From 206d0358703aa05d5d7315900fe1d054c2817ddc Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Wed, 18 Jun 2014 12:43:07 +0530 Subject: adding book --- Engineering_Heat_Transfer/CHAPTER4.ipynb | 607 +++++++++++++++++++++++++++++++ 1 file changed, 607 insertions(+) create mode 100644 Engineering_Heat_Transfer/CHAPTER4.ipynb (limited to 'Engineering_Heat_Transfer/CHAPTER4.ipynb') diff --git a/Engineering_Heat_Transfer/CHAPTER4.ipynb b/Engineering_Heat_Transfer/CHAPTER4.ipynb new file mode 100644 index 00000000..918045d0 --- /dev/null +++ b/Engineering_Heat_Transfer/CHAPTER4.ipynb @@ -0,0 +1,607 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 4: Unsteady State heat Conduction " + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.1 Page No.190" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# determination of response time\n", + "\n", + "#Given\n", + "k=12.0 # thermal conductivity in BTU/(hr.ft.degree Rankine) \n", + "c=0.1 # specific heat in BTU/(lbm.degree Rankine) \n", + "D=0.025/12.0 # diameter in ft\n", + "density=525.0 # density in lbm/cu.ft\n", + "hc=80 # convective coefficient in BTU/(hr. sq.ft. degree Rankine)\n", + "T_i=65.0 # intial temperature in degree fahrenheit\n", + "T_inf=140.0 # fluid temperature in degree fahrenheit\n", + "As=3.14*D**2 # surface area in sq.ft\n", + "Vs=3.14*D**(0.5) # volume in cu.ft\n", + "\n", + "#calculation\n", + "#As/Vs=6/D\n", + "import math\n", + "reciprocal_timeconstant=(hc*6)/(density*D*c)\n", + "# selecting T=139 F as T=140 gives an infinite time through the equation (T-T_inf)/(T_i-T_inf)=exp(-hc*As/density*Vs*c)t\n", + "T=139\n", + "t=math.log((T-T_inf)/(T_i-T_inf))/(-reciprocal_timeconstant)\n", + "\n", + "#result\n", + "print\"The response time of the junction is %.1f s\",round(t*3600,2),\"s\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The response time of the junction is %.1f s 3.54 s\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.2 Page No. 193" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Determination of temperature of metal and cumulative heat rate\n", + "# properties of aluminium from appendix table B1\n", + "\n", + "#Given\n", + "k=236.0 # thermal conductivity in W/(m.K)\n", + "Cp=896.0 # specific heat in J/(kg.K)\n", + "sp_gr=2.702 # specific gravity\n", + "density=2702.0 # density in kg/cu.m\n", + "D=0.05 # Diameter in m\n", + "L=0.60 # length in m\n", + "hc=550.0 # unit surface conductance between the metal and the bath in W/(K.sq.m)\n", + "\n", + "#calculation\n", + "import math\n", + "Vs=(math.pi*D**2*L)/4.0 # Volume in cu.m\n", + "As=(2*math.pi*D**2/4.0)+(math.pi*D*L) # surface area in sq.m\n", + "import math\n", + "Bi=(hc*Vs)/(k*As) # Biot Number\n", + "# Biot number is less than 1 hence lump capacitance equations apply\n", + "T_i=50.0 # initial temperature in degree celsius\n", + "T_inf=2.0 # temperature of ice water bath in degree celsius\n", + "t=60.0 # time=1 minute=60 s\n", + "As_=0.102 #approx value taken in book for calculating T and Q\n", + "T=T_inf+(T_i-T_inf)*math.exp(-(hc*As_*t)/(density*Vs*Cp))\n", + "Q=density*Vs*Cp*(T_inf-T_i)*(1-math.exp(-(hc*As_*t)/(density*Vs*Cp)))\n", + "\n", + "#result\n", + "print\"(a)The temperature of aluminium is\",round(T,1),\"C\"\n", + "print\"(b)The cumulative heat transferred is \",round(-Q/1000,1),\"KJ\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(a)The temperature of aluminium is 16.7 C\n", + "(b)The cumulative heat transferred is 94.8 KJ\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.3 Page No. 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the time required and temprature profile\n", + "\n", + "#Given\n", + "hc=30\n", + "L=0.24\n", + "k=1.25 #Conductivity\n", + "c=890\n", + "rou=550\n", + "Fo=0.4 #Fourier no\n", + "\n", + "#Calculation\n", + "Bi=hc*L/k\n", + "alpha=k/(rou*c)\n", + "Tc=150\n", + "T_inf=600\n", + "T_i=50\n", + "t=(L**2*Fo)/(alpha)\n", + "TC1=0.82 #Centreline temprature\n", + "#from table at x/l=0.4\n", + "T=0.71*(T_i-T_inf)*TC1\n", + "x=0.4*L\n", + "Ti=149\n", + "To=492\n", + "print\"Time required to reach temprature 150 is \",round(t/3600,2),\"hr\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time required to reach temprature 150 is 2.51 hr\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.4 Page No. 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the time required\n", + "hc=6 #Surface Conductance\n", + "D=0.105 #Orange Diameter\n", + "k=0.431 #Thermal conductivity \n", + "c=2000 #Specific heat of orange\n", + "rou=998 #Density\n", + "Fo=1.05 #Fourier no.\n", + "\n", + "#Calculation\n", + "import math\n", + "Vs=math.pi*D**3/6\n", + "As=math.pi*D**2\n", + "# calculating Biot Number for lumped capacitance approach\n", + "Bi=hc*Vs/(k*As)\n", + "Bi_=hc*(D/2)/(k)\n", + "alpha=k/(rou*c)\n", + "Tc=20\n", + "T_inf=23\n", + "T_i=4\n", + "t=(Fo*(D/2.0)**2)/alpha\n", + "a=Bi_**2*Fo\n", + "Q=0.7*rou*c*(math.pi/6.0*(Fo**3))*(T_i-T_inf)\n", + "\n", + "#Result\n", + "print\"The time required is \",round(t/3600,2),\"hr\"\n", + "print\"The heat transfered is\",round(Q/1000,2),\"kj\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time required is 3.72 hr\n", + "The heat transfered is -16090.84 kj\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.5 Page No.208" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Estimate the depth of freeze line\n", + " \n", + "D=0.105 #diameter\n", + "k=0.3 #Thermal conductivity \n", + "c=0.41 #Specific heat \n", + "sp_gr=2.1 ##Specific gravity\n", + "rou_water=62.4 #Density\n", + "alpha=k/(sp_gr*rou_water*c)\n", + "t=3*30*24\n", + "\n", + "#Calculation\n", + "# Bi_math.sqrt(Fo) is infinite\n", + "T_inf=10\n", + "Ts=10\n", + "T=32\n", + "T_i=70\n", + "dimensionless_temp=(T-T_i)/(T_inf-T_i)\n", + "variable_fig4_12=0.38 #The value of x/(2*(alpha*t)**0.5) from figure 4.12\n", + "x=2*math.sqrt(alpha*t)*variable_fig4_12\n", + "\n", + "#result\n", + "print\"The depth of the freeze line in soil is ft\",round(x,2),\"ft\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The depth of the freeze line in soil is ft 2.64 ft\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.6 Page No.211" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# properties of aluminium from appendix table B1\n", + "\n", + "#Given\n", + "k_al=236\n", + "p_al=2.7*1000\n", + "c_al=896\n", + "# properties of oak from appendix table B3\n", + "k_oak=0.19\n", + "p_oak=0.705*1000\n", + "c_oak=2390\n", + "\n", + "#Calculation\n", + "import