diff options
author | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
---|---|---|
committer | Jovina Dsouza | 2014-07-07 16:34:28 +0530 |
commit | fffcc90da91b66ee607066d410b57f34024bd1de (patch) | |
tree | 7b8011d61013305e0bf7794a275706abd1fdb0d3 /A_Heat_Transfer_Text_Book/Chapter3.ipynb | |
parent | 299711403e92ffa94a643fbd960c6f879639302c (diff) | |
download | Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.gz Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.tar.bz2 Python-Textbook-Companions-fffcc90da91b66ee607066d410b57f34024bd1de.zip |
adding book
Diffstat (limited to 'A_Heat_Transfer_Text_Book/Chapter3.ipynb')
-rwxr-xr-x | A_Heat_Transfer_Text_Book/Chapter3.ipynb | 227 |
1 files changed, 227 insertions, 0 deletions
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 |