diff options
| author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
| commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
| tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /Heat_Transfer_Applications_for_the_Practicing_Engineer | |
| parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
| download | Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.gz Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.bz2 Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.zip | |
adding book
Diffstat (limited to 'Heat_Transfer_Applications_for_the_Practicing_Engineer')
28 files changed, 11531 insertions, 0 deletions
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 100755 index 00000000..dd27a183 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_03.ipynb @@ -0,0 +1,457 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:18e19bec17b5f55566a2079756ce3fc602f406d3355e580aba8938f8d7833673" + }, + "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": [ + "\n", + " \n", + "#Variable Declaration:\n", + "Q1 = 8.03 #Years(part 1)\n", + "D = 365 #Days in a year\n", + "H = 24 #Hours in a day\n", + "M = 60 #Minutes in an hour\n", + "S = 60 #Seconds in a minute\n", + "Q2 = 150 #Miles per hour(part 2)\n", + "FM = 5280 #Feet in a mile\n", + "YF = 1.0/3.0 #Yard in a feet\n", + "Q3 = 100 #Meter per second square(part 3)\n", + "Cmm = 100 #Centimeter in a meter\n", + "FC = 1.0/30.48 #Feet in a centimeter\n", + "SsMs = 60**2 #Second square in a minute square\n", + "Q4 = 0.03 #Gram per centimeter cube (part 4)\n", + "PG = 1.0/454.0 #Pound in a gram\n", + "CF = (30.48)**3 #Centimeter in a feet\n", + " \n", + "#Calculation:\n", + "A1 = Q1*D*H*M*S #Seconds (s)\n", + "A2 = Q2*FM*YF #Yards per hour (yd/hr)\n", + "A3 = Q3*Cmm*FC*SsMs #Feet per min square (ft/min^2)\n", + "A4 = Q4*PG*CF #Pound per feet cube (lb/ft^3)\n", + " \n", + "#Results:\n", + "print \"1. Seconds in\",Q1,\"year is:\",round(A1/10**8,2),\" x 10**8 s\"\n", + "print \"2. Yards per hour in\",Q2,\"miles per hour is:\",round(A2/10**5,1),\" x 10**5 yd/h\"\n", + "print \"3. Feets per minute square in\",Q3,\"meter per square is:\",round(A3/10**6,3),\" x 10**6 ft/min^2\"\n", + "print \"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\n", + "2. Yards per hour in 150 miles per hour is: 2.6 x 10**5 yd/h\n", + "3. Feets per minute square in 100 meter per square is: 1.181 x 10**6 ft/min^2\n", + "4. 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": [ + "\n", + " \n", + "#Variable Declaration:\n", + "Q1 = 32.2 #Gravitational acceleration (ft/s^2) (part 1)\n", + "CF = 32.2 #Conversion factor (lb.ft/lbf.s^2)\n", + "M = 100 #Mass (lb)\n", + "SA = 3 #Surface area (in^2)\n", + "FsIs = (1.0/12.0)**2 #Feet square in a inch square\n", + "Q2 = 14.7 #Atmospheric pressure (psi) (part 2)\n", + "GP = 35 #Gauge Pressure (psig)\n", + " \n", + "#Caculations:\n", + "F = M*Q1/CF #Force (lbf)\n", + "P = F/SA/FsIs #Pressure at the base (lbf/ft^2)\n", + "Pa = GP+Q2 #Absolute pressure (psia)\n", + " \n", + "#Results:\n", + "print \"1. Pressure at the base is:\",round(P),\"lbf/ft^2\"\n", + "print \"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\n", + "2. 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": [ + "\n", + " \n", + "#Variable Declaration:\n", + "Q1 = 20.0 #Mass (lb) (part 1)\n", + "MH = 1.008 #Molecular weight of H (lb/lbmol)\n", + "MO = 15.999 #Molecular weight of O (lb/lbmol)\n", + "Q2 = 454 #Gram in pound (part 2)\n", + "Q3 = 6.023*10**23 #Avogadro nuber (part 3)\n", + " \n", + "#Calculations:\n", + "Mol = 2*MH+MO #Molecular weight of water (lb/lbmol)\n", + "A1 = Q1/Mol #Pound.moles of water (lbmol)\n", + "A2 = Q1*Q2/Mol #Gram.moles of water (gmol)\n", + "A3 = A2*Q3 #Molecules of water (molecules)\n", + " \n", + "#Results:\n", + "print \"1. Pound.moles of water is:\",round(A1,2),\"lbmol water\"\n", + "print \"2. Gram.moles of water is:\",round(A2),\"gmol water\"\n", + "print \"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\n", + "2. Gram.moles of water is: 504.0 gmol water\n", + "3. 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": [ + "\n", + " \n", + "#Variable declaration:\n", + "SG = 0.92 #Specific gavity of liquid, methanol\n", + "DW = 62.4 #Density of reference substance, water (lb/ft^3)\n", + " \n", + "#Calculation:\n", + "DM = SG*DW #Density of methanol (lb/ft^3)\n", + " \n", + "#Result:\n", + "print \"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": [ + "\n", + " \n", + "#Variable declaration:\n", + "SG = 0.8 #Specific Gravity\n", + "AV = 0.02 #Absolute Viscosity (cP)\n", + "cP = 1 #Viscosity of centipoise (cP)\n", + "VcP = 6.72 * 10**-4 #Pound per feet.sec in a centipoise (lb/ft.s)\n", + "pR = 62.43 #Reference density (lb/ft^3)\n", + " \n", + "#Calculations:\n", + "u = AV*VcP/cP #Viscosity of gas (lb/ft.s)\n", + "p = SG*pR #Density of gas (lb/ft^3)\n", + "v = u/p #Kinematic viscosity of gas (ft^2/s)\n", + " \n", + "#Result:\n", + "print \"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": [ + "\n", + " \n", + "#Variable declaration:\n", + "X = 7.0 #Coordinate X of H2SO4\n", + "Y = 24.8 #Coordinate Y of H2SO4\n", + "S = 45 #Slope\n", + " \n", + "#Calculations:\n", + "#From the figure C.1 we found the intersection of curves mu = 12cP\n", + "mu = 12\n", + " \n", + "#Results:\n", + "print \"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": [ + "\n", + " \n", + "#Variable declaration:\n", + "CpM = 0.61 #Heat capacity of methanol (cal/g.\u00b0C)\n", + "G = 454 #Grams in a pound\n", + "B = 1.0/252.0 #Btu in a calorie\n", + "C = 1.0/1.8 #Degree celsius in a degree fahrenheit\n", + " \n", + "#Calculation:\n", + "Cp = CpM*G*B*C #Heat capacity in English units (Btu/lb.\u00b0F)\n", + " \n", + "#Result:\n", + "print \"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": [ + "\n", + " \n", + "#Variable declaration:\n", + "kM = 0.0512 #Thermal conductivity of methanol (cal/m.s\u00b0C)\n", + "B = 1.0/252.0 #Btu in a calorie\n", + "M = 0.3048 #Meters in a feet\n", + "S = 3600 #Seconds in an hour\n", + "C = 1.0/1.8 #Degree celsius in a degree fahrenheit\n", + " \n", + "#Calculation:\n", + "k = kM*B*M*S*C #Thermal conductivity in English units (Btu/ft.h.\u00b0F)\n", + " \n", + "#Result:\n", + "print \"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": [ + "\n", + " \n", + "#Variable declaration:\n", + "D = 5 #Diameter of pipe (ft)\n", + "V = 10 #Fluid velocity (ft/s)\n", + "p = 50 #Fluid density (lb/ft^3)\n", + "u = 0.65 #Fluid viscosity (lb/ft.s)\n", + "F = 1.0/12.0 #Feet in an inch\n", + "VCp = 6.72*10**-4 #Viscosity of centipoise (lb/ft.s)\n", + " \n", + "#Calculation:\n", + "A = D*V*p*F/u/VCp #Reynolds Number\n", + " \n", + "#Result:\n", + "if(A>2100):\n", + " print \"The Reynolds number is :\",round(A,-2),\"; therefore, the flow is turbulent.\"\n", + "elif(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": [ + "\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.\n", + "z1 = 0 #Intial height from ground level (m)\n", + "z2 = 10 #Final height from ground level (m)\n", + "PE1 = 0 #Initial potential energy at z1 (J)\n", + "m = 1 #Mass of water (kg)\n", + "g = 9.8 #Gravitational acceleration (m/s^2)\n", + "gc = 1 #Conversion factor\n", + " \n", + "#Calculations:\n", + "PE2 = m*(g/gc)*z2 #Final potential energy at z2 (J)\n", + " \n", + "#Result:\n", + "print \"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 100755 index 00000000..3fece5a7 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_04.ipynb @@ -0,0 +1,555 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:da9a4303c6fc07c594d52c82662dd9ab6e23eae788f020d82b784ed3a6873663" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Vx_in = 420 #Entry Velocity in X direction (m/s)\n", + "Vx_out = 0 #Exit Velocity in X direction (m/s)\n", + "Vy_in = 0 #Entry Velocity in Y direction (m/s)\n", + "Vy_out = 420 #Exit Velocity in Y direction (m/s)\n", + "m = 0.15 #Rate of water entrained by the steam (kg/s)\n", + "lb = 1.0/4.46 #Pound force in a newton force\n", + "\n", + "#Calculations:\n", + "Mx_out = m*Vx_out #Rate of change of momentum at entry in x-direction (kg.m)\n", + "Mx_in = m*Vx_in #Rate of change of momentum at exit in x-direction (kg.m)\n", + "My_out = m*Vy_out #Rate of change of momentum at entry in y-direction (kg.m)\n", + "My_in = m*Vy_in #Rate of change of momentum at exit in y-direction (kg.m)\n", + "Fxgc = (Mx_out - Mx_in)*lb #Force in X direction (lbf)\n", + "Fygc = (My_out - My_in)*lb #Force in X direction (lbf)\n", + "\n", + "#Results:\n", + "if Fxgc < 1:\n", + " print \"The x-direction supporting force acting on the 90\u00b0 elbow is :\",round(-Fxgc,1),\" lbf acting toward the left. \"\n", + "else:\n", + " print \"The x-direction supporting force acting on the 90\u00b0 elbow is :\",round(Fxgc,1),\" lbf acting toward the right. \"\n", + "if Fygc < 1:\n", + " print \"The y-direction supporting force acting on the 90\u00b0 elbow is :\",round(-Fygc,1),\" lbf acting downwards. \" \n", + "else:\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. \n", + "The 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": [ + "\n", + "\n", + "from math import sqrt,degrees,atan2\n", + "\n", + "#Variable declaration:\n", + "Fx = -63 #Force component in X direction (N)\n", + "Fy = 63 #Force component in Y direction (N)\n", + "lbf = 0.22481 #Pound-forrce in unit newton (lbf)\n", + "\n", + "#Calculations:\n", + "Fr = sqrt(Fx**2 + Fy**2)*lbf #The resultant supporting force (lbf)\n", + "u = degrees(atan2(Fy,Fx)) #Angle between the positive x axis and the direction of the force (degrees)\n", + "\n", + "#Result: \n", + "if (0<u<90):\n", + " print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201cnortheast\u201d direction.\"\n", + "elif (90<u<180):\n", + " print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201cnorthwest\u201d direction.\"\n", + "elif (180<u<270):\n", + " print \"The supporting force is :\",round(Fr,1),\" lbf acting at\",u,\"\u00b0 i.e in the \u201csouthwest\u201d direction.\"\n", + "elif (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": [ + "\n", + "\n", + "#Variable declaration:\n", + "R1_in = 10000 #Rate of fuel fed into the boiler (lb/h)\n", + "R2_1n = 20000 #Rate of air fed into the boiler (lb/h)\n", + "R3_in = 2000 #Rate of methane fed into the boiler (lb/h)\n", + "\n", + "#Calculations:\n", + "m_in = R1_in + R2_1n + R3_in #Rate of mass in (lb/h)\n", + "m_out = m_in #Rate of mass out (lb/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "E1 = 65 #Efficiency of spray tower (%)\n", + "E2 = 98 #Efficiency of packed column (%)\n", + "m_in = 76 #Mass flow rate of HCl entering the system (lb/h)\n", + "\n", + "#Calculations:\n", + "m1_out = (1 - E1/100.0)*m_in #Mass flow rate of HCl leaving the spray tower (lb/h)\n", + "m2_out = (1 - E2/100.0)*m1_out #Mass flow rate of HCl entering the packed column (lb/h)\n", + "E = (m_in - m2_out)/m_in #Overall fractional efficiency (%)\n", + "\n", + "#Result:\n", + "print \"The mass flow rate of HCl leaving the spray tower is :\",round(m1_out,2),\" lb/h HCL\"\n", + "print \"The mass flow rate of HCl entering the packed column is :\",round(m2_out,3),\" lb/h HCL\"\n", + "print \"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\n", + "The mass flow rate of HCl entering the packed column is : 0.532 lb/h HCL\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "m1 = 1000 #Flowrate data 1 (lb/min)\n", + "m2 = 1000 #Flowrate data 2 (lb/min)\n", + "m4 = 200 #Flowrate data 4 (lb/min)\n", + "\n", + "#Calculations:\n", + "m5 = m1 + m2 - m4 #Flowrate data 5 (lb/min)\n", + "m6 = m2 #Flowrate data 6 (lb/min)\n", + "m = m5 - m6 #Flowrate of water lost in operation (lb/min)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "q1 = 1000.0 #Volumetric flowrate from tank 1 (gal/day)\n", + "q2 = 1000.0 #Volumetric flowrate from tank 2 (gal/day)\n", + "q3 = 2000.0 #Volumetric flowrate from tank 3 (gal/day)\n", + "q4 = 200.0 #Volumetric flowrate from tank 4 (gal/day)\n", + "q5 = 1800.0 #Volumetric flowrate from tank 5 (gal/day)\n", + "q6 = 1000.0 #Volumetric flowrate from tank 6 (gal/day)\n", + "C1 = 4.0 #Phosphate concentration in tank 1 (ppm)\n", + "C2 = 0.0 #Phosphate concentration in tank 2 (ppm)\n", + "C3 = 2.0 #Phosphate concentration in tank 3 (ppm)\n", + "C4 = 20.0 #Phosphate concentration in tank 4 (ppm)\n", + "C5 = 0.0 #Phosphate concentration in tank 5 (ppm)\n", + "C6 = 0.0 #Phosphate concentration in tank 6 (ppm)\n", + "Cf = 120000.0 #conversion factor for water (gal/10**6lb)\n", + "\n", + "#Calculations:\n", + "C1q1 = C1*q1/Cf #Data 1 (lb/day)\n", + "C2q2 = C2*q2/Cf #Data 2 (lb/day)\n", + "C3q3 = C3*q3/Cf #Data 3 (lb/day)\n", + "C4q4 = C4*q4/Cf #Data 4 (lb/day)\n", + "C5q5 = C5*q5/Cf #Data 5 (lb/day)\n", + "C6q6 = C6*q6/Cf #Data 6 (lb/day)\n", + "\n", + "#Results:\n", + "if (((C1q1 + C2q2) == C3q3) and C3q3 == (C4q4 + C5q5) and C5q5 == C6q6 and C2q2 == C6q6):\n", + " print \"The data appear to be consistent .\"\n", + "else:\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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Dz = 3000 #Height (ft)\n", + "V0 = 500000 #Flowrate of water (gal/min)\n", + "n = 30 #Turbine efficiency (%)\n", + "m = 0.3048 #Meters in a feet\n", + "m3 = 0.00378 #Meters-cube in a gallon\n", + "g = 9.8 #Gravitational acceleration (m/s^2)\n", + "gc = 1 #Conversion factor\n", + "MW = 10**(-6) #Megawatt in newton-meter-per-second\n", + "\n", + "#Calculations:\n", + "V1 = (V0*m3)*1000.0/60.0 #The mass flow rate of the water in kilograms/second (kg/s)\n", + "DPE = V1*g*Dz*m/gc*MW #The loss in potential energy (MW)\n", + "AP = n/100.0*DPE #The actual power output (MW)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "n = 111.4 #Flowrate of air stream (lbmol/min)\n", + "H1 = 1170 #Average heat capacity at 200\u00b0F (Btu/lbmol)\n", + "H2 = 4010 #Average heat capacity at 600\u00b0F (Btu/lbmol)\n", + "\n", + "#Calculation:\n", + "Q = n*(H2 - H1) #The heat transfer rate (Btu/min)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "n = 600 #The mass flow rate of fluid (lbmol/min)\n", + "Cp_AV = 0.271 #Heat capacity (Btu/lbmol . \u00b0F)\n", + "T1 = 200 #Initial temperature(\u00b0F)\n", + "T2 = 600 #Final temperature(\u00b0F)\n", + "\n", + "#Calcultaion:\n", + "Q = n*Cp_AV*(T2 - T1) #The required heat rate (Btu/min)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "T_c1 = 20 #Initial cold fluid temperature (\u00b0C)\n", + "T_h1 = 82 #Initial hot fluid temperature (\u00b0C)\n", + "T_h2 = 94 #Final hot fluid temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "T_c2 = (T_h2 - T_h1 + T_c1) #Final cold fluid temperature (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate is:\",round(T_c2),\" \u00b0C\"\n", + "print \"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\n", + "There 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Q = -5.5*10**6 #The heat transferred out from the gas (W)\n", + "Cp = 1090.0 #The average heat capacity of the gas (J/(kg . \u00b0C))\n", + "m = 9.0 #The gas mass flow rate (kg/s)\n", + "T1 = 650 #The gas inlet temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "T2 = Q/(m*Cp)+T1 #The gas outlet temperature (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "n = 3500.0 #Inlet flowrate of water (gal/min)\n", + "Cp_W = 75.4 #Heat capacity of water (J/(gmol . \u00b0C)\n", + "p = 62.4 #Density of water (lb/ft^3)\n", + "M = 24*60.0 #Minutes in a day (min/day)\n", + "G = 7.48 #Gallons in a feet cube (gal/ft^3)\n", + "gm = 454.0 #Grams in a pound (g/lb)\n", + "J = 1054.0 #Joules in a Btu (J/Btu)\n", + "g = 18.0 #Grams in a gmol (g/gmol)\n", + "F = 1.8 #Degree fahrenheit in a degree celcius (\u00b0F)\n", + "Ti = 38.0 #Initial temperature (\u00b0F)\n", + "Tf = 36.2 #Final temperature (\u00b0F)\n", + "\n", + "#Calculations:\n", + "T= Ti-Tf #Temperature loss (\u00b0F)\n", + "m = n*p*M/G #Mass flow rate of water (lb/day)\n", + "Cp = Cp_W*gm/J/g/F #Heat capacity in cosistent units (Btu/(lb.\u00b0F))\n", + "Q = m*Cp*T #Rate of heat flow from water (Btu/day)\n", + "\n", + "#Result:\n", + "print \"The rate of Btu removed from the water per day is :\",round(Q/10**8,2),\" x 10**8 Btu/day .\"\n", + "print \"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 .\n", + "There 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 100755 index 00000000..d5d367e4 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_05.ipynb @@ -0,0 +1,464 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:33ba4a11fa0be53b880079008a0280bb47a9d8531c545010b70c7a3f75197aa0" + }, + "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": [ + "\n", + "#Variable declaration:\n", + "qi = 3500 #Initial volumetric flow rate of gas (acfm)\n", + "Ti = 100.0 #Initial temperature (\u00b0F)\n", + "Tf = 300.0 #Final temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Ti_R = Ti+460 #Initial temperatur in Rankine scale (\u00b0R)\n", + "Tf_R = Tf+460 #Final temperatur in Rankine scale (\u00b0R)\n", + "qf = qi*(Tf_R/Ti_R) #Final volumetric flow rate of gas (acfm)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "qi = 3500 #Initial volumetric flow rate of gas (acfm)\n", + "Pi = 1.0 #Iitial pressure (atm)\n", + "Pf = 3.0 #Final pressure (atm)\n", + "\n", + "#Calculation:\n", + "qf = qi*(Pi/Pf) #Final volumetric flow rate of gas (acfm)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "qi = 3500 #Initial volumetric flow rate of the gas (acfm)\n", + "Pi = 1.0 #Initial pressure (atm)\n", + "Pf = 3.0 #Final pressure (atm)\n", + "Tf = 300.0+460.0 #Final temperature in Rankine scale (\u00b0R)\n", + "Ti = 100.0+460.0 #Initial temperature in Rankine scale (\u00b0R)\n", + "\n", + "#Calculation:\n", + "qf = qi*(Pi/Pf)*(Tf/Ti) #Final volumetric flow rate of the gas (acfm)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "P = 14.7 #Absolute pressure of air (psia)\n", + "MW = 29 #Molecular weight of air (lb/lbmol)\n", + "T = 75+460 #Temperature in Rankine scale (\u00b0R)\n", + "R = 10.73 #Universal gas constant (ft^3.psi/lbmol.\u00b0R)\n", + "\n", + "#Calculation:\n", + "p = P*MW/R/T #Density of air (lb/ft^3)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "n = 1 #Molar flow rate of gas (lbmol/h)\n", + "R = 10.73 #Universal gas constant (ft^3.psi/lbmol.\u00b0R)\n", + "T = 60+460 #Temperature in Rankine scale (\u00b0R)\n", + "P = 14.7 #Absolute pressure of gas (psia)\n", + "\n", + "#Calculation:\n", + "V = n*R*T/P #Volume of gas (ft^3)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "P = 1.2 #Abslute pressure of gas (psia)\n", + "MW = 29 #Molecular weight of gas (g/gmol)\n", + "R = 82.06 #Universal gas constant (atm.cm^3/gmol.K)\n", + "T = 20+273 #Temperature in Kelvin (K)\n", + "\n", + "#Calculation:\n", + "p = P*MW/R/T #Dendity of gas (g/cm^3)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "R = 10.73 #Universal gas constant (psia . ft^3/lbmol .\u00b0R)\n", + "T = 70+460 #Temperature in Rankine scale (\u00b0R)\n", + "v = 10.58 #Specific volume (ft^3/lb)\n", + "P = 14.7 #Absolute pressure (psia)\n", + "\n", + "#Calculation:\n", + "MW = R*T/v/P #Molecular weight of gas (lb/lbmol)\n", + "\n", + "#Result:\n", + "print \"The molecular weight of the gas is :\",round(MW,2),\" lb/lbmol.\"\n", + "print \"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.\n", + "It 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "qs = 30000 #Volumetric flow rate at standard conditions (scfm)\n", + "Ta = 1100+460 #Actual absolute temperature in Rankine scale (\u00b0R)\n", + "Ts = 60+460 #Standard absolute temperature in Rankine scale (\u00b0R)\n", + "\n", + "#Calculation:\n", + "qa = qs*Ta/Ts #Volumetric flow rate at actual conditions (acfm)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "qs = 1000 #Volumetric flow rate at standard conditions (scfm)\n", + "Ta = 300+460 #Actual absolute temperature in Rankine scale (\u00b0R)\n", + "Ts = 70+460 #Standard absolute temperature in Rankine scale (\u00b0R)\n", + "A = 2.0 #Inlet area of stack (ft^2)\n", + "\n", + "#Calculations:\n", + "qa = qs*Ta/Ts #Volumetric flow rate at actual conditions (acfm)\n", + "v = qa/A/60 #Velocity of gas (ft/s)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "qs1 = 5000.0 #Volumetric flow rate of C6H5Cl at standard conditions (scfm)\n", + "qs2 = 3000.0 #Volumetric flow rate of air at standard conditions (scfm)\n", + "Ta = 70+460.0 #Actual absolute temperature in Rankine scale (\u00b0R)\n", + "Ts = 60+460.0 #Standard absolute temperature in Rankine scale (\u00b0R)\n", + "V = 387.0 #Volume occupied by one lbmol of any ideal gas (ft^3)\n", + "M1 = 112.5 #Molecular weight of C6H5Cl (lb/lbmol)\n", + "M2 = 29.0 #Molecular weight of air (lb/lbmol)\n", + "T = 60.0 #Absolute temperature (\u00b0F)\n", + "\n", + "#Calculations:\n", + "qa1 = qs1*(Ta/Ts) #Volumetric flow rate of C6H5Cl at actual conditions (acfm)\n", + "qa2 = qs2*(Ta/Ts) #Volumetric flow rate of air at actual conditions (acfm)\n", + "n1 = qa1/V #Molar flow rate of C6H5Cl (lbmol/min)\n", + "n2 = qa2/V #Molar flow rate of air (lbmol/min)\n", + "m1 = n1*M1*T #Mass flow rate of C6H5Cl (lb/h)\n", + "m2 = n2*M2*T #Mass flow rate of air (lb/h)\n", + "m_in = m1+m2 #Total mass flow rate of both streams entering the oxidizer (lb/h)\n", + "m_out = m_in #Total mass flow rate of both streams exit the cooler (lb/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "p = 0.15 #Partial pressure of SO3 (mm Hg)\n", + "P = 760.0 #Atmospheric pressure (mm Hg)\n", + "m = 10**6 #Particles in a million\n", + "\n", + "#Calculation:\n", + "y = p/P #Mole fraction of SO3\n", + "ppm = y*m #Parts per million of SO3 (ppm)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..e4393f65 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_06.ipynb @@ -0,0 +1,363 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:54e2e4fc776d95c57fe6ec3f8739ba703b59fc814a535a0e713a806e8b2afe6d" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "NPS = 2 #Nominal pipe size (inch)\n", + "SN = 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.\n", + "ID = 2.067\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "NPS = 3 #Nominal pipe size (inch)\n", + "SN = 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.\n", + "ID = 3.068\n", + "OD = 3.5\n", + "WT = 0.216\n", + "PW = 7.58\n", + "\n", + "#Result:\n", + "print \"The inside diameter of steel pipe is :\",ID,\" in\"\n", + "print \"The outside diameter of steel pipe is :\",OD,\" in\"\n", + "print \"The wall thickness of steel pipe is :\",WT,\" in\"\n", + "print \"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\n", + "The outside diameter of steel pipe is : 3.5 in\n", + "The wall thickness of steel pipe is : 0.216 in\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "ID = 0.957 #Inside diameter of pipe (in)\n", + "OD = 1.315 #Outside diameter of pipe (in)\n", + "WT = 0.179 #Wall thickness of pipe (in)\n", + "PW = 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.\n", + "NSP = 1\n", + "SN = 80\n", + "\n", + "#Result:\n", + "print \"The nominal size of the pipe is :\",NSP,\" in.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "#Variable declaration:\n", + "S = 3/4 #Tube size (in)\n", + "BWG = 16 #Birmingham Wire Gauge number (gauge)\n", + "\n", + "#calculation:\n", + "#From table 6.3, we get:\n", + "ID = 0.620 #Internal diameter of tube (in)\n", + "WT = 0.065 #Wall thickness of tube (in)\n", + "OD = ID+2*WT #Outside diameter of tube (in)\n", + "EA = 0.1963 #External area per foot (ft)\n", + "\n", + "#Result:\n", + "print \"The inside diameter is :\",ID,\" in\"\n", + "print \"The wall thickness is :\",WT,\" in\"\n", + "print \"The outside diamater is :\",OD,\" in\"\n", + "print \"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\n", + "The wall thickness is : 0.065 in\n", + "The outside diamater is : 0.75 in\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "a = 1 #Length of cross-section (m)\n", + "b = 0.25 #Width of cross-section (m)\n", + "v = 1*10**-5 #Kinematic viscosity of air (m^2/s)\n", + "Re = 2300.0 #Reynolds Number\n", + "cm = 100 #Cenitmeters in a meter\n", + "\n", + "#Calculation:\n", + "Dh = 2*a*b/(a+b) #Hydraulic diameter of duct (m)\n", + "V = Re*v/Dh*cm #Maximum air velocity (cm/s)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "q = 0.486 #Flow rate of fluid (ft^3/s)\n", + "D = 2.0/12.0 #Diameter of tube in feet (ft)\n", + "pi = 3.14 #Value of pi\n", + "p = 70.0 #Density of fluid (lb/ft^3)\n", + "u = 0.1806 #Viscosity of fluid (lb/ft)\n", + "\n", + "#Calculation:\n", + "V = 4*q/pi/D**2 #Flow velocity (ft/s)\n", + "Re = D*V*p/u #Reynolds Number\n", + "\n", + "#Result:\n", + "if(Re<2100):\n", + " print \"The flow is laminar.\"\n", + "elif(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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 6.12, we have:\n", + "D = 2.0/12.0 #Diameter of pipe in feet (ft)\n", + "Re = 1440.0 #Reynolds number\n", + "\n", + "#Calculation:\n", + "Lc = 0.05*D*Re #Length of pipe (ft)\n", + "\n", + "#Result:\n", + "print \"The pipe length to ensure a fully developed flow is:\",Lc,\" ft.\"\n", + "print \"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.\n", + "This 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "u = 6.72*10**-4 #Viscosity of water (lb/ft.s)\n", + "p = 62.4 #Density of water (lb/ft^3)\n", + "#For laminar flow:\n", + "Re = 2100.0 #Reynolds number\n", + "#From table 6.2, we have:\n", + "D = 2.067/12.0 #Inside diameter of pipe (ft)\n", + "\n", + "#Calculation:\n", + "V = Re*u/D/p #Average velocity of water flowing (ft/s)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..bfa882d7 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_07.ipynb @@ -0,0 +1,412 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:057d6eaa086db820c820c1a5a80f0143b18739ecbe1b8c285287723cf460e4b6" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Q = 3000.0 #The rate of heat flow through the glass window (W)\n", + "L = 0.01 #Thickness of glass window (m)\n", + "A = 3.0 #Area of heat transfer (m^2)\n", + "TC = 10+273 #Temperature at the outside surface (K)\n", + "k = 1.4 #Thermal onductivity of glass (W/m.K)\n", + "\n", + "#Calculation:\n", + "TH = TC+Q*L/k/A #Temperature at the inner surface (K)\n", + "\n", + "#Result:\n", + "print \"The temperature at the inner surface is :\",round(TH,1),\" K\"\n", + "print \"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\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "k = 0.026 #Thermal conductivity of insulating material (Btu/ft.h.\u00b0F)\n", + "L = 1.0 #Thickness of insulating material (ft)\n", + "TC = 70.0 #Temperature on the cold side surface (\u00b0F)\n", + "TH = 210.0 #Temperature on the hot side surface (\u00b0F)\n", + "c = 0.252 #Kilocalorie per hour in a Btu per hour\n", + "m = 0.093 #meter square in a feet square\n", + "\n", + "#Calculation:\n", + "DT = TH-TC #Change in temperature (\u00b0F)\n", + "Q1 = k*DT/L #Rate of heat flux throughthe wall (Btu/f^t2.h.)\n", + "Q2 = Q1*c/m #Rate of heat flux throughthe wall in SI units (kcal/m^2.h)\n", + "\n", + "#Result:\n", + "print \"The rate of heat flux in Btu/ft^2.h is :\",round(Q1,3),\" Btu/ft^2.h .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "TH = 1592.0 #Temperature of inside surface (K)\n", + "TC = 1364.0 #Temperature of outside surface (K)\n", + "H = 3.0 #Height of furnace wall (m)\n", + "W = 1.2 #Width of furnace wall (m)\n", + "L = 0.17 #Thickness furnace wall (m)\n", + "m = 0.0929 #Meter square per second in a feet square per second\n", + "Btu = 3.412 #Btu per hour in a Watt\n", + "Btu2 = 0.3171 #Btu per feet square hour in a watt per meter square\n", + "\n", + "#Calculation:\n", + "Tav = (TH+TC)/2 #Average wall temperature (K)\n", + "#From Table in Appendix:\n", + "p = 2645.0 #Density of material (kg/m^3)\n", + "k = 1.8 #Thermal conductivity (W/m.K)\n", + "Cp = 960.0 #Heat capacity of material (J/kg.K)\n", + "a = k/(p*Cp)/m #Thermal diffusivity (ft^2/s)\n", + "t = (TC-TH)/L #Temperature gradient (\u00b0C/m)\n", + "A = H*W #Heat transfer area (m^2)\n", + "Q1 = k*A*(TH-TC)/L*Btu #Heat transfer rate (Btu/h)\n", + "Q2 = k*(TH-TC)/L*Btu2 #Heat transfer flux (Btu/h.ft^2)\n", + "R = L/(k*A) #Thermal resistance (\u00b0C/W)\n", + "\n", + "#Result:\n", + "print \"The temperature gradient is :\",round(t),\" \u00b0C/m.\"\n", + "print \"The heat transfer rate is :\",round(Q1),\" Btu/h.\"\n", + "print \"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.\n", + "The heat transfer rate is : 29653.0 Btu/h.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "TH = 25.0 #Temperature at inner suface of wall (\u00b0C)\n", + "TC = -15.0 #Temperature at outer suface of wall (\u00b0C)\n", + "L = 0.3 #Thickness of wall (m)\n", + "k = 1.0 #Thermal conductivity of concrete (W/m)\n", + "A = 30.0 #Sueface area of wall (m^2)\n", + "\n", + "#Calculation:\n", + "DT = TH-TC #Driving force for heat transfer (\u00b0C) (part 2)\n", + "R = L/(k*A) #Thermal resistance (\u00b0C/W) (part 3)\n", + "Q = DT/R/10**3 #Heat loss through the wall (kW)\n", + "\n", + "#Result:\n", + "print \"1. Theoretical part.\"\n", + "print \"2. The driving force for heat transfer is :\",DT,\" \u00b0C.\"\n", + "print \"3. The heat loss through the wall is :\",Q,\" kW.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1. Theoretical part.\n", + "2. The driving force for heat transfer is : 40.0 \u00b0C.\n", + "3. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "TC = 27.0 #Inside temperature of walls (\u00b0C)\n", + "TH = 68.7 #Outside temperature of walls (\u00b0C)\n", + "LC = 6*0.0254 #Thickness of concrete (m)\n", + "LB = 8*0.0254 #Thickness of cork-board (m)\n", + "LW = 1*0.0254 #Thickness of wood (m)\n", + "kC = 0.762 #Thermal conductivity of concrete (W/m.K)\n", + "kB = 0.0433 #Thermal conductivity of cork-board (W/m.K)\n", + "kW = 0.151 #Thermal conductivity of wood (W/m.K)\n", + "\n", + "#Calculation:\n", + "RC = LC/kC #Thermal resistance of concrete (K/W)\n", + "RB = LB/kB #Thermal resistance of cork-board (K/W)\n", + "RW = LW/kW #Thermal resistance of wood (K/W)\n", + "Q = (TC-TH)/(RC+RB+RW) #Heat transfer rate across the wall (W)\n", + "T = -(Q*RW-TC) #Interface temperature between wood and cork-board (K)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate across the wall is :\",round(Q,3),\" W.\"\n", + "print \"The interface temperature between wood and cork-board is :\",round(T,1),\" \u00b0C.\"\n", + "print \"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.\n", + "The interface temperature between wood and cork-board is : 28.4 \u00b0C.\n", + "The 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": [ + "\n", + "\n", + "from math import pi, log\n", + "from sympy import symbols\n", + "\n", + "#Variable declaration:\n", + "Z = symbols ('z') #Length of pipe\n", + "D1s = 4.0 #Glass wool inside diameter (in)\n", + "D2s = 8.0 #Glass wool outside diameter (in)\n", + "D1a = 3.0 #Asbestos inside diameter (in)\n", + "D2a = 4.0 #Asbestos outside diameter (in)\n", + "TH = 500.0 #Outer surface temperature of pipe (\u00b0F)\n", + "TC = 100.0 #Outer surface temperature of glass wool (\u00b0F)\n", + "La = 0.5/12.0 #Thickness of asbestos (ft)\n", + "Lb = 2.0/12.0 #Thickness of glss wool (ft)\n", + "ka = 0.120 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\n", + "kb = 0.0317 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\n", + "\n", + "#Calculation:\n", + "Aa = (pi*Z*(D2a-D1a)/12.0)/log(D2a/D1a) #Area of asbestos (ft^2)\n", + "Ab = (pi*Z*(D2s-D1s)/12.0)/log(D2s/D1s) #Area of glass wool (ft^2)\n", + "Q1 = (TH-TC)/(La/(ka*Aa)+Lb/(kb*Ab)) #Steady-state heat transfer per foot of pipe (Btu/h.)\n", + "Q2 = Q1/Z #Factorization of Q/Z (Btu/h.ft)\n", + "\n", + "#Result:\n", + "print \"The steady-state heat transfer per foot of pipe, Z, is :\",round(Q1/Z,1),\" x z Btu/h.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 7.6:\n", + "TH = 500 #Outer surface temperature of pipe (\u00b0F)\n", + "Lb = 2.0/12.0 #Thickness of glss wool (ft)\n", + "kb = 0.0317 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\n", + "Ab = 1.51 #Area of glass wool (ft^2)\n", + "Q = 103.5 #Steady-state heat transfer per foot of pipe (Btu/h.)\n", + "La = 0.5/12.0 #Thickness of asbestos (ft)\n", + "ka = 0.120 #Thermal conductivity of asbestos (Btu/h.ft.\u00b0F)\n", + "Aa = 0.91 #Area of asbestos (ft^2)\n", + "TC = 100 #Outer surface temperature of glass wool (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Ti_b = -((Lb*Q)/(kb*Ab)-TH) #Interfacial temperature of glass wool layer (\u00b0F)\n", + "Ti_a = (Q*La)/(ka*Aa)+TC #Interfacial temperature of asbestos layer (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The interfacial temperature of glass wool layer is :\",round(Ti_b),\" \u00b0F.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "from sympy import cos,symbols,diff,pi\n", + "\n", + "#Variable declaration:\n", + "z,h,k = symbols('z, h, k') #Length, height, thermal conductivity\n", + "T = 100*cos((pi*z)/(2*h)) #Temperature of solid slab\n", + "\n", + "#Calculation:\n", + "DT = diff(T,z) #Temperature at z\n", + "Q = -k*(DT) #Heat flux in slab (Btu/s.ft^2)\n", + "Q1 = Q.subs(z,0) #Heat flux in slab at z = 0 (Btu/s.ft^2)\n", + "Q2 = Q.subs(z,h) #Heat flux in slab at z = h (Btu/s.ft^2)\n", + "\n", + "#Result:\n", + "print \"The heat flux in slab is :\",Q,\" Btu/s.ft^2 .\"\n", + "print \"The heat flux in slab at z = 0 is :\",Q1,\" Btu/s.ft^2 .\"\n", + "print \"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 .\n", + "The heat flux in slab at z = 0 is : 0 Btu/s.ft^2 .\n", + "The 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 100755 index 00000000..51da1dc2 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_08.ipynb @@ -0,0 +1,73 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:f803274ef7ccff0534bae20ab018bc7691f3977ae1b9f41d5ea33a08004d0ffb" + }, + "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": [ + "\n", + "\n", + "from math import pi,sin,e\n", + "\n", + "#Variable declaration:\n", + "k = 9.1 #Thermal coductivity of steel rod (Btu/h.ft.\u00b0F)\n", + "p = 0.29*1728 #Density of steel rod (lb/ft^3)\n", + "Cp = 0.12 #Heat capacity of steel rod (Btu/lb.\u00b0F)\n", + "P = 15+14.7 #Absolute pressure (psia)\n", + "Ta = 71.0 #Initial temperature (\u00b0F)\n", + "L = 20.0/12.0 #Length of rod (ft)\n", + "t = 30.0/60.0 #Time taken (h)\n", + "x = 0.875/12.0 #Length from one of end (ft)\n", + "#From assumption:\n", + "n = 1.0 #First term\n", + "#From tables in Appendix:\n", + "Ts = 249.7 #Saturated steam temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "a = k/(p*Cp) #Thermal diffusivity (ft^2/s)\n", + "T = 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:\n", + "print \"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 100755 index 00000000..cba4bd5f --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_09.ipynb @@ -0,0 +1,612 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:717821d247948d7fb9af37b8a67927013610ff26888719ba58ba9ef00c80f713" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "D = 1.0 #Diamete of vessel (ft)\n", + "L = 1.5 #Length of vessel (ft)\n", + "T1 = 390.0 #Surface temperature of vessel (\u00b0F)\n", + "T2 = 50.0 #Surrounding temperature of vessel (\u00b0F)\n", + "h = 4.0 #Convective heat transfer coefficient (Btu/h.ft.\u00b0F)\n", + "\n", + "#Calculation:\n", + "A = pi*D*L+2*pi*(D/2)**2 #Total heat transfer area (ft^2)\n", + "Q = h*A*(T1-T2) #Rate of heat transfer (Btu/h)\n", + "R = 1/(h*A) #Thermal resistance (\u00b0F.h/Btu)\n", + "\n", + "#Result:\n", + "print \"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:\n", + "R = 0.0398 #Theral resistance (\u00b0F.h/Btu)\n", + "Btu = 3.412 #Btu/h in a watt\n", + "C = 1.8 #Change in degree fahrenheit for a degree change in celsius\n", + "K = 1 #Change in degree celsius for a unit change in Kelvin\n", + "\n", + "#Calculation:\n", + "Rc = R*Btu/C #Thermal resistance in degree cesius per watt (\u00b0C/W)\n", + "Rk = Rc/K #Thermal resistance in Kelvin per watt (K/W)\n", + "\n", + "#Result:\n", + "print \"The thermal resistance in \u00b0C/W is :\",round(Rc,3),\" \u00b0C/W.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "h = 48.0 #Convective heat transfer coefficient (Btu/h.ft.\u00b0F)\n", + "A = 2*1.5 #Total heat transfer area (ft^2)\n", + "Ts = 530.0 #Surface temperature of plate (\u00b0F)\n", + "Tm = 105.0 #Maintained temperature of opposite side of plate (\u00b0F)\n", + "kW = 3.4123*10**3 #Units kW in a Btu/h\n", + "\n", + "#Calculation:\n", + "Q = h*A*(Ts-Tm) #Heat transfer rate in Btu/h (Btu/h)\n", + "Q1 = Q/kW #Heat transfer rate in kW (kW)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate in Btu/h is :\",round(Q),\" Btu/h.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "TS = 10+273 #Outer surface temperature of wall (K)\n", + "Q = 3000.0 #Heat transfer rate (W)\n", + "h = 100.0 #Convection coefficient of air (W/m^2)\n", + "A = 3.0 #Area of glass window (m^2)\n", + "\n", + "#Calculation:\n", + "TM = TS-Q/(h*A) #Bulk temperature of fluid (K)\n", + "\n", + "#Result:\n", + "print \"The bulk temperature of fluid is :\",round(TM),\" K.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "h = 24.0 #Plant operating hour per day (h/day)\n", + "d = 350.0 #Plant operating day per year (day/yr)\n", + "\n", + "#Calculation:\n", + "N = h*d #Operating hours per year (h/yr)\n", + "#From example 9.1:\n", + "Q = 8545.0 #Rate of energy loss (Btu/h)\n", + "Qy = Q*N #Steady-state energy loss yearly (Btu/yr)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from sympy import symbols, integrate\n", + "\n", + "#Variable declaration:\n", + "x = 0.3 #Length from the leading age of the plate (m)\n", + "L = 1.2 #Length of plate (m)\n", + "TS = 58.0 #Surface temperature of plate (\u00b0C)\n", + "Ta = 21.0 #Temperature of flowing air (\u00b0C)\n", + "\n", + "#Calculation:\n", + "hx = 25/x**0.4 #Local heat transfer coefficient at 0.3m (W/m^2.K) (Part 1)\n", + "y = symbols('y') #Length\n", + "hy = 25/y**0.4 #hx at the end of the plate (W/m^2.K)\n", + "h = integrate(hy, (y,0,L))/L #Average heat transfer coefficient (W/m^2.K)\n", + "Q = hx*(TS-Ta) #Heat flux at 0.3m from leading edge of plate (W/m^2)\n", + "hL = 25/L**0.4 #Local heat transfer coefficient at plate end (W/m^2.K) (Part 2) \n", + "r = h/hL #Ratio h/hL at the end of the plate\n", + "\n", + "#Result:\n", + "print \"1. The heat flux at 0.3 m from the leading edge of the plate is :\",round(Q),\" W/m^2.\"\n", + "print \"2. The local heat transfer coefficient at the end of the plate is :\",round(hL,1),\" W/m^2.K.\"\n", + "print \"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.\n", + "2. The local heat transfer coefficient at the end of the plate is : 23.2 W/m^2.K.\n", + "3. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 9.7:\n", + "b = 1.0 #Width of plate (m)\n", + "L = 1.2 #Length of plate (m)\n", + "TS = 58.0 #Surface temperture of plate (\u00b0C)\n", + "Ta = 21.0 #Air flow temperature (\u00b0C)\n", + "h = 38.7 #Average heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "A = b*L #Area for heat transfer for the entire plate (m^2)\n", + "Q = h*A*(TS-Ta) #Rate of heat transfer over the whole length of the plate (W)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from math import pi\n", + "#Variable declaration:\n", + "m = 0.075 #Mass rate of air flow (kg/s)\n", + "D = 0.225 #Diameter of tube (m)\n", + "mu = 208*10**-7 #Dynamic viscosity of fluid (N)\n", + "Pr = 0.71 #Prandtl number\n", + "k = 0.030 #Thermal conductivity of air (W/m.K)\n", + "\n", + "#Calculation:\n", + "Re = 4*m/(pi*D*mu) #Reynolds number\n", + "#From equation 9.26:\n", + "Nu = 0.023*(Re**0.8)*(Pr**0.3) #Nusselt number\n", + "h = (k/D)*Nu #Heat transfer coefficient of air (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "D = 0.902/12.0 #Inside diameter of tube (ft)\n", + "T_in = 60.0 #Temperature water entering the tube (\u00b0F)\n", + "T_out = 70.0 #Temperature water leaving the tube (\u00b0F)\n", + "V = 7.0 #Average wave velocity water (ft/s)\n", + "p = 62.3 #Density of water (lb/ft^3)\n", + "mu = 2.51/3600.0 #Dynamic viscosity of water (lb/ft.s)\n", + "Cp = 1.0 #Viscosity of centipoise (Btu/lb.\u00b0F)\n", + "k = 0.34 #Thermal conductivity of water (Btu/h.ft.\u00b0F)\n", + "\n", + "#Calculation:\n", + "Re = D*V*p/mu #Reynolds Number\n", + "Pr = Cp*mu/k*3600 #Prandtl number\n", + "#From equation 9.26:\n", + "Nu = 0.023*(Re**0.8)*(Pr**0.4) #Nusselt number\n", + "h = (k/D)*Nu #Average film heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "P = 1.0132 * 10**5 #Air pressure (Pa)\n", + "T = 300.0+273.0 #Air temperature (K)\n", + "V = 5.0 #Air flow velocity (m/s)\n", + "D = 2.54/100.0 #Diameter of tube (m)\n", + "R = 287.0 #Gas constant (m^2/s^2.K)\n", + "#From Appendix:\n", + "Pr = 0.713 #Prandtl number of nitrogen\n", + "mu = 1.784*10**(-5) #Dynamic viscosity of nitrogen (kg/m.s)\n", + "k = 0.0262 #Thermal conductivity of nitrogen (W/m.K)\n", + "Cp = 1.041 #Heat capacity of nitrogen (kJ/kg.K)\n", + "\n", + "#Calculation:\n", + "p = P/(R*T) #Density of air\n", + "Re = D*V*p/mu #Reynolds number\n", + "#From table 9.5:\n", + "Nu = 0.023*(Re**0.8)*(Pr**0.3) #Nusselt number\n", + "h = (k/D)*Nu #Heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "T1 = 15.0 #Water entering temperature (\u00b0C)\n", + "T2 = 60.0 #Water leaving temperature (\u00b0C)\n", + "D = 0.022 #Inside diameter of tube (m)\n", + "V = 0.355 #Average water flow velocity (m/s)\n", + "TC = 150.0 #Outside wall temperature (\u00b0C)\n", + "#From Appendix:\n", + "p = 993.0 #Density of water (kg/m^3)\n", + "mu = 0.000683 #Dynamic viscosity of water (kg/m.s)\n", + "Cp = 4.17*10**3 #Heat capacity of water (J/kg.K)\n", + "k = 0.63 #Thermal conductivity of water (W/m.K)\n", + "\n", + "#Calculation:\n", + "Tav1 = (T1+T2)/2.0 #Average bulk temperature of water (\u00b0C)\n", + "Re = D*V*p/mu #Reynolds number\n", + "Pr = Cp*mu/k #Prandtl number\n", + "Tav2 = (Tav1+TC)/2.0 #Fluid's average wall temperature (\u00b0C)\n", + "#From Appendix:\n", + "mu_w = 0.000306 #Dynamic viscosity of fluid at wall (kg/m.s)\n", + "#From Table 9.5:\n", + "h = (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:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 9.7:\n", + "h = 38.7 #Average heat transfer coefficient (W/m^2.K)\n", + "L = 1.2 #Length of plate (m)\n", + "k = 0.025 #Thermal conductivity of air (W/m)\n", + "\n", + "#Calculation:\n", + "Bi = h*L/k #Average Biot number\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from math import pi\n", + "#Variable declaration:\n", + "k = 60.0 #Thermal conductivity of rod (W/m.K)\n", + "p = 7850.0 #Density of rod (kg/m^3)\n", + "Cp = 434.0 #Heat capacity of rod (J/kg.K)\n", + "h = 140.0 #Convection heat transfer coefficient (W/m^2.K)\n", + "D = 0.01 #Diameter of rod (m)\n", + "kf = 0.6 #Thermal conductivity of fluid (W/m.K)\n", + "L = 2.5 #Length of rod (m)\n", + "Ts = 250.0 #Surface temperature of rod (\u00b0C)\n", + "Tf = 25.0 #Fluid temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "#Case 1:\n", + "a = k/(p*Cp) #Thermal diffusivity of bare rod (m^2/s)\n", + "#Case 2:\n", + "Nu = h*D/kf #Nusselt number\n", + "#Case 3:\n", + "Bi = h*D/k #Biot number of bare rod\n", + "#Case 4:\n", + "Q = h*(pi*D*L)*(Ts-Tf) #Heat transferred from rod to fluid (W)\n", + "\n", + "#Result:\n", + "print \"1. The thermal diffusivity of the bare rod is :\",round(a/10**-5,2),\" x 10^-5 m^2/s.\"\n", + "print \"2. The nusselt number is :\",round(Nu,2),\" .\"\n", + "print \"3. The Biot number is :\",round(Bi,4),\" .\"\n", + "print \"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.\n", + "2. The nusselt number is : 2.33 .\n", + "3. The Biot number is : 0.0233 .\n", + "4. 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 100755 index 00000000..f1c9afa8 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_10.ipynb @@ -0,0 +1,514 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:5a160b2a8858072f2a32a8c5f96e1426d911509afe45130ab4e23facf3414ba2" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Gr = 100.0 #Grashof number\n", + "Re = 50.0 #Reynolds number\n", + "\n", + "#Calculation:\n", + "LT = Gr/Re**2 #Measure of influence of convection effect\n", + "\n", + "#Result:\n", + "if (LT<1.0):\n", + " print \"The free convection effects can be neglected.\"\n", + "elif (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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ts = 110.0+273.0 #Surface temperature of plate (K)\n", + "Too = 30.0+273.0 #Ambient air temperature (K)\n", + "L = 3.5 #Height of plate (m)\n", + "g = 9.807 #Gravitational acceleration (m^2/s)\n", + "\n", + "#Calculation:\n", + "Tf = (Ts+Too)/2 #Film temperature (K)\n", + "DT = Ts - Too #Temperature difference between surface and air (K)\n", + "#From appendix:\n", + "v = 2.0*10**-5 #Kinematic viscosity for air (m^2/s)\n", + "k = 0.029 #Thermal conductivity for air (W/m.K)\n", + "Pr = 0.7 #Prandtl number\n", + "B = 1.0/Tf #Coefficient of expansion (K^-1)\n", + "Gr = g*B*DT*L**3/v**2 #Grashof number\n", + "Ra = Gr*Pr #Rayleigh number\n", + "\n", + "#Result:\n", + "print \"The Grashof number is :\",round(Gr/10**11,2),\" x 10^11 .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 10.2:\n", + "Ra = 1.71*10**11 #Rayleigh number\n", + "\n", + "#Result:\n", + "if (Ra>10**9):\n", + " print \"The convection flow category is turbulent.\"\n", + "elif(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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From Table 10.1:\n", + "c = 0.1 #Constant c\n", + "m = 1.0/3.0 #Constant for turbulent free conection\n", + "#From example 10.2:\n", + "Ra = 1.71*10**11 #Rayleigh number\n", + "k = 0.029 #Thermal conductivity (W/m.K)\n", + "L = 3.5 #Thickness of plate (m)\n", + "\n", + "#Calculation:\n", + "Nu = c*Ra**m #Average Nusselt number\n", + "h = Nu*k/L #Average heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import log,e\n", + "\n", + "#Variable declaration:\n", + "Ts = 200.0+460.0 #Surface temperature of pipe (\u00b0R)\n", + "Too = 70.0+460.0 #Air temperature (\u00b0R)\n", + "D = 0.5 #Diameter of pipe (ft)\n", + "R = 0.73 #Universal gas constant (ft^3.atm.R^\u22121.lb.mol^\u22121)\n", + "P = 1.0 #Atmospheric pressure (Pa)\n", + "MW = 29.0 #Molecular weight of fluid (mol)\n", + "#From Appendix:\n", + "mu = 1.28*10**-5 #Absolute viscosity (lb/ft.s)\n", + "k = 0.016/3600.0 #Thermal conductivity (Btu/s.ft.\u00b0F)\n", + "g = 32.174 #Gravitational acceleration (ft/s^2)\n", + "\n", + "#Calculation:\n", + "Tav = (Ts+Too)/2 #Average temperature (\u00b0R)\n", + "v = R*Tav/P #kinematic viscosity (ft^3/lbmol)\n", + "p = MW/v #Air density (lb/ft^3)\n", + "B = 1.0/Tav #Coefficient of expansion (\u00b0R^-1)\n", + "DT = Ts-Too #Temperature difference (\u00b0R)\n", + "Gr = D**3*p**2*g*B*DT/mu**2 #Grashof number\n", + "#From equation 10.5:\n", + "Cp = 0.25 #Air heat capacity (Btu/lb.\u00b0F)\n", + "Pr = Cp*mu/k #Prandtl number\n", + "GrPr = 10**8.24 #Rayleigh number\n", + "#From Holman^(3):\n", + "Nu = 10**(1.5) #Nusselt number\n", + "h = Nu*(k/D)*3600.0 #Air heat transfer film coefficient (Btu/h.ft.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ts = 120.0+460 #Surface temperature of plate (\u00b0R)\n", + "Too = 60.0+460 #Ambient temperature of nitrogen (\u00b0R)\n", + "L = 6 #Height of plate (ft)\n", + "#From Appendix:\n", + "p = 0.0713 #Air density (lb/ft^3)\n", + "k = 0.01514 #Thermal conductivity (Btu/h.ft.\u00b0F)\n", + "v = 16.82*10**-5 #Kinematic viscosity (ft^2/s)\n", + "Pr = 0.713 #Prandtl number\n", + "g = 32.2 #Gravitational acceleration (ft/s^2)\n", + "\n", + "#Calculation:\n", + "Tf = (Ts+Too)/2 #Mean film temperature (\u00b0R)\n", + "B = 1.0/Tf #Coefficient of expansion (\u00b0R^-1)\n", + "Gr = g*B*(Ts-Too)*L**3/v**2 #Grashof number\n", + "Ra = Gr*Pr #Rayleigh number\n", + "#From equation 10.13(Table 10.2) and costants from Table 10.1:\n", + "h = 0.10*(k/L)*Ra**(1.0/3.0) #Free convection heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The free convection heat transfer coefficient is :\",round(h,3),\" Btu/h.ft^2.\u00b0F .\"\n", + "print \"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 .\n", + "There 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example:\n", + "h = 0.675 #Free convection heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "A = 6.0*8.0 #Area of plate (ft^2)\n", + "Ts = 120.0 #Surface temperature of plate (\u00b0F)\n", + "Too = 60.0 #Ambient temperature of nitrogen (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Q = h*A*(Ts-Too) #Heat loss (Btu/h)\n", + "\n", + "#Result:\n", + "print \"The heat loss is :\",round(Q,-1),\" Btu/h .\"\n", + "print \" 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ts = 113.0+273.0 #Surface temperature of bulb (K)\n", + "Too = 31.0+273.0 #Ambient air temperature (K)\n", + "D = 0.06 #Diameter of sphere (m)\n", + "g = 9.8 #Gravitational acceleration (m/s^2)\n", + "\n", + "#Calculation:\n", + "Tf = (Ts+Too)/2 #Mean temperature (K)\n", + "#From Appendix:\n", + "v = (22.38*10**-5)*0.0929 #Kinematic viscosity (m^2/s)\n", + "Pr = 0.70 #Prandtl number\n", + "k = 0.01735*1.729 #Thermal conductivity (W/m.K)\n", + "B = 1.0/(Tf) #Coefficient of expansion (K^-1)\n", + "Gr = g*B*(Ts-Too)*D**3/v**2 #Grashof number\n", + "Ra = Gr*Pr #Rayleigh number\n", + "\n", + "#From equation 10.13:\n", + "h = (k/D)*0.6*Ra**(1.0/4.0) #Heat transferred from bulb (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "#From example 10.9:\n", + "h = 9.01 #Heat transferred from bulb (W/m^2.K)\n", + "D = 0.06 #Diameter of sphere (m)\n", + "Ts = 113.0+273.0 #Surface temperature of bulb (K)\n", + "Too = 31.0+273.0 #Ambient air temperature (K)\n", + "\n", + "#Calculation:\n", + "A = pi*D**2 #Surface area of bulb (m^2)\n", + "Q = h*A*(Ts-Too) #Heat transfer lost by free convection from light bulb (W)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 10.9-10.10:\n", + "Q = 8.36 #Heat transfer lost by free convection from light bulb (W)\n", + "\n", + "#Calculation:\n", + "E = Q/100.0*(100.0) #Percent energy lost by free convection (%)\n", + "\n", + "#Result:\n", + "print \"The percentage of the energy lost by free convection is :\",round(E,2),\" % .\"\n", + "print \"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 % .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "F = 50.0 #Buoyancy flux of gas (m^4/s^3)\n", + "u = 4.0 #wind speed (m/s)\n", + "\n", + "#Calculation:\n", + "xc = 14*F**(5.0/8.0) #Downward distance (m)\n", + "xf = 3.5*xc #distance of transition from first stage of rise to the second stage of rise (m)\n", + "Dh = 1.6*F**(1.0/3.0)*u**-1*xf**(2.0/3.0) #Plume rise (m)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..70c1dfd8 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_11.ipynb @@ -0,0 +1,686 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:063e55263471b05545d8cee123782ea0408a4dbcfbef0baacac009fc80ca4fe1" + }, + "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": [ + "\n", + "\n", + "from sympy import symbols, integrate,oo,exp,pi\n", + "\n", + "#Variable declaration:\n", + "l = symbols('l') #Wavelength (mu.m)\n", + "I = 40*exp(-l**2) #Intensity of radiation (Btu/h.ft^2.mu.m)\n", + "\n", + "#Calculation:\n", + "E = integrate(I, (l,0,oo)).evalf() #Total emissive power (Btu/h.ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "l = 0.25 #Wavelength (mu.m)\n", + "#From equation 11.4:\n", + "lT = 2884 #Product of wavelength and absolute temperature (mu.m.\u00b0R)\n", + "\n", + "#Calculation:\n", + "T = lT/l #Sun's temperature (\u00b0R)\n", + "\n", + "#Result:\n", + "print \"The Sun's temperature is :\",round(T,-2),\" \u00b0R .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "T1 = 1500.0+460.0 #Absolute temperature 1 (\u00b0R)\n", + "T2 = 1000.0+460.0 #Absolute temperature 2 (\u00b0R)\n", + "\n", + "#Calculation:\n", + "X = T1**4/T2**4 #Ratio of quantity of heat transferred\n", + "x = 100*(T1**4-T2**4)/T2**4 #Percentage increase in heat transfer (%)\n", + "\n", + "#Result:\n", + "print \"The ratio of the quantity/rate of heat transferred is :\",round(X,2),\" .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "T1 = 1200.0+460.0 #Absolute temperature of wall 1 (\u00b0R)\n", + "T2 = 800.0+460.0 #Absolute temperature of wall 2 (\u00b0R)\n", + "\n", + "#Calculation:\n", + "#From equation 11.23:\n", + "X = 0.173*((T1/100.0)**4-(T2/100.0)**4) #Heat removed from colder wall (Btu/h.ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "s = 0.173 #Stefan-Boltzmann constant (Btu/h.ft^2.\u00b0R)\n", + "EH = 0.5 #Energy transferred from hotter body (Btu/h.ft^2)\n", + "EC = 0.75 #Energy transferred to colder body (Btu/h.ft^2)\n", + "TH = 1660.0 #Absolute temperature of hotter body (\u00b0R)\n", + "TC = 1260.0 #Absolute temperature of colder body (\u00b0R)\n", + "\n", + "#Calculation:\n", + "E = 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:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 11.6-11.7:\n", + "E1 = 8776.0 #Energy exchange between black bodies (Btu/h.ft^2)\n", + "E2 = 3760.0 #Energy exchange between non-black bodies (Btu/h.ft^2)\n", + "\n", + "#Calculation:\n", + "D = (E1-E2)/E1*100 #Percent difference in energy (%)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "s = 0.173*10**-8 #Stefan-Boltzmann constant (Btu/h.ft^2.\u00b0R)\n", + "TH = 300.0+460.0 #Absolute temperature of external surface (\u00b0R)\n", + "TC = 75.0+460.0 #Absolute temperature of duct (\u00b0R)\n", + "#From Table 6.2:\n", + "AH = 0.622 #External surface area of pipe (ft^2)\n", + "#From Table 11.2:\n", + "EH = 0.44 #Emissivity of oxidized steel\n", + "AC = 4.0*1.0*1.0 #External surface area of duct (ft^2)\n", + "EC = 0.23 #Emissivity of galvanized zinc\n", + "\n", + "#Calculation:\n", + "FE = 1.0/(1.0/EH+((AH/AC)*(1.0/EC-1.0))) #Emissivity correction factor\n", + "Q = FE*AH*s*(TH**4-TC**4) #Net radiation heat transfer (Btu/h.ft)\n", + "\n", + "#Result:\n", + "print \"The net radiation heat transfer is :\",round(Q,2),\" Btu/h.ft^2 .\"\n", + "print \"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 .\n", + "There 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "TH = 140.0+460.0 #Absolute outside temperature of pipe (ft^2)\n", + "TC = 60.0+460.0 #Absolute temperature of surrounding atmosphere (ft^2)\n", + "A = 10.0 #Area of pipe (ft^2)\n", + "E = 0.9 #Emissivity of pipe\n", + "\n", + "#Calculation:\n", + "Q = E*A*0.173*((TH/100.0)**4-(TC/100.0)**4) #Heat loss due to radiation (Btu/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#Froma example 11.10:\n", + "Q = 880.0 #Heat loss due to radiation (Btu/h)\n", + "A = 10.0 #Area of pipe (ft^2)\n", + "TH = 140.0 #Absolute outside temperature of pipe (\u00b0F)\n", + "TC = 60.0 #Absolute temperature of surrounding atmosphere (\u00b0F)\n", + "\n", + "#Calculation:\n", + "hr = Q/(A*(TH-TC)) #Radiation heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "D = 0.0833 #Diameter of tube (ft)\n", + "L = 2.0 #Length of tube (ft)\n", + "h = 2.8 #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "Ta1 = 1500.0+460.0 #Temperature of hot air in furnace (\u00b0R)\n", + "Ta2 = 1350.0+460.0 #Temperature of hot air in the furnace brick walls (\u00b0R)\n", + "Tt = 600.0+460.0 #Surface temperature of tube (\u00b0R)\n", + "E = 0.6 #Surface emissivity of tube\n", + "s = 0.1713*10**-8 #Stefan-Boltzmann constant\n", + "\n", + "\n", + "#Calculation:\n", + "#Case 1:\n", + "A = pi*D*L #Area of tube (ft^2)\n", + "Qc = round(h*A*(Ta1-Tt),-1) #Convection heat transfer from air to tube (Btu/h)\n", + "Qr = round(E*s*A*(Ta2**4-Tt**4),-2) #Radiation feat transfer from wall to tube (Btu/h)\n", + "Q = Qr+Qc #Total heat transfer (Btu/h)\n", + "#Case 2:\n", + "Qp = Qr/Q*100 #Radiation percent \n", + "#Case 3:\n", + "hr = Qr/(A*(Ta2-Tt)) #Radiation heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "#Case 4:\n", + "T = Ta2-Tt #Temperature difference (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"1. The convective heat transferred to the metal tube is :\",Qc,\" Btu/h .\"\n", + "print \" The radiative heat transferred to the metal tube is :\",Qr,\" Btu/h .\"\n", + "print \" The total heat transferred to the metal tube is :\",Q,\" Btu/h .\"\n", + "print \"2. The percent of total heat transferred by radiation is :\",round(Qp,1),\" % .\"\n", + "print \"3. The radiation heat transfer coefficient is :\",round(hr,1),\" Btu/h.ft^2.\u00b0F .\"\n", + "if (T > 200):\n", + " print \"4. The use of the approximation Equation (11.30), hr = 4EsTav^3, is not appropriate.\"\n", + "elif (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 .\n", + "2. The percent of total heat transferred by radiation is : 79.4 % .\n", + "3. The radiation heat transfer coefficient is : 13.0 Btu/h.ft^2.\u00b0F .\n", + "4. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Q = 5.0 #Radiation heat transfer (W)\n", + "E = 1.0 #Emissivity of filament\n", + "s = 5.669*10**-8 #Stefan-Boltzmann constant\n", + "T1 = 900.0+273.0 #Light bulb temperature (K)\n", + "T2 = 150.0+273.0 #Glass bulb temperature (K)\n", + "\n", + "#Calculation:\n", + "A = Q/(E*s*(T1**4-T2**4)) #Surface area of the filament (m^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "T1 = 127.0+273.0 #Surface temperature (K)\n", + "T2 = 20.0+273.0 #Wall temperature (K)\n", + "T3 = 22.0+273.0 #Air temperature (K)\n", + "s = 5.669*10**-8 #Stefan-Boltzmann constant\n", + "e = 0.76 #Surface emissivity of anodized aluminium\n", + "D = 0.06 #Diameter of pipe (m)\n", + "L = 100.0 #Length of pipe (m)\n", + "h = 15.0 #Pipe convective heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "Eb = s*T1**4 #Emissive energy of pipe (W/m^2)\n", + "E = e*Eb #Emissive power from surface of pipe (W/m^2)\n", + "A = pi*D*L #Surface area of pipe (m^2)\n", + "Qc = h*A*(T1-T3) #Convection heat transfer to air (W)\n", + "Qr = e*s*A*(T1**4-T2**4) #Radiation heat transfer rate (W)\n", + "Q = Qc+Qr #Total heat transfer rate (Btu/h)\n", + "Tav = (T1+T2)/2.0 #Average temperature (K)\n", + "hr = 4*e*s*Tav**3 #Radiation heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"The emissive power from surface of pipe is :\",round(E),\" W/m^2 .\"\n", + "print \"The convection heat transfer to air is :\",round(Qc/10**3,1),\" kW .\"\n", + "print \"The radiation heat transfer rate is :\",round(Qr/10**3,1),\" kW \"\n", + "print \"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 .\n", + "The convection heat transfer to air is : 29.7 kW .\n", + "The radiation heat transfer rate is : 14.8 kW \n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 11.14:\n", + "Qc = 15.0 #Convection heat transfer coefficient (W/m^2.K)\n", + "hr = 7.2 #Radiation heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "X = hr/(Qc+hr)*100.0 #Percent heat transfer by radiation (%)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "FV = 1.0 #Correction factor\n", + "#From example 11.9:\n", + "FE = 0.358 #Emissivity correction factor\n", + "TH = 300.0+460.0 #Absolute temperature of external surface (\u00b0R)\n", + "TC = 75.0+460.0 #Absolute temperature of duct (\u00b0R)\n", + "AH = 0.622 #Area of pipe (ft^2)\n", + "s = 0.173*10**-8 #Stefan-Boltzmann constant\n", + "\n", + "#Calculation:\n", + "Q = FV*FE*AH*s*(TH**4-TC**4) #Heat transfer rate (Btu/h.ft)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate is :\",round(Q,2),\" Btu/h.ft\"\n", + "print \"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\n", + "Since, '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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From figure 11.2:\n", + "L = 1.0 #Space between plates (m)\n", + "X = 0.5 #Length of plate (m)\n", + "Y = 2.0 #Width of plate (m)\n", + "s = 5.669*10**-8 #Stefan-Boltzmann constant\n", + "TH = 2000.0+273.0 #Temperature of hotter plate (K)\n", + "TC = 1000.0+273.0 #Temperature of colder plate (K)\n", + "Btu = 0.2934*10**-3 #Btu/h in a KW\n", + "\n", + "#Calculation:\n", + "A = X*Y #Area of plate (m^2)\n", + "Z1 = Y/L #Ratio of width with space\n", + "Z2 = X/L #Ratio of length with space\n", + "#From figure 11.2:\n", + "FV = 0.18 #Correction factor\n", + "FE = 1.0 #Emissivity correction factor\n", + "Q1 = FV*FE*s*A*(TH**4-TC**4) #Net radiant heat exchange between plates (kW)\n", + "Q2 = Q1/Btu #Net radiant heat exchange between plates in Btu/h (Btu/h)\n", + "\n", + "#Result:\n", + "print \"The net radiant heat exchange between plates is :\",round(Q1,-2),\" kW .\"\n", + "print \"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 .\n", + "The 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 100755 index 00000000..dd2d06ef --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_12.ipynb @@ -0,0 +1,483 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:2b2623ff77afaafd7c16fc0720e69537bf598cad89dca70b267d9306235c5968" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "C = 1 #Number of constituents\n", + "P = 1 #Number of phases\n", + "\n", + "#Calculation:\n", + "F = C-P+2 #Number of degrees of freedom\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "\n", + "U1 = 1237.1 #Internnal energy of gas (Btu/lb)\n", + "U2_g = 1112.2 #Internal energy of gas (Btu/lb)\n", + "U2_l = 343.15 #Internal energy of liquid (Btu/lb)\n", + "\n", + "#Calculation:\n", + "Q = 0.5*(U2_g+U2_l)-1*U1 #Heat removed (Btu/lb)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "from sympy import symbols,solve\n", + "\n", + "#Variable declaration:\n", + "T1 = 99.0 #Mean film temperature (\u00b0C)\n", + "T2 = 98.0 #Plate surface temperature (\u00b0C)\n", + "g = 9.807 #Gravitational acceleration (m/s^2)\n", + "#From Appendix:\n", + "T3 = 100.0 #Saturation temperatre (\u00b0C)\n", + "h_vap1 = 970.3 #Latent heat of steam in Btu/lb (Btu/lb)\n", + "h_vap2 = 2.255*10**6 #Latent heat of steam in J/kg (J/kg)\n", + "p_v = 0.577 #Density of steam (kg/m^3)\n", + "p_l = 960.0 #Density of liquid water condensate (kg/m^3)\n", + "mu_l = 2.82*10**-4 #Absolute viscosity of liquid water condensate (kg/m.s)\n", + "k = 0.68 #Thermal conductivity of water (W/m.K)\n", + "#From table 12.2\n", + "Z = 0.4 #Height of rectangular plate (m)\n", + "Pw = 0.2 #Wetted perimeter of rectangular plate (m)\n", + "h = symbols('h') #Average heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "A = Z*Pw #Heat transfer area of plate (m^2)\n", + "R = A/Pw #Ratio A/Pw (m)\n", + "v_l = mu_l/p_l #Kinematic viscosity of liquid water condensate (m^2/s)\n", + "Co1 = (h/k)*(v_l**2/g/(1-p_v/p_l))**(1/3) #Condensation number (in terms of the average heat transfer coefficient)\n", + "Re = 4*h*Z*(T3-T2)/(mu_l*h_vap2) #Reynolds number in terms of the average heat transfer coefficient\n", + "#From equation 12.14:\n", + "CO1 = 0.0077*Re**Z #Co in terms of Reynolds number for flow type 1\n", + "x1 = solve(Co1-CO1,h) #Solving heat transfer coefficient (W/m^2.K)\n", + "h1 =x1[1]; #Average heat transfer coefficient for flow type 1 (W/m^2.K)\n", + "Re1 = Re.subs(h,h1); #Reynolds number for flow type 1\n", + "CO2 = 1.874*Re**(-1/3) #Co in terms of Reynolds number for flow tupe 2\n", + "x2 = solve(Co1-CO2,h) #Solving average heat transfer coefficient for flow type 2 (W/m^2.K)\n", + "h2 = x2[0]; #Average heat transfer coefficient for flow type 2 (W/m^2.K)\n", + "Re2 = Re.subs(h,h2) #Reynolds number for flow type 2\n", + "\n", + "#Result:\n", + "print \"The type of condensation flow type 2 is laminar.\"\n", + "print \"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.\n", + "And 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 12.5:\n", + "Re = 73.9 #Reynolds number\n", + "mu_l = 2.82*10**-4 #Absolute viscosity of liquid water condensate (kg/m.s)\n", + "Pw = 0.2 #Wetted perimeter of rectangular plate (m)\n", + "h = 14700.0 #Heat transfer coefficient (W/m^2.K)\n", + "T_sat = 100.0 #Saturation temperature (\u00b0C)\n", + "Ts = 98.0 #Surface temperature (\u00b0C)\n", + "A = 0.2*0.4 #Heat transfer area of plate (m^2) \n", + "\n", + "#Calculation:\n", + "m1 = Re*mu_l/4.0 #Mass flow rate of condensate (kg/m.s)\n", + "m = Pw*m1 #Mass flow rate of condensate (kg/s)\n", + "Co = (3.038*10**-5)*h #Condensation number\n", + "Q = h*A*(T_sat-Ts) #Heat transfer rate (W)\n", + "\n", + "#Result:\n", + "print \"1. The mass flow rate of condensate is :\",round(m1,4),\" kg/m.s . \"\n", + "print \"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 . \n", + "2. 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": [ + "\n", + "#Variable declaration:\n", + "T_sat = 126.0 #Saturation temperature (\u00b0F)\n", + "T = 64.0 #Surface temperature of tube (\u00b0F)\n", + "g = 32.2 #Gravitational acceleration (ft^2/s)\n", + "D = 4.0/12.0 #Outside diameter of tube (ft)\n", + "\n", + "#Calculation:\n", + "Tf = (T_sat+T)/2.0 #Mean film temperature (\u00b0F)\n", + "#From approximate values of key properties:\n", + "h_vap = 1022.0 #Latent heat of steam (Btu/lb)\n", + "p_v = 0.00576 #Density of steam (lb/ft^3)\n", + "p_l = 62.03 #Density of liquid (lb/ft^3)\n", + "k_l = 0.364 #Thermal conductivity of liquid (Btu/h.ft.\u00b0F)\n", + "mu_l = 4.26*10**-4 #Absolute viscosity of liquid water condensate (lb/ft.s)\n", + "h = 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:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "Qs1 = 9800.0 #Heat flux (W/m^2)\n", + "Ts1 = 102.0 #Original surface temperature (\u00b0C)\n", + "Ts2 = 103.0 #New surface temperature (\u00b0C)\n", + "Tsat = 100.0 #Saturation temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "h1 = Qs1/(Ts1-Tsat) #Original heat transfer coefficient (W/m^2.K)\n", + "DT1 = (Ts1 - Tsat) #Original excess temperature (\u00b0C)\n", + "DT2 = (Ts2 - Tsat) #New excess temperature (\u00b0C)\n", + "n = 0.25 #Value of n for laminar flow\n", + "h2 = h1*(DT2/DT1)**(n) #New heat transfer coefficient (W/m^2.K)\n", + "Qs2 = h2*(Ts2-Tsat) #New heat flux (W/m^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "#From example 12.9:\n", + "Ts1 = 102.0 #Original surface temperature (\u00b0C)\n", + "Ts2 = 103.0 #New surface temperature (\u00b0C)\n", + "Tsat = 100.0 #Saturation temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "DTe1 = (Ts1 - Tsat) #Original excess temperature (\u00b0C)\n", + "DTe2 = (Ts2 - Tsat) #New excess temperature (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"The original excess temperature is: DTe = \",DTe1,\" \u00b0C .\"\n", + "print \"The new excess temperature is: DTe = \",DTe2,\" \u00b0C .\"\n", + "if ((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 .\n", + "The new excess temperature is: DTe = 3.0 \u00b0C .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 12.9:\n", + "Cp = 4127.0 #heat capacity (J/kg . K)\n", + "DTe = 3.0 #New excess temperature (\u00b0C)\n", + "h_vap = 2.26*10**6 #latent heat of vaporization (J/kg)\n", + "\n", + "#Calculation:\n", + "Ja_L = Cp*DTe/h_vap #Liquid Jakob number\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ts = 106.0 #Surface temperature (\u00b0C)\n", + "Tsat = 100.0 #Saturation temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "DTe = Ts-Tsat #Excess temperature (\u00b0C)\n", + "#From table 12.5:\n", + "C1 = 5.56 #Constant C1\n", + "n1 = 3.0 #Constant n1\n", + "C2 = 1040.0 #Constant C2\n", + "n2 = 1.0/3.0 #Constant n2\n", + "P = 1.0 #Absolute pressure (atm)\n", + "Pa = 1.0 #Ambient absolute pressure (atm)\n", + "\n", + "#Calculation:\n", + "h1 = C1*DTe**n1*(P/Pa)**0.4 #Boiling water heat transfer coefficient (W/m^2)\n", + "Qs1 = h1*DTe #Surface flux (W/m^2)\n", + "h2 = C2*DTe**n2*(P/Pa)**0.4 #Second Boiling water heat transfer coefficient (W/m^2)\n", + "Qs2 = h2*DTe #Second Surface flux (W/m^2) \n", + "\n", + "#Result:\n", + " \n", + "if (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 .\"\n", + "elif (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 .\n", + "The 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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "#From example 12.12:\n", + "Qs1 = 11340.0 #Surface flux (W/m^2)\n", + "D = 0.3 #Diameter of electric heater (m)\n", + "\n", + "#Calculation:\n", + "A = pi*(D/2.0)**2 #Surface area of heater (m^2)\n", + "Qs = Qs1*A #Heat transfer rate (W)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..b4812ba2 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_13.ipynb @@ -0,0 +1,425 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c1f753b566ef5be1a578722a730958eae554497c733345943c47490976e91069" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "LR = 7.5/12.0 #Thickness of refractory (ft)\n", + "LI = 3.0/12.0 #Thickness of insulation (ft)\n", + "LS = 0.25/12.0 #Thickness of steel (ft)\n", + "kR = 0.75 #Thermal conductivity of refractory\n", + "kI = 0.08 #Thermal conductivity of insulation\n", + "kS = 26.0 #Thermal conductivity of steel\n", + "TR = 2000.0 #Average surface temperature of the inner face of the refractory (\u00b0F)\n", + "TS = 220.0 #Average surface temperature of the outer face of the steel (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT = TR-TS #Temperature difference (\u00b0F)\n", + "Q = DT/(LR/kR+LI/kI+LS/kS) #Heat loss (Btu/h.ft^2)(here representing Qdot/A)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "LR = 7.5/12.0 #Thickness of refractory (ft)\n", + "kR = 0.75 #Thermal conductivity of refractory\n", + "TR = 2000.0 #Average surface temperature of the inner face of the refractory (\u00b0F)\n", + "Q = 450.0 #Heat loss (Btu/h.ft^2)\n", + "\n", + "#Calculation:\n", + "TI = TR - Q*(LR/kR) #Temperature of the boundary where the refractory meets the insulation (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "QbyA = 70000.0 #Total heat loss (Btu/h)\n", + "Q = 450.0 #Heat loss (Btu/h.ft^2)\n", + "\n", + "#Calculation:\n", + "A = QbyA/Q #Area available for heat transfer (ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "h_out = 390.0 #Enthalpy of the fluid that exits from the evaporator (kJ/kg)\n", + "h_in = 230.0 #Enthalpy of the fluid that enters the unit (kJ/kg)\n", + "\n", + "#Calculation:\n", + "QC = h_out - h_in #Heat absorbed by the evaporator (kJ/kg)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 13.9:\n", + "TS = -10.0+273.0 #Fluid\u2019s saturation temperature expressed in Kelvin (K)\n", + "QC = 160.0 #Heat absorbed by the evaporator (kJ/kg)\n", + "\n", + "#Calcuation:\n", + "DS = QC/TS #Fluid\u2019s change in entropy(kJ/kg.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From figure 13.2:\n", + "h1 = 390.0 #Fluid enthalpy on entering the compressor (kJ/kg)\n", + "h2 = 430.0 #Fluid enthalpy on leaving the compressor (kJ/kg)\n", + "h3 = 230.0 #Fluid enthalpy on leaving the condenser (kJ/kg)\n", + "\n", + "#Calculation:\n", + "QH = h2 - h3 #Heat rejected from the condenser (kJ/kg)\n", + "W_in = h2 - h1 #Change in enthalpy across the compressor (kJ/kg)\n", + "QC = QH - W_in #Heat absorbed by the evaporator (kJ/kg)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 13.11:\n", + "W_in = 40.0 #Change in enthalpy across the compressor (kJ/kg)\n", + "QC = 160.0 #Heat absorbed by the evaporator (kJ/kg)\n", + "\n", + "#Calculation:\n", + "COP = QC/W_in #Refrigerator\u2019s C.O.P.\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "h1 = 548.0 #Steam enthalpy at the entry and exit to the boiler (kJ/kg)\n", + "h2 = 3989.0 #Steam enthalpy at the entry and exit to the turbine (kJ/kg)\n", + "h3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\n", + "QH = 2043.0 #Heat rejected by the condenser (kJ/kg)\n", + "\n", + "#Calculation:\n", + "h4 = h3 - QH #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\n", + "Qb = h2 - h1 #Enthalpy change across the boiler (kJ/kg)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "#From example 13.4:\n", + "h1 = 548.0 #Steam enthalpy at the entry and exit to the boiler (kJ/kg)\n", + "h2 = 3989.0 #Steam enthalpy at the entry and exit to the turbine (kJ/kg)\n", + "h3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\n", + "h4 = 448.0 #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\n", + "Qb = 3441.0 #Enthalpy change across the boiler (kJ/kg)\n", + "\n", + "#Calculation:\n", + "Wt = h2 - h3 #Work produced by the turbine (kJ/kg)\n", + "Wp = h1 - h4 #Work used by the pump (kJ/kg)\n", + "W_net = Wt - Wp #Net work by subtracting the pump work from the turbine work (kJ/kg)\n", + "n_th = W_net/Qb #Thermal efficiency\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From table 13.4:\n", + "x3 = 0.9575 #Mass fraction vapour at point 3\n", + "h3 = 2491.0 #Steam enthalpy at the entry and exit to the pump (kJ/kg)\n", + "s3 = 7.7630 #Entropy at the entry and exit to the pump (kJ/kg.K)\n", + "s4 = 1.4410 #Entropy at the entry and exit to the condenser (kJ/kg.K)\n", + "#From example13.14:\n", + "h4 = 448.0 #Steam enthalpy at the entry and exit to the condenser (kJ/kg)\n", + "\n", + "#Calculation:\n", + "Q_out = h3 - h4 #Heat rejected (kJ/kg)\n", + "DS = s3 - s4 #Process change in entropy (kJ/kg)\n", + "T3 = Q_out/DS #Temperature at point 3 (K)\n", + "\n", + "#Result:\n", + "print \"The temperature at point 3 is :\",round(T3),\" K .\"\n", + "print \"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 .\n", + "Or, 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 100755 index 00000000..7bc0f737 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_14.ipynb @@ -0,0 +1,705 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e3c042bf60cf5ea8673efb2741d5df5f7935daa012691b520a2d1357e838bd6c" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "scfm = 20000.0 #Volumetric flow rate of air at standard conditions (scfm)\n", + "H1 = 1170.0 #Enthalpy at 200\u00b0F (Btu/lbmol)\n", + "H2 = 14970.0 #Enthalpy at 2000\u00b0F (Btu/lbmol)\n", + "Cp = 7.53 #Average heat capacity (Btu/lbmol.\u00b0F)\n", + "T1 = 200.0 #Initial temperature (\u00b0F)\n", + "T2 = 2000.0 #Final temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "n = scfm/359.0 #Flow rate of air in a molar flow rate (lbmol/min)\n", + "DH = H2 - H1 #Change in enthalpy (Btu/lbmol)\n", + "DT = T2 - T1 #Change in temperature (\u00b0F)\n", + "Q1 = n*DH #Heat transfer rate using enthalpy data (Btu/min)\n", + "Q2 = n*Cp*DT #Heat transfer rate using the average heat capacity data (Btu/min)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate using enthalpy data is :\",round(Q1/10**5,2),\" x 10^5 Btu/min.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "n = 1200.0 #Flow rate of air in a molar flow rate (lbmol/min)\n", + "Cp = 0.26 #Average heat capacity (Btu/lbmol.\u00b0F)\n", + "T1 = 200.0 #Initial temperature (\u00b0F)\n", + "T2 = 1200.0 #Final temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT = T2 - T1 #Change in temperature (\u00b0F)\n", + "Q = n*Cp*DT #Required heat rate (Btu/min)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Tc1 = 25.0 #Initial temperature of cold fluid (\u00b0C)\n", + "Th1 = 72.0 #Initial temperature of hot fluid (\u00b0C)\n", + "Th2 = 84.0 #Final temperature of hot fluid (\u00b0C)\n", + "\n", + "#Calculation:\n", + "#From equation 14.2:\n", + "Tc2 = (Th2-Th1)+Tc1 #Final temperature of cold fluid (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"The final temperature of the cold liquid is :\",Tc2,\" \u00b0C .\"\n", + "print \"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 .\n", + "There 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ts = 100.0 #Steam temperature at 1 atm (\u00b0C)\n", + "Tl = 25.0 #Fluid temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "DTlm = Ts - Tl #Log mean temperature difference (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "Ts = 100.0 #Steam temperature at 1 atm (\u00b0C)\n", + "T1 = 25.0 #Initial fluid temperature (\u00b0C)\n", + "T2 = 80.0 #Final fluid temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "DT1 = Ts - T1 #Temperature difference driving force at the fluid entrance (\u00b0C)\n", + "DT2 = Ts - T2 #Temperature driving force at the fluid exit (\u00b0C)\n", + "DTlm = (DT1 - DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"The LMTD is :\",round(DTlm,1),\" \u00b0C .\"\n", + "print \"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 .\n", + "There 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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "T1 = 500.0 #Temperature of hot fluid entering the heat exchanger (\u00b0F)\n", + "T2 = 400.0 #Temperature of hot fluid exiting the heat exchanger (\u00b0F)\n", + "t1 = 120.0 #Temperature of cold fluid entering the heat exchanger (\u00b0F)\n", + "t2 = 310.0 #Temperature of cold fluid exiting the heat exchanger (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T1 - t2 #Temperature difference driving force at the heat exchanger entrance (\u00b0F)\n", + "DT2 = T2 - t1 #Temperature difference driving force at the heat exchanger exit (\u00b0F)\n", + "DTlm = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "m = 8000.0 #Rate of oil flow inside the tube (lb/h)\n", + "Cp = 0.55 #Heat capacity of oil (Btu/lb.\u00b0F)\n", + "T1 = 210.0 #Initial temperature of oil (\u00b0F)\n", + "T2 = 170.0 #Final temperature of oil (\u00b0F)\n", + "t = 60.0 #Tube surface temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT = T2 - T1 #Change in temperature (\u00b0F)\n", + "Q = m*Cp*DT #Heat transferred from the heavy oil (Btu/h)\n", + "DT1 = T1 - t #Temperature difference driving force at the pipe entrance (\u00b0F)\n", + "DT2 = T2 - t #Temperature difference driving force at the pipe exit (\u00b0F)\n", + "DTlm = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate is :\",round(Q),\" Btu/h .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "T1 = 138.0 #Temperature of oil entering the cooler (\u00b0F)\n", + "T2 = 103.0 #Temperature of oil leaving the cooler (\u00b0F)\n", + "t1 = 88.0 #Temperature of coolant entering the cooler (\u00b0F)\n", + "t2 = 98.0 #Temperature of coolant leaving the cooler (\u00b0F)\n", + "\n", + "#Calculation:\n", + "#For counter flow unit:\n", + "DT1 = T1 - t2 #Temperature difference driving force at the cooler entrance (\u00b0F)\n", + "DT2 = T2 - t1 #Temperature difference driving force at the cooler exit (\u00b0F)\n", + "DTlm1 = (DT1 - DT2)/(log(DT1/DT2)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n", + "#For parallel flow unit:\n", + "DT3 = T1 - t1 #Temperature difference driving force at the cooler entrance (\u00b0F)\n", + "DT4 = T2 - t2 #Temperature difference driving force at the cooler exit (\u00b0F)\n", + "DTlm2 = (DT3 - DT4)/(log(DT3/DT4)) #LMTD (driving force) for the heat exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The LMTD for counter-current flow unit is :\",round(DTlm1,1),\" \u00b0F .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "A = 1.0 #Surface area of glass (m^2)\n", + "h1 = 11.0 #Heat transfer coefficient inside room (W/m^2.K)\n", + "L2 = 0.125*0.0254 #Thickness of glass (m)\n", + "k2 = 1.4 #Thermal conductivity of glass (W/m.K)\n", + "h3 = 9.0 #Heat transfer coefficient from window to surrounding cold air (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "R1 = 1.0/(h1*A) #Internal convection resistance (K/W)\n", + "R2 = L2/(k2*A) #Conduction resistance through glass panel (K/W)\n", + "R3 = 1.0/(h3*A) #Outside convection resistance (K/W)\n", + "Rt = R1+R2+R3 #Total thermal resistance (K/W)\n", + "U = 1.0/(A*Rt) #Overall heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Dx = 0.049/12.0 #Thickness of copper plate (ft)\n", + "h1 = 208.0 #Film coefficient of surface one (Btu/h.ft^2.\u00b0F)\n", + "h2 = 10.8 #Film coefficient of surface two (Btu/h.ft^2.\u00b0F)\n", + "k = 220.0 #Thermal conductivity for copper (W/m.K)\n", + "\n", + "#Calculation:\n", + "U = 1.0/(1.0/h1+Dx/k+1.0/h2) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Do = 0.06 #Outside diameter of pipe (m)\n", + "Di = 0.05 #Inside diameter of pipe (m)\n", + "ho = 8.25 #Outside coefficient (W/m^2.K)\n", + "hi = 2000.0 #Inside coefficient (W/m^2.K)\n", + "R = 1.33*10**-4 #Resistance for steel (m^2.K/W)\n", + "\n", + "#Calculation:\n", + "U = 1.0/(Do/(hi*Di)+R+1.0/ho) #Overall heat transfer coefficient (W/m^2.\u00b0K)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import pi,log\n", + "\n", + "#Variable declaration:\n", + "Di = 0.825/12.0 #Pipe inside diameter (ft)\n", + "Do = 1.05/12.0 #Pipe outside diameter (ft)\n", + "Dl = 4.05/12.0 #Insulation thickness (ft)\n", + "l = 1.0 #Pipe length (ft)\n", + "kp = 26.0 #Thermal conductivity of pipe (Btu/h.ft.\u00b0F)\n", + "kl = 0.037 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\n", + "hi = 800.0 #Steam film coefficient (Btu/h.ft^2.\u00b0F)\n", + "ho = 2.5 #Air film coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "ri = Di/2.0 #Pipe inside radius (ft)\n", + "ro = Do/2.0 #Pipe outside radius (ft)\n", + "rl = Dl/2.0 #Insulation radius (ft)\n", + "Ai = pi*Di*l #Inside area of pipe (ft^2)\n", + "Ao = pi*Do*l #Outside area of pipe (ft^2)\n", + "Al = pi*Dl*l #Insulation area of pipe (ft^2)\n", + "A_Plm = (Ao-Ai)/log(Ao/Ai) #Log mean area for steel pipe (ft^2)\n", + "A_Ilm = (Al-Ao)/log(Al/Ao) #Log mean area for insulation (ft^2)\n", + "Ri = 1.0/(hi*Ai) #Air resistance (m^2.K/W)\n", + "Ro = 1.0/(ho*Al) #Steam resistance (m^2.K/W)\n", + "Rp = (ro-ri)/(kp*A_Plm) #Pipe resistance (m^2.K/W)\n", + "Rl = (rl-ro)/(kl*A_Ilm) #Insulation resistance (m^2.K/W)\n", + "U = 1.0/(Ai*(Ri+Rp+Ro+Rl)) #Overall heat coefficient based on the inside area (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "#From example 14.14:\n", + "Di = 0.825/12.0 #Pipe inside diameter (ft)\n", + "L = 1.0 #Pipe length (ft)\n", + "Ui = 0.7492 #Overall heat coefficient (Btu/h.ft^2.\u00b0F)\n", + "Ts = 247.0 #Steam temperature (\u00b0F)\n", + "ta = 60.0 #Air temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Ai = pi*Di*L #Inside area of pipe (ft^2)\n", + "Q = Ui*Ai*(Ts-ta) #Heat transfer rate (Btu/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "hw = 200.0 #Water heat coefficient (Btu/h.ft^2.\u00b0F)\n", + "ho = 50.0 #Oil heat coefficient (Btu/h.ft^2.\u00b0F)\n", + "hf = 1000.0 #Fouling heat coefficient (Btu/h.ft^2.\u00b0F)\n", + "DTlm = 90.0 #Log mean temperature difference (\u00b0F)\n", + "A = 15.0 #Area of wall (ft^2)\n", + "\n", + "#Calculation:\n", + "X = 1.0/hw+1.0/ho+1.0/hf #Equation 14.34 for constant A\n", + "U = 1.0/X #Overall heat coeffocient (Btu/h.ft^2.\u00b0F)\n", + "Q = U*A*DTlm #Heat transfer rate (Btu/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "from sympy import symbols,log,nsolve\n", + "\n", + "\n", + "#Variable declaration:\n", + "T = 80.0 #Pipe surface temperature (\u00b0F)\n", + "t1 = 10.0 #Brine inlet temperature (\u00b0F)\n", + "DT2 = symbols('DT2') #Discharge temperature of the brine solution (\u00b0F)\n", + "m = 20*60 #Flowrate of brine solution (lb/h)\n", + "Cp = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\n", + "U1 = 150 #Overall heat transfer coefficient at brine solution entrance (Btu/h.ft^2.\u00b0F)\n", + "U2 = 140 #Overall heat transfer coefficientat at brine solution exit (Btu/h.ft^2.\u00b0F)\n", + "A = 2.5 #Pipe surface area for heat transfer (ft^2)\n", + "\n", + "#Calculation:\n", + "DT1 = T-t1 #Temperature approach at the pipe entrance (\u00b0F)\n", + "Q = m*Cp*(DT1-DT2) #Energy balance to the brine solution across the full length of the pipe (Btu/h)\n", + "DT1m = (DT1-DT2)/log(DT1/DT2) #Equation for the LMTD\n", + "QQ = A*(U2*DT1-U1*DT2)/log(U2*DT1/U1/DT2) #Equation for the heat transfer rate (Btu/h)\n", + "E = QQ-Q #Energy balance equation\n", + "R = nsolve([E],[DT2],[1.2]) #\n", + "DT = R[0] #Log mean temperature difference\n", + "t2 = T-DT #In discharge temperature of the brine solution (\u00b0F)\n", + "t2c = 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:\n", + "print \"The temperature approach at the brine inlet side is :\",round(DT1,1),\" \u00b0F.\"\n", + "print \"Or, the temperature approach at the brine inlet side is :\",round(DT1/1.8,1),\" \u00b0C.\"\n", + "print \"The exit temperature of the brine solution is :\",round(t2,1),\" \u00b0F.\"\n", + "print \"Or, the exit temperature of the brine solution is :\",round((t2-32)/1.8,1),\" \u00b0C.\"\n", + "print \"The rate of heat transfer is :\",round(_Q_,-1),\" Btu/h.\"\n", + "print \"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.\n", + "Or, the temperature approach at the brine inlet side is : 38.9 \u00b0C.\n", + "The exit temperature of the brine solution is : 28.4 \u00b0F.\n", + "Or, the exit temperature of the brine solution is : -2.0 \u00b0C.\n", + "The rate of heat transfer is : 21830.0 Btu/h.\n", + "Or, 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 100755 index 00000000..8331e128 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_15.ipynb @@ -0,0 +1,815 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9c90790dace23fbdfc3e0ce273ce32b43acc5cfa1b423fc8f4e186650e91a101" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Q = 12000.0 #Heat transfer rate (Btu/h)\n", + "U = 48.0 #Overall heat coefficient (Btu/ft^2.h.\u00b0F)\n", + "DTlm = 50.0 #Log mean temperature difference (\u00b0F)\n", + "\n", + "#Calculation:\n", + "A = Q/(U*DTlm) #Area of exchanger (ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from sympy import symbols,solve,log\n", + "\n", + "#Variable declaration:\n", + "Q = 56760 #Heat transfer rate (Btu/h)\n", + "U = 35.35 #Overall heat coefficient (Btu/ft.h.\u00b0F)\n", + "A = 32.1 #Area of exachanger (ft^2)\n", + "t1 = 63.0 #Outlet cold water temperature (\u00b0F)\n", + "T1 = 164 #Outlet hot water temperature (\u00b0F)\n", + "T2 = 99 #Inlet hot water temperature (\u00b0F)\n", + "t2 = symbols('t2') #Inlet cold water temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DTlm = Q/(U*A) #Log mean temperature difference (\u00b0F)\n", + "dT1 = T1-t1 #Temperature approach at pipe outlet (\u00b0F)\n", + "dT2 = T2-t2 #Temperature approach at pipe inlet (\u00b0F)\n", + "Eq = (dT2-dT1)/log(dT2/dT1)-DTlm\n", + "R = solve(Eq,t2) #Inlet cold water temperature (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "m = 14.6 #Flow rate of water inside the tube (lb/min)\n", + "Cp = 1 #Heat capacity of water (Btu/lb.\u00b0F)\n", + "t2 = 79 #Initial temperature of water (\u00b0F)\n", + "t1 = 63 #Final temperature of water (\u00b0F)\n", + "#From example 15.3:\n", + "Q1 = 56760 #Old heat transfer rate (Btu/h)\n", + "\n", + "#Calculation:\n", + "Q2 = m*Cp*(t2-t1) #New heat transfer rate (Btu/min)\n", + "\n", + "#Result:\n", + "print \"The new heat transfer rate is :\",round(Q2),\" Btu/min.\" \n", + "print \"Or, the new heat transfer rate is :\",round(Q2*60),\" Btu/h.\"\n", + "if (Q1==Q2) :\n", + " print \"This result agree with the Q\u02d9 provided in the problem statement. Shakespeare is wrong, nothing is rotten there.\"\n", + "else :\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.\n", + "Or, the new heat transfer rate is : 14016.0 Btu/h.\n", + "This 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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "T1 = 210.0 #Initial temperature of oil (\u00b0F)\n", + "T2 = 170.0 #Final temperature of oil (\u00b0F)\n", + "T3 = 60.0 #Surface temperature of oil (\u00b0F)\n", + "m = 8000.0 #Flow rate of oil inside tube (lb/h)\n", + "cp = 0.55 #Heat capacity of oil (Btu/lb.\u00b0F)\n", + "U = 63.0 #Overall heat teansfer coefficient (Btu.h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T1-T3 #Temperature difference 1 (\u00b0F)\n", + "DT2 = T2-T3 #Temperature difference 2 (\u00b0F)\n", + "DTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temerature difference (\u00b0F)\n", + "Q = m*cp*(T1-T2) #Heat transferred (Btu/h)\n", + "A = Q/(U*DTlm) #Heat transfer area (ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "T1 = 140.0 #Initial temperature of hot water (\u00b0F)\n", + "T2 = 110.0 #Final temperature of hot water (\u00b0F)\n", + "T3 = 60.0 #Initial temperature of cold water (\u00b0F)\n", + "T4 = 90.0 #Initial temperature of cold water (\u00b0F)\n", + "DTlm2 = 50.0 #Log mean temerature difference for countercurrent flow, a constant (\u00b0F) (part 2)\n", + "m = 100.0*60 #Water flow rate (lb/h)\n", + "cp = 1.0 ##Heat capacity of water (Btu/lb.\u00b0F)\n", + "U = 750.0 #Overall heat teansfer coefficient (Btu.h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T1-T3 #Temperature difference 1 (\u00b0F) (part 1)\n", + "DT2 = T2-T4 #Temperature difference 2 (\u00b0F)\n", + "DTlm1 = (DT1-DT2)/log(DT1/DT2) #Log mean temerature difference (\u00b0F)\n", + "Q = m*cp*(T1-T2) #Heat transferred (Btu/h)\n", + "Ap = Q/(U*DTlm1) #Heat transfer area for parallel flow (ft^2)\n", + "Ac = Q/(U*DTlm2) #Heat transfer area for counter flow (ft^2)\n", + "\n", + "#Result:\n", + "print \"1. The double pipe co-current flow is :\",round(Ap,2),\" ft^2 .\"\n", + "print \"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 .\n", + "1. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi,log\n", + "\n", + "#Variable declaration:\n", + "uC = 3.7*10**-4 \t #Viscosity of benzene (lb/ft.s)\n", + "uH = 2.05*10**-4 \t #Viscosity of water @200 \u00b0F (lb/ft.s)\n", + "u2 = 2.16*10**-4 \t\t\t\t #Viscosity of water @192 \u00b0F (lb/ft.s)\n", + "pC = 54.8 #Density of benzene (lb/ft^3)\n", + "pH = 60.13 #Density of water (lb/ft^3)\n", + "cpC = 0.415 #Specific heat capacity of benzene (Btu/lb.\u00b0F)\n", + "cpH = 1 #Specific heat capacity of water (Btu/lb.\u00b0F)\n", + "sgC = 0.879 \n", + "kC = 0.092 #Thermal conductivity of benzene (Btu/h.ft.\u00b0F)\n", + "kH = 0.392 #Thermal conductivity of water @200 \u00b0F (Btu/h.ft.\u00b0F)\n", + "k2 = 0.390\t\t\t\t\t #Thermal conductivity of water @192 \u00b0F (Btu/h.ft.\u00b0F)\n", + "mC = 2500\t \t\t\t #Flow rate of benzene (lb/s)\n", + "mH = 4000 #Flow rate of water (lb/s)\n", + "Re = 13000 #Reynolds number\n", + "dTc = 120-60\t\t\t\t\t #Difference in temperature heating for benzene\n", + "Tw = 200\t\t\t\t #Temperatperature of hot water (\u00b0F)\n", + "#For 2-inch schedule 40 pipe\n", + "Ai = 0.541 #Inside area of pipe (ft^2/ft)\n", + "Ao = 0.622 #Outside area of pipe (ft^2/ft)\n", + "Di = 2.067 #Inside diameter of pipe (inch)\n", + "Do = 2.375 #Outside diameter of pipe (inch)\n", + "Si = 0.0233 #Inside surface area of pipe (ft^2)\n", + "dXw = 0.128 #Width of pipe (ft)\n", + "\n", + "#For 4-inch schedule 40 pipe\n", + "Dio = 4.026 #Inside diameter of pipe (inch)\n", + "Doi = Do #Outside diameter of pipe (inch)\n", + "kw = 26 \n", + "\n", + "#Calculations:\n", + "def St(Re,Pr):\t\t\t\t #Dittus Boelter equation\n", + "\treturn 0.023*Re**-0.2*Pr**-0.667\n", + "#For inside tubes:\n", + "Dicalc = 4*mC/(Re*pi*uC)/3600 #Inside diameter (ft)\n", + "mHcalc = Re*pi*uH*(Doi+Dio)/4*3600/12 #Mass flow rate of water (lb/h)\n", + "Q = mC*cpC*dTc\t\t\t\t\t #Heat in water (Btu/h)\n", + "dTH = Q/mH #Temperature difference of water (\u00b0F)\n", + "THo = Tw - dTH #Outlet temperature of water (\u00b0F)\n", + "THav = (Tw+THo)/2 #Average temperature of water (\u00b0F) \n", + "#For benzene:\n", + "PrC = cpC*uC/kC*3600 #Prandtl number\n", + "StC = round(St(13000, PrC),5) #Stanton number\n", + "hi = StC*cpC*mC/Si #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "#For water:\n", + "ReH = 4*mH/3600/(pi*u2*(Doi+Dio)/12) #Reynolds number\n", + "PrH = round(cpH*(u2)/k2*3600 ,2) #Prandtl number\n", + "StH = round(St(ReH, PrH),5) #Stanton number\n", + "Sann = round(pi/4*(Dio**2-Doi**2)/144,4) #Surface area of annulus (ft^2)\n", + "ho = round(StH*cpH*mH/Sann) #Heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "#For pipe:\n", + "Dlm = round((Do-Di)/log(Do/Di)*12,3) #Log mean difference in diameter (ft)\n", + "Uo = 1/(Do/Di/hi + dXw*Do/kw/Dlm + 1/ho) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "dTlm = (124.4-80)/log(124.4/80) #Log mean temperature difference (\u00b0F)\n", + "L = Q/(Uo*0.622*dTlm) #Length of pipe (ft)\n", + "\n", + "#Result:\n", + "print '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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import e\n", + "\n", + "#Variable declaration:\n", + "MC = 2000.0 \n", + "mc = 1000.0\n", + "U = 2000.0\n", + "A = 10.0\n", + "T1 = 300.0\n", + "t1 = 60.0\n", + "\n", + "#Calculation:\n", + "B = 1.0/mc \n", + "b = 1.0/MC\n", + "x = B/b\n", + "y = U*(B-b)\n", + "T2 = ((x-y)*T1 + x*(e-y)*t1)/(2*e-1)\n", + "t2 = t1+(T1-T2)/x\n", + "\n", + "#Result:\n", + "print \"T2 = :\",round(T2),\" \u00b0F\"\n", + "print \"t2 = :\",round(t2),\" \u00b0F\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2 = : 114.0 \u00b0F\n", + "t2 = : 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": [ + "\n", + "\n", + "from sympy import symbols,solve,log\n", + "\n", + "#Variable declaration:\n", + "h1 = 1200.0 #Hot film coefficient (Btu/h.ft^2.\u00b0F)\n", + "h2 = 1175.0 #Cold film coefficient (Btu/h.ft^2.\u00b0F)\n", + "L = 200.0 #Length of pipe (ft)\n", + "MC = 30000.0\n", + "mc = 22300.0\n", + "T1 = 300.0 #Inlet temperature of hot fluid in pipe (\u00b0F)\n", + "t1 = 60.0 #Inlet temperature of cold fluid in pipe (\u00b0F)\n", + "T2 = symbols('T2') #Outlet temperature of hot fluid \u00b0F\n", + "t2 = symbols('t2') #Outlet temperature of cold fluid \u00b0F\n", + "#From table 6.2:\n", + "ID = 2.067 #Inside diameter of pipe (in)\n", + "OD = 2.375 #Outside diameter of pipe (in)\n", + "Dx = 0.154 #Thickness of pipe (in)\n", + "Ai = 0.541 #Inside sectional area of pipe (ft^2/ft)\n", + "k = 25.0 #Thermal conductivity of pipe (Btu/h)\n", + "\n", + "#Calculation:\n", + "Ui = 1.0/((1.0/h1) +(Dx/(k*12.0))+(1.0/(h2*(OD/ID)))) #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "Ai1 = Ai*L #Inside area of pipe (ft^3/ft)\n", + "QH = MC*(T1-T2) #Heat transfer rate of hot fluid (Btu/h)\n", + "QC = mc*(t2-t1) #Heat transfer rate of cold fluid (Btu/h)\n", + "t2ht = 195 #t2 by hit and trial\n", + "x = solve(QC-QH,T2)\n", + "T2 = x[0]\n", + "DTlm = (T1-t1-T2+t2)/log((T1-t1)/(T2-t2)) #Log mean temperature difference (\u00b0F)\n", + "Q = Ui*Ai1*DTlm.subs(t2,t2ht) #Total heat transfer rate (Btu/h)\n", + "\n", + "#Result:\n", + "print \"T2 :\", round(T2.subs(t2,t2ht)),\"(\u00b0F)\"\n", + "print \"t2 :\", t2.subs(t2,t2ht),\"(\u00b0F)\"\n", + "print \"Qdot :\", round(Q/10**6) ,\"x 10**6 Btu/h\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "T2 : 200.0 (\u00b0F)\n", + "t2 : 195 (\u00b0F)\n", + "Qdot : 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log,e\n", + "\n", + "#Variable declaration:\n", + "B = 3.33*10**-5\n", + "b = 4.48*10**-5\n", + "#From example 15.11:\n", + "A = 108.2 #Inside area of pipe (ft^3/ft)\n", + "U = 482 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "MC = 30000.0\n", + "mc = 23000.0\n", + "T1 = 300.0 #Inlet temperature of hot fluid in pipe (\u00b0F)\n", + "t1 = 60.0 #Inlet temperature of cold fluid in pipe (\u00b0F)\n", + "\n", + "#Calculation:\n", + "#From equation 15.28:\n", + "T2 = ((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:\n", + "t2 = ((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)\n", + "DT = ((T2-t1)-(T1-t2))/(log((T2-t1)/(T1-t2))) #Log mean difference temperature (\u00b0F)\n", + "Q1 = U*A*DT #Heat transfer rate of hot fluid (Btu/h)\n", + "Q2 = MC*(T1-T2) #Heat transfer rate of cold fluid (Btu/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "from math import log,pi\n", + "\n", + "#Variable declaration:\n", + "Ts = 100.0 #Saturation temperature (\u00b0C)\n", + "t1 = 25.0 #Initial temperature of water (\u00b0C)\n", + "t2 = 73.0 #Final temperature of water (\u00b0C)\n", + "m = 228.0/3600.0 #Mass flow rate of water (kg/s)\n", + "cp = 4174.0 #Heat capacity of water (J/kg.K)\n", + "m_s = 55.0/3600.0 #Mass flow rate of steam (kg/s)\n", + "h_vap = 2.26*10**26 #Latent heat of condensation (J/kg)\n", + "k = 54.0 #Thermal conductivity for 0.5% carbon steel (W/m.K)\n", + "rii = 0.013 #Inner radius of inner pipe of the double pipe heat exchanger (m)\n", + "roi = 0.019 #Outer radius of inner pipe of the double pipe heat exchanger (m)\n", + "Rf = 0.0002 #Fouling factor (m^2.K/W)\n", + "Uc = 0.00045 #Clean overall heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "DT1 = Ts-t1 #Temperature driving force at end 1 (K)\n", + "DT2 = Ts-t2 #Temperature driving force at end 2 (K)\n", + "DTlm = (DT1-DT2)/(log(DT1/DT2)) #Log mean difference temperature (\u00b0C)\n", + "Cw =m*cp #Capacitance rate of water (W/K)\n", + "Q = Cw*(t2-t1) #Heat transfer rate (W)\n", + "Qmax1 = Cw*(Ts-t1) #Maximum heat term from the water stream (W)\n", + "Qmax2 = m_s*h_vap #Maximum heat term from the steam (W)\n", + "E = Q/Qmax1 #Effectiveness\n", + "Lmin = (Q*(log(roi/rii)))/(2*pi*k*(Ts-t1)) #Minimum required length of heat exchanger (m)\n", + "Ud = 1.0/(1.0/Uc+Rf) #Dirty overall heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Result:\n", + "print \"1. The temperature profile of the water and steam along the length of the exchanger is :\",round(DTlm),\" \u00b0C .\"\n", + "print \"2. Effectiveness of energy from steam to heat the water is :\",round(E,3),\" .\"\n", + "print \"3. The minimum length of the heat exchanger is :\",round(Lmin,3),\" m .\"\n", + "print \"4. The dirty overall heat transfer coefficient :\",round(Ud,5),\" W/m^2.K .\"\n", + "print \"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 .\n", + "2. Effectiveness of energy from steam to heat the water is : 0.64 .\n", + "3. The minimum length of the heat exchanger is : 0.189 m .\n", + "4. The dirty overall heat transfer coefficient : 0.00045 W/m^2.K .\n", + "5. 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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "Q = 12700.0 #Heat transfer rate (W)\n", + "Ud = 2220.0 #Dirty overall heat transfer coefficient (W/m^2.K)\n", + "DTlm = 47.0 #Log mean difference temperature (\u00b0C)\n", + "rii = 0.013 #Inner radius of inner pipe of the double pipe heat exchanger (m)\n", + "#Calculation:\n", + "A = Q/(Ud*DTlm) #Heat transfer area (m^2)\n", + "L = A/(2*pi*rii) #Tube length (m)\n", + "\n", + "#Result:\n", + "print \"The heat transfer area is :\",round(A,4),\" m^2 .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ud = 2220.0 #Dirty overall heat transfer coefficient (W/m^2.K)\n", + "A = 0.1217 #Heat transfer area (m^2)\n", + "Cw = 264.0 #Capacitance rate of water (W/K)\n", + "\n", + "#Calculation:\n", + "NTU = (Ud*A)/Cw #Number of transfer units of the exchanger\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Ao = 1.85 #Area of heat exchanger (ft^2)\n", + "\n", + "#Calculation:\n", + "#From figure 15.6:\n", + "y = 0.560*10**-3 #Intercept 1/UoAo (\u00b0F.h/Btu)\n", + "ho = 1.0/(Ao*y) #Thermal conductivity for heat exchanger (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From figure 15.7:\n", + "a = 0.00126\n", + "b = 0.0276\n", + "\n", + "#Calculation:\n", + "ho = 1.0/a #The value of ho (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "Di = 0.902/12.0 #Inside diameter of tube (ft)\n", + "Do = 1.0/12.0 #Outside diameter of tube (ft)\n", + "k = 60.0 #Thermal conductivity of tube (Btu/h.ft^2.\u00b0F) \n", + "\n", + "#Calculation:\n", + "#From example 15.19:\n", + "a = 0.00126\n", + "Dr = (Do - Di)/2.0 #Radial thickness of tube wall (ft)\n", + "Rw = Dr/k #Resistance of wall (Btu/h.\u00b0F)\n", + "ho = 1.0/(a-Rw) #The revised ho (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "a1 = 0.00044 #Term 'a' for U_clean\n", + "a2 = 0.00089 #Term 'a' for U_dirty\n", + "\n", + "#Calculation:\n", + "Rs = a2 - a1 #Resistance associated with the scale\n", + "hs = 1.0/Rs #Scale film coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..af0c69ba --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_16.ipynb @@ -0,0 +1,583 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4d89093b0aef649c7f239bf75c18374320afd98a6f8286cbe2c1fcb242b5a141" + }, + "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": [ + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "#From figure 16.13, for ideal countercurrent heat exchanger:\n", + "T1 = 150.0 #Inlet temperature of hot fluid (\u00b0F)\n", + "T2 = 100.0 #Outet temperature of hot fluid (\u00b0F)\n", + "t1 = 50.0 #Inlet temperature of cold fluid (\u00b0F)\n", + "t2 = 80.0 #Outet temperature of hot fluid (\u00b0F)\n", + "#From figure 16.14, for shell and tube exchanger:\n", + "T_1 = 50.0 #Inlet temperature of cold fluid (\u00b0F)\n", + "T_2 = 80.0 #Outet temperature of hot fluid (\u00b0F)\n", + "t_1 = 150.0 #Inlet temperature of hot fluid (\u00b0F)\n", + "t_2 = 100.0 #Outet temperature of hot fluid (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T1 - t2 #Temperature driving force 1 (\u00b0F)\n", + "DT2 = T2 - t1 #Temperature driving force 1 (\u00b0F)\n", + "DTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0F)\n", + "P = (t2-t1)/(T1 - t1) #Dimensionless ratio P\n", + "R = (T1-T2)/(t2-t1) #Dimensionless ratio R\n", + "#From figure 16.7:\n", + "F = 0.925 #Correction Factor\n", + "DTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The log mean temperature difference for ideal system is :\",round(DTlm1,1),\" \u00b0F .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "from math import log\n", + "5\n", + "#Variable declaration:\n", + "T1 = 400.0 #Temperature of fluid entering the shell (\u00b0F)\n", + "T2 = 250.0 #Temperature of fluid leaving the shell (\u00b0F)\n", + "t1 = 100.0 #Temperature of fluid entering the tube (\u00b0F)\n", + "t2 = 175.0 #Temperature of fluid leaving the tube (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T1 - T2 #Temperature driving force 1 (\u00b0F)\n", + "DT2 = t2 - t1 #Temperature driving force 1 (\u00b0F)\n", + "DTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0F)\n", + "P = (t2-t1)/(T1 - t1) #Dimensionless ratio P\n", + "R = (T1-T2)/(t2-t1) #Dimensionless ratio R\n", + "#From figure 16.8:\n", + "F = 0.985 #Correction factor\n", + "DTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "#Variable declaration:\n", + "#From example 16.5:\n", + "P1 = 0.30 #Dimensionless ratio P\n", + "R1 = 1.67 #Dimensionless ratio R\n", + "#From example 16.6:\n", + "P2 = 0.30 #Dimensionless ratio P\n", + "R2 = 1.67 #Dimensionless ratio R\n", + "\n", + "#Calculation:\n", + "#Applying Equation 16.27:\n", + "F1 = 0.92 #Correction Factor\n", + "#Applying Equation 16.33:\n", + "F2 = 0.985 #Correction Factor\n", + "#From example 16.6:\n", + "LMTD1 = 59.4 #Log mean temperature driving force 1 for ideal countercurrent heat exchanger (\u00b0F)\n", + "LMTD2 = 108.0 #Log mean temperature driving force 2 for ideal countercurrent heat exchanger (\u00b0F)\n", + "DTlm1 = F1*LMTD1 #Log mean temperature driving force 1 for shell and tube exchanger (\u00b0F)\n", + "DTlm2 = F2*LMTD2 #Log mean temperature driving force 2 for shell and tube exchanger (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The log mean temperature difference for real system (in example 16.5) is :\",round(DTlm1,2),\" \u00b0F .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "t2 = 75.0 #Temperature of water leaving the shell (\u00b0C)\n", + "t1 = 35.0 #Temperature of water enteringing the shell (\u00b0C)\n", + "T2 = 75.0 #Temperature of oil leaving the tube (\u00b0C)\n", + "T1 = 110.0 #Temperature of oil entering the tube (\u00b0C)\n", + "m = 1.133 #Mass flowrate of water (kg/s)\n", + "cp = 4180.0 #Heat capacity of water (J/kg.K)\n", + "F = 0.965 #Correction factor\n", + "U = 350.0 #Overall heat transfer coefficient (W/m^2.K)\n", + "\n", + "#Calculation:\n", + "Q = m*cp*(t2-t1) #Heat load (W)\n", + "DT1 = T1-t2 #Temperature driving force 1 (\u00b0C)\n", + "DT2 = T2-t1 #Temperature driving force 2 (\u00b0C)\n", + "DTlm1 = (DT1-DT2)/log(DT1/DT2)+273.0 #Countercurrent log-mean temperature difference (K)\n", + "DTlm2 = F*DTlm1 #Corrected log-mean temperature difference (K)\n", + "A = Q/(U*DTlm2) #Required heat transfer area (m^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "t2 = 84.0 #Temperature of water leaving the tube (\u00b0C)\n", + "t1 = 16.0 #Temperature of water entering the tube (\u00b0C)\n", + "m1 = 10000.0/3600.0 #Mass flowrate of water (kg/s)\n", + "T2 = 94.0 #Temperature of oil leaving the shell (\u00b0C)\n", + "T1 = 160.0 #Temperature of oil entering the shell (\u00b0C)\n", + "\n", + "#Calculation:\n", + "Tw = (t1+t2)/2.0 #Average bulk temperature of water (\u00b0C)\n", + "To = (T1+T2)/2.0 #Average bulk temperature of oil (\u00b0C)\n", + "#From table 16.1:\n", + "p1 = 987.0 #Density of water (kg/m^3)\n", + "cp1 = 4176.0 #Heat capacity of water (J/kg.\u00b0C)\n", + "p2 = 822.0 #Density of oil (kg/m^3)\n", + "Q = m1*cp1*(t2-t1) #Heat load (W)\n", + "cp2 = 4820.0 #Heat capacity of oil (J/kg.\u00b0C)\n", + "m2 = Q/(cp2*(T1-T2)) #Mass flowrate of oil (kg/s)\n", + "DT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\n", + "DT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\n", + "DTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\n", + "P = (t2-t1)/(T1 - t1) #Dimensionless ratio P\n", + "R = (T1-T2)/(t2-t1) #Dimensionless ratio R\n", + "#From figure 16.7:\n", + "F = 0.965 #Correction factor\n", + "DTlm2 = F*DTlm1 #Log mean temperature driving force for 1-4 shell and tube exchanger (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"1. The heat load is :\",round(Q/10**6,3),\" MW .\"\n", + "print \"2. The countercurrent flow log mean temperature difference is :\",round(DTlm1),\" \u00b0C .\"\n", + "print \"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 .\n", + "2. The countercurrent flow log mean temperature difference is : 77.0 \u00b0C .\n", + "3. 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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "#From example 16.10:\n", + "U = 350.0 #Over all heat transfer coefficient (W/m^2.\u00b0C)\n", + "DTlm = 74.3 #Log mean temperature driving force for 1-4 shell and tube exchanger (\u00b0C)\n", + "Q = 788800.0 #Heat load (W)\n", + "Nt = 11.0 #Number of tubes per pass\n", + "Np = 4.0 #Number of passes\n", + "Di = 0.0229 #Inside diameter of tube (m)\n", + "\n", + "#Calculation:\n", + "A = Q/(U*DTlm) #Heat transfer area required for heat exchanger (m^2)\n", + "N = Nt*Np #Total number of tubes\n", + "L = A/(pi*Di*N) #Tube length (m)\n", + "\n", + "#Result:\n", + "print \"The heat transfer area required for the heat exchanger is :\",round(A,2),\" m^2 .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 16.10:\n", + "m1 = 2.778 #Mass flowrate of water (kg/s)\n", + "cp1 = 4176.0 #Heat capacity of water (J/kg.\u00b0C)\n", + "cp2 = 4820.0 #Heat capacity of oil (J/kg.\u00b0C)\n", + "m2 = 2.48 #Mass flowrate of oil (kg/s)\n", + "t2 = 84.0 #Temperature of water leaving the tube (\u00b0C)\n", + "t1 = 16.0 #Temperature of water entering the tube (\u00b0C)\n", + "T2 = 94.0 #Temperature of oil leaving the shell (\u00b0C)\n", + "T1 = 160.0 #Temperature of oil entering the shell (\u00b0C)\n", + "U = 350.0 #Over all heat transfer coefficient (W/m^2.\u00b0C)\n", + "A = 30.33 #Heat transfer area required for heat exchanger (m^2)\n", + "\n", + "#Calculation:\n", + "C1 = m1*cp1 #Capacitance rate of water (W/\u00b0C)\n", + "C2 = m2*cp2 #Capacitance rate of oil (W/\u00b0C)\n", + "Q = C1*(t2-t1) #Heat load of water (W)\n", + "Qmax = C1*(T1-t1) #Maximum heat load of water (W)\n", + "E = Q/Qmax #Effectiveness\n", + "if (C1<C2):\n", + " Cmin = C1 #Minimum capacitance rate (W/\u00b0C)\n", + " Cmax = C2 #Maximum capacitance rate (W/\u00b0C)\n", + "else:\n", + " Cmin = C2 #Minimum capacitance rate (W/\u00b0C)\n", + " Cmax = C1 #Maximum capacitance rate (W/\u00b0C)\n", + "NTU = U*A/Cmin #Number of transfer units\n", + "C = Cmin/Cmax #Capacitance rate ratio\n", + "\n", + "#Result:\n", + "print \"The effectiveness is :\",round(E,3),\".\"\n", + "print \"The number of transfer units is :\",round(NTU,3),\".\"\n", + "print \"The capacitance rate ratio is :\",round(C,3),\".\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The effectiveness is : 0.472 .\n", + "The number of transfer units is : 0.915 .\n", + "The 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": [ + "\n", + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "#From table 16.4:\n", + "Cw = 11680.3 #Capacitance rate of water (W/\u00b0C)\n", + "t2 = 65.0 #Temperature of water leaving the tube (\u00b0C)\n", + "t1 = 20.0 #Temperature of water entering the tube (\u00b0C)\n", + "T2 = 107.3 #Temperature of steam leaving the shell (\u00b0C)\n", + "T1 = 107.3 #Temperature of steam entering the shell (\u00b0C)\n", + "hv = 2.238*10**6 #Latenet heat of condensation for steam (J/kg)\n", + "U = 2000.0 #Overall heat transfer coefficient (W/m^2.\u00b0C)\n", + "\n", + "#Calculation:\n", + "Q = Cw*(t2-t1) #Heat load (W)\n", + "m2 = Q/hv #Steam condensation rate (kg/s)\n", + "DT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\n", + "DT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\n", + "DTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\n", + "F = 1.0 #Correction factor (since, T2 = T1)\n", + "DTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0C)\n", + "A1 = Q/(U*DTlm2) #Heat transfer area using LMTD method (m^2)\n", + "E = (t2-t1)/(T1-t1) #Effectiveness\n", + "#From figure 16.18:\n", + "NTU = 0.7 #Number of transfer units\n", + "A2 = (NTU*Cw)/U #Heat transfer area using E-NTU method (m^2)\n", + "\n", + "#Result:\n", + "print \"The heat transfr area for the exchanger (using LMTD method) is :\",round(A1,2),\" m^2 .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "from math import pi,log\n", + "\n", + "#Variable declaration:\n", + "#From table 16.5:\n", + "t2 = 75.0 #Temperature of water leaving the shell (\u00b0C)\n", + "t1 = 35.0 #Temperature of water entering the shell (\u00b0C\n", + "T2 = 75.0 #Temperature of oil leaving the tube (\u00b0C)\n", + "T1 = 110.0 #Temperature of oil entering the tube (\u00b0C)\n", + "mw = 1.133 #Mass flowrtae of water (kg/s)\n", + "cpw = 4180.0 #Heat capacity of water (J/kg.K)\n", + "cpo = 1900.0 #Heat capacity of oil (J/kg.K)\n", + "p = 850.0 #Density of oil (kg/m^3)\n", + "Di = 0.01905 #Inside diameter of tube (m)\n", + "V = 0.3 #Average velocity of oil flow inside the tube (m/s)\n", + "Np = 2.0 #Number of passes\n", + "Uc = 350.0 #Overall heat transfer coefficient for clean heat exchanger (W/m^2)\n", + "Rf = 0.00027 #Fouling factor (m^2.K/w)\n", + "\n", + "#Calculation:\n", + "Cw = mw*cpw #Water capacitance rate (W/K)\n", + "Q = Cw*(t2-t1) #Heat load (W)\n", + "Co = Q/(T1-T2) #Oil capacitance rate (W/K)\n", + "mo = Co/cpo #Total flowrate of oil (kg/s)\n", + "if (Cw<Co):\n", + " Cmin = Cw #Minimum capacitance rate (W/K)\n", + " Cmax = Co #Maximum capacitance rate (W/K)\n", + "else:\n", + " Cmin = Co #Minimum capacitance rate (W/K)\n", + " Cmax = Cw #Maximum capacitance rate (W/K)\n", + "m_ot = p*V*(pi/4.0)*Di**2 #Oil flowrate per tube (kg/s)\n", + "Nt = mo/m_ot #Number of tubes per pass\n", + "N = Nt*Np #Number of tubes\n", + "DT1 = T2-t1 #Temperature driving force 1 (\u00b0C)\n", + "DT2 = T1-t2 #Temperature driving force 2 (\u00b0C)\n", + "DTlm1 = ((DT1-DT2)/log(DT1/DT2)) #Log mean temperature driving force for ideal countercurrent heat exchanger (\u00b0C)\n", + "P = (t2-t1)/(T1 - t1) #Dimensionless parameter P\n", + "R = (T1-T2)/(t2-t1) #Dimensionless parameter R\n", + "#From figure 16.7:\n", + "F = 0.81 #Correction factor\n", + "DTlm2 = F*DTlm1 #Log mean temperature driving force for shell and tube exchanger (\u00b0C)\n", + "Ud = 1.0/(1.0/Uc+Rf) #Dirty overall heat transfer coefficient (W/m^2.K)\n", + "A = Q/(Ud*DTlm2) #Required heat transfer area (m^2)\n", + "L = A/(N*pi*Di) #Tube length (m)\n", + "\n", + "#Result:\n", + "print \"1. The mass flow rate of the oil is :\",round(mo,2),\" kg/s .\"\n", + "print \"2. The minimum and maximum heat capacity rate is :\",round(Cmin),\" and \",round(Cmax,1),\" W/K .\"\n", + "print \"3. The heat load, Q is :\",round(Q),\" W .\"\n", + "print \"4. The total number of tubes is :\",round(N,-1),\".\"\n", + "print \"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 .\n", + "2. The minimum and maximum heat capacity rate is : 4736.0 and 5412.5 W/K .\n", + "3. The heat load, Q is : 189438.0 W .\n", + "4. The total number of tubes is : 80.0 .\n", + "5. 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": [ + "\n", + "#Variable declaration:\n", + "#From example 16.22:\n", + "t2 = 75.0 #Temperature of water leaving the shell (\u00b0F)\n", + "t1 = 35.0 #Temperature of water entering the shell (\u00b0F)\n", + "T2 = 75.0 #Temperature of oil leaving the tube (\u00b0F)\n", + "T1 = 110.0 #Temperature of oil entering the tube (\u00b0F)\n", + "U = 320.0 #Overall heat transfer coefficient (W/m^2.K)\n", + "A = 19.5 #Required heat transfer area (m^2)\n", + "Cmin = 4736.0 #Minimum capacitance rate (W/K)\n", + "\n", + "#Calculation:\n", + "DT1 = t2-t1 #Actual water temperature change (\u00b0F)\n", + "DT2 = T1 - t1 #Maximum water temperature change (\u00b0F)\n", + "E = DT1/DT2 #Effectiveness\n", + "NTU = (U*A)/Cmin #Number of transfer units\n", + "\n", + "#Result:\n", + "print \"The effectiveness is :\",round(E,3),\".\"\n", + "print \"The NTU is :\",round(NTU,3),\".\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The effectiveness is : 0.533 .\n", + "The 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 100755 index 00000000..4a60ab42 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_17.ipynb @@ -0,0 +1,832 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:475ce8fa81a36fce30309ad0a5a8ca165ef95371bfe5c210d4123243261a9f37" + }, + "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": [ + "\n", + "\n", + "#Variable declaration:\n", + "w1 = 1.5 #Thicknessof fin (in)\n", + "L = 12.0 #Length of fin (in)\n", + "w2 = 0.1 #Thickness of fin(in)\n", + "\n", + "#Calculation:\n", + "Af = 2*w1*L #Face area of fin (in^2)\n", + "At = Af + L*w2 #Total area of fin (in^2)\n", + "\n", + "#Result:\n", + "print \"The face area of the fin is :\",round(Af),\" in^2 .\"\n", + "print \"The face area of the fin is :\",round(Af/12**2,2),\" ft^2 .\"\n", + "print \"The total area of the fin is :\",round(At,1),\" in^2 .\"\n", + "print \"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 .\n", + "The face area of the fin is : 0.25 ft^2 .\n", + "The total area of the fin is : 37.2 in^2 .\n", + "The 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": [ + "\n", + "\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "rf = 6.0/12.0 #Outside radius of fin (ft)\n", + "ro = 4.0/12.0 #Outside radius of pipe (ft)\n", + "t = 0.1/12.0 #Thickness of fin (ft)\n", + "\n", + "#Calculation:\n", + "Af = 2*pi*(rf**2-ro**2) #Face area of fin (ft^2)\n", + "At = Af + 2*pi*rf*t #Total area of fin (ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from math import sqrt\n", + "\n", + "#Variable declaration:\n", + "L = 3.0*0.0254 #Height of fin (m)\n", + "t = 1.0*0.0254 #Thickness of fin (m)\n", + "h = 15.0 #Heat transfer coefficient (W/m^2.K)\n", + "k = 300.0 #Thermal conductivity (W/m.K)\n", + "\n", + "#Calculation:\n", + "Lc = L + t/2.0 #Corrected height of fin (m)\n", + "Ap = Lc*t #Profile area of fin (m^2)\n", + "x = sqrt((Lc**3*h)/(k*Ap)) #x-coordinate of figure 17.3\n", + "#From figure 17.3:\n", + "nf = 98.0 #Fin efficiency\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 17.4:\n", + "X = 0.1246 #X-coordinate of figure 17.3\n", + "\n", + "#Calculation:\n", + "#Applying equation (A) from Table 17.3:\n", + "Y = 4.5128*X**3 - 10.079*X**2 - 31.413*X + 101.47\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from math import sqrt,atan\n", + "\n", + "#Variable declaration:\n", + "w = 0.2/100.0 #Width of fin (m)\n", + "t = 0.2/100.0 #Thickness of fin (m)\n", + "L = 1.0/100.0 #Length of fin (m)\n", + "h = 16.0 #Heat transfer coefficient (W/m^2.K)\n", + "k = 400.0 #Thermal conductivity of fin (W/m.K)\n", + "Tc = 100.0 #Circuit temperature (\u00b0C)\n", + "Ta = 25.0 #Air temperature (\u00b0C)\n", + "\n", + "#Calculation:\n", + "P = 4*w #Fin cross-section parameter (m)\n", + "Ac = w*t #Cross-sectional area of fin (m^2)\n", + "Lc = L+Ac/P #Corrected height of fin (m)\n", + "m = sqrt((h*P)/(k*Ac)) #Location of minimum temperature (m^-1)\n", + "Q = (sqrt(h*P*k*Ac))*(Tc-Ta)*atan(h)*(m*Lc) #Heat transfer from each micro-fin (W)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "from math import sqrt\n", + "#Variable declaration:\n", + "h1 = 13.0 #Air-side heat transfer coefficient (W/m^2.K)\n", + "A = 1.0 #Base wall area (m^2)\n", + "L = 2.5/100 #Length of steel fins (m)\n", + "L2 = 1.5/10**3 #Length of steel wall (m)\n", + "k = 13.0 #Thermal conductivity of fin (W/m.K)\n", + "k1 = 38.0 #Thermal conductivity of steel wall (W/m.K)\n", + "h2 = 260.0 #Water side heat transfer coefficient (W/m^2.K)\n", + "T4 = 19.0 #Air temperature (\u00b0C)\n", + "T1 = 83.0 #Water temperature (\u00b0C)\n", + "t = 1.3/10**3 #Thickness of steel fins (m)\n", + "w = 1.0 #Width of wall (m)\n", + "S = 1.3/100 #Fin pitch(m)\n", + "\n", + "#Calculation:\n", + "R1 = 1/(h1*A) #Air resistance (\u00b0C/W) (part 1)\n", + "R2 = L2/(k1*A) #Conduction resistance (\u00b0C/W)\n", + "R3 = 1/(h2*A) #Water resistance (\u00b0C/W)\n", + "Rt = (R1+R3) #Total resistance (\u00b0C/W) (part 2)\n", + "Q = (T1-T4)/Rt #Total heat transfer (W)\n", + "Nf = 1/S #Number of fins (part 3)\n", + "Lbe = w - Nf*t #Unfinned exposed base surface\n", + "Abe = w*Lbe #Exposed base surface area (m^2)\n", + "Lc = L+t/2 #Corrected length (m)\n", + "Ap = Lc*t #Profile area (m^2)\n", + "Af = 2*w*Lc #Fin surface area (m^2)\n", + "Bi = h1*(t/2)/k1 #Biot number\n", + "a = sqrt(Lc**3*h1/(k*Ap)) #Abscissa of the fin efficiency\n", + "#From figure 17.3:\n", + "nf = 0.88 #Fin efficiency\n", + "Rb = 1/(h1*Abe) #Air thermal resistance of base wall (\u00b0C/W)\n", + "Rf = 1/(h1*Nf*Af*nf) #Air thermal resistance of fins (\u00b0C/W)\n", + "RT1 = 1/(1/Rb+1/Rf) #Total outside resistance of the fin array (\u00b0C/W)\n", + "Rt3 = RT1+R3 #Total resistance on air side fins (\u00b0C/W)\n", + "Qt = (T1-T4)/round(Rt3,5) #Heat transfer rate on air side fins (W)\n", + "I = (Qt/Q - 1)*100 #Percent increase in heat transfer rate to air side fins (W)\n", + "A = sqrt(Lc**3*h2/(k1*Ap)) #Abscissa of the new fin efficiency (part 4)\n", + "#From figure 17.3:\n", + "nf2 = 38.0 #New fin efficiency\n", + "Rb2 = 1/(h2*Abe) #Thermal resistance of base wall (\u00b0C/W)\n", + "Rf2 = 1/(h2*Nf*Af*nf2) #Thermal resistance of fins (\u00b0C/W)\n", + "Rt4 = 1/(1/Rb2+1/Rf2) #Total resistance of the finned surface (\u00b0C/W)\n", + "Rt5 = R1+Rt4 #Total resistance on water side fins (\u00b0C/W)\n", + "QT1 = (T1-T4)/Rt5 #Heat transfer rate on water side fins (W)\n", + "I2 = (QT1/Q - 1)*100 #Percent increase in heat transfer rate to water side fins (W)\n", + "\n", + "#Result:\n", + "if (R2<R1 or R2<R3):\n", + " print \"1. The conduction resistance may be neglected.\"\n", + "else:\n", + " print \"1. The conduction resistance can not be neglected.\"\n", + "print \"2. The rate of heat transfer from water to air is :\",round(Q,1),\" W .\"\n", + "print \"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),\" % .\"\n", + "print \"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),\" % .\"\n", + "print \"____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.\n", + "2. The rate of heat transfer from water to air is : 792.4 W .\n", + "3. The percent increase in steady-state heat transfer rate by adding fins to the air side of the plane wall is : 276.7 % .\n", + "4. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi,sqrt\n", + "\n", + "#Variable declaration:\n", + "Do = 2.5/100 #Outside diameter of tube (m)\n", + "t = 1/10**3 #Thickness of fin (m)\n", + "T = 25 #Fluid temperature (\u00b0C)\n", + "Tb = 170 #Surface temperature (\u00b0C)\n", + "h = 130 #Heat transfer coefficient (W/m^2.K)\n", + "k = 200 #Thermal conductivity of fin (W/m.K)\n", + "rf = 2.75/100 #Outside radius of fin (m)\n", + "\n", + "#Calculation:\n", + "ro = Do/2 #Radius of tube (m)\n", + "Ab = 2*pi*ro*t #Area of the base of the fin (m^2)\n", + "Te = Tb-T #Excess temperature at the base of the fin (K)\n", + "Q1 = h*Ab*Te #Total heat transfer rate without the fin (W)\n", + "Bi = h*(t/2)/k #Biot number\n", + "L = rf-ro #Fin height (m)\n", + "rc = rf+t/2 #Corrected radius (m)\n", + "Lc = L+t/2 #Corrected height (m)\n", + "Ap = Lc*t #Profile area (m^2)\n", + "Af = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)\n", + "Qm = h*Af*Te #Maximum fin heat transfer rate (W)\n", + "A = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency\n", + "C = rf/ro #Curve parameter of fin efficiency\n", + "#From figure 17.4:\n", + "nf = 0.86 #Fin efficiency\n", + "Qf = nf*Qm #Fin heat transfer rate (W)\n", + "R = Te/Qf #Fin resistance (K/W)\n", + "\n", + "#Result:\n", + "print \"1. The heat transfer rate without the fin is :\",round(Q1,2),\" W .\"\n", + "print \"Or, the heat transfer rate without the fin is :\",round(Q1*3.412),\" Btu/h .\"\n", + "print \"2. The corrected length is :\",round(Lc,4),\" m .\"\n", + "print \"3. The outer radius is :\",round(rc,3),\" m .\"\n", + "print \"4. The maximum heat transfer rate from the fin is :\",round(Qm,2),\" W .\" \n", + "print \"5. The fin efficiency is :\",round(nf*100),\" % .\" \n", + "print \"6. The fin heat transfer rate is :\",round(Qf),\" W .\"\n", + "print \"Or, the fin heat transfer rate is :\",round(Qf*3.412),\" Btu/h .\"\n", + "print \"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 .\n", + "Or, the heat transfer rate without the fin is : 5.0 Btu/h .\n", + "2. The corrected length is : 0.0155 m .\n", + "3. The outer radius is : 0.028 m .\n", + "4. The maximum heat transfer rate from the fin is : 74.35 W .\n", + "5. The fin efficiency is : 86.0 % .\n", + "6. The fin heat transfer rate is : 64.0 W .\n", + "Or, the fin heat transfer rate is : 218.0 Btu/h .\n", + "7. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 17.10:\n", + "Qf = 64 #Fin heat transfer rate (W)\n", + "Q1 = 1.48 #Total heat transfer rate without the fin (W)\n", + "\n", + "#Calculation:\n", + "E = Qf/Q1 #Fin effectiveness\n", + "\n", + "#Result:\n", + "print \"The fin effectiveness is :\",round(E,1),\" .\"\n", + "if 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 .\n", + "Hence, 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "w = 1 #Length of tube (m)\n", + "S = 10/10**3 #Fin patch (m)\n", + "#From example 17.10:\n", + "t = 1/10**3 #Thickness of fin (m)\n", + "ro = 0.0125 #Radius of tube (m)\n", + "Af = 3.94*10**-3 #Fin surface area (m^2)\n", + "Tb = 145 #Excess temperature at the base of the fin (K)\n", + "h = 130 #Heat transfer coefficient (W/m^2.K)\n", + "Qf = 64 #Fin heat transfer rate (W)\n", + "\n", + "#Calculation:\n", + "Nf = w/S #Number of fins in tube length\n", + "wb = w-Nf*t #Unfinned base length (m)\n", + "Ab = 2*pi*ro*wb #Unfinned base area (m^2)\n", + "At =Ab+Nf*Af #Total transfer surface area (m^2)\n", + "Qt = h*(2*pi*ro*w*Tb) #Total heat rate without fins (W)\n", + "Qb = h*Ab*Tb #Heat flow rate from the exposed tube base (W)\n", + "Qft = Nf*Qf #Heat flow rate from all the fins (W)\n", + "Qt2 = Qb+Qft #Total heat flow rate (W)\n", + "Qm = h*At*Tb #Maximum heat transfer rate (W)\n", + "no = Qt2/Qm #Overall fin efficiency\n", + "Eo = Qt2/Qt #Overall effectiveness\n", + "Rb = 1/(h*Ab) #Thermal resistance of base (K/W)\n", + "Rf = 1/(h*Nf*Af*no) #Thermal resistance of fins (K/W)\n", + "\n", + "#Result:\n", + "print \"1. The total surface area for heat transfer is :\",round(At,3),\" m^2 .\"\n", + "print \"2. The exposed tube base total heat transfer rate is :\",round(Qb,1),\" W .\"\n", + "print \"Or, the exposed tube base total heat transfer rate is :\",round(Qb*3.412),\" Btu/h .\"\n", + "print \"3. The overall efficiency of the surface is :\",round(no*100,1),\" % .\"\n", + "print \"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 .\n", + "2. The exposed tube base total heat transfer rate is : 1332.4 W .\n", + "Or, the exposed tube base total heat transfer rate is : 4546.0 Btu/h .\n", + "3. The overall efficiency of the surface is : 88.3 % .\n", + "4. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import sqrt\n", + "\n", + "#Variable declaration:\n", + "w = 1 #Width of single of fin (m)\n", + "t = 2/10**3 #Fin base thickness (m)\n", + "l = 6/10**3 #Fin length thickness (m)\n", + "T1 = 250 #Surface temperature (\u00b0C)\n", + "T2 = 20 #Ambient air temperature (\u00b0C)\n", + "h = 40 #Surface convection coefficient (W/m^2.K)\n", + "k = 240 #Thermal conductivity of fin (W/m.K)\n", + "\n", + "#Calculation:\n", + "Ab = t*w #Base area of the fin (m^2)\n", + "Te = T1-T2 #Excess temperature at the base of the fin (K)\n", + "Qw = h*Ab*Te #Heat transfer rate without a fin (W)\n", + "Af = 2*w*(sqrt(l**2-(t/2)**2)) #Fin surface area (m^2)\n", + "Qm = h*Af*Te #Maximum heat transfer rate (m^2)\n", + "Bi = h*(t/2)/k #Biot number\n", + "Lc = l #Corrected length (m)\n", + "Ap = l*t/2 #Profile area (m^2)\n", + "A = sqrt((Lc**3*h)/k*Ap) #Abscissa for the fin efficiency figure\n", + "#From figure 17.4:\n", + "nf = 0.99 #Fin efficiency\n", + "Qf = nf*Qm #Fin heat transfer rate (W)\n", + "R = Te/Qf #Fin thermal resistance (K/W)\n", + "E = Qf/Qw #Fin effectiveness\n", + "\n", + "#Result:\n", + "print \"1. The heat transfer rate without the fin is :\",round(Qw,1),\" W .\"\n", + "print \"2. The maximum heat transfer rate from the fin is :\",round(Qm,-1),\" W .\"\n", + "print \"3. The fin efficiency is :\",round(nf*100),\" % .\"\n", + "print \" The fin thermal resistance is :\",round(R,1),\" \u00b0C/W .\"\n", + "print \" 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 .\n", + "2. The maximum heat transfer rate from the fin is : 110.0 W .\n", + "3. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 17.13:\n", + "Qf = 108.9 #Fin heat transfer rate (W)\n", + "Qw = 18.4 #Total heat transfer rate without the fin (W)\n", + "\n", + "#Calculation:\n", + "E = Qf/Qw #Fin effectiveness\n", + "\n", + "#Result:\n", + "print \"The fin effectiveness is :\",round(E,2),\" .\"\n", + "if 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 .\n", + "Hence, 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi,sqrt\n", + "\n", + "#Variable declaration:\n", + "Do = 50/10**3 #Outside diameter of tube (m)\n", + "t = 4/10**3 #Thickness of fin (m)\n", + "T = 20 #Fluid temperature (\u00b0C)\n", + "Tb = 200 #Surface temperature (\u00b0C)\n", + "h = 40 #Heat transfer coefficient (W/m^2.K)\n", + "k = 240 #Thermal conductivity of fin (W/m.K)\n", + "l = 15/10**3 #Length of fin (m)\n", + "\n", + "#Calculation:\n", + "ro = Do/2 #Radius of tube (m)\n", + "rf = ro+l #Outside radius of fin (m)\n", + "Ab = 2*pi*ro*t #Area of the base of the fin (m^2)\n", + "Te = Tb-T #Excess temperature at the base of the fin (K)\n", + "Q1 = h*Ab*Te #Total heat transfer rate without the fin (W)\n", + "Bi = h*(t/2)/k #Biot number\n", + "L = rf-ro #Fin height (m)\n", + "rc = rf+t/2 #Corrected radius (m)\n", + "Lc = L+t/2 #Corrected height (m)\n", + "Ap = Lc*t #Profile area (m^2)\n", + "Af = 2*pi*(rc**2-ro**2) #Fin surface area (m^2)\n", + "Qm = h*Af*Te #Maximum fin heat transfer rate (W)\n", + "A = sqrt(Lc**3*h/(k*Ap)) #Abscissa of fin efficiency\n", + "C = rf/ro #Curve parameter of fin efficiency\n", + "#From figure 17.4:\n", + "nf = 0.97 #Fin efficiency\n", + "Qf = nf*Qm #Fin heat transfer rate (W)\n", + "R = Te/Qf #Fin resistance (K/W)\n", + "E = Qf/Q1 #Fin effectiveness\n", + "\n", + "#Result:\n", + "print \"The fin efficiency is :\",round(nf*100),\" % .\"\n", + "print \"The fin thermal resistance is :\",round(R,1),\" \u00b0C/W .\"\n", + "print \"The fin effectiveness is :\",round(E,2),\" .\"\n", + "print \"The maximum heat transfer rate from a single fin is :\",round(Qm,2),\" W .\"\n", + "if 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 % .\n", + "The fin thermal resistance is : 3.6 \u00b0C/W .\n", + "The fin effectiveness is : 11.05 .\n", + "The maximum heat transfer rate from a single fin is : 51.53 W .\n", + "Since 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi,sqrt\n", + "\n", + "#Variable declaration:\n", + "Nf = 125 #Array of fins per meter\n", + "w = 1 #Length of fin (m)\n", + "#From example 17.15:\n", + "t = 4/10**3 #Thickness of fin (m)\n", + "Do = 50/10**3 #Outside diameter of tube (m)\n", + "Af = 7.157*10**-3 #Fin surface area (m^2)\n", + "h = 40 #Heat transfer coefficient (W/m^2.K)\n", + "DTb = 180 #Excess temperature at the base of the fin (K)\n", + "Qf = 50 #Fin heat transfer rate (W)\n", + "\n", + "#Calculation:\n", + "ro = Do/2 #Radius of tube (m)\n", + "wb = w-Nf*t #Unfinned exposed base length (m)\n", + "Ab = 2*pi*ro*wb #Area of the base of the fin (m^2)\n", + "At = Ab+Nf*Af #Total heat transfer surface area (m^2)\n", + "Qw = h*(2*pi*ro*w)*DTb #Heat rate without fin (W)\n", + "Qb = h*Ab*DTb #Heat rate from the base (W)\n", + "Qft = Nf*Qf #Heat rate from the fin (W)\n", + "Qt = Qb+Qft #Total heat rate (W)\n", + "Qm = h*At*DTb #Maximum heat transfer rate (W)\n", + "n = Qt/Qm #Overall fin efficiency\n", + "E = Qt/Qw #Overall fin effectiveness\n", + "Rb = 1/(h*Ab) #Thermal resistance of base (\u00b0C/W)\n", + "Rf = 1/(h*Nf*Af*n) #Thermal resistance of fin (\u00b0C/W)\n", + "\n", + "#Result:\n", + "print \"The rate of heat transfer per unit length of tube is :\",round(Qt,1),\" W .\"\n", + "print \"Or, the rate of heat transfer per unit length of tube is :\",round(Qt/10**3,2),\" kW .\"\n", + "print \"The overall fin efficiency is :\",round(n*100,1),\" % .\"\n", + "print \"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 .\n", + "Or, the rate of heat transfer per unit length of tube is : 6.82 kW .\n", + "The overall fin efficiency is : 97.3 % .\n", + "The 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": [ + "\n", + "#Variable declaration:\n", + "print '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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "#From example 17.18:\n", + "T = 250 #Base temperature of fin (\u00b0F)\n", + "h = 15 #Convection coefficient of heat transfer (Btu/h.ft.\u00b0F)\n", + "w = 1 #Base width of fin (ft)\n", + "t = 1 #Thickness of fin (in)\n", + "H = 1/8 #Height of fin (in)\n", + "l = 1 #Length of fin (in)\n", + "Q = 357.2 #Heat transfer rate (Btu/h.ft)\n", + "\n", + "#Calculation:\n", + "A = (l*w+t*w+H*w)/12 #Heat transfer area of fin (ft^2)\n", + "Qm = h*A*(T-70) #Maximum heat transfer rate (Btu/h.ft)\n", + "n = Q/Qm*100 #Fin efficiency\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..61e88510 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_18.ipynb @@ -0,0 +1,658 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4239ad8a8aef7d8398d3c69721341c39df37be6f21def1df2b04a8d6ccbef700" + }, + "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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "T1 = 25 #Temperature of H2SO4 (\u00b0C)\n", + "m = 50+200 #Mass of H2SO4 (lb)\n", + "#From figure 18.2:\n", + "W1 = 50+100 #Weight of H2SO4 (lb)\n", + "W2 = 100 #Weight of H2O (lb)\n", + "\n", + "#Calculation:\n", + "m = W1/(W1+W2)*100 #Percent weight of H2SO4 (%)\n", + "m2 = W1+W2 #Mass of mixture (lb)\n", + "#From fgure 18.2:\n", + "T2 = 140 #Final temperature between the 50% solution and pure H2SO4 at 25\u00b0C (\u00b0F)\n", + "h1 = -86 #Specific heat capacity of H2O (Btu/lb)\n", + "h2 = -121.5 #Specific heat capacity of H2SO4 (Btu/lb)\n", + "Q = m2*(h2-h1) #Heat transferred (Btu)\n", + "\n", + "#Result:\n", + "print \"The final temperature between the 50% solution and pure H2SO4 at 25\u00b0C is :\",round(T2),\" \u00b0F .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "F = 10000 #Mass flow rate of NaOH (lb/h)\n", + "C1 = 10 #Old concentration of NaOH solution (%)\n", + "C2 = 75 #New concentration of NaOH solution (%)\n", + "h1 = 1150 #Enthalpy of saturated steam at 14.7 psia (Btu/lb)\n", + "U = 500 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "T1 = 212 #Absolute temperature of evaporator (\u00b0F)\n", + "T2 = 340 #Saturated steam temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "L = F*(C1/100)/(C2/100) #Flow rate of steam leaving the evaporator (lb/h)\n", + "V = F-L #Overall material balance (lb/h)\n", + "#From figure 18.3:\n", + "hF = 81 #Enthalpy of solution entering the unit (Btu/lb)\n", + "hL = 395 #Enthalpy of the 75% NaOH solution (Btu/lb)\n", + "Q = round(V)*h1+round(L)*hL-F*hF #Evaporator heat required (Btu/h)\n", + "A = Q/(U*(T2-T1)) #Area of the evaporaor (ft^2)\n", + "\n", + "#Result:\n", + "print \"The heat transfer rate required for the evaporator is :\",round(Q,-2),\" Btu/h .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "U1 = 240 #Overall heat transfer coefficient for first effect (Btu/h.ft^2.\u00b0F)\n", + "U2 = 200 #Overall heat transfer coefficient for second effect (Btu/h.ft^2.\u00b0F)\n", + "U3 = 125 #Overall heat transfer coefficient for third effect (Btu/h.ft^2.\u00b0F)\n", + "A1 = 125 #Heating surface area in first effect (ft^3)\n", + "A2 = 150 #Heating surface area in second effect (ft^3)\n", + "A3 = 160 #Heating surface area in third effect (ft^3)\n", + "T1 = 400 #Condensation stream temperature in the first effect (\u00b0F)\n", + "T2 = 120 #Vapor leaving temperature in the first effect (\u00b0F)\n", + "\n", + "#Calculation:\n", + "R1 = 1/(U1*A1) #Resistance across first effect\n", + "R2 = 1/(U2*A2) #Resistance across second effect\n", + "R3 = 1/(U3*A3) #Resistance across third effect\n", + "R = R1+R2+R3 #Total resistance\n", + "DT1 = (R1/R)*(T1-T2) #Temperature drop across the heating surface in the first effect (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "F = 5000 #Mass of soltuion fed in the evaporator (lb)\n", + "xF = 2/100 #Concentration of feed\n", + "xL = 5/100 #Concentration of liquor\n", + "U = 280 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "#From figure 18.1 & 18.3:\n", + "TF = 100 #Feed temperature (\u00b0F)\n", + "TS = 227 #Steam temperature (\u00b0F)\n", + "TV = 212 #Vapour temperature (\u00b0F)\n", + "TL = 212 #Liquor temperature (\u00b0F)\n", + "TC = 227 #Condensate temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "#From steam tables:\n", + "hF = 68 #Enthalpy of feed (Btu/lb)\n", + "hL = 180 #Enthalpy of liquor (Btu/lb)\n", + "hV = 1150 #Enthalpy of vapour (Btu/lb)\n", + "hS = 1156 #Enthalpy of steam (Btu/lb)\n", + "hC = 195 #Enthalpy of condensate (Btu/lb)\n", + "s1 = F*xF #Total solids in feed (lb)\n", + "w = F-s1 #Total water in feed (lb)\n", + "s2 = F*xF #Total solids in liquor (lb)\n", + "L = s2/xL #Total water in liquor (lb)\n", + "V = F-L #Overall balance (lb)\n", + "S = (V*hV+L*hL-F*hF)/(hS-hC) #Mass of steam (lb)\n", + "Q = S*(hS-hC) #Total heat requirement (Btu)\n", + "A = Q/(U*(TS-TL)) #Required surface aea (ft^2)\n", + "\n", + "#Result:\n", + "print \"The mass of vapor produced is :\",round(V),\" lb .\"\n", + "print \"The total mass of steam required is :\",round(S),\" lb .\"\n", + "print \"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 .\n", + "The total mass of steam required is : 3611.0 lb .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "F = 5000 #Mass flow rate of NaOH (lb/h)\n", + "xF = 20/100 #Old concentration of NaOH solution\n", + "TF = 100 #Feed temperature (\u00b0F)\n", + "xL = 40/100 #New concentration of NaOH solution\n", + "xv = 0 #Vapour concentration at x\n", + "yv = 0 #Vapour concentration at y\n", + "T1 = 198 #Boiling temperature of solution in the evaporator (\u00b0F)\n", + "T2 = 125 #Saturated steam temperature (\u00b0F)\n", + "U = 400 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "Ts = 228 #Steam temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "#From steam tables at 228\u00b0F and 5 psig:\n", + "hS = 1156 #Enthalpy of steam (Btu/lb)\n", + "hC = 196 #Enthalpy of condensate (Btu/lb)\n", + "hV = hS-hC #Enthalpy of vapour (Btu/lb)\n", + "Tw = 125.4 #Boiling point of water at 4 in Hg absolute (\u00b0F)\n", + "hS2 = 1116 #Enthalpy of saturated steam at 125\u00b0F (Btu/lb)\n", + "hs = 0.46 #Heat capacity of superheated steam (Btu/lb.\u00b0F)\n", + "#From figure 18.3:\n", + "hF = 55 #Enthalpy of feed (Btu/lb)\n", + "hL = 177 #Enthalpy of liquor (Btu/lb)\n", + "L = F*xF/xL #Mass of liquor (lb)\n", + "V = L #Mass of vapour (lb)\n", + "hV = hS2+hs*(T1-T2) #Enthalpy of vapour leaving the solution (Btu/lb)\n", + "S = (V*hV+L*hL-F*hF)/(hS-hC) #Mass flow rate of steam (lb/h)\n", + "Q = S*(hS-hC) #Total heat requirement (Btu)\n", + "A = Q/(U*(Ts-T1)) #Required heat transfer area (ft^2)\n", + "\n", + "#Result:\n", + "print \"The steam flow rate is :\",round(S,-1),\" lb/h .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "T1 = 2000 #Hot gas temperature (\u00b0F)\n", + "T2 = 550 #Cool gas temperature (\u00b0F)\n", + "T3 = 330 #Steam temperature (\u00b0F)\n", + "T4 = 140 #Water temperature (\u00b0F)\n", + "m = 30000 #Mass flow rate of steam (lb/h)\n", + "cp = 0.279 #Average heat capacity of gas (Btu/lb.\u00b0F)\n", + "N = 800 #Number of boiler tubes\n", + "\n", + "#Calculation:\n", + "DT = (T1-T3)/(T2-T3) #Temperature difference ratio\n", + "Tav = (T1+T2)/2 #Average gas temperature (\u00b0F)\n", + "#From steam tables (Appendix):\n", + "hs = 1187.7 #Steam enthalpy (Btu/lb)\n", + "hw = 107.89 #Water enthalpy (Btu/lb)\n", + "Q = m*(hs-hw) #Heat duty (Btu/h)\n", + "mh = Q/cp*(T1-T2) #Mass flow rate of gas (lb/h)\n", + "x = mh/N #Gas mass flow rate per tube (lb/h)\n", + "#From figure 18.5:\n", + "L = 15 #Length of boiler tubes (ft)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "T1 = 1800 #Hot gas temperature (\u00b0F)\n", + "T2 = 500 #Cool gas temperature (\u00b0F)\n", + "#From steam tables:\n", + "Tw = 312 #Boiling point of water at 80 psia (\u00b0F)\n", + "m1 = 120000 #Mass flow rate of flue gas (lb/h)\n", + "D = 2/12 #Inside diameter of tube (ft)\n", + "cp = 0.26 #Average heat capacity of flue gas (Btu/lb.\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT = (T1-Tw)/(T2-Tw) #Temperature difference ratio\n", + "Tav = (T1+T2)/2 #Average gas temperature (\u00b0F)\n", + "#From figure 18.4:\n", + "x = 150 #Gas mass flow rate per tube (m/N) (lb/h)\n", + "N = m1/x #Number of tubes\n", + "L = 21.5 #Length of tubes (ft)\n", + "A = N*L*D #Total heat transfer area (ft^2)\n", + "Q = m1*cp*(T1-T2) #Heat duty (Btu/h)\n", + "#From steam tables (Appendix):\n", + "hs = 1183.1 #Steam enthalpy at 80 psia (Btu/lb)\n", + "hw = 168.1 #Water enthalpy at 200\u00b0F (Btu/lb)\n", + "m2 = Q/(hs-hw) #Mass flow rate of water (lb/h)\n", + "\n", + "#Result:\n", + "print \"The required heat transfer area is :\",round(A),\" ft^2 .\"\n", + "print \"The tube length is :\",L,\" ft .\"\n", + "print \"The heat duty is :\",round(Q/10**7,2),\" x 10^7 .\"\n", + "print \"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 .\n", + "The tube length is : 21.5 ft .\n", + "The heat duty is : 4.06 x 10^7 .\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "m1 = 144206 #Mass flow rate of flue gas (lb/h)\n", + "cp = 0.3 #Average flue gas heat capacity (Btu/lb.\u00b0F)\n", + "T1 = 2050 #Initial temperature of gas (\u00b0F)\n", + "T2 = 560 #Final temperature of gas (\u00b0F)\n", + "T3 = 70 #Ambient air temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Q = m1*cp*(T1-T2) #Duty rate (Btu/h)\n", + "#From appendix:\n", + "cpa = 0.243 #Average ambient air heat capacity 70\u00b0F (Btu/lb.\u00b0F)\n", + "MW = 29 #Molecular weight of air at 70\u00b0F\n", + "ma = round(Q,-5)/(cpa*(T2-T3)) #Mass of air required (lb/h)\n", + "m2 = round(ma)/MW #Moles of air required (lb mol/h)\n", + "m3 = round(ma)*13.32 #Volume of air required (ft^3/h)\n", + "\n", + "#Result:\n", + "print \"The mass of air required is :\",round(ma,-2),\" lb/h .\"\n", + "print \"The moles of air required is :\",round(m2,-1),\"lb mol/h .\"\n", + "print \"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 .\n", + "The moles of air required is : 18680.0 lb mol/h .\n", + "The 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": [ + "\n", + "#Variable declaration:\n", + "#From example 18.19:\n", + "m1 = 144200 #Mass flow rate of flue gas (lb/h)\n", + "m2 = 541700 #Mass flow rate of air (lb/h)\n", + "R = 0.73 #Universal gas constant (psia.ft^3/lbmol.\u00b0R)\n", + "P = 1 #Absolute pressure (psia)\n", + "T = 1020 #Absolute temperature (\u00b0R)\n", + "MW = 29 #Molecular weight of air\n", + "t = 1.5 #Residence time (s)\n", + "\n", + "#Calculation:\n", + "m = m1+m2 #Total mass flow rate of the gas (lb/h)\n", + "q = m*R*T/(P*MW) #Volumetric flow at 560\u00b0F (ft^3/h)\n", + "V = q*t/3600 #Volume of tank (ft^3)\n", + "\n", + "#Result:\n", + "print \"The total mass flow rate of the gas is :\",round(m,-2),\" lb/h .\"\n", + "print \"The volumetric flow at 560\u00b0F is :\",round(q/10**7,2),\" x 10^7 ft^3/h .\"\n", + "print \"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 .\n", + "The volumetric flow at 560\u00b0F is : 1.76 x 10^7 ft^3/h .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "#Fro example 18.20:\n", + "V = 7335 #Volume of tank (ft^3)\n", + "\n", + "#Calculation:\n", + "D = (4*V/pi)**(1/3) #Diameter of tank (ft)\n", + "H = D #Height of tube (ft)\n", + "\n", + "#Result:\n", + "print \"The diameter of tank is :\",round(H,2),\" ft .\"\n", + "print \"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 .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "m1 = 144206 #Mass flow rate of flue gas (lb/h)\n", + "cp1 = 0.3 #Average heat capacities of the flue gas (Btu/lb\u00b0F)\n", + "cp2 = 0.88 #Average heat capacities of the solid (Btu/lb\u00b0F)\n", + "#From example 18.18:\n", + "T1 = 550 #Initial temperature of gas (\u00b0F)\n", + "T2 = 2050 #Final temperature of gas (\u00b0F)\n", + "T3 = 70 #Initial temperature of solid (\u00b0F)\n", + "T4 = 550-40 #Final temperature of solid (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Dhf = m1*cp1*(T2-T1) #For the flue gas, the enthalpy change for one hour of operation (Btu)\n", + "Dhs = round(Dhf,-4) #For the solids, the enthalpy change for one hour of operation (Btu)\n", + "m2 = Dhs/(cp2*(T4-T3)) #Mass of solid (lb)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log,sqrt,pi\n", + "\n", + "#Variable declaration:\n", + "#From example 18.21:\n", + "m = 144206 #Mass flow rate of flue gas (lb/h)\n", + "cp = 0.3 #Average heat capacities of the flue gas (Btu/lb\u00b0F)\n", + "T1 = 2050 #Initial temperature of gas (\u00b0F)\n", + "T2 = 180 #Final temperature of gas (\u00b0F)\n", + "T3 = 60 #Ambient air temperature (\u00b0F)\n", + "U = 1.5 #Overall heat transfer coefficient for cooler (Btu/h.ft^2.\u00b0F)\n", + "MW = 28.27 #Molecular weight of gas\n", + "R = 379 #Universal gas constant (psia.ft^3/lbmol.\u00b0R)\n", + "v = 60 #Duct or pipe velcity at inlet (2050\u00b0F) (ft/s)\n", + "\n", + "#Calculation:\n", + "Q = m*cp*(T1-T2) #Heat duty (Btu/h)\n", + "DTlm = ((T1-T3)-(T2-T3))/log((T1-T3)/(T2-T3)) #Log-mean temperature difference (\u00b0F)\n", + "A1 = round(Q,-5)/(U*round(DTlm)) #Radiative surface area (ft^2)\n", + "q = m*R*(T1+460)/(T3+460)/MW #Volumetric flow at inlet (ft^3/h)\n", + "A2 = q/(v*3600) #Duct area (ft^2)\n", + "D = sqrt(A2*4/pi) #Duct diameter (ft)\n", + "L = A1/(pi*D) #Length of required heat exchange ducting (ft)\n", + "\n", + "#Result:\n", + "print \" The radiative surface area required is :\",round(A1,-1),\" ft^2 .\"\n", + "print \" 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 100755 index 00000000..3617f433 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_19.ipynb @@ -0,0 +1,861 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e60aa082c3bd1f8f9672b5ca74c6357f1b26803a7fb296a1d0a644ab9efb4ea1" + }, + "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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "H = 2.5 #Height of wall (m)\n", + "W = 4 #Width of wall (m)\n", + "h = 11 #Convective heat transfer coefficient (W/m^2.K)\n", + "T1 = 24 #Outside surface temperature (\u00b0C)\n", + "T3 = -15 #Outside air temperature (\u00b0C)\n", + "L = 7.62/10**3 #Insulation thickness (m)\n", + "k = 0.04 #Thermal conductivity of wool (W/m.K)\n", + "\n", + "#Calculation:\n", + "A = H*W #Heat transfer area (m^2)\n", + "Q = h*A*(T1-T3) #Heat transfer rate (W)\n", + "Ri = L/(k*A) #Insuation resistance (K/W)\n", + "Rc = 1/(h*A) #Convective resitance (K/W)\n", + "R = Ri+Rc #Total resistance (K/W)\n", + "Qt = (T1-T3)/R #Revised heat transfer rate (Btu/h)\n", + " \n", + "#Result:\n", + "print \"1. The heat transfer rate without insulation is :\",round(Q),\" W .\"\n", + "print \"Or, the heat transfer rate without insulation is :\",round(Q*3.412),\" Btu/h .\"\n", + "print \"2. The revised heat transfer rate with insulation is :\",round(Qt),\" W .\"\n", + "print \"Or, the revised heat transfer rate with insulation is :\",round(Qt*3.412),\" Btu/h .\"\n", + "print \"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 .\n", + "Or, the heat transfer rate without insulation is : 14637.0 Btu/h .\n", + "2. The revised heat transfer rate with insulation is : 1386.0 W .\n", + "Or, the revised heat transfer rate with insulation is : 4729.0 Btu/h .\n", + "There 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 19.1:\n", + "T1 = 24 #Outside surface temperature (\u00b0C)\n", + "Ri = 0.0191 #Insulation resistance (K/W)\n", + "Q = 1383 #Revised heat transfer rate (Btu/h)\n", + "\n", + "#Calculation:\n", + "T2 = T1-Q*Ri #Temperature at outer surface of insulation (\u00b0C)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 19.1:\n", + "h = 11 #Convective heat transfer coefficient (W/m^2.K)\n", + "L = 7.62/10**3 #Insulation thickness (m)\n", + "k = 0.04 #Thermal conductivity of wool (W/m.K)\n", + "\n", + "#Calculation:\n", + "Bi = h*L/k #Biot number\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "k = 0.022 #Thermal conductivity of glass wool (Btu/h.ft.\u00b0F)\n", + "T1 = 400 #Inside wall temperature (\u00b0F)\n", + "T2 = 25 #Outside wall temperature (\u00b0C)\n", + "L = 3/12 #Length of insulation cover (ft)\n", + "\n", + "#Calculation:\n", + "T_2 = T2*(9/5)+32 #Outside wall temperature in fahrenheit scale (\u00b0F)\n", + "QbyA = k*(T1-T_2)/L #Heat flux across the wall (Btu/h.ft^2)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "w = 8 #Width of wall (m)\n", + "H = 3 #Height of wall (m)\n", + "h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\n", + "T1 = -18 #Outside surace of wall temperature (\u00b0C)\n", + "T3 = 26 #Surrounding air temperature (\u00b0C)\n", + "l1 = 80/100 #Reduction in cooling load\n", + "k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\n", + "T = 12000 #Units Btu/h in 1 ton of refrigeration\n", + "\n", + "#Calculation:\n", + "A = w*H #Heat transfer area (m^2) (part 1)\n", + "Q1 = h*A*(T1-T3) #Rate of heat flow in the absence of insulation (W)\n", + "Q2 = Q1*3.4123/T #Rate of heat flow in the absence of insulation (ton of refrigeration)\n", + "l2 = 1-l1 #Reduced cooling load (part 2)\n", + "Q3 = l2*Q1 #Heat rate with insulation (W)\n", + "Rt = (T1-T3)/Q3 #Total thermal resistance (\u00b0C/W)\n", + "R2 = 1/(h*A) #Convection thermal resistance (\u00b0C/W)\n", + "R1 = Rt-R2 #Insulation conduction resistance (\u00b0C/W)\n", + "L = R1*k*A #Required insulation thickness (m)\n", + "\n", + "#Result:\n", + "print \"1. The rate of heat flow through the rectangular wall without insulation is :\",round(Q1/10**3,2),\" kW .\"\n", + "print \"Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is :\",round(Q2,1),\" ton of refrigeration .\"\n", + "if (Q1<0):\n", + " print \" The negative sign indicates heat flow from the surrounding air into the cold room.\"\n", + "else :\n", + " print \" The positive sign indicates heat flow from the surrounding air into the cold room.\"\n", + "print \"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 .\n", + "Or, 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.\n", + "2. 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "#From example 19.5:\n", + "Q = -4435.2 #Heat rate with insulation (W)\n", + "R2 = 0.00198 #Convection thermal resistance (\u00b0C/W)\n", + "T3 = 26 #Surrounding air temperature (\u00b0C)\n", + "h = 21 #Convective heat transfer coefficient between the air and the surface (W/m^2.K)\n", + "k = 0.0433 #Thermal conductivity of cork board insulation (W/m.K)\n", + "L = 0.00825 #Required insulation thickness (m)\n", + "\n", + "#Calculation:\n", + "T2 = T3+Q*R2 #Interface temperature (\u00b0C) (part 1)\n", + "Bi = h*L/k #Biot number (part 2)\n", + "\n", + "#Result:\n", + "print \"1. The interface temperature is :\",round(T2,2),\" \u00b0C .\"\n", + "print \"2. The Biot number is :\",round(Bi),\" .\"\n", + "print \"3. Theoretical part.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1. The interface temperature is : 17.22 \u00b0C .\n", + "2. The Biot number is : 4.0 .\n", + "3. 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": [ + "\n", + "from __future__ import division\n", + "from math import pi,log\n", + "\n", + "#Variable declaration:\n", + "D2 = 0.5/10**3 #External diameter of needle (m)\n", + "h3 = 12 #Heat transfer coefficient (W/m^2.K)\n", + "L = 1 #Insulation thickness (m)\n", + "T1 = 95 #Reactant temperature (\u00b0C)\n", + "T3 = 20 #Ambient air temperature (\u00b0C)\n", + "k1 = 16 #Thermal conductivity of needle (W/m.K)\n", + "k3 = 0.0242 #Thermal conductivity of air (W/m.K)\n", + "D3 = 2/10**3 #Diameter of rubber tube (m)\n", + "\n", + "#Calculation:\n", + "r2 = D2/2 #External radius of needle (m)\n", + "r3 = D3/2 #Radius of rubber tube (m)\n", + "Rt1 = 1/(h3*(2*pi*r2*L)) #Thermal resistance (\u00b0C/W)\n", + "Q1 = (T1-T3)/Rt1 #Rate of heat flow in the absence of insulation (W)\n", + "Bi = h3*D2/k1 #Biot number \n", + "Nu = h3*D2/k3 #Nusselt number\n", + "R2 = log(r3/r2) #Thermal resistance of needle (\u00b0C/W)\n", + "R3 = 1/(h3*(2*pi*r3*L)) #Thermal resistance of rubber tube (\u00b0C/W)\n", + "Rt2 = R2+R3 #Total thermal resistance (\u00b0C/W)\n", + "Q2 = (T1-T3)/Rt2 #Rate of heat loss (W)\n", + "\n", + "#Result:\n", + "print \"1. The rate of the heat loss from the hypodermic needle with the rubber insulation is :\",round(Q1,2),\" W .\"\n", + "print \" The rate of the heat loss from the hypodermic needle without the rubber insulation is :\",round(Q2,2),\" W .\"\n", + "print \"2. The Biot number is :\",round(Bi,6),\" .\"\n", + "print \" 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 .\n", + "2. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log, pi\n", + "\n", + "#Variable declaration:\n", + "h = 140 #Convention heat transfer coefficient (W/m^2.K)\n", + "D1 = 10/10**3 #Rod diameter (m)\n", + "L = 2.5 #Rod length (m)\n", + "T1 = 200 #Surface temperature of rod (\u00b0C)\n", + "T2 = 25 #Fluid temperature (\u00b0C)\n", + "k = 1.4 #Thermal conductivity of bakellite (W/m.K)\n", + "l = 55/10**3 #Insulation thickness (m)\n", + "\n", + "#Calculation:\n", + "Q1 = h*pi*D1*L*(T1-T2) #Rate of heat transfer for the bare rod (W) (part 1)\n", + "Bi = 2 #Critical Biot number (part 2)\n", + "D2 = Bi*k/h #Critical diameter associated with the bakelite coating (m)\n", + "r2 = D2/2 #Critical radius associated with the bakelite coating (m)\n", + "r1 = D1/2 #Rod radius (m)\n", + "R1 = log(r2/r1)/(2*pi*k*L) #Insulation conduction resistance (\u00b0C/W)\n", + "R2 = 1/(h*(2*pi*r2*L)) #Convection thermal resistance (\u00b0C/W)\n", + "Rt1 = R1+R2 #Total thermal resistance (\u00b0C/W)\n", + "Qc = (T1-T2)/Rt1 #Heat transfer rate at the critical radius (W)\n", + "r3 = r1+l #New radius associated with the bakelite coating after insulation (m) (part 3)\n", + "R3 = log(r3/r1)/(2*pi*k*L) #Insulation conduction bakelite resistance (\u00b0C/W)\n", + "R4 = 1/(h*(2*pi*r3*L)) #Convection bakelite thermal resistance (\u00b0C/W)\n", + "Rt2 = R3+R4 #Total bakelite thermal resistance (\u00b0C/W)\n", + "Q2 = (T1-T2)/Rt2 #Heat transfer rate at the bakelite critical radius (W)\n", + "Re = ((Q1-Q2)/Q1)*100 #Percent reduction in heat transfer rate relative to the case of a bare rod (%)\n", + "\n", + "#Result:\n", + "print \"1. The rate of heat transfer for the bare rod is :\",round(Q1),\" W .\"\n", + "print \"2. The critical radius associated with the bakelite coating is :\",round(r2*10**3),\" mm.\"\n", + "print \" & the heat transfer rate at the critical radius is :\",round(Qc),\" W .\"\n", + "print \"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 .\n", + "2. 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 .\n", + "3. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi, log\n", + "\n", + "#Variable declaration:\n", + "r1 = 1.1/100 #Inside radius of pipe (m)\n", + "r2 = 1.3/100 #Outside radius of pipe (m)\n", + "r3 = 3.8/100 #Outside radius of asbestos insulation (m)\n", + "L = 1 #Length of tube (m)\n", + "h1 = 190 #Heat transfer coefficient from ethylene glycol to the stainless steel pipe (W/m^2.K)\n", + "k2 = 19 #Thermal conductivity of pipe (W/m.K)\n", + "h2 = 14 #Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K)\n", + "k3 = 0.2 #Thermal conductivity of asbestos (W/m.K)\n", + "T1 = 124 #Hot ethylene glycol temperature (\u00b0C)\n", + "T5 = 2 #Surrounding air temperature (\u00b0C)\n", + "k4 = 0.0242 #Thermal conductivity of air (W/m.K)\n", + "\n", + "#Calculation:\n", + "A1 = 2*pi*r1*L #Inside surface area of pipe (m^2) (part1)\n", + "A2 = 2*pi*r2*L #Outside surface area of pipe (m^2)\n", + "A3 = 2*pi*r3*L #Outside surface area of asbestos insulation (m^2)\n", + "R1 = 1/(h1*A1) #Inside convection resistance (\u00b0C/W)\n", + "R2 = log(r2/r1)/(2*pi*k2*L) #Conduction resistance through the tube (\u00b0C/W)\n", + "R3 = 1/(h2*A2) #Outside convection resistance (\u00b0C/W)\n", + "Rt1 = R1+R2+R3 #Total resistance without insulation (\u00b0C/W)\n", + "Q1 = (T1 - T5)/Rt1 #Heat transfer rate without insulation (W)\n", + "R4 = log(r3/r2)/(2*pi*k3*L) #Conduction resistance associated with the insulation (\u00b0C/W) (part 2)\n", + "R5 = 1/(h2*A3) #Outside convection resistance (\u00b0C/W)\n", + "Rt2 = R1+R2+R4+R5 #Total rsistance with the insulation (\u00b0C/W)\n", + "Q2 = (T1-T5)/Rt2 #Heat transfer rate with the insulation (W)\n", + "U1 = 1/(Rt2*A1) #Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3)\n", + "U3 = 1/(Rt2*A3) #Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4)\n", + "T3 = T1-(R1+R2)*Q2 #Temperature at the steel\u2013insulation interface (\u00b0C) (part 5)\n", + "Bi1 = h2*(2*r3)/k3 #Outside Biot number (part 6)\n", + "Bi2 = h1*(2*r1)/k2 #Inside Biot number\n", + "Nu = h1*(2*r1)/k4 #Nusselt number of the air\n", + "rlm = (r3-r2)/log(r3/r2) #Log mean radius of the insulation (m) (part 7)\n", + "\n", + "#Result:\n", + "print \"1. The rate of heat transfer without insulation is :\",round(Q1,1),\" W.\"\n", + "print \"2. The rate of heat transfer with insulation is :\",round(Q2,1),\" W.\"\n", + "print \"3. The overall heat transfer coefficient based on the inside area of the tube is :\",round(U1,2),\" W/m^2.K .\"\n", + "print \"4. The overall heat transfer coefficient based on the outside area of the insulation is :\",round(U3,1),\" W/m^2.K .\"\n", + "print \"5. The temperature, T3, at the steel\u2013insulation interface is :\",round(T3,1),\" \u00b0C.\"\n", + "print \"6. The inside Biot numbers is :\",round(Bi2,2),\" .\"\n", + "print \" The outside Biot numbers is :\",round(Bi1,2),\" .\"\n", + "print \" The Nusselt number is :\",round(Nu,1),\" .\"\n", + "print \"7. The log mean radius of insulation is :\",round(rlm*100,2),\" cm.\"\n", + "print \"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.\n", + "2. The rate of heat transfer with insulation is : 99.2 W.\n", + "3. The overall heat transfer coefficient based on the inside area of the tube is : 11.76 W/m^2.K .\n", + "4. The overall heat transfer coefficient based on the outside area of the insulation is : 3.4 W/m^2.K .\n", + "5. The temperature, T3, at the steel\u2013insulation interface is : 116.3 \u00b0C.\n", + "6. The inside Biot numbers is : 0.22 .\n", + " The outside Biot numbers is : 5.32 .\n", + " The Nusselt number is : 172.7 .\n", + "7. The log mean radius of insulation is : 2.33 cm.\n", + "There 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi\n", + "\n", + "#Variable declaration:\n", + "h1 = 800 #Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2.\u00b0F)\n", + "h2 = 40 #Heat transfer coefficient for oil outside coil (Btu/h.ft^2.\u00b0F)\n", + "h3 = 40 #Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2.\u00b0F)\n", + "h4 = 2 #Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2.\u00b0F)\n", + "k1 = 0.039 #Thermal conductivity of insulation layer (Btu/h.ft.\u00b0F)\n", + "l1 = 2/12 #Thickness of insulation layer (ft)\n", + "D = 10 #Diameter of tank (ft)\n", + "H = 30 #Height of tank (ft)\n", + "k2 = 224 #Thermal conductivity of copper tube (Btu/h.ft.\u00b0F)\n", + "l2 = (3/4)/12 #Thickness of insulation layer (ft)\n", + "T1 = 120 #Temperature of tank (\u00b0F)\n", + "T2 = 5 #Outdoor temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Uo1 = 1/(1/h3+(l1/k1)+1/h4) #Overall heat transfer coefficient for tank (Btu/h.ft^2.\u00b0F)\n", + "At = pi*(D+2*l1)*H #Surface area of tank (ft^2)\n", + "Q = Uo1*At*(T1-T2) #Heat transfer rate lost from the tank (Btu/h)\n", + "#From table 6.3:\n", + "l2 = 0.049/12 #Thickness of coil (ft)\n", + "A = 0.1963 #Area of 18 guage, 3/4-inch copper tube (ft^2/ft)\n", + "Uo2 = 1/(1/h2+(l2/k2)+1/h1) #Overall heat transfer coefficient for coil (Btu/h.ft^2.\u00b0F)\n", + "#From steam tables:\n", + "Tst = 240 #Temperature for 10 psia (24.7 psia) steam (\u00b0F)\n", + "Ac = Q/(Uo2*(Tst-T1)) #Area of tube (ft^2)\n", + "L = Ac/A #Lengt of tube (ft)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from __future__ import division\n", + "from math import pi, log\n", + "from numpy import array,log as LOG\n", + "\n", + "#Variable declaration:\n", + "#For 1-inch pipe schedule 40:\n", + "Di = 1.049/12 #Inside diameter (ft)\n", + "Do = 1.315/12 #Outside diameter (ft)\n", + "L = 8000 #Length of pipe (ft)\n", + "hi = 2000 #Heat transfer coefficient inside of the pipe (Btu/h.ft^2.\u00b0F)\n", + "ho = 100 #Outside heat transfer coefficient (Btu/h.ft.\u00b0F)\n", + "kl = 0.01 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\n", + "T1 = 240 #Steam temperature (\u00b0F)\n", + "T2 = 20 #Air temperature (\u00b0F)\n", + "k = 24.8 #Thermal conductivity for steel (Btu/h.ft.\u00b0F)\n", + "Dxl = array([3/8,1/2,3/4,1])/12 #thickness(ft)\n", + "amt = array([1.51,3.54,5.54,8.36])/6 #Cost per feet($) \n", + "\n", + "#Calculation:\n", + "D_ = (Do-Di)/log(Do/Di) #Log-mean diameter of the pipe (ft)\n", + "Dl = Do+2*(Dxl) #Insulation thickness (ft)\n", + "D_l = (Dl-Do)/LOG(Dl/Do) #Log mean diameter of pipe (ft)\n", + "Dxw = (Do-Di)/2 #Pipe thickness (ft)\n", + "Rw = Dxw/(k*pi*D_*L) #Wall resistance ((Btu/h.\u00b0F)^-1)\n", + "Ri = 1/(hi*pi*Di*L) #Inside steam convection resistance ((Btu/h.\u00b0F)^-1)\n", + "Rl = Dxl/(kl*pi*D_l*L) #Insulation resistance ((Btu/h.\u00b0F)^-1)\n", + "Ro = 1/(ho*pi*Dl*L) #Outside air convection resistance ((Btu/h.\u00b0F)^-1)\n", + "R = Ri+Rw+Rl+Ro #Total resistance ((Btu/h.\u00b0F)^-1)\n", + "Uo = 1/(R*pi*Dl*L) #Overall outside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "Ui = 1/(R*pi*Di*L) #Overall inside heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "dT = T1-T2\n", + "Ai = pi*Di*L #Inside area (ft^2)\n", + "Q = Ui*Ai*dT #Energy loss (Btu/h)\n", + "def energyPerDollar(Q1,Q2,amt1,amt2):\n", + " return round((Q1-Q2)/(8000*(amt2-amt1)),1)\n", + "\n", + "#Results:\n", + "print \"Energy saved per dollar ingoing from 3/8 to 1/2 inch is :\",energyPerDollar(Q[0],Q[1],amt[0],amt[1]),' Btu/h.$'\n", + "print \"Energy saved per dollar ingoing from 1/2 to 3/4 inch is :\",energyPerDollar(Q[1],Q[2],amt[1],amt[2]),' Btu/h.$'\n", + "print \"Energy saved per dollar ingoing from 3/4 to 1 inch is :\",energyPerDollar(Q[2],Q[3],amt[2],amt[3]),' Btu/h.$'" + ], + "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.$\n", + "Energy saved per dollar ingoing from 1/2 to 3/4 inch is : 18.8 Btu/h.$\n", + "Energy 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "ki = 0.44 #Thermal conductivity of insulation (Btu/h.ft.\u00b0F)\n", + "ho = 1.32 #Air flow coefficient (Btu/h.ft^2.\u00b0F)\n", + "OD = 2 #Outside diameter of pipe (in)\n", + "\n", + "#Calculation:\n", + "rc = (ki/ho)*12 #Outer critical radius of insulation (in)\n", + "ro = OD/2 #Outside radius of pipe (in)\n", + "L = rc-ro #Critical insulation thickness (in)\n", + "\n", + "#Result:\n", + "print \"The outer critical radius of insulation is :\",round(rc),\" in .\"\n", + "if ro<rc:\n", + " print \"Since, ro<rc, the heat loss will increase as insulation is added.\"\n", + "else :\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 .\n", + "Since, 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "Lf = 6/12 #Length of firebrick (ft)\n", + "kf = 0.61 #Thermal conductivity of firebrick (Btu/h.ft.\u00b0F)\n", + "A = 480 #Surface area of wall (ft^2)\n", + "Lw = 8/12 #Length of rock wool (ft)\n", + "kw = 0.023 #Thermal conductivity of rock wool (Btu/h.ft.\u00b0F)\n", + "T1 = 1900 #Temperature of insulation of firebrick (\u00b0F)\n", + "T2 = 140 #Temperature of insulation of rock wool (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Rf = Lf/(kf*A) #Resistance of firebrick (h.\u00b0F/Btu)\n", + "Rw = Lw/(kw*A) #Resistance of rock wool (h.\u00b0F/Btu)\n", + "R = Rf+Rw #Total resitance (h.\u00b0F/Btu)\n", + "Q = (T1-T2)/R #Heat loss through the wall (Btu/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "h1 = 1700 #Steam heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "h2 = 2 #Air heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "A = 1 #Area of base (ft^2) (assumption)\n", + "k1 = 26 #Thermal conductivity of steel (Btu/h.ft.\u00b0F)\n", + "k2 = 218 #Thermal conductivity of copper (Btu/h.ft.\u00b0F)\n", + "t = 0.375 #Thickness of steel sheet (in)\n", + "h3 = 2500 #Increased steam heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "h4 = 12 #Increased air heat-transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "R1 = 1/(h1*A) #Steam resistance (h.\u00b0F/Btu)\n", + "R2 = 1/(h2*A) #Air resistance (h.\u00b0F/Btu)\n", + "R3 = (t/12)/(k1*A) #Steel resistance (h.\u00b0F/Btu)\n", + "Rt1 = R1+R2+R3 #Total resistance (with steel) (h.\u00b0F/Btu)\n", + "R4 = (t/12)/(k2*A) #Copper resistance (h.\u00b0F/Btu) (part 1)\n", + "Rt2 = R1+R2+R4 #Total resistance (with copper) (h.\u00b0F/Btu)\n", + "R5 = 1/(h1*A) #New steam resistance (h.\u00b0F/Btu)\n", + "Rt3 = R5+R2+R3 #Total resistance after increasing the steam coefficient (h.\u00b0F/Btu)\n", + "R6 = 1/(h4*A) #Air resistance (h.\u00b0F/Btu)\n", + "Rt4 = R1+R6+R3 #Total resistance after increasing the air coefficient (h.\u00b0F/Btu)\n", + "\n", + "#Result:\n", + "if (Rt1==Rt2):\n", + " print \"1.The rate of heat transfer is essentially unaffected.\"\n", + "else :\n", + " print \"1. The rate of heat transfer is essentially affected.\"\n", + "if (Rt1==Rt3):\n", + " print \"2. The rate is again unaffected.\"\n", + "else :\n", + " print \"2. The rate is again affected.\"\n", + "if (Rt1==Rt4):\n", + " print \"3. The rate is unaffected for this case.\"\n", + "else :\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.\n", + "2. The rate is again unaffected.\n", + "3. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import pi, log, exp\n", + "\n", + "#Variable declaration:\n", + "rfo = 12/2 #Outside radius of firebrick (ft)\n", + "rfi = 5.167 #Inside radius of firebrick (ft)\n", + "rso = 6.479 #Outside radius of sil-o-cel (ft)\n", + "rsi = 6.063 #Inside radius of fsil-o-cel (ft)\n", + "L = 30 #Length of incinerator (ft)\n", + "kf = 0.608 #Thermal conductivity of firebrick (Btu/h.ft.\u00b0F)\n", + "ks = 0.035 #Thermal conductivity of sil-o-cel (Btu/h.ft.\u00b0F)\n", + "\n", + "#Calculation:\n", + "Rf= log(rfo/rfi)/(2*pi*L*kf) #Resistance of firebrick (h.ft.\u00b0F/Btu)\n", + "Rs= log(rso/rsi)/(2*pi*L*ks) #Resistance of sil-o-cel (h.ft.\u00b0F/Btu)\n", + "R = Rf+Rs #Total resistance (h.ft.\u00b0F/Btu)\n", + "ro = exp(R*(2*pi*L*ks))*rso #New outside radius of sil-o-cel (ft)\n", + "r= ro-rso #Extra thickness (ft)\n", + "\n", + "#Result:\n", + "print \"The extra thickness is :\",round(r,3),\" ft.\"\n", + "print \"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.\n", + "Or, 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 100755 index 00000000..5f8510ce --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_21.ipynb @@ -0,0 +1,167 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:bd26ff767164e3be8b1fcb13ef6fd471083a3c697418421a2ed4fbad1197f805" + }, + "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": [ + "\n", + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "m = 1 #Mass flowrate (lb)\n", + "cP = 1 #Heat capacity (Btu/lb.\u00b0F)\n", + "#From figure 21.3:\n", + "T1 = 300 #Temperature of hot fluid leaving exchanger (\u00b0F)\n", + "T2 = 540 #Temperature of hot fluid entering exchanger (\u00b0F)\n", + "T3 = 60 #Temperature of cold fluid leaving exchanger (\u00b0F)\n", + "T4 = 300 #Temperature of cold fluid entering exchanger (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DSh = m*cP*log((T1+460)/(T2+460)) #Entropy for hot fluid (Btu/\u00b0F)\n", + "DSc = m*cP*log((T4+460)/(T3+460)) #Entropy for cold fluid (Btu/\u00b0F)\n", + "DSa = DSh+DSc #Entropy for one exchanger (Btu/\u00b0F)\n", + "DSt = DSa*2 #Total entropy change (Btu/\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The entropy chage is :\",round(DSt,4),\" Btu/\u00b0F .\"\n", + "if (DSt>0):\n", + " print \"There is a positive entropy change.\"\n", + "else :\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 .\n", + "There 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "#From example 21.1:\n", + "DSh = -0.2744 #Entropy for hot fluid (Btu/\u00b0F)\n", + "DSc = 0.3795 #Entropy for cold fluid (Btu/\u00b0F)\n", + "m = 1 #Mass flowrate (lb)\n", + "cP = 1 #Heat capacity (Btu/lb.\u00b0F)\n", + "#From figure 21.4:\n", + "DT = 0 #Temperature difference driving force (\u00b0F)\n", + "DS_D = 0 #Entropy for D exchanger (Btu/\u00b0F)\n", + "\n", + "#Calculation:\n", + "DS_C = DSh+DSc #Entropy for C exchanger (Btu/\u00b0F)\n", + "DSt = DS_C+DS_D #Total entropy change of exchangers (Btu/\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "#From figure 21.5:\n", + "m = 2 #Mass flowrate (lb)\n", + "cP = 1 #Heat capacity (Btu/lb.\u00b0F)\n", + "DS1 = -0.2744 #Entropy for hot fluid for E exchanger (Btu/\u00b0F)\n", + "T1 = 180 #Temperature cold fluid entering the E exchabger (\u00b0F)\n", + "T2 = 60 #Temperature cold fluid leaving the E exchabger (\u00b0F)\n", + "\n", + "#Calculation:\n", + "DS2 = m*cP*log((T1+460)/(T2+460)) #Entropy for cold fluid for E exchanger (Btu/\u00b0F)\n", + "DS_E = DS1+DS2 #Entropy for E exchanger (Btu/\u00b0F)\n", + "DS_F = DS_E #Entropy for F exchanger (Btu/\u00b0F)\n", + "DSt = DS_F+DS_E #Entropy change in exchangers E and F (Btu/\u00b0F)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..65afae0f --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_22.ipynb @@ -0,0 +1,544 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:02a3e83bf8e73eb737ccd32c864c102b912719b43e71a2afa02159b5cd3f3cf2" + }, + "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": [ + "\n", + "from __future__ import division\n", + "from sympy import symbols,solve\n", + "\n", + "#Variable declaration:\n", + "#From steam tables:\n", + "h1 = 1572 #Enthalpy for super heated steam at (P = 40 atm, T = 1000\u00b0F) (Btu/lb)\n", + "h2 = 1316 #Enthalpy for super heated steam at (P = 20 atm, T = 600\u00b0F) (Btu/lb)\n", + "h3 = 1151 #Enthalpy for saturated steam (Btu/lb)\n", + "h4 = 28.1 #Enthalpy for saturated water (Btu/lb)\n", + "m1 = 1000 #Mass flowrate of steam (lb/h)\n", + "m = symbols('m') #Mass flow rate of steam (lb/h)\n", + "\n", + "#Calculation:\n", + "Dh1 = m1*(h3-h4) #The change in enthalpy for the vaporization of the water stream (Btu/h)\n", + "Dh2 = m*(h1-h2) #The change in enthalpy for the cooling of the water stream (Btu/h)\n", + "x = solve(Dh1-Dh2,m) #Mass flowrate of steam (lb/h)\n", + "m2 = x[0]; #Mass flowrate of steam (lb/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "#From table 22.1:\n", + "QH1 = 12*10**6 #Heat duty for process unit 1 (Btu/h)\n", + "QH2 = 6*10**6 #Heat duty for process unit 2 (Btu/h)\n", + "QH3 = 23.5*10**6 #Heat duty for process unit 3 (Btu/h)\n", + "QH4 = 17*10**6 #Heat duty for process unit 4 (Btu/h)\n", + "QH5 = 31*10**6 #Heat duty for process unit 5 (Btu/h)\n", + "T1 = 90 #Supply water temperature (\u00b0F)\n", + "T2 = 115 #Return water temperature (\u00b0F)\n", + "cP = 1 #Cooling water heat capacity (Btu/(lb.\u00b0F))\n", + "p = 62*0.1337 #Density of water (lb/gal)\n", + "BDR = 5/100 #Blow-down rate\n", + "\n", + "#Calculation:\n", + "QHL = (QH1+QH2+QH3+QH4+QH5)/60 #Heat load (Btu/min)\n", + "DT = T2-T1 #Change in temperature (\u00b0F)\n", + "qCW = round(QHL,-5)/(DT*cP*p) #Required cooling water flowrate (gpm)\n", + "qBD = BDR*qCW #Blow-down flow (gpm)\n", + "\n", + "#Result:\n", + "print \"The total flowrate of cooling water required for the services is :\",round(qCW,-1),\" gpm.\"\n", + "print \"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.\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "Q1 = 10*10**6 #Unit heat duty for process unit 1 (Btu/h)\n", + "Q2 = 8*10**6 #Unit heat duty for process unit 2 (Btu/h)\n", + "Q3 = 12*10**6 #Unit heat duty for process unit 3 (Btu/h)\n", + "Q4 = 20*10**6 #Unit heat duty for process unit 4 (Btu/h)\n", + "hv = 751 #Enthalpy of vaporization for pressure 500 psig (Btu/lb)\n", + "\n", + "#Calculation:\n", + "mB1 = Q1/hv #Mass flowrate of 500 psig steam through unit 1 (lb/h)\n", + "mB2 = Q2/hv #Mass flowrate of 500 psig steam through unit 2 (lb/h)\n", + "mB3 = Q3/hv #Mass flowrate of 500 psig steam through unit 3 (lb/h)\n", + "mB4 = Q4/hv #Mass flowrate of 500 psig steam through unit 4 (lb/h)\n", + "mBT = mB1+mB2+mB3+mB4 #Total steam required (lb/h)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "from sympy import symbols,solve\n", + "from math import log,pi\n", + "\n", + "#Variable declaration:\n", + "po = 53*16.0185 #Density of oil (kg/m^3)\n", + "co = 0.46*4186.7 #Heat capacity of oil (J/kg.\u00b0C)\n", + "muo = 150/1000 #Dynamic viscosity of oil (kg/m.s)\n", + "ko = 0.11*1.7303 #Thermal conductivity of oil (W/m.\u00b0C)\n", + "qo = 28830*4.381*10**-8 #Volumetric flowrate of oil (m^3/s)\n", + "pw = 964 #Density of water (kg/m^3)\n", + "cw = 4204 #Heat capacity of water (J/kg.\u00b0C)\n", + "muw = 0.7/3600*1.4881 #Dynamic viscosity of water (kg/m.s)\n", + "kw = 0.678 #Thermal conductivity of water (W/m.\u00b0C)\n", + "qw = 8406*4.381*10**-8 #Volumetric flowrate of water (m^3/s)\n", + "t1 = 23.5 #Initial temperature of oil (\u00b0C)\n", + "t2 = 27 #Final temperature of oil (\u00b0C)\n", + "T1 = 93 #Water heating temperature of water (\u00b0C)\n", + "T2 = symbols('T2') #Minimum temperature of heating water (\u00b0C)\n", + "A = symbols('A') #Heat transfer area (m^2)\n", + "Uc = 35.4 #Clean heat transfer coefficient (W/m^2.K)\n", + "Rf = 0.0007 #Thermal resistance (m^2.K/W)\n", + "D = 6*0.0254 #Inside diameter of pipe (m)\n", + "\n", + "#Calculation:\n", + "vo = muo/po #Kinematic viscosity of oil (m^2/s)\n", + "mo = po*qo #Mass flowrate of oil (kg/s)\n", + "vw = muw/pw #Kinematic viscosity of (m^2/s)\n", + "mw = pw*qw #Masss flow rate of water (kg/s)\n", + "Q1 = mo*co*(t2-t1) #Duty of exchanger of oil (W)\n", + "T2m = t1 #Lowest possible temperature of the water (\u00b0C) (part 1)\n", + "Qmw = mw*cw*(T1-T2m) #Maximum duty of exchanger of water (W) (part 2)\n", + "Q2 = mw*cw*(T1-T2) #Duty of exchanger of water in terms of T2 (W)\n", + "x = solve(Q1-Q2,T2) #Solving value for T2 (\u00b0C)\n", + "T3 = x[0]; #Minimum temperature of heating water (\u00b0C)\n", + "DT1 = T3-t1 #Inlet temperature difference (\u00b0C)\n", + "DT2 = T1-t2 #Outlet temperature difference (\u00b0C)\n", + "DTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\n", + "Ud1 = 1/Uc+Rf #Dirty heat transfer coefficient (W/m^2.K) (part 3)\n", + "Ud2 = 34.6 #Dirty heat transfer coefficient (W/m^2.\u00b0C)\n", + "Q3 = Ud2*A*DTlm #Duty of exchanger (W) (part 4)\n", + "y = solve(Q1-Q3,A) #Heat transfer area (m^2)\n", + "A1 = y[0]; #Required heat transfer area (m^2)\n", + "L = A1/(pi*D) #Required heat transfer length (m)\n", + "Qmo = mo*co*(T1-t1) #Maximum duty of exchanger of oil (W) (part 5)\n", + "Qm = Qmw #Maximum duty of exchanger (W)\n", + "E = Q1/Qm*100 #Effectiveness (%)\n", + "NTU = Ud2*A1/(mw*cw) #Number of transfer units\n", + "\n", + "#Result:\n", + "print \"1. The lowest possible temperature of the water is :\",T2m,\" \u00b0C .\"\n", + "print \"2. The log mean temperature difference is :\",round(DTlm,2),\" \u00b0C .\"\n", + "print \"3. The overall heat transfer coefficient for the new clean exchanger is :\",round(Ud2,1),\" (W/m^2.\u00b0C .\"\n", + "print \"4. The length of the double pipe heat exchanger is :\",round(L,2),\" m .\"\n", + "print \"5. The effectiveness of the exchanger is :\",round(E,2),\" % .\"\n", + "print \" 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 .\n", + "2. The log mean temperature difference is : 65.33 \u00b0C .\n", + "3. The overall heat transfer coefficient for the new clean exchanger is : 34.6 (W/m^2.\u00b0C .\n", + "4. The length of the double pipe heat exchanger is : 6.68 m .\n", + "5. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log,pi\n", + "\n", + "#Variable declaration:\n", + "#From example 22.9:\n", + "t1 = 23.5 #Initial temperature of oil (\u00b0C)\n", + "t2 = 27 #Final temperature of oil (\u00b0C)\n", + "T1 = 93 #Water heating temperature of water (\u00b0C)\n", + "T2 = 88.16 #Minimum temperature of heating water (\u00b0C)\n", + "U = 34.6 #Overall heat transfer coefficient (W/m^2.\u00b0C)\n", + "Q = 7227.2 #Duty of exchanger (W)\n", + "D = 6*0.0254 #Inside diameter of pipe (m)\n", + "l = 6.68 #Previous heat transfer length (m)\n", + "\n", + "#Calculation:\n", + "DT1 = T1-t1 #Inlet temperature difference (\u00b0C)\n", + "DT2 = T2-t2 #Outlet temperature difference (\u00b0C)\n", + "DTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0C)\n", + "A = Q/(U*DTlm) #Required heat transfer area (m^2)\n", + "L = A/(pi*D) #Required heat transfer length (m)\n", + "\n", + "#Result:\n", + "print \"The length of the parallel pipe heat exchanger is :\",round(L,2),\" m .\"\n", + "if L>l:\n", + " print \"The tube length would increase slightly.\"\n", + "elif 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 .\n", + "The 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from scipy.optimize import fsolve\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "T = 80 #Pipe surface temperature (\u00b0F)\n", + "t1 = 10 #Inlet temperature of brine solution (\u00b0F)\n", + "m = 1200 #mass flowrate of solution (kg/s)\n", + "c = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\n", + "A = 2.5 #Heat transfer area (ft^2)\n", + "U1 = 150 #Overall heat transfer coefficient at temperature approach (Btu/h.ft^2.\u00b0F)\n", + "U2 = 140 #Overall heat transfer coefficient at inlet brine temperature (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "DT1 = T-t1 #Temperature approach at the pipe entrance (\u00b0F)\n", + "\n", + "def 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\n", + "t2 = round(T-fsolve(equation,1),1)\t #The temperature of the brine solution (\u00b0F)\n", + "\n", + "#Results:\n", + "print '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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "m = 1200 #mass flowrate of solution (kg/s)\n", + "c = 0.99 #Heat capacity of brine solution (Btu/lb.\u00b0F)\n", + "DT1 = 70 #Temperature approach at the pipe entrance (\u00b0F)\n", + "DT2 = 51.6 #Temperature difference at the pipe exit (\u00b0F)\n", + "\n", + "#Calculation:\n", + "Q = m*c*(DT1-DT2) #Heat transfer rate (Btu/h)\n", + "DTlm = (DT1-DT2)/log(DT1/DT2) #Log mean temperature difference (\u00b0F)\n", + "\n", + "#Result:\n", + "print \"1. The rate of heat transfer is :\",round(Q,-1),\" Btu/h.\"\n", + "print \"Or, the rate of heat transfer is :\",round(Q/3.412),\" W.\"\n", + "print \"2. The log mean temperature difference is :\",round(DTlm,1),\" \u00b0F.\"\n", + "print \"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.\n", + "Or, the rate of heat transfer is : 6407.0 W.\n", + "2. The log mean temperature difference is : 60.3 \u00b0F.\n", + "Or, 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "Too = 100 #Steam temperature (\u00b0C)\n", + "Ti = 18 #Initial temperature of liquid TCA (\u00b0C)\n", + "Tf = 74 #Final temperature of liquid TCA (\u00b0C)\n", + "t = 180 #Heating time (s)\n", + "p = 87.4 #Density of TCA (lb/ft^3)\n", + "V = 18 #Kinematic viscosity of TCA (m^2/s)\n", + "cp = 0.23 #Heat capacity of TCA (Btu/lb.\u00b0F)\n", + "U = 200 #Overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "ui = Too-Ti #Initial excess temperature (\u00b0C)\n", + "uf = Too-Tf #Final excess temperature (\u00b0C)\n", + "R = log(ui/uf) #Ratio t/r\n", + "r = t/R #Thermal time constant (s)\n", + "A = p*V*cp/(3600*U*r) #Required heating area (ft^3)\n", + "Ti_F = Ti*9/5+32 #Initial temperature in fahrenheit scale (\u00b0F)\n", + "Tf_F = Tf*9/5+32 #Final temperature in fahrenheit scale (\u00b0F)\n", + "Q = p*V*cp*(Tf_F-Ti_F) #Total amount of heat added (Btu)\n", + "\n", + "#Result:\n", + "print \"1. The required surface area of the heating coil is :\",(A,1),\" ft^3 .\"\n", + "print \"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 .\n", + "2. 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": [ + "\n", + "#Variable declaration:\n", + "m1 = 62000 #Mass flowrate of alcohol (lb/h)\n", + "h1 = 365 #Enthalpy of vapour (Btu/lb)\n", + "cp = 1 #Heat capacity of water (Btu/lb.\u00b0F)\n", + "T1 = 85 #Entering temperature of water (\u00b0F)\n", + "T2 = 120 #Exit temperature of water (\u00b0F)\n", + "a1 = 2.11 #Flow area for the shell side (ft^2)\n", + "N = 700 #Total number of tubes\n", + "a2 = 0.546 #Flow area per tube (in^2/tube)\n", + "n = 4 #Number of tube passes\n", + "p = 62.5 #Density of water (lb/ft^3)\n", + "L = 16 #Length of condenser (ft)\n", + "hio = 862.4 #Cooling water inside film coefficient (Btu/h.ft^2.\u00b0F)\n", + "g = 9.8 #Gravitational accleration (m^2/s)\n", + "Rf = 0.003 #Fouling factor (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Calculation:\n", + "Q1 = m1*h1 #Heat loss from alcohol (Btu/h)\n", + "Q2 = Q1 #Heat gained by water (Btu/h)\n", + "DT = T2-T1 #Temperature difference (\u00b0F)\n", + "m2 = Q2/(cp*DT) #Water mass flow rate (lb/h)\n", + "LMTD = ((T2-32)-(T1-32))/log((T2-32)/(T1-32)) #Log mean temperature difference (\u00b0F)\n", + "at = (N*a2)/(144*n) #Total flow area for tube side (ft^2)\n", + "G1 = m1/a1 #Mass velocity of flow in shell side (lb/h.ft^2)\n", + "G2 = m2/at #Mass velocity of flow in tube side (lb/h.ft^2)\n", + "V = G2/(3600*p) #Velocity of water (ft/s)\n", + "G3 = m1/(L*N)**(2/3) #Loading G (lb/h.ft)\n", + "#For alcohol:\n", + "kf = 0.105 #Thermal conductivity (Btu/h.ft.\u00b0F)\n", + "muf = 0.55*2.42 #Dynamic viscosity (lb/ft.h)\n", + "sf = 0.79 #\n", + "pf = sf*p #Density (lb/ft^3)\n", + "h = 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)\n", + "ho = h #Outside heat transfer coefficient of the tube bundle (Btu/h.ft^2.\u00b0F)\n", + "Uc = (hio*ho)/(hio+ho) #Overall heat transfer coefficient for a new (clean) heat exchanger (Btu/h.ft^2.\u00b0F)\n", + "A = N*L*0.2618 #Area for heat transfer (ft^2)\n", + "Ud = Q1/(A*DT) #Design (D) overall heat transfer coefficient (Btu/h.ft^2.\u00b0F)\n", + "Rd = (Uc-Ud)/(Uc*Ud) #Dirt (d) factor (Btu/h.ft^2.\u00b0F)\n", + "\n", + "#Result:\n", + "print \"The dirt (d) factor is :\",round(Rd,4),\" Btu/h.ft^2.\u00b0F .\"\n", + "if (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.\"\n", + "else:\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 .\n", + "Therefore, 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 100755 index 00000000..02544a8f --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_23.ipynb @@ -0,0 +1,106 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ea99fc2be6052cbd9b5accb00ac50cd0a25a5282b964ce8d492c2c528e36d584" + }, + "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": [ + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "Q = 20000 #Fuel input (Btu)\n", + "e = 1 #Energy produced (kW.h)\n", + "Btu = 3412 #Units Btu in 1 kW.h\n", + "\n", + "#Calulation:\n", + "ER = Q/Btu #Energy requirement in 1990 (kW.h)\n", + "E = e/ER*100 #Efficiency of energy conversion (%)\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "ADL1 = 2 #Average daily load (MW)\n", + "R = 25/100 #Reduction in electrical load (%)\n", + "\n", + "#Calculation:\n", + "L = 1-R #New load fraction\n", + "ADL2 = ADL1*L #New average daily load (MW)\n", + "AR = ADL1-ADL2 #Average reduction in electrical load (MW)\n", + "\n", + "#Result:\n", + "print \"The new Average daily load for the plant is :\",ADL2,\" MW.\"\n", + "print \"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.\n", + "The 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 100755 index 00000000..fb821b6b --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_24.ipynb @@ -0,0 +1,387 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:20841f2d078ed8f6b672e93204c42bfe5b2cecc320cf09e962fc7cfcf63372c5" + }, + "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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "fm = 30/100 #Mole fraction of methane\n", + "fe = 50/100 #Mole fraction of ethane\n", + "fp = 20/100 #Mole fraction of pentane\n", + "LFLm = 0.046 #Lower flammability limit for methane\n", + "LFLe = 0.035 #Lower flammability limit for ethane\n", + "LFLp = 0.014 #Lower flammability limit for propane\n", + "UFLm = 0.142 #Upper flammability limit for methane\n", + "UFLe = 0.151 #Upper flammability limit for ethane\n", + "UFLp = 0.078 #Upper flammability limit for propane\n", + "\n", + "#Calculation:\n", + "LFLmix = 1/((fm/LFLm)+(fe/LFLe)+(fp/LFLp)) #Lower flammability limit of gas mixture\n", + "UFLmix = 1/((fm/UFLm)+(fe/UFLe)+(fp/UFLp)) #Upper flammability limit of gas mixture\n", + "\n", + "#Result:\n", + "print \"The upper flammability limit (UFL) of the gas mixture is :\",round(UFLmix*100,2),\" % .\"\n", + "print \"The lower flammability limit (LFL) of the gas mixture is :\",round(LFLmix*100,2),\" % .\"\n", + "print \"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 % .\n", + "The lower flammability limit (LFL) of the gas mixture is : 2.85 % .\n", + "There 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from sympy import nsimplify as changeToFraction\n", + "\n", + "#Variable declaration:\n", + "P_A = 10/100 #Probability that the first tube is defective if the first is replaced\n", + "P_B = 10/100 #Probability that the second tube is defective if the first is replaced\n", + "\n", + "#Calculation:\n", + "P_AB = P_A*P_B #Probability that the two tubes are defective if the first is replaced\n", + "P_B_A = 9/99 #Probability that the second tube is defective if the first tube is not replaced\n", + "Pd_AB = P_A*P_B_A #Probability that both tubes are defective if the first tube is not replaced\n", + "\n", + "#Result:\n", + "print \"The probability that both tubes are defective if :\"\n", + "print \"(a) the first is replaced before the second is drawn is :\",changeToFraction(P_AB),\" .\"\n", + "print \"(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": [ + "\n", + "\n", + "from sympy import symbols, integrate, exp\n", + "\n", + "#Variable declaration:\n", + "X = symbols('X') #Range of X\n", + "Px = 1.7*(exp(-1.7*X)) #Probability distribution function\n", + "\n", + "#Calculation:\n", + "P = integrate(Px, (X,2,6)) #Probability that X will have a value between 2 and 6\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import factorial\n", + "\n", + "#Variable Declaration:\n", + "n = 20 #Total number of components\n", + "p = 0.1 #Probability of success\n", + "\n", + "#Calculations:\n", + "def 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:\n", + "print \"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": [ + "\n", + "\n", + "from sympy import symbols, integrate, exp\n", + "\n", + "#Variable declaration:\n", + "a = 1.3*10**-3 #Constant a\n", + "B = 0.77 #Constant B\n", + "t = symbols('t') #Time (h)\n", + "Ft = a*B*t**(B-1)*(exp(-a*t**B)) #Pdf for heat exchanger tube\n", + "Pt = integrate(Ft, (t,0,1000)) #Probability that a heat exchanger will fail within 100 hours\n", + "\n", + "#Result:\n", + "print \"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": [ + "\n", + "\n", + "from __future__ import division\n", + "from scipy.stats import norm as f\n", + "\n", + "#Variable declaration:\n", + "m = 0.4008 #Mean(inch)\n", + "s = 0.0004 #Standard Deviation(inch)\n", + "UL = 0.4000+0.001 #Upper Limit\n", + "LL = 0.4000-0.001 #Upper Limit\n", + "\n", + "#Calculation:\n", + "Ps = f.cdf(UL,m,s)-f.cdf(LL,m,s)#Probability of meeting specs\n", + "Pd = 1-Ps #Probability of defect\n", + "\n", + "#Results:\n", + "print 'Probability of meeting specifications:',round(Ps*100,2),'%'\n", + "print 'Probability of Defect:',round(Pd*100,2),'%'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Probability of meeting specifications: 69.15 %\n", + "Probability 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from numpy import array,minimum\n", + "from math import sqrt\n", + "\n", + "#variable Declaration:\n", + "mTa = array([100]*10) #Mean weeks for thermometer failure(A)\n", + "mTb = array([90]*10) #Mean weeks for thermometer failure(B)\n", + "mTc = array([80]*10) #Mean weeks for thermometer failure(C)\n", + "sTa = 30 #Standard deviation (weeks) for thermometer failure(A)\n", + "sTb = 20 #Standard deviation (weeks) for thermometer failure(B)\n", + "sTc = 10 #Standard deviation (weeks) for thermometer failure(C)\n", + "Ra = 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\n", + "Rb = 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\n", + "Rc = 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\n", + "Za = 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\n", + "Zb = 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\n", + "Zc = 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:\n", + "Ta = mTa+sTa*Za\n", + "Tb = mTb+sTb*Zb\n", + "Tc = mTc+sTc*Zc\n", + "Ts = minimum(Ta,Tb)\n", + "Ts = minimum(Ts,Tc)\n", + "m = array([sum(Ts)/len(Ts)]*10)\n", + "s = sqrt(sum((Ts-m)**2)/(len(Ts)-1))\n", + "\n", + "#Results:\n", + "print '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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "t = 273 #Standard temperature (K)\n", + "v = 0.0224 #Volume of air occupied by 1 gmol of ideal gas (m^3)\n", + "V = 1100 #Volume of heat exchanger (m^3)\n", + "T = 22+273 #Temperature of heat exchanger (K)\n", + "x1 = 0.75 #gmols of hydrocarbon leaking from the exchanger (gmol)\n", + "\n", + "#Calculation:\n", + "n = V*(1/v)*(t/T) #Total number of gmols of air in the room (gmol)\n", + "xHC = (x1/(n+x1))*10**6 #The mole fraction of hydrocarbon in the room (ppm)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..64b80ece --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_26.ipynb @@ -0,0 +1,167 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:4c0513ab1606f42cc2658f4efc57e03793ddba6911c2c7915b3159fc7138c6b8" + }, + "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": [ + "\n", + "from __future__ import division\n", + "from sympy import symbols,solve,log\n", + "from math import exp\n", + "\n", + "#Variable Declaration:\n", + "A,B,r,C = symbols('A B r C');\n", + "\n", + "#Calculation:\n", + "res = solve([A + B*log(2)-log(3),A + B*log(4)-log(12)],[A,B])\n", + "A = round(float(res[A]),4)\n", + "B = round(float(res[B]))\n", + "kA = round(exp(A),2)\n", + "a = B\n", + "\n", + "#Result:\n", + "print '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": [ + "\n", + "import scipy.stats as f\n", + "\n", + "#Variable Declaration:\n", + "T = [-40,-20,0,10,12,30,40,50,60,80,100,150,200,250,300,400,500]\n", + "u = [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:\n", + "B,A,r_value, p_value, std_err = f.linregress(T,u)\n", + "\n", + "#Results:\n", + "print 'The value of A in regression model is:',round(A,4)\n", + "print '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\n", + "The 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": [ + "\n", + "from __future__ import division\n", + "from scipy.optimize import fmin_cobyla as optimize\n", + "\n", + "\n", + "#Variable Declaration:\n", + "def f(x):\t\n", + " return -2.0*x[0] - 1.6*x[1]\n", + "\n", + "def c1(x):\n", + " return 16820 - x[0]\n", + "\n", + "def c2(x):\n", + " return 1152 - x[1]\n", + "\n", + "def c3(x):\n", + " return 1500 - 0.08*x[0] - 0.11*x[1]\n", + "\n", + "def c4(x):\n", + " return 6000 - 0.29*x[0] - 0.54*x[1]\n", + "\n", + "def c5(x):\n", + " return 11000 - 0.63*x[0] - 0.35*x[1]\n", + "\n", + "def c6(x):\n", + " return x[0]\n", + "\n", + "def c7(x):\n", + " return x[1]\n", + "\n", + "#Calculation\n", + "X = optimize(f,[16820,1152],[c1,c2,c3,c4,c5,c6], disp = 0)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..c5e5dea1 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_27.ipynb @@ -0,0 +1,544 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:9cb92879a3edfd07367a11876840b47bf3060ee079039238aa0305841473df9a" + }, + "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": [ + "\n", + "#Variable declaration:\n", + "i = 0.03375 #Rate of interest (%)\n", + "n = 9 #Years to the end of life (yr)\n", + "P = 60000 #Cost of exchanger ($)\n", + "L = 500 #Salvage value ($)\n", + "x = 5 #Time after 5 years (yr)\n", + "\n", + "#Calculation:\n", + "SFDF = i/((1+i)**n-1) #Sinking fund depreciation factor\n", + "UAP = (P-L)*SFDF #Uniform annual payment ($)\n", + "B = P-((P-L)/n)*x #Appraisal value after 5 years ($)\n", + "\n", + "#Result:\n", + "print \"1. The uniform annual payment made into the fund at the of the year is : $\",round(UAP),\" .\"\n", + "print \"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 .\n", + "2. 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": [ + "\n", + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "C = 150000 #Capital cost ($)\n", + "i = 7/100 #Interest rate\n", + "n = 5 #Time (yr)\n", + "OC = 15000 #Operating cost ($)\n", + "A = 75000 #Annual cost for the old process ($)\n", + "\n", + "#Calculation:\n", + "CRF = (i*(1+i)**n)/((1+i)**n-1) #Capital recovery factor\n", + "IC = CRF*C #Initial cost ($)\n", + "AC = IC+OC #Total annualized cost ($)\n", + "\n", + "#Result:\n", + "print \"The annualized cost for the new heating system is : $\",round(AC),\" .\"\n", + "if (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.\"\n", + "else :\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 .\n", + "Since this cost is lower than the annual cost of $75,000 for the old process, the proposed plan should be implemented.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "ILLUSTRATIVE EXAMPLE 27.7, Page number: 577" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from __future__ import division\n", + "\n", + "#Variable declaration:\n", + "i = 12/100 #Intersest rate\n", + "n = 12 #Lifetime period (yr)\n", + "CC = 2625000 #Capital cost ($)\n", + "IC = 1575000 #Installation cost ($)\n", + "#From table 27.3:\n", + "Ic1 = 2000000 #Income credit for double pipe ($/yr)\n", + "Ic2 = 2500000 #Income credit for Shell-and-tube ($/yr)\n", + "AC1 = 1728000 #Total annual cost for double pipe ($/yr)\n", + "AC2 = 2080000 #Total annual cost for Shell-and-tube ($/yr)\n", + "\n", + "#Calculation:\n", + "CRF = i/(1-(1+i)**-n) #Capital recovery factor\n", + "DPc = (CC+IC)*CRF #Annual capital and installation costs for the DP unit ($/yr)\n", + "STc = (CC+IC)*CRF #Annual capital and installation costs for the ST unit ($/yr)\n", + "DPp = Ic1-AC1 #Profit for the DP unit ($/yr)\n", + "STp = Ic2-AC2 #Profit for the ST unit ($/yr)\n", + "\n", + "#Result:\n", + "print \"The profit for the shell-and-tube unit is : $\",round(DPp),\"/yr .\"\n", + "print \"The profit for the double pipe unit is : $\",round(STp),\"/yr .\"\n", + "if (STp>DPp):\n", + " print \"A shell-and-tube heat exchanger should therefore be selected based on the above economic analysis.\"\n", + "else :\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 .\n", + "The profit for the double pipe unit is : $ 420000.0 /yr .\n", + "A 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from math import log\n", + "\n", + "#Variable declaration:\n", + "m = 50000 #Mass flowrate of the organic fluid (lb/h)\n", + "cP = 0.6 #The heat capacity of the organic liquid (Btu/lb.\u00b0F)\n", + "T1 = 150 #Initial temperature of organic fluid (\u00b0F)\n", + "T2 = 330 #Final temperature of organic fluid (\u00b0F)\n", + "Ts1 = 358 #Saturation temperature for 150 psia (\u00b0F)\n", + "Ts2 = 417 #Saturation temperature for 300 psia (\u00b0F)\n", + "L1 = 863.6 #Latent heat for 150 psia (Btu/lb)\n", + "L2 = 809 #Latent heat for 300 psia (Btu/lb)\n", + "c1 = 5.20/1000 #Cost for 150 psia ($/lb)\n", + "c2 = 5.75/1000 #Cost for 300 psia ($/lb)\n", + "CI1 = 230 #Cost index in 1998 \n", + "CI2 = 360 #Cost index in 2011\n", + "IF = 3.29 #Installation factor\n", + "PF1 = 1.15 #Pressure factors for 100 to 200 psig\n", + "PF2 = 1.20 #Pressure factors for 200 to 300 psig\n", + "OP = 90/100 #Plant on-stream operation factor\n", + "h = 365*24 #Hours in a year (h)\n", + "\n", + "#Calculation:\n", + "Q = m*cP*(T2-T1) #Overall heta duty (Btu/h)\n", + "DT1 = Ts1-T1 #Temperature driving force 1 for 150 psia (\u00b0F)\n", + "DT2 = Ts1-T2 #Temperature driving force 2 for 150 psia (\u00b0F)\n", + "LMTD1 = (DT1-DT2)/log(DT1/DT2) #Log-mean temperature difference for 150 psia (\u00b0F)\n", + "DT3 = Ts2-T1 #Temperature driving force 1 for 300 psia (\u00b0F)\n", + "DT4 = Ts2-T2 #Temperature driving force 2 for 300 psia (\u00b0F)\n", + "LMTD2 = (DT3-DT4)/log(DT3/DT4) #Log-mean temperature difference for 1300 psia (\u00b0F)\n", + "A1 = Q/(138*LMTD1) #Required heat transfer area for 150 psia (ft^2)\n", + "A2 = Q/(138*LMTD2) #Required heat transfer area for 300 psia (ft^2)\n", + "BC1 = 117*A1**0.65 #Base cost for 150 psia ($)\n", + "BC2 = 117*A2**0.65 #Base cost for 13000 psia ($)\n", + "C1 = BC1*(CI2/CI1)*IF*PF1 #Capital cost for 150 psia ($)\n", + "C2 = BC2*(CI2/CI1)*IF*PF2 #Capital cost for 300 psia ($)\n", + "S1 = Q*(h*OP)/L1 #Steam requirement for 150 psia (lb/yr)\n", + "S2 = Q*(h*OP)/L2 #Steam requirement for 300 psia (lb/yr)\n", + "SC1 = S1*c1 #Annual steam cost for 150 psia ($/yr)\n", + "SC2 = S2*c2 #Annual steam cost for 300 psia ($/yr)\n", + "\n", + "#Result:\n", + "print \"1. The capital cost for 150 psia is : $\",round(C1,-3),\" .\"\n", + "print \" The capital cost for 300 psia is : $\",round(C2,-3),\" .\"\n", + "print \"2. The annual steam cost for 150 psia is : $\",round(SC1,-3),\"/yr .\"\n", + "print \" The annual steam cost for 300 psia is : $\",round(SC2,-3),\"/yr .\"\n", + "if (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.\"\n", + "elif (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 .\n", + "2. The annual steam cost for 150 psia is : $ 256000.0 /yr .\n", + " The annual steam cost for 300 psia is : $ 303000.0 /yr .\n", + "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.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "ILLUSTRATIVE EXAMPLE 27.9, Page number: 581" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "from __future__ import division\n", + "from sympy import symbols,solve\n", + "from scipy.optimize import fsolve\n", + "\n", + "#Variable declaration:\n", + "TCC_TB = 2500000 #Total capital cost ($)\n", + "R_TB = 3600000 #R_TBevenue generated from the facility ($)\n", + "AOC_TB = 1200000 #Annual operating costs ($)\n", + "TCC_FB = 3500000 #Total capital cost ($)\n", + "R_FB = 5300000 #R_TBevenue generated from the facility ($)\n", + "AOC_FB = 1400000 #Annual operating costs ($)\n", + "n = 10 \t#Time of facility (yr)\n", + "\n", + "#Calculation:\n", + "D = 0.1*TCC_TB #Depriciation ($)\n", + "WC = 0.1*TCC_TB #Working capital ($)\n", + "TI = R_TB-AOC_TB-D #Taxable income ($)\n", + "IT = 0.5*TI #Income tax to be paid ($)\n", + "A = R_TB-AOC_TB-IT #After-tax cash flow ($)\n", + "def eqTB(i):\n", + "\tx = (((1+i)**n-1)/(i*(1+i)**n))*A + (1/(1+i)**n)*WC #Equation for computing rate of return for TB unit\n", + "\ty = WC + 0.5*TCC_TB + 0.5*TCC_TB*(1+i)**1 #Equation for computing rate of return for TB unit\n", + "\treturn x-y\n", + "iTB = round(fsolve(eqTB,0.8)*100,1) #Rate of return for TB unit (%)\n", + "\n", + "D = 0.1*TCC_FB #Depriciation ($)\n", + "WC = 0.1*TCC_FB #Working capital ($)\n", + "TI = R_FB-AOC_FB-D #Taxable income ($)\n", + "IT = 0.5*TI #Income tax to be paid ($)\n", + "A = R_FB-AOC_FB-IT #After-tax cash flow ($)\n", + "\n", + "def 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\n", + "iFB = round(fsolve(eqFB,0.8)*100,1) #Rate of return for FB unit (%)\n", + "\n", + "#Results:\n", + "print \"The rate of return for TB unit is:\",round(iTB),\" %.\"\n", + "print \"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 %.\n", + "The 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": [ + "\n", + "\n", + "#Variable declaration:\n", + "f = 100000 #Flow rate of flue gas (acfm)\n", + "i = 0.1 #Interest rate\n", + "#From table 27.4:\n", + "#For finned preheater:\n", + "ac1 = 3.1 #Equipment cost ($/acfm)\n", + "ac2 = 0.8 #Installation cost ($/acfm)\n", + "ac3 = 0.06 #Operating cost ($/acfm-yr)\n", + "ac4 = 14000 #Maintenance cost ($/yr)\n", + "an = 20 #Lifetime (yr)\n", + "#For 4-pass preheater:\n", + "bc1 = 1.9 #Equipment cost ($/acfm)\n", + "bc2 = 1.4 #Installation cost ($/acfm)\n", + "bc3 = 0.06 #Operating cost for ($/acfm-yr)\n", + "bc4 = 28000 #Maintenance cost ($/yr)\n", + "bn = 15 #Lifetime of (yr)\n", + "#For 2-pass preheater:\n", + "cc1 = 2.5 #Equipment cost ($/acfm)\n", + "cc2 = 1.0 #Installation cost ($/acfm)\n", + "cc3 = 0.095 #Operating cost for ($/acfm-yr)\n", + "cc4 = 9500 #Maintenance cost for ($/yr)\n", + "cn = 20 #Lifetime of (yr)\n", + "\n", + "#Calculation:\n", + "#For Finned preheater:\n", + "aEC = f*ac1 #Total equipment cost ($)\n", + "aIC = f*ac2 #Total installation cost ($)\n", + "aOC = f*ac3 #Total operating cost ($)\n", + "aMC = f*ac4 #Total maintenance cost ($)\n", + "aCRF = (i*(1+i)**an)/((1+i)**an-1) #Capital recovery factor\n", + "aAEC = aEC*aCRF #Equipment annual cost ($/yr)\n", + "aAIC = aIC*aCRF #Installation annual cost($/yr)\n", + "aAOC = ac3*f #Annual operating cost ($)\n", + "aAMC = ac4 #Annual maintenance cost ($)\n", + "aTAC = aAEC+aAIC+aAOC+aAMC #Total annual cost ($)\n", + "\n", + "#For 4-pass preheater:\n", + "bEC = f*bc1 #Total equipment cost ($)\n", + "bIC = f*bc2 #Total installation cost ($)\n", + "bOC = f*bc3 #Total operating cost ($)\n", + "bMC = f*bc4 #Total maintenance cost ($)\n", + "bCRF = (i*(1+i)**bn)/((1+i)**bn-1) #Capital recovery factor\n", + "bAEC = bEC*bCRF #Equipment annual cost ($/yr)\n", + "bAIC = bIC*bCRF #Installation annual cost($/yr)\n", + "bAOC = bc3*f #Annual operating cost ($)\n", + "bAMC = bc4 #Annual maintenance cost ($)\n", + "bTAC = bAEC+bAIC+bAOC+bAMC #Total annual cost ($)\n", + "#For 2-pass preheater:\n", + "cEC = f*cc1 #Total equipment cost ($)\n", + "cIC = f*cc2 #Total installation cost ($)\n", + "cOC = f*cc3 #Total operating cost ($)\n", + "cMC = f*cc4 #Total maintenance cost ($)\n", + "cCRF = (i*(1+i)**cn)/((1+i)**cn-1) #Capital recovery factor\n", + "cAEC = cEC*cCRF #Equipment annual cost ($/yr)\n", + "cAIC = cIC*cCRF #Installation annual cost($/yr)\n", + "cAOC = cc3*f #Annual operating cost ($)\n", + "cAMC = cc4 #Annual maintenance cost ($)\n", + "cTAC = cAEC+cAIC+cAOC+cAMC #Total annual cost ($)\n", + "\n", + "#Result:\n", + "print \"Total annual cost for finned preheater is : $\",round(aTAC),\" .\"\n", + "print \"Total annual cost for 4-pass preheater is : $\",round(bTAC),\" .\"\n", + "print \"Total annual cost for 2-pass preheater is : $\",round(cTAC),\" .\"\n", + "if (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.\"\n", + "elif (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.\"\n", + "elif (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 .\n", + "Total annual cost for 4-pass preheater is : $ 77386.0 .\n", + "Total annual cost for 2-pass preheater is : $ 60111.0 .\n", + "According 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": [ + "\n", + "from __future__ import division\n", + "from numpy import roots\n", + "from math import sqrt\n", + "\n", + "#Variable declaration:\n", + "TH = 500 #Hot stream temperature at exchanger 1 (\u00b0F)\n", + "tc = 100 #Cold stream temperature at exchanger 2 (\u00b0F)\n", + "A = 10 #Constant A\n", + "B1 = 100000 #Constant B1\n", + "B2 = 4000 #Constant B2\n", + "B3 = 400000 #Constant B3\n", + "\n", + "#Calculations:\n", + "#It forms equation fo form t^2 - t(Th-tc) +tcTH +B/A\n", + "t1 = roots([1, -(TH+tc),(tc*TH + B1/A) ]); #Roots\n", + "tmax1 = TH - sqrt(B1/A) #Upon maximising profit\n", + "t2 = roots([1, -(TH+tc),(tc*TH + B2/A) ]); #Roots\n", + "tmax2 = TH - sqrt(B2/A) #Upon maximising profit\n", + "t3 = roots([1, -(TH+tc),(tc*TH + B3/A) ]); #Roots\n", + "tmax3 = TH - sqrt(B3/A) #Upon maximising profit\n", + "\n", + "#Results:\n", + "print 'tBE for case 1: ',round(t1[0]),'\u00b0F',round(t1[1]),'\u00b0F'\n", + "print 'tmax1:', round(tmax1),'\u00b0F'\n", + "print 'tBE for case 2: ',round(t2[0]),'\u00b0F',round(t2[1]),'\u00b0F'\n", + "print 'tmax1:', round(tmax2),'\u00b0F'\n", + "print 'tBE for case 1: ',round(t3[0]),'\u00b0F',round(t3[1]),'\u00b0F'\n", + "print 'tmax1:', round(tmax3),'\u00b0F'" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "tBE for case 1: 473.0 \u00b0F 127.0 \u00b0F\n", + "tmax1: 400.0 \u00b0F\n", + "tBE for case 2: 499.0 \u00b0F 101.0 \u00b0F\n", + "tmax1: 480.0 \u00b0F\n", + "tBE for case 1: 300.0 \u00b0F 300.0 \u00b0F\n", + "tmax1: 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": [ + "\n", + "\n", + "from __future__ import division\n", + "from scipy.optimize import fmin_cobyla as optimize\n", + "\n", + "#Key:\n", + "#f(x) : Objective Function\n", + "#ci(x)'s : Constraints\n", + "\n", + "#Variable Declaration:\n", + "def f(x):\t\n", + " return -1.70*x[0] - 2*x[1]\n", + "\n", + "def c1(x):\n", + " return 8000 - x[0]\n", + "\n", + "def c2(x):\n", + " return 6000 - x[1]\n", + "\n", + "def c3(x):\n", + " return 12000 - 0.75*x[0] - 0.40*x[1]\n", + "\n", + "def c4(x):\n", + " return 6000 - 0.60*x[0] - 0.25*x[1]\n", + "\n", + "def c5(x):\n", + " return x[0]\n", + "\n", + "def c6(x):\n", + " return x[1]\n", + "\n", + "#Calculation\n", + "X = optimize(f,[7000,6000],[c1,c2,c3,c4,c5,c6], disp = 0)\n", + "\n", + "#Result:\n", + "print \"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 100755 index 00000000..9031d510 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/Chapter_28.ipynb @@ -0,0 +1,108 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:14824d48a431915b3dac3068270b7ce0d86b9e98bf4622dc413ebf717903e290" + }, + "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": [ + "\n", + "\n", + "hm1 = 50000 #Mass flowrate (lb/h)\n", + "hcP1 = 0.65 #Heat capacity (Btu/lb.\u00b0F)\n", + "hTi1 = 70 #Inlet temperature (\u00b0F)\n", + "hTo1 = 300 #Outlet temperature (\u00b0F)\n", + "#For stream 2 to be heated:\n", + "hm2 = 60000 #Mass flowrate (lb/h)\n", + "hcP2 = 0.58 #Heat capacity (Btu/lb.\u00b0F)\n", + "hTi2 = 120 #Inlet temperature (\u00b0F)\n", + "hTo2 = 310 #Outlet temperature (\u00b0F)\n", + "#For stream 3 to be heated:\n", + "hm3 = 80000 #Mass flowrate (lb/h)\n", + "hcP3 = 0.78 #Heat capacity (Btu/lb.\u00b0F)\n", + "hTi3 = 90 #Inlet temperature (\u00b0F)\n", + "hTo3 = 250 #Outlet temperature (\u00b0F)\n", + "#From table 28.4:\n", + "#For stream 1 to be cooled:\n", + "cm1 = 60000 #Mass flowrate (lb/h)\n", + "ccP1 = 0.70 #Heat capacity (Btu/lb.\u00b0F)\n", + "cTi1 = 420 #Inlet temperature (\u00b0F)\n", + "cTo1 = 120 #Outlet temperature (\u00b0F)\n", + "#For stream 2 to be cooled:\n", + "cm2 = 40000 #Mass flowrate (lb/h)\n", + "ccP2 = 0.52 #Heat capacity (Btu/lb.\u00b0F)\n", + "cTi2 = 300 #Inlet temperature (\u00b0F)\n", + "cTo2 = 100 #Outlet temperature (\u00b0F)\n", + "#For stream 3 to be cooled:\n", + "cm3 = 35000 #Mass flowrate (lb/h)\n", + "ccP3 = 0.60 #Heat capacity (Btu/lb.\u00b0F)\n", + "cTi3 = 240 #Inlet temperature (\u00b0F)\n", + "cTo3 = 90 #Outlet temperature (\u00b0F)\n", + "\n", + "#Calculation:\n", + "H1 = hm1*hcP1*(hTo1-hTi1) #Heating duty for stream 1 (Btu/h)\n", + "H2 = hm2*hcP2*(hTo2-hTi2) #Heating duty for stream 2 (Btu/h)\n", + "H3 = hm3*hcP3*(hTo3-hTi3) #Heating duty for stream 1 (Btu/h)\n", + "H = H1+H2+H3 #Total heating duty (Btu/h)\n", + "C1 = cm1*ccP1*(cTi1-cTo1) #Cooling duty for stream 1 (Btu/h)\n", + "C2 = cm2*ccP2*(cTi2-cTo2) #Cooling duty for stream 2 (Btu/h)\n", + "C3 = cm3*ccP3*(cTi3-cTo3) #Cooling duty for stream 1 (Btu/h)\n", + "C = C1+C2+C3 #Total Cooling duty (Btu/h)\n", + "\n", + "#Result:\n", + "print \"Table: Duty Requirements.\"\n", + "print \"Stream Duty, Btu/h\"\n", + "print \"1 \",round(H1)\n", + "print \"2 \",round(H2)\n", + "print \"3 \",round(H3)\n", + "print \"4 \",round(C1)\n", + "print \"5 \",round(C2)\n", + "print \"6 \",round(C3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Table: Duty Requirements.\n", + "Stream Duty, Btu/h\n", + "1 7475000.0\n", + "2 6612000.0\n", + "3 9984000.0\n", + "4 12600000.0\n", + "5 4160000.0\n", + "6 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 100755 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 100755 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 100755 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 100755 index 00000000..427304c9 --- /dev/null +++ b/Heat_Transfer_Applications_for_the_Practicing_Engineer/screenshots/shell_heat.png |
