Thermodynamics Lecture
Series
Ideal Rankine Cycle –The
Practical Cycle
Applied Sciences Education Research
Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
email: drjjlanita@hotmail.com
http://www5.uitm.edu.my/faculties/fsg/drjj1.html
Steam Power Plant
Example: A steam power cycle.
Combustion
Products
Steam
Turbine
Fuel
Air
Pump
Mechanical Energy
to Generator
Heat
Exchanger
Cooling Water
System Boundary
for Thermodynamic
Analysis
Second Law
Working fluid:
Water
High T Res., TH
Furnace
qin = qH
Purpose:
Produce work,
Wout, out
Steam Power Plant
qout = qL
Low T Res., TL
Water from river
An Energy-Flow diagram for a SPP
net,out
Second Law – Dream Engine
What is the maximum performance of
real engines if it can never achieve
100%??
Carnot
Cycle
P -  diagram for a Carnot (ideal) power plant
P, kPa
qin
1
2
desired output  net ,out


required input
qin
rev
4
 qin  qout 

 
 qin
 rev
3
qout
, m3/kg
Second Law – Will a Process Happen
Carnot Principles
• For heat engines in contact with the same hot
and cold reservoir
P1: 1 = 2 = 3 (Equality)
real  rev
P2: real < rev (Inequality)
Consequence
 qL 
TL (K)

 
;
 qH  rev TH (K)
rev
 qL
 1  
 qH

TL (K)
  1 
TH (K)
 rev
Processes satisfying Carnot Principles
obeys the Second Law of
Second Law – Will a Process Happen
Clausius Inequality :
• Sum of Q/T in a cyclic process must be zero
for reversible processes and negative for real
processes
q
kJ
Q
kJ
 0,
 0,
T
K
T
kg  K
Q
 0 , reversible
Q
T
 0 , real
T
Q
 0 , impossible
T





Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Isolated systems
FIGURE 6-6
The entropy
change of an
isolated
system is the
sum of the
entropy
changes of its
components,
and is never
less than
zero.
6-3
Entropy – Quantifying Disorder
Increase of Entropy Principle – closed system
The entropy of an isolated (closed and
adiabatic) system undergoing any process,
will always increase.
Sisolated  S heat  S gen  S sys  S surr  0
For pure
Ssys  m( s2  s1 )
substance
:
Qin  Qout surr
Ssurr 
and
Tsurr
Then
S gen  m( s2  s1 ) 
 
Qnet ,in
T
surr
Surrounding
System
Entropy – Quantifying Disorder
Entropy Balance – for any general system
For any system undergoing any process,
Energy must be conserved (Ein – Eout = Esys)
Mass must be conserved (min – mout = msys)
Entropy will always be generated except for
reversible processes
Entropy balance is (Sin – Sout + Sgen = Ssys)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Entropy Transfer
FIGURE 6-61
Mechanisms of
entropy transfer
for a general
system.
6-18
Entropy – Quantifying Disorder
Entropy Balance –Steady-flow device




Q in  Q out  W in  W out


  
  
  m    m  , k W

 out 
 in

S in  S out  S gen  Ssys  0
Then:

S gen

S gen


  S heat  S mass


Q out 

 
Tout




So , S gen  S out  S in



   S heat  S mass
out 

Q in 

m s 

 
exit Tin 





in

m s 
inlet

Entropy – Quantifying Disorder
Entropy Balance –Steady-flow device

Turbine:




Q in  Q out  W in  W out  mexit  inlet , kW



Assume adiabatic,
where minlet  mexit  m


kemass = 0,
0  0  0  W out  mh4  h3 , kW
pemass = 0





Q
Q
In,3
Entropy S gen  out  in  m 4 s  m3 s , kW
4
3
T
T
K
Balance
out
in

S gen

kW
 0  0  ms4  s3 ,
K
Out
Entropy – Quantifying Disorder
Entropy Balance –Steady-flow device
Mixing Chamber:




Q in  Q out  W in  W out  




 
 m   

exit



 
, kW
 m  

inlet

Qin  Qout  W in  W out  m3 h3  m2 h2  m1 h1 , kW

S gen





Qout Q in
kW 3


 m3 s3  m2 s2  m1 s1 ,
Tout
Tin
K


where minlet  mexit
1
2
Vapor Cycle
Steam Power Plant
External combustion
Fuel (qH) from nuclear reactors, natural gas, charcoal
Working fluid is H2O
Cheap, easily available & high enthalpy of
vaporization hfg
Cycle is closed thermodynamic cycle
Alternates between liquid and gas phase
Can Carnot cycle be used for representing real
SPP??
Aim: To decrease ratio of TL/TH
Vapor Cycle – Carnot Cycle
Efficiency of a Carnot Cycle SPP
rev
rev
TL
15  273
1 
1 
 0.55
TH
374  273
TL
15  273
1 
1 
 0.627
TH
500  273
Vapor Cycle –Carnot Cycle
Impracticalities of Carnot Cycle
T, C
qin = qH
Tcrit
TH
TL
qout = qL
s1 = s2
s3 = s4
 Isothermal expansion: TH
limited to only Tcrit for H2O.
 High moisture at turbine exit
 Not economical to design
pump to work in 2-phase (end
of Isothermal compression)
 No assurance can get same x
for every cycle (end of
Isothermal compression)
s, kJ/kgK
Vapor Cycle – Alternate Carnot Cycle
Impracticalities of Alternate Carnot Cycle
T, C
qin = qH
TH
Tcrit
TL
Still Problematic
Isothermal expansion but at
variable pressure
Pump to very high pressure
Can the boiler sustain the high
P?
qout = qL
s1 = s2
s3 = s4
s, kJ/kgK
Vapor Cycle – Ideal Rankine Cycle
Overcoming Impracticalities of Carnot Cycle
Superheat the H2O at a constant pressure (isobaric
expansion)
 Can easily achieve desired TH higher than Tcrit.
 reduces moisture content at turbine exit
Remove all excess heat at condenser
 Phase is sat. liquid at condenser exit, hence need only
a pump to increase pressure
 Quality is zero for every cycle at condenser exit
(pump inlet)
Vapor Cycle – Ideal Rankine Cycle
Working fluid:
Water
High T Res., TH
Furnace
qin = qH
Boiler
Pump
Turbin
e
in
out
Condenser
qin - qout = out - in
qout = qL
Low T Res., TL
Water from river
qin - qout = net,out
A Schematic diagram for a Steam Power Plant
Vapor Cycle – Ideal Rankine Cycle
T- s diagram for an Ideal Rankine Cycle
T, C
3
TH
boiler
Tcrit q = q
in
H
Tsat@P2
TL= Tsat@P4
in
pump
turbine
PH
PL
out
2
1 q =q
out
L
s1 = s2
4
condenser s3 = s4
s, kJ/kgK
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-2
The simple
ideal Rankine
cycle.
9-2
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
qin = qH
In,2
Out,3
Assume ke =0, pe =0 for
the moving mass, kJ/kg
Boiler
qin – qout+ in – out = qout – qin, kJ/kg
qin – 0 + 0 – 0 = hexit – hinlet, kJ/kg
qin = h3 – h2, kJ/kg


Qin = m(h3 – h2), kJ
Q in  mh3  h2 , kW
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
Assume ke =0, pe =0 for
the moving mass, kJ/kg
qin – qout+ in – out = qout – qin, kJ/kg
0 – qout + 0 – 0 = hexit – hinlet
- qout = h1 – h4,
So, qout = h4 – h1, kJ/kg
In,4
Out,1
Condenser
qout = qL
Qout = m(h4 – h1), kJ


Q out  mh4  h1 , kW
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
In,3
Turbin
e
qin – qout+ in – out = qout – qin, kJ/kg
0 – 0 + 0 – out = hexit – hinlet, kJ/kg
- out = h4 – h3, kJ/kg

Out,4
So, out = h3 – h4, kJ/kg
Wout = m(h3 – h4), kJ

out
W out  mh3  h4 , kW
Assume ke =0, pe =0 for
the moving mass, kJ/kg
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
qin – qout+ in – out = qout – qin, kJ/kg
Pump
Out,2
in
0 – 0 + in – 0 = hexit – hinlet, kJ/kg
in = h2 – h1, kJ/kg
For
reversibl
e pumps
where
So, Win = m(h2 – h1), kJ
2
2
2
1
1
1
In,1
 pump ,in   Pd  dP  0    dP
 pump ,in   P2  P1   h2  h1
 2  1   f @ P1


W in  mh2  h1 , kW
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
Efficiency


qin  qout
h3  h2  h4  h1 


qin
h3  h2
 net ,out
qin
 net ,out
qin

 out   in
qin
h3  h4  h2  h1 

h3  h2
h3  h4  h2  h1
 
h3  h2
Vapor Cycle – Ideal Rankine Cycle
T- s diagram for an Ideal Rankine Cycle
T, C
Note that P1 = P4
s1 = sf@P1
h1 = hf@P1
3
TH
Tcrit
Tsat@P2
TL= Tsat@P4
in
pump
boiler
qin = qH
turbine
PH
s3 =
s@P3,T3
h3 = h@P3,T3
PL
out
2
s4 = [sf +xsfg]@P4 = s3
x
1 q =q
out
L
4
s3  sf @ P 4
sfg @ P4
h4 = [hf +xhfg]@P4
s, kJ/kgK
s1 = s 2
condenser
h2 = h1 +2(P2 – P1); where
s3 = s 4
 2  1   f @ P1
Vapor Cycle – Ideal Rankine Cycle
Energy Analysis
Increasing Efficiency
Must increase net,out = qin – qout
Increase area under process cycle
Decrease condenser pressure; P1=P4
Pmin > Psat@Tcooling+10 deg C
Superheat
T3 limited to metullargical strength of boiler
Increase boiler pressure; P2=P3
Will decrease quality (an increase in moisture).
Minimum x is 89.6%.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lowering Condenser Pressure
FIGURE 9-6
The effect of
lowering the
condenser
pressure on the
ideal Rankine
cycle.
9-4
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Superheating Steam
FIGURE 9-7
The effect of
superheating
the steam to
higher
temperatures
on the ideal
Rankine cycle.
9-5
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Increasing Boiler Pressure
FIGURE 9-8
The effect of
increasing the
boiler pressure on
the ideal Rankine
cycle.
9-6
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-10
T-s diagrams
of the three
cycles
discussed in
Example 9–3.
9-8
Vapor Cycle – Reheat Rankine Cycle
High T Reservoir, TH
qin = qH
3
Boiler
Pump
in
Hig
hP
turb
ine
2
4
qreheat
out,2
Lo
wP
turb
ine
5
1
out,1
Condenser
qout = qL
Low T Reservoir,
6
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 9-11
The ideal reheat
Rankine cycle.
9-9
Vapor Cycle – Reheat Rankine Cycle
Reheating increases  and reduces moisture in
turbine
T, C
TH
Tcrit
qprimary = h3-h2
Tsat@P3
Tsat@P4
qreheat = h5-h4
3
5
P4 = P5
P3 
out
out, II
P6 = P1
4
2
TL= Tsat@P1
in
1
s1 = s2
qout = h6-h1
6
s3 = s4 s5 = s6
s, kJ/kgK
Vapor Cycle – Reheat Rankine Cycle
Energy Analysis
q in = qprimary + qreheat = h3 - h2 + h5 - h4
qout = h6-h1
net,out = out,1 + out,2 - in = h3 - h4 + h5 - h6 – h2 + h1


 net ,out
qin
 net ,out
qin

qin  qout h3  h2  h5  h4  h6  h1 


qin
h3  h2  h5  h4
 out1   out 2  in
qin
h3  h4  h5  h6  h2  h1

h3  h2  h5  h4
Vapor Cycle – Reheat Rankine Cycle
Energy Analysis
where
s6 = [sf +xsfg]@P6. Use x = 0.896 and s5 = s6
h6 = [hf +xhfg]@P6
Knowing s5 and T5, P5 needs to be estimated
(usually approximately a quarter of P3 to ensure x
is around 89%. On the property table, choose P5 so
that the entropy is lower than s5 above. Then can
find h5 = h@P5,T5.
Vapor Cycle – Reheat Rankine Cycle
Energy Analysis
where
s1 = sf@P1
h1 = hf@P1
s3 = s@P3,T3 = s4.
h2 = h1 +2(P2 – P1); where
h3 = h@P3,T3
 2  1   f @ P1
P5 = P4.
From P4 and s4, lookup for h4 in the table. If not
found, then do interpolation.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Supercritical Rankine Cycle
FIGURE 9-9
A supercritical
Rankine
cycle.
9-7