# Chapter 7 :- Exergy Analysis¶

## Example 7.1 Page no-279

In :
# 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.'

-> The specific exergy of the gas is  368.912 kJ/kg.

## Example 7.2 Page no-280

In :
# 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–mass assembly is ',round(deltaEiso,2),'kJ.'

-> 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–mass assembly is  -96.2 kJ.

## Example 7.3 Page no-287

In :
# 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.'

-> 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.

## Example 7.4 Page no-289

In :
# 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.'

-> 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.

## Example 7.5 Page no-295

In :
# 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.'

-> 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.

## Example 7.6 Page no-296

In :
# 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.'

-> 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.

## Example 7.7 Page no-299

In :
# 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 '• Rate of exergy destruction',Ed,'kJ/kg.'

-> 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 .
• Rate of exergy destruction 148.484885714 kJ/kg.

## Example 7.8 Page no-300

In :
# 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 '• Rate of exergy out'
print '-> power developed',1772.8-netREout-Eddot-EdDot,'kJ/kg.'
print '-> water stream  ',netREout
print '• Rate of exergy destruction'
print '-> heat-recovery steam generator',Eddot,'kJ/kg'
print '-> turbine',EdDot

-> balance sheet
- Net rate of exergy in: 1775.34276 kJ/kg.
-> Disposition of the exergy:
• Rate of exergy out
-> power developed 874.25724 kJ/kg.
-> water stream   210.180672
• Rate of exergy destruction
-> heat-recovery steam generator 366.018792 kJ/kg
-> turbine 322.343296

## Example 7.9 Page no-302

In :
# 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)

-> 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

## Example 7.10 Page no-313

In :
# 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.'

-> 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.