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diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter1.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter1.ipynb new file mode 100644 index 00000000..f0595b96 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter1.ipynb @@ -0,0 +1,272 @@ +{ + "metadata": { + "name": "chapter1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 1:Properties of Fluids" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.1 Page No5 " + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nw=500 #mass of liquid in N\n\ng=9.806 #gravity in m/s*2\n\ng1=3.5 #gravity in m/s*2\n\ng2=18.0 #gravity in m/s*2\n\n\n#Solution\n\nm=w/g #mass\n\nprint \"a)m=\",round(m,2),\"kg\"\n\nw1=m*g1 #weight\n\nw2=m*g2\n\nprint \"b)w1=\",round(w1,2),\"N\"\n\nprint \"w2=\",round(w2,1),\"N\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "a)m= 50.99 kg\nb)w1= 178.46 N\nw2= 917.8 N\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.2 Page No5" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nweight=400 # weight in N\n\ngravity= 9.806 # gravity in m/s^2\n\nF=800 # force in n\n\ng=1.6 # gravity in m/s^2\n\n#Solution\n\nm=round(weight/gravity,3)\n\n#mass is constant and does not change with location.By Newton's second law F=m*a\"\n\na=F/m\n\n#acceleration is independent of g.Hence both on the earth ,as well as on moon\"\n\nprint \"a=\",round(a,3),\"m/s^2\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "a= 19.612 m/s^2\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.3 Page No5" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nRD=0.8 #relative density\n\nrho_w=998 #density of water at 20 degree C in kg/m^3\n\ng=9.81 #gravity in m/s^2\n\nv=2.3 #viscosity in centistoke\n\n#calculation\n\nUnitWeight=((rho_w*RD)*g)/1000 #Unit Weight \n\nprint \"(i) gamma=\",round(UnitWeight,3),\"kN/m^3\"\n\nv=2.3/(10**6) #viscosity in m^2/s\n\nrho=round(rho_w*RD,1) #dymanic viscosity \n\nmu=v*rho\n\nprint \"(ii) mu=\"'%4.3E' % mu,\"Pa.s.\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i) gamma= 7.832 kN/m^3\n(ii) mu=1.836E-03 Pa.s.\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.15 Page No9" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nsigma=0.073 #surface Tension at air-water interface in N/m\n\nd=0.01 #diameter of air bubble in mm\n\n#calculation\n\n#air bubble has only one surface\n\nR=(d/2)/1000\n\ndelta_p=(2*sigma)/R #Pressure difference in N/m^2\n\nprint \"delta_p=\",delta_p/1000,\"kPa\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "delta_p= 29.2 kPa\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.16 Page No9" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nsigma=0.088 #surface Tension at soap-air interface in N/m\n\nd=3 #diameter of air bubble in cm\n\n#Solution\n\n#soap bubble has two interfaces\n\nR=0.03/2 #in m\n\ndelta_p=(4*sigma)/R #in N/m^2\n\nprint \"delta_p=\",round(delta_p,2),\"N/m^2 above atmospheric pressure\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "delta_p= 23.47 N/m^2 above atmospheric pressure\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.17 Page No9" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nd1=6 #diameter in mm\n\nd2=16 #diameter in mm\n\nsigma=0.073 #surface tension of water in N/m\n\ng=9.81 #gravity\n\nrho=998 #density\n\n#Solution\n\n#assume angle of contact theta=0 degree\n\nR1=0.006/2 #radius in m\n\nR2=0.016/2 #raidus in m\n\nr=(1/R1)-(1/R2)\n\nh=(2*sigma*r)/(rho*g) #difference in water level\n\nprint \"h=\",round(h,4)*1000,\"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "h= 3.1 mm\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.18 Page No9" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#variable initialisation\n\nd=3E-3 #diameter of tube\n\nsigma=0.48 #surface tension in N/m\n\na=130 #angle of contact in degree C\n\nrho=13600 #density of liquid in kg/m^3\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nR=d/2\n\ngamma=rho*g\n\ne=math.radians(a) #converting degree into radians\n\nc=math.cos(e) #cosine value\n\nh=(2*sigma*c)/(gamma*R)\n\nh1=round(h*1000,2)\n\nprint \"Therefore,there is a capillary depression of\",-(h1),\"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Therefore,there is a capillary depression of 3.08 mm\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.19 Page No10" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nd=0.005 #diamter of grain in mm\n\nsigma=0.073 #surface tension of air-water interface in N/m\n\ng= 9.81 #gravity in m/s^2\n\nde=998 #density\n\n#solution\n\nR=(d/2)/1000\n\nga=g*de\n\n#by assuming theta = 0 degree\n\ndel_h=(2*sigma)/(ga*R) #height of water rise\n\nprint \"Answer given in the book is incorrect.It's given as 5.95\"\n\nprint \"correct Answer is\"\n\nprint \"delta_h=\",round(del_h,2),\"m\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Answer given in the book is incorrect.It's given as 5.95\ncorrect Answer is\ndelta_h= 5.97 m\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.22 Page No10" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nM=28.97 #molecular weight of air\n\np=120E+3 #in abs\n\nT=60 #Temperature in degree C\n\n#solution\n\nR=math.ceil(8312/28.97) #gas constant for air\n\nT=T+273 #Temperature in K\n\nrho=round(p/(R*T),3)\n\nprint \"(i)density of air=\",rho,\"kg/m^3\"\n\nM1=44\n\nR=int(math.ceil(8312/44)) # gas constant for co2\n\na=R*T\n\nrho=p/a\n\nprint \"(ii)density of co2=\",round(rho,3),\"kg/m^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i)density of air= 1.256 kg/m^3\n(ii)density of co2= 1.907 kg/m^3\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.24 Page No11" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\np=100 #compression rate in abs\n\np1=80 #compression rate in abs\n\n#solution\n\n#In isothermal change K=p\n\nKA=p #bulk modulus \n\nprint \"KA=\",KA,\"kPa\"\n\n#In adiabatic change K=kp\n\nk=1.4 #constant for gases\n\nKB=int(k*p1) #bulk modulus\n\nprint \"KB=\",KB,\"kPa\"\n\nprint \"KA<KB,gas A is more compressible than gas B,in the notified situation\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "KA= 100 kPa\nKB= 112 kPa\nKA<KB,gas A is more compressible than gas B,in the notified situation\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.26 Page No11" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math \n\n#variable initialisation\n\nk=1.43E+9 #bulk modulus of elasticity of kerosene in Pa\n\nr=0.804 #relative density \n\n#solution\n\nrho=round(r*998,1) #density of kerosene in kg/m^3\n\nC=math.ceil(math.sqrt(k/rho))\n\nprint \"velocity of sound C=\",int(C),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "velocity of sound C= 1335 m/s\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 1.28 Page No11" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#variable initialisation\n\nR=287 #gas constant\n\nT=80 #temperature in degree C\n\n#solution\n\nT=273+80 #temperature in K\n\nk=1.4 #for air\n\nC=round(math.sqrt(k*R*T),1) #velocity of sound at 80 degree\n\nprint \"C=\",C,\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 376.6 m/s\n" + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter10.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter10.ipynb new file mode 100644 index 00000000..4340223a --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter10.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "chapter10" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 10:Turbulent Pipe Flow" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.1 Page No216" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD=2 #diameter in cm\n\nv=0.0098 #viscosity of water\n\nRe=2000 #Critical Reynolds number\n\nrho=998 #density in kg/m^3\n\n#solution\n\nV=Re*v/10000/(D/100) #Calculating V\n\nL=(math.pi/4)*((D/100)**2)*V\n\nprint \"The required Largest Discharge\",round(L*1000*60,3),\"L/min\"\n\nf=64/Re\n\nu=V*(math.sqrt(f/8)) #Boundary shear stress\n\ntou=rho*(u**2)\n\nprint \"tou=\",round(tou,4),\"Pa\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required Largest Discharge 1.847 L/min\ntou= 0.0383 Pa\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 10.5 Page No217" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nf=0.035 #friction factor\n\nRe=1E+5 #Reynolds number\n\nd=10 #diameter in cm\n\nEs=0.80 #roughness of pipe\n\n#solution\n\nro=(d/2)/100 #Radius in m\n\ndelta=round((65.6*ro)/(Re*math.sqrt(f))*1000,3) #Thickness of lamina sublayer\n\nprint \"delta=\",delta,\"mm\" \n\nEs_delta=Es/delta #Calculating Es_delta\n\nprint \"Es_delta=\",round(Es_delta,2)\n\nprint \"This value being less than 6.0 and greater than 0.25,the pipe flow is in transition regime\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "delta= 0.175 mm\nEs_delta= 4.57\nThis value being less than 6.0 and greater than 0.25,the pipe flow is in transition regime\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter12.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter12.ipynb new file mode 100644 index 00000000..3f6328f7 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter12.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "chapter12" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 12:Flow in Open Channels" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.1 Page No277" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#variable initialisation\n\nB=2.5 #width in m\n\ny=1.2 #depth in m\n\ns_o=0.0036 #Bed slope\n\nga=998\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nA=B*y #area\n\nP=B+2*y #Wetted Perimeter\n\nR=A/P #Hydraulic radius\n\ntou=ga*g*R*s_o #average shear stress\n\nprint \"tou=\",round(tou,2),\"Pa\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "tou= 21.58 Pa\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.3 Page No278" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nw=10 #width in m\n\nh=1.5 # horizantal slope\n\nv=1 #vertical slope\n\nd=3 #depth in m\n\nn=0.015 #mannings n\n\nQ=100\n\n#solution\n\nA=(w+(h*d))*d #Area\n\nP=round((w+(2*d)*math.sqrt((h**2)+v)),3) #wetted perimeter\n\nR=A/P #hydraulic radius\n\n#By mannings formula\n\nSo=((Q*n)/(A*(math.pow(R,2/3))))**2\n\nprint \"So=\",'%0.3E' % So", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "So= 4.451E-04\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.4 Page No278" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nw=5 #width in m\n\nd=2.3 #depthin m\n\nf=0.02 #friction factor\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nA=w*d #area\n\nP=w+(2*d) #Wetted perimeter\n\nR=round(A/P ,3) #Hydraulic radius\n\nC=round(math.sqrt((8*g)/f),1) #value of chezy's C\n\nprint \"C=\",C\n\nn=(math.pow(R,1/6))\n\nprint \"n=\",round(n/C,4)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 62.6\nn= 0.0165\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.5 Page No278" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable iniatialization\n\nB=3 #width in m\n\nh=1 #horizantal slope\n\nm=1 #vertical slope\n\nSo=0.0036 #bottom slope of channel\n\ny=1.25 #depth in m\n\nQ=15 #discharge in m^3/s\n\n#solution\n\nA=(B+(h*y))*y #area\n\nV=round(Q/A,3) #velocity\n\nP=round(B+(2*y*(math.sqrt((m**2)+1))),4) #wetted perimeter\n\nR=round(A/P,4)\n \nC=V/(math.sqrt(R*So)) #By chezy formula\n\nprint \"C=\",round(C,1)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 52.2\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 12.45 Page No293" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\ny1=0.5 #depth of flow before jump\n\ny2=3.0 #depth of flow after jump\n\ng=9.81 #gravity in m/s^2\n\nga=9790 #specific weight\n\n#solution\n\n#sequent depths related to critical depth as,\n\nyc=((y1)*(y2)*(y1+y2))/2 #Critical depth\n\nprint \"(i)yc=\",round(math.pow(yc,1/3),3),\"m\"\n\n#(q**2)/g=yc**3\n\nq=round(math.sqrt(yc*g),3)\n\nE_L=round(((y2-y1)**3)/(4*y1*y2),3) #Head loss\n\nP=ga*q*E_L #Power lost per metre width\n\nprint \"(ii)P=\",round(P/1000,1),\"kW\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i)yc= 1.379 m\n(ii)P= 129.4 kW\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter13.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter13.ipynb new file mode 100644 index 00000000..4931c306 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter13.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "chapter13" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 13:Flow Measurement" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.2 Page No321" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nd=25 #diameter im mm\n\nH=5.5 #head in m\n\nQ=3 #rate in L/s\n\nx=1.5 #horizontal distance\n\ny=0.12 #vertical distance\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nCv=round(x/(math.sqrt(4*y*H)),3) #Coefficient of velocity\n\nprint \"Cv=\",Cv \n\nQ=Q/1000 \n\na=((math.pi)/4)*((d/1000)**2)\n\nCd=round(Q/(a*(math.sqrt(2*g*H))),3) #Coefficient of contadiction\n\nprint \"Cd=\",Cd\n\nCc=Cd/Cv #Coefficient of discharge\n\nprint \"Cc=\",round(Cc,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Cv= 0.923\nCd= 0.588\nCc= 0.637\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.5 Page No322" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD=10 #diameter in cm\n\nV1=2.5 #Velocity in pipe in m/s\n\nP1=50 #pressure in kPa\n\nCv=0.98 #coefficient of velocity\n\nga=9.79 #Relative density\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nH=round((P1/ga)+((V1**2)/(2*g)),3) #Total head\n\nVj=round(Cv*(math.sqrt(2*g*H)),2) #Velocity of jet\n\nprint \"Vj=\",Vj,\"m/s\"\n\nA1=(math.pi/4)*((D/100)**2) #Discharge\n\naj=(A1*V1)/Vj\n\nDj=math.sqrt(aj*4/(math.pi))\n\nprint \"Dj=\",round((Dj*100),2),\"cm\"\n\nHl=((1/(Cv**2))-1)*((Vj**2)/(2*g)) #Headloss\n\nprint \"HL=\",round(Hl,3),\"m\" ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Vj= 10.11 m/s\nDj= 4.97 cm\nHL= 0.215 m\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.8 Page No323" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nQ=20 #Discharge in L/s\n\ng=9.81 #gravity in m/s^2\n\nKf=0.99 #flow coefficient of nozzle\n\nD=3.0 #diameter in cm\n\nga=0.70*9.79 #relative density\n\n#solution\n\nA2=math.pi/4*(D/100)**2 #calculating area\n\nQ=Q/10000 #converting into m^3/s\n\ndel_H=((Q/(Kf*A2))**2)/(2*g) #Head difference across nozzle\n\n#For a Horizontal nozzle Z1-Z2=0\n\ndel_P=ga*del_H #calculating del_p\n\nprint \"del_p=\", round(del_P,3),\"kPa\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "del_p= 2.853 kPa\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.18 Page No328" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD2=10 #diameter in cm\n\nD1=20 #diameter in cm\n\nSm=13.6 #relative density constant\n\nSp=0.9 #relative density\n\nCd=0.99 #coefficient of discharge\n\ng=9.81 #gravity in m/s^2\n\ny=9 #manometer reading in cm\n\n#solution\n\n#for a differential monometer\n\nr=(D1/2)/100 #radius\n\nA2=(math.pi/4)*((r)**2) #area\n\nD=round(math.sqrt(1-(D2/D1)**4),3) \n\ndelh=(Sm/Sp)-1 #calculating delh\n\nQ=round(((Cd*A2)/D)*(math.sqrt(2*g*(delh*(y/100)))),2) #calculating flow\n\nprint \"Q=\",Q*1000,\"L/s\"\n\n#when Q=50L/s\n\nQ=50/1000\n\ns=(((Cd*A2)/D)*(math.sqrt(2*g*(delh))))\n\ny=round((Q/s)**2,2) #manometer reading\n\nprint \"y=\",int(y*100),\"cm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Q= 40.0 L/s\ny= 14 cm\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.22 Page No330" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD1=20 #diameter in cm\n\nCd=0.985 #coefficient of discharge\n\nSm=0.6 #relative density of liquid\n\nSp=1.0 #relative density\n\ny=15 #U-tube reading in cm\n\nD2=10 #diameter in cm\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\n#from the inverted differential manometer\n\ndelh=(y/100)*(1-(Sm/Sp))\n\n#By the venturimeter equation,\n\nA2=(math.pi/4)*(((D1/100)/2)**2) #area\n\ns=math.sqrt((1-((D2/D1)**4)))\n\nQ=((Cd*A2)/s)*(math.sqrt((2*g*delh)))\n\nprint \"Q=\",round(Q*1000,2),\"L/s\"\n\nHLi=(1-(Cd**2))*delh #Head loss in the intel section\n\nprint \"HLi=\",(round(HLi,4)*1000),\"mm of water\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Q= 8.67 L/s\nHLi= 1.8 mm of water\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.24 Page No331" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nSp=0.85 #Relative density\n\ny=4 #pitot static tube reading in cm\n\nC=0.99 #coefficient of pitot tube\n\nSm=13.6 #relative density\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\n#For the differential monometer\n\ndel_h=(y/100)*((Sm/Sp)-1) #del_h\n\n#For the pitot tube\n\nVo=C*(math.sqrt(2*g*del_h)) #Velocity\n\nprint \"Velocity at M=\",round(Vo,3),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Velocity at M= 3.397 m/s\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.26 Page No332" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nPs=3.0 #Stagnation Pressure in kPa\n\nPo=-3.0 #Static pressure in kPa\n\nrho=1.20 #mass density in kg/m^3\n\nC=0.98 #instrument coefficient\n\n#solution\n\n#In a pitot static tube,\n\n#Velocity of flow\n\nVo=C*math.sqrt(((2*(Ps-Po))/rho)*1000)\n\nprint \"Vo=\",int(Vo),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Vo= 98 m/s\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.27 Page No332" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nD=30 #diameter in cm\n\nPs=-10 #static pressure in cm\n\nPsn=1.0 #stagnation pressure in N/cm^2\n\nC=0.98 #coefficient of tube\n \nga=9790 #density\n\ng=9.81 #gravity\n\n#solution\n\nPo=(Ps/100)*13.6 #Calculating Po/ga\n\nPs=Psn*(10**4)/ga #Calculating ps/ga\n\ndel_h=(Ps-Po)\n\nVm=C*math.sqrt(2*g*del_h) #Centreline velocity\n\nV=0.85*Vm #Mean velocity\n\nQ=(math.pi/4)*((D/100)**2)*V #Discharge\n\nprint \"Q=\",round(Q,3),\"m^3/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Q= 0.402 m^3/s\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.31 Page No333" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nL=0.40 #Width in m\n\nQ=25 #discharge in L/s\n\nH1=10 #head in cm\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\n#By the weir formula,\n\nCd=(Q/1000)/((2/3)*(math.sqrt(2*g))*L*(math.pow((H1/100),3/2)))\n\nprint \"Cd=\",round(Cd,3) #Coefficient of discharge", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Cd= 0.669\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 13.35 Page No334" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nL=2.5 #length in m\n \nCd=0.62 #Coefficient of discharge\n\nH1=0.7 #head in m\n\nt=0.15 #thickness in m\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nL=L-2*t #crest length\n\nn=2+(2*2) #no of end contractions\n\nLe=L-(0.1*n*H1) #effective crest length\n\n#The discharge from Francis Formula by neglecting the velocity of approach,\n\nQ=(2/3)*(Cd)*(math.sqrt(2*g))*(Le)*(math.pow(H1,3/2)) \n\nprint \"Q=\",round(Q,2),\"m^3/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Q= 1.91 m^3/s\n" + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter14.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter14.ipynb new file mode 100644 index 00000000..90cd2965 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter14.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "chapter14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 14:Unsteady Flow" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14.11 Page No358" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\n#bulk modulus of elasticity for water in MPa\n\nk_w=2.22E+3 \n\nk_c=1.50E+3\n\nk_g=9.58E+2\n\nrho=998 #density in kg/m^3\n\nRD_c=0.8 #relative density of crude oil\n\nRD_g=0.68 #relative density of gasolene\n\n#solution\n\nC=math.sqrt((k_w*10**6)/rho) #for water\n\nprint \"C=\",round(C,1),\"m/s\"\n\nC=math.sqrt((k_c*10**6)/(RD_c*rho)) #for crude oil\n\nprint \"C=\",round(C,1),\"m/s\"\n\nC=math.sqrt((k_g*10**6)/(RD_g*rho)) #for gasolene\n\nprint \"C=\",int(C),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 1491.5 m/s\nC= 1370.7 m/s\nC= 1188 m/s\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14.12 Page No358" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nE=2.07E+5 #pressure in steel pipe\n\nK=1.43E+3 #Bulk modulus in MPa\n\nt=2 #thickness in cm\n\nD=2.5 #diameter in m\n\nRD=0.80 #relative density\n\nrho=998 #density in m/s^2\n\n#solution\n\nD=D*100 #converting D to cm\n\nE=E*(10**6) #converting MPa to Pa\n\nK=K*(10**6) #converting MPa to Pa\n\n#velocity of pressure wave\n\nC=(math.sqrt(K/(rho*RD)))*(math.pow(1/(1+((D/t)*(K/E))),1/2))\n\nprint \"C=\",round(C,1),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 980.4 m/s\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14.14 Page No358" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nE=1.0E+11 #modulus of elasticity\n\nK=2.0E+9 #Bulk modulus in Pa\n\nt=1.25 #thickness in mm\n\nD=0.9 #diameter in m\n\nrho=998 #density in kg/m^3\n\nL=1000 #length in m\n\nV=2.60 #velocity in m/s\n\ng=9.81 #gravity in m/s^2\n\n#solution\n\nD=D*100 #converting D to cm\n\n#velocity of pressure wave\n\nC=(math.sqrt(K/(rho)))*(math.pow(1/(1+((D/t)*(K/E))),1/2))\n\n#with Pressure rise,\n\ndel_V=0-V #calculating del_V\n\nh_h=-(C*del_V)/g #Water Wammer Head\n\nprint \"h_W=\",round(h_h,1),\"m\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "h_W= 240.2 m\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14.16 Page No359" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nE=2.10E+11 #modulus of elasticity\n\nK=1.43E+9 #Bulk modulus in Pa\n\nt=80 #thickness in mm\n\nD=80 #diameter in cm\n\nRD=0.8 #relative density for kerosene\n\nrho=998 #density in m/s^2\n\nL=1000 #length in m\n\n#solution\n\nD=D*10 #converting D to mm\n\n#velocity of pressure wave\n\nC=(math.sqrt(K/(rho*RD)))*(math.pow(1/(1+((D/t)*(K/E))),1/2))\n\nTo=(2*L)/C\n\nprint \"Hence maximum time for a sudden closure is\",round(To,3),\"s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Hence maximum time for a sudden closure is 1.544 s\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 14.20 Page No360" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nK=2.2E+3 #bulk modulus in MPa\n\nrho=998 #density in kg/m^3\n\nL=3500 #length in m\n\nVo=0.8 #velocity in m/s\n\nT=4.0 #closure in s\n\n#solution\n\nC=round(math.sqrt((K*10**6)/rho),1) #velocity of pressure wave\n\nTo=(2*L)/C #critical time\n\n#T=4.0 s,the closure is rapid.\n\nPh=(rho*C*Vo)/(10**6) #water hammer pressure\n\nprint \"Ph=\",round(Ph,3),\"MPa\"\n\n#length of the pipe affected by peak pressure\n\nXo=L-((C*T)/2)\n\nprint \"Xo=\",Xo,\"m\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ph= 1.185 MPa\nXo= 530.6 m\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter15.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter15.ipynb new file mode 100644 index 00000000..4c41f511 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter15.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "chapter15" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 15:Compressible Flow" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15.1 Page No376" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nMg=44 #molecular weight of carbondioxide\n\n#solution\n\nR=8314/Mg #gas constant for carbondioxide\n\nprint \"R=\",int(round(R,1)),\"N.m/(Kg.K)\"\n\nMg=32 #molecular weight of oxygen\n\nR=8314/Mg #gas constant for oxygen\n\nprint \"R=\",int(math.ceil(round(R,1))),\"N.m/(Kg.K)\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 189 N.m/(Kg.K)\nR= 260 N.m/(Kg.K)\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15.2 Page No376" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nMg=28.96 #molecular weight of air\n\nk=1.4 #constant for air \n\n#solution\n\nR=int(8314/Mg) #in J/(kg.K)\n\nCp=(k/(k-1.0))*R\n\nprint \"Cp=\",Cp,\"J/(Kg.K)\"\n\nCv=R/(k-1) #in J/(kg.K)\n\nprint \"Cv=\",Cv,\"J/(Kg.K)\"\n\n#1 Kcal = 4187 J\n\nprint \"In heat units\"\n\nCp=Cp/4187\n\nprint \"Cp=\",round(Cp,3),\"Kcal/(Kg.K)\" \n\nCv=Cv/4187\n\nprint \"Cv=\",round(Cv,3),\"Kcal/(Kg.K)\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Cp= 1004.5 J/(Kg.K)\nCv= 717.5 J/(Kg.K)\nIn heat units\nCp= 0.24 Kcal/(Kg.K)\nCv= 0.171 Kcal/(Kg.K)\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15.7 Page No378" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nk=1.40 #constant for gases\n\nR=260 #gas constant in J/(Kg.K)\n\nT=25 #temperature in degree C\n\n#solution\n\nT=273 +T #converting into K\n\nC=math.sqrt(k*R*T) #Speed of sound\n\nprint \"C=\",round(C,1),\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 329.4 m/s\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15.8 Page No378" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nM=1.5 #Mach number\n\np=89.89 #Pressure in kPa\n\nrho=1.112 #density in kg/m^3\n\nk=1.4 #constant for gases\n\n#solution\n\n#At 1000 m altitude,\n\nC=math.sqrt((k*(p*1000))/rho)\n\nV=C*M #Using Mach number\n\nV=(V*3600)/1000 #Converting V from m/s to Km/h\n\nprint \"V=\",int(math.ceil(round(V,1))),\"Km/h\"\n\n#At 10,000 m altitude\n\np=26.42 #Pressure in kPa\n\nrho=0.4125 #density in kg/m^3\n\nC=math.sqrt((k*(p*1000))/rho)\n\nV=C*M\n\nV=(V*3600)/1000 #Converting V from m/s to Km/h\n\nprint \"V=\",int(V),\"Km/h\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 1817 Km/h\nV= 1617 Km/h\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 15.16 Page No381" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nTo=35 #temperature in degree C\n\nR=287 #gas constant in (J/Kg.K)\n\npo=250 #pressure in kPa\n\nk=1.4 #constant for gases\n\nV1=200 #velocity in m/s\n\n#solution\n\nTo=273+To #converting into K\n\nCp=(k*R)/(k-1)\n\nT1=int(To-((V1**2)/(2*Cp)))\n\n#The Mach number M1,at the exit is,\n\nM1=math.sqrt(((To/T1)-1)/0.2)\n\nprint \"M1=\",round(M1,3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "M1= 0.589\n" + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter16.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter16.ipynb new file mode 100644 index 00000000..6fea7c1c --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter16.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "chapter16" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 16:Fluid Flow Machines" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16.17 Page No411" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialization\n\nH=500 #Head in m\n\nCv=0.98 #specific heat\n\ng=9.81 #gravity in m/s^2\n\nfi=0.45 #assumed fi value\n\neta_o=0.85 #assumed eta value\n\nd=18 #diameter in cm\n\ngamma=9.79 #specific weight\n\nN=420 \n\n#solution\n\nV1=Cv*(math.sqrt(2*g*H)) #Velocity\n\nd=d/100 #diameter in m\n\nQ=(math.pi/4)*(d**2)*V1 #Discharge\n\nP=eta_o*gamma*Q*H #Power developed\n\nNs=(N*math.sqrt(P))/(math.pow(H,5/4)) #Specific speed\n\nprint \"Ns=\",int(Ns)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ns= 18\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16.19 Page No412" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nN1=100 #speed of turbine in rpm\n\nH1=30 #head on turbine in m\n\nH2=18 #head reduced in m\n\nP1=8000 #p in kW\n\n#solution\n\n#For geometrically similar turbines,the unit speed,Nu=N/math.sqrt(H)\n\nN2=round(N1*(math.sqrt(H2/H1)),2) #speed \n\nprint \"N2=\",N2,\"rpm\"\n\nP2=P1*(math.pow((H2/H1),3/2)) #power developed\n\nprint \"P2=\",int(P2),\"kW\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "N2= 77.46 rpm\nP2= 3718 kW\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 16.20 Page No412" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nP1=6750 #p in kW\n\nN1=300 #speed in rpm\n\nH1=45 #net head in m\n\nH2=60 #net head under homologus conditions in m\n\nete_o=85 #efficiency in percentage\n\nga=9.81*998 #density in kg/m^3\n\n#solution\n\n#using unit relationships,\n\neta_o=85/100\n\nQ=round(P1/((eta_o)*(ga/1000)*H1),2) \n\nN2=round(N1*(math.sqrt(H2/H1)),1) #revolutions per minute\n\nprint \"N2=\",N2,\"rpm\"\n\nQ1=18.03 #Q value\n\nQ2=Q1*(math.sqrt(H2/H1)) #Discharge\n\nprint \"Q2=\",round(Q2,2),\"m^3/s\"\n\nP2=P1*(math.pow((H2/H1),3/2)) #brake power\n\nprint \"P2=\",int(round(P2,5)),\"kW\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "N2= 346.4 rpm\nQ2= 20.82 m^3/s\nP2= 10392 kW\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter2.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter2.ipynb new file mode 100644 index 00000000..7f060e84 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter2.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "chapter2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 2:Fluid Statics" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.1 Page No22" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#variable initialisation\n\nh1=10 #column water in cm\n\nh2=5 #column of oil in cm\n\nrd=0.75 #relative density of oil\n\nh3=2 #column of mercury in cm\n\n#solution\n\nga=9790 #density in N/m^3\n\np=ga*h1/100 #pressure of water column\n\nprint \"(i)For water Column,p=\",int(p),\"N/m^2\"\n\nga_o=rd*ga \n\np_o=(ga_o*h2)/100 #pressure of oil column\n\nprint \"(ii)For Oil Column, p=\",round(p_o,2),\"N/m^2\"\n\nga_m=13.6*ga\n\np_m=ga_m*h3/100 #pressure of mercury column\n\nprint \"(iii)for Mercury Column,p=\",round(p_m,1),\"N/m^2\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i)For water Column,p= 979 N/m^2\n(ii)For Oil Column, p= 367.13 N/m^2\n(iii)for Mercury Column,p= 2662.9 N/m^2\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.37 Page No38" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#variable initialisation\n\nD=2.5 #diameter in m\n\nrd=0.25 #relative density \n\np=1000 #pressure in kPA\n\nf_a=120 #allowable stress in MPa\n\n#solution\n\n#Hoop stress in cylinder,sigma=(p*D)/(2*t)\n\n#sigma=f_a=allowable stress\n\nt=(p*D)/(2*f_a*1000)\n\nprint \"A Thickness of\",round(t*1000,2),\"mm can therefore be used\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "A Thickness of 10.42 mm can therefore be used\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 2.42 Page No40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#variable initialisation\n\nW=0.2 #weight in N\n\nD=5 #diameter in mm\n\n#solution\n\n#(i)distance between marking of RD of 1.0 and 0.95\n\nh1=(W/(9790*((math.pi/4)*(D*D)/(1000*1000))))*((1/0.95)-1)\n\nprint \"h1=\",round(h1*1000,2),\"mm\"\n\n#(ii)distance between marking of RD of 1.0 and 1.05\n\nh2=(W/(9790*((math.pi/4)*(D*D)/(1000*1000))))*((1/1.05)-1)\n\nprint \"h2=\",round(h2*1000,2),\"mm\"\n \nprint \"h2 will be below the marking corresponding to relative density of 1.0\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "h1= 54.76 mm\nh2= -49.54 mm\nh2 will be below the marking corresponding to relative density of 1.0\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter6.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter6.ipynb new file mode 100644 index 00000000..7bc6260b --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter6.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "chapter6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 6:Dimensional Analysis and Similitude" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.17 Page No148" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nrho_p=917 #density in kg/m^3\n\nmu_p=0.29 #viscosity in Pa.s\n\nL_p=15.0 #diameter in cm\n\nV_p=2 #velocity of pipe\n\nrho_m=998 #density in kg/m^3\n\nL_m=1.0 #diameter in cm\n\nmu_m=1.31E-3 #viscosity in Pa.s\n\n#solution\n\n#Reynold's similarity law of applicable,referring oil with p and water with m\n\nm=round((mu_m/mu_p)*(1)/((L_m/L_p)*(rho_m)/(rho_p)),4)\n\nV_m=V_p*m #velocity of water flow\n\nprint \"V_m=\",V_m,\"m/s\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V_m= 0.1246 m/s\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.18 Page No148" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nV_p=60 #velocity in Km/h\n\nF_m=250 #model drag\n\nmu_r=1 #viscosity\n\nrho_r=1 #density in in kg/m^3\n\n#solution\n\n#Reynold's similarity law is applicable\n\n#V_r=vr/L_r\n\n#If vr=1,then V_r=1/L_r\n\nL_r=6 \n\nV_m=((V_p*L_r)*1000)/3600 #V_m in m/s\n\n#using force ratio,Fm/Fp=(mu_r**2)/rho_r\n\n#Fm/Fp=1.0\n\nFp=F_m/((mu_r*mu_r))/rho_r #drag required\n\nprint \"Fp=\",int(Fp),\"N(same as in the model)\"\n\nV_p=(V_p*1000)/(3600) #V_m in m/s\n\nPp=Fp*V_p #power to overcome drag in the prototype\n\nprint \"Pp=\",math.ceil(Pp)/1000,\"kW\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Fp= 250 N(same as in the model)\nPp= 4.167 kW\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.21 Page No149" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#variable initialisation\n\nD_p=1.5 #Diameter in m\n\nrd=0.9 #relative density\n\nv_p=0.03 #viscosity in stoke\n\nQ=3.0 #rate\n\nD_m=0.15 #diameter2 in m\n\nV_m=0.01 #viscosity2 in stoke \n\n#solution\n\n#Reynolds number must be the same in the model and prototype for similar pipe flows\n\nV_p=round((Q)/((math.pi/4)*(1.5*1.5)),4)\n\nV_m=V_p*(D_p/D_m)*(V_m/v_p) \n\nprint \"V_m=\",V_m,\"m/s\"\n\nQ_m=(math.pi/4)*(D_m**2)*(V_m) #Discharge in the model\n\nprint \"Q_m=\",round(Q_m,1),\"m^3/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V_m= 5.659 m/s\nQ_m= 0.1 m^3/s\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.25 Page No151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nL_r=1/200 #horizontal scale\n\nh_r=1/40 #vertical scale\n\nQ_p=152 #normal discharge in m^3/s\n\ny_p=2.0 #depth in metres\n\nB_p=90 #width in metres\n\nn_p=0.025 #roughness value\n\n#solution\n\nL_r=1/200\n\nh_r=1/40\n\nh=math.sqrt(h_r)\n \nQ_m=Q_p*L_r*h*h_r*10 #Discharge \n\nprint \"(i)Qm=\",round(Q_m,2),\"m^3/s\"\n\ny_m=y_p*h_r #Depth\n\nprint \"(ii) ym=\",y_m,\"m\"\n\nB_m=B_p*L_r #Width\n\nprint \"(iii)Bm=\",B_m/10,\"m\"\n \na=math.pow(h_r,2/3) #Manning's n_m\n\nb=math.sqrt(L_r)*L_r\n\nn_m=((n_p)*(math.pow(h_r,2/3)))/(math.pow(L_r,1/2))\n\nprint \"(iv) n_m=\",round(n_m,2)\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i)Qm= 0.03 m^3/s\n(ii) ym= 0.05 m\n(iii)Bm= 0.045 m\n(iv) n_m= 0.03\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 6.27 Page No151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nL_r=1/250 #horizontal scale\n\nh_r=1/25 #vertical scale\n\nS_p=0.0002 #prototype slope\n\nV_m=0.50 #velocity in m/s\n\nQ_m=0.02 #discharge in m^3/s\n\n#solution\n\nh_r=1/25\n\nL_r=1/250\n\nS_r=h_r/L_r\n\nS_m=round((S_p)*(S_r),3) #Slope ratio \n\nprint \"(i)S_m=\",S_m\n\nV_r=math.sqrt(h_r)\n\nV_p=V_m/V_r #Velocity ratio\n\nprint \"(ii)V_p=\", V_p,\"m/s\"\n\nQ_p=(L_r)*(math.sqrt(h_r))*(h_r)\n\nQ_p=(Q_m)/(Q_p) #Discharge ratio\n\nprint \"(iii)Q_p=\",int(round(Q_p,1)),\"m^3/s\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(i)S_m= 0.002\n(ii)V_p= 2.5 m/s\n(iii)Q_p= 625 m^3/s\n" + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter7.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter7.ipynb new file mode 100644 index 00000000..b76b5c41 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter7.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "chapter7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 7:Laminar Flow" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.9 Page No164" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nR=2000 #maximum Reynolds number\n\nrho=950 #density in kg/m^3\n\nmu=8E-2 #Viscosity in Pa.s\n\nL=200 #length in metre\n\nD=50 #diameter in cm\n\n#solution\n \nD=15/100 #diameter in m\n\n#Re=(V*D)/v\n\nV=R*(mu/rho)/D\n\n#by,Hagen-Poiseuille equation,\n\nh_f=(32*mu*V*L)/((rho*9.81)*(D**2)) #headloss\n\nprint \"Maximum difference in oil surface elevations h_f=\",round(h_f,3),\"m\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Maximum difference in oil surface elevations h_f= 2.742 m\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.10 Page No164" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD=0.075 #Diamter in m\n\nmu=0.1 #visocity of oil in N.s/m^2\n\nP=5.4 #power in kW\n\nrho=0.90*998 #relative density\n\nE=60 #efficiency in %\n\nL=1000 #length in m\n\n#solution\n\nP=P*(E/100) #power spent in fluid friction\n\nQ=round(math.sqrt(((P*1000)*(math.pi)*(D**4))/(128*mu*L)),5) #quantity\n\nprint \"Quantity of oil=\",round(Q*1000*60,3),\"L/min\"\n\nV=round((Q)/((math.pi/4)*(D**2)),3) #Velocity in m/s\n\nRe=(V*D*rho)/mu #Reynolds number\n\nprint \"Reynolds number=\",int (Re)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Quantity of oil= 301.2 L/min\nReynolds number= 765\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 7.19 Page No168" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nB=0.12 #thickness in mm\n\nL=50 #length in cm\n\nh_f=5.0 #headloss in m\n\nga=9790 #Specific weight\n\nmu=(998)*(0.01E-4) #viscosity \n\n#solution\n\nV=((h_f)*(ga)*((B/1000)**2))/(12*mu*(L/100)) #calculating V in m/s\n\nq=(1*V*(B/1000))*60*1000 #Discharge per metre width\n\nprint round(q,3),\"L/min per metre width of crack\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "0.848 L/min per metre width of crack\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter8.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter8.ipynb new file mode 100644 index 00000000..1eff42bc --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter8.ipynb @@ -0,0 +1,43 @@ +{ + "metadata": { + "name": "chapter8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 8.8 Page No183" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nU=3.0 #Velocity in m/s\n\nv=1.45E-5 #velocity of air in m^2/s\n\nrho_a=1.2 #density in kg/m^3\n\nRe=5E+5 #critical reynolds number\n\nw=1.5 #width in m\n\n#solution\n\nL=round((Re*v)/U,3) #Maximum Length of plate\n\nprint \"L=\",L,\"m\"\n\nC_df=(1.328)/(math.sqrt(Re)) #length of plate in a laminar sublayer\n\nF_D=C_df*(w*L)*((2*(U**2))/2) #Drag force in one side of the plate\n\nprint \"F_D=\",round(F_D/100*60,5),\"N\"\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "L= 2.417 m\nF_D= 0.03677 N\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "code", + "collapsed": false, + "input": "", + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter9.ipynb b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter9.ipynb new file mode 100644 index 00000000..9c92b722 --- /dev/null +++ b/1000_solved_Problems_in_Fluid_Mechanics_includes_Hydraulic_machines_by_K.Subramanya/chapter9.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "chapter9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 9:Drag and Lift on Immersed Bodies" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.2 Page No198" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD=20 #diameter in cm\n\nrho_s=2.8*998 #relative density\n\nrho_w=998 #relative density of water in kg/m^3\n\nv=1E-6 #velocity in m^2/s\n\ng=9.81 #gravity constant in m/s^2\n\nrho_f=998\n\n#solution\n\n#assume Re>3*E+5.then\n\nC_D=0.20\n\n#At terminal velocity V-0t,submerged weight=Drag\n\nD=D/100 #converting D in m\n\nV_ot=round(math.sqrt((4/3)*(D/C_D)*((g)*((rho_s/rho_f)-1))),3)\n\nRe=(V_ot*(D/100))/v #calculating Re\n\n#from Re,its clear,Re>3*10^5\n\nprint \"Hence the initial assumption is correct and V_ot=\",V_ot,\"m/s\"\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Hence the initial assumption is correct and V_ot= 4.852 m/s\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.3 Page No199" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nrho_f=998 #density of water in kg/m^3\n\nmu=1E-3 #coefficient of viscosity in Pa.s\n\ng=9.81 #gravity in m/s^2\n\nga_s=2.65*998 #particles RD\n\nga_f=998 #relative density in kg/m^3\n\n#solution\n\n#Stoke's law is valid upto Re=1.0\n\n#For maximum size particles that obey stoke's law,\n\n#Using fall velocity and stoke's law,\n\nV_ot=(1/18)*g*(mu/rho_f)*((ga_s/ga_f)-1) #terminal velocity\n\nV_o=round(math.pow(V_ot,1/3)*1000,2) #taking cube root\n\nprint \"V_ot=\",V_o,\"mm/s\"\n\nD=(mu)/(rho_f*V_o/1000) #size\n\nprint \"D=\",round(D*1000,4),\"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V_ot= 9.66 mm/s\nD= 0.1037 mm\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.4 Page No199" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#variable initialisation\n\nga_s=2.60*998 #relative density of sphere\n\nD=2.0 #Diameter in mm\n\nv=1.25 #velocity in cm/s\n\nga_f=917 #density of oil in kg/m^3 \n\ng=9.81 #gravity in m/s^2\n\n#solution\n\n#Assuming validity of Stoke's law,\n\n#using Fall Velocity equation,\n\nmu=round((((D/1000)**2)*((ga_s*g)-(ga_f*g)))/(18*(v/100)),3)\n\nprint \"coefficient of dynamic viscosity mu=\",mu,\"Pa.s\"\n\n#Reynold's number,\n\nRe=round((ga_f*(v/100)*(D/1000))/(mu),3)\n\nprint Re,\" < 1.0\"\n\nprint \"Hence the assumed Stoke's law is valid\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "coefficient of dynamic viscosity mu= 0.293 Pa.s\n0.078 < 1.0\nHence the assumed Stoke's law is valid\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.5 Page No199" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nD=0.2 #diameter in mm\n\nrho_f=1.20 #density of air in kg/m^3\n\nrho_s=998 #density of water in kg/m^3\n\ng=9.81 #gravity in m/s^2\n\nC_D=4.20 #Drag coefficient\n\nD1=2.0 #Diameter in mm\n\nC_D1=0.517 #Drag coefficient\n\n#solution\n\nD=D/1000 #converting D to m\n\nprint \"For 0.2 mm rain drop:\"\n\nV_ot=round(math.sqrt((((4*g*D)/(3*C_D))*((rho_s/rho_f)-1))),2)\n\nprint \"V_ot=\",V_ot,\"m/s\"\n\nprint \"For 2.0 mm rain drop:\"\n\nD1=D1/1000 #converting D to m\n\nV_ot=round(math.sqrt((((4*g*D1)/(3*C_D1))*((rho_s/rho_f)-1))),2)\n\nprint \"V_ot=\",V_ot,\"m/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "For 0.2 mm rain drop:\nV_ot= 0.72 m/s\nFor 2.0 mm rain drop:\nV_ot= 6.48 m/s\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.6 Page No199" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\nV_o=60 #velocity in Km/h\n\nC_d=0.35 #Drag coefficient\n\nC_d1=0.30 #reduced Drag coefficient \n\nA=1.6 #area in m^2\n\nrho=1.2 \n\n#solution\n\nV_o=(V_o*1000)/(60*60) #converting V_o into m/s\n\n#power required to overcome wind resistance by the car\n\nF_D=round(C_d * A * rho * ((V_o**2)/2),2)\n\np=F_D*V_o #power \n\nprint \"a)power=\",round(p/1000,3),\"kW\"\n\nV_o=(p*2)/(C_d1*A*rho) #Speed\n\nV=math.pow(V_o,1/3)*18/5 #taking cuberoot\n\nprint \"b)Vo=\",round(V,2),\"Km/h\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "a)power= 1.556 kW\nb)Vo= 63.16 Km/h\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Example 9.9 Page No201" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#variable initialisation\n\ng=9.81 #gravity in m/s^2\n\nh=2 #height in m\n\nF_D=1000 #total load in N\n\nrho=1.2 #density in kg/m^3\n\n#solution\n\nC_d=1.33 #for the hemisphere with concave frontal surface\n\nV_ot=round(math.sqrt(2*g*h),2) #Terminal Velocity\n\nD=F_D*(4/3.14)*(2/(C_d*rho*(V_ot**2)))#minimum size\n \nprint \"D=\",round(math.sqrt(D),2),\"m,say 6.5 m\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "D= 6.38 m,say 6.5 m\n" + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
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"metadata": { + "name": "chapter1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 1:Basic Electricity and Algebra" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.1, Page no:2" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of current in toaster\n\n#Initialization\n\nV=120 #voltage in V\n\nR=12 #Resistance in Ohm\n\n#calculation\n\nI=V/R\n\nprint \"I=\",I,\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "I= 10 A\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.3, Page no:3" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n\n#cal of current in toaster\n\n#initialization\n\nR=12 #Resistance in Ohm\n\nV=100 #voltage in V\n\n\n#calculation\n\nI=V/R #ohm's Law\n\nprint \"When V=100 V,I=\",round(I,1),\"A\"\n\nV=150 #Voltage in V\n\n#calculation\n\nI=V/R\n\nprint \"When V=150 V,I=\",I,\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "When V=100 V,I= 8.3 A\nWhen V=150 V,I= 12.5 A\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.2, Page no:2" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of current in toaster\n\n#initialization\n\nR=10 #Resistance in Ohm\n\nV=120 #voltage in V\n\n#calculation\n\nI=V/R #ohm's Law\n\nprint \"When R=10 ohm,I=\",I,\"A\"\n\nR=15 #Resistance in Ohm\n\n#calculation\n\nI=V/R\n\nprint \"When R=15 Ohm, I=\",I,\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "When R=10 ohm,I= 12 A\nWhen R=15 Ohm, I= 8 A\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.10, Page no:7" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of potential difference across the terminals\n\n#Initialization\n\nI=2 #Current in A\n\nR=8 #Resistance in V\n\n#calculation\n\nV=I*R #ohm's Law\n\nprint \"V=\",V,\"V\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 16 V\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.11, Page no:7" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Resistance of Electric heater\n\n#initialization\n\nV=120 #Voltage in V\n\nI=25 #Current in A\n\n#calculation\n\nR=V/I #Ohm's Law\n\nprint \"R=\",R,\"Ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 4.8 Ohm\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:1.16, Page no:9" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of unknown resistance\n\n#initialization\n\n#Resistances in Ohm\n\nR1=10\n\nR2=1000\n\nR3=26\n\n#calculation\n\nRx=(R2*R3)/R1 #Wheatstone Bridge Formula\n\nprint \"Rx=\",int(Rx),\"Ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Rx= 2600 Ohm\n" + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter10.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter10.ipynb new file mode 100644 index 00000000..f7dd8391 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter10.ipynb @@ -0,0 +1,146 @@ +{ + "metadata": { + "name": "chapter10" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 10:Inductance" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.1, Page no:115\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of average self-induced emf\n\n#Initialization\n\nL=25 #Inductance in mH\n\ndel_I=0.2 #Current in A\n\ndel_t=0.01 #Time in s\n\n#Calculation\n\nL=L/1000 #Inductance in H\n\nV=L*(del_I/del_t)\n\nprint \"V=\",V,\"V\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 0.5 V\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.3, Page no:116" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of inductance\n\n#Initialization\n\nN=500 #No. of Turns\n\nmu_o=1.26E-6 #Magnetic permeability of solenoid in air\n\nA=20 #Area in cm^2\n\nl=10 #Length in cm\n\n\n\n#calculation\n\nl=10/100 #convert length in m\n\nA=20/10000 #Convert area in m^2\n\nL= ((mu_o)*(N*N)*(A))/(l) #Formula for inductance of solenoid\n\nL=L*1000 #convert inductance in mH\n\nprint \"L=\",L,\"mH\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "L= 6.3 mH\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.4, Page no:116\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of turns of wire\n\n#Initialization\n\nL=0.178 #Inductance in mH\n\nl=20 #Length of solenoid in cm\n\nd=2 #Diameter in cm\n\nmu_o=1.26E-6 #magnetic permeability in air\n\n#Calculation\n\nL=L/1000 #Convert Inductance in H\n\nl=l/100 #Convert Length in m\n\nd=2/100 #Convert diameter in m\n\nr=d/2 #Cal of radius\n\nA=(math.pi)*(r**2) #Cal of Area in m^2\n\nN=math.sqrt((L*l)/(mu_o*A)) #Calculating No.of Turns\n\nprint \"N= \",int(round(N)),\"turns\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "N= 300 turns\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.5, Page no:116" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of Inductance\n\n#Initialization\n\nD=5 #Diameter in cm\n\nA=1 #Cross-sectional area in cm^&2\n\nmu_o=1.26E-6 #Magnetic permeability in air\n\nmu=400*mu_o #Permeability of iron given\n\nN=1000 #No. of turns\n\n#Calculation\n\nl=math.pi*(D/100) #Calculating length in m\n\nA=A/10000 #Convert Area in m^2\n\nL=(mu*(N**2)*A)/(l) #inductance Formula\n\nprint \"L=\",round(L,2),\"H\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "L= 0.32 H\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.6, Page no:117\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of magnetic potential energy\n\n#initialization\n\nL=20 #inductance in mH\n\nI=0.2 #Current in A\n\n#Calculation\n\nL=L/1000 #Convert Inductance in H\n\nW=(1/2)*(L*I**2) #Formula for energy in current-carrying conductor\n\nprint \"W=\",'%4.0E' % W,\"J\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "W= 4E-04 J\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:10.7, Page no:117" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of Current\n\n#initialization\n\nL=20 #Inductance in mH\n\nW=1 #Energy in J\n\n#Calculation\n\nL=L/1000 #convert Inductance in H\n\nI=math.sqrt((2*W)/L)\n\nprint \"I=\",int(I),\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "I= 10 A\n" + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter11.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter11.ipynb new file mode 100644 index 00000000..96f879c2 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter11.ipynb @@ -0,0 +1,167 @@ +{ + "metadata": { + "name": "chapter11" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 11:Capacitance" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.1, Page no:123" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of charge on capacitors plates\n\n#initialization\n\nC=200 #Capacitance in pF\n\nV=100 #Voltage in V\n\n#Calculation\n\nC=200E-12 #Convert Capacitance in F\n\nQ=C*V #Capacitance formula\n\nprint \"Q=\",Q,\"C\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Q= 2e-08 C\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.2, Page no:123" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Capacitance\n\n#Initialization\n\nQ=5E-4 #Charge in C\n\nV=300 #Voltage in V\n\n#Calculation\n\nC=Q/V #Capacitance Formula \n\nC=C*(10**6) #Convert Capacitance in muF\n \nprint \"C=\",round(C,2),\"muF\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Capacitance is 1.67 muF\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.3, Page no:124" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Capacitance\n\n#Initialization\n\nEo=8.85E-12 #Permittivity in F/m\n\nA=5 #Area in cm^2\n\nd=0.1 #distance in mm\n\n#Calculation\n\nK=1 #dielectric constant in air\n\nA=A/100 #Convert Area in m^2\n\nd=d/1000 #Convert diameter in m\n\nC=(K*Eo*(A**2))/(d)\n\nC=int(C*(10**12)) #Convert Capacitance in pF\n\nprint \"(a)C=\",C,\"pF\"\n\nK=6 #Permittivity for mica given\n\nC=K*C\n\nprint \"(b)C=\",C,\"pF\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a)C= 221 pF\n(b)C= 1326 pF\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.4, Page no:124" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of dielectric constant of Benzene\n\n#Initialization\n\nC1=2 #Capacitance in air in muF\n\nC2=4.6 #Capacitance in Benzene\n\nK1=1 #Dielectric constant for air\n\n#Calculation\n\nK2=K1*(C2/C1)\n\nprint \"K2=\",K2", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "K2= 2.3\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.7, Page no:125" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of voltage and charge\n\n#Initialization\n\nC=100 #Capacitance in muF\n\nW=50 #Energy in J\n\n#Calculation\n \nC=C*(10**-6) #Convert Capacitance in F \n \nV=math.sqrt((2*W)/C)\n\nprint \"(a)V=\",int(V),\"V\"\n\nQ=C*V\nprint \"(b)Q=\",Q,\"C\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a)V= 1000 V\n(b)Q= 0.1 C\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.8, Page no:126" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Equivalent Capacitance\n\n#Initialization\n\n#Capacitance in muF\n\nC1=1\n\nC2=2\n\nC3=3\n\n#Calculation\n\nC=1/((1/C1)+(1/C2)+(1/C3)) #Formula for capacitors in series\n\nprint \"C=\",round(C,3),\"muF\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 0.545 muF\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:11.9, Page no:126" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Equivalent Capacitance\n\n#Initialization\n\n#Capacitance in muF\n\nC1=2\n\nC2=3\n\n#Calculation\n\nC=(C1*C2)/(C1+C2) #Formula for Capacitors in series\n\nprint \"C=\",C,\"muF\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "C= 1.2 muF\n" + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter12.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter12.ipynb new file mode 100644 index 00000000..4f8354c9 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter12.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "chapter12" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 12:Trigonometry and Vectors" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:12.1, Page no:134" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Expressing the degree in radians\n\n#Initialization\n\nd=8 #Given degree\n\n#Calculation\n\ntheta=d*0.01745 #Converting degree to radians\n\nprint \"theta=\",round(theta,2),\"rad\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "theta= 0.14 rad\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:12.15, Page no:138" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of length of hypotenuse\n\n#initialization\n\na=6 #Length of one side in in\n\nb=12 #Length of other side in in\n\n#Calculation\n\nc=math.sqrt((a**2)+(b**2))\n\nprint \"c=\",round(c,1),\"in\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "c= 13.4 in\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:12.16, Page no:138" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of length of other side\n\n#initialization\n\nc=8 #Length of hypotenuse in m\n\nb=6 #Length of other side in m\n\n#Calculation\n\na=math.sqrt((c**2)-(b**2))\n\nprint \" a=\",round(a,1),\"m\"\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " a= 5.3 m\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter13.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter13.ipynb new file mode 100644 index 00000000..9d4aac22 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter13.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "chapter13" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 13:Alternating Current" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.2, Page no:150" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nfrom __future__ import division\n\n#Cal of Frequency,Velocity,Amplitude\n\n#Initialization\n\nT=4 #Time period in s\n\nLambda=30 #wavelength in m\n\nf=0.25 #Frequency in Hz\n\ntot_range=2 #in m\n\n#Calculation\n\nf=1/T\n\nv=f*Lambda\n\nprint \"(a) f=\",f,\"Hz\"\n\nprint \"(b) v=\",v,\"m/s\"\n\nprint \"(c) The amplitude is half the total range,hence A=\",int(tot_range/2),\"m\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) f= 0.25 Hz\n(b) v= 7.5 m/s\n(c) The amplitude is half the total range,hence A= 1 m\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.3, Page no:150" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Frequency\n\n#Initialization\n\nv=25 #Wave velocity in cm/s\n\nLambda=0.1 #Wavelength in mm\n\n#Calculation\n\nLambda=Lambda/1000 #Convert Wavelength in m\n\nv=v/100 #Convert wave velocity in m/s\n\nf=v/Lambda\n\nprint \"f=\",int(f),\"Hz\"\n\n\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "f= 2500 Hz\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.4, Page no:151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Wavelength\n\n#Initialization\n\nv=5020 #Velocity in ft/s\n\nf=256 #Frequency in Hz\n\n#Calculation\n\nLambda=v/f\n\nprint \"Lambda=\",round(Lambda,1),\"ft\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Lambda= 19.6 ft\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.5, Page no:151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "\nfrom __future__ import division\n\n#Cal of Frequency\n\n#Initialization\n\nLambda=3.2 #Wavelength in cm\n\nc=3E+8 #Velocity of light in m/s\n\n#Calculation\n\nLambda=Lambda/100 #Convert Wavelength in m\n\nf=c/Lambda\n\nprint \"f=\",'%0.1E' % f,\"Hz\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "f= 9.4E+09 Hz\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.7, Page no:152" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Effective ac potential difference\n\n#Initialization\n\nVmax=300 #Potential difference in V\n\n#Calculation\n\nVeff=0.707*(Vmax)\n\nprint \"Veff=\",int(Veff),\"V\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Veff= 212 V\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.8, Page no:152" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Effective current\n\n#Initialization\n\nImax=10 #Current in A\n\nR=20 #Resistance in Ohm \n\n#Calculation\n\nIeff=0.707*Imax\n\nP=(Ieff**2)*R\n\nprint \"Ieff=\",Ieff,\"A\"\n\nprint \"P=\",int(round(P)),\"W\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Ieff= 7.07 A\nP= 1000 W\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.9, Page no:153" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Secondary voltage and current\n\n#Initialization\n\nN1=100 #No.of turns in primary winding\n\nN2=500 #No. of turns in secondary winding\n\nV1=120 #Primary voltage in V\n\nI1=3 #Primary current in A\n\n#Calculation\n\nV2=(N2/N1)*V1\n\nI2=(N1/N2)*I1\n\nprint \"V2=\",int(V2),\"V\"\n\nprint \"I2=\",I2,\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V2= 600 V\nI2= 0.6 A\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:13.10, Page no:154" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Initialization\n\nV1=5000 #primary voltage in V\n\nV2=240 #Secondary voltage in V\n\nP=10 #Power in kW\n\n#Cal of Ratio of turns\n\nN1_N2=V1/V2\n\nP=P*1000 #Convert power in W\n\n#Cal of Maximum current\n\nI2=P/V2\n\nprint \"(a) N1_N2=\",round(N1_N2,1)\n\nprint \"(b) I2=\",round(I2,1),\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) N1_N2= 20.8\n(b) I2= 41.7 A\n" + } + ], + "prompt_number": 9 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter14.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter14.ipynb new file mode 100644 index 00000000..dd9e9a8d --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter14.ipynb @@ -0,0 +1,104 @@ +{ + "metadata": { + "name": "chapter14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 14:Series AC Circuits" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:14.2, Page no:160" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Initialization\n\nC=10 #Capacitance in muF\n\nV=15 #Voltage in V\n\nf=5 #Frequency in kHz\n\n#Cal of reactance of capacitor\n\nf=f*1000\n\nC=C*(10**(-6))\n\nXc=round(1/(2*math.pi*f*C),2)\n\nprint \"(a) Xc=\",Xc,\"Ohm\"\n\n#Cal of Current\n\nI=V/Xc\n\nprint \"(b) I=\",round(I,2),\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) Xc= 3.18 Ohm\n(b) I= 4.72 A\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:14.4, Page no:160" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Initializtion\n\nV=24 #Voltage in V\n\nf=60 #Frequency in Hz\n\nL=0.3 #Inductance in H\n\n#Cal of Reactance\n\nXL=2*math.pi*f*L\n\nprint \"(a) XL=\",int(XL),\"Ohm\"\n\n#Cal of Current\n\nI=V/XL\n\nprint \"(b) I=\",round(I,2),\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) XL= 113 Ohm\n(b) I= 0.21 A\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:14.5, Page no:160" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of Inductance\n\n#Initialization\n\nf=500 #Frequency in Hz\n\nXL=80 #Reactance in Ohm\n\n#Calculation\n\nL=XL/(2*math.pi*f)\n\nL=L*1000 #Convert L in mH\n\nprint \"L=\",round(L,1),\"mH\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "L= 25.5 mH\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:14.11, Page no:167" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Initialization\n\nL=5 #inductance in mH\n\nC=5 #Capacitance in pF\n\nR=5 #Resistance in Ohm\n\nV=5E-4\n\n#Cal of frequency\n\nL=L/1000 #Convert L in H\n\nC=C/(10**12) #Convert C in F\n\n\nfo=1/(2*math.pi*(math.sqrt(L*C))) #Frequency\n\nfo=fo/1000 #Convert fo in kHz\n\nprint \"(a) fo=\",int(math.ceil(fo)),\"kHz\"\n\n#Cal of Current\n\n#At Resonance XL=XC and Z=R\n\nI=V/R\n\nI=I*1000\n\nprint \"(b) I=\",I,\"mA\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) fo= 1007 kHz\n(b) I= 0.1 mA\n" + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter15.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter15.ipynb new file mode 100644 index 00000000..75c1a0fc --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter15.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "chapter15" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 15:Parallel AC Circuits" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:15.12, Page no:183" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#Cal of resonance frquency\n\n#Initialization\n\nL=180 #Inductance in muH\n\nC=100 #Capacitance in pF\n\n#Calculation\n\nL=L*(10**-6)\n\nC=C*(10**-12)\n\nfo=1/((2*math.pi)*(math.sqrt(L*C)))\n\nfo=fo/10**6 #Convert fo in MHz\n\nprint \"fo=\",round(fo,1),\"MHz\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "fo= 1.2 MHz\n" + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter2.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter2.ipynb new file mode 100644 index 00000000..ce638a24 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter2.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "chapter2" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "chapter:2 Fractions,Decimals,and Percentage" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:2.10, Page no:17" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Initialisation\n\nI=3.7 #Current in ampere\n\nt=1.4 #time in hours\n \n#Calculation \n\nE=I*t #Energy \n\nprint \"E=\",E,\"Ah\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "E= 5.18 Ah\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter3.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter3.ipynb new file mode 100644 index 00000000..1467bcbc --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter3.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "chapter3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 3:Power and Energy" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.3,Page no:24" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of Highest voltage drop\n\n#initialization\n\nP=10 #Power in W\n\nI=0.04 #Current in A\n\n#calculation\n\nV=P/I\n\nprint \"V=\",int(V),\"V\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 250 V\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.5,Page no:24" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of power dissipated as heat\n\n#initialization\n\nR=50 #Resistance on Ohm\n\nI=2 #Current in A\n\n#calculation\n\nP=(I**2)*R #Power Equation\n\nprint \"P=\",P,\"W\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "P= 200 W\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.6,Page no:24" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Resistance\n\n#Initialization\n\nP=3000 #Power in W\n\nV=240 #Voltage in V\n\n#calculation\n\nR=(V**2)/P #Another form of power equation\n\nprint \"R=\",R,\"Ohm\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 19.2 Ohm\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.7,Page no:25" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of area of circle\n\n#initialization\n\nr=6 #Radius in inches\n\npi=3.14 #Pi constant Value\n\n#calculation\n\nA=pi*r**2 #Area of Circle Euation in square inches\n\nprint \"A=\",int(A),\"in^2\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "A= 113 in^2\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.8,Page no:25" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#Cal of Volume of a spherical tank\n\n#Initialization\n\nr=10 #Radius in m\n\npi=3.14 #Pi value constant\n\n\n#Calculation\n\nV=math.ceil((4*pi*r**3)/3) #Equation of volume of sphere\n\nprint \"V=\",int(V),\"m^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 4187 m^3\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.11,Page no:26" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Operating voltage\n\n#Initialization\n\nR=18 #resistance in ohm\n\nP=2 #Power in W\n\nimport math\n\n#calculation\n\nV=math.sqrt(P*R)\n\nprint \"V=\",int(V),\"V\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "V= 6 V\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.12,Page no:26" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\nimport math\n\n#cal of Maximum current\n\n#initialization\n\nR=50 #Resistance in Ohm\n\nP=10 #Power in W\n\n#calculation\n\nI=math.sqrt(P/R)\n\nprint \"I=\",round(I,3),\"A\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "I= 0.447 A\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.13,Page no:27" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#cal of radius of wire\n\n#initialization\n\nA=5 #Area in mm^2\n\npi=3.14 #pi constant value\n\nimport math\n\n#calculation\n\nr=math.sqrt(A/pi)\n\nprint \"r=\",round(r,2),\"mm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "r= 1.26 mm\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.15,Page no:28" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#cal of Energy\n\n#initialization\n\nV=240 #Voltage in V\n\nI=15 #Current in A\n\nt=45 #Time in minutes\n\n\n\n#Calculation\n\nP=(I*V)/1000 #Power in kW\n\nt=t/60 #convert time to hours\n\nW=P*t #Work done in kWh\n\nprint \"W=\",round(W,1),\"kWh\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "W= 2.7 kWh\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:3.16,Page no:28" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "import math\n\n#cal of Cost per week\n\n#Initialization\n\nP=75 #Power in W\n\n#Calculation\n\nt=7*24 #Number of hours in a week\n\nP=75/1000 #Power in kWh\n\nW=P*t #Energy used per week\n\nC=12.6*0.09 #Cost at $0.09 per kWh\n\nprint \"Cost per week W =\",\"$\",round(C,2)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Cost per week W = $ 1.13\n" + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter4.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter4.ipynb new file mode 100644 index 00000000..a5fc44d3 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter4.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "chapter4" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 4:Powers of 10 and Logarithms" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:4.17, Page no:37" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Inductive Reactance\n\n#Iniatialization\n\nL=5 #Inductance in mH\n\nf=10 #Frequency in kHz\n\n\n#Calculation\n\nL=5/1000 #Convert Inductance in H\n\nf=10*1000 #Convert Frequency in Hz\n\nXL=2*3.14*f*L #Inductance Formula\n\nprint \"The inductive Reactance of 5mH inductor is\",int(XL),\"Ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The inductive Reactance of 5mH inductor is 314 Ohm\n" + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter5.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter5.ipynb new file mode 100644 index 00000000..bd329287 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter5.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "chapter5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 5:Resistance and Wire Size" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.1, Page no: 46" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Length of wire\n\n#Initialization\n\nL1=20 #Length in ft\n\nR1=13 #Resistance in Ohm\n\nR2=8 #Resistance in Ohm\n\n\n#Calculation\n\nL2=(R2*L1)/R1\n\nprint \"L2=\",round(L2,1),\"ft\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "L2= 12.3 ft\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no.:5.2, Page no: 47" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Resistance\n\n#Initialization\n\nR1=2 #Resistance in Ohm\n\nD1=0.04 #Diameter in inches\n\nD2=0.10 #Diameter in inches\n\n\n\n\n#Cal of Resistance when diameter is 0.10 in\n\nR2=(D1/D2)**2*R1\n\nprint \"(a) The resistance when diameter is 0.10 in is\",R2,\"Ohm\"\n\n\n\n\n#Cal of Resistance when diameter is 0.01 in \n\nD2=0.01 #Diameter in inches\n\nR2=(D1/D2)**2*R1\n\nprint \"(b) The resistance when diameter is 0.01 in is\",int(R2),\"Ohm\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "(a) The resistance when diameter is 0.10 in is 0.32 Ohm\n(b) The resistance when diameter is 0.01 in is 32 Ohm\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.4, Page no:47" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Cross-sectional area\n\n#Initialization\n\nD=0.06408 #Diameter in inches\n\n#Calculation\n\nD=0.06408/0.001 #Convert Diameter in mils\n\nA=D**2\n\nprint \"A_cmil=\",int(A),\"cmil\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "A_cmil= 4106 cmil\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.5, Page no:48" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Comparing the Resistances of wires\n\n#Initialization\n\nA10=10380 #Area in cmil\n\nA14=4106 #Area in cmil\n\nfrom decimal import *\n\n#Calculation\n\nR14_R10=A10/A14\n\nprint \"The no. 14 wire has\",round(R14_R10,2),\"times as much resistance as the no. 10 wire\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The no. 14 wire has 2.53 times as much resistance as the no. 10 wire\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.6, Page no:48" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Resistance\n\n#Initialization\n\nD_mil=0.080 #Diameter in inches\n\nl=1500 #Length in ft\n\nrho=10.4 #Specific resistance in Ohm\n\n#Calculation\n\nD_mil=0.080/0.001 #Diameter in mils\n\nA_cmil=D_mil*D_mil #Area in cmil\n\nR=(rho*l)/A_cmil #Specific Resistance formula\n\nprint \"R=\",round(R,2),\"ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 2.44 ohm\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.7, Page no:48" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Length of Nichrome wire\n\n#Initialization\n\nD_mil=20 #Diameter in inches\n\nrho=600 #Specific resistance in Ohm\n\nR=5 #Resistance in ohm\n\n\n\n#Calculation\n\nA_cmil=D_mil*D_mil #Area in cmil\n\nl=(R*A_cmil)/rho #Specific Resistance formula\n\nprint \"l=\",round(l,2),\"ft\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "l= 3.33 ft\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.8, Page no:48" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of minimum diameter of wire\n\n#initialization\n\nl=400 #Length in ft\n\nR=1.5 #Resistance in ohm\n\nrho=10.4 #Specific resistance for copper wire in cmil/ft\n\nimport math\n\n#Calculation\n\nA_cmil=(rho*l)/R\n\nD_mil=math.sqrt(A_cmil)\n\nprint \"D_mil=\",round(D_mil,1),\"mils\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "D_mil= 52.7 mils\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.14, Page no: 51" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Maximum length\n\n#Initialization\n\nA=6530 #Area in cmil\n\nV=5 #Voltage in V\n\nI=15 #Current in A\n\n#Calculation\n\nl=(A*V)/(10.4*I)\n\nprint \"l=\",int(l),\"ft\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "l= 209 ft\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.15, Page no: 52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Resistance\n\n#Initialization\n\nl=80 #length in m\n\nA=2.5 #Area in mm^2\n\nrho=0.0175 #Specific Resistance of copper wire in mm^2/m\n\n#Calculation\n\nR=(rho*l)/A\n\nprint \"R=\",R,\"Ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 0.56 Ohm\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.16, Page no: 52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Length of wire\n\n#Iniatialization\n\nA=0.1 #Area in mm^2\n\nR=3 #Resistance in Ohm\n\nrho=0.0175 #Specific Resistance of copper wire in mm^2/m\n\n#Calculation\n\nl=(R*A)/rho\n\nprint \"l= \",int(l),\"m\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "l= 17 m\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:5.17, Page no: 52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of cross-sectional area\n\n#Initialization\n\nl=50 #Length in m\n\nR=0.2 #Resistance in ohm\n\nrho=0.0175 #Specific Resistance of copper wire in mm^2/m\n\n#Calculation\n\nA=(rho*l)/R\n\nprint \"A=\",round(A,1),\"mm^2\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "A= 4.4 mm^2\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter6.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter6.ipynb new file mode 100644 index 00000000..aec8aee8 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter6.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "chapter6" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 6:Series Circuits" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:6.1, Page no:54" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Resistance\n\n#Initialization\n\n#Resistances in ohm connected in series\n\nR1=5\n\nR2=5\n\nR3=5\n\n#Calculation\n\nR=R1+R2+R3 #Formula for series resistors\n\nprint \"The Equivalent resistance is\",R,\"ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Equivalent resistance is 15 ohm\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:6.7, Page no:57" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "#Cal of Current\n\n#initialization\n\nV_emf=1.5 #Voltage in V\n\nR=0.4 #Resistance in Ohm\n\nr=0.05 #internal resistance in Ohm\n\n#Calculation\n\nI=V_emf/(R+r) #Formula for current with emf and internal resistance\n\nprint \"The Current in the circuit is\",round(I,2),\"A\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Current in the circuit is 3.33 A\n" + } + ], + "prompt_number": 2 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter7.ipynb b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter7.ipynb new file mode 100644 index 00000000..4aaec8eb --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/chapter7.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "chapter7" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 7:Parallel Circuits" + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": "Problem no:7.5, Page no:67\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "from __future__ import division\n\n#Cal of Equivalent resistance\n\n#Initialization\n\n#Resistors connected in parallel in Ohm\n\nR1=5\n\nR2=10\n\nfrom decimal import *\n\n#Calculation\n\nR=(R1*R2)/(R1+R2)\n\nprint \"R=\",round(R,2),\"ohm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "R= 3.33 ohm\n" + } + ], + "prompt_number": 1 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:13:43.png b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:13:43.png Binary files differnew file mode 100644 index 00000000..3d0f8e9d --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:13:43.png diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:16:31.png b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:16:31.png Binary files differnew file mode 100644 index 00000000..858fc655 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:16:31.png diff --git a/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:17:27.png b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:17:27.png Binary files differnew file mode 100644 index 00000000..05059d00 --- /dev/null +++ b/Basic_Mathematics_for_Electricity_and_Electronics_by_Arthur_Beiser/screenshots/Screenshot_from_2016-01-14_16:17:27.png diff --git a/Electrical_And_Electronics_Engineering_Materials_by_J._B._Gupta/README.txt b/Electrical_And_Electronics_Engineering_Materials_by_J._B._Gupta/README.txt new file mode 100644 index 00000000..bd63408b --- /dev/null +++ b/Electrical_And_Electronics_Engineering_Materials_by_J._B._Gupta/README.txt @@ -0,0 +1,10 @@ +Contributed By: Suhaib Alam +Course: btech +College/Institute/Organization: Phonics Group of Institutions +Department/Designation: ME +Book Title: Electrical And Electronics Engineering Materials +Author: J. B. Gupta +Publisher: S.k. Katariya & Sons, New Delhi +Year of publication: 2010 +Isbn: 81-89757-13-x +Edition: 3
\ No newline at end of file diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter1.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter1.ipynb new file mode 100644 index 00000000..9816da2d --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter1.ipynb @@ -0,0 +1,800 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 1 : Wave Optics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1, Page Number 14" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of sheet= 1178*10**-8 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5890*10**-8 #wavelength\n", + "myu=1.6 #refractive index\n", + "m=12 #order of the fringe\n", + "\n", + "#Calculations\n", + "t=(lamda*m)/(myu-1)/10**-6\n", + "\n", + "#Result\n", + "print\"Thickness of sheet= %i*10**-8 cm\" %t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2, Page Number 15" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of sheet= 1071.45*10**-8 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5893*10**-8 #wavelength\n", + "myu=1.55 #refractive index\n", + "n=10 #order of the fringe\n", + "\n", + "#Calculations\n", + "t=(lamda*n)/(myu-1)/(10**-3)*10**3\n", + "\n", + "#Result\n", + "print\"Thickness of sheet= %0.2f*10**-8 cm\" %t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3, Page Number 15" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angular position of first minima is= 0.16\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5640*10**-8 #wavelength\n", + "d=0.01 #distnce between slits\n", + "n=0 #first minimum\n", + "\n", + "#Calculations\n", + "theta=(n+(1/2))*(lamda/d)\n", + "\n", + "#Result\n", + "print\"Angular position of first minima is= %0.2f\" %math.degrees(theta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4, Page Number 16" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "minimum thickness of the film t= 3.927*10**-5 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5890*10**-8 #wavelength\n", + "myu=1.5 #refractive index of glass\n", + "n=1 #first minimum\n", + "r=60 #angle in degrees\n", + "\n", + "#Calculations\n", + "t=(n*lamda)/(2*myu*0.5)/10**-5\n", + "\n", + "#Result\n", + "print\"minimum thickness of the film t= %1.3f*10**-5 cm\" %t #The answer provided in the textbook is incorrect" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5, Page Number 16" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)For the first order wavelength= 9.58*10**-5 cm\n", + "(ii)For the second order wavelength= 4.79*10**-5 cm\n", + "(iii)For the third order wavelength= 3.19*10**-5 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "i=35 #incident angle in degrees\n", + "myu=1.33 #refractive index \n", + "n=1 #first minimum\n", + "t=4*10**-5 #thickness\n", + "\n", + "#Calculations\n", + "cos_r=0.90\n", + "lamda1=2*myu*t*cos_r/10**-5 #for first order n=1\n", + "lamda2=(2*myu*t*cos_r)/2/10**-5 #for second order n=2\n", + "lamda3=(2*myu*t*cos_r)/3/10**-5 #for third order n=3\n", + "\n", + "#Result\n", + "print\"(i)For the first order wavelength= %1.2f*10**-5 cm\" %lamda1 #The answer provided in the textbook is incorrect\n", + "print\"(ii)For the second order wavelength= %1.2f*10**-5 cm\" %lamda2\n", + "print\"(iii)For the third order wavelength= %1.2f*10**-5 cm\" %lamda3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6, Page Number 16" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Thickness of the film t= 3.927*10**-4 mm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5890*10**-10 #wavelength\n", + "myu=1.5 #refractive index of glass\n", + "n=1 #first minimum\n", + "r=60 #angle in degrees\n", + "\n", + "#Calculations\n", + "t=(n*lamda)/(2*myu*0.5)/10**-7\n", + "\n", + "#Result\n", + "print\"Thickness of the film t= %1.3f*10**-4 mm\" %t " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 7, Page Number 17" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thickness of the film t= 3.132*10**-4 mm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5890*10**-10 #wavelength\n", + "myu=1.33 #refractive index of glass\n", + "n=1 #first minimum\n", + "r=45 #angle in degrees\n", + "cos_r=0.707\n", + "\n", + "#Calculations\n", + "t=(n*lamda)/(2*myu*cos_r)/10**-7\n", + "\n", + "#Result\n", + "print\"thickness of the film t= %1.3f*10**-4 mm\" %t " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 8, Page Number 17" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thickness of the film t= 1.2*10**-5 m\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5.1*10**-7 #wavelength\n", + "myu=4/3 #refractive index of glass\n", + "r=45 #angle in degrees\n", + "sin_i=4/5\n", + "n=50\n", + "\n", + "#Calculations\n", + "sin_r=sin_i/myu\n", + "cos_r=(1-sin_r**2)**0.5\n", + "t=(n*lamda)/(2*myu*cos_r)/10**-5\n", + "\n", + "#Result\n", + "print\"thickness of the film t= %1.1f*10**-5 m\" %t " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 9, Page Number 18" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "order of the dark ring n1= 80\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "n2=20 #order of dark ring\n", + "lamda=5890*10**-8 #wavelenght\n", + "\n", + "#Calculations\n", + "n1=n2*4\n", + "\n", + "#Result\n", + "print\"order of the dark ring n1= %i\" %n1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 10 , Page Number 18" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radius of curvature= 99.8 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "Dm=0.590 #diameter of ring 15\n", + "Dn=0.336 #diameter of ring 5\n", + "lamda=5890*10**-8\n", + "m=15\n", + "n=5\n", + "\n", + "#Calculation\n", + "R=(Dm**2-Dn**2)/(4*lamda*(m-n))\n", + "\n", + "#Result\n", + "print\"Radius of curvature= %0.1f cm\" %R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 11 , Page Number 19" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "diameter of the nth ring= 0.2546 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda1=6000*10**-8 \n", + "lamda2=4500*10**-8\n", + "R=90 #Radius of curvature\n", + "n=3\n", + "\n", + "#Calculation\n", + "dn=math.sqrt(4*n*lamda1*R)\n", + "\n", + "#Result\n", + "print\"diameter of the nth ring= %0.4f cm\" %dn\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 12, Page Number 19" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radius of curvature= 105.93 cms\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "Dm=0.50 #diameter of ring 1\n", + "lamda=5900*10**-8 #wavelenght\n", + "m=10\n", + "\n", + "#Calculation\n", + "R=(Dm**2)/(4*lamda*m)/10**2*10**2\n", + "\n", + "#Result\n", + "print\"Radius of curvature= %0.2f cms\" %R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 13, Page Number 20" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refractive index of liquid= 1.31 \n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "dn=0.3 #diameter of ring 5\n", + "lamda=5.895*10**-5 #wavelenght\n", + "R=100\n", + "n=5\n", + "\n", + "#Calculation\n", + "myu=(4*R*n*lamda)/dn**2\n", + "\n", + "#Result\n", + "print\"Refractive index of liquid= %0.2f \" %myu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 14, Page Number 20" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refractive index of liquid= 1.361 \n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "D2=1.40 \n", + "D1=1.20 \n", + "\n", + "#Calculation\n", + "myu=(D2/D1)**2\n", + "\n", + "#Result\n", + "print\"Refractive index of liquid= %1.3f \" %myu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 15, Page Number 20" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Radius of curvature= 139 cm\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "myu=4/3\n", + "Dn=0.5 #diameter of 10th ring\n", + "lamda=6*10**-5\n", + "n=10\n", + "\n", + "#Calculation\n", + "R=(myu*Dn**2)/(4*n*lamda)\n", + "\n", + "#Result\n", + "print\"Radius of curvature= %1.0f cm\" %R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 16, Page Number 21" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "refractive index of liquid= 1.441 \n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "myu=4/3\n", + "Dn=0.5 #diameter of 10th ring\n", + "lamda=5895*10**-8\n", + "n=6\n", + "R=100\n", + "r=0.15\n", + "\n", + "#Calculation\n", + "myu=(((2*n)-1)*lamda*R)/(2*r**2)\n", + "\n", + "#Result\n", + "print\"refractive index of liquid= %1.3f \" %myu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 17, Page Number 21" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "wavelength of liquid used= 5880 Armstrong\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "D15=5.90*10**-3 #diameter of 15th ring\n", + "D5=3.36*10**-3 #diameter of 5th ring\n", + "m=10\n", + "R=100\n", + "\n", + "#Calculation\n", + "lamda=(D15**2-D5**2)/(4*m*R)/10**-9*10**3\n", + "\n", + "#Result\n", + "print\"wavelength of liquid used= %i Armstrong\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 18, Page Number 21" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "refractive index of liquid= 1.331\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "D1=1.50\n", + "D2=1.30\n", + "\n", + "#Calculation\n", + "myu=(D1/D2)**2\n", + "\n", + "#Result\n", + "print\"refractive index of liquid= %1.3f\" %myu" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 19 , Page Number 22" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)radius of curvature R= 4.583051 m\n", + "(ii)Thickness of air film t= 2.95*10**-6 m\n" + ] + } + ], + "source": [ + "#importing modules \n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=5.9*10**-7\n", + "r=5.2*10**-3 #radius of ring\n", + "n=10\n", + "\n", + "#Calculation\n", + "R=(r**2)/(n*lamda)\n", + "t=(n*lamda)/2/10**-6\n", + "\n", + "#Result\n", + "print\"(i)radius of curvature R= %f m\" %R\n", + "print\"(ii)Thickness of air film t= %1.2f*10**-6 m\" %t" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter10.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter10.ipynb new file mode 100644 index 00000000..4944ce31 --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter10.ipynb @@ -0,0 +1,355 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 : Crystallography and Crystal Imperfections" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 178" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "lattice parameter a= 4.075*10**-8 cm\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "rho=10.6 #density of material\n", + "n=4 #No. of atoms/cell\n", + "A=108 #Atomic weigth\n", + "No=6.023*10**23 #Avagadro's No.\n", + "\n", + "#Calculations\n", + "a=((n*A)/(No*rho))**(1/3)/10**-8\n", + "\n", + "#Result\n", + "print\"lattice parameter a= %1.3f*10**-8 cm\" %a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 179" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "lattice parameter a= 2.87*10**-10 m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "rho=7.86*10**3 #density of material\n", + "n=2 #No. of atoms/cell\n", + "A=55.85 #Atomic weigth\n", + "No=6.023*10**26 #Avagadro's No.\n", + "\n", + "#Calculations\n", + "a=((n*A)/(No*rho))**(1/3)/10**-10\n", + "\n", + "#Result\n", + "print\"lattice parameter a= %1.2f*10**-10 m\" %a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 179" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Distance between like atoms a= 5.628*10**-10 m\n", + "Distance between adjacent atoms a/2= 2.81*10**-10 m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "M=58.5 #formula weigth\n", + "rho=2180 #density of material\n", + "n=4 #No. of atoms/cell \n", + "No=6.02*10**26 #Avagadro's No.\n", + "\n", + "#Calculations\n", + "a=((n*M)/(No*rho))**(1/3)/10**-10/2\n", + "\n", + "#Result\n", + "print\"Distance between like atoms a= %1.3f*10**-10 m\" %(2*a)\n", + "print\"Distance between adjacent atoms a/2= %1.2f*10**-10 m\" %a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 179" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lattice constant a= 5.64*10**-10 m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "M=58.45 #formula weigth\n", + "rho=2.17*10**3 #density of material\n", + "n=4 #No. of atoms/cell \n", + "No=6*10**26 #Avagadro's No.\n", + "\n", + "#Calculations\n", + "a=((n*M)/(No*rho))**(1/3)/10**-10\n", + "\n", + "#Result\n", + "print\"Lattice constant a= %1.2f*10**-10 m\" %a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5 , Page number 180" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lattice constant a= 3.615 Armstrong\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "r=1.278 #atomic weigth \n", + "No=6.02*10**26 #Avagadro's No.\n", + "\n", + "#Calculations\n", + "a=2*math.sqrt(2)*r\n", + "\n", + "#Result\n", + "print\"Lattice constant a= %1.3f\" %a,\"Armstrong\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6 , Page number 180" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Density of the material= 8.94 gm/cc\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "r=1.278 #atomic weigth \n", + "N0=6.02*10**23 #Avagadro's No.