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| author | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
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| committer | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
| commit | 206d0358703aa05d5d7315900fe1d054c2817ddc (patch) | |
| tree | f2403e29f3aded0caf7a2434ea50dd507f6545e2 /Heat_Transfer_Applications_for_the_Practicing_Engineer_ | |
| parent | c6f0d6aeb95beaf41e4b679e78bb42c4ffe45a40 (diff) | |
| download | Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.tar.gz Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.tar.bz2 Python-Textbook-Companions-206d0358703aa05d5d7315900fe1d054c2817ddc.zip | |
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diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_03.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_03.ipynb new file mode 100644 index 00000000..0ee01bdd --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_03.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "Chapter 03" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 3: Process Variables" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.2, Page number: 17" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Convert the following:\n1. 8.03 yr to seconds (s)\n2. 150 miles/h to yards/h\n3. 100.0 m/s2 to ft/min2\n4. 0.03 g/cm3 to lb/ft3\n'''\n \n#Variable Declaration:\nQ1 = 8.03 #Years(part 1)\nD = 365 #Days in a year\nH = 24 #Hours in a day\nM = 60 #Minutes in an hour\nS = 60 #Seconds in a minute\nQ2 = 150 #Miles per hour(part 2)\nFM = 5280 #Feet in a mile\nYF = 1.0/3.0 #Yard in a feet\nQ3 = 100 #Meter per second square(part 3)\nCmm = 100 #Centimeter in a meter\nFC = 1.0/30.48 #Feet in a centimeter\nSsMs = 60**2 #Second square in a minute square\nQ4 = 0.03 #Gram per centimeter cube (part 4)\nPG = 1.0/454.0 #Pound in a gram\nCF = (30.48)**3 #Centimeter in a feet\n \n#Calculation:\nA1 = Q1*D*H*M*S #Seconds (s)\nA2 = Q2*FM*YF #Yards per hour (yd/hr)\nA3 = Q3*Cmm*FC*SsMs #Feet per min square (ft/min^2)\nA4 = Q4*PG*CF #Pound per feet cube (lb/ft^3)\n \n#Results:\nprint \"1. Seconds in\",Q1,\"year is:\",round(A1/10**8,2),\" x 10**8 s\"\nprint \"2. Yards per hour in\",Q2,\"miles per hour is:\",round(A2/10**5,1),\" x 10**5 yd/h\"\nprint \"3. Feets per minute square in\",Q3,\"meter per square is:\",round(A3/10**6,3),\" x 10**6 ft/min^2\"\nprint \"4. Pounds per feet cube in\",Q4,\"gram per centimeter cube is:\",round(A4),\"lb/ft^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. Seconds in 8.03 year is: 2.53 x 10**8 s\n2. Yards per hour in 150 miles per hour is: 2.6 x 10**5 yd/h\n3. Feets per minute square in 100 meter per square is: 1.181 x 10**6 ft/min^2\n4. Pounds per feet cube in 0.03 gram per centimeter cube is: 2.0 lb/ft^3\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.3, Page number: 21" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the following pressure calculations.\n1. A liquid weighing 100 lb held in a cylindrical column with a base area of 3 in2 exerts\nhow much pressure at the base in lbf/ft2?\n2. If a pressure reading is 35 psig (pounds per square inch gauge), what is the absolute\npressure?\n'''\n \n#Variable Declaration:\nQ1 = 32.2 #Gravitational acceleration (ft/s^2) (part 1)\nCF = 32.2 #Conversion factor (lb.ft/lbf.s^2)\nM = 100 #Mass (lb)\nSA = 3 #Surface area (in^2)\nFsIs = (1.0/12.0)**2 #Feet square in a inch square\nQ2 = 14.7 #Atmospheric pressure (psi) (part 2)\nGP = 35 #Gauge Pressure (psig)\n \n#Caculations:\nF = M*Q1/CF #Force (lbf)\nP = F/SA/FsIs #Pressure at the base (lbf/ft^2)\nPa = GP+Q2 #Absolute pressure (psia)\n \n#Results:\nprint \"1. Pressure at the base is:\",round(P),\"lbf/ft^2\"\nprint \"2. Absolute pressure is:\",round(Pa,1),\"psia\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. Pressure at the base is: 4800.0 lbf/ft^2\n2. Absolute pressure is: 49.7 psia\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.4, Page number: 23" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If a 55-gal tank contains 20.0 lb of water,\n1. How many pound . moles of water does it contain?\n2. How many gram . moles does it contain?\n3. How many molecules does it contain?\n'''\n \n#Variable Declaration:\nQ1 = 20.0 #Mass (lb) (part 1)\nMH = 1.008 #Molecular weight of H (lb/lbmol)\nMO = 15.999 #Molecular weight of O (lb/lbmol)\nQ2 = 454 #Gram in pound (part 2)\nQ3 = 6.023*10**23 #Avogadro nuber (part 3)\n \n#Calculations:\nMol = 2*MH+MO #Molecular weight of water (lb/lbmol)\nA1 = Q1/Mol #Pound.moles of water (lbmol)\nA2 = Q1*Q2/Mol #Gram.moles of water (gmol)\nA3 = A2*Q3 #Molecules of water (molecules)\n \n#Results:\nprint \"1. Pound.moles of water is:\",round(A1,2),\"lbmol water\"\nprint \"2. Gram.moles of water is:\",round(A2),\"gmol water\"\nprint \"3. Molecules of water is:\",round(A3/10**26,3),\" x 10**26 molecules\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. Pound.moles of water is: 1.11 lbmol water\n2. Gram.moles of water is: 504.0 gmol water\n3. Molecules of water is: 3.036 x 10**26 molecules\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.5, Page number: 25" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is given:\nSpecific gravity of liquid (methanol) = 0.92 (at 60\u00b0F)\nDensity of reference substance (water) = 62.4 lb/ft^3 (at 60\u00b0F)\nDetermine the density of methanol in lb/ft^3.\n'''\n \n#Variable declaration:\nSG = 0.92 #Specific gavity of liquid, methanol\nDW = 62.4 #Density of reference substance, water (lb/ft^3)\n \n#Calculation:\nDM = SG*DW #Density of methanol (lb/ft^3)\n \n#Result:\nprint \"Density of methanol is:\",round(DM,1),\"lb/ft^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Density of methanol is: 57.4 lb/ft^3\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.6, Page number: 27\n" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the kinematic viscosity of a gas, if the specific gravity and absolute viscosity are 0.8 and\n0.02 cP, respectively?\n'''\n \n#Variable declaration:\nSG = 0.8 #Specific Gravity\nAV = 0.02 #Absolute Viscosity (cP)\ncP = 1 #Viscosity of centipoise (cP)\nVcP = 6.72 * 10**-4 #Pound per feet.sec in a centipoise (lb/ft.s)\npR = 62.43 #Reference density (lb/ft^3)\n \n#Calculations:\nu = AV*VcP/cP #Viscosity of gas (lb/ft.s)\np = SG*pR #Density of gas (lb/ft^3)\nv = u/p #Kinematic viscosity of gas (ft^2/s)\n \n#Result:\nprint \"Kinematic viscosity of gas is:\",round(v/10**-7,3),\"x 10**-7 ft^2/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Kinematic viscosity of gas is: 2.691 x 10**-7 ft^2/s\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.7, Page number: 27" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''To illustrate the use of a monogram, calculate the absolute viscosity of a 98% sulfuric acid\nsolution at 45\u00b0C.\n'''\n \n#Variable declaration:\nX = 7.0 #Coordinate X of H2SO4\nY = 24.8 #Coordinate Y of H2SO4\nS = 45 #Slope\n \n#Calculations:\n#From the figure C.1 we found the intersection of curves mu = 12cP\nmu = 12\n \n#Results:\nprint \"Absolute viscosity of a 98% sulfuric acid solution at 45\u00b0 C is :\",mu*10**-2,\" g/cm.s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Absolute viscosity of a 98% sulfuric acid solution at 45\u00b0 C is : 0.12 g/cm.s\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.8, Page number: 28" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following is given:\nHeat capacity of methanol = 0.61 cal/g.\u00b0C (at 60\u00b0F)\nConvert the heat capacities to English units.\n'''\n \n#Variable declaration:\nCpM = 0.61 #Heat capacity of methanol (cal/g.\u00b0C)\nG = 454 #Grams in a pound\nB = 1.0/252.0 #Btu in a calorie\nC = 1.0/1.8 #Degree celsius in a degree fahrenheit\n \n#Calculation:\nCp = CpM*G*B*C #Heat capacity in English units (Btu/lb.\u00b0F)\n \n#Result:\nprint \"Heat capacity in English units is: \",round(Cp,2),\" Btu/lb.\u00b0F\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat capacity in English units is: 0.61 Btu/lb.\u00b0F\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.9, Page number: 29" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following data is given:\nThermal conductivity of methanol = 0.0512 cal/m.s.\u00b0C(at 60\u00b0F)\nConvert the thermal conductivity to English units.\n'''\n \n#Variable declaration:\nkM = 0.0512 #Thermal conductivity of methanol (cal/m.s\u00b0C)\nB = 1.0/252.0 #Btu in a calorie\nM = 0.3048 #Meters in a feet\nS = 3600 #Seconds in an hour\nC = 1.0/1.8 #Degree celsius in a degree fahrenheit\n \n#Calculation:\nk = kM*B*M*S*C #Thermal conductivity in English units (Btu/ft.h.\u00b0F)\n \n#Result:\nprint \"Thermal coductivity in English units is:\",round(k,3),\"Btu/ft.h.\u00b0F\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal coductivity in English units is: 0.124 Btu/ft.h.\u00b0F\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.11, Page number: 31" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the Reynolds number for a fluid flowing through a 5-inch diameter pipe at 10 fps\n(feet per second) with a density of 50 lb/ft3 and a viscosity of 0.65 cP. Is the flow turbulent\nor laminar?\n'''\n \n#Variable declaration:\nD = 5 #Diameter of pipe (ft)\nV = 10 #Fluid velocity (ft/s)\np = 50 #Fluid density (lb/ft^3)\nu = 0.65 #Fluid viscosity (lb/ft.s)\nF = 1.0/12.0 #Feet in an inch\nVCp = 6.72*10**-4 #Viscosity of centipoise (lb/ft.s)\n \n#Calculation:\nA = D*V*p*F/u/VCp #Reynolds Number\n \n#Result:\nif(A>2100):\n print \"The Reynolds number is :\",round(A,-2),\"; therefore, the flow is turbulent.\"\nelif(A<2100):\n print \"The Reynolds number is :\",round(A,-2),\"; therefore, the flow is not turbulent.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Reynolds number is : 477000.0 ; therefore, the flow is turbulent.\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 3.12, Page number: 32" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''As part of a heat transfer course, a young environmental engineering major has been requested to\ndetermine the potential energy of water before it flows over a waterfall 10 meters in height above\nground level conditions.\n'''\n#Variable declaration:\n#For the problem at hand, take as a basis 1 kilogram of water and assume the potential energy to be zero at ground level conditions.\nz1 = 0 #Intial height from ground level (m)\nz2 = 10 #Final height from ground level (m)\nPE1 = 0 #Initial potential energy at z1 (J)\nm = 1 #Mass of water (kg)\ng = 9.8 #Gravitational acceleration (m/s^2)\ngc = 1 #Conversion factor\n \n#Calculations:\nPE2 = m*(g/gc)*z2 #Final potential energy at z2 (J)\n \n#Result:\nprint \"The potential energy of water is :\",PE2,\"J \"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The potential energy of water is : 98.0 J \n" + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_04.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_04.ipynb new file mode 100644 index 00000000..7d7ce512 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_04.ipynb @@ -0,0 +1,272 @@ +{ + "metadata": { + "name": "Chapter 04" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 4: The Conservation Law for Momentum" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.1, Page number: 39" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A 10-cm-diameter horizontal line carries saturated steam at 420 m/s. Water is entrained by the\nsteam at a rate of 0.15 kg/s. The line has a 90\u00b0 bend. Calculate the force components in the horizontal\nand vertical directions required to hold the bend in place due to the entrained water.\n'''\n\n#Variable declaration:\nVx_in = 420 #Entry Velocity in X direction (m/s)\nVx_out = 0 #Exit Velocity in X direction (m/s)\nVy_in = 0 #Entry Velocity in Y direction (m/s)\nVy_out = 420 #Exit Velocity in Y direction (m/s)\nm = 0.15 #Rate of water entrained by the steam (kg/s)\nlb = 1.0/4.46 #Pound force in a newton force\n\n#Calculations:\nMx_out = m*Vx_out #Rate of change of momentum at entry in x-direction (kg.m)\nMx_in = m*Vx_in #Rate of change of momentum at exit in x-direction (kg.m)\nMy_out = m*Vy_out #Rate of change of momentum at entry in y-direction (kg.m)\nMy_in = m*Vy_in #Rate of change of momentum at exit in y-direction (kg.m)\nFxgc = (Mx_out - Mx_in)*lb #Force in X direction (lbf)\nFygc = (My_out - My_in)*lb #Force in X direction (lbf)\n\n#Results:\nif Fxgc < 1:\n print \"The x-direction supporting force acting on the 90\u00b0 elbow is :\",round(-Fxgc,1),\" lbf acting toward the left. \"\nelse:\n print \"The x-direction supporting force acting on the 90\u00b0 elbow is :\",round(Fxgc,1),\" lbf acting toward the right. \"\nif Fygc < 1:\n print \"The y-direction supporting force acting on the 90\u00b0 elbow is :\",round(-Fygc,1),\" lbf acting downwards. \" \nelse:\n print \"The y-direction supporting force acting on the 90\u00b0 elbow is :\",round(Fygc,1),\" lbf acting upwards. \"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The x-direction supporting force acting on the 90\u00b0 elbow is : 14.1 lbf acting toward the left. \nThe y-direction supporting force acting on the 90\u00b0 elbow is : 14.1 lbf acting upwards. \n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.2 Page number: 40" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 4.1. Calculate the magnitude and direction of the resultant force.\n'''\n\nfrom math import sqrt,degrees,atan2\n\n#Variable declaration:\nFx = -63 #Force component in X direction (N)\nFy = 63 #Force component in Y direction (N)\nlbf = 0.22481 #Pound-forrce in unit newton (lbf)\n\n#Calculations:\nFr = sqrt(Fx**2 + Fy**2)*lbf #The resultant supporting force (lbf)\nu = degrees(atan2(Fy,Fx)) #Angle between the positive x axis and the direction of the force (degrees)\n\n#Result: \nif (0<u<90):\n print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201cnortheast\u201d direction.\"\nelif (90<u<180):\n print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201cnorthwest\u201d direction.\"\nelif (180<u<270):\n print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201csouthwest\u201d direction.\"\nelif (270<u<360):\n print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201csoutheast\u201d direction.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The supporting force is : 20.0 lbf acting at 135.0 \u00b0 i.e in the \u201cnorthwest\u201d direction.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.3, Page number: 42" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Fuel is fed into a boiler at a rate of 10,000 lb/h in the presence of 20,000 lb/h of air. Due to the\nlow heating value of the fuel, 2000 lb/h of methane is added to assist in the combustion of the\nfuel. At what rate (lb/h) do the product gases exit the incinerator?\n'''\n\n#Variable declaration:\nR1_in = 10000 #Rate of fuel fed into the boiler (lb/h)\nR2_1n = 20000 #Rate of air fed into the boiler (lb/h)\nR3_in = 2000 #Rate of methane fed into the boiler (lb/h)\n\n#Calculations:\nm_in = R1_in + R2_1n + R3_in #Rate of mass in (lb/h)\nm_out = m_in #Rate of mass out (lb/h)\n\n#Result:\nprint \"The rate of the product gases exit from the incinerator is :\",round(m_in),\" lb/h\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of the product gases exit from the incinerator is : 32000.0 lb/h\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.4, Page number: 42" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A proposed heat exchanger design requires that a packed column and a spray tower are to be used\nin series for the removal of HCl from a gas feed stream. The spray tower is operating at an efficiency\nof 65% and the packed column at an efficiency of 98%. Calculate the mass flow rate of HCl leaving the spray tower, the mass flow rate of HCl entering the packed tower, and the overall\nfractional efficiency of the removal system if 76.0 lb of HCl enters the system every hour.\n'''\n\n#Variable declaration:\nE1 = 65 #Efficiency of spray tower (%)\nE2 = 98 #Efficiency of packed column (%)\nm_in = 76 #Mass flow rate of HCl entering the system (lb/h)\n\n#Calculations:\nm1_out = (1 - E1/100.0)*m_in #Mass flow rate of HCl leaving the spray tower (lb/h)\nm2_out = (1 - E2/100.0)*m1_out #Mass flow rate of HCl entering the packed column (lb/h)\nE = (m_in - m2_out)/m_in #Overall fractional efficiency (%)\n\n#Result:\nprint \"The mass flow rate of HCl leaving the spray tower is :\",round(m1_out,2),\" lb/h HCL\"\nprint \"The mass flow rate of HCl entering the packed column is :\",round(m2_out,3),\" lb/h HCL\"\nprint \"The overall fractional efficiency is :\",round(E*100,1),\" %\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The mass flow rate of HCl leaving the spray tower is : 26.6 lb/h HCL\nThe mass flow rate of HCl entering the packed column is : 0.532 lb/h HCL\nThe overall fractional efficiency is : 99.3 %\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.5, Page number: 43" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the flow diagram in Figure 4.3 for a wastewater treatment system. The following\nflowrate data are given:\nm\u02d91 = 1000 lb/min\nm\u02d92 = 1000 lb/min\nm\u02d94 = 200 lb/min\nCalculate the amount of water lost by evaporation in the operation, m\u02d9 .\n'''\n\n#Variable declaration:\nm1 = 1000 #Flowrate data 1 (lb/min)\nm2 = 1000 #Flowrate data 2 (lb/min)\nm4 = 200 #Flowrate data 4 (lb/min)\n\n#Calculations:\nm5 = m1 + m2 - m4 #Flowrate data 5 (lb/min)\nm6 = m2 #Flowrate data 6 (lb/min)\nm = m5 - m6 #Flowrate of water lost in operation (lb/min)\n\n#Result:\nprint \"The amount of water lost by evaporation in the operation is\", round(m),\" lb/min\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The amount of water lost by evaporation in the operation is 800.0 lb/min\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.6, Page number: 44" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the system shown in Figure 4.4. The following volumetric flowrate and phosphate concentration\n(volume basis) data have been provided by the plant manager. Are the data correct\nand/or consistent?\nq1 = 1000 gal/day\nq2 = 1000 gal/day\nq3 = 2000 gal/day\nq4 = 200 gal/day\nq5 = 1800 gal/day\nq6 = 1000 gal/day\nC1 = 4 ppm\nC2 = 0 ppm\nC3 = 2 ppm\nC4 = 20 ppm\nC5 = 0 ppm\nC6 = 0 ppm\n'''\n\n#Variable declaration:\nq1 = 1000.0 #Volumetric flowrate from tank 1 (gal/day)\nq2 = 1000.0 #Volumetric flowrate from tank 2 (gal/day)\nq3 = 2000.0 #Volumetric flowrate from tank 3 (gal/day)\nq4 = 200.0 #Volumetric flowrate from tank 4 (gal/day)\nq5 = 1800.0 #Volumetric flowrate from tank 5 (gal/day)\nq6 = 1000.0 #Volumetric flowrate from tank 6 (gal/day)\nC1 = 4.0 #Phosphate concentration in tank 1 (ppm)\nC2 = 0.0 #Phosphate concentration in tank 2 (ppm)\nC3 = 2.0 #Phosphate concentration in tank 3 (ppm)\nC4 = 20.0 #Phosphate concentration in tank 4 (ppm)\nC5 = 0.0 #Phosphate concentration in tank 5 (ppm)\nC6 = 0.0 #Phosphate concentration in tank 6 (ppm)\nCf = 120000.0 #conversion factor for water (gal/10**6lb)\n\n#Calculations:\nC1q1 = C1*q1/Cf #Data 1 (lb/day)\nC2q2 = C2*q2/Cf #Data 2 (lb/day)\nC3q3 = C3*q3/Cf #Data 3 (lb/day)\nC4q4 = C4*q4/Cf #Data 4 (lb/day)\nC5q5 = C5*q5/Cf #Data 5 (lb/day)\nC6q6 = C6*q6/Cf #Data 6 (lb/day)\n\n#Results:\nif (((C1q1 + C2q2) == C3q3) and C3q3 == (C4q4 + C5q5) and C5q5 == C6q6 and C2q2 == C6q6):\n print \"The data appear to be consistent .\"\nelse:\n print \"The data appear to be inconsistent .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The data appear to be consistent .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.7, Page number: 48" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A lake is located at the top of a mountain. A power plant has been constructed at the bottom of\nthe mountain. The potential energy of the water traveling downhill can be used to spin turbines\nand generate electricity. This is the operating mode in the daytime during peak electrical\ndemand. At night, when demand is reduced, the water is pumped back up the mountain. The\noperation is shown in Figure 4.5.\nUsing the method of power \u201cproduction\u201d described above, determine how much power\n(Watts) is generated by the lake located at an elevation of 3000 ft above the power plant. The\nflowrate of water is 500,000 gpm. The turbine efficiency is 30%. Neglect friction effects.\nNote: This programmed-instructional problem is a modified and edited version (with permission)\nof an illustrative example prepared by Marie Gillman, a graduate mechanical engineering\nstudent at Manhattan College.\n'''\n\n#Variable declaration:\nDz = 3000 #Height (ft)\nV0 = 500000 #Flowrate of water (gal/min)\nn = 30 #Turbine efficiency (%)\nm = 0.3048 #Meters in a feet\nm3 = 0.00378 #Meters-cube in a gallon\ng = 9.8 #Gravitational acceleration (m/s^2)\ngc = 1 #Conversion factor\nMW = 10**(-6) #Megawatt in newton-meter-per-second\n\n#Calculations:\nV1 = (V0*m3)*1000.0/60.0 #The mass flow rate of the water in kilograms/second (kg/s)\nDPE = V1*g*Dz*m/gc*MW #The loss in potential energy (MW)\nAP = n/100.0*DPE #The actual power output (MW)\n\n#Result:\nprint \"The power generated by the lake located is :\",round(AP,1),\" MW\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The power generated by the lake located is : 84.7 MW\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.8, Page number: 50" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If 111.4 lbmol/min of an air stream is heated from 200\u00b0F to 600\u00b0F, calculate the heat transfer\nrate required to bring about this change in temperature. Use the following enthalpy and average\nheat capacity data:\nH200\u00b0F = 1170 Btu/lbmol\nH600\u00b0F = 4010 Btu/lbmol\n'''\n\n#Variable declaration:\nn = 111.4 #Flowrate of air stream (lbmol/min)\nH1 = 1170 #Average heat capacity at 200\u00b0F (Btu/lbmol)\nH2 = 4010 #Average heat capacity at 600\u00b0F (Btu/lbmol)\n\n#Calculation:\nQ = n*(H2 - H1) #The heat transfer rate (Btu/min)\n\n#Result:\nprint \"The heat transfer rate required is:\",round(Q/10**5,2),\" x 10**5 Btu/min\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate required is: 3.16 x 10**5 Btu/min\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.9, Page number: 50" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Given the mass flow rate of a fluid and its heat capacity, determine the required heat rate to\nchange the fluid from one temperature to another. Data are provided below:\nn = 600 lbmol/min\nCp,AV = 0.271 Btu/lbmol . \u00b0F (over the 200\u2013600\u00b0F range)\nT1 = 200\u00b0F\nT2 = 600\u00b0F\n'''\n\n#Variable declaration:\nn = 600 #The mass flow rate of fluid (lbmol/min)\nCp_AV = 0.271 #Heat capacity (Btu/lbmol . \u00b0F)\nT1 = 200 #Initial temperature(\u00b0F)\nT2 = 600 #Final temperature(\u00b0F)\n\n#Calcultaion:\nQ = n*Cp_AV*(T2 - T1) #The required heat rate (Btu/min)\n\n#Result:\nprint \"The required heat rate is :\",round(Q,-2),\" Btu/min\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required heat rate is : 65000.0 Btu/min\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.10, Page number: 51" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Obtain the heat transfer rate in an exchanger if equal mass liquid flow rates are used. The hot\nfluid is cooled from 94 \u00b0C to 82 \u00b0C while the cold fluid is initially at 20 \u00b0C.\n'''\n\n#Variable declaration:\nT_c1 = 20 #Initial cold fluid temperature (\u00b0C)\nT_h1 = 82 #Initial hot fluid temperature (\u00b0C)\nT_h2 = 94 #Final hot fluid temperature (\u00b0C)\n\n#Calculation:\nT_c2 = (T_h2 - T_h1 + T_c1) #Final cold fluid temperature (\u00b0C)\n\n#Result:\nprint \"The heat transfer rate is:\",round(T_c2),\" \u00b0C\"\nprint \"There is a printing mistake in book regarding unit of the final result.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate is: 32.0 \u00b0C\nThere is a printing mistake in book regarding unit of the final result.\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.11, Page number: 51" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''As a gas flows through a cooler, 5.5MW of heat is transferred from the gas. The average heat\ncapacity of the gas is 1090 J/(kg . \u00b0C), the gas mass flow rate, m\u02d9 , is 9 kg/s and the gas inlet temperature,\nT1, is 650 \u00b0C. For this example, kinetic and potential energy effects are again neglected.\nFurthermore, there is no shaft work. Determine the gas outlet temperature.\n'''\n\n#Variable declaration:\nQ = -5.5*10**6 #The heat transferred out from the gas (W)\nCp = 1090.0 #The average heat capacity of the gas (J/(kg . \u00b0C))\nm = 9.0 #The gas mass flow rate (kg/s)\nT1 = 650 #The gas inlet temperature (\u00b0C)\n\n#Calculation:\nT2 = Q/(m*Cp)+T1 #The gas outlet temperature (\u00b0C)\n\n#Result:\nprint \"The gas outlet temperature is :\",round(T2),\" \u00b0C\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The gas outlet temperature is : 89.0 \u00b0C\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 4.12, Page number: 52" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heat pump takes in 3500 gpm of water at a temperature of 38 \u00b0F and discharges it back to the\nlake at 36.2 \u00b0F. How many Btu are removed from thewater per day [Cp for H2O = 75.4 J/(gmol .\u00b0C),\np = 62.4 lb/ft^3]?\n '''\n\n#Variable declaration:\nn = 3500.0 #Inlet flowrate of water (gal/min)\nCp_W = 75.4 #Heat capacity of water (J/(gmol . \u00b0C)\np = 62.4 #Density of water (lb/ft^3)\nM = 24*60.0 #Minutes in a day (min/day)\nG = 7.48 #Gallons in a feet cube (gal/ft^3)\ngm = 454.0 #Grams in a pound (g/lb)\nJ = 1054.0 #Joules in a Btu (J/Btu)\ng = 18.0 #Grams in a gmol (g/gmol)\nF = 1.8 #Degree fahrenheit in a degree celcius (\u00b0F)\nTi = 38.0 #Initial temperature (\u00b0F)\nTf = 36.2 #Final temperature (\u00b0F)\n\n#Calculations:\nT= Ti-Tf #Temperature loss (\u00b0F)\nm = n*p*M/G #Mass flow rate of water (lb/day)\nCp = Cp_W*gm/J/g/F #Heat capacity in cosistent units (Btu/(lb.\u00b0F))\nQ = m*Cp*T #Rate of heat flow from water (Btu/day)\n\n#Result:\nprint \"The rate of Btu removed from the water per day is :\",round(Q/10**8,2),\" x 10**8 Btu/day .\"\nprint \"There is a calculation mistake in the book regarding the final result.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of Btu removed from the water per day is : 0.76 x 10**8 Btu/day .\nThere is a calculation mistake in the book regarding the final result.\n" + } + ], + "prompt_number": 15 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_05.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_05.ipynb new file mode 100644 index 00000000..8cc3b0eb --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_05.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "Chapter 05" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 5: Gas Laws" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.1, Page number: 56" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the final ( f ) volumetric flow rate of a gas that is heated at constant pressure in a heat\nexchanger from 100 to 300\u00b0F if its initial (i) flow is 3500 actual cubic feet per minute (acfm).\n'''\n\n#Variable declaration:\nqi = 3500 #Initial volumetric flow rate of gas (acfm)\nTi = 100.0 #Initial temperature (\u00b0F)\nTf = 300.0 #Final temperature (\u00b0F)\n\n#Calculation:\nTi_R = Ti+460 #Initial temperatur in Rankine scale (\u00b0R)\nTf_R = Tf+460 #Final temperatur in Rankine scale (\u00b0R)\nqf = qi*(Tf_R/Ti_R) #Final volumetric flow rate of gas (acfm)\n\n#Result:\nprint \"The final volumetric flow rate of gas is :\",round(qf),\" acfm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The final volumetric flow rate of gas is : 4750.0 acfm\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.2, Page number: 57" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the volumetric flow rate of the gas (100\u00b0F, 1 atm) in the previous example if it is compressed\nisothermally (constant temperature) to 3 atm?\n'''\n\n#Variable declaration:\nqi = 3500 #Initial volumetric flow rate of gas (acfm)\nPi = 1.0 #Iitial pressure (atm)\nPf = 3.0 #Final pressure (atm)\n\n#Calculation:\nqf = qi*(Pi/Pf) #Final volumetric flow rate of gas (acfm)\n\n#Result:\nprint \"The volumetric flow rate of the gas (100\u00b0F, 1 atm) is:\",round(qf),\" acfm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The volumetric flow rate of the gas (100\u00b0F, 1 atm) is: 1167.0 acfm\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.3, Page number: 57" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the volumetric flow rate of the gas in the previous example if the final temperature\nis 300\u00b0F.\n'''\n\n#Variable declaration:\nqi = 3500 #Initial volumetric flow rate of the gas (acfm)\nPi = 1.0 #Initial pressure (atm)\nPf = 3.0 #Final pressure (atm)\nTf = 300.0+460.0 #Final temperature in Rankine scale (\u00b0R)\nTi = 100.0+460.0 #Initial temperature in Rankine scale (\u00b0R)\n\n#Calculation:\nqf = qi*(Pi/Pf)*(Tf/Ti) #Final volumetric flow rate of the gas (acfm)\n\n#Result:\nprint \"The volumetric flow rate of the gas at 300\u00b0F temperature is :\",round(qf),\" acfm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The volumetric flow rate of the gas at 300\u00b0F temperature is : 1583.0 acfm\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.4, Page number: 59" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the density of air at 75 \u00b0F and 14.7 psia? The molecular weight of air is approximately 29.\n'''\n\n#Variable declaration:\nP = 14.7 #Absolute pressure of air (psia)\nMW = 29 #Molecular weight of air (lb/lbmol)\nT = 75+460 #Temperature in Rankine scale (\u00b0R)\nR = 10.73 #Universal gas constant (ft^3.psi/lbmol.\u00b0R)\n\n#Calculation:\np = P*MW/R/T #Density of air (lb/ft^3)\n\n#Result:\nprint \"The density of air at 75\u00b0F and 14.7 psia is :\",round(p,4),\" lb/ft^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The density of air at 75\u00b0F and 14.7 psia is : 0.0743 lb/ft^3\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.5, Page number: 59" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the volume (in ft3) of 1.0 lbmol of any ideal gas at 60\u00b0F and 14.7 psia.\n'''\n\n#Variable declaration:\nn = 1 #Molar flow rate of gas (lbmol/h)\nR = 10.73 #Universal gas constant (ft^3.psi/lbmol.\u00b0R)\nT = 60+460 #Temperature in Rankine scale (\u00b0R)\nP = 14.7 #Absolute pressure of gas (psia)\n\n#Calculation:\nV = n*R*T/P #Volume of gas (ft^3)\n\n#Result:\nprint \"The volume of given ideal gas is :\",round(V,1),\" ft^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The volume of given ideal gas is : 379.6 ft^3\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.6, Page number: 59" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the density of a gas (MW = 29) in g/cm3 at 20\u00b0C and 1.2 atm using the ideal gas law.\n'''\n\n#Variable declaration:\nP = 1.2 #Abslute pressure of gas (psia)\nMW = 29 #Molecular weight of gas (g/gmol)\nR = 82.06 #Universal gas constant (atm.cm^3/gmol.K)\nT = 20+273 #Temperature in Kelvin (K)\n\n#Calculation:\np = P*MW/R/T #Dendity of gas (g/cm^3)\n\n#Result:\nprint \"The density of given gas is :\",round(p,5),\" g/cm^3\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The density of given gas is : 0.00145 g/cm^3\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.7, Page number: 60" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A certain pure-component two-element ideal gas has a specific volume v of 10.58 ft3/lb at 70\u00b0F\nand 14.7 psia. Determine the molecular weight of the gas and state its name. (Hint: The gas is\nacidic and soluble in water.)\n'''\n\n#Variable declaration:\nR = 10.73 #Universal gas constant (psia . ft^3/lbmol .\u00b0R)\nT = 70+460 #Temperature in Rankine scale (\u00b0R)\nv = 10.58 #Specific volume (ft^3/lb)\nP = 14.7 #Absolute pressure (psia)\n\n#Calculation:\nMW = R*T/v/P #Molecular weight of gas (lb/lbmol)\n\n#Result:\nprint \"The molecular weight of the gas is :\",round(MW,2),\" lb/lbmol.\"\nprint \"It appears that the gas is HCl (i.e., hydrogen chloride).\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The molecular weight of the gas is : 36.57 lb/lbmol.\nIt appears that the gas is HCl (i.e., hydrogen chloride).\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.8, Page number: 61" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Data from a heat exchanger indicate a volumetric flow rate of 30,000 scfm (60\u00b0F, 1 atm). If the\noperating temperature and pressure of the unit are 1100\u00b0F and 1 atm, respectively, calculate the\nflow rate in actual cubic feet per minute (acfm).\n'''\n\n#Variable declaration:\nqs = 30000 #Volumetric flow rate at standard conditions (scfm)\nTa = 1100+460 #Actual absolute temperature in Rankine scale (\u00b0R)\nTs = 60+460 #Standard absolute temperature in Rankine scale (\u00b0R)\n\n#Calculation:\nqa = qs*Ta/Ts #Volumetric flow rate at actual conditions (acfm)\n\n#Result:\nprint \"The volumetric flow rate in actual cubic feet per minute is :\",round(qa),\" acfm\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The volumetric flow rate in actual cubic feet per minute is : 90000.0 acfm\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.9, Page number: 62" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The exhaust gas flow rate from a condenser is 1000 scfm. All of the gas is vented through a small\nstack that has an inlet area of 2.0 ft^2. The exhaust gas temperature is 300\u00b0F. What is the velocity\nof the gas through the stack inlet in feet per second? Assume standard conditions to be 70\u00b0F and\n1.0 atm. Neglect the pressure drop across the stack.\n'''\n\n#Variable declaration:\nqs = 1000 #Volumetric flow rate at standard conditions (scfm)\nTa = 300+460 #Actual absolute temperature in Rankine scale (\u00b0R)\nTs = 70+460 #Standard absolute temperature in Rankine scale (\u00b0R)\nA = 2.0 #Inlet area of stack (ft^2)\n\n#Calculations:\nqa = qs*Ta/Ts #Volumetric flow rate at actual conditions (acfm)\nv = qa/A/60 #Velocity of gas (ft/s)\n\n#Result:\nprint \"The velocity of the gas through the stack inlet is :\",round(v),\" ft/s\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The velocity of the gas through the stack inlet is : 12.0 ft/s\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.10, Page number: 62" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''C6H5Cl is fed into a thermal oxidizer at a rate of 5000 scfm (60\u00b0F, 1 atm) and is combusted in the\npresence of air fed at a rate of 3000 scfm (60\u00b0F, 1 atm). Both streams enter the oxidizer at 70\u00b0F.\nThe products are then cooled from 20008F and exit the cooler at 180\u00b0F. At what rate (lb/h) do the\nproducts exit the cooler? The molecular weight of C6H5Cl is 112.5; the molecular weight of\nair is 29.\n'''\n\n#Variable declaration:\nqs1 = 5000.0 #Volumetric flow rate of C6H5Cl at standard conditions (scfm)\nqs2 = 3000.0 #Volumetric flow rate of air at standard conditions (scfm)\nTa = 70+460.0 #Actual absolute temperature in Rankine scale (\u00b0R)\nTs = 60+460.0 #Standard absolute temperature in Rankine scale (\u00b0R)\nV = 387.0 #Volume occupied by one lbmol of any ideal gas (ft^3)\nM1 = 112.5 #Molecular weight of C6H5Cl (lb/lbmol)\nM2 = 29.0 #Molecular weight of air (lb/lbmol)\nT = 60.0 #Absolute temperature (\u00b0F)\n\n#Calculations:\nqa1 = qs1*(Ta/Ts) #Volumetric flow rate of C6H5Cl at actual conditions (acfm)\nqa2 = qs2*(Ta/Ts) #Volumetric flow rate of air at actual conditions (acfm)\nn1 = qa1/V #Molar flow rate of C6H5Cl (lbmol/min)\nn2 = qa2/V #Molar flow rate of air (lbmol/min)\nm1 = n1*M1*T #Mass flow rate of C6H5Cl (lb/h)\nm2 = n2*M2*T #Mass flow rate of air (lb/h)\nm_in = m1+m2 #Total mass flow rate of both streams entering the oxidizer (lb/h)\nm_out = m_in #Total mass flow rate of both streams exit the cooler (lb/h)\n\n#Result:\nprint \"The rate of the products exit the cooler is :\",round(m_out),\" lb/h\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of the products exit the cooler is : 102634.0 lb/h\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 5.11, Page number: 64" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The exhaust to the atmosphere from a thermal device has a SO3 concentration expressed as\n0.15 mm Hg partial pressure. Calculate the parts per million of SO3 in the exhaust.\n'''\n\n#Variable declaration:\np = 0.15 #Partial pressure of SO3 (mm Hg)\nP = 760.0 #Atmospheric pressure (mm Hg)\nm = 10**6 #Particles in a million\n\n#Calculation:\ny = p/P #Mole fraction of SO3\nppm = y*m #Parts per million of SO3 (ppm)\n\n#Result:\nprint \"The parts per million of SO3 in the exhaust is :\",round(ppm),\" ppm .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The parts per million of SO3 in the exhaust is : 197.0 ppm .\n" + } + ], + "prompt_number": 12 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_06.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_06.ipynb new file mode 100644 index 00000000..dfc904b2 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_06.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "Chapter 06" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 6: Heat Exchanger Pipes and Tubes" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.1, Page number: 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Obtain the inside diameter in inches of 2 inch schedule 40 nominal steel pipe.\n'''\n\n#Variable declaration:\nNPS = 2 #Nominal pipe size (inch)\nSN = 40 #Schedule number\n\n#Calculation:\n#From Table 6.2, we obtain that the inside diameter of steel pipe is ID = 2.067 in.\nID = 2.067\n\n#Result:\nprint \"The inside diameter of steel pipe is :\",ID,\" in.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The inside diameter of steel pipe is : 2.067 in.\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.2, Page number: 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine the inside diameter (ID), outside diameter (OD), wall thickness, and pipe weight\n(in lb/ft) of 3 inch schedule 40 steel pipe.\n'''\n\n#Variable declaration:\nNPS = 3 #Nominal pipe size (inch)\nSN = 40 #Schedule number\n\n#Calculation:\n#From Table 6.2, we obtain that the inside diameter of steel pipe is ID = 3.068 in, outside diameter OD = 3.5 in, wal thickness WT = 0.216 in, and pipe weight PW = 7.58 lb/ft.\nID = 3.068\nOD = 3.5\nWT = 0.216\nPW = 7.58\n\n#Result:\nprint \"The inside diameter of steel pipe is :\",ID,\" in\"\nprint \"The outside diameter of steel pipe is :\",OD,\" in\"\nprint \"The wall thickness of steel pipe is :\",WT,\" in\"\nprint \"The weight of steel pipe is :\",PW,\" lb/ft.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The inside diameter of steel pipe is : 3.068 in\nThe outside diameter of steel pipe is : 3.5 in\nThe wall thickness of steel pipe is : 0.216 in\nThe weight of steel pipe is : 7.58 lb/ft.\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.3, Page number: 73" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is provided for a steel pipe:\nID = 0.957 in\nOD = 1.315 in\nWall thickness = 0.179 in\nPipe weight = 2.17 lb/ft\nDetermine the nominal size and schedule number of the pipe.\n'''\n\n#Variable declaration:\nID = 0.957 #Inside diameter of pipe (in)\nOD = 1.315 #Outside diameter of pipe (in)\nWT = 0.179 #Wall thickness of pipe (in)\nPW = 2.17 #Weight of pipe (lb/ft)\n\n#Calculation:\n#From Table 6.2, it indicates that the steel pipe is 1 inch schedule 80.\nNSP = 1\nSN = 80\n\n#Result:\nprint \"The nominal size of the pipe is :\",NSP,\" in.\"\nprint \"The schedule number of the pipe is:\",SN,\".\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The nominal size of the pipe is : 1 in.\nThe schedule number of the pipe is: 80 .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.4, Page number: 75" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine the inside diameter, wall thickness, outside diameter, and external area per foot (EA)\nof a 3/4 inch 16 BWG tube.\n'''\n\n#Variable declaration:\nS = 3/4 #Tube size (in)\nBWG = 16 #Birmingham Wire Gauge number (gauge)\n\n#calculation:\n#From table 6.3, we get:\nID = 0.620 #Internal diameter of tube (in)\nWT = 0.065 #Wall thickness of tube (in)\nOD = ID+2*WT #Outside diameter of tube (in)\nEA = 0.1963 #External area per foot (ft)\n\n#Result:\nprint \"The inside diameter is :\",ID,\" in\"\nprint \"The wall thickness is :\",WT,\" in\"\nprint \"The outside diamater is :\",OD,\" in\"\nprint \"The external area per foot per foot :\",EA,\" ft.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The inside diameter is : 0.62 in\nThe wall thickness is : 0.065 in\nThe outside diamater is : 0.75 in\nThe external area per foot per foot : 0.1963 ft.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.11, Page number: 81" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An air-conditioning duct has a rectangular cross-section of 1 m by 0.25 m. If the kinematic viscosity\nof the air is approximately 1 x 10**-5 m2/s, determine the maximum air velocity before\nthe flow becomes turbulent. Assume the critical Reynolds number is 2300.\n'''\n\n#Variable declaration:\na = 1 #Length of cross-section (m)\nb = 0.25 #Width of cross-section (m)\nv = 1*10**-5 #Kinematic viscosity of air (m^2/s)\nRe = 2300.0 #Reynolds Number\ncm = 100 #Cenitmeters in a meter\n\n#Calculation:\nDh = 2*a*b/(a+b) #Hydraulic diameter of duct (m)\nV = Re*v/Dh*cm #Maximum air velocity (cm/s)\n\n#Result:\nprint \"The maximum air velocity before the flow becomes turbulent is :\",round(V,1),\" cm/s.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The maximum air velocity before the flow becomes turbulent is : 5.8 cm/s.\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.12, Page number: 82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A circular 2-inch diameter horizontal tube contains a cold liquid fluid of density and viscosity\n70 lb/ft3 and 0.1806 lb/ft . s, respectively. If, the flow rate is 0.486 ft3/s, determine if the flow\nis laminar.\n'''\n\n#Variable declaration:\nq = 0.486 #Flow rate of fluid (ft^3/s)\nD = 2.0/12.0 #Diameter of tube in feet (ft)\npi = 3.14 #Value of pi\np = 70.0 #Density of fluid (lb/ft^3)\nu = 0.1806 #Viscosity of fluid (lb/ft)\n\n#Calculation:\nV = 4*q/pi/D**2 #Flow velocity (ft/s)\nRe = D*V*p/u #Reynolds Number\n\n#Result:\nif(Re<2100):\n print \"The flow is laminar.\"\nelif(Re>2100):\n print \"The flow is turbulant.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The flow is laminar.\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.13, Page number: 82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 6.12. How long must the pipe be to ensure a fully developed flow?'''\n\n#Variable declaration:\n#From example 6.12, we have:\nD = 2.0/12.0 #Diameter of pipe in feet (ft)\nRe = 1440.0 #Reynolds number\n\n#Calculation:\nLc = 0.05*D*Re #Length of pipe (ft)\n\n#Result:\nprint \"The pipe length to ensure a fully developed flow is:\",Lc,\" ft.\"\nprint \"This is an abnormally long calming length for a pipe (or tube) in a heat exchanger.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The pipe length to ensure a fully developed flow is: 12.0 ft.\nThis is an abnormally long calming length for a pipe (or tube) in a heat exchanger.\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 6.14, Page number: 82" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the average velocity of water flowing in a 2-inch schedule 40 standard pipe at 608F for\nwhich the flow is viscous (laminar).\n'''\n\n#Variable declaration:\nu = 6.72*10**-4 #Viscosity of water (lb/ft.s)\np = 62.4 #Density of water (lb/ft^3)\n#For laminar flow:\nRe = 2100.0 #Reynolds number\n#From table 6.2, we have:\nD = 2.067/12.0 #Inside diameter of pipe (ft)\n\n#Calculation:\nV = Re*u/D/p #Average velocity of water flowing (ft/s)\n\n#Result:\nprint \"The average velocity of water flowing is:\",round(V,2),\" ft/s.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The average velocity of water flowing is: 0.13 ft/s.\n" + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_07.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_07.ipynb new file mode 100644 index 00000000..939dc2b5 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_07.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "Chapter 07" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 7: Steady-State Heat Conduction" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.1, Page number: 93" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A glass window is 1 cm thick and has an area of 3 m^2. The temperature at the outer surface is\n10\u00b0C. The glass has a conductivity of 1.4 W/m . K. The heat transfer rate is 3 kW. Calculate\nthe temperature at the inner surface.\n'''\n\n#Variable declaration:\nQ = 3000.0 #The rate of heat flow through the glass window (W)\nL = 0.01 #Thickness of glass window (m)\nA = 3.0 #Area of heat transfer (m^2)\nTC = 10+273 #Temperature at the outside surface (K)\nk = 1.4 #Thermal onductivity of glass (W/m.K)\n\n#Calculation:\nTH = TC+Q*L/k/A #Temperature at the inner surface (K)\n\n#Result:\nprint \"The temperature at the inner surface is :\",round(TH,1),\" K\"\nprint \"The temperature at the inner surface is :\",round(TH-273,1),\" \u00b0C\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature at the inner surface is : 290.1 K\nThe temperature at the inner surface is : 17.1 \u00b0C\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.2, Page number: 94" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A new 1 ft thick insulating material was recently tested for heat resistant properties. The data\nrecorded temperatures of 70\u00b0F and 210\u00b0F on the cold and hot sides, respectively. If the thermal\nconductivity of the insulating material is 0.026 Btu/ft . h . \u00b0F, calculate the rate of the heat flux,\nQ/A, through the wall in Btu/ft^2 . h. Resolve the problem in SI units.\n'''\n\n#Variable declaration:\nk = 0.026 #Thermal conductivity of insulating material (Btu/ft.h.\u00b0F)\nL = 1.0 #Thickness of insulating material (ft)\nTC = 70.0 #Temperature on the cold side surface (\u00b0F)\nTH = 210.0 #Temperature on the hot side surface (\u00b0F)\nc = 0.252 #Kilocalorie per hour in a Btu per hour\nm = 0.093 #meter square in a feet square\n\n#Calculation:\nDT = TH-TC #Change in temperature (\u00b0F)\nQ1 = k*DT/L #Rate of heat flux throughthe wall (Btu/f^t2.h.)\nQ2 = Q1*c/m #Rate of heat flux throughthe wall in SI units (kcal/m^2.h)\n\n#Result:\nprint \"The rate of heat flux in Btu/ft^2.h is :\",round(Q1,3),\" Btu/ft^2.h .\"\nprint \"The rate of heat flux in SI units is :\",round(Q2,3),\" kcal/m^2.h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of heat flux in Btu/ft^2.h is : 3.64 Btu/ft^2.h .\nThe rate of heat flux in SI units is : 9.863 kcal/m^2.h .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.3, Page number: 94" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is provided. A rectangular furnace wall is made of fire clay (height,\nH = 3 m, width, W = 1.2 m, thickness, L = 0.17 m). Temperature of inside surface (area =\nH x W), TH, is 1592 K, and of the outside surface (area = H x W), TC, is 1364 K.\nDetermine the temperature gradient, the heat transfer rate, and the heat transfer flux.\n'''\n\n#Variable declaration:\nTH = 1592.0 #Temperature of inside surface (K)\nTC = 1364.0 #Temperature of outside surface (K)\nH = 3.0 #Height of furnace wall (m)\nW = 1.2 #Width of furnace wall (m)\nL = 0.17 #Thickness furnace wall (m)\nm = 0.0929 #Meter square per second in a feet square per second\nBtu = 3.412 #Btu per hour in a Watt\nBtu2 = 0.3171 #Btu per feet square hour in a watt per meter square\n\n#Calculation:\nTav = (TH+TC)/2 #Average wall temperature (K)\n#From Table in Appendix:\np = 2645.0 #Density of material (kg/m^3)\nk = 1.8 #Thermal conductivity (W/m.K)\nCp = 960.0 #Heat capacity of material (J/kg.K)\na = k/(p*Cp)/m #Thermal diffusivity (ft^2/s)\nt = (TC-TH)/L #Temperature gradient (\u00b0C/m)\nA = H*W #Heat transfer area (m^2)\nQ1 = k*A*(TH-TC)/L*Btu #Heat transfer rate (Btu/h)\nQ2 = k*(TH-TC)/L*Btu2 #Heat transfer flux (Btu/h.ft^2)\nR = L/(k*A) #Thermal resistance (\u00b0C/W)\n\n#Result:\nprint \"The temperature gradient is :\",round(t),\" \u00b0C/m.\"\nprint \"The heat transfer rate is :\",round(Q1),\" Btu/h.\"\nprint \"The heat transfer flux is :\",round(Q2,1),\" Btu/h.ft^2.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature gradient is : -1341.0 \u00b0C/m.\nThe heat transfer rate is : 29653.0 Btu/h.\nThe heat transfer flux is : 765.5 Btu/h.ft^2.\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.4, Page number: 96" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A concrete wall has a surface area of 30 m^2 and is 0.30 m thick. It separates warm room air from\ncold ambient air. The inner surface of the wall is known to be at a temperature of 25\u00b0C, while the\nouter surface is at -15\u00b0C. The thermal conductivity of the concrete is 1.0 W/m . K.\n1. Describe the conditions that must be satisfied in order for the temperature distribution in\nthe wall to be linear.\n2. What is the driving force for the transfer of heat?\n3. What is the heat loss through the wall?\n'''\n\n#Variable declaration:\nTH = 25.0 #Temperature at inner suface of wall (\u00b0C)\nTC = -15.0 #Temperature at outer suface of wall (\u00b0C)\nL = 0.3 #Thickness of wall (m)\nk = 1.0 #Thermal conductivity of concrete (W/m)\nA = 30.0 #Sueface area of wall (m^2)\n\n#Calculation:\nDT = TH-TC #Driving force for heat transfer (\u00b0C) (part 2)\nR = L/(k*A) #Thermal resistance (\u00b0C/W) (part 3)\nQ = DT/R/10**3 #Heat loss through the wall (kW)\n\n#Result:\nprint \"1. Theoretical part.\"\nprint \"2. The driving force for heat transfer is :\",DT,\" \u00b0C.\"\nprint \"3. The heat loss through the wall is :\",Q,\" kW.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. Theoretical part.\n2. The driving force for heat transfer is : 40.0 \u00b0C.\n3. The heat loss through the wall is : 4.0 kW.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.5, Page number: 97" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The inside and outside temperatures of the walls of a room are 27\u00b0C and 68.7\u00b0C, respectively.\nThe room walls consist of 6-inch concrete (C), 8-inch cork-board (B), and 1-inch wood (W) with\ncorresponding thermal conductivities of 0.762, 0.0433, and 0.151W/m . K, respectively.\nCalculate the heat transfer rate across the wall and determine the temperature at the interface\nbetween the wood and the cork-board.\n'''\n\n#Variable declaration:\nTC = 27.0 #Inside temperature of walls (\u00b0C)\nTH = 68.7 #Outside temperature of walls (\u00b0C)\nLC = 6*0.0254 #Thickness of concrete (m)\nLB = 8*0.0254 #Thickness of cork-board (m)\nLW = 1*0.0254 #Thickness of wood (m)\nkC = 0.762 #Thermal conductivity of concrete (W/m.K)\nkB = 0.0433 #Thermal conductivity of cork-board (W/m.K)\nkW = 0.151 #Thermal conductivity of wood (W/m.K)\n\n#Calculation:\nRC = LC/kC #Thermal resistance of concrete (K/W)\nRB = LB/kB #Thermal resistance of cork-board (K/W)\nRW = LW/kW #Thermal resistance of wood (K/W)\nQ = (TC-TH)/(RC+RB+RW) #Heat transfer rate across the wall (W)\nT = -(Q*RW-TC) #Interface temperature between wood and cork-board (K)\n\n#Result:\nprint \"The heat transfer rate across the wall is :\",round(Q,3),\" W.\"\nprint \"The interface temperature between wood and cork-board is :\",round(T,1),\" \u00b0C.\"\nprint \"The interface temperature between wood and cork-board is :\",round(T+273,1),\" K.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate across the wall is : -8.239 W.\nThe interface temperature between wood and cork-board is : 28.4 \u00b0C.\nThe interface temperature between wood and cork-board is : 301.4 K.\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.6, Page number: 98" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A 3-inch outside diameter steel pipe is covered with a 1/2-inch layer of asbestos, (a), which in turn\nis covered with a 2-inch layer of glass wool, (b). Determine the steady-state heat transfer per foot\nof pipe, Z, if the pipe outer surface temperature is 500\u00b0F and the glass wool outer temperature\nis 100\u00b0F.\nAssume an asbestos\u2013glass wool interfacial temperature of 300\u00b0F and an average asbestos (a)\ntemperature of 200\u00b0F and glass wool (b) temperature of 400\u00b0F. Based on a literature review,\nasbestos and glass wool thermal conductivity values have been estimated to be:\nk(a)(200\u00b0F) = 0.120 Btu=h . ft . \u00b0F\nk(b)(400\u00b0F) = 0.0317 Btu=h . ft . \u00b0F\n'''\n\nfrom math import pi, log\nfrom sympy import symbols\n\n#Variable declaration:\nZ = symbols ('z') #Length of pipe\nD1s = 4.0 #Glass wool inside diameter (in)\nD2s = 8.0 #Glass wool outside diameter (in)\nD1a = 3.0 #Asbestos inside diameter (in)\nD2a = 4.0 #Asbestos outside diameter (in)\nTH = 500.0 #Outer surface temperature of pipe (\u00b0F)\nTC = 100.0 #Outer surface temperature of glass wool (\u00b0F)\nLa = 0.5/12.0 #Thickness of asbestos (ft)\nLb = 2.0/12.0 #Thickness of glss wool (ft)\nka = 0.120 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\nkb = 0.0317 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\n\n#Calculation:\nAa = (pi*Z*(D2a-D1a)/12.0)/log(D2a/D1a) #Area of asbestos (ft^2)\nAb = (pi*Z*(D2s-D1s)/12.0)/log(D2s/D1s) #Area of glass wool (ft^2)\nQ1 = (TH-TC)/(La/(ka*Aa)+Lb/(kb*Ab)) #Steady-state heat transfer per foot of pipe (Btu/h.)\nQ2 = Q1/Z #Factorization of Q/Z (Btu/h.ft)\n\n#Result:\nprint \"The steady-state heat transfer per foot of pipe, Z, is :\",round(Q1/Z,1),\" x z Btu/h.\"\nprint \"The steady-state heat transfer factorizating out Z is :\",round(Q2,1),\" Btu/h.ft.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The steady-state heat transfer per foot of pipe, Z, is : 103.6 x z Btu/h.\nThe steady-state heat transfer factorizating out Z is : 103.6 Btu/h.ft.\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.7, Page number: 99" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example. Calculate the outer asbestos temperature, Ti (i.e., the interfacial\ntemperature between the asbestos and glass wool).\n'''\n\n#Variable declaration:\n#From example 7.6:\nTH = 500 #Outer surface temperature of pipe (\u00b0F)\nLb = 2.0/12.0 #Thickness of glss wool (ft)\nkb = 0.0317 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\nAb = 1.51 #Area of glass wool (ft^2)\nQ = 103.5 #Steady-state heat transfer per foot of pipe (Btu/h.)\nLa = 0.5/12.0 #Thickness of asbestos (ft)\nka = 0.120 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\nAa = 0.91 #Area of asbestos (ft^2)\nTC = 100 #Outer surface temperature of glass wool (\u00b0F)\n\n#Calculation:\nTi_b = -((Lb*Q)/(kb*Ab)-TH) #Interfacial temperature of glass wool layer (\u00b0F)\nTi_a = (Q*La)/(ka*Aa)+TC #Interfacial temperature of asbestos layer (\u00b0F)\n\n#Result:\nprint \"The interfacial temperature of glass wool layer is :\",round(Ti_b),\" \u00b0F.\"\nprint \"The interfacial temperature of asbestos layer is :\",round(Ti_a,1),\" \u00b0F.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The interfacial temperature of glass wool layer is : 140.0 \u00b0F.\nThe interfacial temperature of asbestos layer is : 139.5 \u00b0F.\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 7.8, Page number: 100" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The temperature of the solid slab pictured in Figure 7.4 is given by the expression\nT = 100cos(pz/2h)\nCalculate the heat flux at z = 0 and z = h in terms of the thermal conductivity and h. Comment\non the results.\n'''\n\nfrom sympy import cos,symbols,diff,pi\n\n#Variable declaration:\nz,h,k = symbols('z, h, k') #Length, height, thermal conductivity\nT = 100*cos((pi*z)/(2*h)) #Temperature of solid slab\n\n#Calculation:\nDT = diff(T,z) #Temperature at z\nQ = -k*(DT) #Heat flux in slab (Btu/s.ft^2)\nQ1 = Q.subs(z,0) #Heat flux in slab at z = 0 (Btu/s.ft^2)\nQ2 = Q.subs(z,h) #Heat flux in slab at z = h (Btu/s.ft^2)\n\n#Result:\nprint \"The heat flux in slab is :\",Q,\" Btu/s.ft^2 .\"\nprint \"The heat flux in slab at z = 0 is :\",Q1,\" Btu/s.ft^2 .\"\nprint \"The heat flux in slab at z = h is :5\",Q2,\" Btu/s.ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat flux in slab is : 50*pi*k*sin(pi*z/(2*h))/h Btu/s.ft^2 .\nThe heat flux in slab at z = 0 is : 0 Btu/s.ft^2 .\nThe heat flux in slab at z = h is :5 50*pi*k/h Btu/s.ft^2 .\n" + } + ], + "prompt_number": 8 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_08.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_08.ipynb new file mode 100644 index 00000000..cb9b01d8 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_08.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "Chapter 08" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 8: Unsteady-State Heat Conduction" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 8.4, Page number: 122" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 8.3. If TS represents saturated steam at 15 psig, the initial temperature\nTA is 71\u00b0F, and the rod is 20 inches in length of stainless steel with k = 9.1 Btu/h . ft . \u00b0F,\nr = 0.29 lb/in3, and cp = 0.12 Btu/lb . \u00b0F, calculate the temperature 0.875 inches from one of\nthe ends after 30 minutes. Note: This unit was designed by the author and is located in the Unit\nOperations Laboratory of Manhattan College.\n'''\n\nfrom math import pi,sin,e\n\n#Variable declaration:\nk = 9.1 #Thermal coductivity of steel rod (Btu/h.ft.\u00b0F)\np = 0.29*1728 #Density of steel rod (lb/ft^3)\nCp = 0.12 #Heat capacity of steel rod (Btu/lb.\u00b0F)\nP = 15+14.7 #Absolute pressure (psia)\nTa = 71.0 #Initial temperature (\u00b0F)\nL = 20.0/12.0 #Length of rod (ft)\nt = 30.0/60.0 #Time taken (h)\nx = 0.875/12.0 #Length from one of end (ft)\n#From assumption:\nn = 1.0 #First term\n#From tables in Appendix:\nTs = 249.7 #Saturated steam temperature (\u00b0F)\n\n#Calculation:\na = k/(p*Cp) #Thermal diffusivity (ft^2/s)\nT = Ts+(Ta-Ts)*(((n+1)*(-1)**2 + 1 )/pi)*e**((-a*((n*pi)/L)**2)*t)*sin((n*pi*x)/L) #Temperature 0.875 inches from one of the ends after 30 minutes (\u00b0F)\n\n#Result:\nprint \"The temperature 0.875 inches from one of the ends after 30 minutes is :\",round(T),\" \u00b0F.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature 0.875 inches from one of the ends after 30 minutes is : 232.0 \u00b0F.\n" + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_09.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_09.ipynb new file mode 100644 index 00000000..f6fe87e6 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_09.ipynb @@ -0,0 +1,293 @@ +{ + "metadata": { + "name": "Chapter 09" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 9: Forced Convection" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.1, Page number: 135" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider a closed cylindrical reactor vessel of diameter D = 1 ft, and length L = 1.5 ft. The surface\ntemperature of the vessel, T1, and the surrounding temperature, T2, are 390\u00b0F and 50\u00b0F,\nrespectively. The convective heat transfer coefficient, h, between the vessel wall and surrounding\nfluid is 4.0 Btu/h . ft . \u00b0F. Calculate the thermal resistance in \u00b0F . h/Btu.\n'''\n\n#Variable declaration:\nD = 1.0 #Diamete of vessel (ft)\nL = 1.5 #Length of vessel (ft)\nT1 = 390.0 #Surface temperature of vessel (\u00b0F)\nT2 = 50.0 #Surrounding temperature of vessel (\u00b0F)\nh = 4.0 #Convective heat transfer coefficient (Btu/h.ft.\u00b0F)\n\n#Calculation:\nA = pi*D*L+2*pi*(D/2)**2 #Total heat transfer area (ft^2)\nQ = h*A*(T1-T2) #Rate of heat transfer (Btu/h)\nR = 1/(h*A) #Thermal resistance (\u00b0F.h/Btu)\n\n#Result:\nprint \"The thermal resistance of vessel wal is :\",round(R,4),\" \u00b0F.h/Btu .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The thermal resistance of vessel wal is : 0.0398 \u00b0F.h/Btu .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.2, Page number: 135" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Referring to the previous example, convert the resistance to K/W and \u00b0C/W.\n'''\n\n#Variable declaration:\n#From example 9.1:\nR = 0.0398 #Theral resistance (\u00b0F.h/Btu)\nBtu = 3.412 #Btu/h in a watt\nC = 1.8 #Change in degree fahrenheit for a degree change in celsius\nK = 1 #Change in degree celsius for a unit change in Kelvin\n\n#Calculation:\nRc = R*Btu/C #Thermal resistance in degree cesius per watt (\u00b0C/W)\nRk = Rc/K #Thermal resistance in Kelvin per watt (K/W)\n\n#Result:\nprint \"The thermal resistance in \u00b0C/W is :\",round(Rc,3),\" \u00b0C/W.\"\nprint \"The thermal resistance in K/W is :\",round(Rk,3),\" K/W.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The thermal resistance in \u00b0C/W is : 0.075 \u00b0C/W.\nThe thermal resistance in K/W is : 0.075 K/W.\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.3, Page number: 136" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Hot gas at 530\u00b0F flows over a flat plate of dimensions 2 ft by 1.5 ft. The convection heat transfer\ncoefficient between the plate and the gas is 48 Btu/ft2 . h . \u00b0F. Determine the heat transfer rate in\nBtu/h and kW from the air to one side of the plate when the plate is maintained at 105\u00b0F.\n'''\n\n#Variable declaration:\nh = 48.0 #Convective heat transfer coefficient (Btu/h.ft.\u00b0F)\nA = 2*1.5 #Total heat transfer area (ft^2)\nTs = 530.0 #Surface temperature of plate (\u00b0F)\nTm = 105.0 #Maintained temperature of opposite side of plate (\u00b0F)\nkW = 3.4123*10**3 #Units kW in a Btu/h\n\n#Calculation:\nQ = h*A*(Ts-Tm) #Heat transfer rate in Btu/h (Btu/h)\nQ1 = Q/kW #Heat transfer rate in kW (kW)\n\n#Result:\nprint \"The heat transfer rate in Btu/h is :\",round(Q),\" Btu/h.\"\nprint \"The heat transfer rate in kW is :\",round(Q1,2),\" kW.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate in Btu/h is : 61200.0 Btu/h.\nThe heat transfer rate in kW is : 17.94 kW.\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.4, Page number: 136" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The glass window shown in Figure 9.3 of area 3.0 m2 has a temperature at the outer surface of\n10\u00b0C. The glass has conductivity of 1.4W/m . K. The convection coefficient (heat transfer\ncoefficient) of the air is 100W/m2 . K. The heat transfer is 3.0 kW. Calculate the bulk temperature\nof the fluid.\n'''\n\n#Variable declaration:\nTS = 10+273 #Outer surface temperature of wall (K)\nQ = 3000.0 #Heat transfer rate (W)\nh = 100.0 #Convection coefficient of air (W/m^2)\nA = 3.0 #Area of glass window (m^2)\n\n#Calculation:\nTM = TS-Q/(h*A) #Bulk temperature of fluid (K)\n\n#Result:\nprint \"The bulk temperature of fluid is :\",round(TM),\" K.\"\nprint \"The bulk temperature of fluid is :\",round(TM-273),\" \u00b0C.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The bulk temperature of fluid is : 273.0 K.\nThe bulk temperature of fluid is : 0.0 \u00b0C.\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.5, Page number: 137" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 9.1. If the convective heat transfer coefficient, h, between the\nvessel wall and the surrounding fluid is constant at 4.0 Btu/h . ft2 . \u00b0F and the plant operates\n24 h/day, 350 days/yr, calculate the steady-state energy loss in Btu/yr.\n'''\n\n#Variable declaration:\nh = 24.0 #Plant operating hour per day (h/day)\nd = 350.0 #Plant operating day per year (day/yr)\n\n#Calculation:\nN = h*d #Operating hours per year (h/yr)\n#From example 9.1:\nQ = 8545.0 #Rate of energy loss (Btu/h)\nQy = Q*N #Steady-state energy loss yearly (Btu/yr)\n\n#Result:\nprint \"The yearly steady-state energy loss is :\",round(Qy/10**7,2),\" x 10^7 Btu/yr.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The yearly steady-state energy loss is : 7.18 x 10^7 Btu/yr.\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.7, Page number: 148" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''For a flow of air over a horizontal flat plate, the local heat transfer coefficient, hx, is given by the\nequation\nhx = 25/x**0.4, W/m^2 . K\nwhere hx is the local heat transfer coefficient at a distance x from the leading edge of the plate and\nx is the distance in meters. The critical Reynolds number, Recr, which is the Reynolds number at\nwhich the flow is no longer laminar, is 500,000.\nConsider the flow of air at T1 = 21\u00b0C (cp = 1004.8 J/kg . K, v = 1.5 x 10**(-5) m^2/s,\nk = 0.025W/m . K, Pr = 0.7), at a velocity of 3 m/s, over a flat plate. The plate has a thermal conductivity k = 33W/m . K,\nsurface temperature, TS = 58\u00b0C, width, b = 1 m, and length,L = 1.2 m. Calculate\n1. the heat flux at 0.3 m from the leading edge of the plate\n2. the local heat transfer coefficient at the end of the plate\n3. the ratio h/hx at the end of the plate\n'''\nfrom sympy import symbols, integrate\n\n#Variable declaration:\nx = 0.3 #Length from the leading age of the plate (m)\nL = 1.2 #Length of plate (m)\nTS = 58.0 #Surface temperature of plate (\u00b0C)\nTa = 21.0 #Temperature of flowing air (\u00b0C)\n\n#Calculation:\nhx = 25/x**0.4 #Local heat transfer coefficient at 0.3m (W/m^2.K) (Part 1)\ny = symbols('y') #Length\nhy = 25/y**0.4 #hx at the end of the plate (W/m^2.K)\nh = integrate(hy, (y,0,L))/L #Average heat transfer coefficient (W/m^2.K)\nQ = hx*(TS-Ta) #Heat flux at 0.3m from leading edge of plate (W/m^2)\nhL = 25/L**0.4 #Local heat transfer coefficient at plate end (W/m^2.K) (Part 2) \nr = h/hL #Ratio h/hL at the end of the plate\n\n#Result:\nprint \"1. The heat flux at 0.3 m from the leading edge of the plate is :\",round(Q),\" W/m^2.\"\nprint \"2. The local heat transfer coefficient at the end of the plate is :\",round(hL,1),\" W/m^2.K.\"\nprint \"3. The ratio h/hL at the end of plate is :\",round(r,3),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The heat flux at 0.3 m from the leading edge of the plate is : 1497.0 W/m^2.\n2. The local heat transfer coefficient at the end of the plate is : 23.2 W/m^2.K.\n3. The ratio h/hL at the end of plate is : 1.667 .\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.8, Page number: 150" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example. Calculate the rate of heat transfer over the whole length of the\nplate.\n'''\n\n#Variable declaration:\n#From example 9.7:\nb = 1.0 #Width of plate (m)\nL = 1.2 #Length of plate (m)\nTS = 58.0 #Surface temperture of plate (\u00b0C)\nTa = 21.0 #Air flow temperature (\u00b0C)\nh = 38.7 #Average heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nA = b*L #Area for heat transfer for the entire plate (m^2)\nQ = h*A*(TS-Ta) #Rate of heat transfer over the whole length of the plate (W)\n\n#Result:\nprint \"The rate of heat transfer over the whole length of the plate is :\",round(Q,-1),\" W.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of heat transfer over the whole length of the plate is : 1720.0 W.\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.9, Page number: 150" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Air with a mass rate of 0.075 kg/s flows through a tube of diameter D = 0.225 m. The air enters\nat 100\u00b0C and, after a distance of L = 5 m, cools to 70\u00b0C. Determine the heat transfer coefficient\nof the air. The properties of air at 85\u00b0C are approximately, cp = 1010 J/kg . K, k = 0.030\nW/m . K, mu = 20\u00b0 x 10**-7 N . s/m2, and Pr = 0.71.\n'''\nfrom math import pi\n#Variable declaration:\nm = 0.075 #Mass rate of air flow (kg/s)\nD = 0.225 #Diameter of tube (m)\nmu = 208*10**-7 #Dynamic viscosity of fluid (N)\nPr = 0.71 #Prandtl number\nk = 0.030 #Thermal conductivity of air (W/m.K)\n\n#Calculation:\nRe = 4*m/(pi*D*mu) #Reynolds number\n#From equation 9.26:\nNu = 0.023*(Re**0.8)*(Pr**0.3) #Nusselt number\nh = (k/D)*Nu #Heat transfer coefficient of air (W/m^2.K)\n\n#Result:\nprint \"The Heat transfer coefficient of air is :\",round(h,2),\" W/m^2.K.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Heat transfer coefficient of air is : 7.76 W/m^2.K.\n" + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.10, Page number: 150" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the average film heat transfer coefficient (Btu/h . ft^2 . \u00b0F) on the water side of a single\npass steam condenser. The tubes are 0.902 inch inside diameter, and the cooling water enters\nat 60\u00b0F and leaves at 70\u00b0F. Employ the Dittus\u2013Boelter equation and assume the average\nwater velocity is 7 ft/s. Pertinent physical properties of water at an average temperature of\n65\u00b0F are:\np = 62.3 lb/ft^3\nmu = 2.51 lb/ft . h\nCp = 1.0 Btu/lb . \u00b0F\nk = 0.340 Btu/h . ft . \u00b0F\n'''\n\n#Variable declaration:\nD = 0.902/12.0 #Inside diameter of tube (ft)\nT_in = 60.0 #Temperature water entering the tube (\u00b0F)\nT_out = 70.0 #Temperature water leaving the tube (\u00b0F)\nV = 7.0 #Average wave velocity water (ft/s)\np = 62.3 #Density of water (lb/ft^3)\nmu = 2.51/3600.0 #Dynamic viscosity of water (lb/ft.s)\nCp = 1.0 #Viscosity of centipoise (Btu/lb.\u00b0F)\nk = 0.34 #Thermal conductivity of water (Btu/h.ft.\u00b0F)\n\n#Calculation:\nRe = D*V*p/mu #Reynolds Number\nPr = Cp*mu/k*3600 #Prandtl number\n#From equation 9.26:\nNu = 0.023*(Re**0.8)*(Pr**0.4) #Nusselt number\nh = (k/D)*Nu #Average film heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The required average film heat transfer coefficient is :\",round(h),\" Btu/h.ft^2.\u00b0F.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required average film heat transfer coefficient is : 1265.0 Btu/h.ft^2.\u00b0F.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.11, Page number: 151" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Air at 1 atm and 300\u00b0C is cooled as it flows at a velocity 5.0 m/s through a tube with a diameter\nof 2.54 cm. Calculate the heat transfer coefficient if a constant heat flux condition is maintained\nat the wall and the wall temperature is 20\u00b0C above the temperature along the entire length of\nthe tube.\n'''\n\n#Variable declaration:\nP = 1.0132 * 10**5 #Air pressure (Pa)\nT = 300.0+273.0 #Air temperature (K)\nV = 5.0 #Air flow velocity (m/s)\nD = 2.54/100.0 #Diameter of tube (m)\nR = 287.0 #Gas constant (m^2/s^2.K)\n#From Appendix:\nPr = 0.713 #Prandtl number of nitrogen\nmu = 1.784*10**(-5) #Dynamic viscosity of nitrogen (kg/m.s)\nk = 0.0262 #Thermal conductivity of nitrogen (W/m.K)\nCp = 1.041 #Heat capacity of nitrogen (kJ/kg.K)\n\n#Calculation:\np = P/(R*T) #Density of air\nRe = D*V*p/mu #Reynolds number\n#From table 9.5:\nNu = 0.023*(Re**0.8)*(Pr**0.3) #Nusselt number\nh = (k/D)*Nu #Heat transfer coefficient (W/m^2.K)\n\n#Result:\nprint \"The required Heat transfer coefficient is :\",round(h,2),\" W/m^2.K.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required Heat transfer coefficient is : 17.57 W/m^2.K.\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.12, Page number: 152" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Water flows with an average velocity of 0.355 m/s through a long copper tube (inside\ndiameter = 2.2 cm) in a heat exchanger. The water is heated by steam condensing at 150\u00b0C\non the outside of the tube. Water enters at 15\u00b0C and leaves at 60\u00b0C. Determine the heat transfer\ncoefficient, h, for the water. (Adapted from Griskey.^(5))\n'''\n\n#Variable declaration:\nT1 = 15.0 #Water entering temperature (\u00b0C)\nT2 = 60.0 #Water leaving temperature (\u00b0C)\nD = 0.022 #Inside diameter of tube (m)\nV = 0.355 #Average water flow velocity (m/s)\nTC = 150.0 #Outside wall temperature (\u00b0C)\n#From Appendix:\np = 993.0 #Density of water (kg/m^3)\nmu = 0.000683 #Dynamic viscosity of water (kg/m.s)\nCp = 4.17*10**3 #Heat capacity of water (J/kg.K)\nk = 0.63 #Thermal conductivity of water (W/m.K)\n\n#Calculation:\nTav1 = (T1+T2)/2.0 #Average bulk temperature of water (\u00b0C)\nRe = D*V*p/mu #Reynolds number\nPr = Cp*mu/k #Prandtl number\nTav2 = (Tav1+TC)/2.0 #Fluid's average wall temperature (\u00b0C)\n#From Appendix:\nmu_w = 0.000306 #Dynamic viscosity of fluid at wall (kg/m.s)\n#From Table 9.5:\nh = (k/D)*0.027*Re**0.8*Pr**0.33*(mu/mu_w)**0.14 #Heat transfer coefficient for water (W/m^2.K)\n\n#Result:\nprint \"The heat transfer coefficient for water is :\",round(h,1),\" W/m^2.K.\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer coefficient for water is : 2497.3 W/m^2.K.\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.13, Page number: 153" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 9.7. Calculate the average Biot number.\n'''\n\n#Variable declaration:\n#From example 9.7:\nh = 38.7 #Average heat transfer coefficient (W/m^2.K)\nL = 1.2 #Length of plate (m)\nk = 0.025 #Thermal conductivity of air (W/m)\n\n#Calculation:\nBi = h*L/k #Average Biot number\n\n#Result:\nprint \"The average Biot number is :\",round(Bi),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The average Biot number is : 1858.0 .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 9.14, Page number: 154" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The surface temperature Ts of a circular conducting rod is maintained at 250\u00b0C by the passage of\nan electric current. The rod diameter is 10 mm, the length is 2.5 m, the thermal conductivity is 60\nW/m . K, the density is 7850 kg/m3, and the heat capacity is 434 J/kg . K. The rod is in a fluid\nat temperature Tf = 25\u00b0C, and the convection heat transfer coefficient is 140 W/m2 . K. The\nthermal conductivity of the fluid is 0.6W/m . K.\n1. What is the thermal diffusivity of the bare rod?\n2. What is the Nusselt number of the fluid in contact with the bare rod?\n3. What is the Biot number of the bare rod?\n4. Calculate the heat transferred from the rod to the fluid.\n'''\nfrom math import pi\n#Variable declaration:\nk = 60.0 #Thermal conductivity of rod (W/m.K)\np = 7850.0 #Density of rod (kg/m^3)\nCp = 434.0 #Heat capacity of rod (J/kg.K)\nh = 140.0 #Convection heat transfer coefficient (W/m^2.K)\nD = 0.01 #Diameter of rod (m)\nkf = 0.6 #Thermal conductivity of fluid (W/m.K)\nL = 2.5 #Length of rod (m)\nTs = 250.0 #Surface temperature of rod (\u00b0C)\nTf = 25.0 #Fluid temperature (\u00b0C)\n\n#Calculation:\n#Case 1:\na = k/(p*Cp) #Thermal diffusivity of bare rod (m^2/s)\n#Case 2:\nNu = h*D/kf #Nusselt number\n#Case 3:\nBi = h*D/k #Biot number of bare rod\n#Case 4:\nQ = h*(pi*D*L)*(Ts-Tf) #Heat transferred from rod to fluid (W)\n\n#Result:\nprint \"1. The thermal diffusivity of the bare rod is :\",round(a/10**-5,2),\" x 10^-5 m^2/s.\"\nprint \"2. The nusselt number is :\",round(Nu,2),\" .\"\nprint \"3. The Biot number is :\",round(Bi,4),\" .\"\nprint \"4. The heat transferred from the rod to the fluid is :\",round(Q),\" W.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The thermal diffusivity of the bare rod is : 1.76 x 10^-5 m^2/s.\n2. The nusselt number is : 2.33 .\n3. The Biot number is : 0.0233 .\n4. The heat transferred from the rod to the fluid is : 2474.0 W.\n" + } + ], + "prompt_number": 14 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_10.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_10.ipynb new file mode 100644 index 00000000..76eae3ba --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_10.ipynb @@ -0,0 +1,251 @@ +{ + "metadata": { + "name": "Chapter 10" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 10: Free Convection" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.1, Page number: 163" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The Grashof and Reynolds numbers for a system involved in a heat transfer process are approximately\n100 and 50, respectively. Can free convection effects be neglected.\n'''\n\n#Variable declaration:\nGr = 100.0 #Grashof number\nRe = 50.0 #Reynolds number\n\n#Calculation:\nLT = Gr/Re**2 #Measure of influence of convection effect\n\n#Result:\nif (LT<1.0):\n print \"The free convection effects can be neglected.\"\nelif (LT>1.0):\n print \"The free convection effects can not be neglected.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The free convection effects can be neglected.\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.2, Page number: 166" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The heat flux rate incident on a vertical flat plate at 110\u00b0C is 800W/m2. The plate is 2 m wide\nand 3.5 m high and is well insulated on the back side. The ambient air temperature is 30\u00b0C. All\nthe incident radiation (800W/m2) on the plate is absorbed and dissipated by free convection to\nthe ambient air at 30\u00b0C. Determine the Grashof and Rayleigh numbers.\n'''\n\n#Variable declaration:\nTs = 110.0+273.0 #Surface temperature of plate (K)\nToo = 30.0+273.0 #Ambient air temperature (K)\nL = 3.5 #Height of plate (m)\ng = 9.807 #Gravitational acceleration (m^2/s)\n\n#Calculation:\nTf = (Ts+Too)/2 #Film temperature (K)\nDT = Ts - Too #Temperature difference between surface and air (K)\n#From appendix:\nv = 2.0*10**-5 #Kinematic viscosity for air (m^2/s)\nk = 0.029 #Thermal conductivity for air (W/m.K)\nPr = 0.7 #Prandtl number\nB = 1.0/Tf #Coefficient of expansion (K^-1)\nGr = g*B*DT*L**3/v**2 #Grashof number\nRa = Gr*Pr #Rayleigh number\n\n#Result:\nprint \"The Grashof number is :\",round(Gr/10**11,2),\" x 10^11 .\"\nprint \"The Rayleigh number is :\",round(Ra/10**11,2),\" x 10^11 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Grashof number is : 2.45 x 10^11 .\nThe Rayleigh number is : 1.72 x 10^11 .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.3, Page number: 166" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 10.2. Determine the type of natural convection (flow regime).\n'''\n\n#Variable declaration:\n#From example 10.2:\nRa = 1.71*10**11 #Rayleigh number\n\n#Result:\nif (Ra>10**9):\n print \"The convection flow category is turbulent.\"\nelif(Ra<10**9):\n print \"The convection flow category is laminar.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The convection flow category is turbulent.\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.4, Page number: 167" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 10.2. Determine the average heat transfer coefficient.\n'''\n\n#Variable declaration:\n#From Table 10.1:\nc = 0.1 #Constant c\nm = 1.0/3.0 #Constant for turbulent free conection\n#From example 10.2:\nRa = 1.71*10**11 #Rayleigh number\nk = 0.029 #Thermal conductivity (W/m.K)\nL = 3.5 #Thickness of plate (m)\n\n#Calculation:\nNu = c*Ra**m #Average Nusselt number\nh = Nu*k/L #Average heat transfer coefficient (W/m^2.K)\n\n#Result:\nprint \"The average heat transfer coefficient is :\",round(h,1),\" W/m^2.K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The average heat transfer coefficient is : 4.6 W/m^2.K .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.6, Page number: 167" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the air heat transfer film coefficient for a horizontal 6 inch diameter pipe whose surface\ntemperature is 200\u00b0F in a room containing air at 70\u00b0F.\n'''\n\nfrom math import log,e\n\n#Variable declaration:\nTs = 200.0+460.0 #Surface temperature of pipe (\u00b0R)\nToo = 70.0+460.0 #Air temperature (\u00b0R)\nD = 0.5 #Diameter of pipe (ft)\nR = 0.73 #Universal gas constant (ft^3.atm.R^\u22121.lb.mol^\u22121)\nP = 1.0 #Atmospheric pressure (Pa)\nMW = 29.0 #Molecular weight of fluid (mol)\n#From Appendix:\nmu = 1.28*10**-5 #Absolute viscosity (lb/ft.s)\nk = 0.016/3600.0 #Thermal conductivity (Btu/s.ft.\u00b0F)\ng = 32.174 #Gravitational acceleration (ft/s^2)\n\n#Calculation:\nTav = (Ts+Too)/2 #Average temperature (\u00b0R)\nv = R*Tav/P #kinematic viscosity (ft^3/lbmol)\np = MW/v #Air density (lb/ft^3)\nB = 1.0/Tav #Coefficient of expansion (\u00b0R^-1)\nDT = Ts-Too #Temperature difference (\u00b0R)\nGr = D**3*p**2*g*B*DT/mu**2 #Grashof number\n#From equation 10.5:\nCp = 0.25 #Air heat capacity (Btu/lb.\u00b0F)\nPr = Cp*mu/k #Prandtl number\nGrPr = 10**8.24 #Rayleigh number\n#From Holman^(3):\nNu = 10**(1.5) #Nusselt number\nh = Nu*(k/D)*3600.0 #Air heat transfer film coefficient (Btu/h.ft.\u00b0F)\n\n#Result:\nprint \"The required air heat transfer film coefficient is :\",round(h,2),\" Btu/h.ft.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required air heat transfer film coefficient is : 1.01 Btu/h.ft.\u00b0F .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.7, Page number: 168" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the free convection heat transfer coefficient for a plate 6 ft high and 8 ft wide at 120\u00b0F\nthat is exposed to nitrogen at 60\u00b0F.\n'''\n\n#Variable declaration:\nTs = 120.0+460 #Surface temperature of plate (\u00b0R)\nToo = 60.0+460 #Ambient temperature of nitrogen (\u00b0R)\nL = 6 #Height of plate (ft)\n#From Appendix:\np = 0.0713 #Air density (lb/ft^3)\nk = 0.01514 #Thermal conductivity (Btu/h.ft.\u00b0F)\nv = 16.82*10**-5 #Kinematic viscosity (ft^2/s)\nPr = 0.713 #Prandtl number\ng = 32.2 #Gravitational acceleration (ft/s^2)\n\n#Calculation:\nTf = (Ts+Too)/2 #Mean film temperature (\u00b0R)\nB = 1.0/Tf #Coefficient of expansion (\u00b0R^-1)\nGr = g*B*(Ts-Too)*L**3/v**2 #Grashof number\nRa = Gr*Pr #Rayleigh number\n#From equation 10.13(Table 10.2) and costants from Table 10.1:\nh = 0.10*(k/L)*Ra**(1.0/3.0) #Free convection heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The free convection heat transfer coefficient is :\",round(h,3),\" Btu/h.ft^2.\u00b0F .\"\nprint \"There is a calculation mistake in the book for calculating 'Gr', so, value of 'h' alters from that given.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The free convection heat transfer coefficient is : 0.675 Btu/h.ft^2.\u00b0F .\nThere is a calculation mistake in the book for calculating 'Gr', so, value of 'h' alters from that given.\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.8, Page number: 169" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the heat loss in the previous example.\n'''\n\n#Variable declaration:\n#From example:\nh = 0.675 #Free convection heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nA = 6.0*8.0 #Area of plate (ft^2)\nTs = 120.0 #Surface temperature of plate (\u00b0F)\nToo = 60.0 #Ambient temperature of nitrogen (\u00b0F)\n\n#Calculation:\nQ = h*A*(Ts-Too) #Heat loss (Btu/h)\n\n#Result:\nprint \"The heat loss is :\",round(Q,-1),\" Btu/h .\"\nprint \" The 'h' obtained in the previous example differs, therefore, 'Q' obtained here also fiffers from that given in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat loss is : 1940.0 Btu/h .\n The 'h' obtained in the previous example differs, therefore, 'Q' obtained here also fiffers from that given in book.\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.9, Page number: 169" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the heat transfer from a 100-W light bulb at 113\u00b0C to 31\u00b0C ambient air. Approximate\nthe bulb as a 120-mm-diameter sphere.\n'''\n\n#Variable declaration:\nTs = 113.0+273.0 #Surface temperature of bulb (K)\nToo = 31.0+273.0 #Ambient air temperature (K)\nD = 0.06 #Diameter of sphere (m)\ng = 9.8 #Gravitational acceleration (m/s^2)\n\n#Calculation:\nTf = (Ts+Too)/2 #Mean temperature (K)\n#From Appendix:\nv = (22.38*10**-5)*0.0929 #Kinematic viscosity (m^2/s)\nPr = 0.70 #Prandtl number\nk = 0.01735*1.729 #Thermal conductivity (W/m.K)\nB = 1.0/(Tf) #Coefficient of expansion (K^-1)\nGr = g*B*(Ts-Too)*D**3/v**2 #Grashof number\nRa = Gr*Pr #Rayleigh number\n\n#From equation 10.13:\nh = (k/D)*0.6*Ra**(1.0/4.0) #Heat transferred from bulb (W/m^2.K)\n\n#Result:\nprint \"The heat transferred from bulb to air is :\",round(h,2),\" W/m^2.K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transferred from bulb to air is : 9.01 W/m^2.K .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.10, Page number: 170" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 10.9. Calculate the heat transfer lost by free convection from the\nlight bulb.\n'''\n\nfrom math import pi\n\n#Variable declaration:\n#From example 10.9:\nh = 9.01 #Heat transferred from bulb (W/m^2.K)\nD = 0.06 #Diameter of sphere (m)\nTs = 113.0+273.0 #Surface temperature of bulb (K)\nToo = 31.0+273.0 #Ambient air temperature (K)\n\n#Calculation:\nA = pi*D**2 #Surface area of bulb (m^2)\nQ = h*A*(Ts-Too) #Heat transfer lost by free convection from light bulb (W)\n\n#Result:\nprint \"The heat transfer lost by free convection from light bulb is :\",round(Q,2),\" W .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer lost by free convection from light bulb is : 8.36 W .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.11, Page number: 170" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''With reference to Illustrative Examples 10.9\u009610.10, what percentage of the energy is lost by free\nconvection.\n'''\n\n#Variable declaration:\n#From example 10.9-10.10:\nQ = 8.36 #Heat transfer lost by free convection from light bulb (W)\n\n#Calculation:\nE = Q/100.0*(100.0) #Percent energy lost by free convection (%)\n\n#Result:\nprint \"The percentage of the energy lost by free convection is :\",round(E,2),\" % .\"\nprint \"The energy lost fraction is :\",round(E/100.0,4),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The percentage of the energy lost by free convection is : 8.36 % .\nThe energy lost fraction is : 0.0836 .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 10.13, Page number: 175" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If a waste source emits a gas with a buoyancy flux of 50 m^4/s^3, and the wind speed averages 4\nm/s, find the plume rise at a distance of 750 m downward from a stack that is 50 m high under\nunstable atmospheric conditions. Several plume rise equations are available. Use the equation\nproposed by Briggs.\nBriggs(5) used the following equations to calculate the plume rise.\nDh = 1.6F^1/3u^-1x^2/3 ; x < xf (10.23)\n= 1.6F^1/3u^-1xf^2/3 ; if x < xf\nxc = 14F^5/8 ; when F < 55m^4/s^3\n= 34F^2/5 ; when F < 55m^4/s^3\nxf = 3.5xc\nwhere Dh = plume rise, m\nF = buoyancy flux, m^4/s^3 =3 .7 x 10^-5 Q\u02d9H\nu = wind speed, m/s\nx* = downward distance, m\nxf = distance of transition from first stage of rise to the second stage of rise, m\nQ\u02d9H = heat emission rate, kcal/s\nIf the term Q\u02d9H is not available, the term F may be estimated by\nF = (g/pi)q(TS - T)/TS (10.24)\nwhere g = gravity term 9.8 m/s^2\nq = stack gas volumetric flowrate, m^3/s (actual conditions)\nTS, T = stack gas and ambient air temperature, K, respectively\n'''\n\n#Variable declaration:\nF = 50.0 #Buoyancy flux of gas (m^4/s^3)\nu = 4.0 #wind speed (m/s)\n\n#Calculation:\nxc = 14*F**(5.0/8.0) #Downward distance (m)\nxf = 3.5*xc #distance of transition from first stage of rise to the second stage of rise (m)\nDh = 1.6*F**(1.0/3.0)*u**-1*xf**(2.0/3.0) #Plume rise (m)\n\n#Result:\nprint \"The plume rise is :\",round(Dh),\" m .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The plume rise is : 101.0 m .\n" + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_11.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_11.ipynb new file mode 100644 index 00000000..485669cf --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_11.ipynb @@ -0,0 +1,335 @@ +{ + "metadata": { + "name": "Chapter 11" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 11: Radiation" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.3, Page number: 181" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The intensity of radiation as a function of wavelength (l = mu*m) is specified as\nI = 40e^-l^2 ; Btu/h.ft^2.mu.m\nCalculate the total emissive power.\n'''\n\nfrom sympy import symbols, integrate,oo,exp,pi\n\n#Variable declaration:\nl = symbols('l') #Wavelength (mu.m)\nI = 40*exp(-l**2) #Intensity of radiation (Btu/h.ft^2.mu.m)\n\n#Calculation:\nE = integrate(I, (l,0,oo)).evalf() #Total emissive power (Btu/h.ft^2)\n\n#Result:\nprint \"The total emissive power is :\",round(E,1),\" Btu/h.ft^2 .\" ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total emissive power is : 35.4 Btu/h.ft^2 .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.4, Page number: 182" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Estimate the Sun\u2019s temperature. Employ equation (11.4).\n'''\n\n#Variable declaration:\nl = 0.25 #Wavelength (mu.m)\n#From equation 11.4:\nlT = 2884 #Product of wavelength and absolute temperature (mu.m.\u00b0R)\n\n#Calculation:\nT = lT/l #Sun's temperature (\u00b0R)\n\n#Result:\nprint \"The Sun's temperature is :\",round(T,-2),\" \u00b0R .\"\nprint \"The Sun's temperature in fahrenheit scale is :\",round(T-460,-3),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Sun's temperature is : 11500.0 \u00b0R .\nThe Sun's temperature in fahrenheit scale is : 11000.0 \u00b0F .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.5, Page number: 188" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Estimate the increase in heat transferred by radiation of a black body at 1500\u00b0F relative to one at\n1000\u00b0F.\n'''\n\n#Variable declaration:\nT1 = 1500.0+460.0 #Absolute temperature 1 (\u00b0R)\nT2 = 1000.0+460.0 #Absolute temperature 2 (\u00b0R)\n\n#Calculation:\nX = T1**4/T2**4 #Ratio of quantity of heat transferred\nx = 100*(T1**4-T2**4)/T2**4 #Percentage increase in heat transfer (%)\n\n#Result:\nprint \"The ratio of the quantity/rate of heat transferred is :\",round(X,2),\" .\"\nprint \"The percentage increase in heat transfer is :\",round(x),\"%\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The ratio of the quantity/rate of heat transferred is : 3.25 .\nThe percentage increase in heat transfer is : 225.0 %\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.6, Page number: 189" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Two large walls are required to be maintained at constant temperatures of 800\u00b0F and 1200\u00b0F.\nAssuming the walls are black bodies, how much heat must be removed from the colder wall\nto maintain a steady-state, constant temperature?\n'''\n\n#Variable declaration:\nT1 = 1200.0+460.0 #Absolute temperature of wall 1 (\u00b0R)\nT2 = 800.0+460.0 #Absolute temperature of wall 2 (\u00b0R)\n\n#Calculation:\n#From equation 11.23:\nX = 0.173*((T1/100.0)**4-(T2/100.0)**4) #Heat removed from colder wall (Btu/h.ft^2)\n\n#Result:\nprint \"The heat removed from the colder wall to maintain a steady-state is :\",round(X),\" Btu/h.ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat removed from the colder wall to maintain a steady-state is : 8776.0 Btu/h.ft^2 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.7, Page number: 190" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If the two bodies from Illustrative Example 11.6 have emissivities of 0.5 and 0.75, respectively,\nwhat is the net energy exchange (per unit area)? Assume that the temperatures remain constant at\n1660\u00b0R and 1260\u00b0R, and the two bodies are of infinite size.\n'''\n\n#Variable declaration:\ns = 0.173 #Stefan-Boltzmann constant (Btu/h.ft^2.\u00b0R)\nEH = 0.5 #Energy transferred from hotter body (Btu/h.ft^2)\nEC = 0.75 #Energy transferred to colder body (Btu/h.ft^2)\nTH = 1660.0 #Absolute temperature of hotter body (\u00b0R)\nTC = 1260.0 #Absolute temperature of colder body (\u00b0R)\n\n#Calculation:\nE = s*((TH/100.0)**4-(TC/100.0)**4)/((1.0/EH)+(1.0/EC)-1.0) #Net energy exchange per unit area (Btu/h.ft^2)\n\n#Result:\nprint \"The net energy exchange per unit area is :\",round(E,-1),\" Btu/h.ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The net energy exchange per unit area is : 3760.0 Btu/h.ft^2 .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.8, Page number: 191" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Compare and discuss the results of the last two illustrative examples.\n'''\n\n#Variable declaration:\n#From example 11.6-11.7:\nE1 = 8776.0 #Energy exchange between black bodies (Btu/h.ft^2)\nE2 = 3760.0 #Energy exchange between non-black bodies (Btu/h.ft^2)\n\n#Calculation:\nD = (E1-E2)/E1*100 #Percent difference in energy (%)\n\n#Result:\nprint \"The percent difference relative to the black body is:\",round(D,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The percent difference relative to the black body is: 57.2 % .\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.9, Page number: 192" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the radiation from a 2-inch IPS cast iron pipe (assume polished) carrying steam at\n300\u00b0F and passing through the center of a 1 ft x 1 ft galvanized zinc duct at 75\u00b0F and whose\noutside is insulated.\n'''\n\n#Variable declaration:\ns = 0.173*10**-8 #Stefan-Boltzmann constant (Btu/h.ft^2.\u00b0R)\nTH = 300.0+460.0 #Absolute temperature of external surface (\u00b0R)\nTC = 75.0+460.0 #Absolute temperature of duct (\u00b0R)\n#From Table 6.2:\nAH = 0.622 #External surface area of pipe (ft^2)\n#From Table 11.2:\nEH = 0.44 #Emissivity of oxidized steel\nAC = 4.0*1.0*1.0 #External surface area of duct (ft^2)\nEC = 0.23 #Emissivity of galvanized zinc\n\n#Calculation:\nFE = 1.0/(1.0/EH+((AH/AC)*(1.0/EC-1.0))) #Emissivity correction factor\nQ = FE*AH*s*(TH**4-TC**4) #Net radiation heat transfer (Btu/h.ft)\n\n#Result:\nprint \"The net radiation heat transfer is :\",round(Q,2),\" Btu/h.ft^2 .\"\nprint \"There is a calculation error in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The net radiation heat transfer is : 96.96 Btu/h.ft^2 .\nThere is a calculation error in book.\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.10, Page number: 193" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The outside temperature of a 10 ft2 hot insulated pipe is 140\u00b0F and the surrounding atmosphere\nis 60\u00b0F. The heat loss by free convection and radiation is 13,020 Btu/h, and the combined coefficient\nof heat transfer is estimated to be 2.10 Btu/h.ft^2.\u00b0F . How much of the heat loss is due to\nradiation? Assume the pipe emissivity is approximately 0.9.\n'''\n\n#Variable declaration:\nTH = 140.0+460.0 #Absolute outside temperature of pipe (ft^2)\nTC = 60.0+460.0 #Absolute temperature of surrounding atmosphere (ft^2)\nA = 10.0 #Area of pipe (ft^2)\nE = 0.9 #Emissivity of pipe\n\n#Calculation:\nQ = E*A*0.173*((TH/100.0)**4-(TC/100.0)**4) #Heat loss due to radiation (Btu/h)\n\n#Result:\nprint \"The heat loss due to radiation is :\",round(Q,-1),\" Btu/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat loss due to radiation is : 880.0 Btu/h .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.11, Page number: 193" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''With reference to Illustrative Example 11.10, calculate the radiation heat transfer coefficient, hr.\n'''\n\n#Variable declaration:\n#Froma example 11.10:\nQ = 880.0 #Heat loss due to radiation (Btu/h)\nA = 10.0 #Area of pipe (ft^2)\nTH = 140.0 #Absolute outside temperature of pipe (\u00b0F)\nTC = 60.0 #Absolute temperature of surrounding atmosphere (\u00b0F)\n\n#Calculation:\nhr = Q/(A*(TH-TC)) #Radiation heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The radiation heat transfer coefficient is :\",round(hr,1),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The radiation heat transfer coefficient is : 1.1 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.12, Page number: 194" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A small oxidized horizontal metal tube is placed in a very large furnace enclosure with firebrick\nwalls. The metal tube has an outside diameter of 1 inch, a length of 2 ft, a surface emissivity of\n0.6, and its surface is maintained at 600\u00b0F. The hot air in the furnace is at 1500\u00b0F and the furnace\nbrick walls are at 1350\u00b0F. The convection heat transfer coefficient for the horizontal tube is 2.8\nBtu/h.ft2.\u00b0F . Calculate\n1. The convective, radiative, and total heat transferred to the metal tube;\n2. The percent of total heat transferred by radiation;\n3. The radiation heat transfer coefficient;\n4. Is it appropriate to use the approximate equation presented in Equation (11.30) for (3)?\n'''\n\nfrom math import pi\n\n#Variable declaration:\nD = 0.0833 #Diameter of tube (ft)\nL = 2.0 #Length of tube (ft)\nh = 2.8 #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nTa1 = 1500.0+460.0 #Temperature of hot air in furnace (\u00b0R)\nTa2 = 1350.0+460.0 #Temperature of hot air in the furnace brick walls (\u00b0R)\nTt = 600.0+460.0 #Surface temperature of tube (\u00b0R)\nE = 0.6 #Surface emissivity of tube\ns = 0.1713*10**-8 #Stefan-Boltzmann constant\n\n\n#Calculation:\n#Case 1:\nA = pi*D*L #Area of tube (ft^2)\nQc = round(h*A*(Ta1-Tt),-1) #Convection heat transfer from air to tube (Btu/h)\nQr = round(E*s*A*(Ta2**4-Tt**4),-2) #Radiation feat transfer from wall to tube (Btu/h)\nQ = Qr+Qc #Total heat transfer (Btu/h)\n#Case 2:\nQp = Qr/Q*100 #Radiation percent \n#Case 3:\nhr = Qr/(A*(Ta2-Tt)) #Radiation heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n#Case 4:\nT = Ta2-Tt #Temperature difference (\u00b0F)\n\n#Result:\nprint \"1. The convective heat transferred to the metal tube is :\",Qc,\" Btu/h .\"\nprint \" The radiative heat transferred to the metal tube is :\",Qr,\" Btu/h .\"\nprint \" The total heat transferred to the metal tube is :\",Q,\" Btu/h .\"\nprint \"2. The percent of total heat transferred by radiation is :\",round(Qp,1),\" % .\"\nprint \"3. The radiation heat transfer coefficient is :\",round(hr,1),\" Btu/h.ft^2.\u00b0F .\"\nif (T > 200):\n print \"4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is not appropriate.\"\nelif (T < 200):\n print \"4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is appropriate.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The convective heat transferred to the metal tube is : 1320.0 Btu/h .\n The radiative heat transferred to the metal tube is : 5100.0 Btu/h .\n The total heat transferred to the metal tube is : 6420.0 Btu/h .\n2. The percent of total heat transferred by radiation is : 79.4 % .\n3. The radiation heat transfer coefficient is : 13.0 Btu/h.ft^2.\u00b0F .\n4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is not appropriate.\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.13, Page number: 194" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The filament of a light bulb is at a temperature of 900\u00b0C and emits 5W of heat toward the glass\nbulb. The interior of a light bulb can be considered a vacuum and the temperature of the glass\nbulb is 150\u00b0C. Ignoring heat transfer to the room and assuming the emissivity of the filament is\n1.0, calculate the surface area of the filament in cm^2.\n'''\n\n#Variable declaration:\nQ = 5.0 #Radiation heat transfer (W)\nE = 1.0 #Emissivity of filament\ns = 5.669*10**-8 #Stefan-Boltzmann constant\nT1 = 900.0+273.0 #Light bulb temperature (K)\nT2 = 150.0+273.0 #Glass bulb temperature (K)\n\n#Calculation:\nA = Q/(E*s*(T1**4-T2**4)) #Surface area of the filament (m^2)\n\n#Result:\nprint \"The surface area of the filament is :\",round(A*10**4,2), \"cm^2\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The surface area of the filament is : 0.47 cm^2\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.14, Page number: 195" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A system consists of an uninsulated steam pipe made of anodized aluminum with a diameter\nD = 0.06 m and a length L = 100 m. The surface temperature is T1 = 127\u00b0C and the surface\nemissivity of anodized aluminum is 1 = 0.76. The pipe is in a large room with a wall temperature\nT2 = 20\u00b0C. The air in the room is at a temperature T3 = 22\u00b0C. The pipe convective heat\ntransfer coefficient is h = 15 W/m^2 . K.\nEstimate the emissive power, the total heat transfer by convection and radiation, and the radiation\nheat transfer coefficient. Assume steady-state operation, constant properties, and a room\nsurface area much larger than the pipe surface area.\n'''\n\nfrom math import pi\n\n#Variable declaration:\nT1 = 127.0+273.0 #Surface temperature (K)\nT2 = 20.0+273.0 #Wall temperature (K)\nT3 = 22.0+273.0 #Air temperature (K)\ns = 5.669*10**-8 #Stefan-Boltzmann constant\ne = 0.76 #Surface emissivity of anodized aluminium\nD = 0.06 #Diameter of pipe (m)\nL = 100.0 #Length of pipe (m)\nh = 15.0 #Pipe convective heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nEb = s*T1**4 #Emissive energy of pipe (W/m^2)\nE = e*Eb #Emissive power from surface of pipe (W/m^2)\nA = pi*D*L #Surface area of pipe (m^2)\nQc = h*A*(T1-T3) #Convection heat transfer to air (W)\nQr = e*s*A*(T1**4-T2**4) #Radiation heat transfer rate (W)\nQ = Qc+Qr #Total heat transfer rate (Btu/h)\nTav = (T1+T2)/2.0 #Average temperature (K)\nhr = 4*e*s*Tav**3 #Radiation heat transfer coefficient (W/m^2.K)\n\n#Result:\nprint \"The emissive power from surface of pipe is :\",round(E),\" W/m^2 .\"\nprint \"The convection heat transfer to air is :\",round(Qc/10**3,1),\" kW .\"\nprint \"The radiation heat transfer rate is :\",round(Qr/10**3,1),\" kW \"\nprint \"The radiation heat transfer coefficient is :\",round(hr,1),\" W/m^2.K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The emissive power from surface of pipe is : 1103.0 W/m^2 .\nThe convection heat transfer to air is : 29.7 kW .\nThe radiation heat transfer rate is : 14.8 kW \nThe radiation heat transfer coefficient is : 7.2 W/m^2.K .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.15, Page number: 196" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 11.14. How much does radiation contribute to the total heat\ntransfer?\n'''\n\n#Variable declaration:\n#From example 11.14:\nQc = 15.0 #Convection heat transfer coefficient (W/m^2.K)\nhr = 7.2 #Radiation heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nX = hr/(Qc+hr)*100.0 #Percent heat transfer by radiation (%)\n\n#Result:\nprint \"The percent heat transfer by radiation is :\",round(X,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The percent heat transfer by radiation is : 32.4 % .\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.16, Page number: 200" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 11.9. Calculate the heat transfer rate if Fv = 1.0.\n'''\n\n#Variable declaration:\nFV = 1.0 #Correction factor\n#From example 11.9:\nFE = 0.358 #Emissivity correction factor\nTH = 300.0+460.0 #Absolute temperature of external surface (\u00b0R)\nTC = 75.0+460.0 #Absolute temperature of duct (\u00b0R)\nAH = 0.622 #Area of pipe (ft^2)\ns = 0.173*10**-8 #Stefan-Boltzmann constant\n\n#Calculation:\nQ = FV*FE*AH*s*(TH**4-TC**4) #Heat transfer rate (Btu/h.ft)\n\n#Result:\nprint \"The heat transfer rate is :\",round(Q,2),\" Btu/h.ft\"\nprint \"Since, 'Q' obtained in (11.9) is 96.96 Btu/h.ft, the solution does not match with book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate is : 96.96 Btu/h.ft\nSince, 'Q' obtained in (11.9) is 96.96 Btu/h.ft, the solution does not match with book.\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 11.17, Page number: 200" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Two parallel rectangular black plates 0.5 m by 2.0 m are spaced 1.0 m apart. One plate is maintained\nat 1000\u00b0C and the other at 2000\u00b0C. What is the net radiant heat exchange between the two\nplates?\n'''\n\n#Variable declaration:\n#From figure 11.2:\nL = 1.0 #Space between plates (m)\nX = 0.5 #Length of plate (m)\nY = 2.0 #Width of plate (m)\ns = 5.669*10**-8 #Stefan-Boltzmann constant\nTH = 2000.0+273.0 #Temperature of hotter plate (K)\nTC = 1000.0+273.0 #Temperature of colder plate (K)\nBtu = 0.2934*10**-3 #Btu/h in a KW\n\n#Calculation:\nA = X*Y #Area of plate (m^2)\nZ1 = Y/L #Ratio of width with space\nZ2 = X/L #Ratio of length with space\n#From figure 11.2:\nFV = 0.18 #Correction factor\nFE = 1.0 #Emissivity correction factor\nQ1 = FV*FE*s*A*(TH**4-TC**4) #Net radiant heat exchange between plates (kW)\nQ2 = Q1/Btu #Net radiant heat exchange between plates in Btu/h (Btu/h)\n\n#Result:\nprint \"The net radiant heat exchange between plates is :\",round(Q1,-2),\" kW .\"\nprint \"The net radiant heat exchange between plates in Btu/h is :\",round(Q2/10**8,2),\" x 10^8 Btu/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The net radiant heat exchange between plates is : 245600.0 kW .\nThe net radiant heat exchange between plates in Btu/h is : 8.37 x 10^8 Btu/h .\n" + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_12.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_12.ipynb new file mode 100644 index 00000000..fab7925c --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_12.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "Chapter 12" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 12: Condensation and Boiling" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.2, Page number: 206" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the number of degrees of freedom for a one-component, one-phase system.\n'''\n\n#Variable declaration:\nC = 1 #Number of constituents\nP = 1 #Number of phases\n\n#Calculation:\nF = C-P+2 #Number of degrees of freedom\n\n#Result:\nprint \"The number of degrees of freedom is :\",F,\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The number of degrees of freedom is : 2 .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.4, Page number: 209" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous illustrative example. How much heat is removed from the system during\nthis process?\n'''\n\n#Variable declaration:\n#From steam tables:\nU1 = 1237.1 #Internnal energy of gas (Btu/lb)\nU2_g = 1112.2 #Internal energy of gas (Btu/lb)\nU2_l = 343.15 #Internal energy of liquid (Btu/lb)\n\n#Calculation:\nQ = 0.5*(U2_g+U2_l)-1*U1 #Heat removed (Btu/lb)\n\n#Result:\nprint \"Heat removed from the system during the process is :\",round(Q,1),\" Btu/lb .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Heat removed from the system during the process is : -509.4 Btu/lb .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.5, Page number: 212" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A vertical rectangular plate is 0.2 m wide and 0.4 m high. One surface of the plate is exposed to\nsaturated steam at atmospheric pressure. The plate surface is maintained at 98\u00b0C. Filmwise condensation\nmay be assumed. Calculate the type of condensation (flow type) and the condensation\nheat transfer coefficient. Assume steady-state, constant properties, an isotropic homogeneous\nsurface, and that the properties of the water condensate may be evaluated at a film temperature\nof 99\u00b0C. The thermal conductivity of water may be assumed equal to 0.68W/m . K.\n'''\nfrom __future__ import division\nfrom sympy import symbols,solve\n\n#Variable declaration:\nT1 = 99.0 #Mean film temperature (\u00b0C)\nT2 = 98.0 #Plate surface temperature (\u00b0C)\ng = 9.807 #Gravitational acceleration (m/s^2)\n#From Appendix:\nT3 = 100.0 #Saturation temperatre (\u00b0C)\nh_vap1 = 970.3 #Latent heat of steam in Btu/lb (Btu/lb)\nh_vap2 = 2.255*10**6 #Latent heat of steam in J/kg (J/kg)\np_v = 0.577 #Density of steam (kg/m^3)\np_l = 960.0 #Density of liquid water condensate (kg/m^3)\nmu_l = 2.82*10**-4 #Absolute viscosity of liquid water condensate (kg/m.s)\nk = 0.68 #Thermal conductivity of water (W/m.K)\n#From table 12.2\nZ = 0.4 #Height of rectangular plate (m)\nPw = 0.2 #Wetted perimeter of rectangular plate (m)\nh = symbols('h') #Average heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nA = Z*Pw #Heat transfer area of plate (m^2)\nR = A/Pw #Ratio A/Pw (m)\nv_l = mu_l/p_l #Kinematic viscosity of liquid water condensate (m^2/s)\nCo1 = (h/k)*(v_l**2/g/(1-p_v/p_l))**(1/3) #Condensation number (in terms of the average heat transfer coefficient)\nRe = 4*h*Z*(T3-T2)/(mu_l*h_vap2) #Reynolds number in terms of the average heat transfer coefficient\n#From equation 12.14:\nCO1 = 0.0077*Re**Z #Co in terms of Reynolds number for flow type 1\nx1 = solve(Co1-CO1,h) #Solving heat transfer coefficient (W/m^2.K)\nh1 =x1[1]; #Average heat transfer coefficient for flow type 1 (W/m^2.K)\nRe1 = Re.subs(h,h1); #Reynolds number for flow type 1\nCO2 = 1.874*Re**(-1/3) #Co in terms of Reynolds number for flow tupe 2\nx2 = solve(Co1-CO2,h) #Solving average heat transfer coefficient for flow type 2 (W/m^2.K)\nh2 = x2[0]; #Average heat transfer coefficient for flow type 2 (W/m^2.K)\nRe2 = Re.subs(h,h2) #Reynolds number for flow type 2\n\n#Result:\nprint \"The type of condensation flow type 2 is laminar.\"\nprint \"And the condensation heat transfer coefficient is :\",round(h2,-1),\" W/m^2.K .\"\n", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The type of condensation flow type 2 is laminar.\nAnd the condensation heat transfer coefficient is : 14700.0 W/m^2.K .\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.6, Page number: 214" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 12.5. Calculate\n1. the mass flow rate of condensate, m\u02d91\n2. the heat transfer rate Q\u02d9 .\n'''\n\n#Variable declaration:\n#From example 12.5:\nRe = 73.9 #Reynolds number\nmu_l = 2.82*10**-4 #Absolute viscosity of liquid water condensate (kg/m.s)\nPw = 0.2 #Wetted perimeter of rectangular plate (m)\nh = 14700.0 #Heat transfer coefficient (W/m^2.K)\nT_sat = 100.0 #Saturation temperature (\u00b0C)\nTs = 98.0 #Surface temperature (\u00b0C)\nA = 0.2*0.4 #Heat transfer area of plate (m^2) \n\n#Calculation:\nm1 = Re*mu_l/4.0 #Mass flow rate of condensate (kg/m.s)\nm = Pw*m1 #Mass flow rate of condensate (kg/s)\nCo = (3.038*10**-5)*h #Condensation number\nQ = h*A*(T_sat-Ts) #Heat transfer rate (W)\n\n#Result:\nprint \"1. The mass flow rate of condensate is :\",round(m1,4),\" kg/m.s . \"\nprint \"2. The heat transfer rate is :\",round(Q/10**3,2),\" kW . \"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The mass flow rate of condensate is : 0.0052 kg/m.s . \n2. The heat transfer rate is : 2.35 kW . \n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.7, Page number: 215" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A horizontal 4-inch OD tube is surrounded by saturated steam at 2.0 psia. The tube is maintained\nat 64\u00b0F. What is the average heat-transfer coefficient? Assume laminar flow.\n'''\n\n#Variable declaration:\nT_sat = 126.0 #Saturation temperature (\u00b0F)\nT = 64.0 #Surface temperature of tube (\u00b0F)\ng = 32.2 #Gravitational acceleration (ft^2/s)\nD = 4.0/12.0 #Outside diameter of tube (ft)\n\n#Calculation:\nTf = (T_sat+T)/2.0 #Mean film temperature (\u00b0F)\n#From approximate values of key properties:\nh_vap = 1022.0 #Latent heat of steam (Btu/lb)\np_v = 0.00576 #Density of steam (lb/ft^3)\np_l = 62.03 #Density of liquid (lb/ft^3)\nk_l = 0.364 #Thermal conductivity of liquid (Btu/h.ft.\u00b0F)\nmu_l = 4.26*10**-4 #Absolute viscosity of liquid water condensate (lb/ft.s)\nh = 0.725*((p_l*(p_l-p_v)*g*h_vap*k_l**3)/(mu_l*D*(T_sat-T)/3600.0))**(1.0/4.0) #Average heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The average heat transfer coefficient is :\",round(h,1),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The average heat transfer coefficient is : 911.4 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.9, Page number: 222" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Boiling water at 100\u00b0C and 1 atm is supplied with heat from a surface at 102\u00b0C. The heat flux is\n9\u00b000W/m2. If the surface temperature is raised to 103\u00b0C, determine the new heat flux.\nAt 1 atm, the saturation temperature of the water is 100\u00b0C. The liquid water properties (SI\nunits) are: density = 957.9 kg/m3, heat capacity = 4217 J/kg . K, latent heat of vaporization =\n2.26 \u0005 106 J/kg. Assume steady-state operation, constant properties, and laminar conditions.\n'''\n\n#Variable declaration:\nQs1 = 9800.0 #Heat flux (W/m^2)\nTs1 = 102.0 #Original surface temperature (\u00b0C)\nTs2 = 103.0 #New surface temperature (\u00b0C)\nTsat = 100.0 #Saturation temperature (\u00b0C)\n\n#Calculation:\nh1 = Qs1/(Ts1-Tsat) #Original heat transfer coefficient (W/m^2.K)\nDT1 = (Ts1 - Tsat) #Original excess temperature (\u00b0C)\nDT2 = (Ts2 - Tsat) #New excess temperature (\u00b0C)\nn = 0.25 #Value of n for laminar flow\nh2 = h1*(DT2/DT1)**(n) #New heat transfer coefficient (W/m^2.K)\nQs2 = h2*(Ts2-Tsat) #New heat flux (W/m^2)\n\n#Result:\nprint \"The new heat flux is :\",round(Qs2),\" W/m^2.K . \"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The new heat flux is : 16268.0 W/m^2.K . \n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.10, Page number: 223" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''With respect to Illustrative Example 12.9, determine the mechanism of boiling heat transfer.\nIs the earlier assumed boiling mechanism still valid?\n'''\n\n#Variable declaration:\n#From example 12.9:\nTs1 = 102.0 #Original surface temperature (\u00b0C)\nTs2 = 103.0 #New surface temperature (\u00b0C)\nTsat = 100.0 #Saturation temperature (\u00b0C)\n\n#Calculation:\nDTe1 = (Ts1 - Tsat) #Original excess temperature (\u00b0C)\nDTe2 = (Ts2 - Tsat) #New excess temperature (\u00b0C)\n\n#Result:\nprint \"The original excess temperature is: DTe = \",DTe1,\" \u00b0C .\"\nprint \"The new excess temperature is: DTe = \",DTe2,\" \u00b0C .\"\nif ((DTe1 < 5) and (DTe2 < 5)):\n print \"The assumption of the free convection mechanism is valid since DTe < 5\u00b0C.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The original excess temperature is: DTe = 2.0 \u00b0C .\nThe new excess temperature is: DTe = 3.0 \u00b0C .\nThe assumption of the free convection mechanism is valid since DTe < 5\u00b0C.\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.11, Page number: 223" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 12.9 and calculate the liquid Jakob number.\n'''\n\n#Variable declaration:\n#From example 12.9:\nCp = 4127.0 #heat capacity (J/kg . K)\nDTe = 3.0 #New excess temperature (\u00b0C)\nh_vap = 2.26*10**6 #latent heat of vaporization (J/kg)\n\n#Calculation:\nJa_L = Cp*DTe/h_vap #Liquid Jakob number\n\n#Result:\nprint \"The liquid Jakob number is :\",round(Ja_L,5),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The liquid Jakob number is : 0.00548 .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.12, Page number: 223" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An electric heater 0.3 m in diameter at 106\u00b0C is used to heat water. The water is at 1 atm and\n100\u00b0C. Determine\n1. the boiling regime, and\n2. The heat transfer coefficient.\nAssume steady-state operation and constant properties.\n'''\n\n#Variable declaration:\nTs = 106.0 #Surface temperature (\u00b0C)\nTsat = 100.0 #Saturation temperature (\u00b0C)\n\n#Calculation:\nDTe = Ts-Tsat #Excess temperature (\u00b0C)\n#From table 12.5:\nC1 = 5.56 #Constant C1\nn1 = 3.0 #Constant n1\nC2 = 1040.0 #Constant C2\nn2 = 1.0/3.0 #Constant n2\nP = 1.0 #Absolute pressure (atm)\nPa = 1.0 #Ambient absolute pressure (atm)\n\n#Calculation:\nh1 = C1*DTe**n1*(P/Pa)**0.4 #Boiling water heat transfer coefficient (W/m^2)\nQs1 = h1*DTe #Surface flux (W/m^2)\nh2 = C2*DTe**n2*(P/Pa)**0.4 #Second Boiling water heat transfer coefficient (W/m^2)\nQs2 = h2*DTe #Second Surface flux (W/m^2) \n\n#Result:\n \nif (Qs1/10**3 > 15.8 and Qs1/10**3 < 236):\n print \"The boiling regime is :\",round(Qs1/10**3,1),\" kW/m^2 .\"\n print \"The heat transfer coefficient is :\",round(h1), \" W/m^2 .\"\nelif (Qs1/10**3 < 15.8):\n print \"The boiling regime is :\",round(Qs2/10**3,2),\" kW/m^2 .\"\n print \"The heat transfer coefficient is :\",round(h2), \" W/m^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The boiling regime is : 11.34 kW/m^2 .\nThe heat transfer coefficient is : 1890.0 W/m^2 .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 12.13, Page number: 224" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example. Calculate the rate of heat transfer.\n'''\n\nfrom math import pi\n\n#Variable declaration:\n#From example 12.12:\nQs1 = 11340.0 #Surface flux (W/m^2)\nD = 0.3 #Diameter of electric heater (m)\n\n#Calculation:\nA = pi*(D/2.0)**2 #Surface area of heater (m^2)\nQs = Qs1*A #Heat transfer rate (W)\n\n#Result:\nprint \"The rate of heat transfer is :\",round(Qs),\" W .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of heat transfer is : 802.0 W .\n" + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_13.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_13.ipynb new file mode 100644 index 00000000..3f81a28b --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_13.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "Chapter 13" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 13: Refrigeration and Cryogenics" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.1, Page number: 237" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The vertical flat walls of the combustion chamber of a furnace consist of 7.5 in of refractory,\nkR = 0.75; 3 in of insulation, kI = 0.08; and 0.25 in of steel, kS = 26.0. The units of k are\nBtu/h.ft.\u00b0F. The average surface temperature of the inner face of the refractory and the outer\nface of the steel are 2000\u00b0F and 220\u00b0F, respectively.\nCalculate the heat loss expressed as Btu/h.ft^2.\n'''\n\n#Variable declaration:\nLR = 7.5/12.0 #Thickness of refractory (ft)\nLI = 3.0/12.0 #Thickness of insulation (ft)\nLS = 0.25/12.0 #Thickness of steel (ft)\nkR = 0.75 #Thermal conductivity of refractory\nkI = 0.08 #Thermal conductivity of insulation\nkS = 26.0 #Thermal conductivity of steel\nTR = 2000.0 #Average surface temperature of the inner face of the refractory (\u00b0F)\nTS = 220.0 #Average surface temperature of the outer face of the steel (\u00b0F)\n\n#Calculation:\nDT = TR-TS #Temperature difference (\u00b0F)\nQ = DT/(LR/kR+LI/kI+LS/kS) #Heat loss (Btu/h.ft^2)(here representing Qdot/A)\n\n#Result:\nprint \"The heat loss is :\",round(Q),\" Btu/h.ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat loss is : 450.0 Btu/h.ft^2 .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.2, Page number: 239" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''For the combustion chamber of the furnace in the previous example, calculate the temperature of\nthe boundary where the refractory meets the insulation, TI.\n'''\n\n#Variable declaration:\nLR = 7.5/12.0 #Thickness of refractory (ft)\nkR = 0.75 #Thermal conductivity of refractory\nTR = 2000.0 #Average surface temperature of the inner face of the refractory (\u00b0F)\nQ = 450.0 #Heat loss (Btu/h.ft^2)\n\n#Calculation:\nTI = TR - Q*(LR/kR) #Temperature of the boundary where the refractory meets the insulation (\u00b0F)\n\n#Result:\nprint \"The temperature of the boundary where the refractory meets the insulation is :\",round(TI),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature of the boundary where the refractory meets the insulation is : 1625.0 \u00b0F .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.3, Page number: 239" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Using the scenario described in Illustrative Examples 13.1 and 13.2, calculate the area available\nfor heat transfer if the total heat loss is 70,000 Btu/h.\n'''\n\n#Variable declaration:\nQbyA = 70000.0 #Total heat loss (Btu/h)\nQ = 450.0 #Heat loss (Btu/h.ft^2)\n\n#Calculation:\nA = QbyA/Q #Area available for heat transfer (ft^2)\n\n#Result:\nprint \"The area available for heat transfer is :\",round(A,1),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The area available for heat transfer is : 155.6 ft^2 .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.9, Page number: 245" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A refrigerator\u2019s evaporator is at 0.2 MPa and has a working fluid that enters the unit as liquid\nwith an enthalpy, hin, of 230 kJ/kg. If the refrigerator\u2019s working fluid (see Table 13.3) exits\nthe evaporator as dry and saturated vapor, calculate the heat absorbed by the evaporator, QC.\nTable 13.3 Fluid Data for Illustrative Example 13.9\nP, MPa Tsat, \u00b0C hf, kJ/kg hg, kJ/kg, h at 20\u00b0C superheated, kJ/kg\n0.2 210 190 390 410\n0.6 20 230 410 430\n'''\n\n#Variable declaration:\nh_out = 390.0 #Enthalpy of the fluid that exits from the evaporator (kJ/kg)\nh_in = 230.0 #Enthalpy of the fluid that enters the unit (kJ/kg)\n\n#Calculation:\nQC = h_out - h_in #Heat absorbed by the evaporator (kJ/kg)\n\n#Result:\nprint \"The heat absorbed by the evaporator is :\",round(QC),\" kJ/kg .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat absorbed by the evaporator is : 160.0 kJ/kg .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.10, Page number: 246" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''As a follow-up to Illustrative Example 13.9, what is the fluid\u2019s change in entropy across the\nevaporator?\n'''\n\n#Variable declaration:\n#From example 13.9:\nTS = -10.0+273.0 #Fluid\u2019s saturation temperature expressed in Kelvin (K)\nQC = 160.0 #Heat absorbed by the evaporator (kJ/kg)\n\n#Calcuation:\nDS = QC/TS #Fluid\u2019s change in entropy(kJ/kg.K)\n\n#Result:\nprint \"The fluid's change in entropy across the evaporator is :\",round(DS,2),\" kJ/kg.K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fluid's change in entropy across the evaporator is : 0.61 kJ/kg.K .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.11, Page number: 247" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The working fluid in a refrigerator enters a compressor at dry-saturated conditions at a pressure\nof 0.2 MPa and exits the compressor 20\u00b0C superheated at a pressure of 0.6 MPa. Given the fluid\ndata from Illustrative Example 13.9, and given that the fluid leaves the condenser wet-saturated,\nwhat is the heat absorbed by the evaporator of the refrigerator?\n'''\n\n#Variable declaration:\n#From figure 13.2:\nh1 = 390.0 #Fluid enthalpy on entering the compressor (kJ/kg)\nh2 = 430.0 #Fluid enthalpy on leaving the compressor (kJ/kg)\nh3 = 230.0 #Fluid enthalpy on leaving the condenser (kJ/kg)\n\n#Calculation:\nQH = h2 - h3 #Heat rejected from the condenser (kJ/kg)\nW_in = h2 - h1 #Change in enthalpy across the compressor (kJ/kg)\nQC = QH - W_in #Heat absorbed by the evaporator (kJ/kg)\n\n#Result:\nprint \"The heat absorbed by the evaporator of the refrigerator is :\",round(QC),\" kJ/kg .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat absorbed by the evaporator of the refrigerator is : 160.0 kJ/kg .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.12, Page number: 248" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''From Illustrative Example 13.11, what is the refrigerator\u2019s C.O.P.?\n'''\n\n#Variable declaration:\n#From example 13.11:\nW_in = 40.0 #Change in enthalpy across the compressor (kJ/kg)\nQC = 160.0 #Heat absorbed by the evaporator (kJ/kg)\n\n#Calculation:\nCOP = QC/W_in #Refrigerator\u2019s C.O.P.\n\n#Result:\nprint \"the refrigerator's C.O.P. is :\",round(COP),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "the refrigerator's C.O.P. is : 4.0 .\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.13, Page number: 250" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A Rankine cycle has a T-S diagram as shown in Figure 13.4, and rejects 2043 kJ/kg of heat\nduring the constant pressure process between points 3 and 4. The data provided in Table 13.4\nis also associated with this cycle. Calculate the enthalpy change across the boiler.\nProperties associated with the condenser are as follows:\nP = 0.1235 bar\nhf = 209 kJ/kg\nhg = 2592 kJ/kg\nsf = 0.7038 kJ/kg . K\nsg = 8.0763 kJ/kg . K\n\nTable 13.4 T-S Data for Illustrative Example 13.13\nPoint x h, kJ/kg s, kJ/kg.K\n1 548\n2 3989 7.5190\n3 0.9575 2491 7.7630\n4 1.4410\n'''\n\n#Variable declaration:\nh1 = 548.0 #Steam enthalpy at the entry and exit to the boiler (kJ/kg)\nh2 = 3989.0 #Steam enthalpy at the entry and exit to the turbine (kJ/kg)\nh3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\nQH = 2043.0 #Heat rejected by the condenser (kJ/kg)\n\n#Calculation:\nh4 = h3 - QH #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\nQb = h2 - h1 #Enthalpy change across the boiler (kJ/kg)\n\n#Result:\nprint \"The enthalpy change across the boiler is :\",round(Qb),\" kJ/kg .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The enthalpy change across the boiler is : 3441.0 kJ/kg .\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.14, Page number: 251" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the thermal efficiency of the cycle in Illustrative Example 13.13?\n'''\n\n#Variable declaration:\n#From example 13.4:\nh1 = 548.0 #Steam enthalpy at the entry and exit to the boiler (kJ/kg)\nh2 = 3989.0 #Steam enthalpy at the entry and exit to the turbine (kJ/kg)\nh3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\nh4 = 448.0 #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\nQb = 3441.0 #Enthalpy change across the boiler (kJ/kg)\n\n#Calculation:\nWt = h2 - h3 #Work produced by the turbine (kJ/kg)\nWp = h1 - h4 #Work used by the pump (kJ/kg)\nW_net = Wt - Wp #Net work by subtracting the pump work from the turbine work (kJ/kg)\nn_th = W_net/Qb #Thermal efficiency\n\n#Result:\nprint \"The thermal efficiency is :\",round(n_th*100,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The thermal efficiency is : 40.6 % .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 13.15, Page number: 252" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Using the Rankine cycle T-s diagram, in Figure 13.4, and the data from Illustrative Example\n13.13, calculate the temperature at point 3.\n'''\n\n#Variable declaration:\n#From table 13.4:\nx3 = 0.9575 #Mass fraction vapour at point 3\nh3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\ns3 = 7.7630 #Entropy at the entry and exit to the pump (kJ/kg.K)\ns4 = 1.4410 #Entropy at the entry and exit to the condenser (kJ/kg.K)\n#From example13.14:\nh4 = 448.0 #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\n\n#Calculation:\nQ_out = h3 - h4 #Heat rejected (kJ/kg)\nDS = s3 - s4 #Process change in entropy (kJ/kg)\nT3 = Q_out/DS #Temperature at point 3 (K)\n\n#Result:\nprint \"The temperature at point 3 is :\",round(T3),\" K .\"\nprint \"Or, the temperature at point 3 is :\",round(T3-273),\" \u00b0C .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature at point 3 is : 323.0 K .\nOr, the temperature at point 3 is : 50.0 \u00b0C .\n" + } + ], + "prompt_number": 16 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_14.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_14.ipynb new file mode 100644 index 00000000..8d13ffd1 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_14.ipynb @@ -0,0 +1,335 @@ +{ + "metadata": { + "name": "Chapter 14" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 14: Introduction to Heat Exchangers" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.1, Page number: 259" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If 20,000 scfm (32\u00b0F, 1 atm) of an air stream is heated from 200\u00b0F to 2000\u00b0F, calculate the heat\ntransfer rate required to bring about this change in temperature. Use the following enthalpy and\naverage heat capacity data:\nH_200\u00b0F = 1170 Btu/lbmol\nH_2000\u00b0F = 14,970 Btu/lbmol\nCp,av = 7.53 Btu/lbmol.\u00b0F (over the 200-2000\u00b0F range)\n'''\n\n#Variable declaration:\nscfm = 20000.0 #Volumetric flow rate of air at standard conditions (scfm)\nH1 = 1170.0 #Enthalpy at 200\u00b0F (Btu/lbmol)\nH2 = 14970.0 #Enthalpy at 2000\u00b0F (Btu/lbmol)\nCp = 7.53 #Average heat capacity (Btu/lbmol.\u00b0F)\nT1 = 200.0 #Initial temperature (\u00b0F)\nT2 = 2000.0 #Final temperature (\u00b0F)\n\n#Calculation:\nn = scfm/359.0 #Flow rate of air in a molar flow rate (lbmol/min)\nDH = H2 - H1 #Change in enthalpy (Btu/lbmol)\nDT = T2 - T1 #Change in temperature (\u00b0F)\nQ1 = n*DH #Heat transfer rate using enthalpy data (Btu/min)\nQ2 = n*Cp*DT #Heat transfer rate using the average heat capacity data (Btu/min)\n\n#Result:\nprint \"The heat transfer rate using enthalpy data is :\",round(Q1/10**5,2),\" x 10^5 Btu/min.\"\nprint \"The heat transfer rate using the average heat capacity data is :\",round(Q2/10**5,2),\" x 10^5 Btu/min.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate using enthalpy data is : 7.69 x 10^5 Btu/min.\nThe heat transfer rate using the average heat capacity data is : 7.55 x 10^5 Btu/min.\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.2, Page number: 259" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Given the mass flow rate of a gas stream and its heat capacity, determine the required heat rate to\nchange the gas stream from one temperature to another. Data are provided below:\nn = 1200 lbmol/min\nCP,av = 0.26 Btu/lbmol.\u00b0F (over the 200-2000\u00b0F range)\nT1 = 200\u00b0F\nT2 = 1200\u00b0F\n'''\n\n#Variable declaration:\nn = 1200.0 #Flow rate of air in a molar flow rate (lbmol/min)\nCp = 0.26 #Average heat capacity (Btu/lbmol.\u00b0F)\nT1 = 200.0 #Initial temperature (\u00b0F)\nT2 = 1200.0 #Final temperature (\u00b0F)\n\n#Calculation:\nDT = T2 - T1 #Change in temperature (\u00b0F)\nQ = n*Cp*DT #Required heat rate (Btu/min)\n\n#Result:\nprint \"The required heat rate is :\",round(Q/10**5,2),\" x 10^5 Btu/min .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required heat rate is : 3.12 x 10^5 Btu/min .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.3, Page number: 260" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Obtain the heat transfer rate in an exchanger if equal similar liquid flow rates are used. The hot\nfluid is cooled from 84\u00b0C to 72\u00b0C while the cold fluid is initially at 25\u00b0C.\n'''\n\n#Variable declaration:\nTc1 = 25.0 #Initial temperature of cold fluid (\u00b0C)\nTh1 = 72.0 #Initial temperature of hot fluid (\u00b0C)\nTh2 = 84.0 #Final temperature of hot fluid (\u00b0C)\n\n#Calculation:\n#From equation 14.2:\nTc2 = (Th2-Th1)+Tc1 #Final temperature of cold fluid (\u00b0C)\n\n#Result:\nprint \"The final temperature of the cold liquid is :\",Tc2,\" \u00b0C .\"\nprint \"There is a printing mistake in unit of final temperature in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The final temperature of the cold liquid is : 37.0 \u00b0C .\nThere is a printing mistake in unit of final temperature in book.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.4, Page number: 265" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Steam at 1 atm outside a pipe is heating a fluid inside a pipe whose temperature may be assumed\nuniform at 25\u00b0C. Calculate the LMTD.\n'''\n\n#Variable declaration:\nTs = 100.0 #Steam temperature at 1 atm (\u00b0C)\nTl = 25.0 #Fluid temperature (\u00b0C)\n\n#Calculation:\nDTlm = Ts - Tl #Log mean temperature difference (\u00b0C)\n\n#Result:\nprint \"The LMTD is :\",DTlm,\" \u00b0C .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The LMTD is : 75.0 \u00b0C .\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.5, Page number: 265" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Repeat the calculation in the previous example if the fluid is heated from 25\u00b0C and 80\u00b0C\n'''\n\nfrom math import log\n\n#Variable declaration:\nTs = 100.0 #Steam temperature at 1 atm (\u00b0C)\nT1 = 25.0 #Initial fluid temperature (\u00b0C)\nT2 = 80.0 #Final fluid temperature (\u00b0C)\n\n#Calculation:\nDT1 = Ts - T1 #Temperature difference driving force at the fluid entrance (\u00b0C)\nDT2 = Ts - T2 #Temperature driving force at the fluid exit (\u00b0C)\nDTlm = (DT1 - DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\n\n#Result:\nprint \"The LMTD is :\",round(DTlm,1),\" \u00b0C .\"\nprint \"There is a calculation mistake regarding final result in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The LMTD is : 41.6 \u00b0C .\nThere is a calculation mistake regarding final result in book.\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.6, Page number: 266" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A liquid-to-liquid counterflow heat exchanger is used to heat a cold fluid from 120\u00b0F to 310\u00b0F.\nAssuming that the hot fluid enters at 500\u00b0F and leaves at 400\u00b0F, calculate the log mean temperature\ndifference (driving force) for the heat exchanger.\n'''\n\nfrom math import log\n\n#Variable declaration:\nT1 = 500.0 #Temperature of hot fluid entering the heat exchanger (\u00b0F)\nT2 = 400.0 #Temperature of hot fluid exiting the heat exchanger (\u00b0F)\nt1 = 120.0 #Temperature of cold fluid entering the heat exchanger (\u00b0F)\nt2 = 310.0 #Temperature of cold fluid exiting the heat exchanger (\u00b0F)\n\n#Calculation:\nDT1 = T1 - t2 #Temperature difference driving force at the heat exchanger entrance (\u00b0F)\nDT2 = T2 - t1 #Temperature difference driving force at the heat exchanger exit (\u00b0F)\nDTlm = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n\n#Result:\nprint \"The LMTD (driving force) for the heat exchanger is :\",round(DTlm),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The LMTD (driving force) for the heat exchanger is : 232.0 \u00b0F .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.7, Page number: 267" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heavy hydrocarbon oil with heat capacity, cp = 0.55 Btu/lb . \u00b0F, is being cooled from T1 =\n210\u00b0F to T2 = 170\u00b0F. The oil flows inside a tube at a rate of 8000 lb/h and the tube surface temperature\nis maintained at 60\u00b0F. Calculate the heat transfer rate and the LMTD in \u00b0F.\n'''\n\nfrom math import log\n\n#Variable declaration:\nm = 8000.0 #Rate of oil flow inside the tube (lb/h)\nCp = 0.55 #Heat capacity of oil (Btu/lb.\u00b0F)\nT1 = 210.0 #Initial temperature of oil (\u00b0F)\nT2 = 170.0 #Final temperature of oil (\u00b0F)\nt = 60.0 #Tube surface temperature (\u00b0F)\n\n#Calculation:\nDT = T2 - T1 #Change in temperature (\u00b0F)\nQ = m*Cp*DT #Heat transferred from the heavy oil (Btu/h)\nDT1 = T1 - t #Temperature difference driving force at the pipe entrance (\u00b0F)\nDT2 = T2 - t #Temperature difference driving force at the pipe exit (\u00b0F)\nDTlm = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n\n#Result:\nprint \"The heat transfer rate is :\",round(Q),\" Btu/h .\"\nprint \"The LMTD for the heat exchanger is :\",round(DTlm),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate is : -176000.0 Btu/h .\nThe LMTD for the heat exchanger is : 129.0 \u00b0F .\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.8, Page number: 267" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An oil cooler is to operate with an inlet of 138\u00b0F and outlet of 103\u00b0F, and the coolant entering at\n88\u00b0F is to be allowed to rise to 98\u00b0F. What is the LMTD for this unit, if it is considered as\n1. counter-current flow unit, and\n2. a parallel flow unit.\n'''\n\nfrom math import log\n\n#Variable declaration:\nT1 = 138.0 #Temperature of oil entering the cooler (\u00b0F)\nT2 = 103.0 #Temperature of oil leaving the cooler (\u00b0F)\nt1 = 88.0 #Temperature of coolant entering the cooler (\u00b0F)\nt2 = 98.0 #Temperature of coolant leaving the cooler (\u00b0F)\n\n#Calculation:\n#For counter flow unit:\nDT1 = T1 - t2 #Temperature difference driving force at the cooler entrance (\u00b0F)\nDT2 = T2 - t1 #Temperature difference driving force at the cooler exit (\u00b0F)\nDTlm1 = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n#For parallel flow unit:\nDT3 = T1 - t1 #Temperature difference driving force at the cooler entrance (\u00b0F)\nDT4 = T2 - t2 #Temperature difference driving force at the cooler exit (\u00b0F)\nDTlm2 = (DT3 - DT4)/(log(DT3/DT4)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n\n#Result:\nprint \"The LMTD for counter-current flow unit is :\",round(DTlm1,1),\" \u00b0F .\"\nprint \"The LMTD for parallel flow unit is :\",round(DTlm2,1),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The LMTD for counter-current flow unit is : 25.5 \u00b0F .\nThe LMTD for parallel flow unit is : 19.5 \u00b0F .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.10, Page number: 272" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Arectangular plane window glass panel is mounted on a house. The glass is 0.125 inch thick and\nhas a surface area of 1.0 m2; its thermal conductivity, k2, is 1.4W/m . K. The inside house temperature,\nT1, and the outside air temperature, t4, are 25\u00b0C and 214\u00b0C, respectively. The heat\ntransfer coefficient inside the room, h1, is 11.0 W/m^2 . K and the heat transfer coefficient\nfrom the window to the surrounding cold air, h3, is 9.0 W/m^2 . K. Calculate the overall heat\ntransfer coefficient in W/m^2 . K.\n'''\n\n#Variable declaration:\nA = 1.0 #Surface area of glass (m^2)\nh1 = 11.0 #Heat transfer coefficient inside room (W/m^2.K)\nL2 = 0.125*0.0254 #Thickness of glass (m)\nk2 = 1.4 #Thermal conductivity of glass (W/m.K)\nh3 = 9.0 #Heat transfer coefficient from window to surrounding cold air (W/m^2.K)\n\n#Calculation:\nR1 = 1.0/(h1*A) #Internal convection resistance (K/W)\nR2 = L2/(k2*A) #Conduction resistance through glass panel (K/W)\nR3 = 1.0/(h3*A) #Outside convection resistance (K/W)\nRt = R1+R2+R3 #Total thermal resistance (K/W)\nU = 1.0/(A*Rt) #Overall heat transfer coefficient (W/m^2.K)\n\n#Result:\nprint \"The overall heat transfer coefficient is :\",round(U,1),\" W/m^2.K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The overall heat transfer coefficient is : 4.9 W/m^2.K .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.11, Page number: 273" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heat exchanger wall consists of a copper plate 0.049 inch thick. If the two surface film coefficients\nare 208 and 10.8 Btu/h.ft^2.\u00b0F, respectively, calculate the overall heat transfer coefficient\nin Btu/h.ft^2.\u00b0F.\n'''\n\n#Variable declaration:\nDx = 0.049/12.0 #Thickness of copper plate (ft)\nh1 = 208.0 #Film coefficient of surface one (Btu/h.ft^2.\u00b0F)\nh2 = 10.8 #Film coefficient of surface two (Btu/h.ft^2.\u00b0F)\nk = 220.0 #Thermal conductivity for copper (W/m.K)\n\n#Calculation:\nU = 1.0/(1.0/h1+Dx/k+1.0/h2) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The overall heat transfer coefficient is :\",round(U,2),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The overall heat transfer coefficient is : 10.26 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.12, Page number: 274" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A coolant flows through a steel pipe (inside diameter of 0.05 m, outside diameter of 0.06 m) at a\nvelocity of 0.25 m/s. What is the overall heat transfer coefficient for the system based on the\npipe\u2019s outside surface area if the inside and outside coefficients are 2000 W/m^2 . K and 8.25\nW/m^2 . K, respectively. Assume the \u201cresistance\u201d term for the steel, that is,\nR = Do x ln(Do/Di)/2k\nis 1.33 x 10^-4 m^2.K/W .\n'''\n\n#Variable declaration:\nDo = 0.06 #Outside diameter of pipe (m)\nDi = 0.05 #Inside diameter of pipe (m)\nho = 8.25 #Outside coefficient (W/m^2.K)\nhi = 2000.0 #Inside coefficient (W/m^2.K)\nR = 1.33*10**-4 #Resistance for steel (m^2.K/W)\n\n#Calculation:\nU = 1.0/(Do/(hi*Di)+R+1.0/ho) #Overall heat transfer coefficient (W/m^2.\u00b0K)\n\n#Result:\nprint \"The overall heat transfer coefficient is :\",round(U,2),\" W/m^2.\u00b0K .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The overall heat transfer coefficient is : 8.2 W/m^2.\u00b0K .\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.14, Page number: 274" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Steam at 247\u00b0F is flowing through a pipe exposed to air but covered with 1.5 inch thick insulation.\nThe following data is provided:\n1. Pipe diameter, inside = 0.825 inch\n2. Pipe diameter, outside = 1.05 inch\n3. Surrounding air temperature = 60\u00b0F\n4. Thermal conductivity, pipe = 26 Btu/h . ft . \u00b0F\n5. Thermal conductivity, insulation = 0.037 Btu/h . ft . \u00b0F\n6. Steam film coefficient = 800 Btu/h . ft^2 . \u00b0F\n7. Air film coefficient = 2.5 Btu/h . ft^2 . \u00b0F\nCalculate the overall heat transfer coefficient based on the inside area of the pipe.\n'''\n\nfrom math import pi,log\n\n#Variable declaration:\nDi = 0.825/12.0 #Pipe inside diameter (ft)\nDo = 1.05/12.0 #Pipe outside diameter (ft)\nDl = 4.05/12.0 #Insulation thickness (ft)\nl = 1.0 #Pipe length (ft)\nkp = 26.0 #Thermal conductivity of pipe (Btu/h.ft.\u00b0F)\nkl = 0.037 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\nhi = 800.0 #Steam film coefficient (Btu/h.ft^2.\u00b0F)\nho = 2.5 #Air film coefficient (Btu/h.ft^2.\u00b0F)\n\n#Calculation:\nri = Di/2.0 #Pipe inside radius (ft)\nro = Do/2.0 #Pipe outside radius (ft)\nrl = Dl/2.0 #Insulation radius (ft)\nAi = pi*Di*l #Inside area of pipe (ft^2)\nAo = pi*Do*l #Outside area of pipe (ft^2)\nAl = pi*Dl*l #Insulation area of pipe (ft^2)\nA_Plm = (Ao-Ai)/log(Ao/Ai) #Log mean area for steel pipe (ft^2)\nA_Ilm = (Al-Ao)/log(Al/Ao) #Log mean area for insulation (ft^2)\nRi = 1.0/(hi*Ai) #Air resistance (m^2.K/W)\nRo = 1.0/(ho*Al) #Steam resistance (m^2.K/W)\nRp = (ro-ri)/(kp*A_Plm) #Pipe resistance (m^2.K/W)\nRl = (rl-ro)/(kl*A_Ilm) #Insulation resistance (m^2.K/W)\nU = 1.0/(Ai*(Ri+Rp+Ro+Rl)) #Overall heat coefficient based on the inside area (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The overall heat transfer coefficient based on the inside area of the pipe is :\",round(U,3),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The overall heat transfer coefficient based on the inside area of the pipe is : 0.748 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.15, Page number: 275" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the rate of heat transfer in the previous example.\n'''\n\nfrom math import pi\n\n#Variable declaration:\n#From example 14.14:\nDi = 0.825/12.0 #Pipe inside diameter (ft)\nL = 1.0 #Pipe length (ft)\nUi = 0.7492 #Overall heat coefficient (Btu/h.ft^2.\u00b0F)\nTs = 247.0 #Steam temperature (\u00b0F)\nta = 60.0 #Air temperature (\u00b0F)\n\n#Calculation:\nAi = pi*Di*L #Inside area of pipe (ft^2)\nQ = Ui*Ai*(Ts-ta) #Heat transfer rate (Btu/h)\n\n#Result:\nprint \"The heat transfer rate is :\",round(Q,1),\" Btu/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate is : 30.3 Btu/h .\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.16, Page number: 276" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is provided for a heat exchanger:\nhw(water) = 200 Btu/h.ft^2.\u00b0F\nho(oil) = 50 Btu/h.ft^2.\u00b0F\nhf (fouling) = 1000 Btu/h.ft^2.\u00b0F\nDTlm = 90\u00b0F\nA = 15.0 ft^2\nCalculate the heat transfer rate for the exchanger if the wall resistance can be neglected.\n'''\n\n#Variable declaration:\nhw = 200.0 #Water heat coefficient (Btu/h.ft^2.\u00b0F)\nho = 50.0 #Oil heat coefficient (Btu/h.ft^2.\u00b0F)\nhf = 1000.0 #Fouling heat coefficient (Btu/h.ft^2.\u00b0F)\nDTlm = 90.0 #Log mean temperature difference (\u00b0F)\nA = 15.0 #Area of wall (ft^2)\n\n#Calculation:\nX = 1.0/hw+1.0/ho+1.0/hf #Equation 14.34 for constant A\nU = 1.0/X #Overall heat coeffocient (Btu/h.ft^2.\u00b0F)\nQ = U*A*DTlm #Heat transfer rate (Btu/h)\n\n#Result:\nprint \"The heat transfer rate is :\",round(Q,-1),\" Btu/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate is : 51920.0 Btu/h .\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 14.17, Page number: 277" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A brine solution at 10\u00b0F in a food processing plant is heated by flowing through a heated pipe.\nThe pipe surface is maintained at 80\u00b0F. The pipe surface area for heat transfer is 2.5 ft^2. The\nbrine solution (with a density of 62.4 lb/ft^3 and a heat capacity of 0.99 Btu/lb.\u00b0F) flows at a\nrate of 20 lb/min. The overall heat transfer coefficient varies linearly with the temperature\napproach, with values of 150 Btu/h.ft^2.\u00b0F at the brine solution entrance (where the brine\ntemperature is 10\u00b0F) and 140 Btu/h.ft^2.\u00b0F at the brine solution exit. Determine the temperature\napproach at the brine inlet side, the exit temperature of the brine solution, and the rate of heat\ntransfer, Q .\n''' \n\nfrom __future__ import division\nfrom sympy import symbols,log,nsolve\n\n\n#Variable declaration:\nT = 80.0 #Pipe surface temperature (\u00b0F)\nt1 = 10.0 #Brine inlet temperature (\u00b0F)\nDT2 = symbols('DT2') #Discharge temperature of the brine solution (\u00b0F)\nm = 20*60 #Flowrate of brine solution (lb/h)\nCp = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\nU1 = 150 #Overall heat transfer coefficient at brine solution entrance (Btu/h.ft^2.\u00b0F)\nU2 = 140 #Overall heat transfer coefficientat at brine solution exit (Btu/h.ft^2.\u00b0F)\nA = 2.5 #Pipe surface area for heat transfer (ft^2)\n\n#Calculation:\nDT1 = T-t1 #Temperature approach at the pipe entrance (\u00b0F)\nQ = m*Cp*(DT1-DT2) #Energy balance to the brine solution across the full length of the pipe (Btu/h)\nDT1m = (DT1-DT2)/log(DT1/DT2) #Equation for the LMTD\nQQ = A*(U2*DT1-U1*DT2)/log(U2*DT1/U1/DT2) #Equation for the heat transfer rate (Btu/h)\nE = QQ-Q #Energy balance equation\nR = nsolve([E],[DT2],[1.2]) #\nDT = R[0] #Log mean temperature difference\nt2 = T-DT #In discharge temperature of the brine solution (\u00b0F)\nt2c = 5/9*(t2-32) #In discharge temperature of the brine solution in \u00b0C (c/5 = (F-32)/9)\n_Q_ = Q.subs(DT2,DT) #Heat transfer rate (Btu/h)\n\n#Result:\nprint \"The temperature approach at the brine inlet side is :\",round(DT1,1),\" \u00b0F.\"\nprint \"Or, the temperature approach at the brine inlet side is :\",round(DT1/1.8,1),\" \u00b0C.\"\nprint \"The exit temperature of the brine solution is :\",round(t2,1),\" \u00b0F.\"\nprint \"Or, the exit temperature of the brine solution is :\",round((t2-32)/1.8,1),\" \u00b0C.\"\nprint \"The rate of heat transfer is :\",round(_Q_,-1),\" Btu/h.\"\nprint \"Or, the rate of heat transfer is :\",round(_Q_/3.412,-2),\" W.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature approach at the brine inlet side is : 70.0 \u00b0F.\nOr, the temperature approach at the brine inlet side is : 38.9 \u00b0C.\nThe exit temperature of the brine solution is : 28.4 \u00b0F.\nOr, the exit temperature of the brine solution is : -2.0 \u00b0C.\nThe rate of heat transfer is : 21830.0 Btu/h.\nOr, the rate of heat transfer is : 6400.0 W.\n" + } + ], + "prompt_number": 18 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_15.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_15.ipynb new file mode 100644 index 00000000..cbfe94ac --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_15.ipynb @@ -0,0 +1,356 @@ +{ + "metadata": { + "name": "Chapter 15" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 15: Double Pipe Heat Exchangers" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.2, Page number: 290" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following data is provided for a double pipe heat exchanger\nQ = 12,000 Btu/h\nU = 48.0 Btu/ft^2.h.\u00b0F\nDTlm = 50\u00b0F\nCalculate the area of the exchanger.\n'''\n\n#Variable declaration:\nQ = 12000.0 #Heat transfer rate (Btu/h)\nU = 48.0 #Overall heat coefficient (Btu/ft^2.h.\u00b0F)\nDTlm = 50.0 #Log mean temperature difference (\u00b0F)\n\n#Calculation:\nA = Q/(U*DTlm) #Area of exchanger (ft^2)\n\n#Result:\nprint \"The area of the exchanger is :\",round(A),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The area of the exchanger is : 5.0 ft^2 .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.3, Page number: 291" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the inlet cold water temperature flowing at a rate of 14.6 lb/min in a double pipe heat\nexchanger given the following data. Assume co-current operation.\nt1 = tc,i = 63\u00b0F\nT1 = TH,i = 164\u00b0F\nT2 = TH,o = 99\u00b0F\nU = 35.35 Btu/ft.h.\u00b0F\nQ = 56,760 Btu/h\nA = 32.1 ft^2\n'''\n\nfrom sympy import symbols,solve,log\n\n#Variable declaration:\nQ = 56760 #Heat transfer rate (Btu/h)\nU = 35.35 #Overall heat coefficient (Btu/ft.h.\u00b0F)\nA = 32.1 #Area of exachanger (ft^2)\nt1 = 63.0 #Outlet cold water temperature (\u00b0F)\nT1 = 164 #Outlet hot water temperature (\u00b0F)\nT2 = 99 #Inlet hot water temperature (\u00b0F)\nt2 = symbols('t2') #Inlet cold water temperature (\u00b0F)\n\n#Calculation:\nDTlm = Q/(U*A) #Log mean temperature difference (\u00b0F)\ndT1 = T1-t1 #Temperature approach at pipe outlet (\u00b0F)\ndT2 = T2-t2 #Temperature approach at pipe inlet (\u00b0F)\nEq = (dT2-dT1)/log(dT2/dT1)-DTlm\nR = solve(Eq,t2) #Inlet cold water temperature (\u00b0F)\n\n#Result:\nprint \"The inlet cold water temperature is : \",round(R[0]),\" \u00b0F.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The inlet cold water temperature is : 79.0 \u00b0F.\n" + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.4, Page number: 292" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''As Shakespeare once said: \u201cSomething is rotten in the state of . . .\u201d Comment on whether the\ninformation provided in the previous example is \u201cconsistent.\u201d\n'''\n\n#Variable declaration:\nm = 14.6 #Flow rate of water inside the tube (lb/min)\nCp = 1 #Heat capacity of water (Btu/lb.\u00b0F)\nt2 = 79 #Initial temperature of water (\u00b0F)\nt1 = 63 #Final temperature of water (\u00b0F)\n#From example 15.3:\nQ1 = 56760 #Old heat transfer rate (Btu/h)\n\n#Calculation:\nQ2 = m*Cp*(t2-t1) #New heat transfer rate (Btu/min)\n\n#Result:\nprint \"The new heat transfer rate is :\",round(Q2),\" Btu/min.\" \nprint \"Or, the new heat transfer rate is :\",round(Q2*60),\" Btu/h.\"\nif (Q1==Q2) :\n print \"This result agree with the Q\u02d9 provided in the problem statement. Shakespeare is wrong, nothing is rotten there.\"\nelse :\n print \"This result does not agree with the Q\u02d9 provided in the problem statement. Shakespeare is right, something is indeed rotten.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The new heat transfer rate is : 234.0 Btu/min.\nOr, the new heat transfer rate is : 14016.0 Btu/h.\nThis result does not agree with the Q\u02d9 provided in the problem statement. Shakespeare is right, something is indeed rotten.\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.5, Page number: 292" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heavy hydrocarbon oil with a heat capacity, cp = 0.55 Btu/lb.\u00b0F, is being cooled in a double\npipe heat exchanger from T1 = 210\u00b0F to T2 = 170\u00b0F. The oil flows inside a tube at a rate of\n8000 lb/h and the tube surface temperature is maintained at 60\u00b0F. The overall heat transfer\ncoefficient U = 63 Btu.h.ft^2.\u00b0F. Calculate the required heat transfer area, A, in ft^2.\n'''\n\nfrom math import log\n\n#Variable declaration:\nT1 = 210.0 #Initial temperature of oil (\u00b0F)\nT2 = 170.0 #Final temperature of oil (\u00b0F)\nT3 = 60.0 #Surface temperature of oil (\u00b0F)\nm = 8000.0 #Flow rate of oil inside tube (lb/h)\ncp = 0.55 #Heat capacity of oil (Btu/lb.\u00b0F)\nU = 63.0 #Overall heat teansfer coefficient (Btu.h.ft^2.\u00b0F)\n\n#Calculation:\nDT1 = T1-T3 #Temperature difference 1 (\u00b0F)\nDT2 = T2-T3 #Temperature difference 2 (\u00b0F)\nDTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temerature difference (\u00b0F)\nQ = m*cp*(T1-T2) #Heat transferred (Btu/h)\nA = Q/(U*DTlm) #Heat transfer area (ft^2)\n\n#Result:\nprint \"The required heat transfer area is :\",round(A,2),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required heat transfer area is : 21.66 ft^2 .\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.6, Page number: 293" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Two variations of the design of a heat exchanger have been proposed. The unit is to cool a hot\nwater stream from 140\u00b0F to 110\u00b0F through indirect contact with a cold water stream that is being\nheated from 60\u00b0F to 90\u00b0F. The water flow rate is 100 lb/min and the overall heat transfer coefficient\nmay be assumed equal to 750 Btu/h.ft^2.\u00b0F. Calculate the area requirements for the following\ntwo exchanger conditions:\n1. double pipe co-current flow\n2. double pipe countercurrent flow\n'''\n\nfrom math import log\n\n#Variable declaration:\nT1 = 140.0 #Initial temperature of hot water (\u00b0F)\nT2 = 110.0 #Final temperature of hot water (\u00b0F)\nT3 = 60.0 #Initial temperature of cold water (\u00b0F)\nT4 = 90.0 #Initial temperature of cold water (\u00b0F)\nDTlm2 = 50.0 #Log mean temerature difference for countercurrent flow, a constant (\u00b0F) (part 2)\nm = 100.0*60 #Water flow rate (lb/h)\ncp = 1.0 ##Heat capacity of water (Btu/lb.\u00b0F)\nU = 750.0 #Overall heat teansfer coefficient (Btu.h.ft^2.\u00b0F)\n\n#Calculation:\nDT1 = T1-T3 #Temperature difference 1 (\u00b0F) (part 1)\nDT2 = T2-T4 #Temperature difference 2 (\u00b0F)\nDTlm1 = (DT1-DT2)/log(DT1/DT2) #Log mean temerature difference (\u00b0F)\nQ = m*cp*(T1-T2) #Heat transferred (Btu/h)\nAp = Q/(U*DTlm1) #Heat transfer area for parallel flow (ft^2)\nAc = Q/(U*DTlm2) #Heat transfer area for counter flow (ft^2)\n\n#Result:\nprint \"1. The double pipe co-current flow is :\",round(Ap,2),\" ft^2 .\"\nprint \"1. The double pipe countercurrent flow is :\",round(Ac,2),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The double pipe co-current flow is : 5.55 ft^2 .\n1. The double pipe countercurrent flow is : 4.8 ft^2 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.8, Page number: 294" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The double pipe heat exchanger shown in Figure 15.4 is used for heating 2500 lb/h benzene\n\tfrom 60\u00b0F to 120\u00b0F. Hot water at 200\u00b0F from a boiler is available for heating purposes in\n\tamounts up to 4000 lb/h. Use schedule 40 brass pipe and an integral number of 15 foot long\n\tsections are to be employed in the design of the exchanger.\n\tThe design is to be based on achieving a Reynolds number of 13,000 (turbulent flow) in both\n\tthe inner pipe and annular region. As an additional constrain because of pressure drop consideration,\n\tthe width of the annular region must be at least equal to one fourth the outside diameter of\n\tthe inner pipe.\n\tNote that one can obtainQ\u02d9 by calculating the duty of the benzene stream, and a heat balance\n\twill yield the exit temperature of the water, Tw. However, the area or the overall heat transfer\n\tcoefficient is not known. Obtaining U becomes more complicated since information on flow\n\tvariables or a pipe diameter is not given.\n\tData is provided below for this problem, but can be estimated or obtained from Tables in the\n\tAppendix:\n\tCenzene at 90\u00b0F:\n\tmu = 0.55 cP = 3.70 x 10^-4 lb/ft.s\n\tSG = 0.879 (at 68.8\u00b0F)\n\tr = 0.879(62.4) = 54.8 lb/ft^3\n\tcp = 0.415 Ctu/lb.\u00b0F\n\tk = 0.092 Ctu/h.ft.\u00b0F (at 86\u00b0F)\n\tHater at 200\u00b0F.\n\tmu = 0.305 cP = 2.05 x 10^-4 lb/ft.s\n\tr = 60.13 lb/ft^3\n\tcp = 1.0 Btu/lb.\u00b0F\n\tk = 0.392 Btu/h.ft.\u00b0F\n'''\n\nfrom __future__ import division\nfrom math import pi,log\n\n#Variable declaration:\nuC = 3.7*10**-4 \t #Viscosity of benzene (lb/ft.s)\nuH = 2.05*10**-4 \t #Viscosity of water @200 \u00b0F (lb/ft.s)\nu2 = 2.16*10**-4 \t\t\t\t #Viscosity of water @192 \u00b0F (lb/ft.s)\npC = 54.8 #Density of benzene (lb/ft^3)\npH = 60.13 #Density of water (lb/ft^3)\ncpC = 0.415 #Specific heat capacity of benzene (Btu/lb.\u00b0F)\ncpH = 1 #Specific heat capacity of water (Btu/lb.\u00b0F)\nsgC = 0.879 \nkC = 0.092 #Thermal conductivity of benzene (Btu/h.ft.\u00b0F)\nkH = 0.392 #Thermal conductivity of water @200 \u00b0F (Btu/h.ft.\u00b0F)\nk2 = 0.390\t\t\t\t\t #Thermal conductivity of water @192 \u00b0F (Btu/h.ft.\u00b0F)\nmC = 2500\t \t\t\t #Flow rate of benzene (lb/s)\nmH = 4000 #Flow rate of water (lb/s)\nRe = 13000 #Reynolds number\ndTc = 120-60\t\t\t\t\t #Difference in temperature heating for benzene\nTw = 200\t\t\t\t #Temperatperature of hot water (\u00b0F)\n#For 2-inch schedule 40 pipe\nAi = 0.541 #Inside area of pipe (ft^2/ft)\nAo = 0.622 #Outside area of pipe (ft^2/ft)\nDi = 2.067 #Inside diameter of pipe (inch)\nDo = 2.375 #Outside diameter of pipe (inch)\nSi = 0.0233 #Inside surface area of pipe (ft^2)\ndXw = 0.128 #Width of pipe (ft)\n\n#For 4-inch schedule 40 pipe\nDio = 4.026 #Inside diameter of pipe (inch)\nDoi = Do #Outside diameter of pipe (inch)\nkw = 26 \n\n#Calculations:\ndef St(Re,Pr):\t\t\t\t #Dittus Boelter equation\n\treturn 0.023*Re**-0.2*Pr**-0.667\n#For inside tubes:\nDicalc = 4*mC/(Re*pi*uC)/3600 #Inside diameter (ft)\nmHcalc = Re*pi*uH*(Doi+Dio)/4*3600/12 #Mass flow rate of water (lb/h)\nQ = mC*cpC*dTc\t\t\t\t\t #Heat in water (Btu/h)\ndTH = Q/mH #Temperature difference of water (\u00b0F)\nTHo = Tw - dTH #Outlet temperature of water (\u00b0F)\nTHav = (Tw+THo)/2 #Average temperature of water (\u00b0F) \n#For benzene:\nPrC = cpC*uC/kC*3600 #Prandtl number\nStC = round(St(13000, PrC),5) #Stanton number\nhi = StC*cpC*mC/Si #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n#For water:\nReH = 4*mH/3600/(pi*u2*(Doi+Dio)/12) #Reynolds number\nPrH = round(cpH*(u2)/k2*3600 ,2) #Prandtl number\nStH = round(St(ReH, PrH),5) #Stanton number\nSann = round(pi/4*(Dio**2-Doi**2)/144,4) #Surface area of annulus (ft^2)\nho = round(StH*cpH*mH/Sann) #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n#For pipe:\nDlm = round((Do-Di)/log(Do/Di)*12,3) #Log mean difference in diameter (ft)\nUo = 1/(Do/Di/hi + dXw*Do/kw/Dlm + 1/ho) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\ndTlm = (124.4-80)/log(124.4/80) #Log mean temperature difference (\u00b0F)\nL = Q/(Uo*0.622*dTlm) #Length of pipe (ft)\n\n#Result:\nprint 'The required length of pipe: ',round(L,1), 'ft'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required length of pipe: 31.4 ft\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.10, Page number: 300" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate T2 and t2 in (consistent units)\nMC = m_HcpH = 2000\nmc = m_CcpC = 1000\nU = 2000\nA = 10\nT1 = 300\nt1 = 60\nCalculate the exit temperatures if the flow is countercurrent by employing Equation (15.32).(7)\n'''\n\nfrom __future__ import division\nfrom math import e\n\n#Variable declaration:\nMC = 2000.0 \nmc = 1000.0\nU = 2000.0\nA = 10.0\nT1 = 300.0\nt1 = 60.0\n\n#Calculation:\nB = 1.0/mc \nb = 1.0/MC\nx = B/b\ny = U*(B-b)\nT2 = ((x-y)*T1 + x*(e-y)*t1)/(2*e-1)\nt2 = t1+(T1-T2)/x\n\n#Result:\nprint \"T2 = :\",round(T2),\" \u00b0F\"\nprint \"t2 = :\",round(t2),\" \u00b0F\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "T2 = : 114.0 \u00b0F\nt2 = : 153.0 \u00b0F\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.11, Page number: 301" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is provided for a co-current flow double pipe heat exchanger. The\npipe consists of 200 ft of 2-inch schedule 40 pipe, k = 25 Btu/h.ft.\u00b0F. The hot and cold\nfilm coefficients are 1200 and 1175 Btu/h.ft^2.\u00b0F, respectively, In addition,\nMC = m_HcpH = 30,000 Btu/h.\u00b0F\nmc = m_CcpC = 22,300 Btu/h.\u00b0F\nT1 = 300 \u00b0F\nt1 = 60 \u00b0F\nCalculate T2 and t2.\n'''\n\nfrom sympy import symbols,solve,log\n\n#Variable declaration:\nh1 = 1200.0 #Hot film coefficient (Btu/h.ft^2.\u00b0F)\nh2 = 1175.0 #Cold film coefficient (Btu/h.ft^2.\u00b0F)\nL = 200.0 #Length of pipe (ft)\nMC = 30000.0\nmc = 22300.0\nT1 = 300.0 #Inlet temperature of hot fluid in pipe (\u00b0F)\nt1 = 60.0 #Inlet temperature of cold fluid in pipe (\u00b0F)\nT2 = symbols('T2') #Outlet temperature of hot fluid \u00b0F\nt2 = symbols('t2') #Outlet temperature of cold fluid \u00b0F\n#From table 6.2:\nID = 2.067 #Inside diameter of pipe (in)\nOD = 2.375 #Outside diameter of pipe (in)\nDx = 0.154 #Thickness of pipe (in)\nAi = 0.541 #Inside sectional area of pipe (ft^2/ft)\nk = 25.0 #Thermal conductivity of pipe (Btu/h)\n\n#Calculation:\nUi = 1.0/((1.0/h1) +(Dx/(k*12.0))+(1.0/(h2*(OD/ID)))) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nAi1 = Ai*L #Inside area of pipe (ft^3/ft)\nQH = MC*(T1-T2) #Heat transfer rate of hot fluid (Btu/h)\nQC = mc*(t2-t1) #Heat transfer rate of cold fluid (Btu/h)\nt2ht = 195 #t2 by hit and trial\nx = solve(QC-QH,T2)\nT2 = x[0]\nDTlm = (T1-t1-T2+t2)/log((T1-t1)/(T2-t2)) #Log mean temperature difference (\u00b0F)\nQ = Ui*Ai1*DTlm.subs(t2,t2ht) #Total heat transfer rate (Btu/h)\n\n#Result:\nprint \"T2 :\", round(T2.subs(t2,t2ht)),\"(\u00b0F)\"\nprint \"t2 :\", t2.subs(t2,t2ht),\"(\u00b0F)\"\nprint \"Qdot :\", round(Q/10**6) ,\"x 10**6 Btu/h\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "T2 : 200.0 (\u00b0F)\nt2 : 195 (\u00b0F)\nQdot : 3.0 x 10**6 Btu/h\n" + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.12, Page number: 302" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Solve the previous example employing the equations above provided by Famularo.(7) Also calculate\nthe discharge temperature, the LMTD, and Q\u02d9 for the countercurrent case.\n'''\n\nfrom __future__ import division\nfrom math import log,e\n\n#Variable declaration:\nB = 3.33*10**-5\nb = 4.48*10**-5\n#From example 15.11:\nA = 108.2 #Inside area of pipe (ft^3/ft)\nU = 482 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nMC = 30000.0\nmc = 23000.0\nT1 = 300.0 #Inlet temperature of hot fluid in pipe (\u00b0F)\nt1 = 60.0 #Inlet temperature of cold fluid in pipe (\u00b0F)\n\n#Calculation:\n#From equation 15.28:\nT2 = ((B/b)*(e**(U*A*(B-b))-1)*t1+T1*(B/b-1))/((B/b)*e**(U*A*(B-b))-1) #Outlet temperature of hot fluid (\u00b0F)\n#From equation 15.32:\nt2 = ((b/B)*(e**(U*A*(b-B))-1)*T1+t1*(b/B-1))/((b/B)*e**(U*A*(b-B))-1) #Outlet temperature of cold fluid (\u00b0F)\nDT = ((T2-t1)-(T1-t2))/(log((T2-t1)/(T1-t2))) #Log mean difference temperature (\u00b0F)\nQ1 = U*A*DT #Heat transfer rate of hot fluid (Btu/h)\nQ2 = MC*(T1-T2) #Heat transfer rate of cold fluid (Btu/h)\n\n#Result:\nprint \"The heat load is :\",round(Q2,-3),\" Btu/h.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat load is : 4078000.0 Btu/h.\n" + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.14, Page number: 305" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A double pipe heat exchanger is employed to heat 228 kg/h of water from 25 \u00b0C to 73 \u00b0C.\nWater is on the tube side and the flow is countercurrent. The heating fluid is saturated steam\nat 1 atm, with 55 kg/h of steam condensing on the annular side. No subcooling of condensed\nwater takes place. The inner pipe of the double pipe heat exchanger has an inner radius (ri,i)\nof 1.3 cm and an outer radius (ro,i) of 1.9 cm. The inner radius of the outer tube (ri,o) is 3.2\ncm. Both tubes are 0.5 % carbon steel. The overall heat transfer coefficient for the new heat\nexchanger, Uclean, was originally determined to be 3994 W/m^2.K, based on the inside area of\nthe inner pipe. After a period of use, the fouling factor, Rf, was estimated to be 0.0002 m^2.K/W.\nDetermine:\n1. the temperature profile of the water and steam along the length of the exchanger,\n2. whether the steam has enough energy to heat the water,\n3. the minimum length of the heat exchanger, and\n4. the dirty overall heat transfer coefficient.\nWater properties at the average bulk temperature (25 \u00fe 73)/2 = 49\u00b0C are: heat capacity,\ncp = 4174 J/kg.K; density, r = 993 kg/m^3; absolute viscosity, m = 6.82 x 10^-4 kg/m.s ;\nkinematic viscosity, n = 6.87 x 10^-7 m2/s, and the Prandtl number, Pr = 4.53. The properties\nof water at the \u201caverage\u201d film temperature (100 + 49)/2 = 74.9 \u00b0C are: heat capacity, cp = 4190\nJ/kg.K; density, r = 975 kg/m^3; absolute viscosity, m = 3.8 x 10^-4 kg/m.s; kinematic viscosity,\nn = 3.89 x 10^-7 m^2/s, and the Prandtl number, Pr = 2.53.\nFor saturated steam at 100\u00b0C: latent heat of condensation = 2.26 x 10^6 J/kg, vapor\ndensity = 0.596 kg/m^3. The thermal conductivity for 0.5% carbon steel at an average temperature\nof (100 + 25)/2 = 62.5 \u00b0C is k = 54 W/m.K\nAssumptions include steady-state operation, incompressible fluids, Newtonian fluids,\nconstant properties, homogeneous isotropic materials, and there are negligible heat losses.\nNotations 1 and 2 denote the water inlet and outlet, respectively, and s the stream.\n'''\n\nfrom __future__ import division\nfrom math import log,pi\n\n#Variable declaration:\nTs = 100.0 #Saturation temperature (\u00b0C)\nt1 = 25.0 #Initial temperature of water (\u00b0C)\nt2 = 73.0 #Final temperature of water (\u00b0C)\nm = 228.0/3600.0 #Mass flow rate of water (kg/s)\ncp = 4174.0 #Heat capacity of water (J/kg.K)\nm_s = 55.0/3600.0 #Mass flow rate of steam (kg/s)\nh_vap = 2.26*10**26 #Latent heat of condensation (J/kg)\nk = 54.0 #Thermal conductivity for 0.5% carbon steel (W/m.K)\nrii = 0.013 #Inner radius of inner pipe of the double pipe heat exchanger (m)\nroi = 0.019 #Outer radius of inner pipe of the double pipe heat exchanger (m)\nRf = 0.0002 #Fouling factor (m^2.K/W)\nUc = 0.00045 #Clean overall heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nDT1 = Ts-t1 #Temperature driving force at end 1 (K)\nDT2 = Ts-t2 #Temperature driving force at end 2 (K)\nDTlm = (DT1-DT2)/(log(DT1/DT2)) #Log mean difference temperature (\u00b0C)\nCw =m*cp #Capacitance rate of water (W/K)\nQ = Cw*(t2-t1) #Heat transfer rate (W)\nQmax1 = Cw*(Ts-t1) #Maximum heat term from the water stream (W)\nQmax2 = m_s*h_vap #Maximum heat term from the steam (W)\nE = Q/Qmax1 #Effectiveness\nLmin = (Q*(log(roi/rii)))/(2*pi*k*(Ts-t1)) #Minimum required length of heat exchanger (m)\nUd = 1.0/(1.0/Uc+Rf) #Dirty overall heat transfer coefficient (W/m^2.K)\n\n#Result:\nprint \"1. The temperature profile of the water and steam along the length of the exchanger is :\",round(DTlm),\" \u00b0C .\"\nprint \"2. Effectiveness of energy from steam to heat the water is :\",round(E,3),\" .\"\nprint \"3. The minimum length of the heat exchanger is :\",round(Lmin,3),\" m .\"\nprint \"4. The dirty overall heat transfer coefficient :\",round(Ud,5),\" W/m^2.K .\"\nprint \"5. U_dirty: \", round(1/Ud,-1),\" W/m^2.K\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The temperature profile of the water and steam along the length of the exchanger is : 47.0 \u00b0C .\n2. Effectiveness of energy from steam to heat the water is : 0.64 .\n3. The minimum length of the heat exchanger is : 0.189 m .\n4. The dirty overall heat transfer coefficient : 0.00045 W/m^2.K .\n5. U_dirty: 2220.0 W/m^2.K\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.15, Page number: 308" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example. Calculate the area and the length of the heat exchanger.'''\n\nfrom math import pi\n\n#Variable declaration:\nQ = 12700.0 #Heat transfer rate (W)\nUd = 2220.0 #Dirty overall heat transfer coefficient (W/m^2.K)\nDTlm = 47.0 #Log mean difference temperature (\u00b0C)\nrii = 0.013 #Inner radius of inner pipe of the double pipe heat exchanger (m)\n#Calculation:\nA = Q/(Ud*DTlm) #Heat transfer area (m^2)\nL = A/(2*pi*rii) #Tube length (m)\n\n#Result:\nprint \"The heat transfer area is :\",round(A,4),\" m^2 .\"\nprint \"The length of the heat exchanger is :\",round(L,2),\" m .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer area is : 0.1217 m^2 .\nThe length of the heat exchanger is : 1.49 m .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.16, Page number: 308" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''With reference to Illustrative Example 15.14, calculate the number of transfer units (NTU) of the\nexchanger.\n'''\n\n#Variable declaration:\nUd = 2220.0 #Dirty overall heat transfer coefficient (W/m^2.K)\nA = 0.1217 #Heat transfer area (m^2)\nCw = 264.0 #Capacitance rate of water (W/K)\n\n#Calculation:\nNTU = (Ud*A)/Cw #Number of transfer units of the exchanger\n\n#Result:\nprint \"The number of transfer units (NTU) of the exchanger is :\",round(NTU,2),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The number of transfer units (NTU) of the exchanger is : 1.02 .\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.18, Page number: 309" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''In 1975, Arthur Andrews(10)\u2014achemical engineering senior at Manhattan College\u2014conducted\na double pipe heat exchanger experiment in the Unit Operations Laboratory. Some of the\nTable 15.4 Wilson Method Information: Illustrative Example 15.18\nRun l/Uo Ao, \u00b0F.h/Btu T, \u00b0F Average temperature l/V, h/ft\n1 1.2176 x 10^-3 124.5 2.29 x 10^-4\n2 0.9454 x 10^-3 125 1.65 x 10^-4\n3 0.9366 x 10^-3 129 1.29 x 10^-4\n4 0.8618 x 10^-3 121.5 1.117 x 10^-4\n5 0.7966 x 10^-3 122 0.962 x 10^-4\nco-current flow data and calculations submitted are presented in Table 15.4. Employing\nAndrew\u2019s data and Wilson\u2019s method, estimate ho for the exchanger fluid. For this lab unit\n(see Figure 15.1), Ao = 1.85 ft^2.\n'''\n\n#Variable declaration:\nAo = 1.85 #Area of heat exchanger (ft^2)\n\n#Calculation:\n#From figure 15.6:\ny = 0.560*10**-3 #Intercept 1/UoAo (\u00b0F.h/Btu)\nho = 1.0/(Ao*y) #Thermal conductivity for heat exchanger (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"Thermal conductivity for the heat exchanger is :\",round(ho),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal conductivity for the heat exchanger is : 965.0 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.19, Page number: 310" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Another experiment(11) on the double pipe heat exchanger in the Unit Operation Laboratory at\nManhattan College produced the following data and calculated results for U (overall heat transfer\ncoefficient) and m\u02d9C (inside cold water flow rate); see Table 15.5.\nTable 15.5 Wilson Experiment Data and Results\nU, Btu/h.ft^2.\u00b0F m\u02d9C, lb/h m_C^-0.8\n687.98 500.97 144.49\n740.74 1001.94 251.58\n748.10 1502.90 347.97\n757.08 2003.87 438.02\n756.31 2204.26 472.73\nUsing the Wilson method, estimate the outside fluid film coefficient. Once again, neglect the\neffects of scale and pipe wall.\n'''\n\n#Variable declaration:\n#From figure 15.7:\na = 0.00126\nb = 0.0276\n\n#Calculation:\nho = 1.0/a #The value of ho (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"Thermal conductivity is :\",round(ho),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Thermal conductivity is : 794.0 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.20, Page number: 311" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous illustrative example. Perform the same calculation but include the resistance\nassociated with the tube (Admiralty metal) if the inside and outside diameters are 1.00 and\n0.902 in, respectively; k for the tube is 60 Btu/h.ft^2.\u00b0F.\nTable 15.6 Wilson Experiment Data and Results\nU, Btu/h.ft^2.\u00b0F 1/U, h.ft^2.\u00b0F m\u02d9C, lb/h m_C^0.8 m_C^-0.8\n687.98 0.0014535 500.97 144.49 0.00692071\n740.74 0.00135 1001.94 251.58 0.003974904\n748.10 0.0013367 1502.90 347.97 0.002873796\n757.08 0.0013209 2003.87 438.02 0.002282992\n756.31 0.0013222 2204.26 472.73 0.002115387\n'''\n\n#Variable declaration:\nDi = 0.902/12.0 #Inside diameter of tube (ft)\nDo = 1.0/12.0 #Outside diameter of tube (ft)\nk = 60.0 #Thermal conductivity of tube (Btu/h.ft^2.\u00b0F) \n\n#Calculation:\n#From example 15.19:\na = 0.00126\nDr = (Do - Di)/2.0 #Radial thickness of tube wall (ft)\nRw = Dr/k #Resistance of wall (Btu/h.\u00b0F)\nho = 1.0/(a-Rw) #The revised ho (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The revised ho is :\",round(ho),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The revised ho is : 839.0 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 15.21, Page number: 312" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Data from experiments based on Wilson\u2019s method produced the following results.\nClean tube:\n1/Uclean = 0.00044 + 0.0050/(m\u02d9)0.8\n\nTube with scale:\n1/Udirty = 0.00089 +0.0050/(m\u02d9)0.8\nCalculate the scale film coefficient if the wall resistance can be neglected.\n'''\n\n#Variable declaration:\na1 = 0.00044 #Term 'a' for U_clean\na2 = 0.00089 #Term 'a' for U_dirty\n\n#Calculation:\nRs = a2 - a1 #Resistance associated with the scale\nhs = 1.0/Rs #Scale film coefficient (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The scale film coefficient neglecting the wall resistance is:\",round(hs),\" Btu/h.ft^2.\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The scale film coefficient neglecting the wall resistance is: 2222.0 Btu/h.ft^2.\u00b0F .\n" + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_16.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_16.ipynb new file mode 100644 index 00000000..ed13cc7c --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_16.ipynb @@ -0,0 +1,230 @@ +{ + "metadata": { + "name": "Chapter 16" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 16: Shell and Tube Heat Exchangers" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.5, Page number: 334" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the shell and tube heat exchanger in Figure 16.13. Calculate the log mean temperature\ndriving force. Also perform the calculation for the exchanger described in Figure 16.14.\n'''\n\nfrom math import log\n\n#Variable declaration:\n#From figure 16.13, for ideal countercurrent heat exchanger:\nT1 = 150.0 #Inlet temperature of hot fluid (\u00b0F)\nT2 = 100.0 #Outet temperature of hot fluid (\u00b0F)\nt1 = 50.0 #Inlet temperature of cold fluid (\u00b0F)\nt2 = 80.0 #Outet temperature of hot fluid (\u00b0F)\n#From figure 16.14, for shell and tube exchanger:\nT_1 = 50.0 #Inlet temperature of cold fluid (\u00b0F)\nT_2 = 80.0 #Outet temperature of hot fluid (\u00b0F)\nt_1 = 150.0 #Inlet temperature of hot fluid (\u00b0F)\nt_2 = 100.0 #Outet temperature of hot fluid (\u00b0F)\n\n#Calculation:\nDT1 = T1 - t2 #Temperature driving force 1 (\u00b0F)\nDT2 = T2 - t1 #Temperature driving force 1 (\u00b0F)\nDTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0F)\nP = (t2-t1)/(T1 - t1) #Dimensionless ratio P\nR = (T1-T2)/(t2-t1) #Dimensionless ratio R\n#From figure 16.7:\nF = 0.925 #Correction Factor\nDTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0F)\n\n#Result:\nprint \"The log mean temperature difference for ideal system is :\",round(DTlm1,1),\" \u00b0F .\"\nprint \"The log mean temperature difference for real system is :\",round(DTlm2,2),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The log mean temperature difference for ideal system is : 59.4 \u00b0F .\nThe log mean temperature difference for real system is : 54.98 \u00b0F .\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.6, Page number: 335" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A shell and tube heat exchanger having two shell passes and four tube passes is being used\nfor cooling. The shell-side fluid enters at 400\u00b0F and leaves at 250\u00b0F and the tube-side fluid\nenters at 100\u00b0F and leaves at 175\u00b0F. What is the log mean temperature difference between the\nhot fluid and the cold fluid?\n'''\n\nfrom math import log\n5\n#Variable declaration:\nT1 = 400.0 #Temperature of fluid entering the shell (\u00b0F)\nT2 = 250.0 #Temperature of fluid leaving the shell (\u00b0F)\nt1 = 100.0 #Temperature of fluid entering the tube (\u00b0F)\nt2 = 175.0 #Temperature of fluid leaving the tube (\u00b0F)\n\n#Calculation:\nDT1 = T1 - T2 #Temperature driving force 1 (\u00b0F)\nDT2 = t2 - t1 #Temperature driving force 1 (\u00b0F)\nDTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0F)\nP = (t2-t1)/(T1 - t1) #Dimensionless ratio P\nR = (T1-T2)/(t2-t1) #Dimensionless ratio R\n#From figure 16.8:\nF = 0.985 #Correction factor\nDTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0F)\n\n#Result:\nprint \"The log mean temperature difference between the hot fluid and the cold fluid is :\",round(DTlm2,1),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The log mean temperature difference between the hot fluid and the cold fluid is : 106.6 \u00b0F .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.7, Page number: 336" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Recalculate the results of the two previous examples using the equation upon which the graphs\nare based.\n'''\n#Variable declaration:\n#From example 16.5:\nP1 = 0.30 #Dimensionless ratio P\nR1 = 1.67 #Dimensionless ratio R\n#From example 16.6:\nP2 = 0.30 #Dimensionless ratio P\nR2 = 1.67 #Dimensionless ratio R\n\n#Calculation:\n#Applying Equation 16.27:\nF1 = 0.92 #Correction Factor\n#Applying Equation 16.33:\nF2 = 0.985 #Correction Factor\n#From example 16.6:\nLMTD1 = 59.4 #Log mean temperature driving force 1 for ideal countercurrent heat exchanger (\u00b0F)\nLMTD2 = 108.0 #Log mean temperature driving force 2 for ideal countercurrent heat exchanger (\u00b0F)\nDTlm1 = F1*LMTD1 #Log mean temperature driving force 1 for shell and tube exchanger (\u00b0F)\nDTlm2 = F2*LMTD2 #Log mean temperature driving force 2 for shell and tube exchanger (\u00b0F)\n\n#Result:\nprint \"The log mean temperature difference for real system (in example 16.5) is :\",round(DTlm1,2),\" \u00b0F .\"\nprint \"The log mean temperature difference for real system (in example 16.6) is :\",round(DTlm2,1),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The log mean temperature difference for real system (in example 16.5) is : 54.65 \u00b0F .\nThe log mean temperature difference for real system (in example 16.6) is : 106.4 \u00b0F .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.8, Page number: 337" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A shell and tube heat exchanger has one pass on the shell side and two passes on the tube side\n(i.e., a 1\u20132 shell and tube heat exchanger). It is being used for oil cooling. The oil flows in the\ntube side. It enters at 110\u00b0C and leaves at 75\u00b0C. The shell-side fluid is water at a flowrate of 1.133\nkg/s, entering at 35\u00b0C and leaving at 75\u00b0C. The heat capacity of the water is 4180 J/kg.K. The\noverall heat-transfer coefficient for the heat exchanger is 350W/m^2.K. The geometry factor F\nhas been previously estimated to be 0.965. Calculate the heat-transfer area requirement for this\nunit in square meters.(9)\n'''\n\nfrom math import log\n\n#Variable declaration:\nt2 = 75.0 #Temperature of water leaving the shell (\u00b0C)\nt1 = 35.0 #Temperature of water enteringing the shell (\u00b0C)\nT2 = 75.0 #Temperature of oil leaving the tube (\u00b0C)\nT1 = 110.0 #Temperature of oil entering the tube (\u00b0C)\nm = 1.133 #Mass flowrate of water (kg/s)\ncp = 4180.0 #Heat capacity of water (J/kg.K)\nF = 0.965 #Correction factor\nU = 350.0 #Overall heat transfer coefficient (W/m^2.K)\n\n#Calculation:\nQ = m*cp*(t2-t1) #Heat load (W)\nDT1 = T1-t2 #Temperature driving force 1 (\u00b0C)\nDT2 = T2-t1 #Temperature driving force 2 (\u00b0C)\nDTlm1 = (DT1-DT2)/log(DT1/DT2)+273.0 #Countercurrent log-mean temperature difference (K)\nDTlm2 = F*DTlm1 #Corrected log-mean temperature difference (K)\nA = Q/(U*DTlm2) #Required heat transfer area (m^2)\n\n#Result:\nprint \"The required heat-transfer area is :\",round(A,3),\" m^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required heat-transfer area is : 1.807 m^2 .\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.10, Page number: 338" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A shell and tube heat exchanger is to be designed for heating 10,000 kg/h of water from 16 \u00b0C to\n84 \u00b0C by a new high heat capacity hot engine oil flowing through the shell. The oil makes a single\nshell pass, entering at 160 \u00b0C and leaving at 94 \u00b0C. Thewater flows through 11 brass tubes of 22.9\nmm inside diameter and 25.4 mm outside diameter with each tube making four passes through\nthe shell. The overall heat transfer coefficient (including the fouling resistance effect) is approximately\n350 W/m^2.\u00b0C and the thermal conductivity of brass is 137 W/m.\u00b0C. Calculate\n1. the heat load, in MW,\n2. the countercurrent flow log mean temperature difference, and\n3. the F correction factor and the corrected log mean temperature difference.\n'''\n\nfrom math import log\n\n#Variable declaration:\nt2 = 84.0 #Temperature of water leaving the tube (\u00b0C)\nt1 = 16.0 #Temperature of water entering the tube (\u00b0C)\nm1 = 10000.0/3600.0 #Mass flowrate of water (kg/s)\nT2 = 94.0 #Temperature of oil leaving the shell (\u00b0C)\nT1 = 160.0 #Temperature of oil entering the shell (\u00b0C)\n\n#Calculation:\nTw = (t1+t2)/2.0 #Average bulk temperature of water (\u00b0C)\nTo = (T1+T2)/2.0 #Average bulk temperature of oil (\u00b0C)\n#From table 16.1:\np1 = 987.0 #Density of water (kg/m^3)\ncp1 = 4176.0 #Heat capacity of water (J/kg.\u00b0C)\np2 = 822.0 #Density of oil (kg/m^3)\nQ = m1*cp1*(t2-t1) #Heat load (W)\ncp2 = 4820.0 #Heat capacity of oil (J/kg.\u00b0C)\nm2 = Q/(cp2*(T1-T2)) #Mass flowrate of oil (kg/s)\nDT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\nDT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\nDTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\nP = (t2-t1)/(T1 - t1) #Dimensionless ratio P\nR = (T1-T2)/(t2-t1) #Dimensionless ratio R\n#From figure 16.7:\nF = 0.965 #Correction factor\nDTlm2 = F*DTlm1 #Log mean temperature driving force for 1-4 shell and tube exchanger (\u00b0C)\n\n#Result:\nprint \"1. The heat load is :\",round(Q/10**6,3),\" MW .\"\nprint \"2. The countercurrent flow log mean temperature difference is :\",round(DTlm1),\" \u00b0C .\"\nprint \"3. The F correction factor and the corrected log mean temperature difference is :\",round(DTlm2,1),\" \u00b0C .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The heat load is : 0.789 MW .\n2. The countercurrent flow log mean temperature difference is : 77.0 \u00b0C .\n3. The F correction factor and the corrected log mean temperature difference is : 74.3 \u00b0C .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.11, Page number: 340" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example. Calculate the area and the length of the tubes required for this\nheat exchanger.\n'''\n\nfrom math import pi\n\n#Variable declaration:\n#From example 16.10:\nU = 350.0 #Over all heat transfer coefficient (W/m^2.\u00b0C)\nDTlm = 74.3 #Log mean temperature driving force for 1-4 shell and tube exchanger (\u00b0C)\nQ = 788800.0 #Heat load (W)\nNt = 11.0 #Number of tubes per pass\nNp = 4.0 #Number of passes\nDi = 0.0229 #Inside diameter of tube (m)\n\n#Calculation:\nA = Q/(U*DTlm) #Heat transfer area required for heat exchanger (m^2)\nN = Nt*Np #Total number of tubes\nL = A/(pi*Di*N) #Tube length (m)\n\n#Result:\nprint \"The heat transfer area required for the heat exchanger is :\",round(A,2),\" m^2 .\"\nprint \"The length of the tubes required for the heat exchanger is :\",round(L*3.28,1),\" ft .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer area required for the heat exchanger is : 30.33 m^2 .\nThe length of the tubes required for the heat exchanger is : 31.4 ft .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.18, Page number: 349" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 16.10. Determine the effectiveness and number of transfer units.'''\n\n#Variable declaration:\n#From example 16.10:\nm1 = 2.778 #Mass flowrate of water (kg/s)\ncp1 = 4176.0 #Heat capacity of water (J/kg.\u00b0C)\ncp2 = 4820.0 #Heat capacity of oil (J/kg.\u00b0C)\nm2 = 2.48 #Mass flowrate of oil (kg/s)\nt2 = 84.0 #Temperature of water leaving the tube (\u00b0C)\nt1 = 16.0 #Temperature of water entering the tube (\u00b0C)\nT2 = 94.0 #Temperature of oil leaving the shell (\u00b0C)\nT1 = 160.0 #Temperature of oil entering the shell (\u00b0C)\nU = 350.0 #Over all heat transfer coefficient (W/m^2.\u00b0C)\nA = 30.33 #Heat transfer area required for heat exchanger (m^2)\n\n#Calculation:\nC1 = m1*cp1 #Capacitance rate of water (W/\u00b0C)\nC2 = m2*cp2 #Capacitance rate of oil (W/\u00b0C)\nQ = C1*(t2-t1) #Heat load of water (W)\nQmax = C1*(T1-t1) #Maximum heat load of water (W)\nE = Q/Qmax #Effectiveness\nif (C1<C2):\n Cmin = C1 #Minimum capacitance rate (W/\u00b0C)\n Cmax = C2 #Maximum capacitance rate (W/\u00b0C)\nelse:\n Cmin = C2 #Minimum capacitance rate (W/\u00b0C)\n Cmax = C1 #Maximum capacitance rate (W/\u00b0C)\nNTU = U*A/Cmin #Number of transfer units\nC = Cmin/Cmax #Capacitance rate ratio\n\n#Result:\nprint \"The effectiveness is :\",round(E,3),\".\"\nprint \"The number of transfer units is :\",round(NTU,3),\".\"\nprint \"The capacitance rate ratio is :\",round(C,3),\".\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The effectiveness is : 0.472 .\nThe number of transfer units is : 0.915 .\nThe capacitance rate ratio is : 0.97 .\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.19, Page number: 351" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The feedwater heater for a boiler supplies 10,000 kg/h of water at 65\u00b0C. The raw feedwater\nenters the tubes with an inlet temperature of 20\u00b0C and is to be heated in a single-shell, twotube\npass heat exchanger by condensing steam at 1.30 bars. Water is on the tube side and\nsteam is on the shell side. The overall heat transfer coefficient is 2000W/m^2.K. Determine\nthe required heat transfer area for the exchanger. Use both the LMTD method and the E-NTU\napproach to obtain the results.\n'''\n\nfrom math import log\n\n#Variable declaration:\n#From table 16.4:\nCw = 11680.3 #Capacitance rate of water (W/\u00b0C)\nt2 = 65.0 #Temperature of water leaving the tube (\u00b0C)\nt1 = 20.0 #Temperature of water entering the tube (\u00b0C)\nT2 = 107.3 #Temperature of steam leaving the shell (\u00b0C)\nT1 = 107.3 #Temperature of steam entering the shell (\u00b0C)\nhv = 2.238*10**6 #Latenet heat of condensation for steam (J/kg)\nU = 2000.0 #Overall heat transfer coefficient (W/m^2.\u00b0C)\n\n#Calculation:\nQ = Cw*(t2-t1) #Heat load (W)\nm2 = Q/hv #Steam condensation rate (kg/s)\nDT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\nDT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\nDTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\nF = 1.0 #Correction factor (since, T2 = T1)\nDTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0C)\nA1 = Q/(U*DTlm2) #Heat transfer area using LMTD method (m^2)\nE = (t2-t1)/(T1-t1) #Effectiveness\n#From figure 16.18:\nNTU = 0.7 #Number of transfer units\nA2 = (NTU*Cw)/U #Heat transfer area using E-NTU method (m^2)\n\n#Result:\nprint \"The heat transfr area for the exchanger (using LMTD method) is :\",round(A1,2),\" m^2 .\"\nprint \"The heat transfr area for the exchanger (using E-NTU method) is :\",round(A2,1),\" m^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfr area for the exchanger (using LMTD method) is : 4.23 m^2 .\nThe heat transfr area for the exchanger (using E-NTU method) is : 4.1 m^2 .\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.21, Page number: 353" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A shell and tube heat exchanger has one pass on the shell side and two passes on the tube side\n(i.e., 1\u20132 shell and tube heat exchanger). It is being used for oil cooling. The oil flows in the\ntube side. It enters at 110\u00b0C and leaves at 75\u00b0C. The shell-side fluid is water at a flow rate of\n1.133 kg/s, entering at 35\u00b0C and leaving at 75\u00b0C. The oil flows in the tubes at an average\nvelocity of 0.3 m/s. The tube inside diameter is 3/4 inch or 0.01905 m. The overall heat transfer\ncoefficient for a clean heat exchanger is 350W/m^2.K. The fouling factor, Rf, is 0.00027 m2 .\nK/W. The oil properties are: p_oil = 850 kg/m3 and c_oil = 1900 J/kg.K. Calculate:\n1. the mass flow rate of the oil\n2. the minimum and maximum heat capacity rate\n3. the heat load, Q\u02d9\n4. the total number of tubes\n5. the tube length.\n'''\n\nfrom math import pi,log\n\n#Variable declaration:\n#From table 16.5:\nt2 = 75.0 #Temperature of water leaving the shell (\u00b0C)\nt1 = 35.0 #Temperature of water entering the shell (\u00b0C\nT2 = 75.0 #Temperature of oil leaving the tube (\u00b0C)\nT1 = 110.0 #Temperature of oil entering the tube (\u00b0C)\nmw = 1.133 #Mass flowrtae of water (kg/s)\ncpw = 4180.0 #Heat capacity of water (J/kg.K)\ncpo = 1900.0 #Heat capacity of oil (J/kg.K)\np = 850.0 #Density of oil (kg/m^3)\nDi = 0.01905 #Inside diameter of tube (m)\nV = 0.3 #Average velocity of oil flow inside the tube (m/s)\nNp = 2.0 #Number of passes\nUc = 350.0 #Overall heat transfer coefficient for clean heat exchanger (W/m^2)\nRf = 0.00027 #Fouling factor (m^2.K/w)\n\n#Calculation:\nCw = mw*cpw #Water capacitance rate (W/K)\nQ = Cw*(t2-t1) #Heat load (W)\nCo = Q/(T1-T2) #Oil capacitance rate (W/K)\nmo = Co/cpo #Total flowrate of oil (kg/s)\nif (Cw<Co):\n Cmin = Cw #Minimum capacitance rate (W/K)\n Cmax = Co #Maximum capacitance rate (W/K)\nelse:\n Cmin = Co #Minimum capacitance rate (W/K)\n Cmax = Cw #Maximum capacitance rate (W/K)\nm_ot = p*V*(pi/4.0)*Di**2 #Oil flowrate per tube (kg/s)\nNt = mo/m_ot #Number of tubes per pass\nN = Nt*Np #Number of tubes\nDT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\nDT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\nDTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\nP = (t2-t1)/(T1 - t1) #Dimensionless parameter P\nR = (T1-T2)/(t2-t1) #Dimensionless parameter R\n#From figure 16.7:\nF = 0.81 #Correction factor\nDTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0C)\nUd = 1.0/(1.0/Uc+Rf) #Dirty overall heat transfer coefficient (W/m^2.K)\nA = Q/(Ud*DTlm2) #Required heat transfer area (m^2)\nL = A/(N*pi*Di) #Tube length (m)\n\n#Result:\nprint \"1. The mass flow rate of the oil is :\",round(mo,2),\" kg/s .\"\nprint \"2. The minimum and maximum heat capacity rate is :\",round(Cmin),\" and \",round(Cmax,1),\" W/K .\"\nprint \"3. The heat load, Q is :\",round(Q),\" W .\"\nprint \"4. The total number of tubes is :\",round(N,-1),\".\"\nprint \"5. The tube length is :\",round(L,1),\" m .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The mass flow rate of the oil is : 2.85 kg/s .\n2. The minimum and maximum heat capacity rate is : 4736.0 and 5412.5 W/K .\n3. The heat load, Q is : 189438.0 W .\n4. The total number of tubes is : 80.0 .\n5. The tube length is : 4.2 m .\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 16.22, Page number: 356" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine the effectiveness and NTU for the previous example.'''\n\n#Variable declaration:\n#From example 16.22:\nt2 = 75.0 #Temperature of water leaving the shell (\u00b0F)\nt1 = 35.0 #Temperature of water entering the shell (\u00b0F)\nT2 = 75.0 #Temperature of oil leaving the tube (\u00b0F)\nT1 = 110.0 #Temperature of oil entering the tube (\u00b0F)\nU = 320.0 #Overall heat transfer coefficient (W/m^2.K)\nA = 19.5 #Required heat transfer area (m^2)\nCmin = 4736.0 #Minimum capacitance rate (W/K)\n\n#Calculation:\nDT1 = t2-t1 #Actual water temperature change (\u00b0F)\nDT2 = T1 - t1 #Maximum water temperature change (\u00b0F)\nE = DT1/DT2 #Effectiveness\nNTU = (U*A)/Cmin #Number of transfer units\n\n#Result:\nprint \"The effectiveness is :\",round(E,3),\".\"\nprint \"The NTU is :\",round(NTU,3),\".\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The effectiveness is : 0.533 .\nThe NTU is : 1.318 .\n" + } + ], + "prompt_number": 22 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_17.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_17.ipynb new file mode 100644 index 00000000..16924e48 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_17.ipynb @@ -0,0 +1,335 @@ +{ + "metadata": { + "name": "Chapter 17" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 17: Fins and Extended Surfaces" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.1, Page number: 358" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the longitudinal (rectangular) fin pictured in Figure 17.1. Estimate the fin face area,\nneglecting the (top) area contribution associated with the fin thickness if w = 1 ft and L =\n1.5 in. The fin thickness is 0.1 in. Also calculate the total area of the fin.\n'''\n\n#Variable declaration:\nw1 = 1.5 #Thicknessof fin (in)\nL = 12.0 #Length of fin (in)\nw2 = 0.1 #Thickness of fin(in)\n\n#Calculation:\nAf = 2*w1*L #Face area of fin (in^2)\nAt = Af + L*w2 #Total area of fin (in^2)\n\n#Result:\nprint \"The face area of the fin is :\",round(Af),\" in^2 .\"\nprint \"The face area of the fin is :\",round(Af/12**2,2),\" ft^2 .\"\nprint \"The total area of the fin is :\",round(At,1),\" in^2 .\"\nprint \"The total area of the fin is :\",round(At/12**2,3),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The face area of the fin is : 36.0 in^2 .\nThe face area of the fin is : 0.25 ft^2 .\nThe total area of the fin is : 37.2 in^2 .\nThe total area of the fin is : 0.258 ft^2 .\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.3, Page number: 359" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Figure 17.2. Estimate the fin area, neglecting the area contribution associated with the\nfin thickness if r0 = 4.0 in, rf = 6.0 in, and t = 0.1 in.\n'''\n\nfrom math import pi\n\n#Variable declaration:\nrf = 6.0/12.0 #Outside radius of fin (ft)\nro = 4.0/12.0 #Outside radius of pipe (ft)\nt = 0.1/12.0 #Thickness of fin (ft)\n\n#Calculation:\nAf = 2*pi*(rf**2-ro**2) #Face area of fin (ft^2)\nAt = Af + 2*pi*rf*t #Total area of fin (ft^2)\n\n#Result:\nprint \"The total fin area is :\",round(At,3),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total fin area is : 0.899 ft^2 .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.4, Page number: 364" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following information is provided for a straight rectangular fin: h = 15 W/m^2.K,\nk = 300 W/m.K, L = 3 in, and t = 1 in. Estimate the fin efficiency.\n'''\n\nfrom math import sqrt\n\n#Variable declaration:\nL = 3.0*0.0254 #Height of fin (m)\nt = 1.0*0.0254 #Thickness of fin (m)\nh = 15.0 #Heat transfer coefficient (W/m^2.K)\nk = 300.0 #Thermal conductivity (W/m.K)\n\n#Calculation:\nLc = L + t/2.0 #Corrected height of fin (m)\nAp = Lc*t #Profile area of fin (m^2)\nx = sqrt((Lc**3*h)/(k*Ap)) #x-coordinate of figure 17.3\n#From figure 17.3:\nnf = 98.0 #Fin efficiency\n\n#Result:\nprint \"The fin efficiency is :\",nf,\"% .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin efficiency is : 98.0 % .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.5, Page number: 365" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Estimate the fin efficiency in the previous example using the equation developed by\nBarkwill et al.(3)\n'''\n\n#Variable declaration:\n#From example 17.4:\nX = 0.1246 #X-coordinate of figure 17.3\n\n#Calculation:\n#Applying equation (A) from Table 17.3:\nY = 4.5128*X**3 - 10.079*X**2 - 31.413*X + 101.47\n\n#Result:\nprint \"The fin efficiency is :\",round(Y,1),\"% .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin efficiency is : 97.4 % .\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.6, Page number: 365" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A set of micro-fins is designed to cool an electronic circuit. Each micro-fin has a square\ncross-section of 0.2 cm by 0.2 cm and a length of 1 cm. The conductivity of the fin material\nis 400 W/m.K and the air heat transfer coefficient is 16 W/m^2.K. The circuit temperature\nis 100\u00b0C and the air temperature is 25\u00b0C. Calculate the heat transfer from each micro-fin in W.\n'''\n\nfrom math import sqrt,atan\n\n#Variable declaration:\nw = 0.2/100.0 #Width of fin (m)\nt = 0.2/100.0 #Thickness of fin (m)\nL = 1.0/100.0 #Length of fin (m)\nh = 16.0 #Heat transfer coefficient (W/m^2.K)\nk = 400.0 #Thermal conductivity of fin (W/m.K)\nTc = 100.0 #Circuit temperature (\u00b0C)\nTa = 25.0 #Air temperature (\u00b0C)\n\n#Calculation:\nP = 4*w #Fin cross-section parameter (m)\nAc = w*t #Cross-sectional area of fin (m^2)\nLc = L+Ac/P #Corrected height of fin (m)\nm = sqrt((h*P)/(k*Ac)) #Location of minimum temperature (m^-1)\nQ = (sqrt(h*P*k*Ac))*(Tc-Ta)*atan(h)*(m*Lc) #Heat transfer from each micro-fin (W)\n\n#Result:\nprint \"The heat transfer from each micro-fin is :\",round(Q,2),\" W .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer from each micro-fin is : 0.15 W .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.8, Page number: 366" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Air and water are separated by a 1.5 mm plane wall made of steel (k = 38 W/m.K; density, r =\n7753 kg/m^3; heat capacity, cp = 486 J/kg.K). The air temperature, T4, is 19\u00b0C, and the water\ntemperature, T4, is 83\u00b0C. Denote the temperature at the air\u2013wall interface T1 and let T3 be the\ntemperature at the wall\u2013water interface. The air-side heat transfer coefficient, h1, is 13 W/m^2.K\nand the water side heat transfer coefficient, h3, is 260 W/m^2.K. Assume an area of the wall that\nis 1 m high and 1 m wide as a basis.\n1. Show whether the conduction resistance may be neglected.\n2. What is the rate of heat transfer from water to air?\nTo increase the rate of heat transfer, it is proposed to add steel fins to the wall. These straight\nrectangular steel fins will be 2.5 cm long, 1.3 mm thick, and will be spaced such that the fin\npitch, S, is 1.3 cm between centers.\n3. Calculate the percent increase in steady-state heat transfer rate that can be realized by\nadding fins to the air side of the plane wall.\n4. Calculate the percent increase in steady-state heat transfer rate that can be realized by\nadding fins to the water side of the plane wall.\n'''\nfrom __future__ import division\nfrom math import sqrt\n#Variable declaration:\nh1 = 13.0 #Air-side heat transfer coefficient (W/m^2.K)\nA = 1.0 #Base wall area (m^2)\nL = 2.5/100 #Length of steel fins (m)\nL2 = 1.5/10**3 #Length of steel wall (m)\nk = 13.0 #Thermal conductivity of fin (W/m.K)\nk1 = 38.0 #Thermal conductivity of steel wall (W/m.K)\nh2 = 260.0 #Water side heat transfer coefficient (W/m^2.K)\nT4 = 19.0 #Air temperature (\u00b0C)\nT1 = 83.0 #Water temperature (\u00b0C)\nt = 1.3/10**3 #Thickness of steel fins (m)\nw = 1.0 #Width of wall (m)\nS = 1.3/100 #Fin pitch(m)\n\n#Calculation:\nR1 = 1/(h1*A) #Air resistance (\u00b0C/W) (part 1)\nR2 = L2/(k1*A) #Conduction resistance (\u00b0C/W)\nR3 = 1/(h2*A) #Water resistance (\u00b0C/W)\nRt = (R1+R3) #Total resistance (\u00b0C/W) (part 2)\nQ = (T1-T4)/Rt #Total heat transfer (W)\nNf = 1/S #Number of fins (part 3)\nLbe = w - Nf*t #Unfinned exposed base surface\nAbe = w*Lbe #Exposed base surface area (m^2)\nLc = L+t/2 #Corrected length (m)\nAp = Lc*t #Profile area (m^2)\nAf = 2*w*Lc #Fin surface area (m^2)\nBi = h1*(t/2)/k1 #Biot number\na = sqrt(Lc**3*h1/(k*Ap)) #Abscissa of the fin efficiency\n#From figure 17.3:\nnf = 0.88 #Fin efficiency\nRb = 1/(h1*Abe) #Air thermal resistance of base wall (\u00b0C/W)\nRf = 1/(h1*Nf*Af*nf) #Air thermal resistance of fins (\u00b0C/W)\nRT1 = 1/(1/Rb+1/Rf) #Total outside resistance of the fin array (\u00b0C/W)\nRt3 = RT1+R3 #Total resistance on air side fins (\u00b0C/W)\nQt = (T1-T4)/round(Rt3,5) #Heat transfer rate on air side fins (W)\nI = (Qt/Q - 1)*100 #Percent increase in heat transfer rate to air side fins (W)\nA = sqrt(Lc**3*h2/(k1*Ap)) #Abscissa of the new fin efficiency (part 4)\n#From figure 17.3:\nnf2 = 38.0 #New fin efficiency\nRb2 = 1/(h2*Abe) #Thermal resistance of base wall (\u00b0C/W)\nRf2 = 1/(h2*Nf*Af*nf2) #Thermal resistance of fins (\u00b0C/W)\nRt4 = 1/(1/Rb2+1/Rf2) #Total resistance of the finned surface (\u00b0C/W)\nRt5 = R1+Rt4 #Total resistance on water side fins (\u00b0C/W)\nQT1 = (T1-T4)/Rt5 #Heat transfer rate on water side fins (W)\nI2 = (QT1/Q - 1)*100 #Percent increase in heat transfer rate to water side fins (W)\n\n#Result:\nif (R2<R1 or R2<R3):\n print \"1. The conduction resistance may be neglected.\"\nelse:\n print \"1. The conduction resistance can not be neglected.\"\nprint \"2. The rate of heat transfer from water to air is :\",round(Q,1),\" W .\"\nprint \"3. The percent increase in steady-state heat transfer rate by adding fins to the air side of the plane wall is :\",round(I,1),\" % .\"\nprint \"4. The percent increase in steady-state heat transfer rate by adding fins to the water side of the plane wall is :\",round(I2,1),\" % .\"\nprint \"____There is a calculation mistake in book in calculating Qt(83-19/0.0214 = 2999), hence slight differences in answer______\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The conduction resistance may be neglected.\n2. The rate of heat transfer from water to air is : 792.4 W .\n3. The percent increase in steady-state heat transfer rate by adding fins to the air side of the plane wall is : 276.7 % .\n4. The percent increase in steady-state heat transfer rate by adding fins to the water side of the plane wall is : 5.0 % .\n____There is a calculation mistake in book in calculating Qt(83-19/0.0214 = 2999), hence slight differences in answer______\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.10, Page number: 369" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A circular tube has an outside diameter of 2.5 cm and a surface tempeature, Tb, of 170\u00b0C. An\nannular aluminum fin of rectangular profile is attached to the tube. The fin has an outside\nradius, rf, of 2.75 cm, a thickness, t, of 1 mm, and a thermal conductivity, k, of 200 W/m . K.\nThe surrounding fluid is at a temperature Too = 25\u00b0C and the associated heat transfer coefficient,\nh, is 130 W/m^2.K. Calculate the heat transfer rate without the fin,Q\u02d9w/o,f, the corrected\nlength, Lc, the outer radius, r2c, the maximum heat transfer rate from the fin,Q\u02d9f,max, the fin efficiency,\nhf, the fin heat transfer rate, qf, and the fin thermal resistance, Rt,f.\n'''\n\nfrom __future__ import division\nfrom math import pi,sqrt\n\n#Variable declaration:\nDo = 2.5/100 #Outside diameter of tube (m)\nt = 1/10**3 #Thickness of fin (m)\nT = 25 #Fluid temperature (\u00b0C)\nTb = 170 #Surface temperature (\u00b0C)\nh = 130 #Heat transfer coefficient (W/m^2.K)\nk = 200 #Thermal conductivity of fin (W/m.K)\nrf = 2.75/100 #Outside radius of fin (m)\n\n#Calculation:\nro = Do/2 #Radius of tube (m)\nAb = 2*pi*ro*t #Area of the base of the fin (m^2)\nTe = Tb-T #Excess temperature at the base of the fin (K)\nQ1 = h*Ab*Te #Total heat transfer rate without the fin (W)\nBi = h*(t/2)/k #Biot number\nL = rf-ro #Fin height (m)\nrc = rf+t/2 #Corrected radius (m)\nLc = L+t/2 #Corrected height (m)\nAp = Lc*t #Profile area (m^2)\nAf = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)\nQm = h*Af*Te #Maximum fin heat transfer rate (W)\nA = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency\nC = rf/ro #Curve parameter of fin efficiency\n#From figure 17.4:\nnf = 0.86 #Fin efficiency\nQf = nf*Qm #Fin heat transfer rate (W)\nR = Te/Qf #Fin resistance (K/W)\n\n#Result:\nprint \"1. The heat transfer rate without the fin is :\",round(Q1,2),\" W .\"\nprint \"Or, the heat transfer rate without the fin is :\",round(Q1*3.412),\" Btu/h .\"\nprint \"2. The corrected length is :\",round(Lc,4),\" m .\"\nprint \"3. The outer radius is :\",round(rc,3),\" m .\"\nprint \"4. The maximum heat transfer rate from the fin is :\",round(Qm,2),\" W .\" \nprint \"5. The fin efficiency is :\",round(nf*100),\" % .\" \nprint \"6. The fin heat transfer rate is :\",round(Qf),\" W .\"\nprint \"Or, the fin heat transfer rate is :\",round(Qf*3.412),\" Btu/h .\"\nprint \"7. The fin thermal resistance is :\",round(R,2),\" K/W .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The heat transfer rate without the fin is : 1.48 W .\nOr, the heat transfer rate without the fin is : 5.0 Btu/h .\n2. The corrected length is : 0.0155 m .\n3. The outer radius is : 0.028 m .\n4. The maximum heat transfer rate from the fin is : 74.35 W .\n5. The fin efficiency is : 86.0 % .\n6. The fin heat transfer rate is : 64.0 W .\nOr, the fin heat transfer rate is : 218.0 Btu/h .\n7. The fin thermal resistance is : 2.27 K/W .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.11, Page number: 370" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 17.10. Calculate the fin effectiveness, 1f, and whether the use of\nthe fin is justified.\n'''\n\n#Variable declaration:\n#From example 17.10:\nQf = 64 #Fin heat transfer rate (W)\nQ1 = 1.48 #Total heat transfer rate without the fin (W)\n\n#Calculation:\nE = Qf/Q1 #Fin effectiveness\n\n#Result:\nprint \"The fin effectiveness is :\",round(E,1),\" .\"\nif E>2:\n print \"Hence, the use of the fin is justified.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin effectiveness is : 43.2 .\nHence, the use of the fin is justified.\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.12, Page number: 370" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''If the tube described in Illustrative Example 17.10 had a length of one meter and fin pitch of\n10 mm, what would be the total surface area for heat transfer, the exposed tube base total\nheat transfer rate,Q\u02d9t, the overall efficiency of the surface, and the overall surface effectiveness?\nNote: This problem is an extension of Illustrative Example 17.10. The information is\nobtained from it and its solution is used in the solution of this problem.\n'''\n\nfrom __future__ import division\nfrom math import pi\n\n#Variable declaration:\nw = 1 #Length of tube (m)\nS = 10/10**3 #Fin patch (m)\n#From example 17.10:\nt = 1/10**3 #Thickness of fin (m)\nro = 0.0125 #Radius of tube (m)\nAf = 3.94*10**-3 #Fin surface area (m^2)\nTb = 145 #Excess temperature at the base of the fin (K)\nh = 130 #Heat transfer coefficient (W/m^2.K)\nQf = 64 #Fin heat transfer rate (W)\n\n#Calculation:\nNf = w/S #Number of fins in tube length\nwb = w-Nf*t #Unfinned base length (m)\nAb = 2*pi*ro*wb #Unfinned base area (m^2)\nAt =Ab+Nf*Af #Total transfer surface area (m^2)\nQt = h*(2*pi*ro*w*Tb) #Total heat rate without fins (W)\nQb = h*Ab*Tb #Heat flow rate from the exposed tube base (W)\nQft = Nf*Qf #Heat flow rate from all the fins (W)\nQt2 = Qb+Qft #Total heat flow rate (W)\nQm = h*At*Tb #Maximum heat transfer rate (W)\nno = Qt2/Qm #Overall fin efficiency\nEo = Qt2/Qt #Overall effectiveness\nRb = 1/(h*Ab) #Thermal resistance of base (K/W)\nRf = 1/(h*Nf*Af*no) #Thermal resistance of fins (K/W)\n\n#Result:\nprint \"1. The total surface area for heat transfer is :\",round(At,3),\" m^2 .\"\nprint \"2. The exposed tube base total heat transfer rate is :\",round(Qb,1),\" W .\"\nprint \"Or, the exposed tube base total heat transfer rate is :\",round(Qb*3.412),\" Btu/h .\"\nprint \"3. The overall efficiency of the surface is :\",round(no*100,1),\" % .\"\nprint \"4. The overall surface effectiveness is :\",round(Eo,2),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The total surface area for heat transfer is : 0.465 m^2 .\n2. The exposed tube base total heat transfer rate is : 1332.4 W .\nOr, the exposed tube base total heat transfer rate is : 4546.0 Btu/h .\n3. The overall efficiency of the surface is : 88.3 % .\n4. The overall surface effectiveness is : 5.22 .\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.13, Page number: 374" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Consider the case of aluminum fins of triangular profile that are attached to a plane wall with a\nsurface temperature is 250\u00b0C. The fin base thickness is 2 mm and its length is 6 mm. The system\nis in ambient air at a temperature of 20\u00b0C and the surface convection coefficient is 40W/m2 . K.\nConsider a 1 m width of a single fin. Determine:\n1. the heat transfer rate without the fin,\n2. the maximum heat transfer rate from the fin, and\n3. the fin efficiency, thermal resistance, and effectiveness.\nProperties of the aluminum may be evaluated at the average temperature, (Tb + T1)/2 =\n(250 + 20)/2 = 135\u00b0C = 408 K, where k \u2243 240 W/m.K.\n'''\n\nfrom __future__ import division\nfrom math import sqrt\n\n#Variable declaration:\nw = 1 #Width of single of fin (m)\nt = 2/10**3 #Fin base thickness (m)\nl = 6/10**3 #Fin length thickness (m)\nT1 = 250 #Surface temperature (\u00b0C)\nT2 = 20 #Ambient air temperature (\u00b0C)\nh = 40 #Surface convection coefficient (W/m^2.K)\nk = 240 #Thermal conductivity of fin (W/m.K)\n\n#Calculation:\nAb = t*w #Base area of the fin (m^2)\nTe = T1-T2 #Excess temperature at the base of the fin (K)\nQw = h*Ab*Te #Heat transfer rate without a fin (W)\nAf = 2*w*(sqrt(l**2-(t/2)**2)) #Fin surface area (m^2)\nQm = h*Af*Te #Maximum heat transfer rate (m^2)\nBi = h*(t/2)/k #Biot number\nLc = l #Corrected length (m)\nAp = l*t/2 #Profile area (m^2)\nA = sqrt((Lc**3*h)/k*Ap) #Abscissa for the fin efficiency figure\n#From figure 17.4:\nnf = 0.99 #Fin efficiency\nQf = nf*Qm #Fin heat transfer rate (W)\nR = Te/Qf #Fin thermal resistance (K/W)\nE = Qf/Qw #Fin effectiveness\n\n#Result:\nprint \"1. The heat transfer rate without the fin is :\",round(Qw,1),\" W .\"\nprint \"2. The maximum heat transfer rate from the fin is :\",round(Qm,-1),\" W .\"\nprint \"3. The fin efficiency is :\",round(nf*100),\" % .\"\nprint \" The fin thermal resistance is :\",round(R,1),\" \u00b0C/W .\"\nprint \" The fin effectiveness is :\",round(E,1),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The heat transfer rate without the fin is : 18.4 W .\n2. The maximum heat transfer rate from the fin is : 110.0 W .\n3. The fin efficiency is : 99.0 % .\n The fin thermal resistance is : 2.1 \u00b0C/W .\n The fin effectiveness is : 5.9 .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.14, Page number: 375" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine whether the use of the fin is justified in the previous example.'''\n\n#Variable declaration:\n#From example 17.13:\nQf = 108.9 #Fin heat transfer rate (W)\nQw = 18.4 #Total heat transfer rate without the fin (W)\n\n#Calculation:\nE = Qf/Qw #Fin effectiveness\n\n#Result:\nprint \"The fin effectiveness is :\",round(E,2),\" .\"\nif E>2:\n print \"Hence, the use of the fin is justified.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin effectiveness is : 5.92 .\nHence, the use of the fin is justified.\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.15, Page number: 375" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Annular aluminum fins of rectangular profile are attached to a circular tube. The outside diameter\nof the tube is 50 mm and the temperature of its outer surface is 200\u00b0C. The fins are 4 mm\nthick and have a length of 15 mm. The system is in ambient air at a temperature of 20\u00b0C and the\nsurface convection coefficient is 40 W/m^2.K. The thermal conductivity of aluminum is\n240 W/m.K. What are the efficiency, thermal resistance, effectiveness, and heat transfer rate\nof a single fin? Is the use of the fin justified?\n'''\n\nfrom __future__ import division\nfrom math import pi,sqrt\n\n#Variable declaration:\nDo = 50/10**3 #Outside diameter of tube (m)\nt = 4/10**3 #Thickness of fin (m)\nT = 20 #Fluid temperature (\u00b0C)\nTb = 200 #Surface temperature (\u00b0C)\nh = 40 #Heat transfer coefficient (W/m^2.K)\nk = 240 #Thermal conductivity of fin (W/m.K)\nl = 15/10**3 #Length of fin (m)\n\n#Calculation:\nro = Do/2 #Radius of tube (m)\nrf = ro+l #Outside radius of fin (m)\nAb = 2*pi*ro*t #Area of the base of the fin (m^2)\nTe = Tb-T #Excess temperature at the base of the fin (K)\nQ1 = h*Ab*Te #Total heat transfer rate without the fin (W)\nBi = h*(t/2)/k #Biot number\nL = rf-ro #Fin height (m)\nrc = rf+t/2 #Corrected radius (m)\nLc = L+t/2 #Corrected height (m)\nAp = Lc*t #Profile area (m^2)\nAf = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)\nQm = h*Af*Te #Maximum fin heat transfer rate (W)\nA = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency\nC = rf/ro #Curve parameter of fin efficiency\n#From figure 17.4:\nnf = 0.97 #Fin efficiency\nQf = nf*Qm #Fin heat transfer rate (W)\nR = Te/Qf #Fin resistance (K/W)\nE = Qf/Q1 #Fin effectiveness\n\n#Result:\nprint \"The fin efficiency is :\",round(nf*100),\" % .\"\nprint \"The fin thermal resistance is :\",round(R,1),\" \u00b0C/W .\"\nprint \"The fin effectiveness is :\",round(E,2),\" .\"\nprint \"The maximum heat transfer rate from a single fin is :\",round(Qm,2),\" W .\"\nif E>2:\n print \"Since Ef = FCP>2, the use of the fin is justified.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin efficiency is : 97.0 % .\nThe fin thermal resistance is : 3.6 \u00b0C/W .\nThe fin effectiveness is : 11.05 .\nThe maximum heat transfer rate from a single fin is : 51.53 W .\nSince Ef = FCP>2, the use of the fin is justified.\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.16, Page number: 376" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''What is the rate of heat transfer per unit length of tube in the previous illustrative example, if\nthere are 125 such fins per meter of tube length? Also calculate the total efficiency and\neffectiveness.\n'''\n\nfrom __future__ import division\nfrom math import pi,sqrt\n\n#Variable declaration:\nNf = 125 #Array of fins per meter\nw = 1 #Length of fin (m)\n#From example 17.15:\nt = 4/10**3 #Thickness of fin (m)\nDo = 50/10**3 #Outside diameter of tube (m)\nAf = 7.157*10**-3 #Fin surface area (m^2)\nh = 40 #Heat transfer coefficient (W/m^2.K)\nDTb = 180 #Excess temperature at the base of the fin (K)\nQf = 50 #Fin heat transfer rate (W)\n\n#Calculation:\nro = Do/2 #Radius of tube (m)\nwb = w-Nf*t #Unfinned exposed base length (m)\nAb = 2*pi*ro*wb #Area of the base of the fin (m^2)\nAt = Ab+Nf*Af #Total heat transfer surface area (m^2)\nQw = h*(2*pi*ro*w)*DTb #Heat rate without fin (W)\nQb = h*Ab*DTb #Heat rate from the base (W)\nQft = Nf*Qf #Heat rate from the fin (W)\nQt = Qb+Qft #Total heat rate (W)\nQm = h*At*DTb #Maximum heat transfer rate (W)\nn = Qt/Qm #Overall fin efficiency\nE = Qt/Qw #Overall fin effectiveness\nRb = 1/(h*Ab) #Thermal resistance of base (\u00b0C/W)\nRf = 1/(h*Nf*Af*n) #Thermal resistance of fin (\u00b0C/W)\n\n#Result:\nprint \"The rate of heat transfer per unit length of tube is :\",round(Qt,1),\" W .\"\nprint \"Or, the rate of heat transfer per unit length of tube is :\",round(Qt/10**3,2),\" kW .