{ "metadata": { "name": "Chapter 7" }, "nbformat": 2, "worksheets": [ { "cells": [ { "cell_type": "markdown", "source": [ "# Chapter 7 :- Exergy Analysis" ] }, { "cell_type": "markdown", "source": [ "## Example 7.1 Page no-279" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "v = 2450.00 # volume of gaseous products in cm^3", "P = 7.00 # pressure of gaseous product in bar", "T = 867.00 # temperature of gaseous product in degree celcius", "T0 = 300.00 # in kelvin", "P0 = 1.013 # in bar", "", "# From table A-22", "u = 880.35 # in kj/kg", "u0 = 214.07 # in kj/kg", "s0T = 3.11883 # in kj/kg.k", "s0T0 = 1.70203 # in kj/kg.k", "", "# Calculations", "import math", "e = (u-u0) + (P0*(8.314/28.97)*(((T+273)/P)-(T0/P0))) - T0*(s0T-s0T0-(8.314/28.97)*math.log(P/P0)) # kj/kg", "", "# Results", "print '-> The specific exergy of the gas is ',round(e,3),'kJ/kg.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The specific exergy of the gas is 368.912 kJ/kg." ] } ], "prompt_number": 1 }, { "cell_type": "markdown", "source": [ "## Example 7.2 Page no-280" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "mR = 1.11 # mass of the refrigerant in kg", "T1 = -28.00 # initial temperature of the saturated vapor in degree celcius", "P2 = 1.4 # final pressure of the refrigerant in bar", "T0 = 293.00 # in kelvin", "P0 = 1.00 # in bar", "", "# Part (a)", "# From table A-10", "u1 = 211.29 # in kj/kg", "v1 = 0.2052 # in m^3/kg", "s1 = 0.9411 # in kj/kg.k", "# From table A-12", "u0 = 246.67 # in kj/kg", "v0 = 0.23349 # in m^3/kg", "s0 = 1.0829 # in kj/kg.k", "", "# From table A-12", "u2 = 300.16 # in kj/kg", "s2 = 1.2369 # in kj/kg.k", "v2 = v1", "", "# Calculations ", "E1 = mR*((u1-u0) + P0*(10**5)*(v1-v0)*(10**(-3))-T0*(s1-s0))", "E2 = mR*((u2-u0) + P0*(10**5)*(v2-v0)*(10**(-3))-T0*(s2-s0))", "", "# Results for Part A", "print '-> Part(a) The initial exergy is ',round(E1,2),'kJ.'", "print '-> The final exergy is ',round(E2,2),'kJ.'", "print '-> The change in exergy of the refrigerant is ',round(E2-E1,2),'kJ.'", "", "", "# Part (b)", "# Calculations", "deltaU = mR*(u2-u1)", "# From energy balance", "deltaPE = -deltaU", "# With the assumption::The only significant changes of state are experienced by the refrigerant and the suspended mass. For the refrigerant, ", "# there is no change in kinetic or potential energy. For the suspended mass, there is no change in kinetic or internal energy. Elevation is ", "# the only intensive property of the suspended mass that changes", "deltaE = deltaPE", "", "# Results for part b", "print '-> Part(b)The change in exergy of the suspended mass is ',round(deltaE,3),'kJ.'", "", "", "# Part(c)", "# Calculations", "deltaEiso = (E2-E1) + deltaE", "", "# Results", "print '-> Part(c)The change in exergy of an isolated system of the vessel and pulley\u2013mass assembly is ',round(deltaEiso,2),'kJ.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Part(a) The initial exergy is 3.71 kJ.", "-> The final exergy is 6.15 kJ.", "-> The change in exergy of the refrigerant is 2.44 kJ.", "-> Part(b)The change in exergy of the suspended mass is -98.646 kJ.", "-> Part(c)The change in exergy of an isolated system of the vessel and pulley\u2013mass assembly is -96.2 kJ." ] } ], "prompt_number": 2 }, { "cell_type": "markdown", "source": [ "##Example 7.3 Page no-287" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given :-", "T = 373.15 # initial temperature of saturated liquid in kelvin", "T0 = 293.15 # in kelvin", "P0 = 1.014 # in bar", "", "# Part(a)", "# From table A-2", "ug = 2506.5 # in kj/kg", "uf = 418.94 # in kj/kg", "vg = 1.673 # in m^3/kg", "vf = 1.0435*(10**(-3)) # in m^3/kg", "sg = 7.3549 # in kj/kg.k", "sf = 1.3069 # in kj/kg.k", "", "", "# Calculations ", "# Energy transfer accompanying work", "etaw = 0 # since p = p0", "# Exergy transfer