{
"metadata": {
"name": "",
"signature": "sha256:e3c042bf60cf5ea8673efb2741d5df5f7935daa012691b520a2d1357e838bd6c"
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"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": {}
}
]
}