{ "metadata": { "name": "Chapter 9" }, "nbformat": 2, "worksheets": [ { "cells": [ { "cell_type": "markdown", "source": [ "# Chapter 9 :- Gas Power Systems" ] }, { "cell_type": "markdown", "source": [ "## Example 9.1 Page no-378" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T1 = 300.00 # The temperature at the beginning of the compression process in kelvin", "p1 = 1.00 # the pressure at the beginning of the compression process in bar", "r = 8.00 # compression ratio", "V1 = 560.00 # the volume at the beginning of the compression process in cm^3", "T3 = 2000.00 # maximum temperature during the cycle in kelvin", "", "# Part(a)", "# At T1 = 300k,table A-22 gives", "u1 = 214.07 # in kj/kg", "vr1 = 621.2 ", "# Interpolating with vr2 in Table A-22, we get", "T2 = 673.00 # in kelvin", "u2 = 491.2 # in kj/kg", "# At T3 \u0003= 2000 K, Table A-22 gives", "u3 = 1678.7 # in kj/kg", "vr3 = 2.776", "# Interpolating in Table A-22 with vr4 gives", "T4 = 1043 # in kelvin", "u4 = 795.8 # in kj/kg", "", "# Calculations", "# For the isentropic compression Process 1\u20132", "vr2 = vr1/r", "# With the ideal gas equation of state", "p2 = p1*(T2/T1)*(r) # in bars", "# Since Process 2\u20133 occurs at constant volume, the ideal gas equation of state gives", "p3 = p2*(T3/T2) # in bars", "# For the isentropic expansion process 3\u20134", "vr4 = vr3*(r)", "# The ideal gas equation of state applied at states 1 and 4 gives", "p4 = p1*(T4/T1) # in bars", "", "# Results", "print '-> At state1, the pressure is:',p1,'bar.'", "print '-> At state1, the temperature is ',T1,'kelvin.'", "print '-> At state2, the pressure is:',round(p2,3),'bar.'", "print '-> At state2, the temperature is',T2,'kelvin.'", "print '-> At state3, the pressure is:',round(p3,3),'bar.'", "print '-> At state3, the temperature is',T3,'kelvin.'", "print '-> At state4, the pressure is:',round(p4,4),'bar.'", "print '-> At state4, the temperature is',T4,'kelvin.'", "", "# Part(b)", "eta = 1-(u4-u1)/(u3-u2) # thermal efficiency", "# Result", "print '-> The thermal efficiency is:',round(eta,2)", "", "# Part(c)", "R = 8.314 # universal gas constant, in SI units", "M = 28.97 # molar mass of air in grams", "# Calculations", "m = ((p1*V1)/((R/M)*T1))*10**-6*10**5*10**-3 # mass of the air in kg", "Wcycle = m*((u3-u4)-(u2-u1)) # the net work per cycle in KJ", "mep = (Wcycle/(V1*(1-1/r)))*10**6*10**3*10**-5 # in bars", "", "# Result", "print '-> The mean effective pressure, is:',round(mep,4),'atm..'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> At state1, the pressure is: 1.0 bar.", "-> At state1, the temperature is 300.0 kelvin.", "-> At state2, the pressure is: 17.947 bar.", "-> At state2, the temperature is 673.0 kelvin.", "-> At state3, the pressure is: 53.333 bar.", "-> At state3, the temperature is 2000.0 kelvin.", "-> At state4, the pressure is: 3.4767 bar.", "-> At state4, the temperature is 1043 kelvin.", "-> The thermal efficiency is: 0.51", "-> The mean effective pressure, is: 8.0411 atm.." ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "source": [ "## Example 9.2 Page no-383" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given :-", "r = 18.00 # compression ratio", "T1 = 300.00 # temperature at the beginning of the compression process in kelvin", "p1 = 0.1 # pressure at the beginning of the compression process in MPa", "rc = 2.00 # cutoff ratio", "", "# Part(a)", "# With T1 =\u0003 300 K, Table A-22 gives", "u1 = 214.07 # in kj/kg", "vr1 = 621.2 ", "# Interpolating in Table A-22, we get", "T2 = 898.3 # in kelvin ", "h2 = 930.98 # in kj/kg", "# From Table A-22,", "h3 = 1999.1 # in kj/kg", "vr3 = 3.97", "", "# Interpolating in Table A-22 with vr4, we get", "u4 = 664.3 # in kj/kg", "T4 = 887.7 # in