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| -rw-r--r-- | Heat_Transfer_by_K._A._Gavhane/Chapter_6.ipynb | 1098 | ||||
| -rw-r--r-- | Heat_Transfer_by_K._A._Gavhane/README.txt | 10 | ||||
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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
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