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diff --git a/Textbook_Of_Heat_Transfer/Chapter_7_Heat_Exchangers.ipynb b/Textbook_Of_Heat_Transfer/Chapter_7_Heat_Exchangers.ipynb new file mode 100755 index 00000000..9fec8619 --- /dev/null +++ b/Textbook_Of_Heat_Transfer/Chapter_7_Heat_Exchangers.ipynb @@ -0,0 +1,440 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c8c0ee88164cfcd21e5e0ef3ff65a90467a8a7878be11fb5b83841227f683525"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 7: Heat Exchangers"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.1, Page no:285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "h = 2000 ; #W/m^2 K\n",
+ "#From Table 7.1\n",
+ "Uf = 0.0001 ; #fouling factor, m^2K/W\n",
+ "\n",
+ "#calculations\n",
+ "hf = 1/(1/ h+ Uf );\n",
+ "p = (h-hf)/h *100;\n",
+ "\n",
+ "#result\n",
+ "print\"Heat transfer coefficient including the effect of foulung =\",round(hf,4),\"W/m^2 K\";\n",
+ "print\"Percentage reduction =\",round(p,4);"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Heat transfer coefficient including the effect of foulung = 1666.6667 W/m^2 K\n",
+ "Percentage reduction = 16.6667\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.2, Page no:294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "m = 1000 ; #kg/h\n",
+ "Thi = 50 ; #C\n",
+ "The = 40 ; #C\n",
+ "Tci = 35 ; #C\n",
+ "Tce = 40 ; #C\n",
+ "U = 1000 ; #OHTC, W/m^2 K\n",
+ "#From fig 7.15,\n",
+ "F =0.91 ;\n",
+ "#Again from fig 7.15,\n",
+ "F =0.91 ;\n",
+ "\n",
+ "#calculations\n",
+ "#Using Eqn 7.5.25\n",
+ "q = m /3600*4174*( Thi - The ) ; #W\n",
+ "deltaT = (( Thi - Tce ) -(The - Tci ))/math.log (( Thi -Tce)/( The -Tci )); #C\n",
+ "#T1 = Th and T2 = Tc\n",
+ "R = (Thi - The )/( Tce - Tci ) ;\n",
+ "S = (Tce - Tci )/( Thi - Tci ) ;\n",
+ "#Alternatively, taking T1 = Tc and T2 = Th\n",
+ "R2 = (Tci - Tce )/( The - Thi );\n",
+ "S2 = (The - Thi )/( Tci - Thi );\n",
+ "A = q/(U*F* deltaT );\n",
+ "\n",
+ "#result\n",
+ "print\"delta T =\",round(deltaT,4);\n",
+ "print\"\\nTaking T1 = Th and T2 = Tc\";\n",
+ "print\"R=\",round(R,4), \"S=\",round(S,4);\n",
+ "print\"Hence, F =\",round(F,4);\n",
+ "print\"\\nTaking T1 = Tc and T2 = Th\";\n",
+ "print\"R2=\",round(R2,4),\"S2=\",round(S2,4);\n",
+ "print\"Hence, F =\",round(F,4);\n",
+ "print\"\\nArea =\",round(A,4),\"m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "delta T = 7.2135\n",
+ "\n",
+ "Taking T1 = Th and T2 = Tc\n",
+ "R= 2.0 S= 0.3333\n",
+ "Hence, F = 0.91\n",
+ "\n",
+ "Taking T1 = Tc and T2 = Th\n",
+ "R2= 0.5 S2= 0.6667\n",
+ "Hence, F = 0.91\n",
+ "\n",
+ "Area = 1.7663 m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.3, Page no:295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "#Because of change of phase , Thi = The\n",
+ "Thi = 100 ; #[C], Saturated steam\n",
+ "The = 100 ; #[C], Condensed steam\n",
+ "Tci = 30 ; #[C], Cooling water inlet\n",
+ "Tce = 70 ; #[C], cooling water outlet\n",
+ "#From fig 7.16\n",
+ "F = 1;\n",
+ "\n",
+ "#calculations\n",
+ "R = (Thi - The )/( Tce - Tci ) ;\n",
+ "S = (Tce - Tci )/( Thi - Tci ) ;\n",
+ "#For counter flow arrangement\n",
+ "Tmcounter = (( Thi - Tce ) -(The - Tci ))/math.log (( Thi - Tce )/(The - Tci )); #For counter flow arrangement\n",
+ "#Therefore\n",
+ "Tm = F* Tmcounter ;\n",
+ "\n",
+ "#result\n",
+ "print\"Mean Temperaature Difference =\",round(Tm,4),\"C\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mean Temperaature Difference = 47.2089 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4(a), Page no:302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "#Using Mean Temperature Difference approach\n",
+ "mhot = 10 ; #kg/min\n",
+ "mcold = 25 ; #kg/min\n",
+ "hh = 1600 ; #[W/m^2 K], Heat transfer coefficient on hot side\n",
+ "hc = 1600 ; #[W/m^2 K], Heat transfer coefficient on cold side\n",
+ "Thi = 70 ; #C\n",
+ "Tci = 25 ; #C\n",
+ "The = 50 ; #C\n",
+ "\n",
+ "#calculations\n",
+ "#Heat Transfer Rate, q\n",
+ "q = mhot /60*4179*( Thi - The ); #W\n",
+ "#Heat gained by cold water = heat lost by the hot water\n",
+ "Tce = 25 + q *1/( mcold /60*4174) ; #C\n",
+ "#Using equation 7.5.13\n",
+ "Tm = (( Thi - Tci ) -(The - Tce ))/math.log (( Thi -Tci)/( The -Tce)); #C\n",
+ "U = 1/(1/ hh + 1/ hc); #W/m^2 K\n",
+ "A = q/(U*Tm); #Area, [m^2]\n",
+ "\n",
+ "#result\n",
+ "print\"(a)\";\n",
+ "print\"Mean Temperature Difference =\",round(Tm,4),\"C\";\n",
+ "print\"Area of Heat Exchanger =\",round(A,4),\"m^2\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(a)\n",
+ "Mean Temperature Difference = 28.7569 C\n",
+ "Area of Heat Exchanger = 0.6055 m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.4(b) , Page no:302"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "#Using Mean Temperature Difference approach\n",
+ "mhot = 10 ; #kg/min\n",
+ "mcold = 25 ; #kg/min\n",
+ "hh = 1600 ; #[W/m^2 K], Heat transfer coefficient on hot side\n",
+ "hc = 1600 ; #[W/m^2 K], Heat transfer coefficient on cold side\n",
+ "Thi = 70 ; #C\n",
+ "Tci = 25 ; #C\n",
+ "The = 50 ; #C\n",
+ "mhot1 = 20 ; #kg/min\n",
+ "\n",
+ "#calculations\n",
+ "#Heat Transfer Rate, q\n",
+ "q = mhot/60*4179*( Thi - The ); #W\n",
+ "#Heat gained by cold water = heat lost by the hot water\n",
+ "Tce = 25 + q *1/( mcold /60*4174) ;\n",
+ "# Using equation 7.5.13\n",
+ "Tm = ((( Thi - Tci ) -(The - Tce ))/math.log (( Thi -Tci)/( The -Tce))); #C\n",
+ "U = 1/(1/ hh + 1/ hc); #W/m^2 K\n",
+ "A = q/(U*Tm); #Area, [m^2]\n",
+ "#Flow rate on hot side i.e. 'hh' is doubled\n",
+ "hh1 = 1600*2**0.8 ; #W/m^2 K\n",
+ "U1 = 1/(1/ hh1 + 1/ hc); #W/m^2 K\n",
+ "mhCph = mhot1 /60*4179 ; #W/K\n",
+ "mcCpc = mcold /60*4174 ; #W/K\n",
+ "#Therefore\n",
+ "C = mhCph / mcCpc ;\n",
+ "ntu = U1*A/ mhCph ;\n",
+ "e = (1 - math.exp ( -(1+C)*ntu) )/(1+ C) ;\n",
+ "#Therefore (Thi - The)/(Thi - Tci) = e , we get\n",
+ "The = Thi - e*( Thi - Tci );\n",
+ "#Equating the heat lost by water to heat gained by cold water , we get\n",
+ "Tce = Tci + ( mhCph *( Thi - The ))/ mcCpc ;\n",
+ "\n",
+ "#result\n",
+ "print\"(b)\\nNTU =\",round(ntu,4);\n",
+ "print\"Exit temperature of cold and hot stream are\",round(Tce,4),\"C and\",round(The,4),\"C respectively.\";"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(b)\n",
+ "NTU = 0.4418\n",
+ "Exit temperature of cold and hot stream are 35.981 C and 56.2901 C respectively.\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 7.5 , Page no:304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from __future__ import division\n",
+ "\n",
+ "#Variable declaration\n",
+ "mc = 2000 ; # [kg/h]\n",
+ "Tce = 40 ; # [C]\n",
+ "Tci = 15 ; # [C]\n",
+ "Thi = 80 ; # [C]\n",
+ "U = 50 ; # OHTC, [W/m**2 K]\n",
+ "A = 10 ; # Area, [m**2]\n",
+ "\n",
+ "#Calculations\n",
+ "# Using effective NTU method\n",
+ "# Assuming m_c*C_pc = (m*C_p)s\n",
+ "NTU = U*A/(mc*1005/3600);\n",
+ "e = (Tce-Tci)/(Thi-Tci);\n",
+ "# From fig 7.23, no value of C is found corresponding to the above values, hence assumption was wrong.\n",
+ "# So, m_h*C_ph must be equal to (m*C_p)s, proceeding by trail and error method\n",
+ "\n",
+ "mh_1 = 200\n",
+ "NTU_1 = U*A/(mh_1*1.161);\n",
+ "#Corresponding Values of C and e from fig 7.23\n",
+ "C = .416;\n",
+ "e = .78;\n",
+ "#From Equation 7.6.2 Page 297\n",
+ "The_1 = Thi - e*(Thi-Tci)\n",
+ "#From Heat Balance\n",
+ "The2_1 = Thi - mc*1005/3600*(Tce-Tci)/(mh_1*1.161);\n",
+ "\n",
+ "mh_2 = 250\n",
+ "NTU_2 = U*A/(mh_2*1.161);\n",
+ "#Corresponding Values of C and e from fig 7.23\n",
+ "C = .520;\n",
+ "e = .69;\n",
+ "#From Equation 7.6.2 Page 297\n",
+ "The_2 = Thi - e*(Thi-Tci)\n",
+ "#From Heat Balance\n",
+ "The2_2 = Thi - mc*1005/3600*(Tce-Tci)/(mh_2*1.161);\n",
+ "\n",
+ "mh_3 = 300\n",
+ "NTU_3 = U*A/(mh_3*1.161);\n",
+ "#Corresponding Values of C and e from fig 7.23\n",
+ "C = .624;\n",
+ "e = .625;\n",
+ "#From Equation 7.6.2 Page 297\n",
+ "The_3 = Thi - e*(Thi-Tci)\n",
+ "#From Heat Balance\n",
+ "The2_3 = Thi - mc*1005/3600*(Tce-Tci)/(mh_3*1.161);\n",
+ "\n",
+ "mh_4 = 350\n",
+ "NTU_4 = U*A/(mh_4*1.161);#Corresponding Values of C and e from fig 7.23\n",
+ "C = .728;\n",
+ "e = .57;\n",
+ "#From Equation 7.6.2 Page 297\n",
+ "The_4 = Thi - e*(Thi-Tci)\n",
+ "#From Heat Balance\n",
+ "The2_4 = Thi - mc*1005/3600*(Tce-Tci)/(mh_4*1.161);\n",
+ "\n",
+ "mh_5 = 400\n",
+ "NTU_5 = U*A/(mh_5*1.161);\n",
+ "#Corresponding Values of C and e from fig 7.23\n",
+ "C = .832;\n",
+ "e = .51;\n",
+ "#From Equation 7.6.2 Page 297\n",
+ "The_5 = Thi - e*(Thi-Tci)\n",
+ "#From Heat Balance\n",
+ "The2_5 = Thi - mc*1005/3600*(Tce-Tci)/(mh_5*1.161);\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print (\"m_h(kg/h) \\t NTU \\t\\t C \\t\\t e \\t T_he(C) \\t\\t T_he(C)(Heat Balance)\");\n",
+ "print mh_1,\"\\t\\t\",round(NTU_1,3),\"\\t\\t\",round(C,3),\"\\t\\t\",round(e,2),\"\\t\\t \",round(The_1,1),\"\\t\\t\\t\",round(The2_1,1);\n",
+ "print mh_2,\"\\t\\t\",round(NTU_2,3),\"\\t\\t\",round(C,3),\"\\t\\t\",round(e,2),\"\\t\\t \",round(The_2,1),\"\\t\\t\\t\",round(The2_2,1);\n",
+ "print mh_3,\"\\t\\t\",round(NTU_3,3),\"\\t\\t\",round(C,3),\"\\t\\t\",round(e,2),\"\\t\\t \",round(The_3,1),\"\\t\\t\\t\",round(The2_3,1);\n",
+ "print mh_4,\"\\t\\t\",round(NTU_4,3),\"\\t\\t\",round(C,3),\"\\t\\t\",round(e,2),\"\\t\\t \",round(The_4,1),\"\\t\\t\\t\",round(The2_4,1);\n",
+ "print mh_5,\"\\t\\t\",round(NTU_5,3),\"\\t\\t\",round(C,3),\"\\t\\t\",round(e,2),\"\\t\\t \",round(The_5,1),\"\\t\\t\\t\",round(The2_5,1);\n",
+ "\n",
+ "#Graph\n",
+ "mh=[200,250,300,350,400];\n",
+ "The=[29.3,35.2,39.4,43,46.9];\n",
+ "The2=[19.9,31.9,39.9,45.7,50];\n",
+ "\n",
+ "import matplotlib.pyplot as plt\n",
+ "import numpy as np\n",
+ "\n",
+ "from pylab import *\n",
+ "figure(1); # For Plotting y-x Diagram\n",
+ "%matplotlib inline\n",
+ "\n",
+ "plt.plot (mh,The,'b',mh,The2,'r',[295,295,200],[0,39.2,39.2],'k');\n",
+ "plt.title (\"mh vs The\");\n",
+ "plt.xlabel(\" mh \");\n",
+ "plt.ylabel(\" The \");"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "m_h(kg/h) \t NTU \t\t C \t\t e \t T_he(C) \t\t T_he(C)(Heat Balance)\n",
+ "200 \t\t2.153 \t\t0.832 \t\t0.51 \t\t 29.3 \t\t\t19.9\n",
+ "250 \t\t1.723 \t\t0.832 \t\t0.51 \t\t 35.2 \t\t\t31.9\n",
+ "300 \t\t1.436 \t\t0.832 \t\t0.51 \t\t 39.4 \t\t\t39.9\n",
+ "350 \t\t1.23 \t\t0.832 \t\t0.51 \t\t 43.0 \t\t\t45.6\n",
+ "400 \t\t1.077 \t\t0.832 \t\t0.51 \t\t 46.9 \t\t\t49.9\n"
+ ]
+ },
+ {
+ "metadata": {},
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x4f73e30>"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |