Chapter 8: Exergy: A Measure of Work Potential¶

Example 8-1 ,Page No.426¶

In :
from math import pi

# given data
D=12.0;#diameter of rotor in m
V=10.0;#average velocity in m/s

# density of air at 25C & 1atm
p=1.18;

#calculations
ke=(V**2)/2/1000;#factor of 1000 for converting J into kJ
m=p*pi*(D**2)*V/4.0;
MP=m*(ke);
print'Maximum power %f kW'%round(MP,1)
Maximum power 66.700000 kW

Example 8-2 ,Page No.427¶

In :
#given values
TH=2000.0;#temperature of large furnace in R
T0=77+460.0;#temperature of enviroment in R
Qin=3000.0;#heat rate in Btu/sec

#calculation
nth=1-(T0/TH);
Wmax=nth*Qin;
Wmax=round(Wmax)
print'the rate of energy flow %i Btu/s'%round(Wmax)
the rate of energy flow 2195 Btu/s

Example 8-3 ,Page No.429¶

In :
#given data
Tsink=300.0;#temperature of sink in K
Tsource=1200.0;#temperature of source in K
Qin=500.0;#heat rate in kJ/s
Wuout=180;#power output in W

#calculations
Wrev=(1-Tsink/Tsource)*Qin;
print'The reversible power %i kW'%round(Wrev);
I=Wrev-Wuout;
print'the irreversiblity rate %i kW'%round(I)
The reversible power 375 kW
the irreversiblity rate 195 kW

Example 8-4 ,Page No.429¶

In :
import math
from pylab import *

#given data
m=500.0;#mass of iron block in kg
T1=473.0;#temperature of iron block in K
T0=300.0;#temperature of surrondings in K

#from Table A-3
cavg=0.45;

#calculations
def intgrnd1(T):
return ((1-T0/T)*(-m*cavg))#intergrant
Qtotal=m*cavg*(T1-T0);
print'The reversible power %i kJ'%round(Wrev);
Wu=0;
I=Wrev-Wu;
print'the irreversiblity rate %i kJ'%round(I)
The reversible power 8191 kJ
the irreversiblity rate 8191 kJ

Example 8-5 ,Page No.431¶

In :
#given data
Wrev=8191.0;#reversible work from Ex - 8.4
Wtotal=38925.0;#total work is equivalent to total heat lost to heat engine from Ex - 8.4
TL=278.0;#outdoor temperature in K
TH=300.0;#temperature of house in K

#calculations
Wrm=Wtotal-Wrev;#work remaining
COPHP=1/(1-TL/TH);
Wd=COPHP*Wrev;#work delivered
PS=Wd+Wrm;
PS=round(PS/1000);#factor of 1000 for converting kJ into MJ
print'Maximum amount of heat that can be supplied %i MJ'%PS
Maximum amount of heat that can be supplied 142 MJ

Example 8-6 ,Page No.434¶

In :
#given data
COP=1.0;# the efficiency that the dealer is referring to is the first law efficiency, meaning that for each unit of electric energy (work) consumed, the heater will supply the house with 1 unit of energy (heat). That is,the advertised heater has a COP of 1
TL=283.0;#outdoor temperature in K
TH=294.0;#indoor temperature in K

#calculations
COPHP=1/(1-TL/TH);
nII=COP/COPHP;
print'the second law efficiency %f'%round(nII,3)
the second law efficiency 0.037000

Example 8-7 ,Page No.438¶

In :
from math import log

#given data
P1=1000.0;#pressure of compressed air in KPa
V=200.0;#volumne of rigid tank in m^3
T1=300.0;#temperature of enviroment in K
T0=T1;#state of air in tank
P0=100.0;#pressure of enviroment in KPa

#constants used
R=0.287;#in kPa m^3/kg K

#calculations
m1=P1*V/(R*T1);
O1=R*T0*(P0/P1-1)+R*T0*log(P1/P0);# O refers to exergy
X1=m1*O1/1000;#factor of 1000 for converting kJ into MJ
print'work obtained %i MJ'%round(X1)
work obtained 281 MJ

Example 8-8 ,Page No.439¶

In :
#given data
T0=20+273;#temperature of enviroment in K
P1=0.14;#intail pressure in MPa
T1=-10;#intail temperature in K
P2=0.8;#final pressure in MPa
T2=50;#final temperature in K

#the properties of refrigerant
#at inlet
h1=246.36;
s1=0.9724;
#at outlet
h2=286.69;
s2=0.9802;
dO=h2-h1-T0*(s2-s1);# O refers to exergy
print'the exergy change of the refrigerant %i kJ/kg'%round(dO)
wmin=dO;
print'the minimum work input that needs to be supplied is %i kJ/kg'%wmin
the exergy change of the refrigerant 38 kJ/kg
the minimum work input that needs to be supplied is 38 kJ/kg

Example 8-10 ,Page No.447¶

In :
#given values
Q=1035.0;#rate of heat transfer in W
T0=273.0;#outdoor temperature in C
Tin=293.0;#inner wall surface temperature in K
Tout=278.0;#outer wall surface temperature in K
T1=300.0;#indoor temperature in C

#calculations
#Xin - Xout - Xdestroyed = dX/dt
Xdestroyed=Q*(1-T0/Tin)-Q*(1-T0/Tout);
Xdestroyed=round(Xdestroyed);
print'the rate of exergy destroyed %i W'%round(Xdestroyed);
#the total rate of exergy destroyed
Xdestroyed=Q*(1-T0/T1)-Q*(1-T0/T0);
print'the total of exergy destroyed %f W'%round(Xdestroyed,1);
the rate of exergy destroyed 52 W
the total of exergy destroyed 93.100000 W

Example 8-11 ,Page No.448¶

In :
#given data
m=0.05;#mass of steam in kg
P1=1000;#intial pressure in MPa
T1=300+273;#intial temperature in K
P2=200;#final pressure in MPa
T2=150+273;#final temperature in K
P0=100;#pressure of surroundings in KPa
T0=25+273;#temperature of surroundings in K
Qout=2;#heat losses in kJ

#from Table A-6 & A-4
u1=2793.7;
v1=0.25799;
s1=7.1246;
u2=2577.1;
v2=0.95986;
s2=7.2810;
u0=104.83;
v0=0.00103;
s0=0.3672;

#calculations
X1=m*(u1-u0-T0*(s1-s0)+P0*(v1-v0));
X2=m*(u2-u0-T0*(s2-s0)+P0*(v2-v0));
print'exergy of intial state %i kJ'%round(X1);
print'exergy of final state %f kJ'%round(X2,1);
dX=X2-X1;
print'exergy change in system %f kJ'%round(dX,1);
Wout=-Qout-m*(u2-u1);
Wu=Wout-P0*m*(v2-v1);
Xdestroyed=X1-X2-Wu;
print'the exergy destroyed %f kJ'%round(Xdestroyed,1);
nII=Wu/(X1-X2);
print'second law efficiency of this process is %f'%round(nII,3)
exergy of intial state 35 kJ
exergy of final state 25.400000 kJ
exergy change in system -9.700000 kJ
the exergy destroyed 4.300000 kJ
second law efficiency of this process is 0.551000

Example 8-12 ,Page No.451¶

In :
import math
from pylab import *

#given data
m=2.0;#mass of air in lbm
T0=70+460.0;#intial temperature in R
P1=20.0;#intial pressure in psia
T1=70+460.0;#temperature of surroundings in R
T2=130+460.0;#final temperature in R

#constants used
Cv=0.172;#in Btu/lbm - F

#calculations
Xdestroyed=T0*m*Cv*log(T2/T1);
print'exergy destroyed %f Btu'%round(Xdestroyed,1);
def intgrnd1(T):
return ((1-T0/T)*m*Cv)#intergrant
print'the reversible work %f Btu'%round(Wrev,1)
exergy destroyed 19.600000 Btu
the reversible work 1.100000 Btu

Example 8-13 ,Page No.453¶

In :
#given data
T0=20+273.0;#temperature of surrounding in K
P0=100.0;#pressure of surrounding in kPa
Tiw=30+273.0;#temperature of water in K
mw=100.0;#mass of water in kg
Tii=350+273.0;#temperature of block in K
mi=5.0;#mass of block in kg

#constants used(Table A-3)
cw=4.18;#specific heat of water in kJ/kg C
ci=0.45;#specific heat of iron in kJ/kg C

#calculations
Tfk=(mi*ci*Tii+mw*cw*Tiw)/(mw*cw+mi*ci);
Tfc=Tfk-273;#in C
print'the final equilibrium temperature %f C'%round(Tfc,1);
X1i=mi*ci*(Tii-T0-T0*log(Tii/T0));
X1w=mw*cw*(Tiw-T0-T0*log(Tiw/T0));
X1t=X1i+X1w;#total exergy
print'intial exergy of combined systems %i kJ'%round(X1t);
X2i=mi*ci*(Tfk-T0-T0*log(Tfk/T0));
X2w=mw*cw*(Tfk-T0-T0*log(Tfk/T0));
X2t=X2i+X2w;#total exergy
print'intial exergy of combined systems %f kJ'%round(X2t,1);
Xdestroyed=X1t-X2t;
print'the wasted work %f kJ'%round(Xdestroyed,1)
the final equilibrium temperature 31.700000 C
intial exergy of combined systems 315 kJ
intial exergy of combined systems 95.800000 kJ
the wasted work 219.100000 kJ

Example 8-14 ,Page No.455¶

In :
#given data
TR=1200.0;#temperature of furnace
T0=300;#temperature of surrounding in K
P0=100;#pressure of surrounding in kPa
Tsys=400.0;#temperature of argon in K
P1=350;#temperature of argon in K
V1=0.01;#intail volumne in m^3
V2=2*V1;#final volumne

#calculations
W=P1*V1*log(V2/V1);
Wsurr=P0*(V2-V1);
Wu=W-Wsurr;
print'the useful work output %f kJ'%round(Wu,2);
# Qin - W = m*Cv*dT, Since dt=0
Q=W;
Sgen=Q/Tsys-Q/TR;
Xdestroyed=T0*Sgen;
print'the exergy destroyed %f kJ/K'%round(Xdestroyed,2);
Wrev=T0*Q/Tsys-Wsurr+(1-T0/TR)*Q;
print'the reversible work is done in the process %f kJ'%round(Wrev,2);
the useful work output 1.430000 kJ
the exergy destroyed 1.210000 kJ/K
the reversible work is done in the process 2.640000 kJ

Example 8-15 ,Page No.460¶

In :
#calculation error in textbook in part - b which changes all the following answers

#given data
m=8.0;#mass flow rate in kg/s
T0=298.0;#temperature of surrounding in K
P0=100.0;#pressure of surrounding in kPa
P1=3000.0;#inlet pressure in kPa
T1=450.0;#inlet temperature in K
P2=200.0;#outlet pressure in kPa
T2=150.0;#outlet temperature in K
Qout=300.0;#heat rate in kW

#from Table A-6 and A-4
h1=3344.9;
s1=7.0856;
h2=2769.1;
s2=7.2810;
h0=104.83;
s0=0.3672;

#calculations
# Ein = Eout
Wout=m*(h1-h2)-Qout;
Wout=round(Wout)
print'the actual power output %i kW'%Wout
# Xin = Xout
Wrev=m*(h1-h2)-(T0*(s1-s2));
Wrev=round(Wrev);
print'the maximum possible work output %i kW'%Wrev;
nII=Wout/Wrev;
print'second law efficiency is %f'%round(nII,3);
Xdestroyed=Wrev-Wout;
print'the exergy destroyed %i kW'%round(Xdestroyed);
X1=h1-h0-T0*(s1-s0);
print'the exergy of the steam at inlet conditions %i kJ/kg'%round(X1);
the actual power output 4306 kW
the maximum possible work output 4665 kW
second law efficiency is 0.923000
the exergy destroyed 359 kW
the exergy of the steam at inlet conditions 1238 kJ/kg

Example 8-16 ,Page No.462¶

In :
#given data
T0=70+460;#temperature of surroundings in R
T1=50;#temperature of water in F
T2=240;#temperature of steam in F
T3=130;#outlet temperature in F
#as dicussed in example 7-20
m1=300;#mass flow rate of water in lbm/min
m2=22.7;#mass flow rate of steam in lbm/min
m3=322.7;#outlet mass flow rate  in lbm/min

#from steam tables
h1=18.07;
s1=0.03609;
h2=1162.3;
s2=1.7406;
h3=97.99;
s3=0.18174;

#calculations
Wrev=m1*(h1-T0*s1)+m2*(h2-T0*s2)-m3*(h3-T0*s3);
Wrev=round(Wrev);
print'the reversible power %i Btu/min'%round(Wrev)
Xdestroyed=Wrev;
print'the rate of exergy destruction %i Btu/min'%Xdestroyed
the reversible power 4588 Btu/min
the rate of exergy destruction 4588 Btu/min

Example 8-17 ,Page No.463¶

In :
#given data
V=200.0;#volumne of rigid tank in m^3
P1=100.0;#intial surroundings in kPa
P2=1000.0;#final pressure in kPa
P0=100.0;#pressure of surroundings in kPa
T=300.0;#temperature of surroundings in K

#constants used
R=0.287;#in kPa m^3/kg K

#calculations
#Xin - Xout = Xdestroyed = X2 - X1
m2=P2*V/(R*T);
X2=R*T*(log(P2/P0)+P0/P2-1);
Wrev=m2*X2/1000;
Wrev=round(Wrev);
print'Work requirement %i MJ'%Wrev
Work requirement 281 MJ