{ "metadata": { "name": "", "signature": "sha256:6310b6690e48041fd88dd8277353623ea538798dfeeeb9a3819409de065b2175" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter10:IRREVERSIBILITY AND AVAILABILITY" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex10.1:pg-386" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#example 1\n", "#Calculating reversible work\n", "\n", "#Form the Steam Tables,the inlet and the exit state properties are \n", "hi=171.95 #initial specific heat of enthalpy in kJ/kg\n", "si=0.5705 #initial specific entropy in kJ/kg-K\n", "se=2.1341 #final specific entropy in kJ/kg-K\n", "he=765.34 #final specific heat of enthalpy in kJ/kg-K\n", "m=5 #mass flow rate of feedwater in kg/s\n", "q1=900/m #heat added by one of the sources in kJ/kg\n", "q2=he-hi-q1 #second heat transfer in kJ/kg\n", "To=25+273.3 #Temp. of the surroundings in K\n", "T1=100+273.2 #temp. of reservoir of one of the source in K\n", "T2=200+273.2 #temp. of reservoir of second source in K\n", "wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg\n", "print\"\\n Hence, the irreversibility is\",round(wrev,1),\"kJ/kg\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Hence, the irreversibility is 62.0 kJ/kg\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex10.2:pg-387" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#example 2\n", "#Calculating reversible work\n", "import math\n", "#Form the Steam Tables,the inlet and the exit state properties are\n", "hi=298.6 #initial specific heat of enthalpy in kJ/kg\n", "si=6.8631 #initial specific entropy in kJ/kg-K\n", "se=7.4664 #final specific entropy in kJ/kg-K\n", "he=544.7 #final specific heat of enthalpy in kJ/kg-K\n", "q=-50 #heat lost to surroundings in kJ/kg\n", "w=hi-he+q #work in kJ/kg\n", "To=25+273.2 #Temp. of the surroundings in K\n", "P1=100 #Pressure of ambient air in kPa\n", "P2=1000 #Final pressure of air after compression in kPa\n", "R=0.287 #Universal gas constant in kJ/kg-K\n", "wrev=To*(se-si-R*math.log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg\n", "i=wrev-w #irreversibility in kJ/kg\n", "print\"\\n Hence, the irreversibility is\",round(i,1),\"kJ/kg\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Hence, the irreversibility is 32.8 kJ/kg\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex10.3:pg-390" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#example 3\n", "#Calculating reversible work and irreversibility\n", "\n", "#Form the Steam Tables at state 1\n", "u1=1243.5 #initial specific internal energy in kJ/kg\n", "s1=4.4819 #initial specific entropy in kJ/kg-K\n", "v1=28.895 #initial specific volume in m^3/kg\n", "v2=2*v1 #final specific volume in kg/m^3\n", "u2=u1 #initial specific internal energy in kJ/kg\n", "#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg\n", "T2=9.1+273.2 #final temp. in K\n", "x2=0.513 #quality in final state\n", "s2=4.644 #final specific entropy in kJ/kg\n", "V1=1 #volume of part of A in m^3\n", "m=V1/v1 #mass flow rate in kg/s\n", "To=20+273.2 #Room temperature in K\n", "Wrev=To*m*(s2-s1) #reversible work in kJ\n", "I=Wrev #irreversibility of the process\n", "print\"\\n The irreversibility is \",round(I,3),\"kJ\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " The irreversibility is 1.645 kJ\n" ] } ], "prompt_number": 5 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }