) \n",
+ "S = 10 #Stability factor\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IB = IC/beta #Base current (in Ampere)\n",
+ "IE = IB + IC #Emitter current (in Ampere)\n",
+ "Rth = beta*S*RE/(1 + beta - S) - RE #Thevenin's eq. resistance (in ohm)\n",
+ "Vth = IB*Rth + VBE + IE*RE #Thevenin's eq. voltage (in volts) \n",
+ "R1 = Rth*VCC/Vth #Resitance1 (in ohm)\n",
+ "R2 = R1*Vth/(VCC - Vth) #Resistance2 (in ohm) \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Rth : \",round(Rth),\"ohm.\"\n",
+ "print \"IB : \",round(IB*10**6),\"micro-A.\"\n",
+ "print \"IE : \",round(IC*10**3,2),\"mA.\"\n",
+ "print \"Vth : \",round(Vth,1),\"V.\"\n",
+ "print \"R1 : \",round(R1*10**-3,1),\"kilo-ohm.\"\n",
+ "print \"R2 : \",round(R2*10**-3,1),\"kilo-ohm.\"\n",
+ "\n",
+ "#Slight variations due to higher precision."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.39 , Page Number 266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 94,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Following is the graph of ac and dc load lines.\n"
+ ]
+ },
+ {
+ "data": {
+ "text/plain": [
+ "<matplotlib.text.Text at 0x858b350>"
+ ]
+ },
+ "execution_count": 94,
+ "metadata": {},
+ "output_type": "execute_result"
+ },
+ {
+ "data": {
+ "image/png": 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+ "text/plain": [
+ "<matplotlib.figure.Figure at 0x8721e70>"
+ ]
+ },
+ "metadata": {},
+ "output_type": "display_data"
+ }
+ ],
+ "source": [
+ "import math\n",
+ "import numpy\n",
+ "%matplotlib inline\n",
+ "from matplotlib.pyplot import plot,title,xlabel,ylabel,ylim,xlim,annotate\n",
+ "\n",
+ "#Variables\n",
+ "\n",
+ "RC = 3.0 * 10**3 #Collector resistance (in ohm)\n",
+ "RL = 12.0 * 10**3 #Load resistance (in ohm)\n",
+ "R1 = 16.0 * 10**3 #Resitance1 (in ohm)\n",
+ "R2 = 4.0 * 10**3 #Resistance2 (in ohm)\n",
+ "RE = 2.0 * 10**3 #Emitter Resistance (in ohm)\n",
+ "VCEcutoff = VCC = 20 #Collector supply voltage (in volts)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "IC = VCC/(RC + RE) #Collector current(in Ampere) \n",
+ "VE = VCC*R2/(R1 + R2) #Voltage across R2 (in volts)\n",
+ "IE = VE/RE #Emitter current (in Ampere)\n",
+ "ICQ = IE #IC at Q-point (in Ampere)\n",
+ "VCEQ = VCC - ICQ*(RC + RE) #VCE at Q-point (in Ampere)\n",
+ "Rac = RC*RL/(RC + RL) #AC resistance (in ohm)\n",
+ "ICsat = ICQ + VCEQ/Rac #IC saturation (in Ampere)\n",
+ "VCEoff = VCEQ + ICQ*Rac #VCE cut-off for ac load line (in volts)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Following is the graph of ac and dc load lines.\"\n",
+ "\n",
+ "#Graph\n",
+ "\n",
+ "x = numpy.linspace(0,20,100)\n",
+ "x2 = numpy.linspace(0,14.8,100)\n",
+ "y1 = numpy.linspace(0,2,100)\n",
+ "x1 = numpy.linspace(0,10,100)\n",
+ "plot(x,4-4/20.0*x,'b')\n",
+ "plot(x2,6.17-6.17/14.8*x2,'r')\n",
+ "plot(x1,2+x1-x1,'--',color='g')\n",
+ "plot(10+y1-y1,y1,'--',color='g')\n",
+ "annotate('Q',xy=(10,2.2))\n",
+ "annotate('DC load line',xy=(14,1.3))\n",
+ "annotate('AC load line',xy=(5.5,4))\n",
+ "xlim(0,20)\n",
+ "ylim(0,7)\n",
+ "title(\"DC and AC Load lines\")\n",
+ "xlabel(\"-VCE in Volts->\")\n",
+ "ylabel(\"-IC in mA->\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##Example 7.40 , Page Number 268"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 95,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Input resistance : 1.667 kilo-ohm.\n",
+ "Current gain : 80.0 .\n",
+ "AC load : 4.0 kilo-ohm.\n",
+ "Voltage gain : 192.0 .\n",
+ "Power gain : 15360.0 .\n"
+ ]
+ }
+ ],
+ "source": [
+ "#Variables\n",
+ "\n",
+ "dVBE = 0.025 #Change in VBE (in volts)\n",
+ "dIB = 15.0 * 10**-6 #Change in base current (in Ampere)\n",
+ "dIC = 1.2 * 10**-3 #Change in collector current (in Ampere)\n",
+ "RC = 6.0 * 10**3 #Collector resistance (in ohm)\n",
+ "RL = 12.0 * 10**3 #Load resistance (in ohm)\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "Rin = dVBE/dIB #Input resistance (in ohm)\n",
+ "beta = dIC/dIB #Current gain in CE\n",
+ "Rac = RC*RL/(RC+RL) #AC load (in ohm)\n",
+ "Av = beta*Rac/Rin #Voltage gain \n",
+ "Ap = beta*beta*Rac/Rin #Power gain \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"Input resistance : \",round(Rin*10**-3,3),\"kilo-ohm.\"\n",
+ "print \"Current gain : \",beta,\".\"\n",
+ "print \"AC load : \",Rac*10**-3,\"kilo-ohm.\"\n",
+ "print \"Voltage gain : \",Av,\".\"\n",
+ "print \"Power gain : \",Ap,\".\""
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.10"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
|
