# Chapter 8 :- Vapour Power Systems¶

## Example 8.1 Page no-331

In [1]:
# Given:-
p1 = 8.0                                            # pressure of saturated vapor entering the turbine in MPa
p3 = 0.008                                          # pressure of saturated liquid exiting the condenser in MPa
Wcycledot = 100.00                                  # the net power output of the cycle in MW

# Analysis
# From table A-3
h1 = 2758.0                                         # in kj/kg
s1 = 5.7432                                         # in kj/kg.k
s2 = s1
sf = 0.5926                                         # in kj/kg.k
sg = 8.2287                                         # in kj/kg.k
hf = 173.88                                         # in kj/kg
hfg = 2403.1                                        # in kj/kg
v3 = 1.0084e-3                                      # in m^3/kg

# State 3 is saturated liquid at 0.008 MPa, so
h3 = 173.88                                         # in kj/kg

# Calculations
x2 = (s2-sf)/(sg-sf)                                # quality at state 2
h2 = hf + x2*hfg
p4 = p1
h4 = h3 + v3*(p4-p3)*10**6*10**-3                   # in kj/kg

# Part(a)
#Mass and energy rate balances for control volumes around the turbine and pump give, respectively
wtdot = h1 - h2
wpdot = h4-h3

# The rate of heat transfer to the working fluid as it passes through the boiler is determined using mass and energy rate balances as
qindot = h1-h4

eta = (wtdot-wpdot)/qindot                           # thermal efficiency)

# Result for part a
print '-> The thermal efficiency for the cycle is ',round(eta,2)

# Part(b)
bwr = wpdot/wtdot                                    # back work ratio

# Result
print '-> The back work ratio is ',bwr

# Part(c)
mdot = (Wcycledot*10**3*3600)/((h1-h2)-(h4-h3))       # mass flow rate in kg/h

# Result
print '-> The mass flow rate of the steam is',round(mdot,2),'kg/h .'

# Part(d)
Qindot = mdot*qindot/(3600*10**3)                     # in MW

# Results
print '-> The rate of heat transfer,Qindot , into the working fluid as it passes through the boiler, is',round(Qindot,2),'MW.'

# Part(e)
Qoutdot = mdot*(h2-h3)/(3600*10**3)                   # in MW

# Results
print '-> The rate of heat transfer,Qoutdot from the condensing steam as it passes through the condenser, is',round(Qoutdot,2),'MW.'

# Part(f)
# From table A-2
hcwout= 146.68                                       # in kj/kg
hcwin= 62.99                                         # in kj/kg
mcwdot= (Qoutdot*10**3*3600)/(hcwout-hcwin)          # in kg/h

# Results
print '-> The mass flow rate of the condenser cooling water is',round(mcwdot,2),'kg/ h.'

-> The thermal efficiency for the cycle is  0.37
-> The back work ratio is  0.00836692570976
-> The mass flow rate of the steam is 376902.57 kg/h .
-> The rate of heat transfer,Qindot , into the working fluid as it passes through the boiler, is 269.7 MW.
-> The rate of heat transfer,Qoutdot from the condensing steam as it passes through the condenser, is 169.7 MW.
-> The mass flow rate of the condenser cooling water is 7299844.18 kg/ h.

## Example 8.2 Page no-338

In [2]:
# Given:-
etat= .85                                   # given that the turbine and the pump each have an isentropic efficiency of 85%
# Analysis
# State 1 is the same as in Example 8.1, so
h1 = 2758.0                                 # in kj/kg
s1 = 5.7432                                 # in kj/kg.k
# From example 8.1
h1 = 2758.0                                 # in kj/kg
h2s = 1794.8                                # in kj/kg
# State 3 is the same as in Example 8.1, so
h3 = 173.88                                 # in kj/kg

# Calculations
h2 = h1 - etat*(h1-h2s)                     # in kj/kg
wpdot = 8.06/etat                           # where the value 8.06 is obtained from example 8.1

h4 = h3 + wpdot

# Part(a)
eta = ((h1-h2)-(h4-h3))/(h1-h4)             # thermal efficiency

# Result for part (a)
print '-> Thermal efficiency is:  ',round(eta,3)

# Part(b)
Wcycledot = 100                             # given,a net power output of 100 MW
# Calculations
mdot = (Wcycledot*(10**3)*3600)/((h1-h2)-(h4-h3))
# Result for part (b)
print '-> The mass flow rate of steam, in kg/h, for a net power output of 100 MW is   ',round(mdot,3),'kg/h.'

# Part(c)
Qindot = mdot*(h1-h4)/(3600 * 10**3)
# Result
print '-> The rate of heat transfer Qindot into the working fluid as it passes through the boiler, is   ',round(Qindot,3),'MW.'

# Part(d)
Qoutdot = mdot*(h2-h3)/(3600*10**3)
# Result
print '-> The rate of heat transfer  Qoutdotfrom the condensing steam as it passes through the condenser, is   ',round(Qoutdot,3),'MW.'

# Part(e)
# From table A-2
hcwout = 146.68                             # in kj/kg
hcwin = 62.99                               # in kj/kg
mcwdot = (Qoutdot*10**3*3600)/(hcwout-hcwin)
# Result
print '-> The mass flow rate of the condenser cooling water, is:    ',round(mcwdot,3),'kg/h.'

-> Thermal efficiency is:   0.314
-> The mass flow rate of steam, in kg/h, for a net power output of 100 MW is    444863.139 kg/h.
-> The rate of heat transfer Qindot into the working fluid as it passes through the boiler, is    318.156 MW.
-> The rate of heat transfer  Qoutdotfrom the condensing steam as it passes through the condenser, is    218.156 MW.
-> The mass flow rate of the condenser cooling water, is:     9384172.373 kg/h.

## Example 8.3 Page no-341

In [3]:
# Given:-
T1 = 480.0                        # temperature of steam entering the first stage turbine in degree celcius
p1 = 8.0                          # pressure of steam entering the first stage turbine in MPa
p2 = 0.7                          # pressure of steam exiting the first stage turbine in MPa
T3 = 440.0                        # temperature of steam before entering the second stage turbine
Pcond = 0.008                     # condenser pressure in MPa
Wcycledot = 100.0                 # the net power output in MW

# Analysis
# From table A-4
h1 = 3348.4                       # in kj/kg
s1 = 6.6586                       # in kj/kg.k
s2 = s1                           # isentropic expansion through the first-stage turbine
# From table A-3
sf = 1.9922                       # in kj/kg.k
sg = 6.708                        # in kj/kg.k
hf = 697.22                       # in kj/kg
hfg = 2066.3                      # in kj/kg

# Calculations
x2 = (s2-sf)/(sg-sf)
h2 = hf + x2*hfg
# State 3 is superheated vapor with p3 = 0.7 MPa and T3=  440C, so from Table A-4
h3 = 3353.3                       # in kj/kg
s3 = 7.7571                       # in kj/kg.k
s4 = s3                           # isentropic expansion through the second-stage turbine
# For determing quality at state 4,from table A-3
sf = 0.5926                       # in kj/kg.k
sg = 8.2287                       # in kj/kg.k
hf = 173.88                       # in kj/kg
hfg = 2403.1                      # in kj/kg

# Calculations
x4 = (s4-sf)/(sg-sf)
h4 = hf + x4*hfg

# State 5 is saturated liquid at 0.008 MPa, so
h5 = 173.88
# The state at the pump exit is the same as in Example 8.1, so
h6 = 181.94

# Part(a)
eta = ((h1-h2)+(h3-h4)-(h6-h5))/((h1-h6)+(h3-h2))
# Result
print '-> The thermal efficiency of the cycle is:',round(eta,2)

# Part(b)
mdot = (Wcycledot*3600*10**3)/((h1-h2)+(h3-h4)-(h6-h5))
print '-> The mass flow rate of steam, is:',round(mdot,2),'kg/h.'