kelvin", "", "# Calculations", "# Since Process 2\u20133 occurs at constant pressure, the ideal gas equation of state gives", "T3 = rc*T2 # in kelvin", "# With the ideal gas equation of state", "p2 = p1*(T2/T1)*(r) # in MPa", "p3 = p2", "# For the isentropic compression process 1\u20132", "vr2 = vr1/r", "# For the isentropic expansion process 3\u20134", "vr4 = (r/rc)*vr3", "# The ideal gas equation of state applied at states 1 and 4 gives", "p4 = p1*(T4/T1) # in MPa", "", "# Results", "print '-> At state1, the pressure is:',round(p1,2),'bar.'", "print '-> At state1, the temperature is',round(T1,2),'kelvin.'", "print '-> At state2, the pressure in bar is:',round(p2,2),'bar.'", "print '-> At state2, the temperature is ',round(T2,2),'kelvin.'", "print '-> At state3, the pressure in bar is:',round(p3,2),'bar.'", "print '-> At state3, the temperature is',round(T3,2),'kelvin.'", "print '-> At state4, the pressure is:',round(p4,2),'bar.'", "print '-> At state4, the temperature is',round(T4,2),'kelvin.'", "", "# Part(b)", "eta = 1- (u4-u1)/(h3-h2)", "print '-> The thermal efficiency is:',round(eta,2)", "", "# Part(c)", "R = 8.314 # universal gas constant, in SI units", "M = 28.97 # molar mass of air in grams", "", "# Calculations", "wcycle = (h3-h2)-(u4-u1) # The net work of the cycle in kj/kg", "v1 = ((R/M)*T1/p1)/10**3 # The specific volume at state 1 in m^3/kg", "mep = (wcycle/(v1*(1-1/r)))*10**3*10**-6 # in MPa", "", "# Results", "print '-> The mean effective pressure, is:',round(mep,2),'MPa.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> At state1, the pressure is: 0.1 bar.", "-> At state1, the temperature is 300.0 kelvin.", "-> At state2, the pressure in bar is: 5.39 bar.", "-> At state2, the temperature is 898.3 kelvin.", "-> At state3, the pressure in bar is: 5.39 bar.", "-> At state3, the temperature is 1796.6 kelvin.", "-> At state4, the pressure is: 0.3 bar.", "-> At state4, the temperature is 887.7 kelvin.", "-> The thermal efficiency is: 0.58", "-> The mean effective pressure, is: 0.76 MPa." ] } ], "prompt_number": 2 }, { "cell_type": "markdown", "source": [ "## Example 9.3 Page no-386" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given :-", "T1 = 300.00 # beginning temperature in kelvin", "p1 = 0.1 # beginning pressure in MPa", "r = 18.00 # compression ratio", "pr = 1.5 # The pressure ratio for the constant volume part of the heating process", "vr = 1.2 # The volume ratio for the constant pressure part of the heating process", "", "# Analysis", "# States 1 and 2 are the same as in Example 9.2, so ", "u1 = 214.07 # in kj/kg", "T2 = 898.3 # in kelvin", "u2 = 673.2 # in kj/kg", "", "# Interpolating in Table A-22, we get", "h3 = 1452.6 # in kj/kg", "u3 = 1065.8 # in kj/kg", "", "# From Table A-22,", "h4 = 1778.3 # in kj/kg", "vr4 = 5.609", "", "# Interpolating in Table A-22, we get", "u5 = 475.96 # in kj/kg", "", "# Calculations", "# Since Process 2\u20133 occurs at constant volume, the ideal gas equation of state reduces to give", "T3 = pr*T2 # in kelvin", "# Since Process 3\u20134 occurs at constant pressure, the ideal gas equation of state reduces to give", "T4 = vr*T3 # in kelvin", "# Process 4\u20135 is an isentropic expansion, so", "vr5 = vr4*r/vr", "", "# Part(a)", "eta = 1-(u5-u1)/((u3-u2)+(h4-h3))", "# Result", "print '-> The thermal efficiency is:',round(eta,2)", "", "# Part(b)", "# The specific volume at state 1 is evaluated in Example 9.2 as", "v1 = 0.861 # in m^3/kg", "mep = (((u3-u2)+(h4-h3)-(u5-u1))/(v1*(1-1/r)))*10**3*10**-6 # in MPa", "", "# Result", "print '-> The mean effective pressure, is:',round(mep,2),'MPa.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is: 0.64", "-> The mean effective pressure, is: 0.56 MPa." ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "source": [ "##Example 9.4 Page no-392" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T1 = 300.00 # in kelvin", "AV = 5.00 # volumetric flow rate in m^3/s", "p1 = 100.00 # in kpa", "pr = 10.00 # compressor pressure ratio", "T3 = 1400.00 # turbine inlet temperature in kelvin", "", "# Analysis", "# At state 1, the temperature is 300 K. From Table A-22,", "h1 = 300.19 # in kj/kg", "pr1 = 1.386", "", "", "# Interpolating in Table A-22,", "h2 = 579.9 # in kj/kg", "# From Table A-22", "h3 = 1515.4 # in kj/kg", "pr3 = 450.5", "", "# Interpolating in Table A-22, we get", "h4 = 808.5 # in kj/kg", "", "# calculations", "pr2 = pr*pr1", "pr4 = pr3*1/pr", "", "", "# Part(a)", "eta = ((h3-h4)-(h2-h1))/(h3-h2) # thermal efficiency", "# Result", "print '-> The thermal efficiency is:',round(eta,4)", "", "# Part(b)", "bwr = (h2-h1)/(h3-h4) # back work ratio", "# Result", "print '-> The back work ratio is:',round(bwr,4)", "", "# Part(c)", "R = 8.314 # universal gas constant, in SI units", "M = 28.97 # molar mass of air in grams", "# Calculations", "mdot = AV*p1/((R/M)*T1) # mass flow rate in kg/s", "Wcycledot = mdot*((h3-h4)-(h2-h1)) # The net power developed", "# Result", "print '-> The net power developed, is:',round(Wcycledot,2),'kW .'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is: 0.4566", "-> The back work ratio is: 0.3957", "-> The net power developed, is: 2480.89 kW ." ] } ], "prompt_number": 4 }, { "cell_type": "markdown", "source": [ "## Example 9.6 Page no-398" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "etat = 0.8 # turbine efficiency", "etac = 0.8 # compressor efficiency", "# Part(a)", "wtdots = 706.9 # The value of wtdots is determined in the solution to Example 9.4 as 706.9 kJ/kg", "wcdots = 279.7 # The value of wcdots is determined in the solution to Example 9.4 as 279.7 kJ/kg", "h1 = 300.19 # h1 is from the solution to Example 9.4, in kj/kg", "h3 = 1515.4 # h3 is from the solution to Example 9.4, in kj/kg", "", "# Calculations", "# The turbine work per unit of mass is", "wtdot = etat*wtdots # in kj/kg", "# For the compressor, the work per unit of mass is", "wcdot = wcdots/etac # in kj/kg", "h2 = h1 + wcdot # in kj/kg", "qindot = h3-h2 # The heat transfer to the working fluid per unit of mass flow in kj/kg", "eta = (wtdot-wcdot)/qindot # thermal efficiency", "", "# Results", "print '-> The thermal efficiency is:',round(eta,2)", "", "# Part(b)", "bwr = wcdot/wtdot # back work ratio", "# Result", "print '-> The back work ratio is:',round(bwr,2)", "", "# Part(c)", "mdot = 5.807 # in kg/s, from example 9.4", "Wcycledot = mdot*(wtdot-wcdot) # The net power developed by the cycle in kw", "# Result", "print '-> The net power developed, is: ',round(Wcycledot,2),'kW.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is: 0.25", "-> The back work ratio is: 0.62", "-> The net power developed, is: 1253.7 kW." ] } ], "prompt_number": 5 }, { "cell_type": "markdown", "source": [ "## Example 9.7 Page no-401" ] }, { "cell_type": "code", "collapsed": false, "input": [ "%matplotlib inline" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "", "Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].", "For more information, type 'help(pylab)'." ] } ], "prompt_number": 6 }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "# Part(a)", "etareg = 0.8 # regenerator effectiveness of 80%.", "# From example 9.4", "h1 = 300.19 # in kj/kg", "h2 = 579.9 # in kj/kg", "h3 = 1515.4 # in kj/kg", "h4 = 808.5 # in kj/kg", "", "# Calculations", "hx = etareg*(h4-h2)+h2 # in kj/kg", "eta = ((h3-h4)- (h2-h1))/(h3-hx) # thermal efficiency", "# Result", "print '-> The thermal efficiency is: '", "print round(eta,2)", "", "# Part(b)", "etareg = []", "x = []", "eta = []", "from numpy import linspace ", "from pylab import plot, show , xlabel, ylabel", "etareg = linspace(0,0.8,50)", "for i in range (0,50):", " x.append(i)", " eta.append(i)", " x[i] = (etareg[i]*(h4-h2))+h2 ", " eta[i] = ((h3-h4)- (h2-h1))/(h3-x[i])", "", "plot(etareg,eta)", "xlabel('Regenerator effectiveness')", "ylabel('Thermal efficiency')", "show()" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is: ", "0.57" ] }, { "output_type": "display_data", "png": 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rrwHt2gGffCKt/0REdDM1dl9aLBYUFxejpKSk2ldxcbGsiyclJaFLly7w8PDA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\n" } ], "prompt_number": 7 }, { "cell_type": "markdown", "source": [ "## Example 9.8 Page no-406" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "# Analysis", "# States 1, 2, and 3 are the same as in Example 9.4:", "h1 = 300.19 # in kj/kg", "h2 = 579.9 # in kj/kg", "h3 = 1515.4 # in kj/kg", "# The temperature at state b is the same as at state 3, so ", "hb = h3", "", "pa = 300.00 # in kpa", "p3 = 1000.00 # in kpa", "# From table A-22", "pr3 = 450.5", "", "# Interpolating in Table A-22, we get", "ha = 1095.9 # in kj/kg", "p4 = 100.00 # in kpa", "pb = 300.00 # in kpa", "# Interpolating in Table A-22, we obtain", "h4 = 1127.6 # in kj/kg", "", "# Calculions", "pra = pr3*(pa/p3)", "prb = pra", "pr4 = prb*(p4/pb)", "# Since the regenerator effectiveness is 100%,", "hx = h4", "eta = ((h3-ha)+(hb-h4)-(h2-h1))/((h3-hx)+(hb-ha)) # thermal efficiency", "", "# Result", "print '-> The thermal efficiency is:'", "print round(eta,2)" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is:", "0.65" ] } ], "prompt_number": 8 }, { "cell_type": "markdown", "source": [ "## Example 9.9 Page no-408" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T1 = 300.00 # in kelvin", "p1 = 100.00 # in kpa", "p2 = 1000.00 # in kpa", "p3 = p2", "pc = 300.00 # in kpa", "pd = 300.00 # in kpa", "Td = 300.00 # in kelvin", "", "", "# Part(a)", "# From table A-22", "prd = 1.386", "# Interpolating in Table A-22, we get", "T2 = 422 # in kelvin", "h2 = 423.8 # in kj/kg", "# Calculations", "pr2 = prd*(p2/pd)", "# Result", "print '-> The temperature at the exit of the second compressor stage is:',round(T2,2),'kelvin.'", "", "# Part(b)", "# From Table A-22 at T1 \u0003= 300", "h1 = 300.19 # in kj/kg", "# Since Td \u0003= T1,", "hd = 300.19 # in kj/kg", "# with pr data from Table A-22 together", "pr1 = 1.386", "# Interpolating in Table A-22, we obtain", "hc = 411.3 # in kj/kg", "# Calculations", "prc = pr1*(pc/p1)", "wcdot = (hc-h1)+(h2-hd) # The total compressor work per unit of mass in kj/kg", "# Result", "print '-> The total compressor work input per unit of mass flow is: ',round(wcdot,2),'kJ/kg'", "", "# Part(c)", "# Interpolating in Table A-22, we get", "T3 = 574 # in kelvin", "h3 = 579.9 # in kj/kg", "# Calculations", "pr3 = pr1*(p3/p1)", "wcdot = h3-h1 # The work input for a single stage of compression in kj/kg", "# Results", "print '-> For a single stage of compression, the temperature at the exit state is: ',round(T3,2),'kelvin'", "print '-> For a single stage of compression, the work input is: ',round(wcdot,2),'kJ.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The temperature at the exit of the second compressor stage is: 422.0 kelvin.", "-> The total compressor work input per unit of mass flow is: 234.72 kJ/kg", "-> For a single stage of compression, the temperature at the exit state is: 574.0 kelvin", "-> For a single stage of compression, the work input is: 279.71 kJ." ] } ], "prompt_number": 9 }, { "cell_type": "markdown", "source": [ "## Example 9.11 Page no-412" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T1 = 300.00 # in kelvin", "p1 = 100.00 # in kpa", "mdot = 5.807 # in kg/s", "p2 = 300.00 # in kpa", "p3 = p2", "p4 = 1000.00 # in kpa", "p5 = p4", "p6 = p4", "T6 = 1400.00 # in kelvin", "T8 = T6", "p7 = 300.00 # in kpa", "p8 = p7", "etac = 0.8 # isentropic efficiency of compressor", "etat = 0.8 # isentropic efficiency of turbine", "etareg = 0.8 # regenerator effectiveness", "# Analysis", "# From example 9.9", "h1 = 300.19 # in kj/kg", "h3 = h1 # in kj/kg", "h2s = 411.3 # in kj/kg", "h4s = 423.8 # in kj/kg", "# From example 9.8", "h6 = 1515.4 # in kj/kg", "h8 = h6", "h7s = 1095.9 # in kj/kg", "h9s = 1127.6 # in kj/kg", "", "# Calculations", "h4 = h3 + (h4s-h3)/etac # in kj/kg", "h2 = h1 + (h2s-h1)/etac # in kj/kg", "h9 = h8-etat*(h8-h9s) # in kj/kg", "h7 = h6-etat*(h6-h7s) # in kj/kg", "h5 = h4+etareg*(h9-h4) # in kj/kg", "", "# Part(a)", "# Calculations", "wtdot = (h6-h7)+(h8-h9) # The total turbine work per unit of mass flow in kj/kg", "wcdot = (h2-h1)+(h4-h3) # The total compressor work input per unit of mass flow in kj/kg", "qindot = (h6-h5)+(h8-h7) # The total heat added per unit of mass flow in kj/kg", "eta = (wtdot-wcdot)/qindot # thermal efficiency", "# Result", "print '-> The thermal efficiency is: '", "print round(eta,2)", "", "# Part(b)", "bwr = wcdot/wtdot # back work ratio", "# Result", "print '-> The back work ratio is:'", "print round(bwr,2)", "", "# Part(c)", "Wcycledot = mdot*(wtdot-wcdot) # net power developed in kw", "# Result", "print '-> The net power developed, is: ',round(Wcycledot,2),'kW.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The thermal efficiency is: ", "0.44", "-> The back work ratio is:", "0.45", "-> The net power developed, is: 2046.62 kW." ] } ], "prompt_number": 10 }, { "cell_type": "markdown", "source": [ "## Example 9.12 Page no-416" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "Ta = 240.00 # in kelvin", "pa = 0.8 # in bar", "Va = 278.00 # in m/s", "PR = 8.00 # pressure ratio across the compressor", "T3 = 1200.00 # in kelvin", "p5 = 0.8 # in bar", "", "# From table A-22", "ha = 240.02 # in kj/kg", "h1 = ha + ((Va**2)/2)*10**-3 # in kj/kg", "# Interpolating in Table A-22 gives", "pr1 = 1.070", "pra = .6355", "", "# Interpolating in Table A-22, we get", "h2 = 505.5 # in kj/kg", "# At state 3 the temperature is given as T3 \u0003= 1200 K. From Table A-22", "h3 = 1277.79 # in kj/kg", "", "", "# Interpolating in Table A-22 with h4, gives", "pr4 = 116.8", "# pr data from table A-22 gives", "pr4 = 116.00", "pr3 = 238.00", "# From table A-22", "h5 = 621.3 # in kj/kg", "", "# The expansion through the nozzle is isentropic to", "p5 = .8 # in bars", "", "# Calculations", "p1 = (pr1/pra)*pa # in bars", "# With the help of assumption, 'The turbine work output equals the work required to drive the compressor.',", "h4 = h3+h1-h2 # in kj/kg", "p2 = PR*p1 # in bars", "# Using assumption 'There is no pressure drop for flow through the combustor', ", "p3 = p2", "p4 = p3*(pr4/pr3) # in bars", "pr5 = pr4*(p5/p4)", "V5 = ((2*(h4-h5)*10**3))**(0.5) # the velocity at the nozzle exit in m/s", "", "# Results", "print '-> The velocity at the nozzle exit in m/s is:'", "print round(V5,2)", "print '-> pa in bars = '", "print round(pa,2)", "print '-> p1 in bars = '", "print round(p1,2)", "print '-> p2 in bars = '", "print round(p2,2)", "print '-> p3 in bars = '", "print round(p3,2)", "print '-> p4 in bars = '", "print round(p4,2)", "print '-> p5 in bars = '", "print round(p5,2)" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The velocity at the nozzle exit in m/s is:", "926.99", "-> pa in bars = ", "0.8", "-> p1 in bars = ", "1.35", "-> p2 in bars = ", "10.78", "-> p3 in bars = ", "10.78", "-> p4 in bars = ", "5.25", "-> p5 in bars = ", "0.8" ] } ], "prompt_number": 11 }, { "cell_type": "markdown", "source": [ "## Example 9.13 Page no-421" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "Wnetdot = 45.00 # in MW", "T1 = 300.00 # in kelvin", "p1 = 100.00 # in kpa", "etac = 0.84 # The isentropic efficiency of the compressor", "T3 = 1400.00 # in kelvin", "p2 = 1200.00 # in kpa", "p3 = p2", "etat = 0.88 # isentropic efficiency of the turbine", "T5 = 400.00 # in kelvin", "p4 = 100.00 # in kpa", "p5 = p4", "T7 = 400.00 # in degree celcius", "p7 = 8.00 # in MPa", "etatw =0.9 # isentropic efficiency of turbine of the vapor cycle", "p8 = 8.00 # in kpa", "p9 = p8", "etap = 0.8 # isentropic efficiency of pump of the vapor cycle", "T0 = 300.00 # in kelvin", "p0 = 100.00 # -in kpa", "", "# Analysis", "# With procedure similar to that used in the examples of chapters 8 and 9,we can determine following property data", "h1 = 300.19 # in kj/kg", "h2 = 669.79 # in kj/kg", "h3 = 1515.42 # in kj/kg", "h4 = 858.02 # in kj/kg", "h5 = 400.98 # in kj/kg", "h6 = 183.96 # in kj/kg", "h7 = 3138.30 # in kj/kg", "h8 = 2104.74 # in kj/kg", "h9 = 173.88 # in kj/kg", "s1 = 1.7020 # in kj/kg.k", "s2 = 2.5088 # in kj/kg.k", "s3 = 3.3620 # in kj/kg.k ", "s4 = 2.7620 # in kj/kg.k", "s5 = 1.9919 # in kj/kg.k", "s6 = 0.5975 # in kj/kg.k", "s7 = 6.3634 # in kj/kg.k", "s8 = 6.7282 # in kj/kg.k", "s9 = 0.5926 # in kj/kg.k", "", "# Part(a)", "# By applying mass and energy rate balances", "# Calculations", "mvdotbymgdot = (h4-h5)/(h7-h6) # ratio of mass flow rates of vapor and air", "mgdot = (Wnetdot*10**3)/(((h3-h4)-(h2-h1)) + mvdotbymgdot*((h7-h8)-(h6-h9))) # mass flow rate of air in kg/s", "mvdot = mvdotbymgdot*mgdot # mass flow rate of vapor in kg/s", "Wgasdot = mgdot*((h3-h4)-(h2-h1))*10**-3 # net power developed by gas turbine in MW", "Wvapdot = mvdot*((h7-h8)-(h6-h9))*10**-3 # net power developed by vapor cycle in MW", "", "# Results", "print '-> Mass flow rate of air is: ',round(mgdot,2),'kg/s.'", "print '-> Mass flow rate of vapor is: ',round(mvdot,2),'kg/s.'", "print '-> Net power developed by gas turbine is: ',round(Wgasdot,2),'MW.'", "print '-> Net power developed by vapor cycle is: ',round(Wvapdot,2),'MW.'", "", "", "# Part(b)", "import math", "# The net rate of exergy increase of the air passing through the combustor is", "Edotf32 = mgdot*(h3-h2-T0*(s3-s2))*10**-3 # in MW", "# The net rate exergy is carried out by the exhaust air stream at 5 is", "Edotf51 = mgdot*(h5-h1-T0*(s5-s1))/10**3 # in MW", "# The net rate exergy is carried out as the water passes through the condenser is", "Edotf89 = mvdot*(h8-h9-T0*(s8-s9))*10**-3 # in MW", "R = 8.314 # universal gas constant, in SI units", "M = 28.97 # molar mass of air in grams", "# The rate of exergy destruction for air turbine is", "Eddott = mgdot*T0*(s4-s3-(R/M)*math.log(p4/p3))/10**3 # in MW", "# The rate of exergy destruction for compressor is", "Eddotc = mgdot*T0*(s2-s1-(R/M)*math.log(p2/p1))/10**3 # in MW", "# The rate of exergy destruction for steam turbine is", "Eddotst = mvdot*T0*(s8-s7)/10**3 # in MW", "# The rate of exergy destruction for pump is", "Eddotp = mvdot*T0*(s6-s9)/10**3 # in MW", "# For heat exchanger", "EddotHE = T0*(mgdot*(s5-s4)+mvdot*(s7-s6))/10**3 # in MW", "", "# Results", "print '-> Balance sheet'", "print 'Net exergy increase of the gas passing'", "print '-> Through the combustor: ',round(Edotf32,2),'MW'", "print 'Disposition of the exergy:'", "print '\u2022 Net power developed'", "print 'gas turbine cycle ',round(Wgasdot,2),'MW'", "print 'vapor cycle ',round(Wvapdot,2),'MW'", "print '\u2022 Net exergy lost'", "print 'with exhaust gas at state 5 ',round(Edotf51,2),'MW'", "print 'from water passing through condenser ',round(Edotf89,2),'MW'", "print '\u2022 Exergy destruction'", "print 'air turbine ',round(Eddott,2),'MW'", "print 'compressor ',round(Eddotc,2),'MW'", "print 'steam turbine ',round(Eddotst,2),'MW'", "print 'pump ',round(Eddotp,2),'MW'", "print 'heat exchanger ',round(EddotHE,2),'MW'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Mass flow rate of air is: 100.87 kg/s.", "-> Mass flow rate of vapor is: 15.6 kg/s.", "-> Net power developed by gas turbine is: 29.03 MW.", "-> Net power developed by vapor cycle is: 15.97 MW.", "-> Balance sheet", "Net exergy increase of the gas passing", "-> Through the combustor: 59.48 MW", "Disposition of the exergy:", "\u2022 Net power developed", "gas turbine cycle 29.03 MW", "vapor cycle 15.97 MW", "\u2022 Net exergy lost", "with exhaust gas at state 5 1.39 MW", "from water passing through condenser 1.41 MW", "\u2022 Exergy destruction", "air turbine 3.42 MW", "compressor 2.83 MW", "steam turbine 1.71 MW", "pump 0.02 MW", "heat exchanger 3.69 MW" ] } ], "prompt_number": 12 }, { "cell_type": "markdown", "source": [ "## Example 9.14 Page no-438" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "Tnot = 360.00 # in kelvin", "pnot = 1.00 # in MPa", "A2 = 0.001 # in m^2", "k = 1.4", "", "# Calculations", "pstarbypnot = (1+(k-1)/2)**(k/(1-k))", "pstar = pstarbypnot*pnot", "", "# Part(a)", "# Since back pressure of 500 kpa is less than critical pressure pstar(528kpa in this case) found above, the nozzle is choked", "# At the exit", "M = 1.00", "p2 = pstar # in MPa", "T2 = Tnot/(1+((k-1)/2)*(M**2)) # exit temperature in kelvin", "R = 8.314 # universal gas constant, in SI units", "Mwt = 28.97 # molar mass of air in grams", "V2 = ((k*(R/Mwt)*T2*10**3)**0.5) # exit velocity in m/s", "mdot = (p2/((R/Mwt)*T2))*A2*V2*10**3 # mass flow rate in kg/s", "", "# Results", "print '-> The exit mach number for back pressure of 500kpa is: '", "print round(M,2)", "print '-> The mass flow rate in kg/s for back pressure of 500kpa is:'", "print round(mdot,2)", "", "# Part(b)", "# Since the back pressure of 784kpa is greater than critical pressure of pstar determined above,the flow throughout the nozzle is subsonic and the exit pressure equals the back pressure,", "p2 = 784.00 # exit pressure in kpa", "# Calculations", "M2 = (((2.00)/(k-1))*(((pnot*10**3)/p2)**((k-1)/k)-1))**0.5 # exit mach number", "T2 = Tnot/(1+((k-1)/2)*(M2**2)) # exit temperature in kelvin", "V2 = M2*((k*(R/Mwt)*10**3*T2)**0.5) # exit velocity in m/s", "mdot2 = (p2/((R/Mwt)*T2))*A2*V2 # mass flow rate in kg/s", "# Results", "print '-> The mass flow rate at the exit for back pressure of 784kpa is: ',round(mdot2,2),'kg/s.'", "print '-> The exit mach number for back pressure of 784 kpa is: '", "print round(M2,2)" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The exit mach number for back pressure of 500kpa is: ", "1.0", "-> The mass flow rate in kg/s for back pressure of 500kpa is:", "2.13", "-> The mass flow rate at the exit for back pressure of 784kpa is: 1.79 kg/s.", "-> The exit mach number for back pressure of 784 kpa is: ", "0.6" ] } ], "prompt_number": 13 }, { "cell_type": "markdown", "source": [ "## Example 9.15 Page no-442" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "# Part(a)", "Mt = 0.7 # mach mumber at the throat", "At = 6.25 # throat area in cm^2", "Ae = 15.00 # exit area in cm^2", "", "# The flow throughout the nozzle, including the exit, is subsonic. Accordingly, with this value for A2byAstar, Table 9.1 gives", "M2 = 0.24", "# For M2 \u0003= 0.24,", "T2byTnot = 0.988", "p2bypnot = 0.959", "k = 1.4", "T0 = 280.00 # in kelvin", "pnot = 6.8 # in bars", "# Calculations", "# With Mt \u0003= 0.7, Table 9.1 gives", "AtbyAstar = 1.09437", "A2byAstar = (Ae/At)*AtbyAstar", "T2 = T2byTnot*T0 # in kelvin", "p2 = p2bypnot*pnot # in bars", "V2 = M2*((k*(8.314/28.97)*T2*10**3)**0.5) # velocity at the exit in m/s", "mdot = (p2/((8.314/28.97)*T2))*Ae*V2*10**-2 # mass flow rate in kg/s", "# Results", "print '-> Part(a) the mass flow rate in kg/s is: '", "print round(mdot,2)", "print '-> The exit pressure in bars is: '", "print round(p2,2)", "print '-> The exit mach number is: '", "print round(M2,2)", "", "# Part(b)", "Mt = 1.00 # mach number at the throat", "# From table 9.1", "M2 = 0.26", "T2byTnot = 0.986 ", "p2bypnot = 0.953", "", "T0 = 280.00 # in kelvin", "pnot = 6.8 # in bars", "# Calculations", "T2 = T2byTnot*T0 # in kelvin", "p2 = p2bypnot*pnot # in bars", "k = 1.4", "V2 = M2*((k*(8314/28.97)*T2)**0.5) # exit velocity in m/s", "mdot = (p2/((8.314/28.97)*T2))*Ae*V2*10**-2 # mass flow rate in kg/s", "# Results", "print '-> Part(b) the mass flow rate is: ',mdot,'kg/s.'", "print '-> The exit pressure is: ',p2,'bars.'", "print '-> The exit mach number is: ',M2", "", "# Part(c)", "# From part (b), the exit Mach number in the present part of the example is", "M2 = 2.4", "# Using this, Table 9.1 gives", "p2bypnot = 0.0684", "pnot = 6.8 # in bars", "# Calculation", "p2 = p2bypnot*pnot # in bars", "# Results", "# Since the nozzle is choked, the mass flow rate is the same as found in part (b).", "print '> Part(c) the mass flow rate is: ',mdot,'kg/s.'", "print '-> The exit pressure is: ',p2,'bars.'", "print '-> The exit mach number is: ',M2", "", "# Part(d)", "# Since a normal shock stands at the exit and the flow upstream of the shock is isentropic, the Mach number Mx and the pressure px correspond to the values found in part (c), ", "Mx = 2.4", "px = 0.465 # in bars", "# Then, from Table 9.2", "My = 0.52", "#py is the exit pressure", "pybypx = 6.5533", "py = px*pybypx", "", "# The pressure downstream of the shock is thus 3.047 bars. This is the exit pressure", "# The mass flow is the same as found in part (b).", "# Results", "print '-> Part(d) the mass flow rate is: ',mdot,'kg/s.'", "print '-> The exit pressure is: ',round(py,3),'bars.'", "print '-> The exit mach number is: ',My", "", "# Part(e)", "# A shock stands in the diverging portion where the area is", "Ax = 12.5 # in cm^2", "# Since a shock occurs, the flow is sonic at the throat, so", "Axstar = 6.25 # in cm^2", "At = Axstar", "# The Mach number Mx can then be found from Table 9.1, by using AxbyAxstar as", "Mx = 2.2", "", "# Results", "# With Mx =\u0003 2.2, the ratio of stagnation pressures is obtained from Table 9.2 as", "pnotybypnotx = 0.62812", "", "# Using this ratio and noting that the flow is subsonic after the shock, Table 9.1 gives", "M2 = 0.43", "# For M2 = 0.43,", "p2bypnoty = 0.88", "# Calculations", "A2byAystar = (Ae/Axstar)*pnotybypnotx", "p2 = p2bypnoty*pnotybypnotx*pnot # in bars", "", "# Results", "# Since the flow is choked, the mass flow rate is the same as that found in part (b).", "print '-> part(e) the mass flow rate is: ',mdot,'kg/s.'", "print '-> the exit pressure is: ',p2,'bars.'", "print '-> the exit mach number is: ',M2" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Part(a) the mass flow rate in kg/s is: ", "0.99", "-> The exit pressure in bars is: ", "6.52", "-> The exit mach number is: ", "0.24", "-> Part(b) the mass flow rate is: 1.06238566635 kg/s.", "-> The exit pressure is: 6.4804 bars.", "-> The exit mach number is: 0.26", "> Part(c) the mass flow rate is: 1.06238566635 kg/s.", "-> The exit pressure is: 0.46512 bars.", "-> The exit mach number is: 2.4", "-> Part(d) the mass flow rate is: 1.06238566635 kg/s.", "-> The exit pressure is: 3.047 bars.", "-> The exit mach number is: 0.52", "-> part(e) the mass flow rate is: 1.06238566635 kg/s.", "-> the exit pressure is: 3.75867008 bars.", "-> the exit mach number is: 0.43" ] } ], "prompt_number": 14 } ] } ] }