{ "metadata": { "name": "", "signature": "sha256:b84dc3021bf00fa69e8fa42a6cbf4b1ad563cdbb8726721608219def960b394a" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 9 : Heat transfer by the combined effect of conduction and convection" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.1 page : 159" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "D = 3./48; # diameter in ft\n", "L = 9./12; # Length of steam vessel in ft\n", "T1 = 210.; # Vessel temperature in degF\n", "T2 = 80.; # Air temperature in degF\n", "th0 = T1-T2; # Temperature difference in degF\n", "h = 1.44; # Assumed heat coefficient in Btu/hr-ft**2-degF\n", "C = math.pi*D; # Circumference of vessel in ft \n", "A = math.pi*D*D/4; # Area of vessel in ft**2\n", "\n", "# Calculations and Results\n", "# For copper\n", "k1 = 219; # Heat conductivity of copper in Btu/hr-ft-degF\n", "m1 = math.sqrt(h*C/(k1*A)); # in /ft\n", "th1 = th0*2/(math.exp(m1*L)+math.exp(-m1*L)); \n", "Tl1 = round(th1+T2); # The temperaure at the free end in degF\n", "print \"Temperature at free end of the copper rod is %d degF \"%(Tl1);\n", "\n", "# For iron\n", "k2 = 36; # heat conductivity of copper in Btu/hr-ft-degF\n", "m2 = math.sqrt(h*C/(k2*A)); # in /ft\n", "th2 = th0*2/(math.exp(m2*L)+math.exp(-m2*L)); \n", "Tl2 = th2+T2; # The temperaure at the free end in degF\n", "print \" Temperature at free end of the iron rod is %.2f degF \"%(Tl2);\n", "\n", "# For glass\n", "k3 = 0.64; # Heat conductivity of copper in Btu/hr-ft-degF\n", "m3 = math.sqrt(h*C/(k3*A)); # in /ft\n", "th3 = th0*2/(math.exp(m3*L)+math.exp(-m3*L));\n", "Tl3 = th3+T2; # The temperaure at the free end in degF\n", "print \" Temperature at free end of the glass rod is %.2f degF \"%(Tl3);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature at free end of the copper rod is 196 degF \n", " Temperature at free end of the iron rod is 151.80 degF \n", " Temperature at free end of the glass rod is 80.03 degF \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.2 page : 163" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "D = 3./48; # diameter in ft\n", "L = 9./12; # Length of steam vessel in ft\n", "T1 = 210.; # Vessel temperature in degF\n", "T2 = 80.; # Air temperature in degF\n", "th0 = T1-T2; # Temperature difference in degF\n", "h = 1.44; # Assumed heat coefficient in Btu/hr-ft**2-degF\n", "C = math.pi*D; # Circumference of vessel in ft \n", "A = math.pi*D*D/4; # Area of vessel in ft**2\n", "\n", "# Calculations \n", "k = 36; # heat conductivity of copper in Btu/hr-ft-degF\n", "m = math.sqrt(h*C/(k*A)); # in /ft\n", "q = k*A*m*th0*(math.exp(m*L)-math.exp(-m*L))/(math.exp(m*L)+math.exp(-m*L)); \n", "# Heat loss by iron rod in Btu/hr\n", "\n", "# Results\n", "print \"The rate of heat loss by iron rod is %.1f Btu/hr\"%(q);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The rate of heat loss by iron rod is 19.2 Btu/hr\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.3 page : 170" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "x = 3./96; # Thickness of plate in ft\n", "k = 220.; # thermal conductivity in Btu/hr-ft-degF\n", "h1 = 480.; # Inner film coefficient in Btu/hr-ft**2-degF\n", "h2 = 1250.; # Outer film coefficient in Btu/hr-ft**2-degF\n", "\n", "# Calculations \n", "U = 1./((1/h1)+(x/k)+(1/h2)); # Overall heat transer coeeficient in Btu-hr-ft**2-degF\n", "\n", "# Results\n", "print \"Overall heat transfer coefficient is %.f Btu/hr-ft**2-degF\"%(U);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Overall heat transfer coefficient is 331 Btu/hr-ft**2-degF\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.4 page : 172" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "r2 = 3./96; # Outer radius in ft\n", "x = 0.1/12; # Thickness of plate in ft\n", "r1 = r2-x; # Outer radius in ft\n", "k = 200.; # thermal conductivity in Btu/hr-ft-degF\n", "h1 = 280.; # Inner film coefficient in Btu/hr-ft**2-degF\n", "h2 = 2000.; # Outer film coefficient in Btu/hr-ft**2-degF\n", "\n", "# Calculations \n", "U = 1/((r2/(h1*r1))+(r2*math.log(r2/r1)/k)+(1/h2)); # Overall heat transer coeeficient in Btu-hr-ft**2-degF\n", "\n", "# Results\n", "print \"Overall heat transfer coefficient is %d Btu/hr-ft**2-degF\"%(U);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Overall heat transfer coefficient is 184 Btu/hr-ft**2-degF\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.5 page : 176" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "Tc1 = 120.; # Inlet cold fluid temperature in degF\n", "Tc2 = 310.; # Outlet cold fluid temperature in degF\n", "Th1 = 500.; # Inlet hot fluid temperature in degF\n", "Th2 = 400.; # Outlet hot fluid temperature in degF\n", "\n", "# Calculations \n", "delt1 = Th2-Tc1; # Maximum temperature difference in degF\n", "delt2 = Th1-Tc2; # Minimum temperature difference in degF\n", "LMTD = (delt1-delt2)/math.log(delt1/delt2); # Log mean temperature difference \n", "\n", "# Results\n", "print \"The log mean temperature difference is %d degF\"%(LMTD) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The log mean temperature difference is 232 degF\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.6 page : 178" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "Tc1 = 120.; # Inlet cold fluid temperature in degF\n", "Tc2 = 310.; # Outlet cold fluid temperature in degF\n", "Th1 = 500.; # Inlet hot fluid temperature in degF\n", "Th2 = 400.; # Outlet hot fluid temperature in degF\n", "\n", "# Calculations \n", "K = (Tc2-Tc1)/(Th2-Tc2); # Temperature ratio\n", "R = (Th1-Th2)/(Tc2-Tc1); # Temperature ratio \n", "delt1 = Th2-Tc1; # Maximum temperature difference in degF\n", "delt2 = Th1-Tc2; # Minimum temperature difference in degF\n", "LMTD = (delt1-delt2)/math.log(delt1/delt2); # Log mean temperature difference\n", "f = 0.99; # Correction factor as seen from figure\n", "LMTDc = round(LMTD*f); # Corrected math.log mean temperature difference\n", "\n", "# Results\n", "print \"Log mean temperature difference is %d degF\"%(LMTDc);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Log mean temperature difference is 230 degF\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.7 page : 181" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "Do = 1./12; # Outside diameter of the condenser in ft\n", "Di = 0.902/12; # Outside diameter of the condenser in ft\n", "Ts = 81.7; # Steam temperature in degF\n", "Tw1 = 61.4; # Water inlet temperature in degF\n", "Tw2 = 69.9; # Water outlet temperature in degF\n", "k = 63.; # Thermal conductivity in Btu/hr-ft-degF\n", "v = 7.; # average velocity in ft/sec\n", "h1 = 1270.; # water side film coefficient i Btu/hr-ft**2-degF\n", "h2 = 1000.; # Steam side film coefficient in Btu/hr-ft**2-degF\n", "\n", "# Calculations \n", "U = 1/((Do/(Di*h1))+(Do*math.log(Do/Di)/(2*k))+(1/h2)); # Heat transfer coefficient\n", "LMTD = ((Ts-Tw1)-(Ts-Tw2))/math.log((Ts-Tw1)/(Ts-Tw2)); # Log mean temperature diff.\n", "m = 731300; # Saturated steam to be handled in lb/hr\n", "L = 1097.4-49.7; # Change in enthalpy in Btu/lb\n", "q = m*L; # Heat required in Btu/hr\n", "A = q/(U*LMTD); # Area of condenser in ft**2\n", "\n", "# Results\n", "print \"The area of steam condenser is %d ft**2\"%(A);\n", "\n", "# book answer is rounded. kindly check." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The area of steam condenser is 94927 ft**2\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.8 page : 182" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "# Variables\n", "Tc1 = 139.7; # Inlet cold fluid temperature in degF\n", "Tc2 = 59.5; # Outlet cold fluid temperature in degF\n", "Th1 = 108.7; # Inlet hot fluid temperature in degF\n", "Th2 = 97.2; # Outlet hot fluid temperature in degF\n", "\n", "# Calculations and Results\n", "delt1 = Tc1-Th2; # Maximum temperature difference in degF\n", "delt2 = Th1-Tc2; # Minimum temperature difference in degF\n", "LMTD = round((delt1-delt2)/math.log(delt1/delt2));\n", "print \" The math.log mean temperature difference is %d degF\"%(LMTD); \n", "\n", "m = 18210.; # Flow rate through tubes\n", "q = m*(Th2-Tc2); # Heat loss in Btu/hr\n", "A = 48.1; # Area in ft**2\n", "U = q/(A*LMTD); # Overall heat transfer coefficient\n", "print \" The overall heat transfer coefficient is %d Btu/hr-ft**2-degF \"%(U);\n", "\n", "\n", "# To calcalute using equations estabilished by correlation\n", "Ts = 113.; # Average tube temperature in degF\n", "Tf = (123.9+Ts)/2; # Film temperature in degF\n", "# At this temperature thermal properties are considered\n", "p1 = 61.7/32.2; # Density in slug/ft**3\n", "u1 = 1.38/32.2; # Vismath.cosity in slug/ft-hr\n", "Cp1 = 1*32.2; # Btu/slug/ft\n", "k1 = 0.366; # Thermal conductivity in Btu/hr-ft-degF\n", "D1 = 0.375/12; # Diameter in ft\n", "v1 = 7610.; # Velocity in ft/sec\n", "Nre1 = v1*D1*p1/u1; # Reynolds number\n", "Npr1 = u1*Cp1/k1; # Prandtls number\n", "Nnu1 = 0.33*Nre1**0.6*Npr1**(1./3); # Nusselt number\n", "h1 = Nnu1*k1/D1; # Heat transfer coefficient\n", "print \" The outer heat transfer coefficient is %d Btu/hr-ft**2-degF \"%(h1);\n", "\n", "# Taking the thermal properties at 78.3 degF\n", "p2 = 62.2/32.2; # Density in slug/ft**3\n", "u2 = 2.13/32.2; # Vismath.cosity in slug/ft-hr\n", "Cp2 = 1*32.2; # Heat capacity in Btu/slug/ft\n", "k2 = 0.348; # Thermal conductivity in Btu/hr-ft-degF\n", "D2 = 0.277/12; # Diameter in ft\n", "v2 = 7140; # Velocity in ft/sec\n", "Nre2 = v2*D2*p2/u2; # Reynolds number\n", "Npr2 = u2*Cp2/k2; # Prandtls number\n", "Nnu2 = 0.023*Nre2**0.8*Npr2**(0.4); # Nusselt number\n", "h2 = Nnu2*k2/D2; # Heat transfer coefficient\n", "print \" The inner heat transfer coefficient is %d Btu/hr-ft**2-degF\"%(h2);\n", "\n", "k3 = 58;\n", "U1 = 1/((D1/(D2*h2))+(D1*math.log(D1/D2)/(2*k3))+(1/h1)); # Heat transfer coefficient \n", "print \" The overall heat transfer coefficient accordind to estabilished correlation is %d Btu/hr-ft**2-degF \"%(U1);\n", "\n", "# note : rounding off error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The math.log mean temperature difference is 46 degF\n", " The overall heat transfer coefficient is 310 Btu/hr-ft**2-degF \n", " The outer heat transfer coefficient is 1567 Btu/hr-ft**2-degF \n", " The inner heat transfer coefficient is 631 Btu/hr-ft**2-degF\n", " The overall heat transfer coefficient accordind to estabilished correlation is 349 Btu/hr-ft**2-degF \n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.9 page : 188" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "To1 = 140.; # inlet temperature of oil in degF\n", "To2 = 90.; # Outlet temperature of oil in degf\n", "Cpo = 0.5; # Specific heat capacity in Btu/lb-degf\n", "Tw1 = 60.; # Inlet tempearture of water in degF\n", "Tw2 = 80.; # Outlet temperature of water in degF\n", "mo = 2000.; # Mass flow rate of oil in lb/hr\n", "\n", "# Calculations \n", "q = mo*Cpo*(To1-To2); # Heat transferred in Btu/hr\n", "Cpw = 1; # Heat capacity of water in Btu/hr\n", "mw = q/(Cpw*(Tw2-Tw1)); # Mass flow rate in lb/hr\n", "E1 = (Tw1-Tw2)/(Tw1-To2); # Effective ratio\n", "# Seeing the effective ratio and mass flow rate ratio, from the graph we get UA\n", "UA = 1.15*mo*Cpo;\n", "\n", "# Results\n", "print \"The product of overall heat transfer and total area is %d Btu/hr-degF\"%(UA);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The product of overall heat transfer and total area is 1150 Btu/hr-degF\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.9 page : 191\n" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "t = 2.; # Thickness of wall in ft\n", "To = 100.; # Initial temperature of wall in degF\n", "Tg = 1000.; # Temperature of hot gases math.exposed in degF\n", "k = 8.; # Thermal conductivity in Btu/hr-ft-degF\n", "p = 162.; # density in lb/ft**-3\n", "Cp = 0.3; # Heat capacity in Btu/lb-degF\n", "h = 1.6; # Heat transfer coefficient in Btu/hr-ft**-2-degF\n", "a = k/(p*Cp); # Thermal diffusivity\n", "\n", "# Considering the values of a and 4at/L**2 and hl/2k, the value of Phis, Phic and Si can be obtained\n", "Phis = 0.37;\n", "Phic = 0.41;\n", "Si = 0.62;\n", "\n", "# Calculations and Results\n", "Ta = Tg+(To-Tg)*Phis; # Temperature of surface in degF\n", "print \"The temperature of surface is %d degF \"%(Ta);\n", "Tc = Tg+(To-Tg)*Phic; # Temperature of center plane in degF\n", "print \"The temperature of surface is %d degF \"%(Tc);\n", "A = 10; # area of wall through which heat is absorbed\n", "q = p*Cp*t*A*Si*(To-Tg); # Heat absorbed in Btu/hr\n", "print \"The heat absorbed by wall is %d Btu\"%(q);\n", "\n", "# note : book answer is rounded off." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The temperature of surface is 667 degF \n", "The temperature of surface is 631 degF \n", "The heat absorbed by wall is -542376 Btu\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.10 page : 189" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "To1 = 160.; # inlet temperature of oil in degF\n", "Cpo = 0.5; # Specific heat capacity in Btu/lb-degf\n", "Tw1 = 60.; # Inlet temperature of water in degF\n", "mo = 1000.; # Mass flow rate of oil in lb/hr\n", "mw = 2500.; # Mass flow rate of water in lb/hr\n", "Cpw = 1.; # Heat capacity of water in Btu/hr\n", "\n", "# Calculations and Results\n", "X = mo*Cpo/(mw*Cpw); # Ratio of flow rates \n", "UA = 1.15*mo*Cpo;\n", "B = UA/mo*Cpo;\n", "\n", "# from the graph, we can locate the point of A and B And corresponding effectiveness ratio\n", "E = 0.86; # Effectiveness ratio\n", "To2 = To1-E*(To1-Tw1); # Outlet temperature of oil in degF\n", "print \"The outlet temperature of oil is %d degF \"%(To2);\n", "\n", "q = mo*Cpo*(To1-To2); # Heat transferred in Btu/hr\n", "Tw2 = Tw1+(q/(mw*Cpw)); # Outlet temperature of oil in degF\n", "print \" The outlet tempearture of water is %.1f degF\"%(Tw2);\n", "\n", "\n", "\n", " \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The outlet temperature of oil is 74 degF \n", " The outlet tempearture of water is 77.2 degF\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 9.11 page : 193" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "from numpy import zeros, nan\n", "\n", "# To compute the temprature distribution\n", "h = 1.; # Heat transfer coefficient in Btu/hr-ft**2-degF\n", "x = 1.; # Assumed thickness in ft\n", "k = 1.; # Thermal conductivity in Btu/hr-ft-degF\n", "N = h*x/k;\n", "t0 = 600.;\n", "t4 = 200.;\n", "t1 = [500, 550, 550, 525, 525, 512.5, 512.5, 512.5, 506.2, 506.2, 506.2, 506.2, 503.1, 503.1];\n", "t2 = [450, 450, 450, 450, 425, 425, 425, 412.5, 412.5, 412.5, 406.3, 406.3, 406.3, 403.1];\n", "t3 = [350, 350, 325, 325 ,325, 325, 312.5, 312.5, 312.5 ,306.3, 306.3, 303.1, 303.1, 303.1];\n", "th1 = zeros(14)\n", "th2 = zeros(14)\n", "th3 = zeros(14)\n", "th1[0] = nan\n", "th2[0] = nan\n", "th3[0] = nan\n", "\n", "# Assumed temperatures in degF for points 1 2 & 3 respectively\n", "for i in range(0,14):\n", " th1[i] = th1[i]+t0+t2[i]-2*t1[i];\n", " th2[i] = th2[i]+t1[i]+t3[i]-2*t2[i];\n", " th3[i] = th3[i]+t2[i]+t4-2*t3[i];\n", " print \"Assuming t1 = %.1f degF t2 = %.1fdegF t3 = %.1fdegF th1 = %.1f degF th2 = %.1f degF th3 = %.1f degF \"%(t1[i],t2[i],t3[i],th1[i],th2[i],th3[i]);\n", "\n", "print \"This way assumption must be continued till all sink strengths are zero\";\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Assuming t1 = 500.0 degF t2 = 450.0degF t3 = 350.0degF th1 = nan degF th2 = nan degF th3 = nan degF \n", "Assuming t1 = 550.0 degF t2 = 450.0degF t3 = 350.0degF th1 = -50.0 degF th2 = 0.0 degF th3 = -50.0 degF \n", "Assuming t1 = 550.0 degF t2 = 450.0degF t3 = 325.0degF th1 = -50.0 degF th2 = -25.0 degF th3 = 0.0 degF \n", "Assuming t1 = 525.0 degF t2 = 450.0degF t3 = 325.0degF th1 = 0.0 degF th2 = -50.0 degF th3 = 0.0 degF \n", "Assuming t1 = 525.0 degF t2 = 425.0degF t3 = 325.0degF th1 = -25.0 degF th2 = 0.0 degF th3 = -25.0 degF \n", "Assuming t1 = 512.5 degF t2 = 425.0degF t3 = 325.0degF th1 = 0.0 degF th2 = -12.5 degF th3 = -25.0 degF \n", "Assuming t1 = 512.5 degF t2 = 425.0degF t3 = 312.5degF th1 = 0.0 degF th2 = -25.0 degF th3 = 0.0 degF \n", "Assuming t1 = 512.5 degF t2 = 412.5degF t3 = 312.5degF th1 = -12.5 degF th2 = 0.0 degF th3 = -12.5 degF \n", "Assuming t1 = 506.2 degF t2 = 412.5degF t3 = 312.5degF th1 = 0.1 degF th2 = -6.3 degF th3 = -12.5 degF \n", "Assuming t1 = 506.2 degF t2 = 412.5degF t3 = 306.3degF th1 = 0.1 degF th2 = -12.5 degF th3 = -0.1 degF \n", "Assuming t1 = 506.2 degF t2 = 406.3degF t3 = 306.3degF th1 = -6.1 degF th2 = -0.1 degF th3 = -6.3 degF \n", "Assuming t1 = 506.2 degF t2 = 406.3degF t3 = 303.1degF th1 = -6.1 degF th2 = -3.3 degF th3 = 0.1 degF \n", "Assuming t1 = 503.1 degF t2 = 406.3degF t3 = 303.1degF th1 = 0.1 degF th2 = -6.4 degF th3 = 0.1 degF \n", "Assuming t1 = 503.1 degF t2 = 403.1degF t3 = 303.1degF th1 = -3.1 degF th2 = 0.0 degF th3 = -3.1 degF \n", "This way assumption must be continued till all sink strengths are zero\n" ] } ], "prompt_number": 22 } ], "metadata": {} } ] }