{
"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": {}
}
]
}