\"\nprint \"The overall fin efficiency is :\",round(n*100,1),\" % .\"\nprint \"The overall fin effectiveness is :\",round(E,2),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of heat transfer per unit length of tube is : 6815.5 W .\nOr, the rate of heat transfer per unit length of tube is : 6.82 kW .\nThe overall fin efficiency is : 97.3 % .\nThe overall fin effectiveness is : 6.03 .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.17, Page number: 377" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A metal fin 1 inch high and 1/8 inch thick has a thermal conductivity, k, of 25 Btu/h.ft.\u00b0F and a\nuniform base temperature of 250\u00b0F. It is exposed to an air stream at 60\u00b0F with a velocity past\nthe fin such that the convection coefficient of heat transfer, h, is 15 Btu/h.ft.\u00b0F. Calculate the\ntemperature at the tip of the fin and the heat transfer from the fin per foot of fin. Solve the problem\nanalytically.\n'''\n#Variable declaration:\nprint 'Analytical Solution'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Analytical Solution\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 17.18, Page number: 379" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Estimate the fin efficiency in the previous example, if the efficiency is defined (as noted earlier)\nas the actual heat transfer divided by the heat rate of the entire fin with the same temperature as its\nbase.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\n#From example 17.18:\nT = 250 #Base temperature of fin (\u00b0F)\nh = 15 #Convection coefficient of heat transfer (Btu/h.ft.\u00b0F)\nw = 1 #Base width of fin (ft)\nt = 1 #Thickness of fin (in)\nH = 1/8 #Height of fin (in)\nl = 1 #Length of fin (in)\nQ = 357.2 #Heat transfer rate (Btu/h.ft)\n\n#Calculation:\nA = (l*w+t*w+H*w)/12 #Heat transfer area of fin (ft^2)\nQm = h*A*(T-70) #Maximum heat transfer rate (Btu/h.ft)\nn = Q/Qm*100 #Fin efficiency\n\n#Result:\nprint \"The fin efficiency is :\",round(n,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The fin efficiency is : 74.7 % .\n" + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_18.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_18.ipynb new file mode 100644 index 00000000..d5d13bdf --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_18.ipynb @@ -0,0 +1,272 @@ +{ + "metadata": { + "name": "Chapter 18" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 18: Other Heat Exchange Equipment" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.2, Page number: 384" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the temperature when 50 lb of pure H2SO4 at 25\u00b0C is mixed adiabatically with 200 lb\nof a 50% by weight aqueous H2SO4 solution at 25\u00b0C. Also calculate the heat effect if the final\nmixture is at 25\u00b0C.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nT1 = 25 #Temperature of H2SO4 (\u00b0C)\nm = 50+200 #Mass of H2SO4 (lb)\n#From figure 18.2:\nW1 = 50+100 #Weight of H2SO4 (lb)\nW2 = 100 #Weight of H2O (lb)\n\n#Calculation:\nm = W1/(W1+W2)*100 #Percent weight of H2SO4 (%)\nm2 = W1+W2 #Mass of mixture (lb)\n#From fgure 18.2:\nT2 = 140 #Final temperature between the 50% solution and pure H2SO4 at 25\u00b0C (\u00b0F)\nh1 = -86 #Specific heat capacity of H2O (Btu/lb)\nh2 = -121.5 #Specific heat capacity of H2SO4 (Btu/lb)\nQ = m2*(h2-h1) #Heat transferred (Btu)\n\n#Result:\nprint \"The final temperature between the 50% solution and pure H2SO4 at 25\u00b0C is :\",round(T2),\" \u00b0F .\"\nprint \"The heat transferred is :\",round(Q),\" Btu .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The final temperature between the 50% solution and pure H2SO4 at 25\u00b0C is : 140.0 \u00b0F .\nThe heat transferred is : -8875.0 Btu .\n" + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.3, Page number: 386" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A single-effect evaporator is to concentrate 10,000 lb/h of a 10% NaOH solution to 75%. The\nfeed enters at 120\u00b0F and the evaporator is to operate at an absolute pressure of 14.7 psi. The 75%\nNaOH solution leaves at the evaporator equilibrium temperature. For what heat transfer rate\n(Btu/h) should the evaporator be designed? Also calculate the area requirement in the evaporator\nif the overall heat transfer coefficient is 500 Btu/h.ft^2.\u00b0F and 103 psig (340\u00b0F) saturated\nsteam is employed in the steam chest. The NaOH\u2013H2O enthalpy-concentration diagram is provided\nin Figure 18.3. The enthalpy of saturated steam at 14.7 psia is approximately 1150 Btu/lb.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nF = 10000 #Mass flow rate of NaOH (lb/h)\nC1 = 10 #Old concentration of NaOH solution (%)\nC2 = 75 #New concentration of NaOH solution (%)\nh1 = 1150 #Enthalpy of saturated steam at 14.7 psia (Btu/lb)\nU = 500 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nT1 = 212 #Absolute temperature of evaporator (\u00b0F)\nT2 = 340 #Saturated steam temperature (\u00b0F)\n\n#Calculation:\nL = F*(C1/100)/(C2/100) #Flow rate of steam leaving the evaporator (lb/h)\nV = F-L #Overall material balance (lb/h)\n#From figure 18.3:\nhF = 81 #Enthalpy of solution entering the unit (Btu/lb)\nhL = 395 #Enthalpy of the 75% NaOH solution (Btu/lb)\nQ = round(V)*h1+round(L)*hL-F*hF #Evaporator heat required (Btu/h)\nA = Q/(U*(T2-T1)) #Area of the evaporaor (ft^2)\n\n#Result:\nprint \"The heat transfer rate required for the evaporator is :\",round(Q,-2),\" Btu/h .\"\nprint \"The area requirement in the evaporator is :\",round(A,1),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat transfer rate required for the evaporator is : 9683600.0 Btu/h .\nThe area requirement in the evaporator is : 151.3 ft^2 .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.4, Page number: 388" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Given the triple-effect evaporator information below, calculate the temperature drop across the\nheating surface in the first effect:\nEffect U, Btu/h.ft^2.\u00b0F A, ft^3\n1 240 125\n2 200 150\n3 125 160\nThe condensing stream temperature in the first effect and the vapor leaving the third effect are\n400\u00b0F and 120\u00b0F, respectively.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nU1 = 240 #Overall heat transfer coefficient for first effect (Btu/h.ft^2.\u00b0F)\nU2 = 200 #Overall heat transfer coefficient for second effect (Btu/h.ft^2.\u00b0F)\nU3 = 125 #Overall heat transfer coefficient for third effect (Btu/h.ft^2.\u00b0F)\nA1 = 125 #Heating surface area in first effect (ft^3)\nA2 = 150 #Heating surface area in second effect (ft^3)\nA3 = 160 #Heating surface area in third effect (ft^3)\nT1 = 400 #Condensation stream temperature in the first effect (\u00b0F)\nT2 = 120 #Vapor leaving temperature in the first effect (\u00b0F)\n\n#Calculation:\nR1 = 1/(U1*A1) #Resistance across first effect\nR2 = 1/(U2*A2) #Resistance across second effect\nR3 = 1/(U3*A3) #Resistance across third effect\nR = R1+R2+R3 #Total resistance\nDT1 = (R1/R)*(T1-T2) #Temperature drop across the heating surface in the first effect (\u00b0F)\n\n#Result:\nprint \"The temperature drop across the heating surface in the first effect is :\",round(DT1),\" \u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature drop across the heating surface in the first effect is : 80.0 \u00b0F .\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.6, Page number: 389" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An evaporator is to be fed with 5000 lb of solution containing 2% solids by weight. The feed,\nF, is at a temperature of 100\u00b0F. It is to be concentrated to a solution of 5% solute by weight in\nan evaporator operating at a pressure of 1 atm in the vapor space. In order to carry out the\nevaporation, the heating surface is supplied with steam at 5 psig (227\u00b0F) and the overall heat\ntransfer coefficient of the evaporator, U, is 280 Btu/h.ft^2.\u00b0F. What is the mass of vapor\nproduced, the total mass of steam required, and the surface area required? Neglect enthalpy\nof solution effects.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nF = 5000 #Mass of soltuion fed in the evaporator (lb)\nxF = 2/100 #Concentration of feed\nxL = 5/100 #Concentration of liquor\nU = 280 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n#From figure 18.1 & 18.3:\nTF = 100 #Feed temperature (\u00b0F)\nTS = 227 #Steam temperature (\u00b0F)\nTV = 212 #Vapour temperature (\u00b0F)\nTL = 212 #Liquor temperature (\u00b0F)\nTC = 227 #Condensate temperature (\u00b0F)\n\n#Calculation:\n#From steam tables:\nhF = 68 #Enthalpy of feed (Btu/lb)\nhL = 180 #Enthalpy of liquor (Btu/lb)\nhV = 1150 #Enthalpy of vapour (Btu/lb)\nhS = 1156 #Enthalpy of steam (Btu/lb)\nhC = 195 #Enthalpy of condensate (Btu/lb)\ns1 = F*xF #Total solids in feed (lb)\nw = F-s1 #Total water in feed (lb)\ns2 = F*xF #Total solids in liquor (lb)\nL = s2/xL #Total water in liquor (lb)\nV = F-L #Overall balance (lb)\nS = (V*hV+L*hL-F*hF)/(hS-hC) #Mass of steam (lb)\nQ = S*(hS-hC) #Total heat requirement (Btu)\nA = Q/(U*(TS-TL)) #Required surface aea (ft^2)\n\n#Result:\nprint \"The mass of vapor produced is :\",round(V),\" lb .\"\nprint \"The total mass of steam required is :\",round(S),\" lb .\"\nprint \"The surface area required is :\",round(A),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The mass of vapor produced is : 3000.0 lb .\nThe total mass of steam required is : 3611.0 lb .\nThe surface area required is : 826.0 ft^2 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.7, Page number: 390" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An evaporator is fed with 5000 lb/h of a 20% solution of sodium hydroxide at 100\u00b0F. This is to\nbe concentrated to a 40% solution. The evaporator is supplied with saturated steam at 5 psig.\nAlthough the unit operates with the vapor space at a pressure of 4 in Hg absolute, the boiling\ntemperature of the solution in the evaporator is 198\u00b0F (due to the superheat created by the\nexposed heating element). The overall heat transfer coefficient is 400 Btu/h.ft^2.\u00b0F.\nCalculate the steam rate and the required heat transfer area.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nF = 5000 #Mass flow rate of NaOH (lb/h)\nxF = 20/100 #Old concentration of NaOH solution\nTF = 100 #Feed temperature (\u00b0F)\nxL = 40/100 #New concentration of NaOH solution\nxv = 0 #Vapour concentration at x\nyv = 0 #Vapour concentration at y\nT1 = 198 #Boiling temperature of solution in the evaporator (\u00b0F)\nT2 = 125 #Saturated steam temperature (\u00b0F)\nU = 400 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nTs = 228 #Steam temperature (\u00b0F)\n\n#Calculation:\n#From steam tables at 228\u00b0F and 5 psig:\nhS = 1156 #Enthalpy of steam (Btu/lb)\nhC = 196 #Enthalpy of condensate (Btu/lb)\nhV = hS-hC #Enthalpy of vapour (Btu/lb)\nTw = 125.4 #Boiling point of water at 4 in Hg absolute (\u00b0F)\nhS2 = 1116 #Enthalpy of saturated steam at 125\u00b0F (Btu/lb)\nhs = 0.46 #Heat capacity of superheated steam (Btu/lb.\u00b0F)\n#From figure 18.3:\nhF = 55 #Enthalpy of feed (Btu/lb)\nhL = 177 #Enthalpy of liquor (Btu/lb)\nL = F*xF/xL #Mass of liquor (lb)\nV = L #Mass of vapour (lb)\nhV = hS2+hs*(T1-T2) #Enthalpy of vapour leaving the solution (Btu/lb)\nS = (V*hV+L*hL-F*hF)/(hS-hC) #Mass flow rate of steam (lb/h)\nQ = S*(hS-hC) #Total heat requirement (Btu)\nA = Q/(U*(Ts-T1)) #Required heat transfer area (ft^2)\n\n#Result:\nprint \"The steam flow rate is :\",round(S,-1),\" lb/h .\"\nprint \"The required heat transfer area is :\",round(A),\" ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The steam flow rate is : 3170.0 lb/h .\nThe required heat transfer area is : 253.0 ft^2 .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.10, Page number: 398" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Using Ganapathy\u2019s method, determine the required \u201clength\u201d of a waste heat boiler to be used to\ncool hot gases (average heat capacity = 0.279 Btu/lb.\u00b0F) from 2000 to 550\u00b0F and generate\n30,000 lb/h of steam at 3308F from water at 140\u00b0F. The boiler contains 800 1.5-inch ID tubes.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nT1 = 2000 #Hot gas temperature (\u00b0F)\nT2 = 550 #Cool gas temperature (\u00b0F)\nT3 = 330 #Steam temperature (\u00b0F)\nT4 = 140 #Water temperature (\u00b0F)\nm = 30000 #Mass flow rate of steam (lb/h)\ncp = 0.279 #Average heat capacity of gas (Btu/lb.\u00b0F)\nN = 800 #Number of boiler tubes\n\n#Calculation:\nDT = (T1-T3)/(T2-T3) #Temperature difference ratio\nTav = (T1+T2)/2 #Average gas temperature (\u00b0F)\n#From steam tables (Appendix):\nhs = 1187.7 #Steam enthalpy (Btu/lb)\nhw = 107.89 #Water enthalpy (Btu/lb)\nQ = m*(hs-hw) #Heat duty (Btu/h)\nmh = Q/cp*(T1-T2) #Mass flow rate of gas (lb/h)\nx = mh/N #Gas mass flow rate per tube (lb/h)\n#From figure 18.5:\nL = 15 #Length of boiler tubes (ft)\n\n#Result:\nprint \"The length of boiler tubes is :\",L,\" ft .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The length of boiler tubes is : 15 ft .\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.12, Page number: 399" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''120,000 lb/h of flue gas from an incinerator is to be cooled from 1800 to 500\u00b0F in a waste heat\nboiler. If 2-inch ID tubes and a flow rate of 150 lb/h through each tube is to be used, estimate the\nrequired heat transfer area, tube length, heat duty, and water mass flow rate. Water at 200\u00b0F is\navailable for the steam generator; saturated steam at 80 psia is needed. The average heat capacity\nof the flue gas is 0.26 Btu/lb.\u00b0F. Use Ganapathy\u2019s method to solve the problem.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nT1 = 1800 #Hot gas temperature (\u00b0F)\nT2 = 500 #Cool gas temperature (\u00b0F)\n#From steam tables:\nTw = 312 #Boiling point of water at 80 psia (\u00b0F)\nm1 = 120000 #Mass flow rate of flue gas (lb/h)\nD = 2/12 #Inside diameter of tube (ft)\ncp = 0.26 #Average heat capacity of flue gas (Btu/lb.\u00b0F)\n\n#Calculation:\nDT = (T1-Tw)/(T2-Tw) #Temperature difference ratio\nTav = (T1+T2)/2 #Average gas temperature (\u00b0F)\n#From figure 18.4:\nx = 150 #Gas mass flow rate per tube (m/N) (lb/h)\nN = m1/x #Number of tubes\nL = 21.5 #Length of tubes (ft)\nA = N*L*D #Total heat transfer area (ft^2)\nQ = m1*cp*(T1-T2) #Heat duty (Btu/h)\n#From steam tables (Appendix):\nhs = 1183.1 #Steam enthalpy at 80 psia (Btu/lb)\nhw = 168.1 #Water enthalpy at 200\u00b0F (Btu/lb)\nm2 = Q/(hs-hw) #Mass flow rate of water (lb/h)\n\n#Result:\nprint \"The required heat transfer area is :\",round(A),\" ft^2 .\"\nprint \"The tube length is :\",L,\" ft .\"\nprint \"The heat duty is :\",round(Q/10**7,2),\" x 10^7 .\"\nprint \"The water mass flow rate is :\",round(m2,-4),\" lb/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The required heat transfer area is : 2867.0 ft^2 .\nThe tube length is : 21.5 ft .\nThe heat duty is : 4.06 x 10^7 .\nThe water mass flow rate is : 40000.0 lb/h .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.18, Page number: 407" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''It is proposed to cool 144,206 lb/h of a hot flue combustion gas by using ambient air at 70\u00b0F.\nCalculate the quantity (mass, mole, and volume basis) of air required to cool the gases from\n2050\u00b0F to an acceptable temperature of 560\u00b0F. Assume an average flue gas heat capacity of\n0.3 Btu/lb.\u00b0F.\n'''\n\n#Variable declaration:\nm1 = 144206 #Mass flow rate of flue gas (lb/h)\ncp = 0.3 #Average flue gas heat capacity (Btu/lb.\u00b0F)\nT1 = 2050 #Initial temperature of gas (\u00b0F)\nT2 = 560 #Final temperature of gas (\u00b0F)\nT3 = 70 #Ambient air temperature (\u00b0F)\n\n#Calculation:\nQ = m1*cp*(T1-T2) #Duty rate (Btu/h)\n#From appendix:\ncpa = 0.243 #Average ambient air heat capacity 70\u00b0F (Btu/lb.\u00b0F)\nMW = 29 #Molecular weight of air at 70\u00b0F\nma = round(Q,-5)/(cpa*(T2-T3)) #Mass of air required (lb/h)\nm2 = round(ma)/MW #Moles of air required (lb mol/h)\nm3 = round(ma)*13.32 #Volume of air required (ft^3/h)\n\n#Result:\nprint \"The mass of air required is :\",round(ma,-2),\" lb/h .\"\nprint \"The moles of air required is :\",round(m2,-1),\"lb mol/h .\"\nprint \"The volume of air required is :\",round(m3,-3),\" ft^3/h .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The mass of air required is : 541700.0 lb/h .\nThe moles of air required is : 18680.0 lb mol/h .\nThe volume of air required is : 7215000.0 ft^3/h .\n" + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.19, Page number: 407" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Design the air quench tank in the previous example if a 1.5-s residence time is required.'''\n\n#Variable declaration:\n#From example 18.19:\nm1 = 144200 #Mass flow rate of flue gas (lb/h)\nm2 = 541700 #Mass flow rate of air (lb/h)\nR = 0.73 #Universal gas constant (psia.ft^3/lbmol.\u00b0R)\nP = 1 #Absolute pressure (psia)\nT = 1020 #Absolute temperature (\u00b0R)\nMW = 29 #Molecular weight of air\nt = 1.5 #Residence time (s)\n\n#Calculation:\nm = m1+m2 #Total mass flow rate of the gas (lb/h)\nq = m*R*T/(P*MW) #Volumetric flow at 560\u00b0F (ft^3/h)\nV = q*t/3600 #Volume of tank (ft^3)\n\n#Result:\nprint \"The total mass flow rate of the gas is :\",round(m,-2),\" lb/h .\"\nprint \"The volumetric flow at 560\u00b0F is :\",round(q/10**7,2),\" x 10^7 ft^3/h .\"\nprint \"The volume of tank is :\",round(V),\" ft^3 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total mass flow rate of the gas is : 685900.0 lb/h .\nThe volumetric flow at 560\u00b0F is : 1.76 x 10^7 ft^3/h .\nThe volume of tank is : 7338.0 ft^3 .\n" + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.20, Page number: 408" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the physical dimensions of the tank in the previous example.'''\n\nfrom __future__ import division\nfrom math import pi\n\n#Variable declaration:\n#Fro example 18.20:\nV = 7335 #Volume of tank (ft^3)\n\n#Calculation:\nD = (4*V/pi)**(1/3) #Diameter of tank (ft)\nH = D #Height of tube (ft)\n\n#Result:\nprint \"The diameter of tank is :\",round(H,2),\" ft .\"\nprint \"The height of tube is :\",round(D,2),\" ft .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The diameter of tank is : 21.06 ft .\nThe height of tube is : 21.06 ft .\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.21, Page number: 408" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''It is proposed to reduce the temperature of 10% of the 144,206 lb/h of flue gas described\nin Illustrative Example 18.18 to 5508F using a solids contact method that operates on a 1-h\ncooling cycle. Assume the average heat capacities of the flue gas and solid to be 0.3 and\n0.88 Btu/lb.\u00b0F, respectively, over the temperature range in question. The initial temperature\nof the solids is 70\u00b0F. Assuming an approach temperature of 40\u00b0F, what mass of solid must be\nprovided in order to cool the flue gas to the required temperature during each hour of operation.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nm1 = 144206 #Mass flow rate of flue gas (lb/h)\ncp1 = 0.3 #Average heat capacities of the flue gas (Btu/lb\u00b0F)\ncp2 = 0.88 #Average heat capacities of the solid (Btu/lb\u00b0F)\n#From example 18.18:\nT1 = 550 #Initial temperature of gas (\u00b0F)\nT2 = 2050 #Final temperature of gas (\u00b0F)\nT3 = 70 #Initial temperature of solid (\u00b0F)\nT4 = 550-40 #Final temperature of solid (\u00b0F)\n\n#Calculation:\nDhf = m1*cp1*(T2-T1) #For the flue gas, the enthalpy change for one hour of operation (Btu)\nDhs = round(Dhf,-4) #For the solids, the enthalpy change for one hour of operation (Btu)\nm2 = Dhs/(cp2*(T4-T3)) #Mass of solid (lb)\n\n#Result:\nprint \"The mass of solid is :\",round(m2),\" lb .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The mass of solid is : 167588.0 lb .\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 18.22, Page number: 409" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''With reference to the previous example, design a radiative heat exchanger to cool the entire\ncombustion gases (MW = 28.27) from 2050 to 180\u00b0F. The ambient air temperature is 60\u00b0F\nand an overall heat transfer coefficient for the cooler of 1.5 Btu/h.ft^2.\u00b0F may be assumed to\napply.\n'''\n\nfrom __future__ import division\nfrom math import log,sqrt,pi\n\n#Variable declaration:\n#From example 18.21:\nm = 144206 #Mass flow rate of flue gas (lb/h)\ncp = 0.3 #Average heat capacities of the flue gas (Btu/lb\u00b0F)\nT1 = 2050 #Initial temperature of gas (\u00b0F)\nT2 = 180 #Final temperature of gas (\u00b0F)\nT3 = 60 #Ambient air temperature (\u00b0F)\nU = 1.5 #Overall heat transfer coefficient for cooler (Btu/h.ft^2.\u00b0F)\nMW = 28.27 #Molecular weight of gas\nR = 379 #Universal gas constant (psia.ft^3/lbmol.\u00b0R)\nv = 60 #Duct or pipe velcity at inlet (2050\u00b0F) (ft/s)\n\n#Calculation:\nQ = m*cp*(T1-T2) #Heat duty (Btu/h)\nDTlm = ((T1-T3)-(T2-T3))/log((T1-T3)/(T2-T3)) #Log-mean temperature difference (\u00b0F)\nA1 = round(Q,-5)/(U*round(DTlm)) #Radiative surface area (ft^2)\nq = m*R*(T1+460)/(T3+460)/MW #Volumetric flow at inlet (ft^3/h)\nA2 = q/(v*3600) #Duct area (ft^2)\nD = sqrt(A2*4/pi) #Duct diameter (ft)\nL = A1/(pi*D) #Length of required heat exchange ducting (ft)\n\n#Result:\nprint \" The radiative surface area required is :\",round(A1,-1),\" ft^2 .\"\nprint \" The length of required heat exchange ducting is :\",round(L),\" ft .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": " The radiative surface area required is : 80980.0 ft^2 .\n The length of required heat exchange ducting is : 3476.0 ft .\n" + } + ], + "prompt_number": 23 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_19.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_19.ipynb new file mode 100644 index 00000000..f5466410 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_19.ipynb @@ -0,0 +1,335 @@ +{ + "metadata": { + "name": "Chapter 19" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 19: Insulation and Refractory" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.1, Page number: 413" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The following data is provided: a rectangular plane room wall, 2.5 m high and 4 m wide, has an\noutside surface temperature, T1 = 24\u00b0C; the outside air temperature is T3 = -15\u00b0C. Calculate\nthe heat transfer rate. The convective heat transfer coefficient between the outside surface and\nthe air is 11W/m^2.K.\nIf loosely packed wool with k = 0.04 W/m.K and a thickness of 7.62 mm (3 in.) is used\nfor insulation on the outer wall, calculate a revised heat transfer rate.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nH = 2.5 #Height of wall (m)\nW = 4 #Width of wall (m)\nh = 11 #Convective heat transfer coefficient (W/m^2.K)\nT1 = 24 #Outside surface temperature (\u00b0C)\nT3 = -15 #Outside air temperature (\u00b0C)\nL = 7.62/10**3 #Insulation thickness (m)\nk = 0.04 #Thermal conductivity of wool (W/m.K)\n\n#Calculation:\nA = H*W #Heat transfer area (m^2)\nQ = h*A*(T1-T3) #Heat transfer rate (W)\nRi = L/(k*A) #Insuation resistance (K/W)\nRc = 1/(h*A) #Convective resitance (K/W)\nR = Ri+Rc #Total resistance (K/W)\nQt = (T1-T3)/R #Revised heat transfer rate (Btu/h)\n \n#Result:\nprint \"1. The heat transfer rate without insulation is :\",round(Q),\" W .\"\nprint \"Or, the heat transfer rate without insulation is :\",round(Q*3.412),\" Btu/h .\"\nprint \"2. The revised heat transfer rate with insulation is :\",round(Qt),\" W .\"\nprint \"Or, the revised heat transfer rate with insulation is :\",round(Qt*3.412),\" Btu/h .\"\nprint \"There is a calculation mistake in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The heat transfer rate without insulation is : 4290.0 W .\nOr, the heat transfer rate without insulation is : 14637.0 Btu/h .\n2. The revised heat transfer rate with insulation is : 1386.0 W .\nOr, the revised heat transfer rate with insulation is : 4729.0 Btu/h .\nThere is a calculation mistake in book.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.2, Page number: 414" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 19.1. Calculate the temperature at the wall-insulation surface.\n'''\n\n#Variable declaration:\n#From example 19.1:\nT1 = 24 #Outside surface temperature (\u00b0C)\nRi = 0.0191 #Insulation resistance (K/W)\nQ = 1383 #Revised heat transfer rate (Btu/h)\n\n#Calculation:\nT2 = T1-Q*Ri #Temperature at outer surface of insulation (\u00b0C)\n\n#Result:\nprint \"The temperature at the outer surface of the insulation is :\",round(T2,1),\" \u00b0C .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature at the outer surface of the insulation is : -2.4 \u00b0C .\n" + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.3, Page number: 415" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the Biot number in the previous example and comment on the results.'''\n\n#Variable declaration:\n#From example 19.1:\nh = 11 #Convective heat transfer coefficient (W/m^2.K)\nL = 7.62/10**3 #Insulation thickness (m)\nk = 0.04 #Thermal conductivity of wool (W/m.K)\n\n#Calculation:\nBi = h*L/k #Biot number\n\n#Result:\nprint \"The Biot nmuber is :\",round(Bi,1),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The Biot nmuber is : 2.1 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.4, Page number: 415" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''One wall of an oven has a 3-inch insulation cover. The temperature on the inside of the wall is at\n400\u00b0F; the temperature on the outside is at 25\u00b0C. What is the heat flux (heat flow rate per unit\narea) across the wall if the insulation is made of glass wool (k = 0.022 Btu/h.ft.\u00b0F)?\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nk = 0.022 #Thermal conductivity of glass wool (Btu/h.ft.\u00b0F)\nT1 = 400 #Inside wall temperature (\u00b0F)\nT2 = 25 #Outside wall temperature (\u00b0C)\nL = 3/12 #Length of insulation cover (ft)\n\n#Calculation:\nT_2 = T2*(9/5)+32 #Outside wall temperature in fahrenheit scale (\u00b0F)\nQbyA = k*(T1-T_2)/L #Heat flux across the wall (Btu/h.ft^2)\n\n#Result:\nprint \"The heat flux across the wall is :\",round(QbyA,1),\" Btu/h.ft^2 .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat flux across the wall is : 28.4 Btu/h.ft^2 .\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.5, Page number: 415" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A cold-storage room has a plane rectangular wall 8 m wide (w) and 3 m high (H). The temperature\nof the outside surface of the wall T1 is -18\u00b0C. The surrounding air temperature T3 is 26\u00b0C.\nThe convective heat transfer coefficient between the air and the surface is 21 W/m^2.K. A layer\nof cork board insulation (thermal conductivity, k = 0.0433 W/m.K) is to be attached to the\noutside wall to reduce the cooling load by 80%.\n1. Calculate the rate of heat flow through the rectangular wall without insulation. Express\nthe answer in tons of refrigeration (1 ton of refrigeration = 12,000 Btu/h). Which direction\nis the heat flowing?\n2. Determine the required thickness of the insulation board.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nw = 8 #Width of wall (m)\nH = 3 #Height of wall (m)\nh = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\nT1 = -18 #Outside surace of wall temperature (\u00b0C)\nT3 = 26 #Surrounding air temperature (\u00b0C)\nl1 = 80/100 #Reduction in cooling load\nk = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\nT = 12000 #Units Btu/h in 1 ton of refrigeration\n\n#Calculation:\nA = w*H #Heat transfer area (m^2) (part 1)\nQ1 = h*A*(T1-T3) #Rate of heat flow in the absence of insulation (W)\nQ2 = Q1*3.4123/T #Rate of heat flow in the absence of insulation (ton of refrigeration)\nl2 = 1-l1 #Reduced cooling load (part 2)\nQ3 = l2*Q1 #Heat rate with insulation (W)\nRt = (T1-T3)/Q3 #Total thermal resistance (\u00b0C/W)\nR2 = 1/(h*A) #Convection thermal resistance (\u00b0C/W)\nR1 = Rt-R2 #Insulation conduction resistance (\u00b0C/W)\nL = R1*k*A #Required insulation thickness (m)\n\n#Result:\nprint \"1. The rate of heat flow through the rectangular wall without insulation is :\",round(Q1/10**3,2),\" kW .\"\nprint \"Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is :\",round(Q2,1),\" ton of refrigeration .\"\nif (Q1<0):\n print \" The negative sign indicates heat flow from the surrounding air into the cold room.\"\nelse :\n print \" The positive sign indicates heat flow from the surrounding air into the cold room.\"\nprint \"2. The required thickness of the insulation board is :\",round(L*10**3,2),\" mm .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of heat flow through the rectangular wall without insulation is : -22.18 kW .\nOr, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is : -6.3 ton of refrigeration .\n The negative sign indicates heat flow from the surrounding air into the cold room.\n2. The required thickness of the insulation board is : 8.25 mm .\n" + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.6, Page number: 417" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the previous example.\n1. Calculate the temperature at the interface between the cork board and the air.\n2. Calculate the Biot number, Bi.\n3. What can one conclude?\n'''\n\n#Variable declaration:\n#From example 19.5:\nQ = -4435.2 #Heat rate with insulation (W)\nR2 = 0.00198 #Convection thermal resistance (\u00b0C/W)\nT3 = 26 #Surrounding air temperature (\u00b0C)\nh = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\nk = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\nL = 0.00825 #Required insulation thickness (m)\n\n#Calculation:\nT2 = T3+Q*R2 #Interface temperature (\u00b0C) (part 1)\nBi = h*L/k #Biot number (part 2)\n\n#Result:\nprint \"1. The interface temperature is :\",round(T2,2),\" \u00b0C .\"\nprint \"2. The Biot number is :\",round(Bi),\" .\"\nprint \"3. Theoretical part.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The interface temperature is : 17.22 \u00b0C .\n2. The Biot number is : 4.0 .\n3. Theoretical part.\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.7, Page number: 417" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A hypodermic needle with an external diameter D1 of 0.50 mm is to be used to transfer a reactant\npreheated to 95\u00b0C in a laboratory reactor. To reduce the heat loss from the transfer line, the hypodermic\nneedle is threaded through the center of a solid rubber insulating tube (thermal conductivity,\nk2 = 0.2 W/m.K) with a diameter of D2 of 2 mm.\n1. Calculate the rate of the heat loss from the hypodermic needle with and without the\nrubber insulation.\n2. Calculate the Biot and Nusselt numbers for the uninsulated needle taking the characteristic\ndimension to be the diameter of the needle.\nThe stainless steel needle has a thermal conductivity, k1, of 16 W/m.K. The ambient air\ntemperature is 20\u00b0C. The thermal conductivity of the air, k3, is 0.0242 W/m.K. The heat transfer\ncoefficient, h3, from the outside surface of the transfer line to the surrounding air is primarily\ndue to natural convection and is approximately equal to 12 W/m^2.K; it may also be assumed\nindependent of the radius or the temperature.\n'''\n\nfrom __future__ import division\nfrom math import pi,log\n\n#Variable declaration:\nD2 = 0.5/10**3 #External diameter of needle (m)\nh3 = 12 #Heat transfer coefficient (W/m^2.K)\nL = 1 #Insulation thickness (m)\nT1 = 95 #Reactant temperature (\u00b0C)\nT3 = 20 #Ambient air temperature (\u00b0C)\nk1 = 16 #Thermal conductivity of needle (W/m.K)\nk3 = 0.0242 #Thermal conductivity of air (W/m.K)\nD3 = 2/10**3 #Diameter of rubber tube (m)\n\n#Calculation:\nr2 = D2/2 #External radius of needle (m)\nr3 = D3/2 #Radius of rubber tube (m)\nRt1 = 1/(h3*(2*pi*r2*L)) #Thermal resistance (\u00b0C/W)\nQ1 = (T1-T3)/Rt1 #Rate of heat flow in the absence of insulation (W)\nBi = h3*D2/k1 #Biot number \nNu = h3*D2/k3 #Nusselt number\nR2 = log(r3/r2) #Thermal resistance of needle (\u00b0C/W)\nR3 = 1/(h3*(2*pi*r3*L)) #Thermal resistance of rubber tube (\u00b0C/W)\nRt2 = R2+R3 #Total thermal resistance (\u00b0C/W)\nQ2 = (T1-T3)/Rt2 #Rate of heat loss (W)\n\n#Result:\nprint \"1. The rate of the heat loss from the hypodermic needle with the rubber insulation is :\",round(Q1,2),\" W .\"\nprint \" The rate of the heat loss from the hypodermic needle without the rubber insulation is :\",round(Q2,2),\" W .\"\nprint \"2. The Biot number is :\",round(Bi,6),\" .\"\nprint \" The nusselt number is :\",round(Nu,3),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of the heat loss from the hypodermic needle with the rubber insulation is : 1.41 W .\n The rate of the heat loss from the hypodermic needle without the rubber insulation is : 5.12 W .\n2. The Biot number is : 0.000375 .\n The nusselt number is : 0.248 .\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.9, Page number: 420" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The surface temperature of a circular conducting rod is maintained at 200\u00b0C (T1) by the passage\nof an electric current. The rod diameter is 10 mm, the length is 2.5 m, the thermal conductivity is\n60 W/m.K, the density is 7850 kg/m^3, and the heat capacity is 434 J/kg.K. A bakelite coating\n(thermal conductivity = 1.4 W/m.K) is applied to the rod. The rod is in a fluid at 25\u00b0C (T3),\nand the convection heat transfer coefficient is 140 W/m^2.K. The thermal conductivity of the\nfluid is 0.6 W/m.K.\n1. Calcualte the rate of heat transfer for the bare rod?\n2. What is the critical radius associated with the bakelite coating? What is the heat transfer\nrate at the critical radius?\n3. If the bakelite insulation thickness is 55 mm, determine the fractional reduction in heat\ntransfer rate relative to the case of a bare rod.\n'''\n\nfrom __future__ import division\nfrom math import log, pi\n\n#Variable declaration:\nh = 140 #Convention heat transfer coefficient (W/m^2.K)\nD1 = 10/10**3 #Rod diameter (m)\nL = 2.5 #Rod length (m)\nT1 = 200 #Surface temperature of rod (\u00b0C)\nT2 = 25 #Fluid temperature (\u00b0C)\nk = 1.4 #Thermal conductivity of bakellite (W/m.K)\nl = 55/10**3 #Insulation thickness (m)\n\n#Calculation:\nQ1 = h*pi*D1*L*(T1-T2) #Rate of heat transfer for the bare rod (W) (part 1)\nBi = 2 #Critical Biot number (part 2)\nD2 = Bi*k/h #Critical diameter associated with the bakelite coating (m)\nr2 = D2/2 #Critical radius associated with the bakelite coating (m)\nr1 = D1/2 #Rod radius (m)\nR1 = log(r2/r1)/(2*pi*k*L) #Insulation conduction resistance (\u00b0C/W)\nR2 = 1/(h*(2*pi*r2*L)) #Convection thermal resistance (\u00b0C/W)\nRt1 = R1+R2 #Total thermal resistance (\u00b0C/W)\nQc = (T1-T2)/Rt1 #Heat transfer rate at the critical radius (W)\nr3 = r1+l #New radius associated with the bakelite coating after insulation (m) (part 3)\nR3 = log(r3/r1)/(2*pi*k*L) #Insulation conduction bakelite resistance (\u00b0C/W)\nR4 = 1/(h*(2*pi*r3*L)) #Convection bakelite thermal resistance (\u00b0C/W)\nRt2 = R3+R4 #Total bakelite thermal resistance (\u00b0C/W)\nQ2 = (T1-T2)/Rt2 #Heat transfer rate at the bakelite critical radius (W)\nRe = ((Q1-Q2)/Q1)*100 #Percent reduction in heat transfer rate relative to the case of a bare rod (%)\n\n#Result:\nprint \"1. The rate of heat transfer for the bare rod is :\",round(Q1),\" W .\"\nprint \"2. The critical radius associated with the bakelite coating is :\",round(r2*10**3),\" mm.\"\nprint \" & the heat transfer rate at the critical radius is :\",round(Qc),\" W .\"\nprint \"3. The fractional reduction in heat transfer rate relative to the case of a bare rod is :\",round(Re,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of heat transfer for the bare rod is : 1924.0 W .\n2. The critical radius associated with the bakelite coating is : 10.0 mm.\n & the heat transfer rate at the critical radius is : 2273.0 W .\n3. The fractional reduction in heat transfer rate relative to the case of a bare rod is : 24.6 % .\n" + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.10, Page number: 421" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A stainless steel tube carries hot ethylene glycol at 124\u00b0C (T1). The surrounding air outside the\ntube is at 2\u00b0C (T5). To reduce the heat losses from the ethylene glycol, the tube is surrounded by\nasbestos insulation. For a 1 m length of the tube, calculate:\n1. the rate of heat transfer without insulation,\n2. the rate of heat transfer with insulation, and\n3. the overall heat transfer coefficient based on the inside area of the tube,\n4. the overall heat transfer coefficient based on the outside area of the insulation,\n5. the temperature, T3, at the steel\u2013insulation interface,\n6. the inside and outside Biot numbers, and the outside Nusselt number, and\n7. the log mean radius of insulation.\nThe following data is provided.\nStainless steel pipe: inside radius, r1 = 1.1 cm; outside radius, r2 = 1.3 cm; thermal conductivity,\nk2 = 19 W/m.K; heat transfer coefficient from ethylene glycol to the stainless\nsteel pipe, h1 = 190 W/m^2.K.\nAsbestos insulation: inside radius, r2 = 1.3 cm; outside radius, r3 = 3.8 cm; thermal conductivity,\nk3 = 0.2 W/m.K.\nThe outside heat transfer coefficient from the air to the surface of the insulation (or of the\npipe, in the case of no insulation), h3 = 14 W/m^2.K. The thermal conductivity, k4, of the air\nis 0.0242 W/m.K.\n'''\n\nfrom __future__ import division\nfrom math import pi, log\n\n#Variable declaration:\nr1 = 1.1/100 #Inside radius of pipe (m)\nr2 = 1.3/100 #Outside radius of pipe (m)\nr3 = 3.8/100 #Outside radius of asbestos insulation (m)\nL = 1 #Length of tube (m)\nh1 = 190 #Heat transfer coefficient from ethylene glycol to the stainless steel pipe (W/m^2.K)\nk2 = 19 #Thermal conductivity of pipe (W/m.K)\nh2 = 14 #Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K)\nk3 = 0.2 #Thermal conductivity of asbestos (W/m.K)\nT1 = 124 #Hot ethylene glycol temperature (\u00b0C)\nT5 = 2 #Surrounding air temperature (\u00b0C)\nk4 = 0.0242 #Thermal conductivity of air (W/m.K)\n\n#Calculation:\nA1 = 2*pi*r1*L #Inside surface area of pipe (m^2) (part1)\nA2 = 2*pi*r2*L #Outside surface area of pipe (m^2)\nA3 = 2*pi*r3*L #Outside surface area of asbestos insulation (m^2)\nR1 = 1/(h1*A1) #Inside convection resistance (\u00b0C/W)\nR2 = log(r2/r1)/(2*pi*k2*L) #Conduction resistance through the tube (\u00b0C/W)\nR3 = 1/(h2*A2) #Outside convection resistance (\u00b0C/W)\nRt1 = R1+R2+R3 #Total resistance without insulation (\u00b0C/W)\nQ1 = (T1 - T5)/Rt1 #Heat transfer rate without insulation (W)\nR4 = log(r3/r2)/(2*pi*k3*L) #Conduction resistance associated with the insulation (\u00b0C/W) (part 2)\nR5 = 1/(h2*A3) #Outside convection resistance (\u00b0C/W)\nRt2 = R1+R2+R4+R5 #Total rsistance with the insulation (\u00b0C/W)\nQ2 = (T1-T5)/Rt2 #Heat transfer rate with the insulation (W)\nU1 = 1/(Rt2*A1) #Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3)\nU3 = 1/(Rt2*A3) #Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4)\nT3 = T1-(R1+R2)*Q2 #Temperature at the steel\u2013insulation interface (\u00b0C) (part 5)\nBi1 = h2*(2*r3)/k3 #Outside Biot number (part 6)\nBi2 = h1*(2*r1)/k2 #Inside Biot number\nNu = h1*(2*r1)/k4 #Nusselt number of the air\nrlm = (r3-r2)/log(r3/r2) #Log mean radius of the insulation (m) (part 7)\n\n#Result:\nprint \"1. The rate of heat transfer without insulation is :\",round(Q1,1),\" W.\"\nprint \"2. The rate of heat transfer with insulation is :\",round(Q2,1),\" W.\"\nprint \"3. The overall heat transfer coefficient based on the inside area of the tube is :\",round(U1,2),\" W/m^2.K .\"\nprint \"4. The overall heat transfer coefficient based on the outside area of the insulation is :\",round(U3,1),\" W/m^2.K .\"\nprint \"5. The temperature, T3, at the steel\u2013insulation interface is :\",round(T3,1),\" \u00b0C.\"\nprint \"6. The inside Biot numbers is :\",round(Bi2,2),\" .\"\nprint \" The outside Biot numbers is :\",round(Bi1,2),\" .\"\nprint \" The Nusselt number is :\",round(Nu,1),\" .\"\nprint \"7. The log mean radius of insulation is :\",round(rlm*100,2),\" cm.\"\nprint \"There is a printing mistake in book for unit in part 7.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of heat transfer without insulation is : 128.1 W.\n2. The rate of heat transfer with insulation is : 99.2 W.\n3. The overall heat transfer coefficient based on the inside area of the tube is : 11.76 W/m^2.K .\n4. The overall heat transfer coefficient based on the outside area of the insulation is : 3.4 W/m^2.K .\n5. The temperature, T3, at the steel\u2013insulation interface is : 116.3 \u00b0C.\n6. The inside Biot numbers is : 0.22 .\n The outside Biot numbers is : 5.32 .\n The Nusselt number is : 172.7 .\n7. The log mean radius of insulation is : 2.33 cm.\nThere is a printing mistake in book for unit in part 7.\n" + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.11, Page number: 424" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A recently developed synthetic oil is stored in a vertical tank 10 feet in diameter and 30 feet\nhigh which is insulated with a 2-inch layer of insulation (k = 0.039 Btu/h.ft.\u00b0F). To prevent\nfreezing of the oil, it is maintained at a temperature of 115\u00b0F by a heating coil consisting of an\n18-gauge, 3/4-inch copper tube (k = 224 Btu/h.ft.\u00b0F) containing saturated steam at 10 psig.\nAssuming that the minimum outdoor temperature is 5\u00b0F and that the oil temperature is uniform,\ncalculate of copper tubing required in feet to maintain the tank at 120\u00b0F in the coldest weather.\nNeglect upper and lower tank surface heat losses. The following additional information is\nprovided:\nSteam condensing inside coil, h = 800 Btu/h.ft^2.\u00b0F\nOil outside coil, h = 40 Btu/h.ft^2\u00b0F\nOil inside tank wall, h = 40 Btu/h.ft^2\u00b0F\nOuter tank wall to ambient air, h = 2.0 Btu/h.ft^2.\u00b0F\n'''\n\nfrom __future__ import division\nfrom math import pi\n\n#Variable declaration:\nh1 = 800 #Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2.\u00b0F)\nh2 = 40 #Heat transfer coefficient for oil outside coil (Btu/h.ft^2.\u00b0F)\nh3 = 40 #Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2.\u00b0F)\nh4 = 2 #Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2.\u00b0F)\nk1 = 0.039 #Thermal conductivity of insulation layer (Btu/h.ft.\u00b0F)\nl1 = 2/12 #Thickness of insulation layer (ft)\nD = 10 #Diameter of tank (ft)\nH = 30 #Height of tank (ft)\nk2 = 224 #Thermal conductivity of copper tube (Btu/h.ft.\u00b0F)\nl2 = (3/4)/12 #Thickness of insulation layer (ft)\nT1 = 120 #Temperature of tank (\u00b0F)\nT2 = 5 #Outdoor temperature (\u00b0F)\n\n#Calculation:\nUo1 = 1/(1/h3+(l1/k1)+1/h4) #Overall heat transfer coefficient for tank (Btu/h.ft^2.\u00b0F)\nAt = pi*(D+2*l1)*H #Surface area of tank (ft^2)\nQ = Uo1*At*(T1-T2) #Heat transfer rate lost from the tank (Btu/h)\n#From table 6.3:\nl2 = 0.049/12 #Thickness of coil (ft)\nA = 0.1963 #Area of 18 guage, 3/4-inch copper tube (ft^2/ft)\nUo2 = 1/(1/h2+(l2/k2)+1/h1) #Overall heat transfer coefficient for coil (Btu/h.ft^2.\u00b0F)\n#From steam tables:\nTst = 240 #Temperature for 10 psia (24.7 psia) steam (\u00b0F)\nAc = Q/(Uo2*(Tst-T1)) #Area of tube (ft^2)\nL = Ac/A #Lengt of tube (ft)\n\n#Result:\nprint \"The length ofcopper tubing required is :\",round(L,1),\" ft .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The length ofcopper tubing required is : 26.0 ft .\n" + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.12, Page number: 426" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Ricci and Theodore (R&T) Consultants have been assigned the job of selecting insulation for all\nthe plant piping at the local power plant. Included in the plant piping are 8000 ft of 1-inch schedule\n40 steel (1% C) pipe carrying steam at 240\u00b0F. It is estimated that the heat transfer coefficient\nfor condensing steam on the inside of the pipe is 2000 Btu/h.ft^2.\u00b0F. The air temperature outside\nof the pipe can drop to 20\u00b0F, and with wind motion the outside heat transfer coefficient can\nbe as high as 100 Btu/h.ft.\u00b0F.\nR&T have decided to use a fiberglass insulation having a thermal conductivity of 0.01\nBtu/h . ft . \u00b0F. It is available in 6 ft lengths in the four thicknesses listed below:\n3/8-inch thick $1.51/6 ft length\n1/2-inch thick $3.54/6 ft length\n3/4-inch thick $5.54/6 ft length\n1-inch thick $8.36/6 ft length\nCalculate the energy saved per dollar of insulation investment in going from 3/8-inch to 1/2-inch\nthick insulation. Repeat this calculation in going from 1/2-inch to 3/4-inch and 3/4-inch to 1-inch.\nExpress the results in units of Btu/h per dollar.\n'''\n\nfrom __future__ import division\nfrom math import pi, log\nfrom numpy import array,log as LOG\n\n#Variable declaration:\n#For 1-inch pipe schedule 40:\nDi = 1.049/12 #Inside diameter (ft)\nDo = 1.315/12 #Outside diameter (ft)\nL = 8000 #Length of pipe (ft)\nhi = 2000 #Heat transfer coefficient inside of the pipe (Btu/h.ft^2.\u00b0F)\nho = 100 #Outside heat transfer coefficient (Btu/h.ft.\u00b0F)\nkl = 0.01 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\nT1 = 240 #Steam temperature (\u00b0F)\nT2 = 20 #Air temperature (\u00b0F)\nk = 24.8 #Thermal conductivity for steel (Btu/h.ft.\u00b0F)\nDxl = array([3/8,1/2,3/4,1])/12 #thickness(ft)\namt = array([1.51,3.54,5.54,8.36])/6 #Cost per feet($) \n\n#Calculation:\nD_ = (Do-Di)/log(Do/Di) #Log-mean diameter of the pipe (ft)\nDl = Do+2*(Dxl) #Insulation thickness (ft)\nD_l = (Dl-Do)/LOG(Dl/Do) #Log mean diameter of pipe (ft)\nDxw = (Do-Di)/2 #Pipe thickness (ft)\nRw = Dxw/(k*pi*D_*L) #Wall resistance ((Btu/h.\u00b0F)^-1)\nRi = 1/(hi*pi*Di*L) #Inside steam convection resistance ((Btu/h.\u00b0F)^-1)\nRl = Dxl/(kl*pi*D_l*L) #Insulation resistance ((Btu/h.\u00b0F)^-1)\nRo = 1/(ho*pi*Dl*L) #Outside air convection resistance ((Btu/h.\u00b0F)^-1)\nR = Ri+Rw+Rl+Ro #Total resistance ((Btu/h.\u00b0F)^-1)\nUo = 1/(R*pi*Dl*L) #Overall outside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nUi = 1/(R*pi*Di*L) #Overall inside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\ndT = T1-T2\nAi = pi*Di*L #Inside area (ft^2)\nQ = Ui*Ai*dT #Energy loss (Btu/h)\ndef energyPerDollar(Q1,Q2,amt1,amt2):\n return round((Q1-Q2)/(8000*(amt2-amt1)),1)\n\n#Results:\nprint \"Energy saved per dollar ingoing from 3/8 to 1/2 inch is :\",energyPerDollar(Q[0],Q[1],amt[0],amt[1]),' Btu/h.$'\nprint \"Energy saved per dollar ingoing from 1/2 to 3/4 inch is :\",energyPerDollar(Q[1],Q[2],amt[1],amt[2]),' Btu/h.$'\nprint \"Energy saved per dollar ingoing from 3/4 to 1 inch is :\",energyPerDollar(Q[2],Q[3],amt[2],amt[3]),' Btu/h.$'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Energy saved per dollar ingoing from 3/8 to 1/2 inch is : 18.2 Btu/h.$\nEnergy saved per dollar ingoing from 1/2 to 3/4 inch is : 18.8 Btu/h.$\nEnergy saved per dollar ingoing from 3/4 to 1 inch is : 6.8 Btu/h.$\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.16, Page number: 434" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the outer critical radius of insulation on a 2.0-inch OD pipe. Assume the air flow coefficient\nto be 1.32 Btu/h.ft^2.\u00b0F and the loosely-packed insulation\u2019s thermal conductivity to be\n0.44 Btu/h.ft.\u00b0F. Comment on the effect of insulation on the heat rate lost from the pipe.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nki = 0.44 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\nho = 1.32 #Air flow coefficient (Btu/h.ft^2.\u00b0F)\nOD = 2 #Outside diameter of pipe (in)\n\n#Calculation:\nrc = (ki/ho)*12 #Outer critical radius of insulation (in)\nro = OD/2 #Outside radius of pipe (in)\nL = rc-ro #Critical insulation thickness (in)\n\n#Result:\nprint \"The outer critical radius of insulation is :\",round(rc),\" in .\"\nif ro<rc:\n print \"Since, ro<rc, the heat loss will increase as insulation is added.\"\nelse :\n print \"Sice, ro>rc, the heat loss will decrease as insulation is added.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The outer critical radius of insulation is : 4.0 in .\nSince, ro<rc, the heat loss will increase as insulation is added.\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.18, Page number: 439" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A flat incinerator wall with a surface area of 480 ft^2 consists of 6 inches of firebrick with a thermal\nconductivity of 0.61 Btu/h.ft.\u00b0F and an 8-inch outer layer of rock wool insulation with a\nthermal conductivity of 0.023 Btu/h.ft.\u00b0F. If the temperature of the insulation of the inside\nface of the firebrick and the outside surface of the rock wool insulation are 1900 and 140\u00b0F,\nrespectively, calculate the following:\n1. The heat loss through the wall in Btu/h.\n2. The temperature of the interface between the firebrick and the rock wool.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nLf = 6/12 #Length of firebrick (ft)\nkf = 0.61 #Thermal conductivity of firebrick (Btu/h.ft.\u00b0F)\nA = 480 #Surface area of wall (ft^2)\nLw = 8/12 #Length of rock wool (ft)\nkw = 0.023 #Thermal conductivity of rock wool (Btu/h.ft.\u00b0F)\nT1 = 1900 #Temperature of insulation of firebrick (\u00b0F)\nT2 = 140 #Temperature of insulation of rock wool (\u00b0F)\n\n#Calculation:\nRf = Lf/(kf*A) #Resistance of firebrick (h.\u00b0F/Btu)\nRw = Lw/(kw*A) #Resistance of rock wool (h.\u00b0F/Btu)\nR = Rf+Rw #Total resitance (h.\u00b0F/Btu)\nQ = (T1-T2)/R #Heat loss through the wall (Btu/h)\n\n#Result:\nprint \"The heat loss through the wall is :\",round(Q),\" Btu/h .\"\n ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The heat loss through the wall is : 28344.0 Btu/h .\n" + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.19, Page number: 440" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Heat is flowing from steam on one side of a 0.375-inch thick vertical steel sheet to air on the\nother side. The steam heat-transfer coefficient is 1700 Btu/h.ft^2.\u00b0F and that of the air is\n2.0 Btu/h.ft^2.\u00b0F. The total temperature difference is 120\u00b0F. How would the rate of heat transfer\nbe affected if:\n1. the wall was copper rather than steel,\n2. by increasing the steam coefficient to 2500, and\n3. by increasing the air coefficient to 12.0?\nNote that the thermal conductivities, k, for steel and copper are 26 and 218 Btu/h.ft.\u00b0F,\nrespectively.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nh1 = 1700 #Steam heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\nh2 = 2 #Air heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\nA = 1 #Area of base (ft^2) (assumption)\nk1 = 26 #Thermal conductivity of steel (Btu/h.ft.\u00b0F)\nk2 = 218 #Thermal conductivity of copper (Btu/h.ft.\u00b0F)\nt = 0.375 #Thickness of steel sheet (in)\nh3 = 2500 #Increased steam heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\nh4 = 12 #Increased air heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Calculation:\nR1 = 1/(h1*A) #Steam resistance (h.\u00b0F/Btu)\nR2 = 1/(h2*A) #Air resistance (h.\u00b0F/Btu)\nR3 = (t/12)/(k1*A) #Steel resistance (h.\u00b0F/Btu)\nRt1 = R1+R2+R3 #Total resistance (with steel) (h.\u00b0F/Btu)\nR4 = (t/12)/(k2*A) #Copper resistance (h.\u00b0F/Btu) (part 1)\nRt2 = R1+R2+R4 #Total resistance (with copper) (h.\u00b0F/Btu)\nR5 = 1/(h1*A) #New steam resistance (h.\u00b0F/Btu)\nRt3 = R5+R2+R3 #Total resistance after increasing the steam coefficient (h.\u00b0F/Btu)\nR6 = 1/(h4*A) #Air resistance (h.\u00b0F/Btu)\nRt4 = R1+R6+R3 #Total resistance after increasing the air coefficient (h.\u00b0F/Btu)\n\n#Result:\nif (Rt1==Rt2):\n print \"1.The rate of heat transfer is essentially unaffected.\"\nelse :\n print \"1. The rate of heat transfer is essentially affected.\"\nif (Rt1==Rt3):\n print \"2. The rate is again unaffected.\"\nelse :\n print \"2. The rate is again affected.\"\nif (Rt1==Rt4):\n print \"3. The rate is unaffected for this case.\"\nelse :\n print \"3. The rate is affected for this case.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of heat transfer is essentially affected.\n2. The rate is again unaffected.\n3. The rate is affected for this case.\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 19.20, Page number: 441" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''An incinerator is 30 ft long, has a 12-ft ID and is constructed of 3/4-inch carbon steel. The inside of\nthe steel shell is protected by 10 in. of firebrick (k = 0.608 Btu/h.ft.\u00b0F) and 5 inches of Sil-o-cel\ninsulation (k = 0.035 Btu/h.ft.\u00b0F) covers the outside. The ambient air temperature is 85\u00b0F\nand the average inside temperature is 1800\u00b0F. The present heat loss through the furnace wall\nis 6% of the heat generated by combustion of a fuel. Calculate the thickness of Sil-o-cel\ninsulation that must be added to cut the losses to 3%.\n'''\n\nfrom __future__ import division\nfrom math import pi, log, exp\n\n#Variable declaration:\nrfo = 12/2 #Outside radius of firebrick (ft)\nrfi = 5.167 #Inside radius of firebrick (ft)\nrso = 6.479 #Outside radius of sil-o-cel (ft)\nrsi = 6.063 #Inside radius of fsil-o-cel (ft)\nL = 30 #Length of incinerator (ft)\nkf = 0.608 #Thermal conductivity of firebrick (Btu/h.ft.\u00b0F)\nks = 0.035 #Thermal conductivity of sil-o-cel (Btu/h.ft.\u00b0F)\n\n#Calculation:\nRf= log(rfo/rfi)/(2*pi*L*kf) #Resistance of firebrick (h.ft.\u00b0F/Btu)\nRs= log(rso/rsi)/(2*pi*L*ks) #Resistance of sil-o-cel (h.ft.\u00b0F/Btu)\nR = Rf+Rs #Total resistance (h.ft.\u00b0F/Btu)\nro = exp(R*(2*pi*L*ks))*rso #New outside radius of sil-o-cel (ft)\nr= ro-rso #Extra thickness (ft)\n\n#Result:\nprint \"The extra thickness is :\",round(r,3),\" ft.\"\nprint \"Or, the extra thickness is :\",round(r*12,2),\" in.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The extra thickness is : 0.504 ft.\nOr, the extra thickness is : 6.05 in.\n" + } + ], + "prompt_number": 28 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_21.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_21.ipynb new file mode 100644 index 00000000..1566d542 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_21.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "Chapter 21" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 21: Entropy Considerations and Analysis" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 21.1, Page number: 460" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Case I, Figure 21.3, above. Calculate the entropy change of the two exchangers. For the\npurposes of analysis, arbitrarily assume cp = l.0 Btu/lb.\u00b0F and mH = mC = l.0 lb\n'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\nm = 1 #Mass flowrate (lb)\ncP = 1 #Heat capacity (Btu/lb.\u00b0F)\n#From figure 21.3:\nT1 = 300 #Temperature of hot fluid leaving exchanger (\u00b0F)\nT2 = 540 #Temperature of hot fluid entering exchanger (\u00b0F)\nT3 = 60 #Temperature of cold fluid leaving exchanger (\u00b0F)\nT4 = 300 #Temperature of cold fluid entering exchanger (\u00b0F)\n\n#Calculation:\nDSh = m*cP*log((T1+460)/(T2+460)) #Entropy for hot fluid (Btu/\u00b0F)\nDSc = m*cP*log((T4+460)/(T3+460)) #Entropy for cold fluid (Btu/\u00b0F)\nDSa = DSh+DSc #Entropy for one exchanger (Btu/\u00b0F)\nDSt = DSa*2 #Total entropy change (Btu/\u00b0F)\n\n#Result:\nprint \"The entropy chage is :\",round(DSt,4),\" Btu/\u00b0F .\"\nif (DSt>0):\n print \"There is a positive entropy change.\"\nelse :\n print \"There is a negative entropy change.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The entropy chage is : 0.2101 Btu/\u00b0F .\nThere is a positive entropy change.\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 21.2, Page number: 461" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Case II, Figure 21.4. Calculate the entropy change of exchangers C and D.'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\n#From example 21.1:\nDSh = -0.2744 #Entropy for hot fluid (Btu/\u00b0F)\nDSc = 0.3795 #Entropy for cold fluid (Btu/\u00b0F)\nm = 1 #Mass flowrate (lb)\ncP = 1 #Heat capacity (Btu/lb.\u00b0F)\n#From figure 21.4:\nDT = 0 #Temperature difference driving force (\u00b0F)\nDS_D = 0 #Entropy for D exchanger (Btu/\u00b0F)\n\n#Calculation:\nDS_C = DSh+DSc #Entropy for C exchanger (Btu/\u00b0F)\nDSt = DS_C+DS_D #Total entropy change of exchangers (Btu/\u00b0F)\n\n#Result:\nprint \"The total entropy change is :\",DSt,\" Btu/\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total entropy change is : 0.1051 Btu/\u00b0F .\n" + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 21.3, Page number: 462" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Case III, Figure 21.5. Calculate the entropy change in exchangers E and F.'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\n#From figure 21.5:\nm = 2 #Mass flowrate (lb)\ncP = 1 #Heat capacity (Btu/lb.\u00b0F)\nDS1 = -0.2744 #Entropy for hot fluid for E exchanger (Btu/\u00b0F)\nT1 = 180 #Temperature cold fluid entering the E exchabger (\u00b0F)\nT2 = 60 #Temperature cold fluid leaving the E exchabger (\u00b0F)\n\n#Calculation:\nDS2 = m*cP*log((T1+460)/(T2+460)) #Entropy for cold fluid for E exchanger (Btu/\u00b0F)\nDS_E = DS1+DS2 #Entropy for E exchanger (Btu/\u00b0F)\nDS_F = DS_E #Entropy for F exchanger (Btu/\u00b0F)\nDSt = DS_F+DS_E #Entropy change in exchangers E and F (Btu/\u00b0F)\n\n#Result:\nprint \"The entropy change in exchangers E and F is :\",round(DSt,4),\" Btu/\u00b0F .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The entropy change in exchangers E and F is : 0.2818 Btu/\u00b0F .\n" + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_22.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_22.ipynb new file mode 100644 index 00000000..92991017 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_22.ipynb @@ -0,0 +1,209 @@ +{ + "metadata": { + "name": "Chapter 22" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 22: Design Principles and Industrial Applications" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.6, Page number: 471" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''It is desired to evaporate 1000 lb/h of 60\u00b0F water at 1 atm at a power plant. Utility superheated\nsteam at 40 atm and 1000\u00b0F is available, but since this steam is to be used elsewhere in the plant,\nit cannot drop below 20 atm and 600\u00b0F. What mass flowrate of the utility steam is required?\nAssume that there is no heat loss in the evaporator.\nFrom the steam tables:\nP = 40 atm, T = 1000\u00b0F, h = 1572 Btu/lb\nP = 20 atm, T = 600\u00b0F, h = 1316 Btu/lb\nFor saturated steam:\nP = 1 atm, h = 1151 Btu/lb\nFor saturated water:\nT = 60\u00b0F, h = 28.1 Btu/lb\n'''\n\nfrom __future__ import division\nfrom sympy import symbols,solve\n\n#Variable declaration:\n#From steam tables:\nh1 = 1572 #Enthalpy for super heated steam at (P = 40 atm, T = 1000\u00b0F) (Btu/lb)\nh2 = 1316 #Enthalpy for super heated steam at (P = 20 atm, T = 600\u00b0F) (Btu/lb)\nh3 = 1151 #Enthalpy for saturated steam (Btu/lb)\nh4 = 28.1 #Enthalpy for saturated water (Btu/lb)\nm1 = 1000 #Mass flowrate of steam (lb/h)\nm = symbols('m') #Mass flow rate of steam (lb/h)\n\n#Calculation:\nDh1 = m1*(h3-h4) #The change in enthalpy for the vaporization of the water stream (Btu/h)\nDh2 = m*(h1-h2) #The change in enthalpy for the cooling of the water stream (Btu/h)\nx = solve(Dh1-Dh2,m) #Mass flowrate of steam (lb/h)\nm2 = x[0]; #Mass flowrate of steam (lb/h)\n\n#Result:\nprint \"The mass flowrate of the utility steam required is :\",round(m2),\" lb/h.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The mass flowrate of the utility steam required is : 4386.0 lb/h.\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.7, Page number: 473" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine the total flowrate of cooling water required for the services listed below if a cooling\ntower system supplies the water at 90\u00b0F with a return temperature of 115\u00b0F. How much fresh\nwater makeup is required if 5% of the return water is sent to \u201cblowdown?\u201d Note that the cooling\nwater heat capacity is 1.00 Btu/(lb.\u00b0F), the heat of vaporization at cooling tower operating\nconditions is 1030 Btu/lb, and the density of water at cooling tower operating conditions is\n62.0 lb/ft^3. Process data is provided in Table 22.1.\nTable 22.1 Data for Illustrative Example 22.7\nProcess unit Heat duty, Btu/h Required\n temperature, \u00b0F\n1 12,000,000 250\n2 6,000,000 200\u2013276\n3 23,500,000 130\u2013175\n4 17,000,000 300\n5 31,500,000 150\u2013225\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\n#From table 22.1:\nQH1 = 12*10**6 #Heat duty for process unit 1 (Btu/h)\nQH2 = 6*10**6 #Heat duty for process unit 2 (Btu/h)\nQH3 = 23.5*10**6 #Heat duty for process unit 3 (Btu/h)\nQH4 = 17*10**6 #Heat duty for process unit 4 (Btu/h)\nQH5 = 31*10**6 #Heat duty for process unit 5 (Btu/h)\nT1 = 90 #Supply water temperature (\u00b0F)\nT2 = 115 #Return water temperature (\u00b0F)\ncP = 1 #Cooling water heat capacity (Btu/(lb.\u00b0F))\np = 62*0.1337 #Density of water (lb/gal)\nBDR = 5/100 #Blow-down rate\n\n#Calculation:\nQHL = (QH1+QH2+QH3+QH4+QH5)/60 #Heat load (Btu/min)\nDT = T2-T1 #Change in temperature (\u00b0F)\nqCW = round(QHL,-5)/(DT*cP*p) #Required cooling water flowrate (gpm)\nqBD = BDR*qCW #Blow-down flow (gpm)\n\n#Result:\nprint \"The total flowrate of cooling water required for the services is :\",round(qCW,-1),\" gpm.\"\nprint \"The required blow-down flow is :\",round(qBD),\" gpm.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total flowrate of cooling water required for the services is : 7240.0 gpm.\nThe required blow-down flow is : 362.0 gpm.\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.8, Page number: 474" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Determine how many pounds per hour of steam are required for the following situation if steam\nis provided at 500 psig, and if steam is provided at both 500 and 75 psig pressures. The plant has\nthe heating requirements given in Table 22.2. Also note the properties of saturated steam in\nTable 22.3.\nTable 22.2 Process Data for Illustrative Example 22.8\nProcess unit Unit heat duty (Q\u02d9 ), Required\n Btu/h temperature, \u00b0F\n1 10,000,000 250\n2 8,000,000 450\n3 12,000,000 400\n4 20,000,000 300\nTable 22.3 Steam Data\nPressure Saturation Enthalpy of vaporization\nprovided, psig temperature, \u00b0F (hvap), Btu/lb\n75 320 894\n500 470 751\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nQ1 = 10*10**6 #Unit heat duty for process unit 1 (Btu/h)\nQ2 = 8*10**6 #Unit heat duty for process unit 2 (Btu/h)\nQ3 = 12*10**6 #Unit heat duty for process unit 3 (Btu/h)\nQ4 = 20*10**6 #Unit heat duty for process unit 4 (Btu/h)\nhv = 751 #Enthalpy of vaporization for pressure 500 psig (Btu/lb)\n\n#Calculation:\nmB1 = Q1/hv #Mass flowrate of 500 psig steam through unit 1 (lb/h)\nmB2 = Q2/hv #Mass flowrate of 500 psig steam through unit 2 (lb/h)\nmB3 = Q3/hv #Mass flowrate of 500 psig steam through unit 3 (lb/h)\nmB4 = Q4/hv #Mass flowrate of 500 psig steam through unit 4 (lb/h)\nmBT = mB1+mB2+mB3+mB4 #Total steam required (lb/h)\n\n#Result:\nprint \"The total steam required is :\", round(mBT,-1),\" lb/h.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The total steam required is : 66580.0 lb/h.\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.9, Page number: 474" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Afeed stream to a distillation column processes of 28,830 gals/day of a light oil, and is fed to the\ncolumn through a 6-inch I.D. pipeline. The temperature of the feed is usually 27\u00b0C. You are\nasked to explore the possibility of using either a countercurrent or a parallel (co-current)\ndouble pipe heat exchanger as an auxiliary heater for the feed stream. The double-pipe exchanger\nis to heat the oil from 23.5\u00b0C to 27\u00b0C.Water at 8406 gals/day and 93\u00b0C is available for heating\nthe oil. The necessary physical properties are provided in Table 22.4. The inner diameter of\nthe pipe forming the annular region is 20.3 cm and the outer surface of that pipe is well insulated.\nFor a counter-current flow arrangement, calculate the following using the SI system:\n1. The lowest temperature the exiting heating water could reach. Is there sufficient energy\navailable from thewater to raise the oil temperature? (If there is not enough energy, there\nis no point in designing the exchanger.)\n2. The log mean temperature difference.\n3. The overall heat transfer coefficient, U, for a new clean exchanger based on the inside\narea is 35.4 W/m^2.K. Correct the U for the fouling of the heat exchanger. Use a\nfouling factor, Rf, of quenching oil, which is 0.0007 m^2.K/W.\n4. The length of the double pipe heat exchanger.\n5. The effectiveness of the exchanger and the NTU.\nTable 22.4 Property Data for Illustrative Example 22.9\nOil (27\u00b0C) Water (93\u00b0C) Pipe\np = 53 lb/ft^3 p = 964 kg/m^3 ID = 6.0 in\ncp = 0.46 Btu/lb.\u00b0F cp = 4204 J/kg.\u00b0C OD= 168 mm\nmu = 150 cP mu = 0.7 lb/h.ft k = 45 W/m.\u00b0C\nk = 0.11 Btu/h.ft.\u00b0F k = 0.678 W/m.\u00b0C\n'''\n\nfrom __future__ import division\nfrom sympy import symbols,solve\nfrom math import log,pi\n\n#Variable declaration:\npo = 53*16.0185 #Density of oil (kg/m^3)\nco = 0.46*4186.7 #Heat capacity of oil (J/kg.\u00b0C)\nmuo = 150/1000 #Dynamic viscosity of oil (kg/m.s)\nko = 0.11*1.7303 #Thermal conductivity of oil (W/m.\u00b0C)\nqo = 28830*4.381*10**-8 #Volumetric flowrate of oil (m^3/s)\npw = 964 #Density of water (kg/m^3)\ncw = 4204 #Heat capacity of water (J/kg.\u00b0C)\nmuw = 0.7/3600*1.4881 #Dynamic viscosity of water (kg/m.s)\nkw = 0.678 #Thermal conductivity of water (W/m.\u00b0C)\nqw = 8406*4.381*10**-8 #Volumetric flowrate of water (m^3/s)\nt1 = 23.5 #Initial temperature of oil (\u00b0C)\nt2 = 27 #Final temperature of oil (\u00b0C)\nT1 = 93 #Water heating temperature of water (\u00b0C)\nT2 = symbols('T2') #Minimum temperature of heating water (\u00b0C)\nA = symbols('A') #Heat transfer area (m^2)\nUc = 35.4 #Clean heat transfer coefficient (W/m^2.K)\nRf = 0.0007 #Thermal resistance (m^2.K/W)\nD = 6*0.0254 #Inside diameter of pipe (m)\n\n#Calculation:\nvo = muo/po #Kinematic viscosity of oil (m^2/s)\nmo = po*qo #Mass flowrate of oil (kg/s)\nvw = muw/pw #Kinematic viscosity of (m^2/s)\nmw = pw*qw #Masss flow rate of water (kg/s)\nQ1 = mo*co*(t2-t1) #Duty of exchanger of oil (W)\nT2m = t1 #Lowest possible temperature of the water (\u00b0C) (part 1)\nQmw = mw*cw*(T1-T2m) #Maximum duty of exchanger of water (W) (part 2)\nQ2 = mw*cw*(T1-T2) #Duty of exchanger of water in terms of T2 (W)\nx = solve(Q1-Q2,T2) #Solving value for T2 (\u00b0C)\nT3 = x[0]; #Minimum temperature of heating water (\u00b0C)\nDT1 = T3-t1 #Inlet temperature difference (\u00b0C)\nDT2 = T1-t2 #Outlet temperature difference (\u00b0C)\nDTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\nUd1 = 1/Uc+Rf #Dirty heat transfer coefficient (W/m^2.K) (part 3)\nUd2 = 34.6 #Dirty heat transfer coefficient (W/m^2.\u00b0C)\nQ3 = Ud2*A*DTlm #Duty of exchanger (W) (part 4)\ny = solve(Q1-Q3,A) #Heat transfer area (m^2)\nA1 = y[0]; #Required heat transfer area (m^2)\nL = A1/(pi*D) #Required heat transfer length (m)\nQmo = mo*co*(T1-t1) #Maximum duty of exchanger of oil (W) (part 5)\nQm = Qmw #Maximum duty of exchanger (W)\nE = Q1/Qm*100 #Effectiveness (%)\nNTU = Ud2*A1/(mw*cw) #Number of transfer units\n\n#Result:\nprint \"1. The lowest possible temperature of the water is :\",T2m,\" \u00b0C .\"\nprint \"2. The log mean temperature difference is :\",round(DTlm,2),\" \u00b0C .\"\nprint \"3. The overall heat transfer coefficient for the new clean exchanger is :\",round(Ud2,1),\" (W/m^2.\u00b0C .\"\nprint \"4. The length of the double pipe heat exchanger is :\",round(L,2),\" m .\"\nprint \"5. The effectiveness of the exchanger is :\",round(E,2),\" % .\"\nprint \" The NTU of the exchanger is :\",round(NTU,4),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The lowest possible temperature of the water is : 23.5 \u00b0C .\n2. The log mean temperature difference is : 65.33 \u00b0C .\n3. The overall heat transfer coefficient for the new clean exchanger is : 34.6 (W/m^2.\u00b0C .\n4. The length of the double pipe heat exchanger is : 6.68 m .\n5. The effectiveness of the exchanger is : 6.97 % .\n The NTU of the exchanger is : 0.0741 .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.10, Page number: 477" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 22.9. How would the answer to Parts 2 and 4 be affected if the flow\nwere parallel?\n'''\n\nfrom __future__ import division\nfrom math import log,pi\n\n#Variable declaration:\n#From example 22.9:\nt1 = 23.5 #Initial temperature of oil (\u00b0C)\nt2 = 27 #Final temperature of oil (\u00b0C)\nT1 = 93 #Water heating temperature of water (\u00b0C)\nT2 = 88.16 #Minimum temperature of heating water (\u00b0C)\nU = 34.6 #Overall heat transfer coefficient (W/m^2.\u00b0C)\nQ = 7227.2 #Duty of exchanger (W)\nD = 6*0.0254 #Inside diameter of pipe (m)\nl = 6.68 #Previous heat transfer length (m)\n\n#Calculation:\nDT1 = T1-t1 #Inlet temperature difference (\u00b0C)\nDT2 = T2-t2 #Outlet temperature difference (\u00b0C)\nDTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\nA = Q/(U*DTlm) #Required heat transfer area (m^2)\nL = A/(pi*D) #Required heat transfer length (m)\n\n#Result:\nprint \"The length of the parallel pipe heat exchanger is :\",round(L,2),\" m .\"\nif L>l:\n print \"The tube length would increase slightly.\"\nelif L<l:\n print \"The tube length would decrease slightly.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The length of the parallel pipe heat exchanger is : 6.69 m .\nThe tube length would increase slightly.\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.12, Page number: 478" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A brine solution at 10\u00b0F is heated in a food processing plant by flowing through a heated\npipe. The pipe surface is maintained at 80\u00b0F. The pipe surface area for heat transfer is 2.5 ft^2.\nThe brine solution (with a density of 62.4 lb/ft^3 and a heat capacity of 0.99 Btu/lb.\u00b0F)\nflows at a rate of 20 lb/min. The overall heat transfer coefficient varies linearly with the temperature\napproach, with values of 150 Btu/h.ft^2.\u00b0F at the brine solution entrance (where the\nbrine temperature is 10\u00b0F) and 140 Btu/h.ft.\u00b0F at the brine solution exit. Determine:\n1. the temperature approach at the brine inlet side\n2. the exit temperature of the brine solution.\n'''\n\nfrom __future__ import division\nfrom scipy.optimize import fsolve\nfrom math import log\n\n#Variable declaration:\nT = 80 #Pipe surface temperature (\u00b0F)\nt1 = 10 #Inlet temperature of brine solution (\u00b0F)\nm = 1200 #mass flowrate of solution (kg/s)\nc = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\nA = 2.5 #Heat transfer area (ft^2)\nU1 = 150 #Overall heat transfer coefficient at temperature approach (Btu/h.ft^2.\u00b0F)\nU2 = 140 #Overall heat transfer coefficient at inlet brine temperature (Btu/h.ft^2.\u00b0F)\n\n#Calculation:\nDT1 = T-t1 #Temperature approach at the pipe entrance (\u00b0F)\n\ndef equation(DT2):\n\tQ1 = m*c*(DT1-DT2) #Energy balance to the brine solution across the full length of the pipe (Btu/h)\n\tDTlm = (DT1-DT2)*log(DT2/DT1) #Log mean temperature difference (\u00b0F)\n\tQ2 = A*(U2*DT1-U1*DT2)/log((U2*DT1)/(U1*DT2)) #Heat transfer rate (Btu/h)\n\treturn Q2-Q1\t\nt2 = round(T-fsolve(equation,1),1)\t #The temperature of the brine solution (\u00b0F)\n\n#Results:\nprint 'The temperature of brine solution is: ',(t2-32)/1.8,'\u00b0C'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The temperature of brine solution is: -2.0 \u00b0C\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.13, Page number: 479" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to Illustrative Example 22.12. Calculate:\n1. the rate of heat transfer, Q\u02d9\n2. the log mean temperature difference.\n'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\nm = 1200 #mass flowrate of solution (kg/s)\nc = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\nDT1 = 70 #Temperature approach at the pipe entrance (\u00b0F)\nDT2 = 51.6 #Temperature difference at the pipe exit (\u00b0F)\n\n#Calculation:\nQ = m*c*(DT1-DT2) #Heat transfer rate (Btu/h)\nDTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0F)\n\n#Result:\nprint \"1. The rate of heat transfer is :\",round(Q,-1),\" Btu/h.\"\nprint \"Or, the rate of heat transfer is :\",round(Q/3.412),\" W.\"\nprint \"2. The log mean temperature difference is :\",round(DTlm,1),\" \u00b0F.\"\nprint \"Or, the log mean temperature difference is :\",round(DTlm/1.8,1),\" \u00b0C.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The rate of heat transfer is : 21860.0 Btu/h.\nOr, the rate of heat transfer is : 6407.0 W.\n2. The log mean temperature difference is : 60.3 \u00b0F.\nOr, the log mean temperature difference is : 33.5 \u00b0C.\n" + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.23, Page number: 484" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Heating or cooling of liquids in batch processes is used in a number of commercial applications.\nSome reasons for using a batch rather than a continuous heat transfer operation include: the\nliquid is not continuously available, the liquid cleaning and regeneration is a significant part\nof the operation, or batch operation is simpler and cheaper.\nFor example, vapor degreasers are widely used for cleaning metal parts. A degreaser consists\nof a tank partially filled with a solvent. The tank is equipped with a heating coil to heat the solvent\nclose to its boiling point. The vapor of the solvent occupies the remaining volume of the\ntank, forming the \u201csolvent vapor zone.\u201d When a metal part is placed in the solvent vapor\nzone, the solvent condenses on the metal part and then drips off, taking contaminants with it.\nFor ease of use, vapor degreasers are often open to the atmosphere. This makes it easier to introduce\nand remove the metal parts. It has been a common practice to use a halogenated hydrocarbon\nfor such cleaning since they are excellent solvents, volatile, and non-flammable;\nhowever, they can be toxic and the open tank of a degreaser can be a significant source of solvent\nemissions or volatile organic components (VOCs).\nWhen an agitated liquid batch, initially at Ti, is heated, the temperature, T, of the liquid at\nany time, t, can be assumed to be uniform if the liquid is well stirred. Three different types of\ncalculations involving the batch heating of liquids are outlined in Table 22.6.\nThe assumptions in these analyses include:\n1. constant properties for the duration of the process,\n2. constant overall heat transfer coefficient, U, for the process,\n3. a constant temperature, T1, of the heating medium,\n\nTable 22.6 Batch Heating of Liquids\nCase Description Calculate\nDesign It is desired to heat the liquid to a The surface area required.\n specified temperature within a\n given time.\n\nSimulation It is desired to heat the liquid for a The final temperature, T.\n specified time; the heat transfer\n surface area is known.\n\nTime The heat transfer surface area and the The batch time, t.\n final temperature are known.\n\n4. a vigorously stirred unit (and therefore a uniform temperature),\n5. negligible heat losses, and\n6. no phase change.\nThe describing equation for these batch systems can be developed. Consider a liquid batch\nthat has a volume, V, density, r, and heat capacity, c. The initial temperature is Ti, and the temperature\nat any time, t, is T. The energy balance on the liquid in the agitated tank simplifies to\n UA(Too - T) = pVcp(dT/dt) (1)\nAn excess temperature, u, can be defined as\n u = T00 - T; ui = T00 - Ti, Ti 2 initial temprature (2)\nSubstituting this expression into Equation (1) and integrating with respect to t yields\n u/ui = e^(-t/r) (3)\nwhere rt is the thermal time constant that is given by\n r = pVcp/(UA) (4)\n Design a degreaser employing the development provided above; i.e., calculate the heating\nsurface area requirement for a vapor degreaser using 1,1,1-trichloroethane, Cl3C-CH3 (TCA)\nas the solvent. The degreaser tank has a rectangular cross section (6 ft x 3 ft) and a height of\n5 ft. Liquid TCA at 18\u00b0C is poured into the tank to a height of 1 ft. The solvent vapor zone\nwill occupy the remaining 4 ft. At time t = 0, saturated steam is passed into the heating coil.\nThe steam temperature is 100\u00b0C. The overall heat transfer coefficient, U, from the steam to\nthe liquid TCA, based on the outside area of the coil, is 200 Btu/h.ft^2.\u00b0F. It is desired to\nheat the liquid TCA from 18\u00b0C to its boiling point (74\u00b0C) without evaporation in 180 seconds.\nThe approximate properties of TCA are: density, p = 87.4 lb/ft3, heat capacity, cp = 0.23 Btu/lb.\u00b0F,\nand viscosity = 0.56 cP. The design should include:\n1. the required surface area of the heating coil\n2. the total heat added to the liquid TCA.\n'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\nToo = 100 #Steam temperature (\u00b0C)\nTi = 18 #Initial temperature of liquid TCA (\u00b0C)\nTf = 74 #Final temperature of liquid TCA (\u00b0C)\nt = 180 #Heating time (s)\np = 87.4 #Density of TCA (lb/ft^3)\nV = 18 #Kinematic viscosity of TCA (m^2/s)\ncp = 0.23 #Heat capacity of TCA (Btu/lb.\u00b0F)\nU = 200 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n\n#Calculation:\nui = Too-Ti #Initial excess temperature (\u00b0C)\nuf = Too-Tf #Final excess temperature (\u00b0C)\nR = log(ui/uf) #Ratio t/r\nr = t/R #Thermal time constant (s)\nA = p*V*cp/(3600*U*r) #Required heating area (ft^3)\nTi_F = Ti*9/5+32 #Initial temperature in fahrenheit scale (\u00b0F)\nTf_F = Tf*9/5+32 #Final temperature in fahrenheit scale (\u00b0F)\nQ = p*V*cp*(Tf_F-Ti_F) #Total amount of heat added (Btu)\n\n#Result:\nprint \"1. The required surface area of the heating coil is :\",(A,1),\" ft^3 .\"\nprint \"2. The total heat added to the liquid TCA is :\",round(Q),\" Btu .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The required surface area of the heating coil is : (3.2068907918330814e-06, 1) ft^3 .\n2. The total heat added to the liquid TCA is : 36473.0 Btu .\n" + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 22.24, Page number: 486" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''62,000 lb/h of pure ethyl alcohol (hvap = 365 Btu/lb; specific gravity, s = 0.79) at 2.0 psig is\nto be condensed by water (r = 62.5 lb/ft^3) entering at 85\u00b0F and exiting at 120\u00b0F. A 1\u20132 horizontal\ncondenser consists of 700(Nt) one-inch outside diameter, 14 BWGtubes, 16-inch long on\na 1.25-inch triangular pitch. There are four (n) tube passes. Assume that the flow is countercurrent\nand that the alcohol is on the shell side. A fouling factor of 0.003 Btu/h.ft^2.\u00b0F is recommended.\nThe cooling water inside film coefficient has been previously determined to be\n862.4 Btu/h.ft^2.\u00b0F. Neglecting any pressure drop considerations, determine if the exchanger\nis suitable. Additional information is provided in Table 22.7.\nTable 22.7 Exchanger Information for Illustrative Example 22.24\nShell Tubes\nID = 39 inch Pitch = 1.25 inch\nBaffle spacing = 39 inch Flow area/tube = 0.546 in2\nFlow area = 2.11 ft^2 ID = 0.834 inch\nCondensation temperature, tc = 173\u00b0F Wall thickness = 0.083 inch\nFor film condensation on horizontal tubes (see also Chapter 12), the average heat transfer\ncoefficient is to be calculated from the equation (consistent units):\nh = 1.51((kf*pf^2gmuf)/(muf^24G))^1/3\nwhere kf = thermal conductivity\npf = fluid density\ng = acceleration due to gravity\nmuf = viscosity of the fluid\nG = loading, G = (m_ alcohol/LNt^(2/3))\nm_alcohol = condensate mass flow\nL = tube length\nNt = number of tubes in bundle\nV = velocity\nFor a clean (unused) tube, the overall heat transfer coefficient can be calculated from the\nfollowing equation:\nUC = Uclean =hioho/(hio + ho)\nwhere UC = Uclean = heat transfer coefficient of a clean (new/unused) tube\nhio = corrected inside heat transfer coefficient to the outside diameter, with\nhio = hi(Di/Do)\nThe dirt fouling factor may be calculated from:\nRD = Rdirty = (Uc - U)/UcU (14.32)\nwhere RD = Rdirty = dirt (fouling) factor\nU = heat transfer coefficient calculated from the actual heat transfer\n'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\nm1 = 62000 #Mass flowrate of alcohol (lb/h)\nh1 = 365 #Enthalpy of vapour (Btu/lb)\ncp = 1 #Heat capacity of water (Btu/lb.\u00b0F)\nT1 = 85 #Entering temperature of water (\u00b0F)\nT2 = 120 #Exit temperature of water (\u00b0F)\na1 = 2.11 #Flow area for the shell side (ft^2)\nN = 700 #Total number of tubes\na2 = 0.546 #Flow area per tube (in^2/tube)\nn = 4 #Number of tube passes\np = 62.5 #Density of water (lb/ft^3)\nL = 16 #Length of condenser (ft)\nhio = 862.4 #Cooling water inside film coefficient (Btu/h.ft^2.\u00b0F)\ng = 9.8 #Gravitational accleration (m^2/s)\nRf = 0.003 #Fouling factor (Btu/h.ft^2.\u00b0F)\n\n#Calculation:\nQ1 = m1*h1 #Heat loss from alcohol (Btu/h)\nQ2 = Q1 #Heat gained by water (Btu/h)\nDT = T2-T1 #Temperature difference (\u00b0F)\nm2 = Q2/(cp*DT) #Water mass flow rate (lb/h)\nLMTD = ((T2-32)-(T1-32))/log((T2-32)/(T1-32)) #Log mean temperature difference (\u00b0F)\nat = (N*a2)/(144*n) #Total flow area for tube side (ft^2)\nG1 = m1/a1 #Mass velocity of flow in shell side (lb/h.ft^2)\nG2 = m2/at #Mass velocity of flow in tube side (lb/h.ft^2)\nV = G2/(3600*p) #Velocity of water (ft/s)\nG3 = m1/(L*N)**(2/3) #Loading G (lb/h.ft)\n#For alcohol:\nkf = 0.105 #Thermal conductivity (Btu/h.ft.\u00b0F)\nmuf = 0.55*2.42 #Dynamic viscosity (lb/ft.h)\nsf = 0.79 #\npf = sf*p #Density (lb/ft^3)\nh = 151*(((kf**3)*(pf**2)*g*muf)/((muf**2)*n*G3))**(1/3) #Heat transfer coefficient for the shell side (Btu/h.ft^2.\u00b0F)\nho = h #Outside heat transfer coefficient of the tube bundle (Btu/h.ft^2.\u00b0F)\nUc = (hio*ho)/(hio+ho) #Overall heat transfer coefficient for a new (clean) heat exchanger (Btu/h.ft^2.\u00b0F)\nA = N*L*0.2618 #Area for heat transfer (ft^2)\nUd = Q1/(A*DT) #Design (D) overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\nRd = (Uc-Ud)/(Uc*Ud) #Dirt (d) factor (Btu/h.ft^2.\u00b0F)\n\n#Result:\nprint \"The dirt (d) factor is :\",round(Rd,4),\" Btu/h.ft^2.\u00b0F .\"\nif (Rd>Rd):\n print \"Therefore, the exchanger as specified is unsuitable for these process conditions since the fouling factor is above the recommended value. Cleaning is recommended.\"\nelse:\n print \"Therefore, the exchanger as specified is suitable for these process conditions since the fouling factor is below the recommended value. Cleaning is not recommended.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The dirt (d) factor is : -0.0157 Btu/h.ft^2.\u00b0F .\nTherefore, the exchanger as specified is suitable for these process conditions since the fouling factor is below the recommended value. Cleaning is not recommended.\n" + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_23.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_23.ipynb new file mode 100644 index 00000000..3ff4626b --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_23.ipynb @@ -0,0 +1,62 @@ +{ + "metadata": { + "name": "Chapter 23" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 23: Environmental Management" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 23.6, Page number: 498" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''In 1900, it took about 20,000 Btu fuel input to produce 1 kW. h of electricity. Estimate the\nefficiency of conversion and compare it with a typical value for today\u2019s power industry.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nQ = 20000 #Fuel input (Btu)\ne = 1 #Energy produced (kW.h)\nBtu = 3412 #Units Btu in 1 kW.h\n\n#Calulation:\nER = Q/Btu #Energy requirement in 1990 (kW.h)\nE = e/ER*100 #Efficiency of energy conversion (%)\n\n#Result:\nprint \"The efficiency of energy conversion is :\",round(E,1),\" % .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The efficiency of energy conversion is : 17.1 % .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 23.7, Page number: 499" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The James David University runs it own coal-fired power plant, consuming Utah bituminous\ncoal with an energy content (in the combustion literature, energy content is defined as the\nlower heating value, LHV) of 25,000 kJ/kg. The coal contains, on average, 1.0 wt% sulfur\nand 1.2 wt% ash (based on the total mass of the coal). The power plant is 35% efficient (indicating\nthat 35% of the energy in the coal is actually converted to electrical energy), and is operated\nat a 2.0-MW average daily load (ADL).\nAssume that the coal is completely burned during combustion, and also that the power plant\ncaptures 99% of the ash and 70% of the sulfur dioxide produced during combustion. After a\nU.S. Environmental Protection Agency (EPA) Green Lights energy audit, James David found\nthat it could install energy-efficient lighting and reduce its average daily electrical generating\nneeds by 25%.\nUsing the information given above, calculate the average reduction in electrical load and the\nnew average daily load for the power plant.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nADL1 = 2 #Average daily load (MW)\nR = 25/100 #Reduction in electrical load (%)\n\n#Calculation:\nL = 1-R #New load fraction\nADL2 = ADL1*L #New average daily load (MW)\nAR = ADL1-ADL2 #Average reduction in electrical load (MW)\n\n#Result:\nprint \"The new Average daily load for the plant is :\",ADL2,\" MW.\"\nprint \"The average reduction in electrical load is :\",AR,\" MW.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The new Average daily load for the plant is : 1.5 MW.\nThe average reduction in electrical load is : 0.5 MW.\n" + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_24.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_24.ipynb new file mode 100644 index 00000000..38c2db87 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_24.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "Chapter 24" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 24: Accident and Emergency Management" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.4, Page number: 514" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Calculate the upper flammability limit (UFL) and the lower flammability limit (LFL) of a gas\nmixture that consists of 30% methane (m), 50% ethane (e), and 20% pentane ( p) by volume.\nEmploy the following equation:\nFL(mixture, n components) = 1/(fi/FLi) ; fi = mole fraction i ; i = 1 to n\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nfm = 30/100 #Mole fraction of methane\nfe = 50/100 #Mole fraction of ethane\nfp = 20/100 #Mole fraction of pentane\nLFLm = 0.046 #Lower flammability limit for methane\nLFLe = 0.035 #Lower flammability limit for ethane\nLFLp = 0.014 #Lower flammability limit for propane\nUFLm = 0.142 #Upper flammability limit for methane\nUFLe = 0.151 #Upper flammability limit for ethane\nUFLp = 0.078 #Upper flammability limit for propane\n\n#Calculation:\nLFLmix = 1/((fm/LFLm)+(fe/LFLe)+(fp/LFLp)) #Lower flammability limit of gas mixture\nUFLmix = 1/((fm/UFLm)+(fe/UFLe)+(fp/UFLp)) #Upper flammability limit of gas mixture\n\n#Result:\nprint \"The upper flammability limit (UFL) of the gas mixture is :\",round(UFLmix*100,2),\" % .\"\nprint \"The lower flammability limit (LFL) of the gas mixture is :\",round(LFLmix*100,2),\" % .\"\nprint \"There is a printing mistake in book.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The upper flammability limit (UFL) of the gas mixture is : 12.52 % .\nThe lower flammability limit (LFL) of the gas mixture is : 2.85 % .\nThere is a printing mistake in book.\n" + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.5, Page number: 514" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Two boiler tubes are drawn in succession from a lot of 100 tubes, of which 10 are defective.\nWhat is the probability that both tubes are defective if (a) the first is replaced before the\nsecond is drawn and (b) the first is not replaced before the second is drawn.\n'''\n\nfrom __future__ import division\nfrom sympy import nsimplify as changeToFraction\n\n#Variable declaration:\nP_A = 10/100 #Probability that the first tube is defective if the first is replaced\nP_B = 10/100 #Probability that the second tube is defective if the first is replaced\n\n#Calculation:\nP_AB = P_A*P_B #Probability that the two tubes are defective if the first is replaced\nP_B_A = 9/99 #Probability that the second tube is defective if the first tube is not replaced\nPd_AB = P_A*P_B_A #Probability that both tubes are defective if the first tube is not replaced\n\n#Result:\nprint \"The probability that both tubes are defective if :\"\nprint \"(a) the first is replaced before the second is drawn is :\",changeToFraction(P_AB),\" .\"\nprint \"(b) the first is not replaced before the second is drawn is :\",changeToFraction(Pd_AB),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The probability that both tubes are defective if :\n(a) the first is replaced before the second is drawn is : 1/100 .\n(b) the first is not replaced before the second is drawn is : 1/110 .\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.6, Page number: 515" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The difference between the magnitude of a large earthquake at a nuclear power plant, on the\nRichter scale, and the threshold value of 3.25, is a random variable X having the following\nprobability distribution function (pdf):\n f(X) = 1.7 exp(-1.7X) ; X > 0\n = 0 ; elsewhere\nFind the probability that X will have a value between 2 and 6; that is, P(2 < X < 6).\n'''\n\nfrom sympy import symbols, integrate, exp\n\n#Variable declaration:\nX = symbols('X') #Range of X\nPx = 1.7*(exp(-1.7*X)) #Probability distribution function\n\n#Calculation:\nP = integrate(Px, (X,2,6)) #Probability that X will have a value between 2 and 6\n\n#Result:\nprint \"The probability that X will have a value between 2 and 6 is :\",round(P,4),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The probability that X will have a value between 2 and 6 is : 0.0333 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.7, Page number: 517" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A coolant sprinkler system in a reactor has 20 independent spray components each of which\nfails with a probability of 0.1. The coolant system is considered to \u201cfail\u201d only if four or more\nof the sprays fail. What is the probability that the sprinkler systems fails?\n'''\n\nfrom __future__ import division\nfrom math import factorial\n\n#Variable Declaration:\nn = 20 #Total number of components\np = 0.1 #Probability of success\n\n#Calculations:\ndef binomial(n,p,x):\n P=0\n for x in range(0,x,1):\n P = P + p**x*(1-p)**(n-x)*factorial(n)/(factorial(x)*factorial(n-x))\n return P\n\n#Results:\nprint \"Probability that the sprinkler system fails :\",round((1-binomial(n,p,4))*100,2),\"%\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Probability that the sprinkler system fails : 13.3 %\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.8, Page number: 518" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Assume the time to failure (in hours), t, of a tube in a heat exchanger has aWeibull distribution\nwith a = 1.3x10^-3 and b = 0.77. Find the probability that a tube in a heat exchanger will fail\nin 1000 hours.\n'''\n\nfrom sympy import symbols, integrate, exp\n\n#Variable declaration:\na = 1.3*10**-3 #Constant a\nB = 0.77 #Constant B\nt = symbols('t') #Time (h)\nFt = a*B*t**(B-1)*(exp(-a*t**B)) #Pdf for heat exchanger tube\nPt = integrate(Ft, (t,0,1000)) #Probability that a heat exchanger will fail within 100 hours\n\n#Result:\nprint \"The probability that a tube in a heat exchanger will fail in 1000 hours is :\",round(Pt,2),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The probability that a tube in a heat exchanger will fail in 1000 hours is : 0.23 .\n" + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.9, Page number: 519" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The measurement of the pitch diameter of an acceptable thread of a boiler tube is normally distributed\nwith mean of 0.4008 inch and standard deviation of 0.0004 inch. The specifications are\ngiven as 0.4000+0.0010 and 0.4000-0.0010. What is the probability of a \u201cdefect\u201d occurring?\n'''\n\nfrom __future__ import division\nfrom scipy.stats import norm as f\n\n#Variable declaration:\nm = 0.4008 #Mean(inch)\ns = 0.0004 #Standard Deviation(inch)\nUL = 0.4000+0.001 #Upper Limit\nLL = 0.4000-0.001 #Upper Limit\n\n#Calculation:\nPs = f.cdf(UL,m,s)-f.cdf(LL,m,s)#Probability of meeting specs\nPd = 1-Ps #Probability of defect\n\n#Results:\nprint 'Probability of meeting specifications:',round(Ps*100,2),'%'\nprint 'Probability of Defect:',round(Pd*100,2),'%'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Probability of meeting specifications: 69.15 %\nProbability of Defect: 30.85 %\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.10, Page number: 522" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Three thermometers (A, B, C) are positioned near the outlet of a heat exchanger. Assume that\nthe individual thermometer component lifetimes are normally distributed with ms and\nstandard deviations provided in Table 24.3.\nUsing the following random numbers, simulate the lifetime (time to failure) of the temperature\nrecording system and estimate its m and standard deviation. The lifetime to failure is\ndefined as the time (in weeks) for one of the thermometers to \u201cfail\u201d.\nMonte Carlo simulation is a procedure for mimicking observations on a random variable that\npermits verification of results that would ordinarily require difficult mathematical calculations or\nextensive experimentation. The method normally uses computer programs called random\nnumber generators. A random number is a number selected from the interval (0, 1) in such a\nway that the probabilities that the number comes from any two subintervals of equal length\nare equal. For example, the probability the number is in the sub-interval (0.1, 0.3) is the\nsame as the probability that the number is in the subinterval (0.5, 0.7). Thus, random numbers\nare observations on a random variable X having a uniform distribution on the interval (0, 1). This\nms that the pdf of X is specified by:\nf(x) = 1; 0 , x , 1\n = 0; elsewhere\nThe above pdf assigns equal probability to subintervals of equal length in the interval (0, 1).\nUsing random number generators, Monte Carlo simulation can generate observed values of a\nrandom variable having any specified pdf. For example, to generate observed values of T, the\ntime to failure, when T is assumed to have a pdf specified by f (t), first use the random\nnumber generator to generate a value of X between 0 and 1. The solution is an observed\nvalue of the random variable T having pdf specified by f (t). Additional information on\nMonte Carlo methods is provided in Chapter 26.\n'''\n\nfrom __future__ import division\nfrom numpy import array,minimum\nfrom math import sqrt\n\n#variable Declaration:\nmTa = array([100]*10) #Mean weeks for thermometer failure(A)\nmTb = array([90]*10) #Mean weeks for thermometer failure(B)\nmTc = array([80]*10) #Mean weeks for thermometer failure(C)\nsTa = 30 #Standard deviation (weeks) for thermometer failure(A)\nsTb = 20 #Standard deviation (weeks) for thermometer failure(B)\nsTc = 10 #Standard deviation (weeks) for thermometer failure(C)\nRa = array([0.52,0.80,0.45,0.68,0.59,0.01,0.50,0.29,0.34,0.46]) #Random No corrosponding to A\nRb = array([0.77,0.54,0.96,0.02,0.73,0.67,0.31,0.34,0.00,0.48]) #Random No corrosponding to B\nRc = array([0.14,0.39,0.06,0.86,0.87,0.90,0.28,0.51,0.56,0.82]) #Random No corrosponding to B\nZa = array([0.05,0.84,-0.13,0.47,0.23,-2.33,0.00,-0.55,-0.41,-0.10]) #Normal variable corrosponding to random No for A\nZb = array([0.74,0.10,1.75,-2.05,0.61,0.44,-0.50,-0.41,-3.90,-0.05]) #Normal variable corrosponding to random No for B\nZc = array([-1.08,-0.28,-1.56,1.08,1.13,1.28,-0.58,0.03,0.15,0.92]) #Normal variable corrosponding to random No for C\n\n#Calculations:\nTa = mTa+sTa*Za\nTb = mTb+sTb*Zb\nTc = mTc+sTc*Zc\nTs = minimum(Ta,Tb)\nTs = minimum(Ts,Tc)\nm = array([sum(Ts)/len(Ts)]*10)\ns = sqrt(sum((Ts-m)**2)/(len(Ts)-1))\n\n#Results:\nprint 'Standard deviation :',round(s,1),\" Weeks\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Standard deviation : 25.9 Weeks\n" + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 24.15, Page number: 531" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heat exchanger is located in a relatively large laboratory with a volume of 1100 m^3 at 22\u00b0C and\n1 atm. The exchanger can leak as much as 0.75 gmol of hydrocarbon (HC) from the flowing\nliquid into the room if the exchanger ruptures. A hydrocarbon mole fraction in the air greater\nthan 425 parts per billion (ppb) constitutes a health and safety hazard.\nSuppose the heat exchanger fails and the maximum amount of HC is released instantaneously.\nAssume the air flow in the room is sufficient to cause the room to behave as a continuously\nstirred tank reactor (CSTR^(17 \u2013 19)); i.e., the air composition is spatially uniform.\nCalculate the ppb of hydrocarbon in the room. Is there a health risk? From a treatment pointof-\nview, what can be done to decrease the environmental hazard or to improve the safety of\nthe exchanger?\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nt = 273 #Standard temperature (K)\nv = 0.0224 #Volume of air occupied by 1 gmol of ideal gas (m^3)\nV = 1100 #Volume of heat exchanger (m^3)\nT = 22+273 #Temperature of heat exchanger (K)\nx1 = 0.75 #gmols of hydrocarbon leaking from the exchanger (gmol)\n\n#Calculation:\nn = V*(1/v)*(t/T) #Total number of gmols of air in the room (gmol)\nxHC = (x1/(n+x1))*10**6 #The mole fraction of hydrocarbon in the room (ppm)\n\n#Result:\nprint \"1. The mole fraction of hydrocarbon in the room is :\",round(xHC*1000,-1),\" ppb .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The mole fraction of hydrocarbon in the room is : 16500.0 ppb .\n" + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_26.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_26.ipynb new file mode 100644 index 00000000..2aff0c90 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_26.ipynb @@ -0,0 +1,83 @@ +{ + "metadata": { + "name": "Chapter 26" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 26: Numerical Methods" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 26.8, Page number: 558" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Species A is undergoing a reaction in a tubular flow unit. The following data (see Table 26.1)\nhave been obtained for the rate, 2rA, versus concentration, CA. Using this data, estimate the\ncoefficient kA and a in the equation below.\n-rA = kACA^a\n'''\nfrom __future__ import division\nfrom sympy import symbols,solve,log\nfrom math import exp\n\n#Variable Declaration:\nA,B,r,C = symbols('A B r C');\n\n#Calculation:\nres = solve([A + B*log(2)-log(3),A + B*log(4)-log(12)],[A,B])\nA = round(float(res[A]),4)\nB = round(float(res[B]))\nkA = round(exp(A),2)\na = B\n\n#Result:\nprint 'The equation for rate of reaction is: ',-r,'=',kA*C**a ", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The equation for rate of reaction is: -r = 0.75*C**2.0\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 26.9, Page number: 559" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The viscosity of air as a function of temperature is provided below in Table 26.2. Assuming a\nlinear relationship, obtain the viscosity as a function of temperature.\n'''\nimport scipy.stats as f\n\n#Variable Declaration:\nT = [-40,-20,0,10,12,30,40,50,60,80,100,150,200,250,300,400,500]\nu = [1.51,1.61,1.71,1.76,1.81,1.86,1.90,1.95,2.00,2.09,2.17,2.38,2.57,2.75,2.93,3.25,3.55]\n\n#Calculations:\nB,A,r_value, p_value, std_err = f.linregress(T,u)\n\n#Results:\nprint 'The value of A in regression model is:',round(A,4)\nprint 'The value of B in regression model is:',round(B,4)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The value of A in regression model is: 1.7484\nThe value of B in regression model is: 0.0038\n" + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 26.11, Page number: 561" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A refinery has two catcrackers that can produce various grades of hydrocarbon products.\nBecause of consumer fuel demand, the production of gasoline, home heating oil and diesel\nmust be limited. This information is provided in Table 26.3.\nThe profit on processing U.S. crude oil is $2.00/gal and on Venezuelan crude is $1.60/gal.\nFind the approximate daily processing rate of the two crudes in order to maximize profits.\n'''\nfrom __future__ import division\nfrom scipy.optimize import fmin_cobyla as optimize\n\n#Key:\n#f(x) : Objective Function\n#ci(x)'s : Constraints\n\n#Variable Declaration:\ndef f(x):\t\n return -2.0*x[0] - 1.6*x[1]\n\ndef c1(x):\n return 16820 - x[0]\n\ndef c2(x):\n return 1152 - x[1]\n\ndef c3(x):\n return 1500 - 0.08*x[0] - 0.11*x[1]\n\ndef c4(x):\n return 6000 - 0.29*x[0] - 0.54*x[1]\n\ndef c5(x):\n return 11000 - 0.63*x[0] - 0.35*x[1]\n\ndef c6(x):\n return x[0]\n\ndef c7(x):\n return x[1]\n\n#Calculation\nX = optimize(f,[16820,1152],[c1,c2,c3,c4,c5,c6], disp = 0)\n\n#Result:\nprint \"Maximum Profit is $\",round(-f(X)), \"/day or $\", -365*f(X), \"/year\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Maximum Profit is $ 35483.0 /day or $ 12951368.0 /year\n" + } + ], + "prompt_number": 7 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_27.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_27.ipynb new file mode 100644 index 00000000..c26d4fbd --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_27.ipynb @@ -0,0 +1,188 @@ +{ + "metadata": { + "name": "Chapter 27" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 27: Economics and Finance" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.5, Page number: 575" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A heat exchanger costing $60,000 has an estimated lifetime of 9 years and a salvage value of\n$500. What uniform annual payment must be made into a fund at the end of the year to replace\nTable 27.1 Comparative Methods of Analysis\nYear Straight-line Double-declining Sum-of-the-year\u2019s digits\n0 1.000 1.000 1.000\n1 0.900 0.800 0.818\n2 0.800 0.640 0.655\n3 0.700 0.512 0.510\n4 0.600 0.410 0.383\n5 0.500 0.328 0.274\n6 0.400 0.262 0.183\n7 0.300 0.210 0.110\n8 0.200 0.168 0.056\n9 0.100 0.134 0.018\n10 0.000 0.108 0.000\nthe exchanger if the fund earns 3.375%? What would be the appraisal value of the exchanger at\nthe end of the fifth year based on straight line depreciation?\n'''\n\n#Variable declaration:\ni = 0.03375 #Rate of interest (%)\nn = 9 #Years to the end of life (yr)\nP = 60000 #Cost of exchanger ($)\nL = 500 #Salvage value ($)\nx = 5 #Time after 5 years (yr)\n\n#Calculation:\nSFDF = i/((1+i)**n-1) #Sinking fund depreciation factor\nUAP = (P-L)*SFDF #Uniform annual payment ($)\nB = P-((P-L)/n)*x #Appraisal value after 5 years ($)\n\n#Result:\nprint \"1. The uniform annual payment made into the fund at the of the year is : $\",round(UAP),\" .\"\nprint \"2. The appraisal value of the exchanger at the end of the fifth year is : $\",round(B),\" .\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The uniform annual payment made into the fund at the of the year is : $ 5768.0 .\n2. The appraisal value of the exchanger at the end of the fifth year is : $ 26945.0 .\n" + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.6, Page number: 576" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''The annual operation costs of an outdated heat exchanger/boiler system is $75,000. Under a\nproposed new design, the installation of a new system will require an initial cost of $150,000\nand an annual operating cost of $15,000 for the first five years. Determine the annualized cost for\nthe new heating system by assuming the system has only five years (n) operational life. The interest\nrate (i) is 7%. The capital recovery factor (CRF) or annual payment of a capital investment\ncan be calculated as follows:\nCRF = (A/P)i,n = i(1 + i)^n/(1 + i)^n - 1 (27.11)\nwhere A is the annual cost and P is the present worth.\nCompare the costs for both the outdated and proposed operations.\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\nC = 150000 #Capital cost ($)\ni = 7/100 #Interest rate\nn = 5 #Time (yr)\nOC = 15000 #Operating cost ($)\nA = 75000 #Annual cost for the old process ($)\n\n#Calculation:\nCRF = (i*(1+i)**n)/((1+i)**n-1) #Capital recovery factor\nIC = CRF*C #Initial cost ($)\nAC = IC+OC #Total annualized cost ($)\n\n#Result:\nprint \"The annualized cost for the new heating system is : $\",round(AC),\" .\"\nif (AC<A):\n print \"Since this cost is lower than the annual cost of $75,000 for the old process, the proposed plan should be implemented.\"\nelse :\n print \"Since this cost is higher than the annual cost of $75,000 for the old process, the proposed plan should not be implemented.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The annualized cost for the new heating system is : $ 51584.0 .\nSince this cost is lower than the annual cost of $75,000 for the old process, the proposed plan should be implemented.\n" + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.7, Page number: 577" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Plans are underway to construct and operate a commercial hazardous waste facility in\nDumpsville in the sate of Egabrag. The company is still undecided as to whether to install a\ndouble pipe or shell-and-tube heat exchanger at the plant site to recover energy. The double\npipe (DP) unit is less expensive to purchase and operate than a comparable shell-and-tube\n(ST) system. However, projected energy recover income from the ST unit is higher since it\nwill handle a larger quantity and different temperature steam.\nBased on economic and financial data provided in Table 27.2, select the heat exchanger or\nthat will yield the higher annual profit.\nCalculations should be based on an interest rate of 12% and a process lifetime of 12 years for\nboth exchangers.\nTable 27.2 Costs/Credits Data\nCosts/credits Double pipe (DP) Shell-and-tube (ST)\nCapital ($) 2,625,000 2,975,000\nInstallation ($) 1,575,000 1,700,000\nOperation ($/yr) 400,000 550,000\nMaintenance ($/yr) 650,000 775,000\nIncome ($/yr) 2,000,000 2,500,000\n'''\n\nfrom __future__ import division\n\n#Variable declaration:\ni = 12/100 #Intersest rate\nn = 12 #Lifetime period (yr)\nCC = 2625000 #Capital cost ($)\nIC = 1575000 #Installation cost ($)\n#From table 27.3:\nIc1 = 2000000 #Income credit for double pipe ($/yr)\nIc2 = 2500000 #Income credit for Shell-and-tube ($/yr)\nAC1 = 1728000 #Total annual cost for double pipe ($/yr)\nAC2 = 2080000 #Total annual cost for Shell-and-tube ($/yr)\n\n#Calculation:\nCRF = i/(1-(1+i)**-n) #Capital recovery factor\nDPc = (CC+IC)*CRF #Annual capital and installation costs for the DP unit ($/yr)\nSTc = (CC+IC)*CRF #Annual capital and installation costs for the ST unit ($/yr)\nDPp = Ic1-AC1 #Profit for the DP unit ($/yr)\nSTp = Ic2-AC2 #Profit for the ST unit ($/yr)\n\n#Result:\nprint \"The profit for the shell-and-tube unit is : $\",round(DPp),\"/yr .\"\nprint \"The profit for the double pipe unit is : $\",round(STp),\"/yr .\"\nif (STp>DPp):\n print \"A shell-and-tube heat exchanger should therefore be selected based on the above economic analysis.\"\nelse :\n print \"A double pipe heat exchanger should therefore be selected based on the above economic analysis.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The profit for the shell-and-tube unit is : $ 272000.0 /yr .\nThe profit for the double pipe unit is : $ 420000.0 /yr .\nA shell-and-tube heat exchanger should therefore be selected based on the above economic analysis.\n" + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.8, Page number: 579" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Based on an outgrowth of a 2011 energy audit study for a new process, it is necessary to heat\n50,000 lb/h of an organic liquid form 150 to 330\u00b0F. The liquid is at a pressure of 135 psia. A\nsimple steam-heated shell-and-tube foating-head carbon steel exchanger is the preferred equipment\nchoice. Steam is available at 150 psia (135 psig) and 300 psia (285 psig). The higher\npressure steam should result in a smaller heat exchanger but the steam will cost more. Which\nsteam choice would be better?\nData:\nThe heat capacity of the organic liquid is 0.6 Btu/lb.\u00b0F.\nThe plant on-stream operation factor is expected to be 90%.\nSteam properties are:\n 150 psia 300 psia\nSaturation temperature, \u00b0F 358.0 417.0\nLatent heat (enthalpy), Btu/lb 863.6 809.0\nCost, $/1000 lb 5.20 5.75\nHeat exchanger cost correlation (1998 basis):\nBase cost (BC) = 117 A^0.65\nInstallation factor (IF) = 3.29\nPressure factors (PF):\n100 to 200 psig = 1.15\n200 to 300 psig = 1.20\nCost indexes (CI):\n1998 = 230\n2011 = 360\nCapital cost (C) = (BC)(IF)(PF)(CI)\n'''\n\nfrom __future__ import division\nfrom math import log\n\n#Variable declaration:\nm = 50000 #Mass flowrate of the organic fluid (lb/h)\ncP = 0.6 #The heat capacity of the organic liquid (Btu/lb.\u00b0F)\nT1 = 150 #Initial temperature of organic fluid (\u00b0F)\nT2 = 330 #Final temperature of organic fluid (\u00b0F)\nTs1 = 358 #Saturation temperature for 150 psia (\u00b0F)\nTs2 = 417 #Saturation temperature for 300 psia (\u00b0F)\nL1 = 863.6 #Latent heat for 150 psia (Btu/lb)\nL2 = 809 #Latent heat for 300 psia (Btu/lb)\nc1 = 5.20/1000 #Cost for 150 psia ($/lb)\nc2 = 5.75/1000 #Cost for 300 psia ($/lb)\nCI1 = 230 #Cost index in 1998 \nCI2 = 360 #Cost index in 2011\nIF = 3.29 #Installation factor\nPF1 = 1.15 #Pressure factors for 100 to 200 psig\nPF2 = 1.20 #Pressure factors for 200 to 300 psig\nOP = 90/100 #Plant on-stream operation factor\nh = 365*24 #Hours in a year (h)\n\n#Calculation:\nQ = m*cP*(T2-T1) #Overall heta duty (Btu/h)\nDT1 = Ts1-T1 #Temperature driving force 1 for 150 psia (\u00b0F)\nDT2 = Ts1-T2 #Temperature driving force 2 for 150 psia (\u00b0F)\nLMTD1 = (DT1-DT2)/log(DT1/DT2) #Log-mean temperature difference for 150 psia (\u00b0F)\nDT3 = Ts2-T1 #Temperature driving force 1 for 300 psia (\u00b0F)\nDT4 = Ts2-T2 #Temperature driving force 2 for 300 psia (\u00b0F)\nLMTD2 = (DT3-DT4)/log(DT3/DT4) #Log-mean temperature difference for 1300 psia (\u00b0F)\nA1 = Q/(138*LMTD1) #Required heat transfer area for 150 psia (ft^2)\nA2 = Q/(138*LMTD2) #Required heat transfer area for 300 psia (ft^2)\nBC1 = 117*A1**0.65 #Base cost for 150 psia ($)\nBC2 = 117*A2**0.65 #Base cost for 13000 psia ($)\nC1 = BC1*(CI2/CI1)*IF*PF1 #Capital cost for 150 psia ($)\nC2 = BC2*(CI2/CI1)*IF*PF2 #Capital cost for 300 psia ($)\nS1 = Q*(h*OP)/L1 #Steam requirement for 150 psia (lb/yr)\nS2 = Q*(h*OP)/L2 #Steam requirement for 300 psia (lb/yr)\nSC1 = S1*c1 #Annual steam cost for 150 psia ($/yr)\nSC2 = S2*c2 #Annual steam cost for 300 psia ($/yr)\n\n#Result:\nprint \"1. The capital cost for 150 psia is : $\",round(C1,-3),\" .\"\nprint \" The capital cost for 300 psia is : $\",round(C2,-3),\" .\"\nprint \"2. The annual steam cost for 150 psia is : $\",round(SC1,-3),\"/yr .\"\nprint \" The annual steam cost for 300 psia is : $\",round(SC2,-3),\"/yr .\"\nif (C1<C2 and SC1>SC2):\n print \"The 300-psia exchanger costs less to purchase and install, but it costs more to operate. Choosing the more expensive, 150-psia exchanger is the obvious choice.\"\nelif (C1>C2 and SC1<SC2):\n print \"The 150-psia exchanger costs less to purchase and install, but it costs more to operate. Choosing the more expensive, 300-psia exchanger is the obvious choice.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "1. The capital cost for 150 psia is : $ 36000.0 .\n The capital cost for 300 psia is : $ 26000.0 .\n2. The annual steam cost for 150 psia is : $ 256000.0 /yr .\n The annual steam cost for 300 psia is : $ 303000.0 /yr .\nThe 150-psia exchanger costs less to purchase and install, but it costs more to operate. Choosing the more expensive, 300-psia exchanger is the obvious choice.\n" + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.9, Page number: 581" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Two small commercial power plant designs are under consideration. The first design involves a\ntraditional boiler (TB) and the second a fluidized fed (FB). For the TB system, the total capital\ncost (TCC_TB) is $2.5 million, the annual operating costs (AOC_TB) are $1.2 million, and the annual\nrevenue generated from the facility (R_TB) is $3.6 million. For the FB system, TCC_TB, AOC_TB and R_TB are\n$3.5, 1.4 and 5.3 million, respectively. Using straight-line depreciation and the discounted cash\nflow method, which unit is more attractive? Assume a 10-yr facility lifetime and a 2 yr construction\nperiod. Note that the solution involves the calculation of the rate of return for each of the two\nproposals.(11)\n'''\n\nfrom __future__ import division\nfrom sympy import symbols,solve\nfrom scipy.optimize import fsolve\n\n#Variable declaration:\nTCC_TB = 2500000 #Total capital cost ($)\nR_TB = 3600000 #R_TBevenue generated from the facility ($)\nAOC_TB = 1200000 #Annual operating costs ($)\nTCC_FB = 3500000 #Total capital cost ($)\nR_FB = 5300000 #R_TBevenue generated from the facility ($)\nAOC_FB = 1400000 #Annual operating costs ($)\nn = 10 \t#Time of facility (yr)\n\n#Calculation:\nD = 0.1*TCC_TB #Depriciation ($)\nWC = 0.1*TCC_TB #Working capital ($)\nTI = R_TB-AOC_TB-D #Taxable income ($)\nIT = 0.5*TI #Income tax to be paid ($)\nA = R_TB-AOC_TB-IT #After-tax cash flow ($)\ndef eqTB(i):\n\tx = (((1+i)**n-1)/(i*(1+i)**n))*A + (1/(1+i)**n)*WC #Equation for computing rate of return for TB unit\n\ty = WC + 0.5*TCC_TB + 0.5*TCC_TB*(1+i)**1 #Equation for computing rate of return for TB unit\n\treturn x-y\niTB = round(fsolve(eqTB,0.8)*100,1) #Rate of return for TB unit (%)\n\nD = 0.1*TCC_FB #Depriciation ($)\nWC = 0.1*TCC_FB #Working capital ($)\nTI = R_FB-AOC_FB-D #Taxable income ($)\nIT = 0.5*TI #Income tax to be paid ($)\nA = R_FB-AOC_FB-IT #After-tax cash flow ($)\n\ndef eqFB(i):\n\tx = (((1+i)**n-1)/(i*(1+i)**n))*A + (1/(1+i)**n)*WC #Equation for computing rate of return for FB unit\n\ty = WC + 0.5*TCC_FB + 0.5*TCC_FB*(1+i)**1 #Equation for computing rate of return for FB unit\n\treturn x-y\niFB = round(fsolve(eqFB,0.8)*100,1) #Rate of return for FB unit (%)\n\n#Results:\nprint \"The rate of return for TB unit is:\",round(iTB),\" %.\"\nprint \"The rate of return for FB unit is:\",round(iFB,1),\" %.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "The rate of return for TB unit is: 40.0 %.\nThe rate of return for FB unit is: 44.8 %.\n" + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.10, Page number: 582" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A stream of 100,000 acfm flue gas from a utility facility is to be cooled in an air preheater. You\nhave been requested to find the best unit to install to cool the flue gas and preheat the combustion\nair feed to the boiler. A reputable vendor has provided information on the cost of three units, as\nwell as installation, operating, and maintenance costs. Table 27.4 summarizes all the data you\nhave collected. Determine what preheater you would select in order to minimize costs on an\nannualized basis.\n'''\n\n#Variable declaration:\nf = 100000 #Flow rate of flue gas (acfm)\ni = 0.1 #Interest rate\n#From table 27.4:\n#For finned preheater:\nac1 = 3.1 #Equipment cost ($/acfm)\nac2 = 0.8 #Installation cost ($/acfm)\nac3 = 0.06 #Operating cost ($/acfm-yr)\nac4 = 14000 #Maintenance cost ($/yr)\nan = 20 #Lifetime (yr)\n#For 4-pass preheater:\nbc1 = 1.9 #Equipment cost ($/acfm)\nbc2 = 1.4 #Installation cost ($/acfm)\nbc3 = 0.06 #Operating cost for ($/acfm-yr)\nbc4 = 28000 #Maintenance cost ($/yr)\nbn = 15 #Lifetime of (yr)\n#For 2-pass preheater:\ncc1 = 2.5 #Equipment cost ($/acfm)\ncc2 = 1.0 #Installation cost ($/acfm)\ncc3 = 0.095 #Operating cost for ($/acfm-yr)\ncc4 = 9500 #Maintenance cost for ($/yr)\ncn = 20 #Lifetime of (yr)\n\n#Calculation:\n#For Finned preheater:\naEC = f*ac1 #Total equipment cost ($)\naIC = f*ac2 #Total installation cost ($)\naOC = f*ac3 #Total operating cost ($)\naMC = f*ac4 #Total maintenance cost ($)\naCRF = (i*(1+i)**an)/((1+i)**an-1) #Capital recovery factor\naAEC = aEC*aCRF #Equipment annual cost ($/yr)\naAIC = aIC*aCRF #Installation annual cost($/yr)\naAOC = ac3*f #Annual operating cost ($)\naAMC = ac4 #Annual maintenance cost ($)\naTAC = aAEC+aAIC+aAOC+aAMC #Total annual cost ($)\n\n#For 4-pass preheater:\nbEC = f*bc1 #Total equipment cost ($)\nbIC = f*bc2 #Total installation cost ($)\nbOC = f*bc3 #Total operating cost ($)\nbMC = f*bc4 #Total maintenance cost ($)\nbCRF = (i*(1+i)**bn)/((1+i)**bn-1) #Capital recovery factor\nbAEC = bEC*bCRF #Equipment annual cost ($/yr)\nbAIC = bIC*bCRF #Installation annual cost($/yr)\nbAOC = bc3*f #Annual operating cost ($)\nbAMC = bc4 #Annual maintenance cost ($)\nbTAC = bAEC+bAIC+bAOC+bAMC #Total annual cost ($)\n#For 2-pass preheater:\ncEC = f*cc1 #Total equipment cost ($)\ncIC = f*cc2 #Total installation cost ($)\ncOC = f*cc3 #Total operating cost ($)\ncMC = f*cc4 #Total maintenance cost ($)\ncCRF = (i*(1+i)**cn)/((1+i)**cn-1) #Capital recovery factor\ncAEC = cEC*cCRF #Equipment annual cost ($/yr)\ncAIC = cIC*cCRF #Installation annual cost($/yr)\ncAOC = cc3*f #Annual operating cost ($)\ncAMC = cc4 #Annual maintenance cost ($)\ncTAC = cAEC+cAIC+cAOC+cAMC #Total annual cost ($)\n\n#Result:\nprint \"Total annual cost for finned preheater is : $\",round(aTAC),\" .\"\nprint \"Total annual cost for 4-pass preheater is : $\",round(bTAC),\" .\"\nprint \"Total annual cost for 2-pass preheater is : $\",round(cTAC),\" .\"\nif (cTAC<aTAC and cTAC<bTAC):\n print \"According to the analysis, the 2-pass exchanger is the most economically attractive device since the annual cost is the lowest.\"\nelif (bTAC<aTAC and bTAC<cTAC):\n print \"According to the analysis, the 4-pass exchanger is the most economically attractive device since the annual cost is the lowest.\"\nelif (aTAC<cTAC and aTAC<bTAC):\n print \"According to the analysis, the finned exchanger is the most economically attractive device since the annual cost is the lowest.\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Total annual cost for finned preheater is : $ 65809.0 .\nTotal annual cost for 4-pass preheater is : $ 77386.0 .\nTotal annual cost for 2-pass preheater is : $ 60111.0 .\nAccording to the analysis, the 2-pass exchanger is the most economically attractive device since the annual cost is the lowest.\n" + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.12, Page number: 584" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Shannon O\u2019Brien, a recent graduate from Manhattan College\u2019s prestigious chemical engineering\nprogramwas given an assignment to design the most cost-effective heat exchanger to recover\nenergy from a hot flue gas at 500\u00b0F. The design is to be based on pre-heating 1008F incoming air\n(to be employed in the boiler) to a temperature that will result in the maximum annual profit to\nthe utility. A line diagram of the proposed countercurrent exchanger is provided in Figure 27.2.\nHaving just completed a heat transfer course with Dr. Flynn and a thermodynamics course\nwith the infamous Dr. Theodore, Shannon realizes that their are two costs that need to be\nconsidered:\n1. the heat exchanger employed for energy recovery, and\n2. the \u201cquality\u201d (from an entropy perspective) of the recovered energy\u2014refer to Chapter\n20 for additional details.\nShe also notes that the higher the discharge temperature of the heated air, t, the smaller will be\nthe temperature difference driving force, and the higher the area requirement of the exchanger\nand the higher the equipment cost. Alternatively, with a higher t, the \u201cquality\u201d of the recovered\nenergy is higher, thus leading to an increase in recovered profits (by reducing fuel costs).\nBased on similar system designs, Ricci Consultants (RC) has provided the following annual\neconomic models:\nRecovered energy profit: A(t - tc); A = $/yr.\u00b0F\nExchange cost: B/(TH - t); B = $.\u00b0F/yr\nFor the above system, RC suggests a value for the coefficients in the cost model be set at:\nA = 10\nB = 100,000\nEmploying the above information, Shannon has been asked to calculate a t that will\n1. provide breakeven operation\n2. maximize profits\nShe is also required to perform the calculation if A = 10, B = 4000, and A = 10, B = 400,000.\nFinally, an analysis of the results is requested.\n'''\nfrom __future__ import division\nfrom numpy import roots\nfrom math import sqrt\n\n#Variable declaration:\nTH = 500 #Hot stream temperature at exchanger 1 (\u00b0F)\ntc = 100 #Cold stream temperature at exchanger 2 (\u00b0F)\nA = 10 #Constant A\nB1 = 100000 #Constant B1\nB2 = 4000 #Constant B2\nB3 = 400000 #Constant B3\n\n#Calculations:\n#It forms equation fo form t^2 - t(Th-tc) +tcTH +B/A\nt1 = roots([1, -(TH+tc),(tc*TH + B1/A) ]); #Roots\ntmax1 = TH - sqrt(B1/A) #Upon maximising profit\nt2 = roots([1, -(TH+tc),(tc*TH + B2/A) ]); #Roots\ntmax2 = TH - sqrt(B2/A) #Upon maximising profit\nt3 = roots([1, -(TH+tc),(tc*TH + B3/A) ]); #Roots\ntmax3 = TH - sqrt(B3/A) #Upon maximising profit\n\n#Results:\nprint 'tBE for case 1: ',round(t1[0]),'\u00b0F',round(t1[1]),'\u00b0F'\nprint 'tmax1:', round(tmax1),'\u00b0F'\nprint 'tBE for case 2: ',round(t2[0]),'\u00b0F',round(t2[1]),'\u00b0F'\nprint 'tmax1:', round(tmax2),'\u00b0F'\nprint 'tBE for case 1: ',round(t3[0]),'\u00b0F',round(t3[1]),'\u00b0F'\nprint 'tmax1:', round(tmax3),'\u00b0F'", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "tBE for case 1: 473.0 \u00b0F 127.0 \u00b0F\ntmax1: 400.0 \u00b0F\ntBE for case 2: 499.0 \u00b0F 101.0 \u00b0F\ntmax1: 480.0 \u00b0F\ntBE for case 1: 300.0 \u00b0F 300.0 \u00b0F\ntmax1: 300.0 \u00b0F\n" + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 27.15, Page number: 588" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''Refer to the two previous examples. Calculate values of qA and qB that will maximize the profit\nfrom this process. Also calculate the annual (365 day basis) profit based on this condition.\n'''\n\nfrom __future__ import division\nfrom scipy.optimize import fmin_cobyla as optimize\n\n#Key:\n#f(x) : Objective Function\n#ci(x)'s : Constraints\n\n#Variable Declaration:\ndef f(x):\t\n return -1.70*x[0] - 2*x[1]\n\ndef c1(x):\n return 8000 - x[0]\n\ndef c2(x):\n return 6000 - x[1]\n\ndef c3(x):\n return 12000 - 0.75*x[0] - 0.40*x[1]\n\ndef c4(x):\n return 6000 - 0.60*x[0] - 0.25*x[1]\n\ndef c5(x):\n return x[0]\n\ndef c6(x):\n return x[1]\n\n#Calculation\nX = optimize(f,[7000,6000],[c1,c2,c3,c4,c5,c6], disp = 0)\n\n#Result:\nprint \"Maximum Profit is $\",-f(X), \"/day or $\", -365*f(X), \"/year\"", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Maximum Profit is $ 24750.0 /day or $ 9033750.0 /year\n" + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_28.ipynb b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_28.ipynb new file mode 100644 index 00000000..ada59276 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/Chapter_28.ipynb @@ -0,0 +1,41 @@ +{ + "metadata": { + "name": "Chapter 28" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": "Chapter 28: Open-Ended Problems" + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": "ILLUSTRATIVE EXAMPLE 28.11, Page number: 600" + }, + { + "cell_type": "code", + "collapsed": false, + "input": "'''A plant has three streams to be heated (see Table 28.3) and three streams to be cooled (see\nTable 28.4). Cooling water (90\u00b0F supply, 155\u00b0F return) and steam (saturated at 250 psia) are\navailable. Note that the saturated steam at 250 psia has a temperature of 401\u00b0F. Calculate the\nheating and cooling duties and indicate what utility (or utilities) should be employed.\n'''\n\n#Variable declaration:\n#From table 28.3:\n#For stream 1 to be heated:\nhm1 = 50000 #Mass flowrate (lb/h)\nhcP1 = 0.65 #Heat capacity (Btu/lb.\u00b0F)\nhTi1 = 70 #Inlet temperature (\u00b0F)\nhTo1 = 300 #Outlet temperature (\u00b0F)\n#For stream 2 to be heated:\nhm2 = 60000 #Mass flowrate (lb/h)\nhcP2 = 0.58 #Heat capacity (Btu/lb.\u00b0F)\nhTi2 = 120 #Inlet temperature (\u00b0F)\nhTo2 = 310 #Outlet temperature (\u00b0F)\n#For stream 3 to be heated:\nhm3 = 80000 #Mass flowrate (lb/h)\nhcP3 = 0.78 #Heat capacity (Btu/lb.\u00b0F)\nhTi3 = 90 #Inlet temperature (\u00b0F)\nhTo3 = 250 #Outlet temperature (\u00b0F)\n#From table 28.4:\n#For stream 1 to be cooled:\ncm1 = 60000 #Mass flowrate (lb/h)\nccP1 = 0.70 #Heat capacity (Btu/lb.\u00b0F)\ncTi1 = 420 #Inlet temperature (\u00b0F)\ncTo1 = 120 #Outlet temperature (\u00b0F)\n#For stream 2 to be cooled:\ncm2 = 40000 #Mass flowrate (lb/h)\nccP2 = 0.52 #Heat capacity (Btu/lb.\u00b0F)\ncTi2 = 300 #Inlet temperature (\u00b0F)\ncTo2 = 100 #Outlet temperature (\u00b0F)\n#For stream 3 to be cooled:\ncm3 = 35000 #Mass flowrate (lb/h)\nccP3 = 0.60 #Heat capacity (Btu/lb.\u00b0F)\ncTi3 = 240 #Inlet temperature (\u00b0F)\ncTo3 = 90 #Outlet temperature (\u00b0F)\n\n#Calculation:\nH1 = hm1*hcP1*(hTo1-hTi1) #Heating duty for stream 1 (Btu/h)\nH2 = hm2*hcP2*(hTo2-hTi2) #Heating duty for stream 2 (Btu/h)\nH3 = hm3*hcP3*(hTo3-hTi3) #Heating duty for stream 1 (Btu/h)\nH = H1+H2+H3 #Total heating duty (Btu/h)\nC1 = cm1*ccP1*(cTi1-cTo1) #Cooling duty for stream 1 (Btu/h)\nC2 = cm2*ccP2*(cTi2-cTo2) #Cooling duty for stream 2 (Btu/h)\nC3 = cm3*ccP3*(cTi3-cTo3) #Cooling duty for stream 1 (Btu/h)\nC = C1+C2+C3 #Total Cooling duty (Btu/h)\n\n#Result:\nprint \"Table: Duty Requirements.\"\nprint \"Stream Duty, Btu/h\"\nprint \"1 \",round(H1)\nprint \"2 \",round(H2)\nprint \"3 \",round(H3)\nprint \"4 \",round(C1)\nprint \"5 \",round(C2)\nprint \"6 \",round(C3)", + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": "Table: Duty Requirements.\nStream Duty, Btu/h\n1 7475000.0\n2 6612000.0\n3 9984000.0\n4 12600000.0\n5 4160000.0\n6 3150000.0\n" + } + ], + "prompt_number": 11 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/README.txt b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/README.txt new file mode 100644 index 00000000..7cc312e6 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/README.txt @@ -0,0 +1,10 @@ +Contributed By: Amitesh Kumar +Course: btech +College/Institute/Organization: Manav Rachna International University +Department/Designation: Electrical & Electronic Engg. +Book Title: Heat Transfer Applications for the Practicing Engineer +Author: Louis Theodore +Publisher: John Wiley & Sons Inc. Publication +Year of publication: 2011 +Isbn: 9780470643723 +Edition: 1st
\ No newline at end of file diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/economics.png b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/economics.png Binary files differnew file mode 100644 index 00000000..e58b6698 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/economics.png diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/forced_convention.png b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/forced_convention.png Binary files differnew file mode 100644 index 00000000..44694e5e --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/forced_convention.png diff --git a/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/shell_heat.png b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/shell_heat.png Binary files differnew file mode 100644 index 00000000..427304c9 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer_/screenshots/shell_heat.png |
