Energy and Exergy Analyses of Power Cycles

1
Energy and Exergy Analyses
of Power Cycles
March/September, 2015
Lecturer:
Luis R. Rojas-Solórzano, Ph.D.
Associate Professor, Dept. of Mechanical Engineering
Nazarbayev University, Republic of Kazakhstan
Luis.rojas@nu.edu.kz
Micro-CV
- Ph.D., 1997 (Carnegie Mellon University-USA)
- Mech.Eng., M.Sc., 1985, 92 (Simon Bolívar
University-Venezuela)
- Member of ASME and Sigma-Xi
- + 25 years teaching Energetic Systems/Fluid
Mechanics
- + 17 years teaching CFD
- + 55 international publications
- + 20 industry technical reports
- + 50 Master and Ph.D. former-current students

CONTENT
1. Introduction. Concepts and Definitions. Pure
Substance and Ideal Gas. Work and Heat. 1st Law of
Thermodynamics. Basic Exercises. (25-min)
2. Second Law of Thermodynamics. Heat Engine
concept. Reversibility. Carnot Engine. Entropy.
Basic Exercises. (20-min)
3. Use of CyclePadTM. Build and Analyze
environments. (30-min)
4. Energy Analysis of Power Cycles. (30-min)
5. Reversible Work and Irreversibility. Exercises.
(15-min)
6. Availability and Exergy Analysis of Power
Cycles. (30-min)
Source:
http://www.lonl.cn
2

• Population growth and development of nations are pushing up global
power demand, making Power Plants of scientific interest.
• 80% of world power generation is based on fossil fuel as primary
energy (89% in Kazakhstan)  important environmental issues.
Note: Kazakhstan has 19.8 GW installed cap. and 15.8 GW available
@2012 and 7.35% T&D losses @ 2011*
• Most Power Plants are currently designed by energetic performance
criteria based only of 1st Law of Thermodynamics  quantitative
analysis only, not qualitative (useful energy loss is not considered)
• Exergetic analysis has been used in recent decades to evaluate,
optimize and improve power plants.
• Exergetic analysis gives a more meaningful assessment of
performance of individual components of power plants; thus, locating
and pinpointing causes of irreversibilities.
http://www.sciencedirect.com/science/article/pii/S1364032110004399
1. Introduction. On Energy and Exergy Analyses.
3
1. Introduction
sources: http://www.reegle.info/policy-and-regulatory-overviews/KZ Based on REEEP Policy Database and (*) Wold Bank

What is Thermodynamics?
Branch of Physics related to Heat and Work, and to all
macroscopic properties of substances related to these two
concepts.
Source: www.ftexploring.com/ energy/first-law_p2.html
1. Introduction to Thermodynamics.
4
1. Introduction

Concepts and definitions
Laws of Thermodynamics:
Zeroth Law (Transitivity of Thermal Equilibrium):
If two thermodynamic systems A and B are in thermal equilibrium, being B and C also in
thermal equilibrium, then A and C are also in thermal equilibrium.
1st Law (Energy Conservation):
The increase in the energy of a closed system is equal to the amount of energy added to
the system by heating, minus the amount lost as work done by the system on its
surroundings.
2nd Law (Entropy):
The total entropy of any isolated thermodynamic system tends to increase over time,
approaching a maximum value.
3rd Law (Absolute Zero Temperature):
When the temperature of a system approaches the absolute zero, all processes stop and
the entropy of the system reaches a minimum value, or zero for a perfect crystalline
substance.
5
1. Introduction. Concepts and Definitions

Thermodynamic System (or Closed System):
Surroundings
System
Constant mass, with possible exchange of heat-work with
surroundings.
Isolated System:
There is not energy exchange (heat-work) with surroundings.
Models: piston-cylinder, rigid container, elastic balloons, etc...
6
Mass
Fixed or moving
boundary (real or virtual)

Open System or Control Volume:
• Interchangeability of mass and energy (heat - work) with
surroundings through control surface.
• Models: pumps, compressors, turbines, boilers, condensers, heat
exchangers, tanks emptying - filling, etc ...
1m
2m
Source: www.ae.su.oz.au/.../ cvanalysis/node35.html
Control Surface
Control Volume
Surroundings
7

Properties of a Pure Substance and Ideal Gas
Pure Substance: substance w/homogeneous chemical
composition (solid - liquid - gas). Well defined Tsat vs. psat .
Example: H2O
Saturation Temperature:
8
T at which a change of phase might
occur, at a given pressure,
denominated Saturation Pressure.
Example: Tsat = 100 °C @ 1 atm for
H2O
Source: http://www.atmos.washington.edu/2003Q3/101/notes/SaturationVaporPressure3.gif
1. Introduction. Pure Substance and Ideal Gas

Saturated liquid: liquid at (p,T)sat
Saturated vapor: gas at (p,T)sat
9Source: http://www.atmos.washington.edu/2003Q3/101/notes/SaturationVaporPressure3.gif
Vapor quality ´x´: gas mass
fraction in saturated mixture
of liquid-gas. x  [0,1]:
x = mg/(mg+mf)

Compressed or Subcooled liquid:
liquid at p > psat @ given T
or
liquid at T < Tsat @ given p
Superheated vapor:
gas at T > Tsat @ given p
or
gas at p < psat @ given T
10

v
T
Properties of saturated liquid-gas mixtures:
t
t
gf
gf
mmixture
m
V
mm
VV
vv 



Vapor (g)
Liquid (f)
fff vmV 
ggg vmV 
fg
g
mm
m
x


fgm vxvxv ).1(. 
 
 fg
fm
vv
vv
x



11Source:
http://www.wiley.com/college/moran/CL_0471465704_S/user/tutorials/tutor
ial2/PvT_Diag2/tv_dome.jpg
Thermodynamic dome

Thermodynamic Tables
Compressed or subcooled liquid
°C
m3/kg kJ/kg kJ/kg-K Tsat
12
v
T
Source:
http://www.wiley.com/college/moran/CL_0471465704_S/us
er/tutorials/tutorial2/PvT_Diag2/tv_dome.jpg

Saturated Liquid-Vapor
13
v
T
Source:

Superheated Vapor
14
v
T
Source:

Properties of a Pure Substance and Ideal Gas
Ideal Gas: gas in which the molecules do
not interact, equivalent to (p, )  0
State Equation:
TRvp 
p: absolute pressure
: molar specific volume
: universal constant
T: absolute temperature
v
R
÷ M (molecular weight)
RTpv 
p: absolute pressure
v: specific volume
R: gas constant
T: absolute temperature
Eg., Air as ideal gas
R = 287J/kg-K = 53,34 BTU/lbm-°R
15

Ideal Gas
Compressibility Diagram (N2)
What are the necessary conditions to approximate a real
gas as an Ideal Gas?
In general:
ZRTpv 
Z  1, if:
p < 400 psia and T ~ 25 °C
or
T > -130 °C and p ~ 1 atm
Error < 1 %
Source: Potter & Somerton, 1993
Source: http://www.chem.queensu.ca/people/faculty/mombourquette/FirstYrChem/GasLaws/
16

Ideal Gas
Special considerations of the model:
• Applies to superheated vapor, within the
recommended (p,T) range, when there are not
available Thermodynamic Tables.
• For a given Ideal Gas:
Boyle-Mariotte (T_ctte) / Charles-Gay-Lussac (p_ctte)_Laws
2
22
1
11
T
vp
T
vp
For Control Volume stations:
For System states:
2
22
1
11
T
Vp
T
Vp

17

Work (W) and Heat (Q)
Work: energy transfer form of a system (open or
closed) to another, which is manifested by
applying a force and generate a
displacement:
xdFW

.
Units:
International System  Nm = Joule = J
Imperial System  lbf-ft
Conversion  1055 J = 778 lbf-ft
Trajectory function!
18
1. Introduction. Work and Heat

Work on/from simple compressible substances*
Source: www.ux1.eiu.edu/~cfadd/ 1360/20Heat/Work.html
pdV
A
dV
ApFdxW  .
 
2
1
V
V
pdVW
12 WWW 
  21WW
19
(*) substances for which the state can be fully determined with 2 independent intensive properties

Wprocess = area under the curve
Wcycle = enclosed area by cycle
Power:
dt
W
W


Units:
International System  J/s = W (Watt)
Imperial System  hp (horse power)
Conversion  1 hp = 0.7455 kW
(+) made by system against
surroundings
(- ) made against the system
WW ,Convention:
20
Work on/from simple compressible substances (cont´d)

Special cases:
• Irrestrict expansion (against absolute vacuum)
p = 0 absp > 0 abs
021 W(1) (2)
• Work made by/against impeller-rotor (Eg., turbine)
dtWW
t
t
2
1
21

21
Work on/from simple compressible substances (cont´d)

Work (W) and Heat (Q)
Heat : form of energy transfer from one system (open
or closed) to another system, which is the
response to a temperature gradient (Conduction-
Convection-Radiation)
Source: http://www.oxfordreference.com/media/images/30740_0.jpg
TA TB
TA > TB
Q
dTQ 
22

Heat: particularities
• Trajectory function, i.e., depends more on
thermodynamic process than on extreme states:
21QQ12
2
1
QQQ  
• Heat flow through boundary:
dt
Q
Q


23

Heat: particularities
• Units:
International System: Calorie = Cal; Cal/hr
1 Cal is the needed amount of heat to raise the temperature of 1g of water, from
14,5 ºC to 15,5 ºC, at 1 atm.
Imperial System: BTU; BTU/hr
1 BTU (British Thermal Unit) quantity of heat that needs to be transfered to 1 lbm
of water to raise its temperature from 39,5 ºF to 40,5 ºF, at 1 atm.
Conversion: 860 kCal = 3412 BTU
(+) to the system
(- ) from the system
Convention:
QQ ,
QQ ,
24

25
The net heat transfer during a thermodynamic cycle is equal
to the net work performed during the same cycle.
1st Law of Thermodynamics
  WQ 
1st Law of Thermodynamics for process 12:
dEWQ  EWQ  2121
Where ‘E’ is the Total Energy corresponding to each state:
UEEE pk 
1. Introduction. First Law of Thermodynamics

26
Total Energy
2
2
1
mCEk 
UEEE pk 
mgHEp  U
Kinetic Energy Potential Energy Internal Energy
The reference level determines the values of E´s
 E is what really matters!

27
piston
fluid
 Q
piston
fluid
1 2
slowly
isobaric
Enthalpy
1st Law:
  pk EEUUWQ  122121
  pk EEUUVVpQ  121221 )(
   1221 VVppdVWy
    pk EEpVUpVUQ  112221

28
piston
fluid
 Q
piston
fluid
1 2
slowly
isobaric
Enthalpy (cont´d.)
1st Law:
    pk EEpVUpVUQ  112221
Definition of Enthalpy: pVUH 
pk EEHHQ  1221

29
Enthalpy: particularities
• Specific Enthalpy: h = H/m
• h for saturated mixtures
  fgm hxxhh  1
• Zero-level of reference, h = 0 kJ/kg for:
Saturated steam @ 0°C (32 °F), 1 atm
Air @ 0 °C (Int. Sys.) or Air @ 0 °F (Imp. Sys.), 1 atm
• hfg represents the Latent Heat of Vaporization

30
Enthalpy: particularities
• h for Ideal Gas. Joule´s Experiment
Source: http://upload.wikimedia.org/wikibooks/en/5/5e/Joule_Engineering_Thermodynamics.png
Air
22 atm 0 atm
0EE;0;0 pk2121  WQ
1st Law: )(0 _ TfUU GasIdeal 
Then, since H=U+pV ⇨ )(_ TfH GasIdeal 

31
Specific Heat
• Cp: Specific heat at constant pressure
• Cv: Specific heat at constant volume
• Cp and Cv for Ideal Gas: since h = f(T) and u = g(T)

dT
dh
Cpo
TCdTCh popogasideal  _

dT
du
Cvo
TCdTCu vovogasideal  _
p
p
T
h
C 





v
v
T
u
C 






32
Analysis by 1st Law for Control Volume (CV)
Mass Conservation for CV
t t + t
mt)cv mt+t)cv
System
Control
Volume
dmi dme dmedmi
 systemttsystemt mm    ecvtticvt dmmdmm  
  
t
dm
Lim
t
dm
Lim
t
mm
Lim e
t
i
t
cvtcvtt
t 







 000

33
Mass Conservation for CV (cont´d)
t t + t
mt)cv mt+t)cv
System
Control
Volume
dmi dme dmedmi
ei
cv
mm
dt
dm
 
  
t
dm
Lim
t
dm
Lim
t
mm
Lim e
t
i
t
cvtcvtt
t 







 000

34
Mass Conservation for CV (cont´d)
AdxdV 
How to calculate the inlet-outlet flow into-from the CV?
A y dVdm 
Adxdm 








  t
dx
ALim
t
dm
Lim
tt

00
ACm 
22211121 CACAmm   
In Steady State regime  Continuity Equation:

35
Energy Conservation (1st Law) for CV
t t + t
Et)cv Et+t)cv
System
Control
Volume
dmi dme dmedmi
pi
vi
Ti
ei
pe
ve
Te
ee
cvW
Q
Recalling, 1st Law for System undergoing process:
  systemsystemsystem EdWQ  (I)

36
Energy Conservation (1st Law) for CV (cont´d)
Additionally:
 cvsystem QQ   (II)
   

 

    
systemtsystemtt E
iicvt
E
eecvttsystemtsystemttsystem dmeEdmeEEEdE 


(III)
  eeeiiicvsystem dmvpdmvpWW   (IV)
gz
c
ue 
2
2
(V)
hpvu  (VI)

37
Energy Conservation (1st Law) for CV (cont´d)
Sustituting (II), (III), (IV), (V) and (VI) in (I):









































  t
dE
gz
c
h
t
dm
Limgz
c
h
t
dm
t
W
t
Q
Lim cv
e
e
e
e
t
i
i
i
icvcv
t 22
2
0
2
0

dt
dE
gz
c
hmgz
c
hmWQ cv
e
e
eei
i
iicvcv 
















22
22

1st Law of Thermodynamics for Control Volume

38
Heat Engine:
Set of components and
equipment operating harmo-
niously in thermodynamic cycle
with heat energy received from
a source at high temperature to
partially convert it into
mechanical energy and
discharge the remnant to a
reservoir at low temperature.
High Temperature (TH)
Low Temperature (TL)
Heat Engine
Power Plants Modeling: The Heat Engine

39
Heat Engine: particularities
Thermal Efficiency ‘T’’:
Represents the ratio between the
cycle-work or power and primary
cycle-energy or heat flow.
H
cycle
Th
Q
W

LHcycle QQW 
1st Law:
H
L
H
LH
Th
Q
Q
Q
QQ


 1
Heat Engine

40
Problem 4. 1st Law for CV in simple-ideal steam cycle
A thermal power plant, based on steam turbine, works with 20 kg/s of
steam, as shown. Neglecting losses in components, find out: (a) Heat
Transfer in the Boiler [MW]; (b) Turbine output Power [MW]; (c) Heat
Transfer in the Condenser [MW]; (d) Pump input power ( )
[MW]; (e) Steam flow velocity at exit of Boiler [m/s]; (f) Thermal Efficiency of
the cycle ( ). (see diagram and table with data)
  112 .. vppmWp  
Boiler
Condenser
Turbine
Pump
Gen.
BoilerQ
CondQpW
tW
  %100x
Q
WW
Boiler
pt

 

1
2
3
4
N T
[°C]
p
[kPa]
x
1 40 10 ---
2 40 10000 ---
3 600 10000 ---
4 --- 10 1
d = 30 cm
Review Exercise on 1st Law:

41
1st Law for CV:
Boiler
Condenser
Turbine
Pump
Gen.
BoilerQ
CondQpW
tW
1
2
3
4
N T
[°C]
p
[kPa]
x h
[kJ/kg]
1 40 10 --- 167,5
2 40 10000 --- 167,5hf
3 600 10000 --- 3625,3
4 --- 10 1 2584,7
CV-1
CV-2
CV-3
CV-4
1st Law on CV´s-1,2,3,4
 
 
 
    %8,29298,0
15,69
2,081,20
)
/9,10])15,0(/[)03837,0)(20(/)/()
2,01000/)1010000)(20(/)()
34,48)6,25845,167)(20()
81,20)3,36256,2584)(20()
15,69)5,1673,3625)(20()
2
33333
112
41
34
23










Boiler
pt
p
Cond
t
Boiler
Q
WW
f
smAvmAmce
MWppmWd
MWhhmQc
MWhhmWb
MWhhmQa










Review Exercise on 1st Law (problem 4, cont´d):

42
2nd Law, ‘Kelvin-Planck Statement’ on Heat Engines:
It is impossible to devise a cyclically operating device, for which
the sole function is to absorb energy in the form of heat from a
single thermal reservoir and to deliver an equivalent amount
of work.
Implies the existence
of an energy sink: T  1
Irreversibilities!
2. Second Law of Thermodynamics. Heat Engine concept
Heat Engine
2. Second Law of Thermodynamics
Note: 1st permits T  1, however, we know that T  1. How to deal
with this incongruence? 2nd Law of Thermodynamics!

43
Reversibility:
Quality of a system of being able to go, from an initial state,
through a process or series of processes in one direction and in
the reverse direction, until the original state, without changing its
initial properties, nor the external environment.
What causes irreversibility in real processes?
• Friction
• Combustion
• Molecular mixing (Eg., water-chlorine, coffee-milk, etc.)
• Heat Transfer between bodies with a finite difference of
temperature
• Irrestrict expansion
• Turbulence
• Etc... (we´ll talk about this later again)
2. Second Law of Thermodynamics. Reversibility

44
Carnot Engine-Cycle (Nicolás Léonard Sadi Carnot 1796-1832)
Virtual heat engine designed to operate in a thermodynamic cycle
based on reversible processes (and, therefore ideals), and even
under such a premise, it satisfies the Second Law of
Thermodynamics.
Postulates of Carnot Engine
1. It is impossible to build a Heat Engine more efficient than Carnot
Engine, operating between the same thermal reservoirs.
2. The efficiency of Carnot Engine does not depend neither on the
working fluid nor on particular features of the design.
3. All reversible heat engines operating between two given thermal
reservoir, have the same efficiency of Carnot Engine operating
between such reservoirs.
2. Second Law of Thermodynamics. Carnot Engine

45
Carnot Engine-Cycle: particularities
• It has the maximum efficiency possible for a heat engine without
violation of the 2nd Law. It serves as a reference for the designer
of power plants.
• p-V diagram (based on Ideal Gas):
Source: http://www.sc.ehu.es/sbweb/fisica/estadistica/carnot/carnot1.gif
Source: http://www.monografias.com/trabajos14/hidro-termodinamica/Image298.gif
Adiabatic
Compression
Adiabatic
Expansion
Isothermal
Expansion @T1
Isothermal
Compression @T2

46
Carnot Engine-Cycle: particularities
Efficiency of Carnot Engine:
21 QQWcycle  1st Law
1
2
1
21
_ 1
Q
Q
Q
QQ
CarnotTh 

 As a Heat Engine
High
Low
CarnotTh
T
T
T
T
 11
1
2
_ Specific only for Carnot Engine
Note: T2 is limited by ambient conditions and T1 is
constrained by fuel properties and materials strength.

47
Problem 14 (Problem 5.3, Potter & Somerton, edition 1993)
An inventor proposes an engine that operates between the 27 °C warm surface
layer of the ocean and a 10 °C layer a few meters down. The inventor claims that
the engine produces 100 kW by pumping 20 kg/s of seawater. Is this possible?
Carnot
Engine
TH = 27 °C
TL = 10 °C
HQ
kWW 100
LQ
A.-




kWK
Kkg
kJ
s
kg
Q
TCmQQ
KCT
H
waterliqHH
o
1421)17(
.
)18,4()20(
1717
max_max
max


%04,70704,0
1421
100
__ 
H
engineproposedTh
Q
W



On the other hand, the most efficient heat engine is Carnot´s:
%67,50567,0
300
283
11_ 
H
L
CarnotTh
T
T
 !
___
impossible
engpropThCarnotTh 
Exercises:

48
Problem 15. (Problem 5.4, Potter & Somerton, edition 1993)
A power utility company desires to use the hot groundwater from a hot spring to
power a heat engine. If the groundwater is at 95 °C, estimate the maximum
power output if a mass flow of 0,2 kg/s. The atmosphere is at 20 °C.
Carnot
Engine
TH = 95 °C
TL = 20 °C
HQ
?W
LQ
A.- The maximumpossible efficiency is Carnot´s between 20-95 °C:
Then, the heat transfer rate from the source can be calculated as:
%38,202038,0
368
293
11_ 
H
L
CarnotTh
T
T



kWK
Kkg
kJ
s
kg
Q
TCmQ
H
waterliquidH
7,62)2095(
.
)18,4()2,0(
_


kWkWQW HCarnotTh 8,12)7,62)(2038,0(_max   
Exercises:

49
Entropy ‘S’
Statistic Thermodynamics: ‘S’ represents the degree of molecular
disorder of a thermodynamic system.
Classical Thermodynamics (macro): ‘S’ is a system property that
represents stored thermal energy unavailable for conversion into
mechanical work.
Entropy ‘S’: mathematical definition
For a cycle composed by reversible processes (eg., Carnot):
0
dS
T
Q
aldifferentiexact
T
Q
T
Q
rev
revrev












(mathematical definition)
2. Second Law of Thermodynamics. Entropy
Note: Adiabatic-reversible processes are isentropic, but the opposite is not
necessarily true!!!

50
Inequality of Clausius
• Inequality of Clausius (consequence of the 2nd Law):
)(0
)(0
0
reversible
leirreversibT
Q






reversibleleirreversib WW  
Heat Engines
• In general, for reversible-irreversible processes:






 T
Q
S
T
Q
dS
reversible
leirreversib

)(
)(

0)(
0_


gssurroundinsystemUniverse
systemisolated
SSS
S
The Entropy of the
Universe always
increases!
eg.,: If the Entropy of a
given system ↓2 units,
then it ↑2(+) units in the
surroundings!

Entropy ‘S’: particularities
• Adiabatic-reversible process  isentropic process (const. S)
• For reversible – irreversible processes, 1st Law leads to:
TdS-pdV = dE (i.e., only depends on extreme states)
• For Ideal Gas Ideal, starting from 1st Law and Equation of State*:
1
2
1
2
12
1
2
1
2
12
lnln
lnln
p
p
R
T
T
Css
or
v
v
R
T
T
Css
po
vo

 And if the process is isentropic (e.g.,
adiabatic-reversible), it leads to*:
  k
k
kk
v
v
p
p
p
p
T
T
v
v
T
T




















2
1
1
2
1
1
2
1
2
1
2
1
1
2
• For Pure Substances: read from tables
• For saturated liquid-gas mixtures: fgm sxxss )1( 
51
(*) assuming constant Specific Heat

Exercises:
Problema 16 (Example 6.8 @ Schaum´s) (Entropy of a System)
Statement:
Solution:
Notice that the entropy of the (closed) system diminishes due to the extraction of
heat from the water. Now, the change of entropy in the surroundings occurs at
constant pressure and temperature. Therefore:
2 kg of superheated steam at 400 ⁰C and 600 kPa is cooled at constant pressure by
transferring heat from a cylinder until the steam is completely condensed. The surroundings
are at 25 ⁰C. Determine the net entropy change of the universe due to this process.
52
    KkJKkJkgssmS
TablesSteam
system /55,11/7086,79316,1.2
_
12 
  
   
 



 KkJ
K
kgkJkg
K
hhm
T
Q
T
Q
S
gssurroundinforsignLawst
gssurroundin /45,17
298
/6,6702,32702
298
)__(_1
2121


  0/90,5/45,1755,11  KkJKkJSSS gssurroundinsystemuniverse

53
Second Law applied to a Control Volume
• Using similar approach as for 1st Law:
t t + t
St)cv St+t)cv
system
control
volume
dmi dme dmedmi
si se
Q
  0_
_____












dingsof Surroun
EntropyChange of
EntropyNetInlet
Volumeof Control
EntropyChange of
SystemClosedbyEntropyofGain
  
01122_ 
gssurroundin
gssurroundin
volumecontrol
T
Q
smsmS

54
01122_ 
gssurroundin
gssurroundin
volumecontrol
T
Q
smsmS
Where the ¨equality¨ is for reversible process, while the ¨inequality¨ for
irreversible ones.
Applying the limit when Δt ➔ 0, we get the instantaneous equation:
01122 
gssurroundin
gssurroundin
cv
T
Q
smsmS



55
• If it is a steady state regime, we have:
Note: Qsurroundings = - Qcv
• Consequences:
• If heat enters the fluid (from surroundings) s2 > s1
• If the process is adiabatic, similarly, due to irreversibilities
s2 > s1. But, we would have s2 = s1 if the process is reversible.
• If we wish to find the production of entropy, then we have:
  012 
gssurroundin
gssurroundin
T
Q
ssm


gssurroundin
gssurroundin
cvproduction
T
Q
smsmSS

  1122

Exercises:
Problem 17 (Example 6.9 @ Schaum´s) (Entropy applied to Control Volume)
Statement:
Solution:
• Steady state + perfectly insulated heat exchanger:
• Mass balance:
• 1st Law:
A preheater is used to preheat water in a power plant cycle, as shown. The superheated
steam is at a temperature of 250 ⁰C and the entering water is subcooled (compressed liquid)
at 45 ⁰C. All pressures are 600 kPa. Calculate the rate of Entropy Production.
56
112233 smsmsmSproduction
 
kW/K638,0
, 33221133


productionS
sThmhmhm


213 mmm  
Steam (2)
0.5 kg/s
Comp. Liquid (1)
4 kg/s
Hot water (3)

3. Use of CyclePadTM
3. Use of CyclePadTM.
• Virtual Laboratory, developed by Qualitative
Reasoning Group, with sponsorship of US-NSF
(National Science Foundation).
Web: http://www.qrg.northwestern.edu/
• Allows us to Build and Analize an ample variety of
thermodynamic systems and cycles in steady state
regime, facilitating the explanation of the reasoning
used to derive the target properties.
57

How does CyclePadTM work ?
• A thermodynamic “device” is a set of components which could: give-
receive heat, do-receive work. The device could be evaluated as a SYSTEM
or as CONTROL VOLUME, which are denominated in CyclePadTM: Closed-
Cycle and Open-Cycle, respectively.
• CyclePadTM permits to:
• Give structure to the design of new devices.
• Analize new or existent designs and verify assumptions (eg., isentropic
processes).
• Develop sensitivity analyses, as for example, to answer the question: how
does the thermal efficiency of a cycle changes as a function of the turbine´s
inlet temperature?
• Perform only steady-state analyses, which are appropriate for Conceptual
Engineering of a given process.
• Work in two environments or modes: Build and Analize. 58

Use of CyclePadTM. Mode BUILD
Equipment-
Components
Nodes-stuff
.- Creation of a new design.
.- Addition of components to
existing designs.
.- Re-label components and
nodes.
.- Manipulation of icons-
components.
.- Removal of nodes.
59

• METER STATIONS permit to:
.- Interact
.- Choose working fluid
.- Introduce model premises
.- Assume numerical values
.- Analize output values
Meter
Stations
• INVESTIGATION
.- Via system of explanation
.- Via Analysis Tool
Online Manual:http://www.qrg.northwestern.edu/projects/NSF/Cyclepad/cpadwtoc.htm 60
Use of CyclePadTM. Mode ANALIZE

Steam Power Plant (Rankine Cycle) and its variants
61
4. Energy Analysis of Power Cycles
http://mae.wvu.edu/~smirnov/mae320/figs/F8-1.jpg

Gas Turbine Power Plant (Brayton-Joule Cycle) and its variants
62
http://www.tva.com/power/images/combturbine.gif

Basic Combined (Brayton-Rankine) Power Plant (watch Virtual Tour)
63
http://www02.abb.com/global/plabb/plabb042.nsf/0/1e9b63f8af2dd7d0c12570cb0042d605/$file/CombinedCycleHeatAndPowerPlantDiagram.jpg

Problem 5 (8.1 @ Schaum´s Engineering Thermodynamics, ed. 1993)
Statement:
Sketch-Diagram:
A steam power plant is designed to operate on a Rankine cycle with a condenser outlet
temperature of 80 ⁰C and boiler outlet temperature of 500 ⁰C. If the pump outlet pressure is
2MPa, calculate the maximum possible thermal efficiency of the cycle. Compare with
efficiency of a Carnot engine operating between the same temperature limits.
64
http://www.powerfromthesun.net/Book/chapter12/chapter12_files/image015.jpg
Boiler
Turbine
Condenser
Exercises

Problem 6 (Problem 8.2 @ Schaum´s*)
Statement:
Sketch:
For the ideal Rankine cycle shown, determine the mass flow rate of steam and the cycle
efficiency.
(*): ¨Theory and Problems of Engineering Thermodynamics by M. Potter & C. Somerton
65
10 kPa
6 MPa
500 ⁰C
Boiler
Turbine
Condenser
Exercises

Problem 7 (Problem 8.5 –modified @ condenser- @ Schaum´s*)
Statement:
An ideal reheat Rankine cycle operates between 8 MPa and 10 kPa with a maximum
temperature of 600 ⁰C, as shown. Two reheat stages, each with a maximum temperature of
600 ⁰C, are to be added at 1 MPa and 100 kPa. Calculate the resulting cycle thermal
efficiency.
Diagram:
66
8 MPa
10 kPa
Exercises

Problem 8 (problem 8.8 @ Schaum´s+)
Statement:
Sketch:
A power plant operates on a reheat-regenerative cycle in which steam at 1000 ⁰F and 2000 psia
enters the turbine. It is reheated at a pressure of 400 psia to 800 ⁰F and has two open feedwater
heaters (OFH*); one using extracted steam at 400 psia and the other using extracted steam at 80
psia. Determine the thermal efficiency of the cycle if the condenser operates at 2 psia.
(*): Open Feedwater Heater
(+): ¨Theory and Problems of Engineering Thermodynamics by M.
Potter & C. Somerton 67http://s3.amazonaws.com/answer-board-
image/3888688f-96f1-43c4-8554-
d7332fe278b0.jpeg
1
2
3
9
3´
7
4
5
6
8
10
3´
3
Exercises

Problem 9 (problem 9.3 @ Schaum´s*)
Statement:
Sketch:
An adiabatic compressor is supplied with 2 kg/s of atmospheric air (1 atm) at 15 ⁰C and delivers
it at 5 MPa. Calculate the efficiency and power input if the exiting temperature is 700 ⁰C.
68
https://ecourses.ou.edu/ebook/thermodynamics/ch07/sec074/media/th070407p.gif
http://upload.wikimedia.org/wikipedia/com
mons/3/38/Axial_compressor-
tech_diagram.jpg
Exercises

Statement:
Sketch:
A gas-turbine power plant is to produce 800 kW of net power by compressing atmospheric air
(1 atm) at 20 ⁰C to 800 kPa. If the maximum temperature is 800 °C, calculate the minimum
mass flow of air (i.e., mass flow without losses). (Hint: use sensitivity tool)
69
http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/fig1BraytonCycleClosed_web.jpg
Exercises

Statement:
Sketch:
Assuming the ideal gas-turbine and regenerator shown, find the Heat-Flow-In Qin and the
Back-Work-ratio. (Hint: Use sensitivity option to find mass flowrate)
70
Air
14.7 psia
80 ⁰F
1200 ⁰F
75 psia
inQ
hpWout 800
Compressor
Regenerator
Combustor
Turbine
s
Exercises

Problem 12 (probems 9.77-78) @ Schaum´s*)
Statement:
Sketch:
A gas-turbine cycle intakes 50 kg/s of air at 100 kPa and 20 ⁰C. It compresses it by a factor of 6 and the
combustor heats it to 900 ⁰C. It then enters the boiler of a simple Rankine cycle power plant that operates on
steam between 8 kPa and 4 MPa. The heat-exchanger boiler (heat recovery steam generator system) outlets
steam at 400 °C and exhaust gases at 300 °C. Determine: (a) Net Power output [MW]; (b) Global thermal
efficiency of the cycle if ηturbine_gas and ηcompressor = 85%; and ηsteam_turbine = 87%.
71
http://sounak4u.weebly.com/vapour--combined-power-cycle.html
Diagram:
Exercises

72
Reversible Work: Is the work associated to a reversible process
between two thermodynamic states undertaken by a system. This is
also the maximum work that can be achieved between such two states.
• As we previously indicated, a reversible process is such that can
be reversed without affecting the system or surroundings.
• The phenomena that originate non-reversible processes are
denominated Irreversibilities and might be for example: friction,
heat transfer due to finite temperature gradient, irrestrict
expansion, molecular mixing, turbulence or combustion. The
work performed by real systems is named: Irreversible or Real
Work (Wirrev or Wreal)
• In heat engines: Wrev > Wirrev
5. Reversible Work and Irreversibility

73
Second Law Efficiency:
Is the ratio between the real work and the reversible work, for turbines
or engines; and viceversa for compressors or pumps; i.e.,
Irreversibility:
It is defined as the difference between the Reversible Work and the
Real Work:
Both parameters, permit to observe deviation of
real case with respect to ideal one.
real
rev
pumpcompressorII
rev
real
engineturbineII
W
W
W
W


__
__


realrev
realrev
wwi
WWI




74
Calculation of the Irreversibility for a Control Volume
• By 1st Law together with 2nd Law, it is possible to demonstrate
that the Irreversibility for a CV is given by:
• Instantaneously, for a CV in steady state regime we have:
• Whilst, the rate of Reversible Work is given by:
  QssTmI o
  12
 

















 122
2
2
21
2
1
1
22
ssTgz
c
hgz
c
hmW orev

  QssTm
dt
dS
TI o
cv
o
  12

Exercises:
Problem 18 (Example 7.1 @ Schaum´s) (Irreversibility of Control Volume)
Statement:
Solution:
An ideal steam turbine is supplied with steam at 12 MPa and 700 ⁰C, and exhausts at 0,6
MPa. Determine: (a) Reversible Work and Irreversibility; (b) If the turbine has an adiabatic
efficiency of 0.88, what is the Reversible Work, Irreversibility and 2nd Law Efficiency?
75
928,0
905
8,839
kJ/kg65,2kJ/kg839,8)-(905w-wi
K)298.15C25T(assumingkJ/kg905)sT-(h-)sT-(hw:Then
C279,4TyK-kJ/kg7,2946spandh
kJ/kg3018,6h)h-(hw:Law1stNow,
)isentropic-(nonkJ/kg8,839
kJ/kg3,954
88,0(b)
kJ/kg0w-wi
kJ/kg954,3)h-(h)sT-(h-)sT-(hw
:handotherOn the
(*)kJ/kg954,3h-hw:Law1st
kJ/kg2904,1hyC225,2TvapordsuperheatekPa600pands
kJ/kg3858,4h;KkJ/kg7,0757ssesSteam_Tabl)(
II
irrevrev
o2o21o1rev
2222
221real
s
irrevrev
212o21o1rev
21real
2222
121











rev
real
real
real
s
real
w
w
w
w
w
w
a






6. Availability and Exergy Analysis of Power Cycles
76
Availability Ψ: Is the maximum reversible work that may be extracted
from a system, which is obtained by taking the system from its current
state to equilibrium with its surroundings:
Therefore, for control volume in steady state regime:
Sub-index ¨o¨ represents the thermodynamic condition of surroundings.
 
 max
max
rev
rev
w
W




   ooo
o
o ssTzzg
cc
hh 

 11
22
1
1
2


77
Exergy:
It appears as a convenient term to describe the Availability in a
compact manner, like:
In few words, the Availability is the change of Exergy from the current
state until the equilibrium with the surroundings (lost of useful energy
between current state and the equilibrium with surroundings).
Exergy can also be defined as the energy that is available to do useful
work in a given state.
Decrease of Exergy Principle:
For an isolated system, the exergy change is:
 oo BBsTgz
c
hB 1
2
2
),0;,0(
0Pr12
reversibleleirreversib
BSTBB Destroyedoductiono



Exercises:
Problem 19 (Example 7.4 @ Schaum´s) (Exergy and Availability)
Statement:
Solution:
How much capacity to do useful work is wasted in the condenser of a power plant which takes
in steam of quality 0,85 and 5 kPa, and delivers saturated liquid at the same pressure?
(Surroundings are at 1 atm and 298 K).
78
 
2)-1betweenedWork wast(UsefulkJ/kg51,6
/0,4717-7,2136298-kJ/kg136,5)-(2197,2:(*)
K-kJ/kg0,4717s;kJ/kg136,5h
:TablesSteamby,ppand0)(xliquidsaturatedis2State
and
K-kJ/kg7,2136s;kJ/kg2197,2h
:TablesSteamusing,pandWith x
(*))()()(
2-1betweenedWork wastUseful
WorkUsefulMaximum
21
2121
22
122
11
11
2211212121
21













kgkJBB
sThsThBBBBBB
tyAvailabili
oooo

Exercises:
Problem 24 (example 7.11.1) @ Wu*
Statement:
Sketch:
An air stream at 1100 K and 500 kPa with mass flow rate of 0.5 kg/s enters a steady-state steady-flow
turbine. The stream leaves the turbine at 500 K and 120 kPa. The environment temperature and pressure
are 290 K and 100 kPa. Find the specific flow exergy of the air at the inlet state and at the exit state. Find
the specific flow exergy change and flow exergy rate change of the air stream.
(*): ¨Thermodynamics And Heat Powered Cycles: A Cognitive Engineering Approach by Chih Wu
79
air
Power
Solution:
Assumptions:
• steady state
• no KE/PE changes

Exercises:
Problem 25 (example 7.11.2) @ Wu*
Statement:
Solution:
A hot water stream at 500 K and 200 kPa with mass flow rate of 0.05 kg/s enters a steady-state steady-
flow heat exchanger and leaves the heat exchanger at 400 K and 200 kPa. A cold water stream at 300 K
and 200 kPa enters the heat exchanger and leaves the heat exchanger at 350 K and 200 kPa. Determine
the rate of flow exergy change of the heat exchanger. The environment temperature and pressure are at
298K and 100 kPa.
80

Exercises:
Problem 25 (example 7.11.2) @ Wu* (cont´d)
Statement:
Solution:
A hot water stream at 500 K and 200 kPa with mass flow rate of 0.05 kg/s enters a steady-state steady-
flow heat exchanger and leaves the heat exchanger at 400 K and 200 kPa. A cold water stream at 300 K
and 200 kPa enters the heat exchanger and leaves the heat exchanger at 350 K and 200 kPa. Determine
the rate of flow exergy change of the heat exchanger. The environment temperature and pressure are at
298K and 100 kPa.
81
Summary of data:
KkgkJskgkJh
KkgkJskgkJh
KkgkJskgkJh
KkgkJskgkJh
skgmm
Q_lineQ_lineskgmm






/04,1;/8,321
/3926,0;/7,112
/16,7;/2720
/62,7;/2924
/05,0
)21(from/0486,0
44
33
22
11
43
21



82
Exergy Effectiveness or Effectiveness of Devices
Use: Evaluate devices such as heaters, coolers, heat exchangers,
throttle valves, etc, that do not involve the production or input of work. It
measures the comparison between the desired output of a non-work
device vs. the exergy input.
Evaluation: Effectiveness ¨ɛII¨ is calculated as:
Example: Given the shown closed heat exchanger, the ¨Effectiveness¨
is calculated as shown.
ExergyInvested
ExergyGained
TransferExergyInput
TransferExergyOutput
II
_
_
__
__

 
 inhotouthothot
incoldoutcoldcold
II
BBm
BBm
__
__







83
Exergy Cycle Efficiency ηex of Power Cycles
Definition: It is the ratio of desirable exergy transfer output and the
required input energy of the cycle (heat input):
• A cycle ηex can be used to see if a real power cycle has a good design
or not.
• ηex addresses the question: to what extent we have used the
available energy? and it´s a complementary rating of the performance
of a real cycle.
InputHeat
WorkNetofTransferExergy
EnergyInput
OutputTransferExergyDesired
ex
_
____
_
___


Exercises:
Problem 26 (example 7.13.1) @ Wu*)
Statement:
Solution:
In a boiler, heat is transferred from the products of combustion to the steam. The temperature of the
products of combustion decreases from 1400 K to 850 K while the pressure remains constant at 100 kPa.
The water enters the boiler at 1000 kPa, 430 K and leaves at 1000 kPa, 530 K with a mass flow rate of 1
kg/s. Determine the effectiveness of the boiler. The ambient temperature and pressure are 298 K and
100kPa.
84

Exercises:
Statement:
Solution:
In a boiler, heat is transferred from the products of combustion to the steam. The temperature of the
products of combustion decreases from 1400 K to 850 K while the pressure remains constant at 100 kPa.
The water enters the boiler at 1000 kPa, 430 K and leaves at 1000 kPa, 530 K with a mass flow rate of 1
kg/s. Determine the effectiveness of the boiler. The ambient temperature and pressure are 298 K and
100kPa.
85
Summary of data:
KkgkJskgkJh
KkgkJskgkJh
skgmm
Water
KkgkJskgkJh
KkgkJskgkJh
skgmm
Air






/95,6;/2957
/86,1;/2,665
/1
:
/47,3;/9,852
/97,3;/1405
/4,15
:
44
33
43
22
11
21


 
 
  
  
4633,0
)47,3*2989,852(97,3*2981405*15,4
86,1*2982,665)95,6*2982957(*1
)()(
)()(
_
_
22111
33443







II
oo
oo
II
sThsThm
sThsThm
ExergyInvestment
ExergyGain





Boiler
Turbine
Condenser
Pump
Exercises:
Statement:
Sketch:
Superheated steam at 10 MPa and 770 K enters the turbine of a Rankine steam power plant operating at
steady state and expands to a condenser pressure of 50 kPa. Assume the efficiencies of the turbine and
pump are 100%. The mass flow rate of the steam is 1 kg/s. The surroundings temperature is 298 K.
Determine the cycle efficiency, the second law cycle efficiency and the exergy cycle efficiency
(Effectiveness) of the power plant.
86http://www.powerfromthesun.net/Book/chapter12/chapter12_files/image015.jpg

Boiler
Turbine
Condenser
Pump
Exercises:
Statement:
Sketch:
Superheated steam at 10 MPa and 770 K enters the turbine of a Rankine steam power plant operating at
steady state and expands to a condenser pressure of 50 kPa. Assume the efficiencies of the turbine and
pump are 100%. The mass flow rate of the steam is 1 kg/s. The surroundings temperature is 298 K.
Determine the cycle efficiency, the second law cycle efficiency and the exergy cycle efficiency
(Effectiveness) of the power plant.
87http://www.powerfromthesun.net/Book/chapter12/chapter12_files/image015.jpg
KkgkJskgkJh
KkgkJskgkJh
KkgkJskgkJh
KkgkJskgkJh
skgm





/59,6;/2289
/59,6;/3366
/09,1;/8,350
/09,1;/5,340
/1
44
33
22
11

  
 
          (35,4%)354,0
2,3015_
___
(66%)66,0
54,0
354,0
(54%)54,0
770
5,354
11_
%)4,35(354,0___
/2,3015_
/7,1066)(__
12124343
II
3
1
23
1243








     pumpturbine
Carnot
th
rev
irrev
th
ssThhssThh
HeatInput
WorkNetTransferExergy
w
w
K
K
T
T
EfficiencyCarnot
EfficThermalCycle
kgkJhhaddedHeat
kgkJhhhhworkspecNet
oo






EXAMPLE: Energy and Exergy Analysis of Thermal Power Plants: A
Review, by S.C. Kaushik, V. Siva Reddy,S.K. Tyagi
Let have a Coal Fire Plant, working with Steam Turbine (Rankine Cycle)
88
Main improved features:
.- Reheating
.- Lower Condenser
pressure
.- Higher Boiler pressure
.- Regeneration
Legend:
B: Boiler
LP: Low Pressure
IP: Intermediate Pressure
HP: High Pressure
T: Turbine
H: Feedwater Heater
C: Condenser
G: Generator

Energy Analysis: Boiler
89

Exergy Analysis: Boiler
90
(b) Second Law efficiency and effectiveness are defined
as:
ɛII, Boiler=
Total Boiler subsystem Second Law Effectiveness is:

Energy Analysis: High Pressure Turbine
91
(a)

Exergy Analysis: High Pressure Turbine
92
(b) Second Law Effectiveness of HPT:
ɛII, HPT =
.
.
.
.
… and so on for the rest of the
components of the plant, until we get the
energy (heat) / exergy heat losses

Energy-Exergy Losses Diagram
93
Total
Energy
Heat
Losses
Total
Exergy
Destruction
Losses
83395.6
Note: largest heat loss in Condenser, but largest exergy destruction in Boiler.
Therefore, it´s critical to optimize combustión process, insulation, etc. in
that component of the plant. Condenser losses are mostly due to the 2nd
Law and physics of the process, but Boiler losses are mostly due to
process inefficiencies.

Comparison of Energy-Exergy Efficiencies in Coal-Fired Plants
94
Exergy-Second Law Effectiveness (%) Energy-First Law Efficiency(%)

CONCLUDING REMARKS
• Exergy analyses are complementary tools to traditional
energy analyses, but provide a tool to measure the quality
of the processes undergoing in devices and equipments.
• Exergy flows (changes) between states, allow us to
measure the level of irreversibilities governing a given
process and help us to develop technology improvements.
• For example, heat exchangers efficiency may be improved
by increasing the area of contact, but added cost may turn
this unbounded solution unviable. In this case, exergy
analysis may provide a tool to track cost/benefit of
potential improvements (not shown in this presentation).
95

96
(luis.rojas@nu.edu.kz)
Dept. of Mechanical Engineering
Nazarbayev University

Energy and Exergy Analyses of Power Cycles

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