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+{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 : Heat and mass transfer in duct flow"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.1 - Page No :497\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Compute and plot the temperature profile.\n",
+ "\n",
+ "import math \n",
+ "\n",
+ "# Variables\n",
+ "# given\n",
+ "K_drywall = 0.28; \t\t\t #[Btu/ft*degF] - thermal conductivity of dry wall\n",
+ "K_fibreglass = 0.024; \t\t #[Btu/ft*degF] - thermal conductivity of fibre glass\n",
+ "K_concrete = 0.5; \t\t\t #[Btu/ft*degF] - thermal conductivity of concrete\n",
+ "T4 = 0.; \t\t #[degF]\n",
+ "T1 = 65.; \t\t\t #[degF]\n",
+ "deltaT = T4-T1; \t \t #[degF]\n",
+ "a = 1.; \t\t #[ft**2] - assuming area of 1 ft**2\n",
+ "deltax1 = 0.5/12; \t\t\t #[ft]\n",
+ "deltax2 = 3.625/12; \t\t #[ft]\n",
+ "deltax3 = 6./12; \t\t\t #[ft]\n",
+ "\n",
+ "# Calculations\n",
+ "R1 = deltax1/(K_drywall*a); \t\t\t #[h*degF/Btu]\n",
+ "R2 = deltax2/(K_fibreglass*a); \t\t\t #[h*degF/Btu]\n",
+ "R3 = deltax3/(K_concrete*a); \t \t\t #[h*degF/Btu]\n",
+ "qx = deltaT/(R1+R2+R3);\n",
+ "q12 = -qx;\n",
+ "q23 = -qx;\n",
+ "q34 = -qx;\n",
+ "deltaT1 = (-q12)*deltax1*(1./(K_drywall*a));\n",
+ "T2 = T1+deltaT1;\n",
+ "deltaT2 = (-q23)*deltax2*(1./(K_fibreglass*a));\n",
+ "T3 = T2+deltaT2;\n",
+ "deltaT3 = (-q34)*deltax3*(1./(K_concrete*a));\n",
+ "T4 = T3+deltaT3;\n",
+ "\n",
+ "# Results\n",
+ "print \" T1 = %.0f F \\\n",
+ "\\n T2 = %.1f F \\\n",
+ "\\n delta T2 = %.2f deg F \\\n",
+ "\\n T3 = %.2f F \\\n",
+ "\\n delta T3 = %.2f deg F \\\n",
+ "\\n T4 = %.0f F\"%(T1,T2,deltaT2,T3,deltaT3,T4);\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " T1 = 65 F \n",
+ " T2 = 64.3 F \n",
+ " delta T2 = -59.56 deg F \n",
+ " T3 = 4.73 F \n",
+ " delta T3 = -4.73 deg F \n",
+ " T4 = 0 F\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.2 - Page No :501\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Find the heat loss and compute the temperature at the steel-magnesia interface.\n",
+ "\n",
+ "# Variables\n",
+ "r1 = (2.067/2.)/(12); \t\t #[ft]\n",
+ "r2 = r1+0.154/12; \t\t\t #[ft]\n",
+ "r3 = r2+3/12.; \t\t\t #[ft]\n",
+ "L = 1.; \t\t\t #[ft]\n",
+ "Ka = 26.; \t\t\t #[Btu/h*ft*degF]\n",
+ "Kb = 0.04; \t\t #[Btu/h*ft*degF]\n",
+ "T1 = 50.; \t\t\t #[degF]\n",
+ "\n",
+ "# Calculations\n",
+ "Ra = (math.log(r2/r1))/(2*math.pi*L*Ka);\n",
+ "Rb = (math.log(r3/r2))/(2*math.pi*L*Kb);\n",
+ "R = Ra+Rb;\n",
+ "deltaT = -18; \t\t\t #[degF] - driving force\n",
+ "Qr = -(deltaT/(R));\n",
+ "deltaT1 = (-Qr)*(Ra);\n",
+ "T2 = T1+deltaT1;\n",
+ "\n",
+ "# Results\n",
+ "print \" Qr = %.3f\"%Qr;\n",
+ "print \" The interface temperature is T2 = %.3f degF\"%(T2);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Qr = 3.589\n",
+ " The interface temperature is T2 = 49.997 degF\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.3 - Page No :502\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Estimate the heat loss for 500 ft of pipe\n",
+ "\n",
+ "# Variables\n",
+ "Ra = 8.502*10**-4; \t\t\t #[h*degF*Btu**-1]\n",
+ "Rb = 5.014; \t\t \t #[h*degF*Btu**-1]\n",
+ "r1 = (2.067/2)/(12.); \t\t\t #[ft]\n",
+ "r2 = r1+0.154/12.; \t\t\t #[ft]\n",
+ "r3 = r2+3/12.; \t\t\t #[ft]\n",
+ "d1 = 2.*r1;\n",
+ "d0 = 2.*r3;\n",
+ "h0 = 25.; \t \t\t #[Btu/h*ft**2*degF]\n",
+ "h1 = 840.; \t\t\t #[Btu/h*ft**2*degF]\n",
+ "L = 1.; \t\t\t #[ft] - considering 1 feet length\n",
+ "\n",
+ "# Calculations\n",
+ "R0 = 1./(h0*math.pi*d0*L);\n",
+ "R1 = 1./(h1*math.pi*d1*L);\n",
+ "R = R0+R1+Ra+Rb;\n",
+ "deltaT = -400; \t\t\t #[degF]\n",
+ "Qr = -(deltaT)/R;\n",
+ "# the heat loss calculated above is the heat loss per foot.therefore for 500 ft\n",
+ "L = 500.;\n",
+ "Qr = Qr*L;\n",
+ "\n",
+ "# Results\n",
+ "print \" the heat loss for a 500 feet pipe is qr = %.2e Btu/h\"%(Qr);\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the heat loss for a 500 feet pipe is qr = 3.97e+04 Btu/h\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.5 - Page No :521\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Estimate the heat transfer coefficient\n",
+ "\n",
+ "# Variables\n",
+ "Nre = 50000.;\n",
+ "d = 0.04; \t\t\t #[m] - diameter of pipe\n",
+ "\n",
+ "# physical properties of water\n",
+ "T1 = 293.15; \t\t\t #[K]\n",
+ "T2 = 303.15; \t\t\t #[K]\n",
+ "T3 = 313.15; \t\t\t #[K]\n",
+ "p1 = 999.; \t\t\t #[kg/m**3] - density of water at temperature T1\n",
+ "p2 = 996.0; \t\t\t #[kg/m**3] - density of water at temperature T2\n",
+ "p3 = 992.1; \t\t\t #[kg/m**3] - density of water at temperature T3\n",
+ "mu1 = 1.001; \t\t\t #[cP] - viscosity of water at temperature T1\n",
+ "mu2 = 0.800; \t\t\t #[cP] - viscosity of water at temperature T2\n",
+ "mu3 = 0.654; \t\t\t #[cP] - viscosity of water at temperature T3\n",
+ "k1 = 0.63; \t\t\t #[W/m*K] - thermal conductivity of water at temperature T1\n",
+ "k2 = 0.618; \t\t\t #[W/m*K] - thermal conductivity of water at temperature T2\n",
+ "k3 = 0.632; \t\t\t #[W/m*K] - thermal conductivity of water at temperature T3\n",
+ "cp1 = 4182.; \t\t\t #[J/kg*K] - heat capacity of water at temperature T1\n",
+ "cp2 = 4178.; \t\t\t #[J/kg*K] - heat capacity of water at temperature T2\n",
+ "cp3 = 4179.; \t\t\t #[J/kg*K] - heat capacity of water at temperature T3\n",
+ "Npr1 = 6.94; \t\t\t # prandtl no. at temperature T1\n",
+ "Npr2 = 5.41; \t\t\t # prandtl no. at temperature T2\n",
+ "Npr3 = 4.32; \t\t\t # prandtl no. at temperature T3\n",
+ "\n",
+ "\n",
+ "# Calculations\n",
+ "# (a) Dittus -Boelter-this correction evalutes all properties at the mean bulk temperature,which is T1\n",
+ "kmb = 0.603\n",
+ "h = (kmb/d)*0.023*((Nre)**(0.8))*((Npr1)**0.4);\n",
+ "\n",
+ "\n",
+ "# Results\n",
+ "print \" a) Dittus -Boelter the heat transfer coefficient is \\nh = %.0f W/m**2*K \\\n",
+ " = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n",
+ "\n",
+ "# (b) Seid_er Tate-this correlation evaluates all the properties save muw at the mean bulk temperature \n",
+ "h = (kmb/d)*(0.027)*((Nre)**0.8)*((Npr1)**(1./3))*((mu1/mu3)**0.14);\n",
+ "print \" b) Seid_er Tate the heat transfer coefficient is \\nh = %.0f W/m**2*K \\\n",
+ " = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n",
+ "\n",
+ "# (c) Sleicher-Rouse equation\n",
+ "a = 0.88-(0.24/(4+Npr3));\n",
+ "b = (1./3)+0.5*math.exp((-0.6)*Npr3);\n",
+ "Nref = Nre*(mu1/mu2)*(p2/p1);\n",
+ "Nnu = 5+0.015*((Nref)**a)*((Npr3)**b);\n",
+ "h = Nnu*(kmb/d);\n",
+ "print \" c) Sleicher-Rouse equation the heat transfer coefficient is \\nh = %.0f W/m**2*K \\\n",
+ " = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n",
+ "\n",
+ "# (d) Colbum Analogy- the j factor for heat transfer is calculated\n",
+ "jh = 0.023*((Nref)**(-0.2));\n",
+ "Nst = jh*((Npr2)**(-2./3));\n",
+ "U = (Nre*mu1*10**-3)/(d*p1);\n",
+ "h = Nst*(p1*cp1*U);\n",
+ "print \" d) Colbum Analogy the heat transfer coefficient is \\nh \\\n",
+ " = %.0f W/m**2*K = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n",
+ "\n",
+ "# (e) Friend-Metzner\n",
+ "f = 0.005227;\n",
+ "Nnu = ((Nre)*(Npr1)*(f/2.)*((mu1/mu3)**0.14))/(1.20+((11.8)*((f/2)**(1./2))*(Npr1-1)*((Npr1)**(-1./3))));\n",
+ "h = Nnu*(kmb/d);\n",
+ "print \" e) Friend-Metzner the heat transfer coefficient is \\nh = %.0f W/m**2*K \\\n",
+ " = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n",
+ "\n",
+ "# (f) Numerical analysis\n",
+ "Nnu = 320.;\n",
+ "h = Nnu*(kmb/d);\n",
+ "print \" f) Numerical analysis the heat transfer coefficient is \\nh = %.0f W/m**2*K \\\n",
+ " = %.0f Btu/ft**2*h**-1*degF\"%(h,h*0.17611);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " a) Dittus -Boelter the heat transfer coefficient is \n",
+ "h = 4322 W/m**2*K = 761 Btu/ft**2*h**-1*degF\n",
+ " b) Seid_er Tate the heat transfer coefficient is \n",
+ "h = 4733 W/m**2*K = 834 Btu/ft**2*h**-1*degF\n",
+ " c) Sleicher-Rouse equation the heat transfer coefficient is \n",
+ "h = 4766 W/m**2*K = 839 Btu/ft**2*h**-1*degF\n",
+ " d) Colbum Analogy the heat transfer coefficient is \n",
+ "h = 4292 W/m**2*K = 756 Btu/ft**2*h**-1*degF\n",
+ " e) Friend-Metzner the heat transfer coefficient is \n",
+ "h = 4713 W/m**2*K = 830 Btu/ft**2*h**-1*degF\n",
+ " f) Numerical analysis the heat transfer coefficient is \n",
+ "h = 4824 W/m**2*K = 850 Btu/ft**2*h**-1*degF\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.6 - Page No :525\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Find the Nusselt number if the fluid temperature is 640 K and the wall temperature is 680 K.\n",
+ "\n",
+ "# Variables\n",
+ "# given\n",
+ "Tw = 680.; \t\t\t #[K] - temperature at the wall\n",
+ "Tb = 640.; \t\t\t #[K] - temperature at the bulk\n",
+ "Tf = (Tw+Tb)/2; \t\t\t #[K]\n",
+ "Nre = 50000.;\n",
+ "vmb = 2.88*10.**-7;\n",
+ "vf = 2.84*10.**-7;\n",
+ "Nref = Nre*(vmb/vf);\n",
+ "k = 27.48;\n",
+ "d = 0.04;\n",
+ "\n",
+ "# Calculation and Results\n",
+ "# from table 11.3 the prandtl no. is\n",
+ "Npr = 8.74*10**-3\n",
+ "\n",
+ "# consmath.tant heat flow\n",
+ "Nnu = 6.3+(0.0167)*((Nref)**0.85)*((Npr)**0.93);\n",
+ "h = Nnu*(k/d);\n",
+ "print \" constant heat flow h = %.0f W/m**2*K = %.0f Btu/ft**2*h*degF\"%(h,round(h*0.17611,-1));\n",
+ "\n",
+ "# constant wall temperature\n",
+ "Nnu = 4.8+0.0156*((Nref)**0.85)*((Npr)**0.93);\n",
+ "h = Nnu*(k/d);\n",
+ "print \" constant wall temperature h = %d W/m**2*K = %d Btu/ft**2*h*degF\"%(h,h*0.17611);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " constant heat flow h = 5723 W/m**2*K = 1010 Btu/ft**2*h*degF\n",
+ " constant wall temperature h = 4600 W/m**2*K = 810 Btu/ft**2*h*degF\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.7 - Page No :536\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Estimate \n",
+ "(a) the duty of the exchanger, \n",
+ "(b) the number of pounds per hour of steam required, and \n",
+ "(c) the length of a double-pipe heat exchanger to perform this task.\n",
+ "'''\n",
+ "\n",
+ "# Variables\n",
+ "di = 0.620; \t\t\t #[inch] - internal diameter\n",
+ "d0 = 0.750; \t\t\t #[inch] - outer diameter\n",
+ "Ai = 0.1623; \t\t\t #[ft**2/ft]\n",
+ "Ao = 0.1963; \t\t\t #[ft**2/ft]\n",
+ "wc = 12*(471.3/0.9425); #[lb/h]\n",
+ "cp = 1.; \t\t\t #[Btu/lbm*degF] - heat capacity of water\n",
+ "Tco = 110.;\n",
+ "Tci = 50.;\n",
+ "\n",
+ "# Calculations\n",
+ "qtotal = wc*cp*(Tco-Tci);\n",
+ "deltaH_coldwater = 3.6*10**5;\n",
+ "deltaH_vapourization = 1179.7-269.59;\n",
+ "wh = deltaH_coldwater/deltaH_vapourization;\n",
+ "hi = 80.; \t\t\t #[Btu/h*ft**2*degF]\n",
+ "ho = 500.; \t\t\t #[Btu/h*ft**2*degF]\n",
+ "km = 26.; \t\t\t #[Btu/h*ft*degF]\n",
+ "Ui = 1./((1./hi)+((Ai*math.log(d0/di))/(2*math.pi*km))+(Ai/(Ao*ho)));\n",
+ "deltaT1 = 300-50.;\n",
+ "deltaT2 = 300-110.;\n",
+ "LMTD = (deltaT1-deltaT2)/(math.log(deltaT1/deltaT2));\n",
+ "A = qtotal/(Ui*LMTD);\n",
+ "L = A/Ai;\n",
+ "\n",
+ "# Results\n",
+ "print \"the length of the heat exchanger is L = %.2f ft\"%(L);\n",
+ "\n",
