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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 3 - Heat exchanger design "
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.3, Page number: 113"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=20; #Entering Temperature of Water, K\n",
"T2=40; #Exit Temperature of water, K\n",
"m=25/60 #Condensation rate of steam, kg/s\n",
"T3=60; #Condensation Temperature,K\n",
"A=12; #area of exchanger, m**2\n",
"h=2358.7*math.pow(10,3); #latent heat, J/kg\n",
"Cp=4174; #Specific heat of water, J/kg K\n",
"\n",
"#Calculations\n",
"U=(m*h)/(A*((T2-T1)/math.log((T3-T1)/(T3-T2))));#Overall heat transfer coefficient, W/(m^2*K)\n",
"Mh=(m*h)/(Cp*(T2-T1));\t\t #Required flow of water, kg/s\n",
"\n",
"#Results\n",
"print \"Overall heat transfer coefficient is :\",round(U,1),\"W/(m^2*K)\\n\"\n",
"print \"Required flow of water is :\",round(Mh,2),\"kg/s\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Overall heat transfer coefficient is : 2838.4 W/(m^2*K)\n",
"\n",
"Required flow of water is : 11.77 kg/s\n",
"\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.4, Page number: 117"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"m=5.795; #flow rate of oil, kg/s\n",
"T1=454; #Entering Temperature of oil, K\n",
"T2=311; #Exit Temperature of oil, K\n",
"T3=305; # Entering Temperature of water, K\n",
"T4=322; #Exit Temperature of water, K\n",
"c=2282; #heat capacity, J/(kg*K)\n",
"U=416; #overall heat transfer coefficient , J/(m**2*K*s)\n",
"F=0.92; #Correction factor for 2 shell and 4 tube-pass exchanger,\n",
"#since R=(T1-T2)/(T4-T3)=8.412 >1, P=(T4-T3)/(T1-T2)=0.114,we can get this value of F by using value of P =R*0.114\n",
"\n",
"#Calculations\n",
"A=(m*c*(T1-T2))/(U*F*((T1-T4-T2+T3)/math.log((T1-T4)/(T2-T3))));#Area for heat exchanger, m^2.\n",
"\n",
"#Results\n",
"print \"Area for heat exchanger is :\",round(A,3),\"m^2\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Area for heat exchanger is : 121.216 m^2\n",
"\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.5, Page number: 112"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=313; #entering temperature of cold water, K\n",
"T2=423; #Entering temperature of hot water, K\n",
"Cc=20000; #heat capacity of cold water, W/K\n",
"Ch=10000; #heat capacity of hot water, W/K\n",
"A=30; #area, m**2\n",
"U=500; #overall heat transfer coefficient, w/(m**2*K)\n",
"e=0.596; #no. of transfer units(NTU)=(U*A)/Ch=1.5, the effectiveness of heat exchanger e can be found by using this value of NTU\n",
"\n",
"#Calculations\n",
"Q=e*Ch*(T2-T1);\t\t\t\t\t\t\t\t#Heat transfer, W\n",
"Q1=Q/1000\t\t\t\t\t \t\t\t#Heat transfer, KW\n",
"Texh=T2-Q/Ch;\t\t\t\t\t\t\t\t#exit hot water temperature, K \n",
"Tn1=Texh-273;\t\t\t\t\t\t\t\t#exit hot water temperature, C\n",
"Texc=T1+Q/Cc\t\t\t\t\t\t\t\t#exit cold water temperature, K\n",
"Tn2=Texc-273;\t\t\t\t\t\t\t\t#exit cold water temperature, C\n",
"\n",
"#Results\n",
"print \"Heat transfer is :\",Q1,\"KW\\n\"\n",
"print \"The exit hot water temperature is:\",Tn1,\"C\\n\"\n",
"print \"The exit cold water temperature is :\",Tn2,\"C\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Heat transfer is : 655.6 KW\n",
"\n",
"The exit hot water temperature is: 84.44 C\n",
"\n",
"The exit cold water temperature is : 72.78 C\n",
"\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 3.6, Page number: 123"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"from __future__ import division\n",
"import math\n",
"\n",
"#Variables\n",
"T1=313; #entering temperature of cold water, K\n",
"T2=423; #Entering temperature of hot water, K\n",
"T3=363; #Exit temperature of hot water, K\n",
"Cc=20000; #heat capacity of cold water, W/K\n",
"Ch=10000; #heat capacity of hot water, W/K\n",
"U=500; #overall heat transfer coefficient, w/(m**2*K)\n",
"\n",
"#Calculations\n",
"T4=T1+(Ch/Cc)*(T2-T3);\t\t \t\t #Exit cold fluid temp. K\n",
"\n",
"e=(T2-T3)/(T2-T1);\t\t\t \t #Effectiveness method\n",
"NTU=1.15;\t\t\t\t\t #No. of transfer unit\n",
"A1=Ch*(NTU)/U; # since NTU=1.15=U*A/Ch, Area can be found by using this formula\n",
"#another way to calculate the area is by using log mean diameter method\n",
"LMTD=(T2-T1-T3+T4)/math.log((T2-T1)/(T3-T4)); #Logarithmic mean temp. difference\n",
"A2=Ch*(T2-T3)/(U*LMTD);\t\t\t\t #Aera by method 2, in meters^2.\n",
"\n",
"#Results\n",
"print \"Area is :\",A1,\"m^2\\n\"\n",
"print \"Area is :\",round(A2,3),\"m^2\\n\"\n",
"print \"There is difference of 1 percent in answers which reflects graph reading inaccuracy.\"\n",
"# we can see that area calulated is same in above 2 methods.\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Area is : 23.0 m^2\n",
"\n",
"Area is : 22.73 m^2\n",
"\n",
"There is difference of 1 percent in answers which reflects graph reading inaccuracy.\n"
]
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}
|