accompanying heat", "Q = 2257 # in kj/kg,obtained from example 6.1", "etah = (1-(T0/T))*Q", "", "# Exergy destruction", "ed = 0 # since the process is accomplished without any irreversibilities", "deltae = ug-uf + P0*(10**5)*(vg-vf)/(10**3)-T0*(sg-sf)", "", "# Results", "print '-> Part(a)the change in exergy is',round(deltae,2),'kJ/kg.'", "print '-> The exergy transfer accompanying work is',round(etaw,2),'kJ/kg.'", "print '-> The exergy transfer accompanying heat is',round(etah,2),'kJ/kg.'", "print '-> The exergy destruction is',round(ed,2),'kJ/kg.'", "", "", "# Part(b)", "Deltae = deltae # since the end states are same ", "Etah = 0 # since process is adiabatic", "# Exergy transfer along work", "W = -2087.56 # in kj/kg from example 6.2", "Etaw = W- P0*(10**5)*(vg-vf)/(10**3)", "# Exergy destruction", "Ed = -(Deltae+Etaw)", "", "# Results", "print '-> Part(b)the change in exergy is ',round(Deltae,2),'kJ/kg.'", "print '-> The exergy transfer accompanying work is',round(Etaw,2),'kJ/kg.'", "print '-> The exergy transfer accompanying heat is',round(Etah,2),'kJ/kg.'", "print '-> The exergy destruction is',round(Ed,2),'kJ/kg.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Part(a)the change in exergy is 484.13 kJ/kg.", "-> The exergy transfer accompanying work is 0.0 kJ/kg.", "-> The exergy transfer accompanying heat is 483.88 kJ/kg.", "-> The exergy destruction is 0.0 kJ/kg.", "-> Part(b)the change in exergy is 484.13 kJ/kg.", "-> The exergy transfer accompanying work is -2257.1 kJ/kg.", "-> The exergy transfer accompanying heat is 0.0 kJ/kg.", "-> The exergy destruction is 1772.97 kJ/kg." ] } ], "prompt_number": 3 }, { "cell_type": "markdown", "source": [ "## Example 7.4 Page no-289" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T0 = 293.00 # in kelvin", "Qdot = -1.2 # in KW, from example 6.4a", "Tb = 300.00 # temperature at the outer surface of the gearbox in kelvin from example 6.4a", "sigmadot = 0.004 # rate of entropy production in KW/k from example 6.4a", "", "# Calculations", "R = -(1-T0/Tb)*Qdot # time rate of exergy transfer accompanying heat", "Eddot = T0*sigmadot # rate of exergy destruction", "", "# Results", "print '-> Balance sheet'", "print '* Rate of exergy in high speed shaft 60Kw' ", "print '-> Disposition of the exergy: Rate of exergy out low-speed shaft 58.8Kw' ", "print '-> Heat transfer is',round(R,3),'kw.'", "print '-> Rate of exergy destruction is',round(Eddot,3),'kw.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Balance sheet", "* Rate of exergy in high speed shaft 60Kw", "-> Disposition of the exergy: Rate of exergy out low-speed shaft 58.8Kw", "-> Heat transfer is 0.028 kw.", "-> Rate of exergy destruction is 1.172 kw." ] } ], "prompt_number": 4 }, { "cell_type": "markdown", "source": [ "##Example 7.5 Page no-295" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "p1 = 3.0 # entry pressure in Mpa", "p2 = 0.5 # exit pressure in Mpa", "T1 = 320.0 # entry temperature in degree celcius", "T0 = 25.0 # in degree celcius", "p0 = 1.0 # in atm", "", "# From table A-4", "h1 = 3043.4 # in kj/kg", "s1 = 6.6245 # in kj/kg.k", "h2 = h1 # from reduction of the steady-state mass and energy rate balances", "s2 = 7.4223 # Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k", "", "# From table A-2", "h0 = 104.89 # in kj/kg", "s0 = 0.3674 # in kj/kg.k", "", "# Calculations ", "ef1 = h1-h0-(T0+273)*(s1-s0) # flow exergy at the inlet", "ef2 = h2-h0-(T0+273)*(s2-s0) # flow exergy at the exit", "# From the steady-state form of the exergy rate balance", "Ed = ef1-ef2 # the exergy destruction per unit of mass flowing is", "", "# Results ", "print '-> The specific flow exergy at the inlet is ',round(ef1,2),'kJ/kg.'", "print '-> The specific flow exergy at the exit is',round(ef2,2),'kJ/kg.'", "print '-> The exergy destruction per unit of mass flowing is',round(Ed,2),'kJ/kg.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The specific flow exergy at the inlet is 1073.89 kJ/kg.", "-> The specific flow exergy at the exit is 836.15 kJ/kg.", "-> The exergy destruction per unit of mass flowing is 237.74 kJ/kg." ] } ], "prompt_number": 5 }, { "cell_type": "markdown", "source": [ "## Example 7.6 Page no-296" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T1 = 610.0 # temperature of the air entering heat exchanger in kelvin", "p1 = 10.0 # pressure of the air entering heat exchanger in bar", "T2 = 860.0 # temperature of the air exiting the heat exchanger in kelvin", "p2 = 9.70 # pressure of the air exiting the heat exchanger in bar", "T3 = 1020.0 # temperature of entering hot combustion gas in kelvin", "p3 = 1.10 # pressure of entering hot combustion gas in bar", "p4 = 1.0 # pressure of exiting hot combustion gas in bar", "mdot = 90.0 # mass flow rate in kg/s", "T0 = 300.0 # in kelvin", "p0 = 1.0 # in bar", "", "# Part (a)", "# From table A-22", "h1 = 617.53 # in kj/kg", "h2 = 888.27 # in kj/kg", "h3 = 1068.89 # in kj/kg", "", "# Calculations", "\"\"\"From reduction of mass and energy rate balances for the control volume at", "steady state \"\"\"", "h4 = h3+h1-h2", "", "# Using interpolation in table A-22 gives", "T4 = 778 # in kelvin", "", "# Results", "print '-> The exit temperature of the combustion gas is',T4,'kelvin.'", "", "# Part(b)", "# From table A-22", "s2 = 2.79783 # in kj/kg.k", "s1 = 2.42644 # in kj/kg.k", "s4 = 2.68769 # in kj/kg.k", "s3 = 2.99034 # in kj/kg.k", "", "# Calculations for part b", "import math", "deltaR = (mdot*((h2-h1)-T0*(s2-s1-(8.314/28.97)*math.log(p2/p1))))/1000", "deltRc = mdot*((h4-h3)-T0*(s4-s3-(8.314/28.97)*math.log(p4/p3)))/1000", "", "# Results for part b", "print '-> The net change in the flow exergy rate from inlet to exit of compressed gas is',round(deltaR,3),'MW.'", "print '-> The net change in the flow exergy rate from inlet to exit of hot combustion gas is',round(deltRc,3),'MW.'", "", "# Part(c)", "#From an exergy rate balance", "Eddot = -deltaR-deltRc", "", "# Results ", "print '-> The rate exergy destroyed, is',round(Eddot,3),'MW.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> The exit temperature of the combustion gas is 778 kelvin.", "-> The net change in the flow exergy rate from inlet to exit of compressed gas is 14.103 MW.", "-> The net change in the flow exergy rate from inlet to exit of hot combustion gas is -16.934 MW.", "-> The rate exergy destroyed, is 2.831 MW." ] } ], "prompt_number": 6 }, { "cell_type": "markdown", "source": [ "##Example 7.7 Page no-299" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "p1 = 30.0 # pressure of entering steam in bar", "t1 = 400.0 # temperature of entering steam in degree celcius", "v1 = 160.0 # velocity of entering steam in m/s", "t2 = 100.0 # temperature of exiting saturated vapor in degree celcius", "v2 = 100.0 # velocity of exiting saturated vapor in m/s", "W = 540.0 # rate of work developed in kj per kg of steam", "Tb = 350.0 # the temperature on the boundary where heat transfer occurs in kelvin", "T0 = 25.0 # in degree celcius", "p0 = 1.0 # in atm", "", "# From table A-4", "h1 = 3230.9 # in kj/kg", "s1 = 6.9212 # in kj/kg.k", "# From table A-2 ", "h2 = 2676.1 # in kj/kg", "s2 = 7.3549 # in kj/kg.k", "# From example 6.6", "Q = -22.6 # in kj/kg", " ", "# Calculations", "DELTAef = (h1-h2)-(T0+273)*(s1-s2)+(v1**2-v2**2)/(2*1000)", "# The net exergy carried in per unit mass of steam flowing in kj/kg", "Eq = (1-(T0+273)/Tb)*(Q) # exergy transfer accompanying heat in kj/kg", "Ed = ((1-(T0+273)/Tb)*(Q))-W+(DELTAef) # The exergy destruction determined by rearranging the steady-state form of the exergy ", " # rate balance", "", "# Results", "print '-> Balance sheet'", "print '-> Net rate of exergy ',DELTAef,'kJ/kg,'", "print '-> Disposition of the exergy:'", "print '* Rate of exergy out'", "print '-> Work',W,'kJ/kg.'", "print '-> Heat transfer',-Eq,'.'", "print '\u2022 Rate of exergy destruction',Ed,'kJ/kg.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Balance sheet", "-> Net rate of exergy 691.8426 kJ/kg,", "-> Disposition of the exergy:", "* Rate of exergy out", "-> Work 540.0 kJ/kg.", "-> Heat transfer 3.35771428571 .", "\u2022 Rate of exergy destruction 148.484885714 kJ/kg." ] } ], "prompt_number": 7 }, { "cell_type": "markdown", "source": [ "## Example 7.8 Page no-300" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "m1dot = 69.78 # in kg/s", "p1 = 1.0 # in bar", "T1 = 478.0 # in kelvin", "T2 = 400.0 # in kelvin", "p2 = 1.0 # in bar", "p3 = 0.275 # in Mpa", "T3 = 38.9 # in degree celcius", "m3dot = 2.08 # in kg/s", "T4 = 180.0 # in degree celcius", "p4 = 0.275 # in Mpa", "p5 = 0.07 # in bar", "x5 = 0.93", "Wcvdot = 876.8 # in kW", "T0 = 298.0 # in kelvin", "", "", "# Part(a)", "# From table A-22", "h1 = 480.35 # in kj/kg", "h2 = 400.97 # in kj/kg", "s1 = 2.173 # in kj/kg", "s2 = 1.992 # in kj/kg", "", "# From table A-2E", "h3 = 162.82 # in kj/kg", "s3 = 0.5598 # in kj/kg.k", "# Using saturation data at 0.07 bars from Table A-3", "h5 = 2403.27 # in kj/kg", "s5 = 7.739 # in kj/kg.k", "#The net rate exergy carried out by the water stream", "", "# From table A-4", "h4 = 2825.0 # in kj/kg", "s4 = 7.2196 # in kj/kg.k", "# Calculations", "import math", "netRE = m1dot*(h1-h2-T0*(s1-s2-(8.314/28.97)*math.log(p1/p2))) # the net rate exergy carried into the control volume", "netREout = m3dot*(h5-h3-T0*(s5-s3))", "# From an exergy rate balance applied to a control volume enclosing the steam generator", "Eddot = netRE + m3dot*(h3-h4-T0*(s3-s4)) # the rate exergy is destroyed in the heat-recovery steam generator", "", "# From an exergy rate balance applied to a control volume enclosing the turbine", "EdDot = -Wcvdot + m3dot*(h4-h5-T0*(s4-s5)) # the rate exergy is destroyed in the tpurbine", "", "# Results ", "print '-> balance sheet'", "print '- Net rate of exergy in:',netRE,'kJ/kg.'", "print '-> Disposition of the exergy:'", "print '\u2022 Rate of exergy out'", "print '-> power developed',1772.8-netREout-Eddot-EdDot,'kJ/kg.'", "print '-> water stream ',netREout", "print '\u2022 Rate of exergy destruction'", "print '-> heat-recovery steam generator',Eddot,'kJ/kg'", "print '-> turbine',EdDot" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> balance sheet", "- Net rate of exergy in: 1775.34276 kJ/kg.", "-> Disposition of the exergy:", "\u2022 Rate of exergy out", "-> power developed 874.25724 kJ/kg.", "-> water stream 210.180672", "\u2022 Rate of exergy destruction", "-> heat-recovery steam generator 366.018792 kJ/kg", "-> turbine 322.343296" ] } ], "prompt_number": 8 }, { "cell_type": "markdown", "source": [ "## Example 7.9 Page no-302" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "T0 = 273.00 # in kelvin", "pricerate = 0.08 # exergy value at $0.08 per kw.h", "", "# From example 6.8", "sigmadotComp = 17.5e-4 # in kw/k", "sigmadotValve = 9.94e-4 # in kw/k", "sigmadotcond = 7.95e-4 # in kw/k", "", "# Calculations", "# The rates of exergy destruction", "EddotComp = T0*sigmadotComp # in kw", "EddotValve = T0*sigmadotValve # in kw", "Eddotcond = T0*sigmadotcond # in kw", "", "mCP = 3.11 # From the solution to Example 6.14, the magnitude of the compressor power in kW", "", "# Results", "print '-> Daily cost in dollars of exergy destruction due to compressor irreversibilities = ',round(EddotComp*pricerate*24,3)", "print '-> Daily cost in dollars of exergy destruction due to irreversibilities in the throttling valve = ',round(EddotValve*pricerate*24,3)", "print '-> Daily cost in dollars of exergy destruction due to irreversibilities in the condenser = ',round(Eddotcond*pricerate*24,3)", "print '-> Daily cost in dollars of electricity to operate compressor = ',round(mCP*pricerate*24,3)" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> Daily cost in dollars of exergy destruction due to compressor irreversibilities = 0.917", "-> Daily cost in dollars of exergy destruction due to irreversibilities in the throttling valve = 0.521", "-> Daily cost in dollars of exergy destruction due to irreversibilities in the condenser = 0.417", "-> Daily cost in dollars of electricity to operate compressor = 5.971" ] } ], "prompt_number": 9 }, { "cell_type": "markdown", "source": [ "## Example 7.10 Page no-313" ] }, { "cell_type": "code", "collapsed": false, "input": [ "", "# Given:-", "EfFdot = 100.00 # exergy rate of fuel entering the boiler in MW", "cF = 1.44 # unit cost of fuel in cents per kw.h", "Zbdot = 1080.00 # the cost of owning and operating boiler in dollars per hour", "Ef1dot = 35.00 # exergy rate of exiting steam from the boiler in MW", "p1 = 50.00 # pressure of exiting steam from the boiler in bar", "T1 = 466.00 # temperature of exiting steam from the boiler in degree celcius", "Ztdot = 92.00 # the cost of owning and operating turbine in dollars per hour", "p2 = 5.00 # pressure of exiting steam from the turbine in bars", "T2 = 205.00 # temperature of exiting steam from the turbine in degree celcius", "m2dot = 26.15 # mass flow rate of exiting steam from the turbine in kg/s", "T0 = 298.00 # in kelvin ", "", "", "# Part(a)", "# From table A-4,", "h1 = 3353.54 # in kj/kg", "h2 = 2865.96 # in kj/kg", "s1 = 6.8773 # in kj/kg.k", "s2 = 7.0806 # in kj/kg.k", "", "# Calculations", "# From assumption,For each control volume,Qcvdot = 0 and kinetic and potential energy effects are negligible,the mass and energy rate ", "# balances for a control volume enclosing the turbine reduce at steady state to give", "Wedot = m2dot *(h1-h2)/1000 # power in MW", "Ef2dot = Ef1dot+m2dot*(h2-h1-T0*(s2-s1))/1000 # the rate exergy exits with the steam in MW", "", "# Results", "print '-> For the turbine,the power is',round(Wedot,2),'MW.'", "print '-> For the turbine,the rate exergy exits with the steam is',round(Ef2dot,2),' MW.'", "", "# Part(b)", "# Calculations", "c1 = cF*(EfFdot/Ef1dot) + ((Zbdot/Ef1dot)/10**3)*100 # unit cost of exiting steam from boiler in cents/Kw.h", "c2 = c1 # Assigning the same unit cost to the steam entering and exiting the turbine", "ce = c1*((Ef1dot-Ef2dot)/Wedot) + ((Ztdot/Wedot)/10**3)*100 # unit cost of power in cents/kw.h", "", "# Results", "print '-> The unit costs of the steam exiting the boiler of exergy is:',round(c1,2),' cents per kw.h.'", "print '-> The unit costs of the steam exiting the turbine of exergy is:',round(c2,2),' cents per kw.h.'", "print '-> Unit cost of power is:',ce,'cents per kw.h.'", "", "# Part(c)", "C2dot = (c2*Ef2dot*10**3)/100 # cost rate for low-pressure steam in dollars per hour", "Cedot = (ce*Wedot*10**3)/100 # cost rate for power in dollars per hour", "", "# Results", "print '-> The cost rate of the steam exiting the turbine is:',round(C2dot,2),' dollars per hour.'", "print '-> The cost rate of the power is: ',round(Cedot,2),' dollars per hour.'" ], "language": "python", "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "-> For the turbine,the power is 12.75 MW.", "-> For the turbine,the rate exergy exits with the steam is 20.67 MW.", "-> The unit costs of the steam exiting the boiler of exergy is: 7.2 cents per kw.h.", "-> The unit costs of the steam exiting the turbine of exergy is: 7.2 cents per kw.h.", "-> Unit cost of power is: 8.81617975223 cents per kw.h.", "-> The cost rate of the steam exiting the turbine is: 1487.92 dollars per hour.", "-> The cost rate of the power is: 1124.08 dollars per hour." ] } ], "prompt_number": 10 } ] } ] }