{ "metadata": { "name": "", "signature": "sha256:a053f39727ace41fe9dd23fd039f18e2237a3b4848b2ddbeb10075aa5411107c" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 10 : Compressor" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.1 Page No : 168" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#Given\n", "V1 = 2.7;#flow rate of CO2 in cubic meter/min\n", "T1 = 273-51;#temperature in K\n", "P1 = 1.0;#initial pressure in Kgf/sq cm\n", "P2 = 10.0;#final pressure in Kgf/sq cm\n", "y = 1.3;#gamma\n", "v1 = 0.41;#specific volume in cubic meter/Kg\n", "H1 = 158.7;# initial enthalpy in Kcal/Kg\n", "H2 = 188.7;#final enthalpy in Kcal/Kg\n", "\n", "#process is isentropic\n", "#To calculate the horsepower required\n", "\n", "#(i)Assuming ideal gas behaviour\n", "#From equation 10.3 (page no 189)\n", "W = (y/(y-1))*(P1*1.03*10**4*V1)*(1-(P2/P1)**((y-1)/y));#work in m Kgf/min\n", "W1 = W/4500.0;\n", "print \"i)The horsepower required is %f hp\"%(W1);\n", "\n", "#(ii)Umath.sing the given data for CO2\n", "#From equation 10.2 (page no 189)\n", "W = -(H2 - H1);#work in Kcal/Kg\n", "M = V1/v1;#Mass rate of gas in Kg/min\n", "W1 = W*M*(427/4500.0);\n", "print \" ii)Compressor work is %f hp\"%(W1);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)The horsepower required is -18.779590 hp\n", " ii)Compressor work is -18.746341 hp\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.2 Page No : 171" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "P1 = 1.0;#Initial pressure in atm\n", "P2 = 29.0;#Final pressure in atm\n", "C = 0.05;#Clearance\n", "y = 1.4;#gamma of air\n", "\n", "#To calculate the volumetric efficiency and the maximum possible pressure that can be attained in a math.single stage\n", "#(i)Calulation of volumetric efficiency\n", "#From equation 10.11 (page no 194)\n", "V_E = 1+C-C*(P2/P1)**(1/y);\n", "print \"i)Volumetric efficiency is %f percent\"%(V_E*100);\n", "\n", "#(ii)Calculation of maximum pressure \n", "V_E = 0;#Minimum efficiency\n", "P2 = P1*(((1+C-V_E)/C)**y);\n", "print \" ii)The maximum possible pressure attained is %f atm\"%(P2);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i)Volumetric efficiency is 49.596143 percent\n", " ii)The maximum possible pressure attained is 70.975263 atm\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.3 Page No : 174" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "V_d = 5.15;#print lacement volume in cubic meter/min\n", "P1 = 1.0;#initial pressure in Kgf/sq cm\n", "P2 = 8.5;#final pressure in Kgf/sq cm\n", "C = 0.06;#Clearance\n", "M_E = 0.8;#Mechenical efficiency\n", "y = 1.31;#gamma\n", "\n", "#To calculate the capacity and the actual horse power of the compressor\n", "v1 = V_d*(1+C-(C*((P2/P1)**(1/y))));\n", "print \"The capacity of the copressor is %f cubic meter/min\"%(v1);\n", "#From equation 10.6 (page no 192)\n", "W = (y/(y-1))*(P1*1*10**4*v1)*(1-(P2/P1)**((y-1)/y));#work in Kgf/min\n", "W1 = W/4500.0;#work in hp\n", "W2 = W1/M_E;\n", "print \" The actual horse power of the compressor is %f hp\"%(W2);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The capacity of the copressor is 3.876154 cubic meter/min\n", " The actual horse power of the compressor is -30.000346 hp\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.4 Page No : 177" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "P1 = 1.0;#Initial pressure in Kgf/sq cm\n", "Pn = 13.0;#Final pressure in Kgf/sq cm\n", "V1 =27.0;#flow rate of gas in cubic meter/min\n", "y = 1.6;#gamma of the gas\n", "n = [1.0,2.0,3.0,4.0,7.0,10.0];#number of stages\n", "print \"No of stages Horse power in hp\";\n", "#To Calculate the theoretical horse power required\n", "W = []\n", "for i in range(0,6):\n", " W.append(n[i]*(y/(y-1))*((P1*10**4)/4500)*V1*(1-(Pn/P1)**((y-1)/(n[i]*y))));\n", " print \" %d\"%(n[i]),\n", " print \" %f\"%(-1*W[i])\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "No of stages Horse power in hp\n", " 1 258.647729\n", " 2 197.623943\n", " 3 181.430407\n", " 4 173.977056\n", " 7 164.971690\n", " 10 161.541416\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 10.5 Page No : 180" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "P1 = 1.0;#Initial pressure in Kgf/sq cm\n", "P4 = 200.0;#Final pressure in Kgf/sq cm\n", "n = 4.0;#no of stages\n", "\n", "#To find out the presure between stages\n", "r = (P4/P1)**(1/n);#Compression ratio\n", "P2 = r*P1;\n", "print \"The pressure after 1st stage is %f Kgf/sq cm\"%(P2);\n", "P3 = r*P2;\n", "print \" The pressure after 2nd stage is %f Kgf/sq cm\"%(P3);\n", "P4 = r*P3;\n", "print \" The pressure after 3rd stage is %f Kgf/sq cm\"%(P4);\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The pressure after 1st stage is 3.760603 Kgf/sq cm\n", " The pressure after 2nd stage is 14.142136 Kgf/sq cm\n", " The pressure after 3rd stage is 53.182959 Kgf/sq cm\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }