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Chapter 1: Introductory concepts and definitions
Thermodynamics
Thermo: Heat
Dynamis: Force
• System
• Surroundings
• Boundary
• Closed System
• Control Volume
• Macroscopic approach: Classical thermodynamics
• Microscopic approach: Statistical thermodynamics
• Property: intensive, extensive
• State
• Process
o Isothermal process
o Adiabatic process
• Cycle
• Equilibrium
• Phase
• Specific Volume
• Density
• Pressure (units)
• Temperature scale
o Kelvin scale
o Rankine scale
Chapter 2: Energy and First Law of Thermodynamics
Work-Energy 2
1
2 22 1
1 ( )2
S
SFds m V V= −∫
Conservation of energy 2 21 1 2
1 12 2
mV mgZ mV mgZ+ = + 2
Expansion /Compression work
Work is a path function
(Depends on process)
Sign convention:
w>0 work done by the system
w<0 work done on the system
A gas in a piston-cylinder undergoes an expansion for which . constantnPV =
Evaluate the work
done by the gas if:
a) n=1.5
b) n=1.0
c) n=0
2
1
v
vw P= ∫ dv Expansion work
c) 2
1
5 3 42 1 2( ) 3*10 (0.2 0.1) 3*10 30KJ
v
v
Nw Pdv P v v m Jm
= = − = − = =∫
b) 2 2
1 1
52
1
0.2ln 3*10 (0.1) ln( ) 20.79KJ0.1
v v
v v
vdvw P dv c PVv v
= = = = =∫ ∫
2
1
1.512 1
2
1 12 1
2 2 1 1
2 2 1 1
0.1) ( ) 3( ) 1.060.2
1
But
17.61
n
n nV
nV
n n
Va P P BarV
c V Vw dV cV n
C PV PV
PV PVW KJn
− −
= = =
−= =
−
= =
−∴ = =
−
∫
Energy transfer by heat:
Q>0 heat transfer to the system
Q<0 heat transfer from the system
Heat transfer modes:
Conduction: xdTQ KAdx
•
= −
Convection: 2 1( )Q hA T T•
= −
Radiation: Stefan-Boltzman Law4 42 1(Q A T Tεσ
•
= − )
1st Last of Thermodynamics for Closed System:
Q W U KE PE= + Δ + Δ + Δ
Energy analysis of cycles:
net netQ W=
(a) Power cycle:
in
in out
in
WQQ Q
Q
η
η
=
−=
(b) Refrigeration cycle:
in
cycle
in
out in
QW
QQ Q
β
β
=
=−
Heater:
out
cycle
QW
γ =
Chapter 3: Evaluating Properties
P-v-T Relation
P=P(T,v)
The graph of such a function is a surface called (P-v-T) surface
Projecting the liquid, two phase, and vapor regions of P-v-T surface onto the temperature-
specific volume Plane results in a (T-v) diagram
Two Phase Flow:
Divide by m
( ) (
(1 )
vap g
tot f g
f g
f f g g
f g
f g
m mx
m m m
V V V
vm v m v m
mf mgv v vm m
v x v xv
= =+
= +
= +
= +
= − +
)
The concept of independent properties:
(1 )
(1 )f g
f g
H V Pvh u Pvu x u x
h x h x
= += += − +
= − +
u
h
Ideal gas model
( )( ) ( )( ) ( )
Pv RTu u Th h T u T Pvh h T u T RT
=== = += = +
Internal energy, enthalpy, and specific heats of ideal gases
For a gas obeying the ideal gas model, specific energy depends only on temperature.
Hence, the specific heat is also a function of temperature alone. That is, vc
( )vduc TdT
= (ideal gas)
This is expressed as an ordinary derivative because depends only onT . u
By separating variables
( )vdu c T dT=
On integration
2
12 1( ) ( ) ( )
T
vTu T u T c T dT− = ∫ (ideal gas)
Similarly, for a gas obeying the ideal gas model, the specific enthalpy depends only on
temperature, so the specific heat defined by Eq. 3.9, is also a function of temperature
alone. That is
pc
( )pdhc TdT
= (ideal gas)
An important relationship between the ideal gas specific heats can be developed by
differentiating the enthalpy equation with respect to temperature
dh du RdT dT
= +
Therefore
( ) ( )p vc T c T R= + (ideal gas)
On a molar basis, this is written as
( ) ( )p vc T c T R= + (ideal gas)
Although each of the two ideal gas specific heats is a function of temperature, the above
equations show that the specific heats differ by just a constant. Knowledge of either
specific heat for a particular gas allows the other to be calculated by using only the gas
constant. The above equations also show that and pc c> v pc c> v , respectively.
For an ideal gas, the specific heat ration, k, is also a function of temperature only
( )( )
p
v
c Tk
c T= (ideal gas)
Since , it follows that . Combining the above two equations results in pc c> v 1k >
v
( )1
c ( )1
pkRc T
kRT
k
=−
=−
Polytropic process of an ideal gas
Recall that a polytropic process of a closed system is described by a pressure-volume
relationship of the form
constantnpV =
Where is a constant. n
For a polytropic process between two states
1 1 2 2n np V p V=
or
2 1
1 2
( )np Vp V
=
The exponent may take on any value from n −∞ to +∞ , depending on the particular
process. When , the process is an isobaric (constant-pressure) process, and when
, the process is an isometric (constant –volume) process.
0n =
n = ±∞
For a polytropic process
22 2 1 1
1( 1
1p V p VpdV n
n)−
= ≠−∫
For any exponent n except . When 1n = 1n = ,
22
1 111
ln ( 1)VpdV p V nV
= =∫
The above equations apply to any gas (or liquid) undergoing a polytropic process.
When the additional idealization of ideal gas behavior is appropriate, further relations
can be derived. Thus, when the ideal gas equation of state is introduced, the following
expressions are obtained, respectively
( 1) / 12 2 1
1 1 2
( ) ( ) (idealgas)n n nT p VT p V
− −=
22 1
1
( ) (ideal gas, n 1)1
mR T TpdVn−
= ≠−∫
22
11
ln (ideal gas, n=1)VpdV mRTV
=∫
For an ideal gas, the case n=1 corresponds to an isothermal (constant-temperature)
process, as can readily be verified. In addition, when the specific heats are constant, the
value of the exponent corresponding to and adiabatic polytropic process of and ideal
gas is the specific heat ratio k
n
Example 3.11 illustrates the use of the closed system energy balance for a system
consisting of an ideal gas undergoing a polytropic process.
Linear Interpolation:
Chapter 4: Control volume analysis
A region of space through which mass flows
Conservation of mass cvi e
dm m mct
• •
= −∑ ∑
Since m VAρ•
=
VAmv
•
=
Then
cv i i e e
i e
dm V A V Adt v v
= −∑ ∑
Most engineering systems can be assumed to be at steady state, meaning that all
properties are unchanged with time.
For a control volume then: 0cvdmdt
=
Therefore eim m• •
=∑ ∑
A stream of water vapor mixes with a liquid water stream to produce a saturated liquid
stream at the exit. The states at the inlets and exit are specified. Mass flow rate and
volumetric flow rate data are given at one inlet and at the exit, respectively.
Determine the mass flow rates at inlet 2 and at the exit, and the velocity . 2V
Analysis:
The principle relations to be employed are the mass rate balance and the expression
. At steady state the mass rate balance becomes: /m VA v•
=
tvdmdt
0
1 2m m m• • •
= + − 3
1
Solving for 2m•
2 3m m m• • •
= −
The mass flow rate is given. The mass flow rate at the exit can be evaluated from the
given volumetric flow rate
1m•
33
3
( )AVmv
•
=
Where , is the specific volume at the exit. In writing this expression, one-dimensional
flow is assumed. From Table A-3, .
3v
3 33 1.108*10 m /kgv −=
Hence
3
3 3 3
0.06 / 54.15kg/s(1.108*10 / )
m smm kg
•
−= =
The mass flow rate at inlet 2 is then
2 3 1 54.15 40 14.15kg/sm m m• • •
= − = − =
For one-dimensional flow at 2, , so 2 2 2 2/m A V v•
=
22 2 /V m v A•
= 2
State 2 is a compressed liquid. The specific volume at this state can be approximated by
2 2( )fv v T= .
From Table A-2 at 40°C, . 3 32 1.0078*10 m /kgv −=
Therefore
3 3 4 2
2 2 2
(14.15kg/s)(1.0078*10 m /kg) 10 cm 5.7m/s25 1
Vcm m
−
= =
Conservation of energy for a control volume
2 2
( ) (2 2
cv i ei ei i e
dE V VQ W m u gz m u gzdt
• • • •
= − + + + − + + )e
( ) ( )cv e e e i i iW W p A V p A V• •
= + −
2 2
( ) (2 2
cv i ecv i ei i i i e e e ecv
dE V VQ W m u p v gz m u p v gzdt
• • • •
= − + + + + − + + + )
2 2
( ) (2 2
cv i ecv i ei i ecv
dE V VQ W m h gz m h gzdt
• • • •
= − + + + − + + )e
2 2
( ) (2 2
cv i ecv i ei i ecv
i e
dE V VQ W m h gz m h gzdt
• • • •
= − + + + − + +∑ ∑ )e
Analyzing control volume at steady state
0cvdEdt
=
2 2
0 ( ) (2 2i e
cv i ei i ecvV VQ W m h gz m h gz
• • • •
= − + + + − + +∑ ∑ )e
•
Since i em m• •
=∑ ∑
Many applications involve one inlet and one exit
Therefore:
i em m m• •
= =
2 21 2
1 2 1 2( )0 [( ) (
2cvcv
V VQ W m h h g z z• • • −
= − + − + + − )]
2 21 2
1 2 1 2( )0 ( ) (
2cvcvQ V VW h h g z z
m m
• •
• •
−= − + − + + − )
2 21 2
1 2 1 20 ( ) (2
V Vq w h h g z z−= − + − + + − )
Or:
2 22 1
2 1 1 2( ) (2
V Vq w h h g z z−= + − + + − )
This is the statement of first law of thermodynamics for controlled volume
Application of the first law to various engineering devices:
Nozzles and diffusers:
2 22 1
2 1 2V Vq h h −
= − +
Turbines:
2 1( )q w h h= + −
Compressors:
2 1( )q w h h= + −
Pump:
2 22 1
2 1 2 1( )2
V Vq w h h g z z−= + − + + −
2 22 1
2 2 2 1 1 1 2 10 (2
V Vw u P v u Pv g z z−= + + − − + + − )
Since water has a low viscosity 2 1 0u u− ≈
And incompressible flow 2 1v v v= =
Then
2 22 1
2 1 2 10 ( ) (2
V Vw v P P g z z−= + − + + − )
)
The above formula is used to find the power used by a pump
2 1(v P P− is known as Pressure head
2 22 1
2V V− is known as Velocity head
2 1(g z z− ) is known as Elevation head
Throttling Devices:
2 22 1
2 1 2 1( ) (2
V Vq w h h g z z−= + − + + − )
There is usually no significant heat transfer with the surrounding and the change in
potential energy from inlet to exit is negligible. With these assumptions, the first law
Of thermodynamics reduces to:
2 21 2
1 22 2V Vh h+ = +
Although velocities may be relatively high in the vicinity of the restriction, measurements
made upstream and downstream of the flow area show in most cases that the change in
the specific kinetic energy of the fluid between these locations can be neglected. With
this further simplification, the last equation reduces to
1 2h h=
System integration (putting it all together):
Heat exchangers:
Devices that transfer energy between fluids at different temperatures by heat transfer
modes such as conduction, convection, and radiation are called heat exchangers. One
common type of heat exchanger is a vessel in which hot and cold streams are mixed
directly as show in figure (a). An open feedwater heater is an example of this type.
Another common type of heat exchanger is one in which a gas or liquid is separated from
another gas or liquid by a wall through which energy is conducted. These heat
exchangers, known as recuperators, take many different forms. Counterflow and parallel
tube-within-a-tube configurations are shown in Figures (b) and (c), respectively. Other
configurations include cross-flow, as in automobile radiators, and multiple-pass shall-
and-tube condensers and evaporators. Figure (d) illustrates a cross-flow heat exchanger.
The only work interaction at the boundary of a control volume enclosing a heat
exchanger is flow work at the places where matter enters and exits, so the term of
the energy rate balance can be set to zero. Although high rates of energy transfer may be
achieved from stream to stream, the heat transfer from the outer surface of the heat
exchanger to the surroundings is often small enough to be neglected. In addition, the
kinetic and potential energies of the flowing streams can often be ignored at the inlets and
exits.
cvW•
Transient analysis:
Chapter 5: The second law of thermodynamics
Although the first law of thermodynamics allows unrestricted convertibility from one
from of energy to another, as long as the overall quantity is conserved, experimental
evidence tells that in certain types of energy conversion, restriction must be placed to the
direction end extent of transformation. For example, it is a fact that although work can be
completely converted into heat, heat cannot be completely converted into work, no matter
how ideal the process may be. For example, rubbing two pieces of sandpaper together,
work used to overcome friction causes an increase in internal energy of the paper and
heat flow to surroundings, but heat addition will not result in doing mechanical work. It
is the second law of thermodynamics that imposes restrictions to the direction and extent
of energy transformation processes. In this chapter we sturdy the second law of
thermodynamics and some consequences of it.
In summary second law provides means for
1. Predicting the direction of processes.
2. Establishing conditions for equilibrium.
3. Determining the best theoretical performance of cycles, engines, and other
devices.
4. Evaluating quantitatively the factors that preclude the attainment of the best
theoretical performance level.
Additional uses of the second law include its roles in
5. Defining a temperature scale independent or the properties of any thermometric
substance.
6. Developing mans for evaluating properties such as u and h in terms of properties
that are more readily obtained experimentally.
It is well known that physical processes in nature process toward equilibrium
spontaneously. Liquids flow from a region of high elevation to one of low elevation;
gases expand from a region of high pressure to one of low pressure; heat flows from a
region of high temperature to one of low temperature; and material diffuses from a region
of high concentration to one of low concentration. A spontaneous process can proceed
only in a particular direction. Energy from an external source is required to reverse such
processes. The second law of thermodynamics epitomizes out experiences with respect
to the unidirectional nature of thermodynamics processes.
Kelvin – Planck statement of the second law (Heat Engine)
Clausius statement of the second law (Heat pump)
Reservoir – A large container that is unaffected in temperature as heat goes into and out
of it.
Reversible process: Is a process that can be reversed and leave no resultant change in
either the system or surroundings. Reversible processes are fictitious, they are ideal
operations that represent the only means by which engineering calculations can be made.
Irreversible process: The system and the surrounding can not be exactly restored to
their initial states. All real processes are irreversible, since they involve dissipative
effects such as friction, heat transfer, etc...
The process is irreversible because of heat transfer to the surroundings. To reverse the
process, heat must be added to the block and an equal amount of mechanical work must
be done, which is impossible.
The most common irreversible processes are:
• Heat transfer across a finite temperature difference
• Unrestrained expansion of a fluid
• Mixing of fluids of different compositions and states
• Solid and fluid friction
• Spontaneous chemical reaction
• Inelastic deformation
• Electric resistance
• Hysteresis effects in magnetization and polarization
Thermodynamic cycles:
1 – Power cycle
Thermal efficiency 1cycle H c c
H H H
W Q Q QQ Q
η −= = = −
Q
2 – Refrigeration/Heat pump cycle
Coefficient of performance
c c
cycle H c
H H
cycle H c
Q QW Q
Q QW Q Q
β
γ
= =Q−
= =−
Maximum thermal efficiency of a power cycle:
For a reversible cycle
( )( )
c c
H H
Q f TQ f T
=
Can use number of temp functions.
Kelvin suggested to use ( )( )
c c
H H
f T Tf T T
==
Therefore c c
H H
Q TQ T
=
max 1 c
H
TT
η = −
Maximum performance of a refrigeration cycle:
max
max
c
H c
H
H c
TT T
TT T
γ
Β =−
=−
Carnot cycle
What is the most efficient heat engine? What heat pump provides the highest coefficient
of performance? All processes in the cycle must be reversible.
Chapter 6: Using Entropy
• Clasius inequality
• Principle of entropy increase
• Entropy is a non conserved property
• Entropy change for pure substances
• Entropy change for ideal gases
• Isentropic process
• Isentropic efficiencies of various engineering devices
Clasius inequality
0QTδ
≤∫ for reversible & irreversible cycles
Sum of all differential heat transfers divided by absolute temp at the boundary.
' cQ W dEcδ δ= + I
Where 'cW W Wδ δ δ= +
Since the cyclic device is a reversible one
'
res
QT T
Qδ δ= II
Eliminating from I & II 'SQ
c resQW T dE
T cδδ = −
Let the system undergo a cycle while the cyclic device undergoes one or more cycles.
c resQW T dE
T cδδ = −∫ ∫ ∫
Since net change of energy during a cycle is zero 0cdE =∫
c resQW T
Tδ
= ∫
For the heat engine, based on Kelvin-Planck statement is a negative quantity and
is positive
cW resT
0QTδ
≤∫
= reversible
< irreversible
Definition of entropy change
A quantity whose cyclic integral is zero depends on the state only and not on the process
path therefore the quantity ( )revQ
Tδ must represent a property in differential form.
( )revQds
Tδ
=
Units of entropy:
KJKg.K
BtuLbm.°R
Integrating both sides:
2
2 1 1( )
rev
Qs s sTδ
Δ = − = ∫
Engineers are often concerned with change in entropy.
The principle of entropy increase:
0QTδ
≤∫ Clasius inequality
2 1
1 2( )
A C
revQ Q
T T0δ δ
+ ≤∫ ∫
2
1 210Q s s
Tδ
+ − ≤∫
Therefore 2
2 1 1
Qs sTδ
− ≥ ∫
Or QdsTδ
≥ statement of second law of thermodynamics
Entropy generation ( 0 irreversible
) 0 reversible0 impossible
ds>⎧⎪=⎨⎪<⎩
Conclusion:
During a thermodynamic process the entropy of the universe can either increase or
remain the same (Reversible process). It is not possible for a process to occur which
involves decrease in entropy of the universe.
( ) ( ) ( )univ system surrS S SΔ = Δ + Δ
The universe is filling up with entropy
1Lbm of saturated R-22 vapor at 155°F is condensed at constant pressure to saturated
liquid at 155°F by transfer of heat to surrounding air at 95°F
112.495 58.04 54.45g fBtuQ h hLbm
= − = − =
54.45( ) 0.09811(95 460)sur
BtuSLbm R
Δ = =+ °
Defined reservoir: no change in temp as heat goes into it.
( ) 0.11036 0.19895 0.08862system f gS s sΔ = − = − = −
( ) ( ) ( ) 0.0095univ system surBtuS S S
Lbm RΔ = Δ + Δ =
°
Find the highest temp of air permissible under 2nd law
ANS: must be a reversible process:
( ) ( ) 0sys surS SΔ + Δ =
54.450.08862 0460T
− ++
=
155T F= °
reversible heat transfer
Entropy change of pure substances:
(1 ) f gs x s x= − + s
Numerical evaluation of entropy:
dQdsT
= for reversible process
dQ=dU+PdV 1st law in differential form for closed system
TdS dU PdV= +
For unit mass I Tds du Pdv→ = +
Units of s: Btu KJ Lbm°R Kg K
h u Pv= +
dh du Pdv vdP= + + plug in equation I
Tds dh vdP= − II
Equations I & II for open, closed, rev systems
Entropy change for ideal gases:
(I) for ideal gasvRTTds c dT dvv
= +
vdT dvds C RT v
= +
2 22 1
1 1
ln lnvT vs s c RT v
− = +
(II) pRTTds c dT dpP
= − for ideal gas
pdT dPds c RT P
= −
2 22 1
1 1
ln lnPT Ps s c RT P
− = −
Isentropic relations for an ideal gas:
2 22 1
1 1
ln lnvT v
s s c RT v
− = +
Since 2 1s s=
2 2
1 1
0 ln lnvT vc RT v
= +
2 2
1 1
0 ln ( ) lnv p vT vc c cT v
= + −
Divide by 2 2
1 1
0 ln ( 1) lnvT vc kT v
→ = + −
2 2
1 1
ln (1 ) lnT vkT v
= −
12 2
1 1
( ) kT vT v
−=
2 22 1
1 1
ln lnpT Ps s c RT P
− = −
2 2
1 1
0 ln ( ) lnp p vT Pc c cT P
= − −
Divide by vc 2 2
1 1
0 ln ( 1) lnT Pk kT P
= − −
2 2
1 1
1ln lnT PkT k
−=
P
12 2
1 1
( )kkT P
T P
−
=
kPv c=
1 1 2 2k kPv P v=
2 1
1 2
( )kP vP v=
Isentropic process = reversible + adiabatic
Enables us to compare the actual performance of engineering devices to the performance
under idealized condition.
2 1s s=
Isentropic efficiencies of various engineering devices:
Turbine:
s s
w Ww W
η•
•= =
Compressor & Pump:
ssw Ww W
η•
•= =
Nozzle:
222
2
( / 2)( / 2)s
VV
η =
More on ideal gas model
For two states having the same specific entropy,
22 2 1 1 2 1
1
( , ) ( , ) ( ) ( ) ln ps T p s T p s T s T Rp
− = ° − ° −
Reduces to
22 1
1
0 ( ) ( ) ln ps T s T Rp
= ° − ° − (I)
The above equation involves four property values: 1 1 2 2, , , and p T p T . If any three are
known, the fourth can be determined. If, for example, the temperature at state 1 and the
pressure ratio 2 / 1p p are known, the temperature at state 2 can be determined from
22 1
1
( ) ( ) ln ps T s T Rp
° = ° + (II)
Since is known, would be obtained from the appropriate table, the value of
would be calculated, and temperature would then be determines by
interpolation. If
1T 1( )s T°
2( )s T° 2T
1p , , and are specified ad the pressure at state 2 is the unknown,
equation I would be solved to obtain
1T 2T
2 12 1
( ) ( )exp s T s Tp pR
° − °⎡ ⎤= ⎢ ⎥⎣ ⎦ (III)
These equations can be used when (or s° s°) data are known, as for the gases of Tables
A-22 and A-23
Air
For the special case of air modeled as an ideal gas the above equation provides the basis
for an alternative approach for relating the temperatures and pressure at two states having
the same specific entropy. To introduce this, rewrite the equation as
[ ][ ]
22
1 1
exp ( ) /exp ( ) /
s T Rpp s T
°=
° R (IV)
The quantity [ ]exp ( ) /s T R° appearing in this expression is solely a function of
temperature, and is given the symbol ( )rp T . A tabulation of rp versus temperature for
air is provided in Tables A-22. In terms of the function pr, the last equation becomes
2 21 2
1 1
(s , air only)r
r
p p sp p
= = (V)
Where 1 2( )r rp p T= and 2 ( )r r 2p p T= . The function rp is sometimes called the relative
pressure. Observe that rp is not truly a pressure, so the name relative pressure is
misleading. Also, be careful not to confuse rp with the reduced pressure of the
compressibility diagram.
A relation between specific volumes and temperature for two states of air having
the same specific entropy can also be developed. With the ideal gas equation of state,
, the ratio of the specific volumes is /v RT p=
2 2
1 2
( )(v RT pv p RT= 1
1
) (VI)
Then, since the two states have the same specific entropy, equation V can be introduced
to give
2 2
1 2
( )( )
r
r
v RT p Tv p T RT
1
1
⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
(VII)
The ratio / ( )rRT p T appearing on the right side of the last equation is solely a function
of temperature, and is given the symbol . Values of for air are tabulated versus
temperature in Tables A-22. In terms of the function , the last equation becomes
( )rv T rv
rv
2 21 2
1 1
( , air only)r
r
v v s sv v= = (VIII)
Where and . The function is sometimes called the relative
volume. Despite the name given to it, is not truly a volume.
1 ( )r rv v T= 1 22 ( )r rv v T= rv
( )rv T
Chapter 8: Cycles
A series of processes in which a system starts at a given state and is returned to that same
state.
• Closed loop cycle: Same working fluid.
• Open loop cycle: New working fluid continuously furnished.
Since in a cycle then 0uΔ =
net netQ W=
Power cycle: Thermal efficiency can not be greater than thermal efficiency of a Carnot
cycle.
Refrigeration cycle: Coefficient of performance.
We will study the application of 1st and 2nd law to analyze the performance of power and
refrigeration cycles.
Vapor power systems – Rankine cycle
turb pumpth
in
w wq
η−
=
bwr = Back work ratio
pump
turb
wbwr
w=
The back work ration is an alternative way of measuring the performance of a Rankine
cycle.
Increasing Rankine cycle efficiency
1 – Lowering the condenser temp
2 – Superheating the steam (before entering turbine)
Note: The boiler and super heater taken together is called steam generator.
3 – Increasing the boiler pressure (while maintaining the peak temp)
4 – Reheat of the steam to maintain the 90% quality of better in turbine to prevent blade
damage.
Principle sources of irreversibilities & losses in turbine and/or pump:
Ideal reheat cycle:
Reheat cycle with turbine irreversibilities:
Regenerative vapor power cycle, open feedwater heater:
Chapter 9: Gas power systems
Internal combustion engine:
Net work for one cycleMean effective pressure = Displacement volume
Air standard Otto cycle:
Cycle analysis:
1-2 isentropic compression
2-3 constant volume heat addition
3-4 isentropic expansion
4-1 constant volume heat rejection
Air standard analysis
34122 1 3
23 413 2 4
,
,
WW u u u um m
Q Qu u u um m
= − = −
= − = −
4
1
(I)
34 123 4 2 1( ) (cycleW W W u u u u
m m m= − = − − − ) (II)
23 413 2 4 1( ) (cycleW Q Q u u u u
m m m= − = − − − ) (III)
3 2 4 1 4
3 2 3 2
( ) ( ) 1u u u u u uu u u u
η − − − −= =
− −1− (IV)
22 1
1
( ) rr r
v vv vv r
= = 1 (V)
44 3
3
( )r rvv v rvv
= = 3r (VI)
Cold air standard analysis
12 1
1 2
( )k kT V rT V
−= = 1− (VII)
1341
3 4
1( )kk
VTT V r
−−= = (VIII)
Effect of compression ratio on performance of Otto cycle
By referring to the T-s diagram of the previous figure, we can conclude that the Otto
cycle thermal efficiency increases as the compression ratio increase. An increase in the
compression ratio changes the cycle from 1-2-3-4-1 to 1-2’-3’4-1. Since the average
temperature of heat addition is greater in the latter cycle and both cycles have the same
heat rejection process, cycle 1-2’-3’-4-1 would have the greater thermal efficiency. The
increase in thermal efficiency with compression ratio is also brought out simply by the
following development on a cold air-standard basis. For constant , Eq. IV becomes vc
4 1
3 2
(1( )
v
v
c T Tc T T
η )−= −
− (IX)
On rearrangement
1 4 1
2 3 2
/ 11 (/ 1
T T TT T T
η )−= −
− (X)
Knowing 14 4
3 3
( ) kT vT v
−= and 11 1
2 2
( ) kT vT v
−=
Since and therefore the right hand sides are equal so the left hand sides
must be equal.
4v v= 1 23v v=
4
3 2
T TT T
= 1 or 34
1 2
TTT T
=
1
2
1 TT
η = − (XI)
Finally, introducing Eq. VII
1
11 krη −= − (XII)
Air Standard Diesel Cycle
Cycle analysis:
1-2 isentropic compression
2-3 constant pressure heat addition
3-4 isentropic expansion
4-1 constant volume heat rejection
In the diesel cycle the heat addition takes place at constant pressure. Accordingly,
Process 2-3 involves both work and heat. The work is given by
3232 3 22(W pdv p v v
m= = −∫ ) (XIII)
The heat added in Process 2-3 can be found by applying the closed system energy
balance
3 2 23 23( )m u u Q W− = −
Introducing Eq. XIII and solving for the heat transfer
233 2 3 2 3 3 2 2 3( ) ( ) ( ) ( )Q u u p v v u pv u pv h h
m= − + − = + − + = − 2 (IXV)
Where the specific enthalpy is introduced to simplify the expression. As in the Otto
cycle, the heat rejected in process 4-1 is given by
414 1
Q u um
= −
The thermal efficiency is the ration of the net work of the cycle to the heat added
41 4 1
23 23 3 2
/ /1 1/ /
cycleW m Q m u uQ m Q m h h
η −= = − = −
− (XV)
As for the Otto cycle, the thermal efficiency from Eq. XV requires values for
or equivalently the temperatures at the principle states of the cycle. Let
us consider next how these temperatures are evaluated. For a given initial temperature
and compression ratio r , the temperature at state 2 can be found using the following
isentropic relationship and data
1 4 2 3, , , and u u h h
1T
rv
32 2
2r c
Vv T rV
= = 2T
1
Where , called the cutoff ratio, has been introduced. Since , the volume
ratio for the isentropic process 3-4 can be expressed as
3 2/cr v v= 4V V=
4 4 2 1 2
3 2 3 2 3 c
V V V V V rV V V V V r
= = = (XVI)
Where the compression ratio and cutoff ratio have been introduces for conciseness.
Using Eq. XVI together with at , the temperature can be determines by
interpolation once is found from the isentropic relationship
r cr
3rv 3T 4T
4rv
44 3
3r r
c
V rv vV r
= = 3rv
In a cold air-standard analysis, the appropriate expression for evaluating is provided
by
2T
12 1
1 2
( )k kT V rT V
1− −= = (constant k)
The temperature is found similarly from 4T
1 134
3 4
( ) ( )k kc
r
V rTT V r
− −= =
Where Eq. XVI has been used to replace the volume ratio.
Effect of compression ratio on performance
1
111( 1)
kc
kc
rr k r
η −
⎡ ⎤−= − ⎢ −⎣ ⎦
⎥ (constant k) (XVII)
Air Standard Brayton Cycle (Gas Turbine Cycle)
Cycle analysis:
1-2 isentropic compression
2-3 constant pressure heat addition
3-4 isentropic expansion
4-1 constant pressure heat rejection
Air standard analysis
3tW h h
m
•
• = − 4 (XVIII)
2cW h h
m
•
• = − 1 (IXX)
3inQ h h
m
•
• = − 2 (XX)
4outQ h h
m
•
• = − 1 (XXI)
3 4 2 1
3 2
( ) (/ /
/
t c
in
h h h hW m W mh hQ m
η• • • •
• •
− − −−= =
−) (XXII)
22 1
1r r
pp pp
= (XXIII)
44 3 3
3 2r r r
pp p pp p
= = 1p (XXIV)
Cold air standard analysis:
14 4
3 3
( )kkT P
T P
−
=
11 1
2 2
( )kkT P
T P
−
=
Effect of compressor pressure ratio on the performance of Brayton cycle
3 4 2 1 4 1
3 2 3 2
( ) ( ) ( )1( ) ( )
p p
p
c T T c T T T Tc T T T T
η− − − −
= =− −
−
1 4 1
2 3 2
/ 11 (/ 1
T T TT T T
η −= −
−)
Knowing 1
4 4
3 3
( )kkT P
T P
−
= and 1
1 1
2 2
( )kkT P
T P
−
=
Since therefore right hand sides are equal and the left hand side must be equal. 4
3 2
P PP P==
1
4
3 2
T TT T
= 1 or 34
1 2
TTT T
=
1
2
1 TT
η = −
( 1) /2 1
11( / ) k kp p
η −= − (k constant) (XXV)
Comparison with Rankine Cycle
The working fluid of Brayton cycle is a gas while the Rankine cycle involves a phase
change (condenser, evap). The pump work is only a small fraction of turbine work,
however in Brayton cycle the compressor work is a large fraction of turbine work this is
why the Rankine cycle has better efficiency. But Brayton cycle is used for Jet engines
because it has a better thrust to weight ratio than Rankine cycle.
T-s Diagram of Ideal Rankine and Brayton Cycles
The Actual Cycle
The deviation of ideal cycle is due to the irreversibilities in turbine and/or compressor.
Effect of Component Efficiencies on Brayton Cycle Performance (Pressure Ratio = 5).
Brayton Cycle With Regenerator
The gases at turbine exit are relatively high (500-600°C). These hot gases are discharged
and not used and represent losses. It is good to recover some of this energy. Such heat
recovery can be accomplished by the use of a ‘regenerator’ or heat exchanger
Hot gases from turbine exhaust are used to heat the cooler air before it enters the
combustion chamber. Therefore less heat (fuel) needs to be supplied in the combustion
chamber to reach a certain max turbine inlet temp.
Disadvantage: Additional ducting and space (extra weight) “not practical for jet engine”.
Effectiveness of a heat exchanger (Regenerator)
2
max 4 2
Actual enthalpy increase of the air flowing through the compressor side of regeneratorMaximum theoretical enthalpy increase
actualq hx heq h h
−= = =
−
2
4 2
'( )''( )
p x
p
c T Te
c T T−
=−
2
4 2
if ' '' (constant specific heat)p p
x
c c
T TeT T
=
−=
−
Example 9.7: Brayton Cycle with Regeneration
Air enters the compressor of an ideal Brayton cycle with regenerator at 100 kPa, 300 K
with a volumetric flow rate of 5 m3/s. The compressor pressure ratio is 10. The turbine
inlet temperature is 1400 K. The regenerator effectiveness is 80%. Determine the thermal
efficiency.
Solution
Known: A regenerative gas turbine operates with air as the working fluid. The
compressor inlet state, turbine inlet temperature, and compressor pressure ratio are
known.
Find: For a regenerator effectiveness of 80%, determine the thermal efficiency. Also
plot the thermal efficiency versus the regenerator effectiveness ranging from 0 to 80%.
Schematic and Given Data:
Assumptions:
1. Each component is analyzed as a control volume at steady state. The control
volumes are shown on the accompanying sketch by dashed lines.
2. The compressor and turbine processes are isentropic.
3. There are no pressure drops for flow through the heat exchangers.
4. The regenerator effectiveness is 80% in part (a).
5. Kinetic and potential energy effects are negligible.
6. The working fluid is air modeled as an ideal gas.
Analysis:
(a) The specific enthalpy values at the numbered states on the T-s diagram are the
same as those in Example 9.4: kg1 300.19 /h kJ= , kg2 579,9 /h kJ= ,
kg , kg . 3 1515.4 /h kJ= =4 808.5 /h kJ
To find the specific enthalpy xh , the regenerator effectiveness is used as follows: By
definition
2
4 2
xreg
h hh h
η −=
−
Solving for xh ,
4 2 2( )
(0.8)(808.5 579.9) 579.9 762.8 /x regh h h h
kJ kg
η= − +
= − + =
With the specific enthalpy values determined above, the thermal efficiency is
3 4 2 1
3
( ) (( / ) ( / )( )( / )
(1515.4 808.5) (579.9 300.19)(1515.4 762.8)
0.568(56.8%)
t c
xin
h h h hW m W mh hQ m
η• • • •
• •
− − −−= =
−
− − −=
−=
)
Gas Turbines for Jet Propulsion:
Gas turbines are particularly suited for aircraft propulsion because of their favorable
power-to-eight ratios. The turbojet engine is commonly used for this purpose. As
illustrated in the figure below, this type of engine consists of three main sections: the
diffuser, the gas generator, and the nozzle. The diffuser placed before the compressor
decelerates the incoming air relative to the engine. A pressure rise known as the ram
effect is associated with this deceleration. The gas generator section consists of a
compressor, combustor, and turbine, with the same functions as the corresponding
components of a stationary gas turbine power plant. In a turbojet engine, the turbine
power output need only be sufficient to drive the compressor and auxiliary equipment,
however. The gases leave the turbine at a pressure significantly greater than atmospheric
and expand through the nozzle to a high velocity before being discharged to the
surroundings. The overall change in the velocity of the gases relative to the engine gives
rise to propulsive force, or thrust.
Turbojet engine with afterburner:
Some turbojets are equipped with an afterburner, as show below. This is essentially a
reheat device in which additional fuel is injected into the gas exiting the turbine and
burned, producing a higher temperature at the nozzle inlet than would be achieved
otherwise. As a consequence, a greater nozzle exit velocity is attained, resulting in
increased thrust.
Cycle analysis
The T-s diagram of the process in an ideal turbojet engine is show on the previous page.
In accordance with the assumptions of an air-standard analysis, the working fluid is air
modeled as an ideal gas. The diffuser, compressor, turbine, and nozzle processes are
isentropic, and the combustor operated as constant pressure.
Process a-1 shows the pressure rise that occurs in the diffuser as the air decelerates
isentropically through this component.
Process 1-2 is an isentropic compression.
Process 2-3 is a constant-pressure heat addition.
Process 3-4 is an isentropic expansion through the turbine during which work is
developed.
Process 4-5 is an isentropic expansion through the nozzle in which the air accelerates and
the pressure decreases.
Owing to irreversibilities in an actual engine, there would be increases in specific entropy
across the diffuser, compressor, turbine, and nozzle. In addition, there would be a
pressure drop through the combustor of an actual engine. The subject of combustion is
discussed in Chap. 13.
In a typical thermodynamic analysis of a turbojet on an air-standard basis the
following quantities might be known: the velocity at the diffuser inlet, the compressor
pressure ratio, and the turbine inlet temperature. The objective of the analysis would be
to determine the velocity at the nozzle exit. Once the nozzle exit velocity is determined,
the thrust is determined by applying Newton’s second law of motion in a form suitable
for a control volume. All principles required for the thermodynamic analysis of turbojet
engines on an air standard basis have been introduces.
Redo example 9.12 pg 441 for homework.
Combined Gas Turbine Vapor Power Cycle
A combined power cycle couples two power cycles such that the energy discharged by
heat transfer from one cycle is used partly or wholly as the input for the other cycle.
The stream exiting the turbine of a gas turbine is at a high temperature. One way this
energy can be used, thereby improving overall fuel utilization, is by a regenerator that
allows the turbine exhaust gas to preheat the air between the compressor and combustor,
which we already discussed.
Another method is the combined cycle. The two power cycles are coupled so that the heat
transfer to the vapor cycle is provided by the gas turbine cycle, called the Topping cycle.
Reheat and Intercooling
Reheat between turbine stages and intercooling between compressor stages provide two
important advantages: The net work output is in creased, and the potential for
regeneration is enhanced. Accordingly, when reheat and intercooling are used together
with regeneration, a substantial improvement in performance can be realized. One
arrangement incorporation reheat, intercooling, and regeneration is shown. This gas
turbine has two stages of compression and two turbine stages. The accompanying T-s
diagram is drawn to indicate irreversibilities in the compressor and turbine stages. The
pressure drops that would occur as the working fluid passes through the intercooler,
regenerator, and combustors are not shown.
Chapter 10: Vapor compression refrigeration systems:
Cycle analysis:
Process 1-2: Isentropic compression of the refrigerant from state 1 to the condenser
pressure at state 2.
Process 2-3: Heat transfer from the refrigerant as it flows at constant pressure through
the condenser. The refrigerant exists as a liquid at state 3.
Process 3-4: Throttling process from state 3 to a two-phase liquid-vapor mixture at
state4.
Process 4-1: Heat transfer to the refrigerant as it flows at constant pressure through the
evaporator to complete the cycle.
1 4inQ h h
m
•
• = −
1 2cW h h
m
•
• = −
2 3outQ h h
m
•
• = −
4 3h h=
/
/in
c
Q m
W mβ
• •
• •=
Therefore:
1 4
1 2
h hh h
β −=
− Which is the coefficient of performance of the ideal cycle
Actual cycle with irreversibilities:
Example:
An automobile air conditioning system uses R-22 as the working fluid. The evaporator
pressure is 80 Lbf/in2, and the condenser outlet pressure is 250 Lbf/in2. The temperature
of the R-22 leaving the evaporator is 5 0F above saturation. Assume an isentropic
compressor with no changes of pressure or temperature in the lines connecting the
components of the cycle. The maximum heat load is 22,000 Btu/h; that is, 22,000 Btu/hr
must be removed to lower the air temperature of the interior of the car to a comfortable
level. Determine the compressor horsepower required and the coefficient of performance
of the cycle. Then repeat the problem for a compressor efficiency of 85 percent.
T-s and p-h diagrams of Vapor Compression Cycle with Nonisentropic Compression
Solution
1 2
1 1 12
1 2 2o 2
2 2
3 2
4 3
80 37.71
Btu42.71 and 80 108.9Lbm
Btu0.2222 and 250Lbm.
Btu 150.8 and 121.3Lbm
Btu at 250 43.46Lbm
Btuand 43.46L
sat
f
LbfP T Fin
LbfT F P hin
Lbfs s PR in
T F h
Lbfh hin
h h
= ∴ = °
= ° = ∴ =
= = =
∴ = ° =
= =
= =
1 4
1 2
bmBtu65.44Lbm
Btu12.40Lbm
5.277
Btu22000 Lbmhr 336.18Btu hr65.94Lbm
1336.18 (12.40 )2545
1.638
Nonisentropic Case:
1.638 10.85
in
comp
in
comp
comp comp
comp
comp
comp
q h h
w h h
qw
m
W m w
Lbm BtuWhr Lbm
W HP
W
β
•
• •
•
•
•
= − =
= − = −
= =
= =
=
=
=
= = .927 The actual work into compressor
1.927 * 2545 4904
4.486
comp
in
comp
HP
BtuWhr
Q
Wβ
•
•
•
= =
= =
Refrigerant properties
Selecting refrigerants