\n", + "no=4 #No. of atoms/cell\n", + "A=63.54 #Atomic weigth\n", + "\n", + "#Calculations\n", + "a=2*math.sqrt(2)*r\n", + "rho=(no*A)/(N0*a**3)/10**-24\n", + "\n", + "#Result\n", + "print\"Density of the material= %1.2f\" %rho,\"gm/cc\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 7 , Page number 180" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Miller indices are ( 3.0 , 2.0 , 2.0 )\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "x=2 #x intercept \n", + "y=3 #y intercept\n", + "z=3 #z intercept\n", + "\n", + "#Calculations \n", + "rx=(1/x)*6 #reciprocal of x intercept\n", + "ry=(1/y)*6 #reciprocal of y intercept\n", + "rz=(1/z)*6 #reciprocal of z intercept\n", + "\n", + "#Result\n", + "print\"Miller indices are (\",rx,\",\",ry,\",\",rz,\")\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 8 , Page number 181" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "inter-planar spacing d= 1.05 Armstrong\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "h=1\n", + "k=1\n", + "l=2\n", + "a=b=2.5\n", + "c=2.6\n", + "\n", + "#Calculations \n", + "d=((h**2/a**2)+(k**2/b**2)+(l**2/c**2))**(-0.5)\n", + "\n", + "#Result\n", + "print\"inter-planar spacing d= %1.2f\" %d,\"Armstrong\"" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter11.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter11.ipynb new file mode 100644 index 00000000..48ce030b --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter11.ipynb @@ -0,0 +1,231 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 : Free Electron Theory Of Metals" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , page number 205 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "root mean square velocity v= 1.17*10**5 m/s\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "k=1.376*10**-23 #Boltzmann's constant in J/K\n", + "T=300 #Temperature\n", + "m=9.11*10**-31 #Mass of electron\n", + "\n", + "#Calculations\n", + "v=math.sqrt((3*k*T)/m)/10**5\n", + "\n", + "#Result\n", + "print\"root mean square velocity v= %1.2f*10**5 m/s\" %v" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , page number 205 " + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mean free path for electron= 6.6*10**-9 m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "sigma=6.8*10**7 #conductivity\n", + "n=8.5*10**28 #number of electrons\n", + "m=9.1*10**-31 #Mass of electron\n", + "e=1.6*10**-19 #charge on electron\n", + "k=1.38*10**-23 #Boltzmann's constant in J/K\n", + "T=300 #temperature in K\n", + "\n", + "#Calculations\n", + "lamda=(2*sigma*math.sqrt(3*m*k*T))/(n*e**2)/10**-9\n", + "\n", + "#Result\n", + "print\"Mean free path for electron= %1.1f*10**-9 m\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , page number 205 " + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relaxation time= 3.97*10**-14 seconds\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "rho=1.54*10**-8 #resistivity\n", + "n=5.8*10**28 #electron density\n", + "e=1.602*10**-19 #charge on electron\n", + "m=9.1*10**-31 #Mass of electron\n", + "\n", + "#Calculations\n", + "tau=m/(n*(e**2)*rho)/10**-14\n", + "\n", + "#Result\n", + "print\"Relaxation time= %1.2f*10**-14 seconds\" %tau" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , page number 206 " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "No. of free electrons per unit volume= 8.395*10**28 electrons per meter**3\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "EF=1.1214*10**-18 #fermi energy in J\n", + "m=9.11*10**-31 #Mass of electron\n", + "h=6.63*10**-34 #planck's constant\n", + "\n", + "#Calculations\n", + "n=((8*m*EF)/(h**2))**(3/2)*(math.pi/3)/10**28\n", + "\n", + "#Result\n", + "print\"No. of free electrons per unit volume= %1.3f*10**28 electrons per meter**3\" %n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5 , page number 206 " + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature= 1261 K\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "import numpy as np\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "fE=0.01 #probability\n", + "delE=8*10**-20 #ev to J\n", + "\n", + "#Calculations\n", + "T=5797/np.log(99)\n", + "\n", + "#Result\n", + "print\"Temperature= %i\" %T,\"K\"" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter12.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter12.ipynb new file mode 100644 index 00000000..c9f7db1e --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter12.ipynb @@ -0,0 +1,321 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 : Semiconductor Physics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1, Page number 228" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistivity = 0.3846 ohm-m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "ni=2.5*10**19 #intrinsic concentration \n", + "myun=0.40 #mobility of electrons \n", + "myup=0.25 #mobility of holes\n", + "e=1.6*10**-19\n", + "\n", + "#Calculations\n", + "sigmai=ni*e*(myun+myup) #conductivity of intrinsic semiconductor\n", + "rhoi=1/sigmai\n", + "\n", + "#Result\n", + "print\"Resistivity =\",round(rhoi,4),\"ohm-m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2, Page number 228" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intrinsic concentration= 5.68*10**18 m**-3\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "myun=0.36 #mobility of electrons \n", + "myup=0.14 #mobility of holes\n", + "e=1.6*10**-19 \n", + "rhoi=2.2 #resistivity\n", + "\n", + "#Calculations\n", + "ni=1/(rhoi*e*(myun+myup))/10**18\n", + "\n", + "#Result\n", + "print\"Intrinsic concentration= %1.2f*10**18 m**-3\" %ni" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3, Page number 228" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Conductivity = 2.4 ohm**-1-m**-1\n", + "resistivity= 0.42 ohm-m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "myun=0.39 #mobility of electrons \n", + "myup=0.21 #mobility of holes\n", + "ni=2.5*10**19 #intrinsic concentration \n", + "e=1.6*10**-19 \n", + "\n", + "#Calculations\n", + "sigmai=ni*e*(myun+myup) #conductivity of intrinsic semiconductor\n", + "rhoi=1/sigmai\n", + "\n", + "#Result\n", + "print\"Conductivity = %1.1f\" %sigmai,\"ohm**-1-m**-1\"\n", + "print\"resistivity= %1.2f\" %rhoi,\"ohm-m\" #the answer provided in the textbook is incorrect" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4, Page number 228" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "concentration of intrinsic charge= 4.83*10**18 /m**3\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "Eg=0.8 #Energy gap width\n", + "T=300\n", + "m=9.1*10**-31 #mass of electron\n", + "k=1.38*10**-23\n", + "h=6.63*10**-34 \n", + "\n", + "#Calculations\n", + "ni=2*((2*22*m*k*T)/(7*h**2))**(3/2)*math.exp((-Eg*1.6*10**-19)/(2*k*T))/10**18\n", + "\n", + "#Result\n", + "print\"concentration of intrinsic charge= %1.2f*10**18 /m**3\" %ni #the answer provided in the textbook is incorrect" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5, Page number 229" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "hole concentration= 19.4*10**21 m**-3\n", + "mobility of holes= 0.0358 m**2 V**-1 s**-1\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "RH=3.22*10**-4 #Hall coefficient\n", + "rho=9.0*10**-3\n", + "e=1.6*10**-19\n", + "\n", + "#Calculations\n", + "p=1/(RH*e)/10**21\n", + "myup=RH/rho\n", + "\n", + "#Result\n", + "print\"hole concentration= %1.1f*10**21 m**-3\" %p\n", + "print\"mobility of holes= %1.4f m**2 V**-1 s**-1\" %myup" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6, Page number 229" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hall voltage VH= 0.183 V\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "RH=3.66*10**-4 #Hall coefficient\n", + "t=10**-3 #thickness\n", + "I=1 #current\n", + "B=0.5 #magnetic induction\n", + "\n", + "#Calculations\n", + "VH=(RH*I*B)/t\n", + "\n", + "#Result\n", + "print\"Hall voltage VH= %1.3f\" %VH,\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 7, Page number 300" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "charge density= 8.33*10**22 /m**3\n", + "mobility= 0.015 m**2 V**-1 s**-1\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "RH=7.5*10**-5 #Hall coefficient\n", + "sigma=200 #conductivity\n", + "e=1.6*10**-19 #electron charge\n", + "\n", + "#Calculations\n", + "n=1/(e*RH)/10**22\n", + "myu=sigma*RH\n", + "\n", + "#Result\n", + "print\"charge density= %1.2f*10**22 /m**3\" %n\n", + "print\"mobility= %0.3f m**2 V**-1 s**-1\" %myu" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter13.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter13.ipynb new file mode 100644 index 00000000..485f024f --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter13.ipynb @@ -0,0 +1,198 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13 : Thin Film Preparation Techniques and their Applications" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 248" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Series Resistance = 500.0 V/m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "delV1=2*10**-3 #milivolts to volts \n", + "delI1=4*10**-6 #microAmpere to Ampere\n", + "\n", + "#Calculations\n", + "Rs=delV1/delI1\n", + "\n", + "#Result\n", + "print\"Series Resistance =\",Rs,\"V/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 248" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in conductivity = 5.0 *10**-3 Ohm**-1-cm**-1\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "I=2*10**-3\n", + "V=1\n", + "\n", + "#Calculations\n", + "rho=(V/I)*2\n", + "delR=rho-800 #change in resitance\n", + "A=1/delR #change in conductance\n", + "\n", + "#Result\n", + "print\"Change in conductivity =\",round(A*10**3),\"*10**-3 Ohm**-1-cm**-1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 248" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Fill factor = 0.42\n", + "Percentage of efficieny = 15.0 %\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "Pmax=21*10**-3 #maximum power output\n", + "Isc=100*10**-3 #short circuit voltage\n", + "Voc=500*10**-3 #open circuit voltage\n", + "Pin=35*10**-3 #Power input\n", + "A=4 #area of solar cell\n", + "\n", + "#Calculations\n", + "Fill_Factor=Pmax/(Isc*Voc)\n", + "n=(Pmax/(Pin*A))*100\n", + "\n", + "#Result\n", + "print\"Fill factor =\",round(Fill_Factor,2)\n", + "print\"Percentage of efficieny =\",n,\"%\" #TextBook has calculation error" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 249" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Isc = 100.0 mA\n", + "Percentage of efficieny = 17.14 %\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "Pmax=18*10**-3 #maximum power output\n", + "F=0.6 #fill factor\n", + "Voc=300*10**-3 #open circuit voltage\n", + "Pin=21*10**-3 #Power input\n", + "A=5 #area of solar cell\n", + "\n", + "#Calculations\n", + "Isc=Pmax/(F*Voc)\n", + "n=(Pmax/(Pin*A))*100\n", + "\n", + "#Result\n", + "print\"Isc =\",Isc*1000,\"mA\"\n", + "print\"Percentage of efficieny =\",round(n,2),\"%\" " + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter15.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter15.ipynb new file mode 100644 index 00000000..a02a528f --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter15.ipynb @@ -0,0 +1,76 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 : Dielectric Material" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 282" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electric field strength = 3.529*10**7 V/m\n", + "Total dipole moment = 33.5*10**-12 C-m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "V=15 #potential difference\n", + "C=6*10**-6 #Capacitance\n", + "epsilon0=8.854*10**-12 #absloute permittivity\n", + "epsilonr=8 #relative permittivity\n", + "A=360*10**-4 #surface Area\n", + "\n", + "#Calculations\n", + "E=(V*C)/(epsilon0*epsilonr*A)/10**7\n", + "T=epsilon0*(epsilonr-1)*V*A/10**-12\n", + "\n", + "#Result\n", + "print\"Electric field strength = %1.3f*10**7 V/m\" %E\n", + "print\"Total dipole moment = %1.1f*10**-12 C-m\" %T\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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter2.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter2.ipynb new file mode 100644 index 00000000..bfcd4dce --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter2.ipynb @@ -0,0 +1,508 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 : Diffraction" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 46" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "width of the slit a= 1.05 micro-m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1 #first minimum\n", + "lamda=6000*10**-10 #waveleght\n", + "theta=math.radians(35) #angle in radians\n", + "\n", + "#Calculations\n", + "a=(m*lamda)/math.sin(theta)/10**-6\n", + "\n", + "#Result\n", + "print\"width of the slit a= %1.2f micro-m\" %a" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 46" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle of first minimum theta= 18.97 degrees\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1 #first minimum\n", + "lamda=6500*10**-10 #waveleght\n", + "a=2*10**-6 #slit width\n", + "\n", + "#Calculations\n", + "theta=math.degrees(math.asin((m*lamda)/a))\n", + "\n", + "#Result\n", + "print\"Angle of first minimum theta= %2.2f degrees\" %theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 46" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of incident light lamda= 5510 Armstrong\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "m=1 #first minimum\n", + "a=90*10**-16 #slit width\n", + "y=6*10**-3 #distance from central maximum\n", + "D=0.98 #Screen distance\n", + "\n", + "#Calculations\n", + "lamda=(y*a)/D/10**-17*10**3\n", + "\n", + "#Result\n", + "print\"Wavelength of incident light lamda= %i Armstrong\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 46" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total linear width= 1.2 cm\n", + "angular position of the minima= 3*10**-3 radian\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "a=2*10**-4 #slit width \n", + "lamda=6*10**-7 #waveleght\n", + "\n", + "#Calculations\n", + "theta=math.asin(lamda/a)\n", + "TLW=4*theta/10**-2\n", + "theta1=math.asin(lamda/a)/10**-3\n", + "\n", + "#Result\n", + "print\"Total linear width= %1.1f cm\" %TLW\n", + "print\"angular position of the minima= %i*10**-3 radian\" %theta1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5 , Page number 47" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "angle between 1st and 2nd order line is 24.95 degrees\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "n1=1\n", + "n2=2 \n", + "lamda=6000*10**-8 #waveleght\n", + "N=6000 #number of lines for diffraction grating\n", + "\n", + "#Calculations\n", + "theta1=math.degrees(math.asin(n1*lamda*N))\n", + "theta2=math.degrees(math.asin(n2*lamda*N))\n", + "\n", + "#Result\n", + "print\"angle between 1st and 2nd order line is %2.2f degrees\" %(theta2-theta1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6 , Page number 47" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "angular separation= 0.06 degrees\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "lamda1=5890*10**-8 #waveleght\n", + "lamda2=5896*10**-8 #waveleght\n", + "N=6000 #number of lines for diffraction grating\n", + "\n", + "#Calculations\n", + "theta1=math.degrees(math.asin(2*lamda1*N))\n", + "theta2=math.degrees(math.asin(2*lamda2*N))\n", + "\n", + "#Result\n", + "print\"angular separation= %2.2f degrees\" %(theta2-theta1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 7 , Page number 48" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "dispersive power in third order spectum= 36.87 degrees\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "lamda=5000*10**-8 #waveleght\n", + "N=4000 #number of lines for diffraction grating\n", + "n=3 #third order \n", + "\n", + "#Calculations\n", + "theta=math.degrees(math.asin(n*lamda*N))\n", + "\n", + "#Result\n", + "print\"dispersive power in third order spectum= %2.2f degrees\" %(theta)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 8 , Page number 48" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "wavelength lamda= 5000 Armstrong\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "N=5000 #number of lines for diffraction grating\n", + "n=2 #second order \n", + "theta2=math.radians(30) #angle in radians\n", + "\n", + "#Calculations\n", + "lamda=math.sin(theta2)/(n*N)/10**-5*10**3\n", + "\n", + "#Result\n", + "print\"wavelength lamda= %1.0f Armstrong\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 9 , Page number 48" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of grating lines= 327.4\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "lamda=5893*10**-8 #wavelenght\n", + "dlamda=6*10**-8\n", + "n=3 #third order\n", + "\n", + "#Calculations\n", + "N=lamda/(n*dlamda)\n", + "\n", + "#Result\n", + "print\"Number of grating lines= %0.1f\" %N" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 10 , Page number 49" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Grating element= 2.511*10**-6 m\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "lamda=6.5*10**-7 #wavelenght\n", + "n=1 #first order\n", + "theta=math.radians(15) #angle in radians\n", + "\n", + "#Calculations\n", + "d=(n*lamda)/math.sin(theta)/10**-6\n", + "\n", + "#Result\n", + "print\"Grating element= %1.3f*10**-6 m\" %d" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 11 , Page number 49" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "lamda= 6656 Armstrong\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "lamda2=4992 #wavelenght\n", + "\n", + "#Calculations\n", + "lamda=(4*lamda2)/3\n", + "\n", + "#Result\n", + "print\"lamda= %i Armstrong\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 12 , Page number 49" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Grating element= 3.24*10**-4 cm\n", + "Number of lines in 1 cm lenght of grating= 3086\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration \n", + "theta=math.radians(30) #angle in radians\n", + "lamda1=5400*10**-10 \n", + "n=3 #third order\n", + "\n", + "#Calculations\n", + "d=(n*lamda1)/math.sin(theta)*10**2/10**-4\n", + "N1=1/d/10**-1*10**3\n", + "\n", + "#Result\n", + "print\"Grating element= %1.2f*10**-4 cm\" %d\n", + "print\"Number of lines in 1 cm lenght of grating= %i\" %N1" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter3.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter3.ipynb new file mode 100644 index 00000000..8338e679 --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter3.ipynb @@ -0,0 +1,305 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Polarization" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 65" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage of light that passes through= 25.0 %\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "theta=math.radians(60) #angle in radians\n", + "\n", + "#Calculations\n", + "Intensityred=100-(1-math.cos(theta)**2)*100\n", + "\n", + "#Result\n", + "print\"Percentage of light that passes through= \",Intensityred,\"%\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 66" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of the two intensities Ie:Io = 3:1\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6000 #wavlenght in Armstrong\n", + "\n", + "#Calculations\n", + "Ie=3/4\n", + "Io=1/4\n", + "Ratio=Ie/Io\n", + "\n", + "#Result\n", + "print\"Ratio of the two intensities Ie:Io = 3:1\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 67" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Angle of polarization= 57.171 degrees\n", + "Angle between reflected and refracted ray= 90 degrees\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "myu=1.55 #refractive index of glass\n", + "\n", + "#Calculations\n", + "theta_p=math.degrees(math.atan(myu))\n", + "theta_r=math.degrees(math.asin(math.sin(math.radians(theta_p))/1.55))\n", + "Total=theta_p+theta_r\n", + "\n", + "#Result\n", + "print\"Angle of polarization= %2.3f degrees\" %theta_p\n", + "print\"Angle between reflected and refracted ray= %i degrees\" %Total" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 67" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thickness of a quarterwave plate= 1.67*10**-5 m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6000*10**-8 #wavelength\n", + "no=1.544 #refractive index of O-ray\n", + "ne=1.553 #refractive index of E-ray\n", + "\n", + "#Calculations\n", + "t=lamda/(4*(ne-no))/10**-3\n", + "\n", + "#Result\n", + "print\"thickness of a quarterwave plate= %1.2f*10**-5 m\" %t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5 , Page number 67" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thickness of a quarterwave plate= 0.001\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6000*10**-8 #wavelength\n", + "no=1.54 #refractive index of O-ray\n", + "ne=1.55 #refractive index of E-ray\n", + "\n", + "#Calculations\n", + "t=lamda/(6*(ne-no))\n", + "\n", + "#Result\n", + "print\"thickness of a quarterwave plate= %1.3f\" %t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6 , Page number 68" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "thickness of a quarterwave plate= 0.003 cm\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6000*10**-8 #wavelength\n", + "no=1.54 #refractive index of O-ray\n", + "ne=1.55 #refractive index of E-ray\n", + "\n", + "#Calculations\n", + "t=lamda/(2*(ne-no))\n", + "\n", + "#Result\n", + "print\"thickness of a quarterwave plate= %1.3f cm\" %t" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 7 , Page number 68" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The solution is of 10%\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "l=2 #length of the tube\n", + "s=60 #specific rotation\n", + "theta=12 #angle of rotation of plane vibration\n", + "\n", + "#Calculations\n", + "C=theta/(l*s)*100\n", + "\n", + "#Result\n", + "print\"The solution is of %i%%\" %C" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter4.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter4.ipynb new file mode 100644 index 00000000..1c009a09 --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter4.ipynb @@ -0,0 +1,153 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Laser and Holography" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 81" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of radiation= 414 nm\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "delE=3*1.6*10**-19 #Energy of laser\n", + "h=6.63*10**-34 #planck's constant\n", + "c=3*10**8 #speed of ligth\n", + "\n", + "#Calculations\n", + "lamda=(h*c)/delE/10**-9\n", + "\n", + "#Result\n", + "print\"Wavelength of radiation= %i nm\" %lamda" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 81" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Energy of radiation= 1.96 eV\n", + "Momentum of electron= 1.048*10**-27 kgm/s\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6328*10**-10 #wavelength\n", + "h=6.63*10**-34 #planck's constant\n", + "c=3*10**8 #speed of ligth\n", + "\n", + "#Calculations\n", + "E=((h*c)/lamda)/1.6*10**19\n", + "Momentum=h/lamda/10**-27\n", + "\n", + "#Result\n", + "print\"Energy of radiation= %1.2f eV\" %E\n", + "print\"Momentum of electron= %1.3f*10**-27 kgm/s\" %Momentum" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 82" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Wavelength of radiation= 338*10**15 photons/sec\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "lamda=6730*10**-18 #wavelength\n", + "h=6.63*10**-34 #planck's constant\n", + "c=3*10**8 #speed of ligth\n", + "P=10**-3 #Power of laser\n", + "\n", + "#Calculations\n", + "n=(P*lamda)/(h*c)/10**5\n", + "\n", + "#Result\n", + "print\"Wavelength of radiation= %i*10**15 photons/sec\" %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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter5.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter5.ipynb new file mode 100644 index 00000000..e08b7c5a --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter5.ipynb @@ -0,0 +1,237 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 : Fibre Optics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 90" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Critical angle= 78.52 degrees\n", + "numerical aperture= 0.298\n", + "acceptance angle= 17.367 degrees\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "n1=1.5 #core refractive index\n", + "n2=1.47 #clad refractive index\n", + "\n", + "#Calculations\n", + "thetac=math.asin(n2/n1)\n", + "NA=(n1**2-n2**2)**0.5\n", + "im=math.asin(NA)\n", + "\n", + "#Result\n", + "print\"Critical angle= %2.2f\" %math.degrees(thetac),\"degrees\"\n", + "print\"numerical aperture= %1.3f\" %NA\n", + "print\"acceptance angle= %2.3f\" %math.degrees(im),\"degrees\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 91" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "refractive index of cladding= 1.587\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "n1=1.6 #core refractive index\n", + "NA=0.2 #Numerical aperture\n", + "\n", + "#Calculations\n", + "NA=(n1**2-NA**2)**0.5\n", + "\n", + "#Result\n", + "print\"refractive index of cladding= %1.3f\" %NA" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 91" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "numerical aperture= 0.252\n", + "Acceptance angle= 14.61 degrees\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "n1=1.6 #core refractive index\n", + "n2=1.58 #clad refractive index\n", + "\n", + "#Calculations\n", + "NA=(n1**2-n2**2)**0.5\n", + "im=math.asin(NA)\n", + "\n", + "#Result\n", + "print\"numerical aperture= %1.3f\" %NA\n", + "print\"Acceptance angle= %2.2f\" %math.degrees(im),\"degrees\" #The answer in the textbook is mathematically incorrect" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 91" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "refractive index of core material= 1.42\n", + "refractive index of cladding material= 1.40\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "delr=12*10**-3 #fractional refractive index change\n", + "NA=0.22 #Numerical aperture\n", + "\n", + "#Calculations\n", + "n1=NA/math.sqrt(2*delr)\n", + "n2=n1-(n1*delr)\n", + "\n", + "#Result\n", + "print\"refractive index of core material= %1.2f\" %n1\n", + "print\"refractive index of cladding material= %1.2f\" %n2" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 5 , Page number 92" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "acceptance angle= 8.6 degrees\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "NA=0.2 #Numerical aperture\n", + "n0=1.33 #refractive index\n", + "n2=1.59 #clad refractive index\n", + "\n", + "#Calculations\n", + "n1=math.sqrt(NA**2+n2**2)\n", + "NA1=math.sqrt(n1**2-n2**2)/n0\n", + "thetac=math.asin(NA1)\n", + "\n", + "#Result\n", + "print\"acceptance angle= %1.1f\" %math.degrees(thetac),\"degrees\"" + ] + } + ], + "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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter9.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter9.ipynb new file mode 100644 index 00000000..f57d114e --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/Chapter9.ipynb @@ -0,0 +1,250 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 : Electromagnetic Theory " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1 , Page number 141" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Amplitude of field E= 30.0 V/m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "P=60 #Power\n", + "r=2 #distance from source\n", + "epsilon0=8.85*10**-12\n", + "C=3*10**2\n", + "\n", + "#Calculations\n", + "E0=math.sqrt((P*2)/(4*math.pi*r**2*C*epsilon0))/1000\n", + "\n", + "#Result\n", + "print\"Amplitude of field E= \",round(E0),\"V/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 2 , Page number 142" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intensity of Beam= 4*10**2 W\n", + "E0= 549 V/m\n", + "B0= 1.83 myu-T\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "P=8*10**-4 #Power\n", + "A=2*10**-6 #cross-sectional Area\n", + "epsilon0=8.85*10**-12\n", + "C=3*10**2\n", + "\n", + "#Calculations\n", + "I=P/A/100\n", + "E0=math.sqrt((2*I)/(C*epsilon0))/100\n", + "B0=E0/C\n", + "\n", + "#Result\n", + "print\"Intensity of Beam= %i*10**2 W\" %I\n", + "print\"E0= %i\" %round(E0),\"V/m\"\n", + "print\"B0= %1.2f\" %B0,\"myu-T\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 3 , Page number 142" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Poynting vector S= 1.59*10**-4 W/m**2\n", + "E0= 0.346 V/m\n", + "B0= 11.5 *10**-10 T\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "E0=9*10**-12\n", + "myu0=4*math.pi*10**-7\n", + "r=10**4 #radius of Hemisphere\n", + "epsilon0=8.85*10**-12\n", + "C=3*10**2\n", + "P=10**5 \n", + "\n", + "#Calculations\n", + "S=P/(2*math.pi*r**2)/10**-4\n", + "E0=math.sqrt((2*S)/(C*epsilon0))/10**5\n", + "B0=E0/C/10**-4\n", + "\n", + "#Result\n", + "print\"Poynting vector S= %1.2f*10**-4 W/m**2\" %S\n", + "print\"E0= %0.3f V/m\" %E0\n", + "print\"B0= %2.1f *10**-10 T\" %B0" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 4 , Page number 143" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intensity of Electric field E= 86.58 V/m\n", + "Intensity of Magnetic field H= 0.23 amp. turn/m\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "myu0=4*math.pi*10**-7\n", + "r=2 #radius of Hemisphere\n", + "epsilon0=8.85*10**-12\n", + "P0=1000 #Power \n", + "\n", + "#Calculations\n", + "E=((P0*math.sqrt(myu0/epsilon0))/(16*math.pi))**(1/2)\n", + "H=P0/(16*math.pi*E)\n", + "\n", + "#Result\n", + "print\"Intensity of Electric field E= %2.2f\" %E,\"V/m\"\n", + "print\"Intensity of Magnetic field H= %1.2f\" %H,\"amp. turn/m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 6 , Page number 143" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Refractive index n= 9\n", + "Velocity of light= 3.33*10**7 m/sec\n" + ] + } + ], + "source": [ + "#importing module\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "E=81\n", + "c=3*10**8 #speed of ligth\n", + "\n", + "#Calculations\n", + "n=math.sqrt(E)\n", + "V=c/n/10**7\n", + "\n", + "#Result\n", + "print\"Refractive index n= %i\" %n\n", + "print\"Velocity of light= %1.2f*10**7 m/sec\" %V" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/chapter16.ipynb b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/chapter16.ipynb new file mode 100644 index 00000000..a62539f7 --- /dev/null +++ b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/chapter16.ipynb @@ -0,0 +1,76 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16 : MAGNETIC MATERIALS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example number 1, Page number 298" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intensity of Magnetization= 5.0 ampere/m\n", + "Flux density in the material= 1.257 weber/m^2\n" + ] + } + ], + "source": [ + "#importing modules\n", + "import math\n", + "from __future__ import division\n", + "\n", + "#Variable declaration\n", + "H=10**6 #Magnetic Field Strength in ampere/m\n", + "x=0.5*10**-5 #Magnetic susceptibility \n", + "mu_0=4*math.pi*10**-7\n", + "\n", + "#Calculatiions\n", + "M=x*H\n", + "B=mu_0*(M+H)\n", + "\n", + "#Result\n", + "print\"Intensity of Magnetization=\",M,\"ampere/m\"\n", + "print\"Flux density in the material=\",round(B,3),\"weber/m^2\"\n", + "\n", + "\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": "2.7.3" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/screenshots/chapter12.png b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/screenshots/chapter12.png Binary files differnew file mode 100644 index 00000000..306b6906 --- /dev/null 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b/Engineering_Physics_by_K.J_Pratap,_G.Prakash_Reddy,_S.Md.Asadullah,_P_Madhusudana_Rao,_B.Bhanu_Prasad_/screenshots/chapter16.png diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter2.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter2.ipynb new file mode 100644 index 00000000..2bd26cb3 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter2.ipynb @@ -0,0 +1,233 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Conduction and Breakdown in Gases" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 64" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the breakdown strength of air for 0.1mm air gap is (kV/cm.) = 43.45\n", + "\n", + "the breakdown strength of air for 20 cm air gap is (kV/cm.) = 25.58\n" + ] + } + ], + "source": [ + "#example 2.1\n", + "#calculation of breakdown strength of air\n", + "\n", + "#given data\n", + "d1=0.1#length(in cm) of the gap\n", + "d2=20#length(in cm) of the gap\n", + "\n", + "#calculation\n", + "#from equation of breakdown strength\n", + "E1=24.22+(6.08/(d1**(1./2)))#for gap d1\n", + "E2=24.22+(6.08/(d2**(1./2)))#for gap d2\n", + "#results\n", + "print 'the breakdown strength of air for 0.1mm air gap is (kV/cm.) = ',round(E1,2)\n", + "print '\\nthe breakdown strength of air for 20 cm air gap is (kV/cm.) = ',round(E2,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 64" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Townsend primary ioniztion coefficient is (/cm torr) = 7.675\n" + ] + } + ], + "source": [ + "#example 2.2\n", + "#calculation of Townsend primary ionization coefficient\n", + "from math import log\n", + "#given data\n", + "d1=0.4#gap distance(in cm)\n", + "d2=0.1#gap distance(in cm)\n", + "I1=5.5*10**-8#value of current(in A)\n", + "I2=5.5*10**-9#value of current(in A)\n", + "\n", + "#calculation\n", + "#from equation of current at anode I=I0*exp(alpha*d)\n", + "alpha=(log(I1/I2))*(1/(d1-d2))\n", + "#results\n", + "print 'Townsend primary ioniztion coefficient is (/cm torr) = ',round(alpha,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 65" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of Townsend secondary ionization coefficient is 9.994e-04\n" + ] + } + ], + "source": [ + "#example 2.3\n", + "#calculation of Townsend secondary ionization coefficient\n", + "from math import exp\n", + "#given data\n", + "d=0.9#gap distance(in cm)\n", + "alpha=7.676#value of alpha\n", + "\n", + "#calculation\n", + "#from condition of breakdown.....gama*exp(alpha*d)=1\n", + "gama=1/(exp(d*alpha))\n", + "#results\n", + "print '%s %.3e' %('the value of Townsend secondary ionization coefficient is ',gama)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 65" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of breakdown voltage of the spark gap is (V) = 5626.0\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#example 2.4\n", + "#calculation of breakdown voltage of a spark gap\n", + "from math import log\n", + "#given data\n", + "A=15.#value of A(in per cm)\n", + "B=360.#value of B(in per cm)\n", + "d=0.1#spark gap(in cm)\n", + "gama=1.5*10**-4#value of gama\n", + "p=760.#value of pressure of gas(in torr)\n", + "\n", + "#calculation\n", + "#from equation of breakdown voltage\n", + "V=(B*p*d)/(log((A*p*d)/(log(1+(1/gama)))))\n", + "\n", + "#results\n", + "print 'the value of breakdown voltage of the spark gap is (V) = ',round(V)\n", + "print 'The answer is a bit different due error in textbook'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 65" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of minimum spark over voltage is (V) = 481.0\n", + "The answer is a bit different due error in textbook\n" + ] + } + ], + "source": [ + "#example 2.5\n", + "#calculation of minimum spark over voltage\n", + "from math import log\n", + "#given data\n", + "A=15.#value of A(in per cm)\n", + "B=360.#value of B(in per cm)\n", + "gama=10**-4#value of gama\n", + "e=2.178#value of constant\n", + "\n", + "#calculation\n", + "Vbmin=(B*e/A)*(log(1+(1/gama)))\n", + "\n", + "#results\n", + "print 'the value of minimum spark over voltage is (V) = ',round(Vbmin)\n", + "print 'The answer is a bit different due error in textbook'" + ] + } + ], + "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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter3.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter3.ipynb new file mode 100644 index 00000000..c7746ecf --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter3.ipynb @@ -0,0 +1,97 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: Conduction and Breakdown in Liquid Dielectrics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 85" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x35e6898>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power law dependence between the gap spacing and the applied voltage of the oil is 24.2 *d^ 0.948\n" + ] + } + ], + "source": [ + "#example 3.1\n", + "#determination of power law dependence between the gap spacing and the applied voltage of the oil\n", + "%matplotlib inline\n", + "from math import log, exp\n", + "from matplotlib import pyplot\n", + "import numpy as np\n", + "#given data\n", + "d1=4.#gap spacing(in mm)\n", + "d2=6.#gap spacing(in mm)\n", + "d3=10.#gap spacing(in mm)\n", + "d4=12.#gap spacing(in mm)\n", + "V1=90.#voltage(in kV) at breakdown\n", + "V2=140.#voltage(in kV) at breakdown\n", + "V3=210.#voltage(in kV) at breakdown\n", + "V4=255.#voltage(in kV) at breakdown\n", + "\n", + "#calculation\n", + "#from the relationship between breakdown voltage and the gap spacing.....V = K*d^n\n", + "#we get n = (log(V)-log(K))/log(d) = slope of line from given data\n", + "n=(log(V4)-log(V1))/(log(d4)-log(d1))\n", + "K=exp(log(V1)-n*log(d1))#Y intercept on the power law dependence graph\n", + "#plotting of graph\n", + "dn=np.linspace(1,20,num=20)\n", + "Vn=K*dn**n\n", + "#results\n", + "pyplot.plot(dn,Vn)\n", + "pyplot.xlabel(\"Gas spacing (mm)\")\n", + "pyplot.ylabel(\"Breakdown voltage (kV)\")\n", + "pyplot.show()\n", + "print 'The power law dependence between the gap spacing and the applied voltage of the oil is ',round(K,1),'*d^',round(n,3)" + ] + } + ], + "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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter4.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter4.ipynb new file mode 100644 index 00000000..f74d0ff9 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter4.ipynb @@ -0,0 +1,168 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Breakdown in Soild Dielectrics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The heat generated in specimen due to dielectric loss is (mW/cm^3) = 0.292\n" + ] + } + ], + "source": [ + "#example 4.1\n", + "#calculation of heat generated in specimen due to dielectric loss\n", + "\n", + "#given data\n", + "epsilonr=4.2#value of the dielectric constant\n", + "tandelta=0.001#value of tandelta\n", + "f=50#value of frequency(in Hz)\n", + "E=50*10**3#value of electric field(in V/cm)\n", + "\n", + "#calculation\n", + "#from equation of dielectric heat loss......H=(E*E*f*epsilonr*tandelta)/(1.8*10^12)\n", + "H=(E*E*f*epsilonr*tandelta)/(1.8*10**12)\n", + "#results\n", + "print 'The heat generated in specimen due to dielectric loss is (mW/cm^3) = ',round(H*10**3,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 123" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the voltage at which an internal discharge can occur is (kV.) = 9.75\n" + ] + } + ], + "source": [ + "#example 4.2\n", + "#calculation of voltage at which an internal discharge can occur\n", + "\n", + "#given data\n", + "d1=1#thickness(in mm) of the internal void\n", + "dt=10#thickness(in mm) of the specimen\n", + "epsilon0=8.89*10**-12#electrical permittivity(in F/m) of free space\n", + "epsilonr=4#relative permittivity of the dielectric\n", + "Vb=3#breakdown strength(in kV/mm) of air\n", + "\n", + "#calculation\n", + "d2=dt-d1\n", + "epsilon1=epsilon0*epsilonr#electrical permittivity(in F/m) of the dielectric\n", + "V1=Vb*d1#voltage at which air void of d1 thickness breaks\n", + "V=(V1*(d1+(epsilon0*d2/epsilon1))/d1)\n", + "#results\n", + "print 'the voltage at which an internal discharge can occur is (kV.) = ',round(V,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 124" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of inner diameter of electrodes in coaxial cylindrical capacitor is (cm) = 11.74\n", + "\n", + "the value of outer diameter of electrodes in coaxial cylindrical capacitor is (cm) = 12.04\n", + "\n", + "the thickness of the insulation is (mm) = 3.0\n" + ] + } + ], + "source": [ + "#example 4.3\n", + "#calculation of the dimensions of electrodes in coaxial cylindrical capacitor\n", + "from math import pi,exp\n", + "#given data\n", + "epsilon0=(36*pi*10**9)**-1#electrical permittivity(in F/m) of free space\n", + "#consider high density polyethylene as the dielectric material\n", + "epsilonr=2.3#relative permittivity of high density polyethylene\n", + "l=0.2#effective length(in m)\n", + "C=1000*10**-12#capacitance(in F) of the capacitor\n", + "V=15#operating voltage(in kV)\n", + "Emax=50#maximum stress(in kV/cm) for breakdown stress 200 kV/cm and factor of safety of 4\n", + "\n", + "#calculation\n", + "#from equation of capacitance of coaxial cylindrical capacitor\n", + "#C=(2*%pi*epsilon0*epsilonr*l)/(lod(d2/d1)).............(1)\n", + "#from equation of Emax occuring near electrodes\n", + "#Emax=V/(r1*(log(r2/r1)))...............................(2)\n", + "#from equation (1) and equation (2),we get\n", + "logr2byr1=(2*pi*epsilon0*epsilonr*l)/C#logd2/d1 = logr2/r1\n", + "r1=V/(Emax*logr2byr1)#from equation (1)\n", + "r2=r1*exp(logr2byr1)\n", + "#results\n", + "print 'the value of inner diameter of electrodes in coaxial cylindrical capacitor is (cm) = ',round(r1,2)\n", + "print '\\nthe value of outer diameter of electrodes in coaxial cylindrical capacitor is (cm) = ',round(r2,2)\n", + "print '\\nthe thickness of the insulation is (mm) = ',round((r2-r1)*10,1)\n", + "\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter6.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter6.ipynb new file mode 100644 index 00000000..f68c1627 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter6.ipynb @@ -0,0 +1,547 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Generation of High voltages and Currents" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 196" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the value of percentage ripple is (percentage) = 4.53\n", + "\n", + "the value of the regulation is (percentage) = 12.4\n", + "\n", + "the optimum number of stages for minimum regulation is 14.0\n" + ] + } + ], + "source": [ + "#example 6.1\n", + "#calculation of percentage ripple,the regulation and the optimum number of stages for minimum regulation in Cockcroft-Walton type voltage multiplier\n", + "from math import sqrt\n", + "#given data\n", + "C=0.05*10**-6#value of capacitance(in F)\n", + "Vmax=125.*10**3#value of supply transformer secondary voltage(in V)\n", + "f=150.#frequency(in Hz)\n", + "I=5.*10**-3#load current(in A)\n", + "nst=8.#number of stages\n", + "\n", + "#calculation\n", + "n=nst*2#number of capacitors\n", + "#from equation of ripple voltage\n", + "deltaV=(I/(f*C))*(n*(n+1)/2)\n", + "perripple=(deltaV*100)/(16*Vmax)\n", + "deltaVn=(I/(f*C))*(((2*nst**3)/3)+(nst*nst/2)-(nst/6))#voltage drop...here n = nst = number of stages\n", + "reg=deltaVn/(2*nst*Vmax)#regulation\n", + "nopt=round(sqrt(Vmax*f*C/I))#optimum number of stages\n", + "#results\n", + "print 'the value of percentage ripple is (percentage) = ',round(perripple,2)\n", + "print '\\nthe value of the regulation is (percentage) = ',round(reg*100,1)\n", + "print '\\nthe optimum number of stages for minimum regulation is ',round(nopt)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 197" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of series inductance is (H) = 3820.0\n", + "\n", + "The value of input voltage to the transformer is (V) = 4.0\n" + ] + } + ], + "source": [ + "#example 6.2\n", + "#calculation of series inductance and input voltage to transformer\n", + "from math import pi\n", + "#given data\n", + "kva=100.*10**3#value of volt-ampere of transformer(in VA)\n", + "V=250.*10**3#value of transformer secondary voltage(in V)\n", + "Vi=400.#value of transformer primary voltage(in V)\n", + "Vc=500.*10**3#voltage(in V)\n", + "Ic=0.4#charging current(in A)\n", + "perX=8.#percentage leakage reactance\n", + "f=50.#value of frequency(in Hz)\n", + "perR1=2.#percentage resistance\n", + "perR2=2.#percentage resistance of inductor\n", + "\n", + "\n", + "#calculation\n", + "I=kva/V#maximum value of current that can be supplied\n", + "Xc=Vc/Ic#reactance of cable\n", + "Xl=(perX*V)/(100*I)#leakage reactance\n", + "adrec=Xc-Xl#additional reactance\n", + "Xadrec=adrec/(2*pi*f)\n", + "perR=perR1+perR2#total resistance\n", + "R=(perR*V)/(100*I)\n", + "VE2=I*R#excitation at secondary\n", + "VE1=VE2*Vi/V#primary voltage\n", + "IkW=(VE1/Vi)*100#input kW\n", + "#results\n", + "print 'The value of series inductance is (H) = ',round(Xadrec)\n", + "print '\\nThe value of input voltage to the transformer is (V) = ',round(IkW)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of series resistance is (ohm) = 420.0\n", + "\n", + "The value of damping resistance is (ohm) = 2981.0\n", + "\n", + "The value of maximum output voltage of the generator is (kV) = 892.02\n", + "The answer is a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#example 6.3\n", + "#calculation of series resistance ,damping resistance and maximum output voltage of the generator\n", + "from math import exp\n", + "#given data\n", + "n=8.#number of stages\n", + "C=0.16*10**-6#value of condenser(in farad)\n", + "Cl=1000.*10**-12#value of load capacitor (in farad)\n", + "t1=1.2*10**-6#time to front(in second)\n", + "t2=50.*10**-6#time to tail(in second)\n", + "Vc=120.*10**3#charging voltage(in V)\n", + "\n", + "#calculation\n", + "C1=C/n#generator capacitance\n", + "C2=Cl#load capacitance\n", + "R1=(t1*(C1+C2))/(3*C1*C2)\n", + "R2=(t2/(0.7*(C1+C2)))-R1\n", + "V=n*Vc#dc charging voltage for n stages\n", + "alpha=1/(R1*C2)\n", + "betaa=1/(R2*C1)\n", + "Vmax=(V*(exp(-alpha*t1)-exp(-betaa*t1)))/(R1*C2*(alpha-betaa))\n", + "#results\n", + "print 'The value of series resistance is (ohm) = ',round(R1)\n", + "print '\\nThe value of damping resistance is (ohm) = ',round(R2)\n", + "print '\\nThe value of maximum output voltage of the generator is (kV) = ',round(-Vmax*10**-3,2)\n", + "print 'The answer is a bit different due to rounding off error in textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 198" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of circuit inductance is (microhenry) = 8.125\n", + "\n", + "The value of dynamic resistance is (ohm) = 0.8694\n", + "\n", + "The value of charging voltage is (kV) = 17.5\n", + "The answer for Charging Voltage is wrong in textbook\n" + ] + } + ], + "source": [ + "#example 6.4\n", + "#calculation of circuit inductance and dynamic resistance\n", + "\n", + "#given data\n", + "alpha=0.0535*10**6#from table\n", + "LC=65.#value of product\n", + "C=8.#value of capacitor (in microfarad)\n", + "Ip=10.#output peak current(in kA)\n", + "t1=8.#time to front(in microsecond)\n", + "\n", + "#calculation\n", + "L=LC/C#inductance(in microhenry)\n", + "Rd=2*(LC*10**-6)*alpha/t1#dynamic resistance\n", + "V=Ip*14./C#charging voltage\n", + "\n", + "#results\n", + "print 'The value of circuit inductance is (microhenry) = ',round(L,3)\n", + "print '\\nThe value of dynamic resistance is (ohm) = ',round(Rd,4)\n", + "print '\\nThe value of charging voltage is (kV) = ',round(V,1)\n", + "print 'The answer for Charging Voltage is wrong in textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 199" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of circuit inductance is (microhenry) = 4.09\n", + "\n", + "The value of dynamic resistance is (ohm) = 1.4302\n", + "\n", + "The value of charging voltage is (kV) = 1.59\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#example 6.5\n", + "#calculation circuit inductance and dynamic resistance\n", + "from math import pi,exp,atan\n", + "#given data\n", + "C=8.*10**-6#value of capacitor (in farad)\n", + "Ip=10.#output peak current(in kA)\n", + "t1=8.*10**-6#time to front(in second)\n", + "t2=20.*10**-6#time to first half cycle(in second)\n", + "V=25.*10**3#charging voltage\n", + "im=10.*10**3#output currennt(in A)\n", + "\n", + "#calculation\n", + "omega=pi/t2\n", + "omegat1=omega*t1\n", + "alpha=omega*(1/atan(omegat1))\n", + "LC=1/((t1**2)+(alpha**2))\n", + "L=LC/C\n", + "R=2*L*alpha \n", + "V=omega*L*10*exp(-alpha*t1)\n", + "#results\n", + "print 'The value of circuit inductance is (microhenry) = ',round(L*10**6,2)\n", + "print '\\nThe value of dynamic resistance is (ohm) = ',round(R,4)\n", + "print '\\nThe value of charging voltage is (kV) = ',round(V,2)\n", + "print 'The answers are a bit different due to rounding off error in textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 199" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The time to front is (microsecond) = 2.19\n", + "\n", + "The time to tail is (microsecond) = 46.7\n" + ] + } + ], + "source": [ + "#example 6.6\n", + "#calculation of front and tail time\n", + "\n", + "#given data\n", + "n=12#number of stages\n", + "C=0.126*10**-6#capacitance(in Farad)\n", + "R1=800#wavefront resistance(in ohm)\n", + "R2=5000#xavetail resistance(in ohm)\n", + "C2=1000*10**-12#load capacitance(in Farad)\n", + "\n", + "\n", + "#calculation\n", + "C1=C/n\n", + "t1=3*R1*(C1*C2)/(C1+C2)\n", + "t2=0.7*(R1+R2)*(C1+C2)\n", + "#results\n", + "print 'The time to front is (microsecond) = ',round(t1*10**6,2)\n", + "print '\\nThe time to tail is (microsecond) = ',round(t2*10**6,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 200" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of highest resonant frequency produced is (kHz) = 8.04714659485\n", + "\n", + "The peak value of output voltage is (kV) - 209.631374545\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#example 6.7\n", + "#calculation of peak value of output voltage and highest resonant frequency produced\n", + "from cmath import sqrt,pi\n", + "#given data\n", + "V=10.*10**3#voltage(in V) at primary winding\n", + "L1=10.*10**-3#inductance(in H)\n", + "L2=200.*10**-3#inductance(in H)\n", + "K=0.6#coefficient of coupling\n", + "C1=2.*10**-6#capacitance(in Farad) on primary side\n", + "C2=1.*10**-9#capacitance(in Farad) on secondary side\n", + "\n", + "#calculation\n", + "M=K*sqrt(L1*L2)\n", + "omega1=1/sqrt(L1*C1)\n", + "sigma=sqrt(1-(K**2))\n", + "omega2=1/sqrt(L2*C2)\n", + "gama2=sqrt(((omega1**2+omega2**2)/2)+sqrt(((omega1**2+omega2**2)/2)-(sigma**2*omega1**2*omega2**2)))\n", + "gama1=sqrt(((omega1**2+omega2**2)/2)-sqrt(((omega1**2+omega2**2)/2)-(sigma**2*omega1**2*omega2**2)))\n", + "fh=gama2/(2*pi)#highest frequency\n", + "V2p=(V*M)/(sigma*L1*L2*C2*(gama2**2-gama1**2))\n", + "#results\n", + "print 'The value of highest resonant frequency produced is (kHz) = ',abs(fh)*10**-3\n", + "print '\\nThe peak value of output voltage is (kV) - ',abs(V2p)*10**-3\n", + "\n", + "#gama1 and gama2 are imaginary numbers....Moreover their magnitudes will also be same....so peak value of output voltage from equation is zero\n", + "print 'The answers are a bit different due to rounding off error in textbook'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 201" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of output voltage is (kV) = 100.0\n", + "correct answer is 100 kV\n" + ] + } + ], + "source": [ + "#example 6.8\n", + "#calculation of output voltage\n", + "from math import sqrt\n", + "#given data\n", + "V1=10.#voltage(in kV) at primary winding \n", + "C1=2.*10**-6#capacitance(in Farad) on primary side\n", + "C2=1.*10**-9#capacitance(in Farad) on secondary side\n", + "pern=5.#energy efficiency(in percentage)\n", + "\n", + "#calculation\n", + "n=pern/100.\n", + "V2=V1*sqrt(n*C1/C2)\n", + "#results\n", + "print 'The value of output voltage is (kV) = ',round(V2,1)\n", + "print 'correct answer is 100 kV'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 201" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of self capacitance is (nF) = 1.126\n", + "\n", + "The value of leakage reactance is (kohm) = 28.0\n" + ] + } + ], + "source": [ + "#example 6.9\n", + "#calculation of self capacitance and leakage reactance\n", + "from math import pi\n", + "#given data\n", + "Vi=350.*10**3#rating(in VA)\n", + "V=350.*10**3#secondary voltage(in V)\n", + "V1=6.6*10**3#primary voltage(in V)\n", + "perV=8.#percentage ratedd voltage\n", + "perR=1.#percentage rise\n", + "f=50.#frequency(in Hz)\n", + "\n", + "#calculation\n", + "I=Vi/V\n", + "Xl=(perV*V)/(100*I)\n", + "I0=perR*V/(100*Xl)\n", + "Xc=((1+(perR/100))*V)/I0\n", + "C=1/(Xc*2*pi*f)\n", + "#results\n", + "print 'The value of self capacitance is (nF) = ',round(C*10**9,3)\n", + "print '\\nThe value of leakage reactance is (kohm) = ',round(Xl*10**-3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 202" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance for 1/50 microsecond voltage is (ohm) = 70.6\n", + "\n", + "The value of inductance for 1/50 microsecond voltage is (microhenry) = 11.6\n", + "\n", + "The value of output voltage is (kV) = 9.88\n", + "\n", + "The value of inductance for 8/20 microsecond voltage is (microhenry) = 65.0\n", + "\n", + "The value of resistance for 8/20 microsecond voltage is (ohm) = 6.955\n", + "\n", + "The peak value of current is (A) = 714.0\n" + ] + } + ], + "source": [ + "#example 6.10\n", + "#calculation of resistance and inductance\n", + "\n", + "#given data\n", + "CR=70.6#value from table\n", + "LC=11.6#value from table\n", + "C=1#capacitance(in microfarad)\n", + "pern=98.8#percentage voltage efficiency\n", + "V=10.#rating(in kV)\n", + "LC2=65.#value from table\n", + "alpha=0.0535#value from table\n", + "\n", + "#calculation\n", + "R=CR/C\n", + "L=LC/C\n", + "Vo=pern*V/100\n", + "L2=LC2/C\n", + "R2=2*L2*alpha\n", + "Ip=V*C/14.\n", + "\n", + "print 'The value of resistance for 1/50 microsecond voltage is (ohm) = ',round(R,1)\n", + "print '\\nThe value of inductance for 1/50 microsecond voltage is (microhenry) = ',round(L,1)\n", + "print '\\nThe value of output voltage is (kV) = ',round(Vo,2)\n", + "print '\\nThe value of inductance for 8/20 microsecond voltage is (microhenry) = ',round(L2)\n", + "print '\\nThe value of resistance for 8/20 microsecond voltage is (ohm) = ',round(R2,3)\n", + "print '\\nThe peak value of current is (A) = ',round(Ip*10**3)\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter7.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter7.ipynb new file mode 100644 index 00000000..9e1c3bce --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter7.ipynb @@ -0,0 +1,380 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Measurement of High Voltages and Currents" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 277" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The capacitance of the generating voltmeter is (pF) = 0.9\n" + ] + } + ], + "source": [ + "#example 7.1\n", + "#calculation of capacitance of generating voltmeter\n", + "from math import pi,sqrt\n", + "#given data\n", + "Irms=2.*10**-6#current(in A)\n", + "V1=20.*10**3#applied voltage(in V)\n", + "V2=200.*10**3#applied voltage(in V)\n", + "rpm=1500.#assume synchronous speed(in rpm) of motor\n", + "\n", + "#calculation\n", + "Cm=Irms*sqrt(2)/(V1*(rpm/60)*2*pi)\n", + "Irmsn=V2*Cm*2*pi*(rpm/60)/sqrt(2)\n", + "#results\n", + "print 'The capacitance of the generating voltmeter is (pF) = ',round(Cm*10**12,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 277" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of Cs is (microfarad) = 1.0\n", + "\n", + "The value of R is 1.0e+07 ohm\n" + ] + } + ], + "source": [ + "#example 7.2\n", + "#Design of a peak reading voltmeter\n", + "\n", + "#given data\n", + "r=1000.#ratio is 1000:1\n", + "V=100.*10**3#read voltage(in V)\n", + "R=10**7#value of resistance(in ohm)\n", + "\n", + "#calculation\n", + "#take range as 0-10 microampere\n", + "Vc2=V/r#voltage at C2 arm\n", + "#Cs * R = 1 to 10 s\n", + "Cs=10./R\n", + "#results\n", + "print 'The value of Cs is (microfarad) = ',round(Cs*10**6)\n", + "print '%s %.1e %s' %('\\nThe value of R is ',R,'ohm')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 278" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The air density factor is 0.9327\n" + ] + } + ], + "source": [ + "#example 7.3\n", + "#calculation of correction factors for atmospheric conditions\n", + "\n", + "#given data\n", + "t=37#temperature(in degree celsius)\n", + "p=750.#atmospheric pressure(in mmHg)\n", + "\n", + "#calculation\n", + "d=p*293./(760*(273+t))\n", + "\t\n", + "#results\n", + "print 'The air density factor is ',round(d,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 278" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The divider ratio is 1238.3\n" + ] + } + ], + "source": [ + "#example 7.4\n", + "#calculation of divider ratio\n", + "\n", + "#given data\n", + "R1=16.*10**3#high voltage arm resistance(in ohm)\n", + "n=16.#number of members\n", + "R=250.#resistance(in ohm) of each member in low voltage arm\n", + "R2dash=75.#terminating resistance(in ohm)\n", + "\n", + "#calculation\n", + "R2=R/n\n", + "a=1+(R1/R2)+(R1/R2dash)\n", + "#results\n", + "print 'The divider ratio is ',round(a,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 278" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of capacitance needed for correct compensation is 1.8e-08 F or 17 nf\n" + ] + } + ], + "source": [ + "#example 7.5\n", + "#calculation of capacitance needed for correct compensation\n", + "\n", + "#given data\n", + "Cgdash=20*10**-12#ground capacitance(in farad)\n", + "n=15.#number of capacitors\n", + "r=120.#resistance(in ohm)\n", + "R2=5.#resistance(in ohm) of LV arm\n", + "\n", + "#calculation\n", + "Ce=(2./3)*n*Cgdash\n", + "R1=n*r/2\n", + "T=R1*Ce/2\n", + "C2=T/R2\n", + "#results\n", + "print '%s %.1e %s %d %s' %('The value of capacitance needed for correct compensation is ',C2,' F or ',C2*10**9,' nf')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 279" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance is (milliohm) = 1.0\n", + "\n", + "The length of shunt is (cm) = 10.0\n", + "\n", + "The radius of shunt is (mm) = 25.6\n", + "\n", + "The thickness of shunt is (mm) = 0.187\n" + ] + } + ], + "source": [ + "#example 7.6\n", + "#calculation of ohmic value of shunt an its dimensions\n", + "from math import sqrt,pi\n", + "#given data\n", + "I=50.*10**3#impulse current (in A)\n", + "Vm=50.#voltage(in V) drop across shunt\n", + "B=10.*10**6#bandwidth(in Hz) of the shunt\n", + "mu0=4.*pi*10**-7#magnetic permeability(in H/m) of free space\n", + "\n", + "#calculation\n", + "R=Vm/I#resistance of shunt\n", + "L0=1.46*R/B\n", + "mu=mu0#in this case ...mu = mu0 * mur ~mu0\n", + "rho=30*10**-8#resistivity(in ohm m) of the tube material\n", + "d=sqrt((1.46*rho)/(mu*B))#thickness of the tube(in m)\n", + "l=10**-1#length(in m) (assume)\n", + "r=(rho*l)/(2*pi*R*d)\n", + "#results\n", + "print 'The value of resistance is (milliohm) = ',round(R*10**3)\n", + "print '\\nThe length of shunt is (cm) = ',round(l*100)\n", + "print '\\nThe radius of shunt is (mm) = ',round(r*10**3,1)\n", + "print '\\nThe thickness of shunt is (mm) = ',round(d*10**3,3)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 280" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of mutual inductance is (nH) = 2.0\n", + "\n", + "The value of resistance is 2.0e+03 ohm\n", + "\n", + "The value of capacitance is (pF) = 1000.0\n" + ] + } + ], + "source": [ + "#example 7.7\n", + "#Estimation of values of mutual inductance,resistance and capacitance\n", + "from math import pi\n", + "#given data\n", + "It=10.*10**3#impulse current(in A)\n", + "Vmt=10.#meter reading(in V) for full scale deflection\n", + "dibydt=10.**11#rate of change of current(in A/s)\n", + "\n", + "#calculation\n", + "MbyCR=Vmt/It\n", + "t=It/dibydt\n", + "f=1/(4.*t)\n", + "omega=2*pi*f\n", + "CR=10*pi/omega\n", + "M=10**-3*CR\n", + "R=2*10**3#assume resistance(in ohm)\n", + "C=CR/R\n", + "#results\n", + "print 'The value of mutual inductance is (nH) = ',round(M*10**9)\n", + "print '%s %.1e %s' %('\\nThe value of resistance is',R,'ohm')\n", + "print '\\nThe value of capacitance is (pF) = ',round(C*10**12)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 281" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of resistance is 2.0e+03 ohm\n", + "\n", + "The value of capacitance is (microfarad) = 0.25\n" + ] + } + ], + "source": [ + "#example 7.8\n", + "#calculation of resistance and capacitance\n", + "from math import pi\n", + "#given data\n", + "t1=8.*10**-6#fronttime(in s)\n", + "t2=20.*10**-6#tailtime(in s)\n", + "\n", + "\n", + "#calculation\n", + "f2=1/t2#frequency corresponding to tail time\n", + "fl=f2/5\n", + "omega=2*pi*fl\n", + "CR=10*pi/omega\n", + "M=10**-3*(1/CR)\n", + "R=2*10**3#assume resistance(in ohm)\n", + "C=CR/R\n", + "\n", + "print '%s %.1e %s' %('The value of resistance is ',R,'ohm')\n", + "print '\\nThe value of capacitance is (microfarad) = ',round(C*10**6,2)\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter8.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter8.ipynb new file mode 100644 index 00000000..10109095 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter8.ipynb @@ -0,0 +1,391 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Overvoltage phenomenon and Insulation coordination in electric power systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 350 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of surge impedance is (ohm) = 374.2\n", + "\n", + "The value of velocity is 3.0e+05 km/s\n", + "\n", + "The time taken by the surge to travel to the other end is (ms) = 1.35\n" + ] + } + ], + "source": [ + "#example 8.1\n", + "#calculation of surge impedance,velocity and time taken by the surge to travel to the other end\n", + "from math import sqrt\n", + "#given data\n", + "L=1.26*10**-3#inductance(in H/km)\n", + "C=0.009*10**-6#capacitance(in F/km)\n", + "l=400.#length(in km) of the transmission line\n", + "\n", + "#calculation\n", + "v=1/sqrt(L*C)\n", + "Xs=sqrt(L/C)\n", + "t=l/v\n", + "#results\n", + "print 'The value of surge impedance is (ohm) = ',round(Xs,1)\n", + "print '%s %.1e %s' %('\\nThe value of velocity is ',v,' km/s')\n", + "print '\\nThe time taken by the surge to travel to the other end is (ms) = ',round(t*10**3,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 350" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of the voltage build up at the junction is (kV) = 893.0\n" + ] + } + ], + "source": [ + "#example 8.2\n", + "#calculation of the voltage build up at the junction\n", + "\n", + "#given data\n", + "Z1=500.#surge impedance(in ohm) of transmission line\n", + "Z2=60.#surge impedance(in ohm) of cable\n", + "e=500.#value of surge(in kV)\n", + "\n", + "#calculation\n", + "tau=(Z1-Z2)/(Z2+Z1)#coefficient of reflection\n", + "Vj=(1+tau)*e\n", + "#results\n", + "print 'The value of the voltage build up at the junction is (kV) = ',round(Vj)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 352" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When wave travels along the cable,the transmitted voltage is (kV) = 374.85\n", + "\n", + "When wave travels along the cable,the reflected voltage is (kV) = 174.85\n", + "\n", + "When wave travels along the cable,the transmitted current is (kA) = 1.002\n", + "\n", + "When wave travels along the cable,the reflected current is (kA) = 6.97\n", + "\n", + "When wave travels along the line,the transmitted voltage is (kV) = 25.15\n", + "\n", + "When wave travels along the line,the reflected voltage is (kV) = -174.85\n", + "\n", + "When wave travels along the line,the transmitted current is (kA) = 1.002\n", + "\n", + "When wave travels along the line,the reflected current is 0.467 kA or 467.0 A\n" + ] + } + ], + "source": [ + "#example 8.5\n", + "#calculation of the transmitted,reflected voltage and current waves\n", + "from math import sqrt\n", + "#given data\n", + "L1=0.189*10**-3#inductance(in H/km) of the cable\n", + "C1=0.3*10**-6#capacitance(in Farad/km) of the cable\n", + "L2=1.26*10**-3#inductance(in H/km) of the overhead line\n", + "C2=0.009*10**-6#capacitance(in Farad/km) of the overhead line\n", + "e=200.*10**3#surge volatge(in kV)\n", + "\n", + "#calculation\n", + "Z1=sqrt(L1/C1)#surge impedance of the cable\n", + "Z2=sqrt(L2/C2)#surge impedance of the line\n", + "tau=(Z2-Z1)/(Z2+Z1)#when wave travels along the cable\n", + "edash=tau*e#reflected wave\n", + "edashdash=(1+tau)*e#transmitted wave\n", + "Idash=edash/Z1#reflected current wave\n", + "Idashdash=edashdash/Z2#transmitted current wave\n", + "Z2n=Z1\n", + "Z1n=Z2\n", + "taun=(Z2n-Z1n)/(Z2n+Z1n)#when wave travels along the line\n", + "edashn=taun*e#reflected wave\n", + "edashdashn=(1+taun)*e#transmitted wave\n", + "Idashdashn=edashdashn/Z2n#transmitted current wave\n", + "Idashn=edashn/Z1n#reflected current wave\n", + "#results\n", + "print 'When wave travels along the cable,the transmitted voltage is (kV) = ',round(edashdash*10**-3,2)\n", + "print '\\nWhen wave travels along the cable,the reflected voltage is (kV) = ',round(edash*10**-3,2)\n", + "print '\\nWhen wave travels along the cable,the transmitted current is (kA) = ',round(Idashdash*10**-3,3)\n", + "print '\\nWhen wave travels along the cable,the reflected current is (kA) = ',round(Idash*10**-3,2) \n", + "print '\\nWhen wave travels along the line,the transmitted voltage is (kV) = ',round(edashdashn*10**-3,2)\n", + "print '\\nWhen wave travels along the line,the reflected voltage is (kV) = ',round(edashn*10**-3,2)\n", + "print '\\nWhen wave travels along the line,the transmitted current is (kA) = ',round(Idashdashn*10**-3,3)\n", + "print '\\nWhen wave travels along the line,the reflected current is ',round(abs(Idashn*10**-3),3),' kA or ',round(abs(Idashn)),'A' \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 353" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of voltage at the receiving end in Bewley lattice diagram is 0.9756 u(t) V\n" + ] + } + ], + "source": [ + "#example 8.6\n", + "#calculation of value of voltage at the receiving end in Bewley lattice diagram\n", + "\n", + "#given data\n", + "alpha=0.8\n", + "\n", + "#calculation\n", + "Vut=2*alpha/(1+alpha**2)\n", + "#results\n", + "print 'The value of voltage at the receiving end in Bewley lattice diagram is ',round(Vut,4),'u(t) V'\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 355" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sparkover voltage for terminated line is (kV) = 347.0\n", + "\n", + "The arrester current for terminated line is (kA) = 5.0\n", + "\n", + "The sparkover voltage for continuous line is (kV) = 294.0\n", + "\n", + "The arrester current for continuous line is (kA) = 2.5\n" + ] + } + ], + "source": [ + "#example 8.7\n", + "#calculation of sparkover voltage and the arrester current\n", + "\n", + "#given data\n", + "Xs=400.#surge impedance(in ohm)\n", + "Xv=1000.#surge voltage(in kV)\n", + "\n", + "#calculation\n", + "#for line terminated\n", + "Iam=2*Xv/Xs#maximum arrester current\n", + "#as Iam = 5 kA from graph Vd = 330 kV\n", + "Vd=330.#sparkover voltage(in kV)\n", + "Vso=Vd+(Vd*5./100)\n", + "#for continuous line\n", + "Iamn=Xv/Xs#maximum arrester current\n", + "#as Iamn = 2.5 kA from graph Vdn = 280 kV\n", + "Vdn=280#sparkover voltage(in kV)\n", + "Vson=Vdn+(Vdn*5./100)\n", + "#results\n", + "print 'The sparkover voltage for terminated line is (kV) = ',round(Vso)\n", + "print '\\nThe arrester current for terminated line is (kA) = ',round(Iam)\n", + "print '\\nThe sparkover voltage for continuous line is (kV) = ',round(Vson)\n", + "print '\\nThe arrester current for continuous line is (kA) = ',round(Iamn,1)\n", + "#values of sparover voltages are\n", + "#for terminated line = 346 kV\n", + "#for continuous line = 294 kV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 356" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Neglecting resistance of line,the rise in voltage at the other end is (kV) = 11.89\n", + "\n", + "Considering all the parameters,the rise in voltage at the other end is (kV) = 94.5\n" + ] + } + ], + "source": [ + "#example 8.8\n", + "#calculation of rise in voltage at the other end\n", + "from cmath import pi,sqrt,cos\n", + "#given data\n", + "R=0.1#resistance(in ohm/km)\n", + "L=1.26*10**-3#inductance(in H/km)\n", + "C=0.009*10**-6#capacitance(in F/km)\n", + "l=400#length(in km) of the line\n", + "V1=230#line voltage(in kV)\n", + "f=50#frequency(in Hz)\n", + "G=0\n", + "\n", + "#calculation\n", + "#Neglecting resistance of line\n", + "V1p=V1/sqrt(3)\n", + "omega=2*pi*f\n", + "Xl=complex(0,omega*L*l)\n", + "Xc=complex(0,-1/(omega*C*l))\n", + "V2=V1p*((1-(Xl/(2*Xc)))-1)\n", + "\n", + "#Considering all the parameters\n", + "omegaL=complex(0,omega*L)\n", + "omegaC=complex(0,omega*C)\n", + "i=l*sqrt((R+omegaL)*(G+omegaC))\n", + "betal=i.imag*l\n", + "V2n=V1p/cos(betal)\n", + "\n", + "print 'Neglecting resistance of line,the rise in voltage at the other end is (kV) = ',round(V2.real,2)\n", + "print '\\nConsidering all the parameters,the rise in voltage at the other end is (kV) = ',round(V2n.real-V1p.real,2)\n", + "\n", + "#By considering all the parameters the rise in voltage at the other end is 94.50 kV\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 357" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The protective margin for lightning impulses is (percentage) = 34.3\n", + "\n", + "The protective margin for switching gears is (percentage) = 27.6\n", + "\n", + "The margin when lightning arrester just sparks is (percentage) = 27.6\n" + ] + } + ], + "source": [ + "#example 8.9\n", + "#working out of insulation coordination\n", + "from math import sqrt\n", + "#given data\n", + "V=220.#voltage(in kV) of substation\n", + "BIL=1050.#value of BIL(in kV)\n", + "BtoS=1.24#ratio of BIL to SIL\n", + "\n", + "#calculation\n", + "Vh=245.#highest voltage(in kV)\n", + "Vg=Vh*sqrt(2.)/sqrt(3)#highest system voltage\n", + "Vs=3*Vg#expected switching voltage(in kV)\n", + "Vfw=760.#impulse sparkover voltage(in kV)\n", + "Vd1=690.#discharge voltage(in kV) for 5 kA\n", + "Vd2=615.#discharge voltage(in kV) for 2 kA\n", + "#SIL = BIL/BtoS = 846 ~ 850 kV\n", + "SIL=850.#value of SIL(in kV)\n", + "Pmlig=(BIL-Vd1)/BIL#protective margin for lightning impulses\n", + "Pmswi=(SIL-Vd2)/SIL#protective margin for switching gears\n", + "Pmspr=(BIL-Vfw)/BIL#margin when lightning arrester just sparks\n", + "#results\n", + "print 'The protective margin for lightning impulses is (percentage) = ',round(Pmlig*100,1)\n", + "print '\\nThe protective margin for switching gears is (percentage) = ',round(Pmswi*100,1)\n", + "print '\\nThe margin when lightning arrester just sparks is (percentage) = ',round(Pmspr*100,1)\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter9.ipynb b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter9.ipynb new file mode 100644 index 00000000..13f9024c --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/Chapter9.ipynb @@ -0,0 +1,452 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Non Destructive testing of materials and electrical apparatus" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 395" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The volume resistivity is (ohmcm) = 1.226e+14\n" + ] + } + ], + "source": [ + "#example 9.1\n", + "#calculation of the volume resistivity\n", + "from math import pi\n", + "#given data\n", + "V=1000.#applied voltage(in V)\n", + "Rs=10**7#standard resistance(in ohm)\n", + "n=3000.#universal shunt ratio\n", + "Ds=33.3#deflection(in cm) for Rs\n", + "D=3.2#deflection(in cm)\n", + "d=10.#diameter(in cm) of the electrodes\n", + "t=2*10**-1#thickness(in cm) of the specimen\n", + "\n", + "#calculation\n", + "G=V/(Rs*n*Ds)#galvanometer sensitivity\n", + "R=V/(D*G)#resistance of the specimen\n", + "r=d/2#radius of the electrodes\n", + "rho=(pi*r**2*R)/t#volume resistivity\n", + "#results\n", + "print '%s %.3e' %('The volume resistivity is (ohmcm) = ',rho)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 395" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistivity of the specimen is (ohmcm) = 1.031e+13\n" + ] + } + ], + "source": [ + "#example 9.2\n", + "#calculation of resistivity of the specimen\n", + "import math\n", + "from math import log\n", + "#given data\n", + "tm=30.#time (in minute)\n", + "ts=20.#time(in second)\n", + "Vn=1000.#voltage(in V) to which the condenser was charged\n", + "V=500.#voltage(in V) fall to\n", + "C=0.1*10**-6#capacitance(in Farad)\n", + "d=10.#diameter(in cm) of the electrodes\n", + "th=2*10**-1#thickness(in cm) of the specimen\n", + "\n", + "#calculation\n", + "t=(tm*60)+ts\n", + "R=t/(C*log(Vn/V))#resistance\n", + "r=d/2#radius of the electrodes\n", + "rho=(math.pi*r**2*R)/th#volume resistivity\n", + "#results\n", + "print '%s %.3e' %('The resistivity of the specimen is (ohmcm) = ',rho)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 396" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The dielectric constant is 4.2\n", + "The complex permittivity(in F/m)is 3.714e-11 -4.456e-14 j\n" + ] + } + ], + "source": [ + "#example 9.3\n", + "#calculation of dielectric constant and complex permittivity of bakelite\n", + "from cmath import pi\n", + "#given data\n", + "C=147*10**-12#capacitance(in Farad)\n", + "Ca=35*10**-12#air capacitance(in Farad)\n", + "tandelta=0.0012\n", + "epsilon0=(36*pi*10**9)**-1#electrical permittivity(in F/m) of free space\n", + "\n", + "\n", + "#calculation\n", + "epsilonr=C/Ca#dielectric constant\n", + "Kdash=epsilonr\n", + "Kdashdash=tandelta*Kdash\n", + "Kim=complex(Kdash,-Kdashdash)\n", + "epsilonast=epsilon0*Kim\n", + "\n", + "print 'The dielectric constant is ',epsilonr\n", + "print '%s %.3e %.3e %s' %('The complex permittivity(in F/m)is ',epsilonast.real,epsilonast.imag,'j')\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 396" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The capacitance is (pF) = 500.0\n", + "\n", + "The value of tandelta of bushing is 0.00125\n" + ] + } + ], + "source": [ + "#example 9.4\n", + "#calculation of capacitance and tandelta of bushing\n", + "from math import pi\n", + "#given data\n", + "R3=3180.#resistance(in ohm)\n", + "R4=636.#resistance(in ohm)\n", + "Cs=100.#standard condenser(in pF)\n", + "f=50.#frequency(in Hz)\n", + "C3=0.00125*10**-6#capacitance(in farad)\n", + "\n", + "#calculation\n", + "omega=2*pi*f\n", + "Cx=R3*Cs/R4#unknown capacitance\n", + "tandelta=omega*C3*R3\n", + "#results\n", + "print 'The capacitance is (pF) = ',Cx\n", + "print '\\nThe value of tandelta of bushing is ',round(tandelta,5)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 396" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The dielectric constant is 2.19\n", + "\n", + "The value of tandelta of the transformer oil is 0.00044\n" + ] + } + ], + "source": [ + "#example 9.5\n", + "#calculation of dielectric constant and tandelta of the transformer oil\n", + "\n", + "#given data\n", + "f=1*10**3#frequency(in Hz)\n", + "C1=504.#capacitance(in pF) for standard condenser and leads\n", + "D1=0.0003#dissipation factor for standard condenser and leads\n", + "C2=525.#capacitance(in pF) for standard condenser in parallel with the empty test cell\n", + "D2=0.00031#dissipation factor for standard condenser in parallel with the empty test cell\n", + "C3=550.#capacitance(in pF) for standard condenser in parallel with the test cell and oil\n", + "D3=0.00075#dissipation factor for standard condenser in parallel with the test cell and oil\n", + "\n", + "#calculation\n", + "Ctc=C2-C1#capacitance of the test cell\n", + "Ctcoil=C3-C1#capacitance of the test cell + oil\n", + "epsilonr=Ctcoil/Ctc#dielectric constant of oil\n", + "deltaDoil=D3-D2#deltaD of oil\n", + "#results\n", + "print 'The dielectric constant is ',round(epsilonr,2)\n", + "print '\\nThe value of tandelta of the transformer oil is ',round(deltaDoil,5)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 397" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The charge transferred from the cavity is (pC) = 0.92\n" + ] + } + ], + "source": [ + "#example 9.6\n", + "#calculation of magnitude of the charge transferred from the cavity\n", + "from math import pi\n", + "#given data\n", + "Vd=0.2#discharge voltage(in V)\n", + "s=1#sensitivity(in pC/V)\n", + "epsilonr=2.5#relative permittivity\n", + "epsilon0=(36*pi*10**9)**-1#electrical permittivity(in F/m) of free space\n", + "d1=1*10**-2#diameter(in m) of the cylindrical disc\n", + "t1=1*10**-2#thickness(in m) of the cylindrical disc\n", + "d2=1*10**-3#diameter(in m) of the cylindrical cavity\n", + "t2=1*10**-3#thickness(in m) of the cylindrical cavity\n", + "\n", + "\n", + "#calculation\n", + "Dm=Vd*s#discharge magnitude\n", + "Ca=epsilon0*(pi*(d2/2)**2)/t2#capacitance of the cavity\n", + "Cb=epsilon0*epsilonr*(pi*(d2/2)**2)/(t1-t2)#capacitance\n", + "qc=((Ca+Cb)/Cb)*Dm\n", + "#results\n", + "print 'The charge transferred from the cavity is (pC) = ',round(qc,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 397" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The dielectric constant is 5.8\n", + "\n", + "The loss factor tandelta is 5e-06\n" + ] + } + ], + "source": [ + "#example 9.7\n", + "#calculation of dielectric constant and loss factor tandelta\n", + "from math import pi\n", + "#given data\n", + "R3=1000./pi#resistance(in ohm) in CD branch\n", + "R4=62.#variable resistance(in ohm)\n", + "Cs=100.*10**-12#standard capacitance(in F)\n", + "epsilon0=8.854*10**-12#electrical permittivity(in F/m) of free space\n", + "f=50.#frequency(in Hz)\n", + "C3=50.*10**-9#variable capacitor(in F)\n", + "d=1.*10**-3#thickness(in m) of sheet\n", + "a=100.*10**-4#electrode effective area(in m**2)\n", + "\n", + "#calculation\n", + "Cx=R3*Cs/R4\n", + "epsilonr=Cx*d/(epsilon0*a)\n", + "omega=2*pi*f\n", + "tandelta=omega*C3*R3*d\n", + "#results\n", + "print 'The dielectric constant is ',round(epsilonr,2)\n", + "print '\\nThe loss factor tandelta is ',round(tandelta,7)\n", + "#In equation of tandelta d is multiplied\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 398" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage across AD branch at balance is (V) = 0.1\n" + ] + } + ], + "source": [ + "#example 9.8\n", + "#calculation of voltage at balance\n", + "from math import pi,sqrt\n", + "#given data\n", + "V=10000#applied voltage(in V)\n", + "R3=1000/pi#resistance(in ohm) in CD branch\n", + "R4=62#variable resistance(in ohm)\n", + "Cs=100*10**-12#standard capacitance(in F)\n", + "f=50#frequency(in Hz)\n", + "C3=50*10**-9#variable capacitor(in F)\n", + "\n", + "#calculation\n", + "Rx=C3*R4/Cs\n", + "Cx=R3*Cs/R4\n", + "omega=2*pi*f\n", + "zx=complex(Rx,-1/(omega*Cx))\n", + "VR4=R4*V/(R4+zx)\n", + "MVR4=sqrt((VR4.real)**2+VR4.imag**2)#magnitude\n", + "#results\n", + "print 'The voltage across AD branch at balance is (V) = ',round(MVR4,1)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 399" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum value of capacitance is (nF) = 11.1\n", + "\n", + "The minimum value of capacitance is (pF) = 10.0\n", + "\n", + "The maximum value of tandelta is 3.87\n", + "\n", + "The minimum value of tandelta is 3.14e-05\n" + ] + } + ], + "source": [ + "#example 9.9\n", + "#calculation of maximum and minimum value of capacitance and tandelta\n", + "from math import pi\n", + "#given data\n", + "R3min=100.#minimum value of R3 resistance(in ohm)\n", + "R3max=11100.#maximum value of R3 resistance(in ohm)\n", + "R4min=100.#minimum value of R4 resistance(in ohm)\n", + "R4max=1000.#maximum value of R4 resistance(in ohm)\n", + "Cs=100.*10**-12#standard capacitance(in farad)\n", + "C3min=1.*10**-9#minimum value of C3 capacitance(in farad)\n", + "C3max=1.11*10**-6#maximum value of C3 capacitance(in farad)\n", + "f=50.#frequency(in Hz)\n", + "\n", + "#calculation\n", + "Cxmax=R3max*Cs/R4min\n", + "Cxmin=R3min*Cs/R4max\n", + "omega=2*pi*f\n", + "tandeltamax=omega*R3max*C3max\n", + "tandeltamin=omega*R3min*C3min\n", + "#results\n", + "print 'The maximum value of capacitance is (nF) = ',round(Cxmax*10**9,1)\n", + "print '\\nThe minimum value of capacitance is (pF) = ',round(Cxmin*10**12)\n", + "print '\\nThe maximum value of tandelta is ',round(tandeltamax,2)\n", + "print '%s %.2e' %('\\nThe minimum value of tandelta is ',tandeltamin)\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter2.png b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter2.png Binary files differnew file mode 100644 index 00000000..78f269d9 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter2.png diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter3.png b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter3.png Binary files differnew file mode 100644 index 00000000..ecdd0b3d --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter3.png diff --git a/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter4.png b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter4.png Binary files differnew file mode 100644 index 00000000..6b19b460 --- /dev/null +++ b/High_Voltage_Engineering_by_V_Kamaraju_,_M_S_Naidu/screenshots/chapter4.png diff --git a/sample_notebooks/SalilKapur/IntroductionConcept_of_Stress.ipynb b/sample_notebooks/SalilKapur/IntroductionConcept_of_Stress.ipynb new file mode 100644 index 00000000..050d69c7 --- /dev/null +++ b/sample_notebooks/SalilKapur/IntroductionConcept_of_Stress.ipynb @@ -0,0 +1,282 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Introduction—Concept of Stress\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.1, Page number 18 " + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Shearing Stress in Pin A = 6790.6 psi\n", + "Case(b): Shearing Stress in Pin C = 7639 psi\n", + "Case(c): Largest Normal Stress in Link ABC = 2286 psi\n", + "Case(d): Average Shearing Stress at B = 171 psi\n", + "Case(e): Bearing Stress in Link at C = 6000 psi\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Fac = 750 #Force on rod AC(lb)\n", + "D = 0.375 #Diameter at the upper junction of rod ABC(in)\n", + "\n", + "\n", + "#Calculation \n", + "#Case(a)\n", + "A=(1/4)*((math.pi)*pow(D,2)) #Area at the upper junction of rod ABC(in^2) \n", + "tA=(Fac/A) #Shearing Stress in Pin A(psi) \n", + "#Case(b) \n", + "Ab=(1/4)*((math.pi)*pow(0.25,2)) #Area at the lower junction of rod ABC(in^2)\n", + "tC=(((1/2)*Fac)/Ab) #Shearing Stress in Pin C(psi)\n", + "#Case(c)\n", + "Anet=(3/8)*(1.25-0.375) #Area of cross section at A(in^2)\n", + "sA=(Fac/Anet) #Largest Normal Stress in Link ABC(psi)\n", + "#Case(d)\n", + "F1=750/2 #Force on each side(lb)\n", + "Ad=(1.25*1.75) #Area at junction B(in^2)\n", + "tB=(F1/Ad) #Average Shearing Stress at B\n", + "#Case(e)\n", + "Ae=0.25*0.25 #Area at point C(in^2)\n", + "sB=(F1/Ae) #Bearing Stress in Link at C\n", + "\n", + "\n", + "#Result\n", + "print('Case(a): Shearing Stress in Pin A = %.1f psi' %tA)\n", + "print('Case(b): Shearing Stress in Pin C = %.f psi' %tC)\n", + "print('Case(c): Largest Normal Stress in Link ABC = %.f psi' %sA)\n", + "print('Case(d): Average Shearing Stress at B = %.f psi' %tB)\n", + "print('Case(e): Bearing Stress in Link at C = %.f psi' %sB)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.2, Page number 19" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Diameter of the bolt = 28 mm\n", + "Case(a): Dimension b at Each End of the Bar = 62 mm\n", + "Case(a): Dimension h of the Bar = 34.300000 mm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "P = 120 #Maximum allowable tension force \n", + "s = 175 #Maximum allowable stress\n", + "t = 100 #Maximum allowable stress\n", + "Sb = 350 #Maximum allowable stress\n", + "\n", + "\n", + "#Calculation\n", + "#Case(a)\n", + "F1=P/2 #Current(A)\n", + "d=math.sqrt(((P/2)*1000)/((22/(4*7))*(100000000))) #Diameter of bolt(m)\n", + "d=d*1000 #Diameter of bolt(mm)\n", + "d=round(d,0) #Rounding of the value of diameter of bolt(mm)\n", + "Ad=(0.020*0.028) #Area of cross section of plate \n", + "tb=((P*1000)/Ad)/(1000000) #Stress between between the 20-mm-thick plate and the 28-mm-diameter bolt\n", + "tb=round(tb,0) #Rounding of the above calculated stress to check if it is less than 350\n", + "a=(P/2)/((0.02)*(175)) #Dimension of cross section of ring \n", + "a=round(a,2) #Rounding dimension of cross section of ring to two decimal places\n", + "b=28 + (2*(a)) #Dimension b at Each End of the Bar\n", + "b=round(b,2) #Rounding the dimension b to two decimal places \n", + "h=(P)/((0.020)*(175)) #Dimension h of the Bar\n", + "h=round(h,1) #Rounding dimension h of bar to 1 decimal place\n", + "\n", + "\n", + "#Result\n", + "print ('Case(a): Diameter of the bolt = %.f mm' %d)\n", + "print ('Case(a): Dimension b at Each End of the Bar = %.f mm' %b)\n", + "print ('Case(a): Dimension h of the Bar = %f mm' %h)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 1.3, Page number 34" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Diameter of the bolt = 16.730000 mm\n", + "Case(a): Dimension b at Each End of the Bar = 22.000000 mm\n", + "Case(a): Dimension h of the Bar = 6.000000 mm\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "Su = 600 #ultimate normal stress(MPa) \n", + "FS = 3.3 #Factor of safety with respect to failure\n", + "tU=350 #Ultimate shearing stress(MPa)\n", + "Cx=40 #X Component of reaction at C(kN)\n", + "Cy=65 #Y Component of reaction at C(kN)\n", + "Smax=300 #Allowable bearing stress of the steel \n", + "\n", + "#Calculation\n", + "C=math.sqrt((math.pow(40,2))+(math.pow(65,2)))\n", + "\n", + "#Case(a)\n", + "P=(15*0.6 + 50*0.3)/(0.6) #Allowable bearing stress of the steel(MPa)\n", + "Sall=(Su/FS) #Allowable Stress(MPa)\n", + "Sall=round(Sall,1) #Rounding Allowable stress to 1 decimal place(MPa)\n", + "Areqa=(P/(Sall*(1000))) #Cross Sectional area(m^2)\n", + "Areqa=round(Areqa,5) #Rounding cross sectional area to 5 decimal places(m^2)\n", + "dAB=math.sqrt(((Areqa)*(4))/(22/7)) #Diameter of AB(m)\n", + "dAB=dAB*1000 #Diameter of AB(mm)\n", + "dAB=round(dAB,2) #Rounding Diameter of AB(mm)\n", + "\n", + "#Case(b)\n", + "tALL=tU/FS #Stress(MPa)\n", + "tALL=round(tALL,1) #Rounding of Stress\n", + "AreqC=((C/2)/tALL) #Cross sectional area(m^2)\n", + "AreqC=AreqC*1000 \n", + "AreqC=round(AreqC,0) #Rounding the cross sectional area\n", + "dC=math.sqrt((4*AreqC)/(22/7)) #Diameter at point C\n", + "dC=round((dC+1),0) #Rounding of the diameter at C\n", + "\n", + "#Case(c)\n", + "\n", + "Areq=((C/2)/Smax) \n", + "Areq=Areq*1000 #Cross sectional area(mm^2)\n", + "t=(Areq/22) #Thickness of the bracket\n", + "t=round(t,0)\n", + "\n", + "#Result\n", + "print ('Case(a): Diameter of the bolt = % f mm' %dAB)\n", + "print ('Case(a): Dimension b at Each End of the Bar = % f mm' %dC)\n", + "print ('Case(a): Dimension h of the Bar = % f mm' %t)" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Example 1.4, Page number 35" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Case(a): Control Rod = 5.263672 kips\n", + "Case(b): Bolt at B = 5.156250 kips\n", + "Case(c): Bolt at D = 6.865179 kips\n", + "Case(d): Bolt at C = 5.238095 kips\n", + "Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= 5.156250 kips\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#Variable declaration\n", + "tU=40 #ultimate tensile stress\n", + "sU=60 #ultimate shearing stress\n", + "FS=3 #Mimnimum factor of safety\n", + "dA=(7/16) #Diameter of bolt at A(in)\n", + "dB=3/8 #Diameter of bolt at B(in) \n", + "dD=3/8 #Diameter of bolt at D(in)\n", + "dC=1/2 #Diameter of bolt at C(in)\n", + "\n", + "\n", + "#Calculation\n", + "Sall=(sU/FS) #Total tensile stress(kips)\n", + "B=Sall*((1/4)*(22/7)*(pow((7/16),2))) #Allowable force in the control rod(kips)\n", + "C1=1.75*(B) #Control Rod(kips)\n", + "tall=(tU/FS) #Total shearing stress\n", + "B=2*(tall*(1/4)*(22/7)*(3/8)*(3/8)) #Allowable magnitude of the force B exerted on the bolt\n", + "C2=1.75*B #Bolt at B(kips) \n", + "D=B #Bolt at D. Since this bolt is the same as bolt B, the allowable force is same(kips) \n", + "C3=2.33*D #Bolt at D(kips)\n", + "C4=2*(tall*(1/4)*(22/7)*(1/2)*(1/2)) #Bolt at C(kips) \n", + "list1=[C1,C2,C3,C4] #Adding all the maximum allowable forces on C(kips) \n", + "\n", + "\n", + "#Result\n", + "print ('Case(a): Control Rod = % f kips' %C1)\n", + "print ('Case(b): Bolt at B = % f kips' %C2)\n", + "print ('Case(c): Bolt at D = % f kips' %C3)\n", + "print ('Case(d): Bolt at C = % f kips' %C4)\n", + "print ('Summary. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the smallest value, namely:= % f kips' %min(list1));" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |
