From fffcc90da91b66ee607066d410b57f34024bd1de Mon Sep 17 00:00:00 2001 From: Jovina Dsouza Date: Mon, 7 Jul 2014 16:34:28 +0530 Subject: adding book --- A_Heat_Transfer_Text_Book/Chapter3.ipynb | 227 +++++++++++++++++++++++++++++++ 1 file changed, 227 insertions(+) create mode 100755 A_Heat_Transfer_Text_Book/Chapter3.ipynb (limited to 'A_Heat_Transfer_Text_Book/Chapter3.ipynb') diff --git a/A_Heat_Transfer_Text_Book/Chapter3.ipynb b/A_Heat_Transfer_Text_Book/Chapter3.ipynb new file mode 100755 index 00000000..6791eb70 --- /dev/null +++ b/A_Heat_Transfer_Text_Book/Chapter3.ipynb @@ -0,0 +1,227 @@ +{ + "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": {} + } + ] +} \ No newline at end of file -- cgit