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diff --git a/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb new file mode 100755 index 00000000..b466521e --- /dev/null +++ b/Chemical_Engineering_Thermodynamics___by_S._Sundaram/ch12.ipynb @@ -0,0 +1,218 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:abd9968b31b6fc770d85971aacc78f09c6ba4e5596d9831c5736d23440171297" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 12 : Refrigeration" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 Page No : 216" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "m = 500.0;#mass of ice produced per hour in Kg\n", + "T1 = 15.0;#Initial temperature of water\n", + "T2 = -5.0;#Final temperature of ice\n", + "Ci = 0.5;#Specific heat of ice in Kcal/Kg deg celsius\n", + "Cw = 1.0;#Specific heat of water in Kcal/Kg deg celsius\n", + "L_f = 79.71;#Latent heat of fusion in Kcal/Kg\n", + "Tf = 0.0;#Frezzing point of ice in deg celsius\n", + "\n", + "#To Calculate the theoretical horse power required\n", + "Q2 = m*(Cw*(T1-Tf)+L_f+Ci*(Tf-T2));#Heat to be extracted per hour in Kcal\n", + "#From equation 12.1 (page no 220)\n", + "COP = (T2+273)/((T1+273)-(T2+273));\n", + "W = Q2/COP;#Work in Kcal/hr\n", + "W1 = W*(427/(60*4500.0));\n", + "print \"The therotical horse power required is %f hp\"%(W1);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The therotical horse power required is 5.736411 hp\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 Page No : 217" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "from scipy.integrate import quad\n", + "\n", + "#Given\n", + "Ta = 298.0;#Initial temperature in K\n", + "Tb = 203.0;#Final temperature in k\n", + "T1 = 298.0;#Water temperature in K\n", + "n = 1.0;#Kgmole of CO2\n", + "#Cp = 5.89+0.0112T ; Specific heat of CO2 in Kcal/Kgmole K\n", + "\n", + "#To Calculate the compressor load\n", + "#From equation 12.2a and b (page no 221)\n", + "def f(T):\n", + " y = ((T1-T)/T)*n*(5.89+0.0112*T);\n", + " return y\n", + "W = quad(f,Ta,Tb)[0];\n", + "print \"The compressor load is %f Kcal/Kgmole\"%(W);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The compressor load is -164.797031 Kcal/Kgmole\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 Page No : 221" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Given\n", + "#Consider the figure 12.4 (page no 226) \n", + "m = 5.0;#tonnes of refrigeration\n", + "T1 = 273-10.0;#temperature of the saturated vapour in K\n", + "T2 = 273+35.0;#temperature of the super heated vapour in K\n", + "T3 = 273+25.0;#temperature of the saturated liquid in K\n", + "T4 = 273+25.0;#temperature of the wet vapour in K\n", + "H1 = 341.8;#enthalpy of the saturated vapour in Kcal/Kg\n", + "H2 = 409.0;#enthalpy of the super heated vapour in Kcal/Kg\n", + "H3 = 350.0;#enthalpy of the saturated liquid in Kcal/Kg\n", + "H4 = 71.3;#enthalpy of the wet vapour in Kcal/Kg\n", + "\n", + "#To Calculate the C.O.P, mass of refrigerant required, compressor horse power required and the C.O.P & compressor horse power for a reversed Carnot cycle\n", + "#(i)Calculation of the C.O.P of the compression cycle\n", + "#From equation 12.6 (page no 226)\n", + "COP = (H1-H4)/(H2-H1);\n", + "print \"i)C.O.P of the compression cycle is %f\"%(COP);\n", + "\n", + "#(ii)Calculation of mass of refrigerant required\n", + "#From equation 12.7 (page no 226)\n", + "M = (m*50.4)/(H1-H4);\n", + "print \" ii)The mass of refrigerant required is %f Kg/mt\"%(M);\n", + "\n", + "#(iii)Calculation of the compressor horse power\n", + "#From equation 12.5 (page no 226)\n", + "C_hp = (H2-H1)*M*(427/4500.0);\n", + "print \" iii)The compressor horse power is %f hp\"%(C_hp);\n", + "\n", + "#(iv)Calculation for reversed Carnot cycle\n", + "#From equation 12.1 (page no 220)\n", + "COP = T1/(T3-T1);\n", + "print \" iv)C.O.P for the reversed Carnot cycle is %f\"%(COP);\n", + "Q2 = m*50.4*(427/4500.0);#in hp\n", + "C_hp = Q2/COP\n", + "print \" Compressor horse power for the reversed Carnot cycle is %f hp\"%(C_hp);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i)C.O.P of the compression cycle is 4.025298\n", + " ii)The mass of refrigerant required is 0.931608 Kg/mt\n", + " iii)The compressor horse power is 5.940430 hp\n", + " iv)C.O.P for the reversed Carnot cycle is 7.514286\n", + " Compressor horse power for the reversed Carnot cycle is 3.182205 hp\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 Page No : 225" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "#Water at 20 deg cel is chilled to 10 deg cel by flash evaporation\n", + "Pv = 0.012;#Vapour pressure of water at 10 deg celsius in Kgf/sq.cm\n", + "H1 = 20.03;#Enthalpy of liquid water at 20 deg cel in Kcal/Kg\n", + "H2 = 10.4;#Enthalpy of liquid water at 10 deg cel in Kcal/Kg\n", + "Hv = 601.6;#Enthalpy of saturated vapour at 10 deg cel in Kcal/kg\n", + "\n", + "#To calculate the pressure in the math.tank and the amount of make up water required\n", + "P = Pv;#pressure in the math.tank = vapour pressure of water\n", + "print \"The pressure in the math.tank is %f Kgf/sq.cm\"%(P);\n", + "#From equation 12.8 (page no 234)\n", + "x = (H1-H2)/(Hv-H2);\n", + "print \" The amount of make up water required is %f Kg\"%(x);\n", + "#end\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pressure in the math.tank is 0.012000 Kgf/sq.cm\n", + " The amount of make up water required is 0.016289 Kg\n" + ] + } + ], + "prompt_number": 5 + } + ], + "metadata": {} + } + ] +}
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