{ "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": {} } ] }