# Part(c)
Qoutdot = (mdot*(h4-h5))/(3600*10**3)
print '-> The rate of heat transfer Qoutdot from the condensing steam as it passes through the condenser, MW is',round(Qoutdot,2),'kg/h.'

-> The thermal efficiency of the cycle is: 0.4
-> The mass flow rate of steam, is: 236344.68 kg/h.
-> The rate of heat transfer Qoutdot from the condensing steam as it passes through the condenser, MW is 148.02 kg/h.

## Example 8.4 Page no-344

In [4]:
%matplotlib inline

Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.
In [5]:
# Given :-
# Part (a)
etat = 0.85                                                                      # given efficiency
# From the solution to Example 8.3, the following specific enthalpy values are known, in kJ/kg
h1 = 3348.4
h2s = 2741.8
h3 = 3353.3
h4s = 2428.5
h5 = 173.88
h6 = 181.94

# Calculations
h2 = h1 - etat*(h1 - h2s)                                                 # The specific enthalpy at the exit of the first-stage turbine in kj/kg
h4 = h3 - etat*(h3-h4s)                                                   # The specific enthalpy at the exit of the second-stage turbine in kj/kg
eta = ((h1-h2)+(h3-h4)-(h6-h5))/((h1-h6)+(h3-h2))

# Result
print '-> The thermal efficiency is:  ',eta

# Part (b)
from numpy import linspace
from pylab import *

h2 = []
h4 = []
y  = []
x = linspace(0.85,1,50)
for i in range(0,50):
h2.append(i)
h4.append(i)
y.append(i)
h2[i] = h1 - x[i]*(h1 - h2s)                                      # The specific enthalpy at the exit of the first-stage turbine in kj/kg
h4[i] = h3 - x[i]*(h3-h4s)                                        # The specific enthalpy at the exit of the second-stage turbine in kj/kg
y[i]  = ((h1-h2[i])+(h3-h4[i])-(h6-h5))/((h1-h6)+(h3-h2[i]))

plot(x,y)
xlabel('isentropic turbine efficiency')
ylabel('cycle thermal efficiency')
show()

-> The thermal efficiency is:   0.350865344714

## Example 8.5 Page no-348

In [6]:
# Given:-
T1 = 480.0                                                                        # temperature of steam entering the turbine in degree celcius
p1 = 8.0                                                                          # pressure of steam entering the turbine in MPa
Pcond = 0.008                                                                     # condenser pressure in MPa
etat = 0.85                                                                       # turbine efficiency
Wcycledot = 100.0                                                                 # net power output of the cycle

# Analysis
# With the help of steam tables
h1 = 3348.4                                                                       # in kj/kg
h2 = 2832.8                                                                       # in kj/kg
s2 = 6.8606                                                                       # in kj/kg.k
h4 = 173.88                                                                       # in kj/kg
# With s3s = s2, the quality at state 3s is x3s=  0.8208; using this, we get
h3s = 2146.3                                                                      # in kj/kg

# Calculations
# The specific enthalpy at state 3 can be determined using the efficiency of the second-stage turbine
h3 = h2 - etat*(h2-h3s)

# State 6 is saturated liquid at 0.7 MPa. Thus,
h6 = 697.22                                                                        # in kj/kg
# For determining specific enthalpies at states 5 and 7 ,we have
p5 = 0.7                                                                           # in MPa
p4 = 0.008                                                                         # in MPa
p7 = 8.0                                                                           # in MPa
p6 = 0.7                                                                           # in MPa
v4 = 1.0084e-3                                                                     # units in m^3/kg,obtained from steam tables
v6 = 1.1080e-3                                                                     # units in m^3/kg,obtained from steam tables

# Calculations
h5 = h4 + v4*(p5-p4)*10**6*10**-3                                                  # in kj/kg
h7 = h6 + v6*(p7-p6)*10**3                                                         # in kj/kg

# Applying mass and energy rate balances to a control volume enclosing the open heater, we find the fraction y of the flow extracted at state 2 from
y = (h6-h5)/(h2-h5)

# Part(a)
wtdot = (h1-h2) + (1-y)*(h2-h3)                                                     # the total turbine work output, units in KJ/Kg
wpdot = (h7-h6) + (1-y)*(h5-h4)                                                     # The total pump work per unit of mass passing through the first-stage turbine,in KJ/kg
qindot = h1 - h7                                                                    # in kj/kg
eta = (wtdot-wpdot)/qindot

# Results
print '-> The thermal efficiency is:',round(eta,2)

# Part(b)
m1dot = (Wcycledot*3600*10**3)/(wtdot-wpdot)

# Results
print '-> The mass flow rate of steam entering the first turbine stage, is:',round(m1dot,2),'kg/h.'

-> The thermal efficiency is: 0.37
-> The mass flow rate of steam entering the first turbine stage, is: 368948.05 kg/h.

## Example 8.6 Page no-352

In [7]:
# Given:-
# Analysis
# State 1 is the same as in Example 8.3, so
h1 = 3348.4                                                                     # in kj/kg
s1 = 6.6586                                                                     # in kj/kg.k
# State 2 is fixed by p2  2.0 MPa and the specific entropy s2, which is the same as that of state 1. Interpolating in Table A-4, we get
h2 = 2963.5                                                                     # in kj/kg
# The state at the exit of the first turbine is the same as at the exit of the first turbine of Example 8.3, so
h3 = 2741.8                                                                     # in kj/kg
# State 4 is superheated vapor at 0.7 MPa, 440C. From Table A-4,
h4 = 3353.3                                                                     # in kj/kg
s4 = 7.7571                                                                     # in kj/kg.k
# Interpolating in table A-4 at p5 = .3MPa and s5 = s4, the enthalpy at state 5 is
h5 = 3101.5                                                                     # in kj/kg
# Using s6 = s4, the quality at state 6 is found to be
x6 = 0.9382
# Using steam tables, for state 6
hf = 173.88                                                                     # in kj/kg
hfg = 2403.1                                                                    # in kj/kg

h6 = hf + x6*hfg

# At the condenser exit, we have
h7 = 173.88                                                                     # in kj/kg
v7 = 1.0084e-3                                                                  # in m^3/kg
p8 = 0.3                                                                        # in MPa
p7 = 0.008                                                                      # in MPa

h8 = h7 + v7*(p8-p7)*10**6*10**-3                                               # The specific enthalpy at the exit of the first pump in kj/kg
# The liquid leaving the open feedwater heater at state 9 is saturated liquid at 0.3 MPa. The specific enthalpy is
h9 = 561.47                                                                     # in kj/kg

# For the exit of the second pump,
v9 = 1.0732e-3                                                                  # in m^3/kg
p10 = 8.0                                                                       # in MPa
p9 = 0.3                                                                        # in MPa
h10 = h9 + v9*(p10-p9)*10**6*10**-3                                             # The specific enthalpy at the exit of the second pump in kj/kg
# The condensate leaving the closed heater is saturated at 2 MPa. From Table A-3,
h12 = 908.79                                                                    # in kj/kg
h13 = h12                                                                       # since The fluid passing through the trap undergoes a throttling process
# For the feedwater exiting the closed heater
hf = 875.1                                                                      # in kj/kg
vf = 1.1646e-3                                                                  # in m^3/kg
p11 = 8.0                                                                       # in MPa
psat = 1.73                                                                     # in MPa
h11 = hf + vf*(p11-psat)*10**6*10**-3                                           # in kj/kg

ydash = (h11-h10)/(h2-h12)                                                      # the fraction of the total flow diverted to the closed heater
ydashdash = ((1-ydash)*h8+ydash*h13-h9)/(h8-h5)                                 # the fraction of the total flow diverted to the open heater

# Part(a)
wt1dot = (h1-h2) + (1-ydash)*(h2-h3)                                            # The work developed by the first turbine per unit of mass entering in kj/kg
wt2dot = (1-ydash)*(h4-h5) + (1-ydash-ydashdash)*(h5-h6)                        # The work developed by the second turbine per unit of mass in kj/kg
wp1dot = (1-ydash-ydashdash)*(h8-h7)                                            # The work for the first pump per unit of mass in kj/kg
wp2dot = h10-h9                                                                 # The work for the second pump per unit of mass in kj/kg
qindot = (h1-h11) + (1-ydash)*(h4-h3)                                           # The total heat added expressed on the basis of a unit of mass entering the first
# turbine
eta = (wt1dot+wt2dot-wp1dot-wp2dot)/qindot                                      # thermal efficiency

# Result
print '-> The thermal efficiency is:  ',round(eta,2)

# Part(b)
Wcycledot = 100.0                                                               # the net power output of the cycle in MW
m1dot = (Wcycledot*3600*10**3)/(wt1dot+wt2dot-wp1dot-wp2dot)

# Result
print '-> The mass flow rate of the steam entering the first turbine, in kg/h is:  ',round(m1dot,2)

-> The thermal efficiency is:   0.43
-> The mass flow rate of the steam entering the first turbine, in kg/h is:   280126.53

## Example 8.7 Page no-360

In [8]:
# Given:-
# Analysis
# The solution to Example 8.2 gives
h1 = 2758                                                                   # in kj/kg
h4 = 183.36                                                                 # in kj/kg
# From table A-22
hi = 1491.44                                                                # in kj/kg
he = 843.98                                                                 # in kj/kg
# Using the conservation of mass principle and energy rate balance, the ratio of mass flow rates of air and water is
# From example 8.2
mdot = 4.449e5                                                              # in kg/h

# Part(a)
T0 = 295                                                                    # in kelvin
# From table A-22
si = 3.34474                                                                # in kj/kg.k
se = 2.74504                                                                # in MW
# Calculation
Rin = madot*(hi-he-T0*(si-se))/(3600*10**3)                                 # The net rate at which exergy is carried into the heat exchanger
# unit by the gaseous stream
# Result
print '-> The net rate at which exergy is carried into the heat exchanger unit by the gas stream, is:',round(Rin,2),'MW '

# Part(b)
# From table A-3
s1 = 5.7432                                                                 # in kj/kg.k
# From interpolation in table A-5 gives
s4 = 0.5957                                                                 # in kj/kg.k
# Calculation
Rout = mdot*(h1-h4-T0*(s1-s4))/(3600*10**3)                                 # in MW
# Result
print '-> The net rate at which exergy is carried from the heat exchanger by the water stream, is:',round(Rout,2),'MW .'

# Part(c)
Eddot = Rin-Rout                                                            # in MW
# Result
print '-> The rate of exergy destruction, in MW is:',round(Eddot,2)

# Part(d)
epsilon = Rout/Rin
# Result
print '-> The exergetic efficiency is: ',round(epsilon,2)

-> The net rate at which exergy is carried into the heat exchanger unit by the gas stream, is: 231.24 MW
-> The net rate at which exergy is carried from the heat exchanger by the water stream, is: 130.52 MW .
-> The rate of exergy destruction, in MW is: 100.72
-> The exergetic efficiency is:  0.56

## Example 8.8 Page no-362

In [9]:
# Given:-
T0 = 295.00                                                                 # in kelvin
P0 = 1.00                                                                   # in atm

# Analysis
# From table A-3
s1 = 5.7432                                                                 # in kj/kg.k
s3 =0.5926                                                                  # in kj/kg.k

# Using h2 = 1939.3 kJ/kg from the solution to Example 8.2, the value of s2 can be determined from Table A-3 as
s2 = 6.2021                                                                 # in kj/kg.k
s4 = 0.5957                                                                 # in kj/kg.k
mdot = 4.449e5                                                              # in kg/h

# Calculations
Eddot = mdot*T0*(s2-s1)/(3600*10**3)                                         # the rate of exergy destruction for the turbine in MW
EddotP = mdot*T0*(s4-s3)/(3600*10**3)                                        # the exergy destruction rate for the pump

# Results
print '-> The rate of exergy destruction for the turbine is: ',round(Eddot,2),'MW.'
# From the solution to Example 8.7, the net rate at which exergy is supplied by the cooling combustion gases is 231.28 MW
print '-> The turbine rate of exergy destruction expressed as a percentage is:  ',round((Eddot/231.28)*100)
# However, since only 69% of the entering fuel exergy remains after the stack loss and combustion exergy destruction are accounted for,
# it can be concluded that
print '-> Percentage of the exergy entering the plant with the fuel destroyed within the turbine is:',round(0.69*(Eddot/231.28)*100,2)
print '-> The exergy destruction rate for the pump in MW is:',round(EddotP,2)
print 'and expressing this as a percentage of the exergy entering the plant as calculated above, we have',round((EddotP/231.28)*69,2)
print '-> The net power output of the vapor power plant of Example 8.2 is 100 MW. Expressing this as a percentage of the rate at which exergy is '
print 'carried into the plant with the fuel, ',round((100/231.28)*69,2)

-> The rate of exergy destruction for the turbine is:  16.73 MW.
-> The turbine rate of exergy destruction expressed as a percentage is:   7.0
-> Percentage of the exergy entering the plant with the fuel destroyed within the turbine is: 4.99
-> The exergy destruction rate for the pump in MW is: 0.11
and expressing this as a percentage of the exergy entering the plant as calculated above, we have 0.03
-> The net power output of the vapor power plant of Example 8.2 is 100 MW. Expressing this as a percentage of the rate at which exergy is
carried into the plant with the fuel,  29.83

## Example 8.9 Page no-364

In [10]:
# Given:-
T0 = 295                                                                # in kelvin
# Analysis
# From solution to Example 8.2.
mcwdot = 9.39e6                                                         # mass flow rate of the cooling water in kg/h

# Part(a)
# With saturated liquid values for specific enthalpy and entropy from Table A-2
he = 146.68                                                             # in kj/kg
hi = 62.99                                                              # in kj/kg
se = 0.5053                                                             # in kj/kg.k
si = 0.2245                                                             # in kj/kg.k
# Calculations
Rout = mcwdot*(he-hi-T0*(se-si))/(3600*10**3)                           # The net rate at which exergy is carried out of the condenser in MW
# Results
print '-> The net rate at which exergy is carried from the condenser by the cooling water, is:',round(Rout,2),'MW.'
print '-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get ',round((Rout/231.28)*69,2),'percent'

# Part(b)
# From table
s3 = 0.5926                                                              # in kj/kg.k
s2 = 6.2021                                                              # in kg/kg.k
mdot = 4.449e5                                                           # in kg/h
# Calculations
Eddot = T0*(mdot*(s3-s2)+mcwdot*(se-si))/(3600*10**3)                    # the rate of exergy destruction for the condenser in MW
# Results
print '-> The rate of exergy destruction for the condenser is:  ',round(Eddot,2),'MW.'
print '-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get,',round((Eddot/231.28)*69,2),'percent'

-> The net rate at which exergy is carried from the condenser by the cooling water, is: 2.23 MW.
-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get  0.66 percent
-> The rate of exergy destruction for the condenser is:   11.56 MW.
-> Expressing this as a percentage of the exergy entering the plant with the fuel, we get, 3.45 percent