+ "# Answer is slightly different becasue of rounding error."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the length of the heat exchanger is L = 145.53 ft\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.8 - Page No :537\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# estimate the outlet temperature of the water.\n",
+ "\n",
+ "# Variables\n",
+ "L = 30.; \t\t\t #[ft] - length\n",
+ "Ai = 0.1623*L;\n",
+ "di = 0.620; \t\t\t #[inch] - internal diameter\n",
+ "d0 = 0.750; \t\t\t #[inch] - outer diameter\n",
+ "Ao = 0.1963*L; \t\t #[ft**2/ft]\n",
+ "wc = 12.*(471.3/0.9425);\n",
+ "cp = 1.; \t\t\t #[Btu/lbm*degF] - heat capacity of water\n",
+ "\n",
+ "# Calculations\n",
+ "deltaH_coldwater = 3.6*10**5;\n",
+ "deltaH_vapourization = 1179.7-269.59;\n",
+ "wh = deltaH_coldwater/deltaH_vapourization;\n",
+ "hi = 80.; \t\t\t #[Btu/h*ft**2*degF]\n",
+ "ho = 500.; \t\t #[Btu/h*ft**2*degF]\n",
+ "km = 26.; \t\t\t #[Btu/h*ft*degF]\n",
+ "Ui = 1./((1./hi)+(((Ai/L)*math.log(d0/di))/(2*math.pi*km))+(Ai/(Ao*ho)));\n",
+ "deltaT1 = 300-50.;\n",
+ "deltaT = deltaT1/(math.exp((Ui*Ai)/(wc*cp)));\n",
+ "Tsat = 300.;\n",
+ "Tc2 = Tsat-deltaT;\n",
+ "\n",
+ "# Results\n",
+ "print \" Therefore, the outlet temperature of the cold fluid_ is Tc2 = %.2f degF\"%(Tc2);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Therefore, the outlet temperature of the cold fluid_ is Tc2 = 63.75 degF\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.9 - Page No :538\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Repeat Example 11.8 if the water side fouling factor is 0.002 h ft\u2019\u201cF Btu\u2019.\n",
+ "\n",
+ "# Variables\n",
+ "Ai = 4.869;\n",
+ "wc = 6000.;\n",
+ "cp = 1.;\n",
+ "Rf = 0.002;\n",
+ "Uclean = 69.685;\n",
+ "\n",
+ "# Calculations\n",
+ "Udirty = 1./(Rf+(1./Uclean));\n",
+ "deltaT1 = 300.-50;\n",
+ "deltaT2 = deltaT1/(math.exp((Udirty*Ai)/(wc*cp)));\n",
+ "Th2 = 300.;\n",
+ "Tc2 = Th2-deltaT2;\n",
+ "\n",
+ "# Results\n",
+ "print \" the outlet temperature is Tc2 = %.1f degF\"%(Tc2);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the outlet temperature is Tc2 = 62.1 degF\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.10 - Page No :544\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Calculate the hydrocarbon flow rate and\n",
+ "the heat exchanger area for the following heat exchangers: (a) parallel double-\n",
+ "pipe; (b) counterflow double-pipe; (c) 1-2 shell-and-tube; and (d) 2-4 shell-tube.\n",
+ "'''\n",
+ "\n",
+ "# Variables\n",
+ "Ui = 325.; \t\t\t #[W/m**2*K] - overall heat transfer coefficient\n",
+ "Thi = 120.; \t\t\t #[degC] - inlet temperature of hydrocarbon\n",
+ "Tho = 65.; \t\t\t #[degC] - outlet temperature of hydrocarbon\n",
+ "Tci = 15.; \t\t\t #[degC] - inlet temperature of water\n",
+ "Tco = 50.; \t\t\t #[degC] - outlet temperture of water\n",
+ "cp = 4184.; \t\t\t #[J/kg*K] - heat capacity of water\n",
+ "ch = 4184.*0.45\t\t\t #[J/kg*K] - heat capacity of hydrocarbon\n",
+ "wc = 1.2; \t\t\t #[kg/sec] - mass flow rate of water\n",
+ "\n",
+ "# Calculation and Results\n",
+ "wh = ((wc*cp)*(Tco-Tci))/((ch)*(Thi-Tho));\n",
+ "qtotal = wc*cp*(Tco-Tci);\n",
+ "\n",
+ "# (a) - parallel double pipe\n",
+ "F = 1.;\n",
+ "Thi = 120.; \t\t\t #[degC] - inlet temperature of hydrocarbon\n",
+ "Tho = 65.; \t\t\t #[degC] - outlet temperature of hydrocarbon\n",
+ "Tci = 15.; \t\t\t #[degC] - inlet temperature of water\n",
+ "Tco = 50.; \t\t\t #[degC] - outlet temperture of water\n",
+ "deltaT1 = Thi-Tci;\n",
+ "deltaT2 = Tho-Tco;\n",
+ "LMTD = (deltaT2-deltaT1)/(math.log(deltaT2/deltaT1));\n",
+ "Ai = qtotal/((Ui*LMTD));\n",
+ "print \" a) parallel double pipe Ai = %.2f m**2\"%(Ai);\n",
+ "\n",
+ "# (b) - counter flow\n",
+ "F = 1.;\n",
+ "Thi = 120.; \t\t\t #[degC] - inlet temperature of hydrocarbon\n",
+ "Tho = 65.; \t\t\t #[degC] - outlet temperature of hydrocarbon\n",
+ "Tco = 15.; \t\t\t #[degC] - inlet temperature of water\n",
+ "Tci = 50.; \t\t\t #[degC] - outlet temperture of water\n",
+ "deltaT1 = Thi-Tci;\n",
+ "deltaT2 = Tho-Tco;\n",
+ "LMTD = (deltaT2-deltaT1)/(math.log(deltaT2/deltaT1));\n",
+ "Ai = qtotal/((Ui*LMTD));\n",
+ "print \" b) counter flow Ai = %.2f m**2\"%(Ai);\n",
+ "\n",
+ "# (c) - 1-2 shell and tube \n",
+ "Thi = 120.; \t\t\t #[degC] - inlet temperature of hydrocarbon\n",
+ "Tho = 65.; \t\t\t #[degC] - outlet temperature of hydrocarbon\n",
+ "Tci = 15.; \t\t\t #[degC] - inlet temperature of water\n",
+ "Tco = 50.; \t\t\t #[degC] - outlet temperture of water\n",
+ "Z = (Thi-Tho)/(Tco-Tci);\n",
+ "nh = (Tco-Tci)/(Thi-Tci);\n",
+ "deltaT1 = Thi-Tco;\n",
+ "deltaT2 = Tho-Tci;\n",
+ "F = 0.92;\n",
+ "LMTD = (F*(deltaT2-deltaT1))/(math.log(deltaT2/deltaT1));\n",
+ "Ai = qtotal/((Ui*LMTD));\n",
+ "print \" c) 1-2 shell and tube Ai = %.2f m**2\"%(Ai);\n",
+ "\n",
+ "# (d) - 2-4 shell and tube\n",
+ "Thi = 120.; \t\t\t #[degC] - inlet temperature of hydrocarbon\n",
+ "Tho = 65.; \t\t\t #[degC] - outlet temperature of hydrocarbon\n",
+ "Tci = 15.; \t\t\t #[degC] - inlet temperature of water\n",
+ "Tco = 50.; \t\t\t #[degC] - outlet temperture of water\n",
+ "Z = (Thi-Tho)/(Tco-Tci);\n",
+ "nh = (Tco-Tci)/(Thi-Tci);\n",
+ "F = 0.975;\n",
+ "LMTD = (F*(deltaT2-deltaT1))/(math.log(deltaT2/deltaT1));\n",
+ "Ai = qtotal/((Ui*LMTD));\n",
+ "print \" d) 2-4 shell and tube Ai = %.2f m**2\"%(Ai);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " a) parallel double pipe Ai = 11.69 m**2\n",
+ " b) counter flow Ai = 9.10 m**2\n",
+ " c) 1-2 shell and tube Ai = 9.89 m**2\n",
+ " d) 2-4 shell and tube Ai = 9.33 m**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
generated by cgit (git 2.34.1) at 2024-12-20 11:32:55 +0000