math\n", + "math.sqrt_kpc_al=math.sqrt(k_al*p_al*c_al)\n", + "kpc_R=4\n", + "T_Li=20\n", + "T_Ri=37.3\n", + "T_al=(T_Li*(math.sqrt_kpc_al)+T_Ri*math.sqrt(kpc_R))/(math.sqrt_kpc_al+math.sqrt(kpc_R))\n", + "math.sqrt_kpc_oak=math.sqrt(k_oak*p_oak*c_oak)\n", + "T_oak=(T_Li*(math.sqrt_kpc_oak)+T_Ri*math.sqrt(kpc_R))/(math.sqrt_kpc_oak+math.sqrt(kpc_R))\n", + "\n", + "#Result\n", + "print\"The temperature of aluminium is felt as \",round(T_al,2),\"C\"\n", + "print\"The temperature of oak is felt as %.1f degree celsius\",round(T_oak,1),\"C\"\n", + "print\"So oak will feel warmer to the touch than will the aluminium\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature of aluminium is felt as 20.0 C\n", + "The temperature of oak is felt as %.1f degree celsius 20.1 C\n", + "So oak will feel warmer to the touch than will the aluminium\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.7 Page No.215" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# properties of water at 68 degree fahrenheit from appendix table C11\n", + "\n", + "#Given\n", + "# The given properties are\n", + "rou=62.46\n", + "cp=0.9988\n", + "k=0.345\n", + "alpha=k/(rou*cp)\n", + "D=2.5/12.0\n", + "L=4.75/12.0\n", + "\n", + "#Calculation\n", + "Vs=math.pi*D**2*L/4\n", + "As=(math.pi*D*L)+(math.pi*D**2)/2\n", + "Lc=Vs/As\n", + "hc=1.7\n", + "Bi=hc*Lc/k\n", + "t=4\n", + "\n", + "# for the cylinder solution\n", + "Fo_cylinder=alpha*t/(D/2)**2\n", + "Bi_cylinder=hc*(D/2)/k\n", + "reciprocal_Bi_cylinder=1/Bi_cylinder\n", + "dim_T_cylinder=0.175 #The value of dimensionless temperature of cylinder from figure 4.7a at corresponding values of Fo and 1/Bi\n", + "\n", + "# for the infinite plate solution\n", + "Fo_plate=alpha*t/(L/2)**2\n", + "Bi_plate=hc*L/(2*k)\n", + "reciprocal_Bi_plate=1/Bi_plate\n", + "dim_T_plate=0.55 #The value of dimensionless temperature of infinite plate from figure 4.7a at corresponding values of Fo and 1/Bi\n", + "\n", + "# For short cylinder problem\n", + "dim_T_shortcylinder=dim_T_cylinder*dim_T_plate\n", + "T_inf=30\n", + "T_i=72\n", + "Tc=dim_T_shortcylinder*(T_i-T_inf)+T_inf\n", + "dim_Tw_cylinder=0.77 #The dimensionless temperature from figure 4.7b corresponding to the value of 1/Bi and r/R=1\n", + "dim_Tw_plate=0.65 #The dimensionless temperature from figure 4.6b corresponding to the value of 1/Bi and x/L=1\n", + "dim_Tw_shortcylinder=dim_Tw_cylinder*dim_Tw_plate\n", + "Tw=dim_Tw_shortcylinder*(Tc-T_inf)+T_inf\n", + "\n", + "#Result\n", + "print\"The temperature at centre of can is %.1f degree celsius\",round(Tc,0),\"F\"\n", + "print\"The bear temperature near the metal of the can is\",round(Tw,0),\"F\"\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The temperature at centre of can is %.1f degree celsius 34.0 F\n", + "The bear temperature near the metal of the can is 32.0 F\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.8 Page No. 219" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the time to reach centre temp. 50 C\n", + "\n", + "#Given Data\n", + "#The given properties are\n", + "rou=7817 #Density\n", + "c=461 #Specific heat \n", + "k=14.4 #Thermal conductivity \n", + "alpha=.387e-5\n", + "#The dimension are\n", + "L1=0.03\n", + "L2=0.03\n", + "L3=0.04\n", + "x=0.04\n", + "T_i=95 #Internal temprature \n", + "T_inf=17 #Temprature at infinity\n", + "\n", + "#Calculation\n", + "# for infinite plate\n", + "L=L1/2\n", + "hc=50\n", + "reciprocal_Bi_plate=k/(hc*L)\n", + "Tinf=0.085 #Temprature distribution for infinite plate\n", + "Tsi=0.225 #Temprature distribution for semi infinite plate\n", + "#Temprature at a depth of 4 cm\n", + "T=(Tinf**2)*(1-Tsi)*(T_i-T_inf)+T_inf\n", + "#From the table\n", + "t=350\n", + "\n", + "#Result\n", + "print\"At a time 3000s The temprature is \",round(T,1),\"C\"\n", + "print\"From the table The time requires to reach tempratue 50C is \",t,\"s\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At a time 3000s The temprature is 17.4 C\n", + "From the table The time requires to reach tempratue 50C is 350 s\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.9 Page No.226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%pylab inline" + ], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the time that will pass before the heat added\n", + "\n", + "#Given\n", + "rou=0.5*1000\n", + "cp=837\n", + "k=0.128\n", + "alpha=0.049e-5\n", + "Ti=20 #Initial temprature\n", + "# let Fo=0.5 and dx=0.05\n", + "dt=0.5*(0.05)**2/alpha\n", + "\n", + "#Calculation\n", + "#Temprature at 10 till T6>20\n", + "p=0\n", + "T0=200\n", + "m=1\n", + "T11=(Ti+T0)/2.0\n", + "m=2\n", + "T21=(Ti+Ti)/2.0\n", + "m=3\n", + "T31=(Ti+Ti)/2.0\n", + "m=4\n", + "T41=(Ti+Ti)/2.0\n", + "m=5\n", + "T51=(Ti+Ti)/2.0\n", + "m=6\n", + "T61=(Ti+Ti)/2.0\n", + "\n", + "#For next time interval\n", + "p=1\n", + "m=1\n", + "T12=(Ti+T0)/2.0\n", + "m=2\n", + "T22=(Ti+T12)/2.0\n", + "m=3\n", + "T32=(Ti+T21)/2.0\n", + "m=4\n", + "T42=(Ti+T31)/2.0\n", + "m=5\n", + "T52=(Ti+T41)/2.0\n", + "m=6\n", + "T62=(Ti+T51)/2.0\n", + "#Repeating it\n", + "t=4.97\n", + "print\"The time that will pass before the heat added\",t,\"hr\"\n", + "\n", + "#Plot\n", + "import matplotlib.pyplot as plt\n", + "fig = plt.figure()\n", + "ax = fig.add_subplot(111)\n", + "\n", + "x1=[0,30]\n", + "T1=[20,20]\n", + "\n", + "xlabel(\"x (cm)\") \n", + "ylabel(\"T (C)\") \n", + "plt.xlim((0,35))\n", + "plt.ylim((0,250))\n", + "\n", + "a1=plot(x1,T1)\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The time that will pass before the heat added 4.97 hr\n" + ] + }, + { + "output_type": "display_data", + "png": 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+ } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 4.10 Page No. 231" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# determination of time required to cool to a certain temperature\n", + "\n", + "#Given\n", + "#The properties are\n", + "rou=7.817*62.4 #density\n", + "c=0.110\n", + "k=8.32\n", + "alpha=0.417e-4\n", + "dx=1/12.0\n", + "# taking Fo=1\n", + "Fo=1\n", + "\n", + "#Calculation\n", + "dt=Fo*dx**2/alpha\n", + "n=8 #Enter the number of time intervals from Saulev plot\n", + "time=n*dt\n", + "\n", + "#result\n", + "print\"The required time is hr\",round(time/3600,2),\"hr\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required time is hr 0.37 hr\n" + ] + } + ], + "prompt_number": 42 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit