1
6 Reversible Processes
Before considering some reversible processes and cycles we summarize once again the
characteristics of a reversible process:
1. A reversible process must be such that, after it has occurred, the system and the
surroundings, can be made to traverse, in the reverse order. All energy transformations
of the original process would be reversed in direction. The form and the magnitude
remains unchanged.
2. The direction of a reversible process can be changed by making infinitesimal changes in the conditions that control it.
3. The process must be a quasi-static (quasi-equilibrium) process, i.e. during the process, the system and the surroundings at all times be in states of equilibrium or
infinitesimally close to states of equilibrium.
4. A reversible process must be free from friction, unrestrained expansion, mixing, heat transfer across a finite temperature difference or inelastic deformation.
Reversible processes are, therefore, purely ideal, limiting cases of actual processes. However,
these are important because they provide the maximum work output from work-producing
devices (such as turbines and expansion engines) and the minimum work input to devices that
absorb work to operate (such as pumps and compressors). The reversible processes are,
therefore, standards of comparison.
6.1 Work Output in Reversible Processes
Reversible processes are characterized by maximum work output. A number of
processes of practical interest consist of one, or a succession of the following processes:
- constant volume (isochoric), constant temperature (isothermal) or constant pressure (isobaric)
- adiabatic (no heat transfer) - polytropic (pressure and volume varies in such a way that pn = const.) or
isentropic (p = const.).
We will consider these in the following section.
6.1.1 Reversible Isothermal Processes
Processes with heat transfer and work output may be performed at constant
temperature. The energy balance for such closed systems for a change of state from point 1 to
2 is given by
U2 - U1 = Q12 + W12 (6.1)
If there is no entropy production in the process then the entropy can change only through heat
transfer over the system boundary. We can then write according to Eq. (5.37):
2
1
12)( TdSQrev
(6.2)
2
For isothermal processes applying equations (6.1), (6.2) and (5.76) we can write
12121212 )()()()( TSUTSUSSTUUWrev (6.3)
or 1212)( AAWrev (6.4)
where A is the Helmholtz free energy. Thus the difference in the Helmholtz free energy gives
the maximum work output for an isothermal reversible process for a closed system.
For open systems and steady state flow follow similar equations considering the
enthalpy balance [neglecting the changes in potential and kinetic energies]
121212 )( HHPQ t (6.5)
Heat transfer for reversible processes
2
1
12)( STdQrev (6.6)
For the isothermal reversible processes follows then the work output
12121212 )()()()( STHSTHSSTHHPrev
t (6.7)
and using equation (5.77) follows the relationship:
1212)( GGPrev
t (6.8)
where G is the Gibbs free energy. Thus the difference in the Gibbs free energy gives the
maximum work output for an isothermal reversible process for open systems and steady state
processes.
In case the work is only the work due to change in volume [compare Equation (5.5)]
then
2
1
12)( pdVWrev
and if the substance could be modelled as ideal gas i.e. U2 = U1 and H2 = H1 then using ideal
gas equation pV = mRT follows:
2
1 1
2
1
212 lnln)(
p
pmRT
V
VmRT
V
mRTdVW rev (6.9)
and
2
1 1
2
1
212 lnln)(
V
VRTm
p
pRTm
p
RTdpmP revt
(6.10)
3
6.1.2 Reversible Adiabatic Processes (Isentropic Processes)
In a reversible process no entropy is produced. If this process is also adiabatic, that is
no heat transfer (also no entropy change due to heat transfer) takes place through the system
boundary then it is called isentropic (constant entropy) process. Heat transfer is a slow
process and hence many technical processes may be modelled as adiabatic processes even
though they are carried at temperatures different from the surroundings. The energy
conversions in turbines, compressors etc. may be modelled quite satisfactorily as adiabatic
processes. If we consider these as reversible then these can be modelled as isentropic. For
such processes simple equations may be derived by using the model of ideal gas.
Writing equations (5.70) and (5.71) for ideal gases the entropy change for a reversible
change of state is given as:
0 dv
RdT
T
cds
igig (6.11)
and 0 dpp
RdT
T
cds
ig
pig (6.12)
If the heat capacities igvc and ig
pc are considered to be independent of temperature then it
follows from Eq. (6.11) and (6.12) that
)(ln)(ln RdTdcig
and )(ln)(ln pRdTdcigp
that is
igc
R
T
T
)(
2
1
1
2 (6.13)
and igpc
R
p
p
T
T)(
1
2
1
2 (6.14)
Using the well known relations
Rcc igigp
and ig
ig
p
c
c
we find
1
2
1
1
2 )(
T
T (6.15)
and 1
1
2
1
2 )(
p
p
T
T (6.16)
From equation (6.15) and (6.16) follows that 2211 pp
i.e. p=constant (6.17) as the condition (relation) for an isentropic process.
4
Simple equations (6.15) and (6.16) are used to calculate the change of state in
reversible adiabatic processes in many practical applications. It should be remembered that
these have been derived assuming ideal gas model and that the heat capacities remain constant
in the temperature range considered. This is an approximation which holds good for many
technical applications.
The isentropic change of state for real fluids cannot be represented by simple
equations like (6.15) and (6.16). However many processes may be approximated fairly well as
polytropic processes for which pvn = const. where 1 < n < . It is then possible to take into
account the condition of constant entropy while interpolating the state properties from
substance property tables (eg. steam table). In this way the isentropic change of state for real
fluids are accessible. For this purpose the so called T-s and h-s diagrams (entropy diagrams)
are used.
6.1.3 Entropy Diagrams
For a reversible process
2
1
Tdsq rev
So, if one constructs a T-s diagram (s = specific entropy) for a simple system (Fig.
6.1), the area under the curve for a reversible process is equal to the heat interaction of 1kg of
material. This is analogous to the work interaction of a reversible process described by the
area under the curve of a p - diagram (see Fig. 6.1). For an irreversible process the areas under the curves in the respective diagrams are larger than the actual heat or work
interactions. Still the T-s diagram is of importance as it gives the highest value for the heat
interaction for a given path.
For reversible process For reversible process
revif
f
i
wpd )( rev
if
f
i
)q(Tds
For irreversible process For irreversible process
if
f
i
wpd iff
i
qTds
Figure 6.1 : p-v and T-s diagrams for reversible and irreversible processes.
i
f
T
s
i
f
p
v
5
The area enclosed by a cycle on a p- diagram is equal to the cyclic integral pd and
the area enclosed on a T-s diagram is equal to Tds . For any given cycle the two areas must be equal (the net work is equal to the net heat)
0pdTdsdu (6.18) However,
Tdsdq (6.19)
and pddw (6.20)
It is useful to indicate on a T-s diagram certain characteristic lines, such as isobars or
isochores. The shape of these lines depends on the nature of the substance. For an ideal gas
the schematic p-v and T-s diagrams are shown in Figure 6.2
Figure 6.2: p- and T-s diagram for an ideal gas
The schematic T-s diagram for a real substance (water) is shown in Figure 6.3. The
bell-shaped curve represents the saturated region. The top of the bell is the critical point. The
left hand side corresponds to the saturated liquid while the right hand side corresponds to the
saturated vapor. Inside the bell (in the liquid-vapor region), constantpressure curves are horizontal and thus coincide with the constant temperature lines, while outside the bell, i.e. in
the superheated region, the shape of a constant pressure curve approaches an exponential
form, as the vapor approaches ideal gas behavior. The constant specific volume lines are not
shown in this diagram. But they have a steeper slope than constant pressure lines. The lines of
constant vapor content (vapor quality) are also shown in the two phase region. In the
superheated vapor region, constant specific enthalpy lines become nearly horizontal as
pressure is reduced. So in this region the enthalpy is determined only by the temperature. The
variation in pressure between states has almost no effect: h = h(T) and the ideal gas model
provides a reasonable approximation. The T-s diagram finds its greatest application in the
analysis of heat and power cycles. The importance lies in the fact that the amount of heat
transferred during a reversible change of state can be determined according to equation (6.2)
or (6.6) as the area under the state points.
p p
v s
T
v
T
s v
s
T
p
6
1200
1100
900
800
700
600
500
400
300
200
100
-2 -1 0 1 2 3 4 5 6 9 10kJ/kgKs
KT
800
x=1
x=0.2
x=
0.4
x=
0.6
x=0.8
2600
1200
1400
K
3000
3200
p,
bar
= 5
00
300
100
10 150
0.1
p = 0,006107 bar
Sublimation area
Tripel line
h, kJ/kg
= 3
500
x=0
Figure 6.3 : T-s diagram for water.
Another useful diagram is the enthalpy-entropy diagram (h-s diagram) shown in
Figure 6.4 for water. This is also known as the Mollier diagram. It displays the lines of
constant pressure and of constant temperature. Here also, the saturation region is represented
by a bell shaped curve. However, it is skewed as compared to that of the T-s diagram, and the
critical point is not located at the top of the bell. The constant-pressure curves in the two
phase region are also constant-temperature lines. These are straight and their slopes are given
by their temperature, since .)( Ts
hp
In the superheated region, the constant pressure curves
approach exponential form, and the constant-temperature lines tend asymptotically to the
horizontal. The h-s diagram for an ideal gas has the same shape as the T-s diagram, because
here the enthalpy is proportional to the temperature, h = const.+ cpT. The data for specific
volume are not usually included. In the two phase region the lines of constant steam content
(quality) appear. This diagram is intended for evaluating properties at superheated vapor
states and for two phase liquid-vapor region. Liquid data are seldom shown. It is used for
calculations involving heating, cooling, expansion or compression. The work of an adiabatic
steady state flow process is given by the vertical distance between the end points of the
process (as h).
7
K
3800
3400
3200
3000
2800
2600
2400
2200
2000
18009.07.06.05.04.0
x=
0,5
0
x=
0,6
0
300
550
500
450
400
350
300
250
200
150
100
50
0,0
50,1
0,2
600
0,0
1
100
150
50
30
12
4,0
1,5
0,51,0
1
0,7
5
20
6,0
2,0
200
x=0,95x=0,90
x=0,80
x=0,85
x=0,75x
=0,70
x=0,6
5
t, C = 650
p,
bar
= 3
80
kJ/kg
kJ/kgK
s
h
Figure 6.4 : h-s diagram for water.
6.2 Steady Flow of Fluids/Nozzles and Diffusers
The flow of fluids is encountered in many engineering problems. A good
understanding of its various aspects is essential for the designing of pipes used for the
transport of fluids; nozzles and flow passages; orifices for fluid metering etc. Pipes are
normally used for transporting fluids. The nozzles and diffusers, on the other hand, are used in
energy conversion processes. The thermal internal energy of a gas which exists under pressure
may be converted to kinetic energy by reducing its pressure in a nozzle (see Fig. 6.5).
Figure 6.5: Conversion of thermal internal energy into kinetic energy in a nozzle.
t1,p1
c1 0
t2,p2
c2 >> 0
8
Similarly by braking a flowing stream in a diffuser the kinetic energy of flow may be
converted for increasing the enthalpy or for increasing the pressure (see Figure 6.6).
Figure 6.6: Conversion of kinetic energy into internal energy in a diffuser.
The flow of fluids is quite complicated but for many practical engineering problems,
the fluid properties may be assumed to be dependent on only one space coordinate (one
dimensional flow). It may be assumed that all properties of flowing fluid are uniform across
each cross section normal to the direction of flow. No real flow is truly one dimensional but
provided there is no sudden change in the area or direction, and that average values of the
properties at any cross-section are used, the one-dimensional treatment yields quite reliable
results.
A diffuser is sometimes also called a diverging nozzle and the nozzle, a converging nozzle.
6.2.1 Adiabatic process with no work
In nozzles and diffusers no shaft work is involved. However, the change of potential
energy may or may not be significant. A diffuser is used to slow down the flow while the
nozzle is used to speed up the flow. The material balance may be written as
mmm 21
and the mass flow rate m is equal to the volume flow rate [velocity(c) x flow area(A)] divided by the specific volume() of the fluid
2
22
1
11
AcAcm (6.21)
or
cAconstant or cA = constant (6.22)
where the specific density = 1/. Equation (6.22) is known as the continuity equation.
The first law for the study of fluid-flow problems then states for steady flow between two
planes 1 and 2:
t2,p2 t1,p1
c1 >> 0 c2 0
9
)()(2
112
2
1
2
21212 zzgcchhq (6.23)
If the process is reversible then 2
1
12 Tdsq and it follows for reversible steady flow
2
1
12
2
1
2
212 )()(2
1zzgcchhTds (6.24)
For a simple fluid we have, from the property relation (Eq. 5.73)
2
1
2
1
12 dphhTds (6.25)
Combining (6.24) and (6.25) we find
2
1
12
2
1
2
2 0)()(2
1zzgccdp (6.26)
If the change in potential energy is negligible the two velocities are connected by
2
1
2
1
2
2 )(2
1dpcc (6.27)
For any given fluid there is a definite relation between p and (Equation of state) and hence dp term may be evaluated. Thus the velocity change may be calculated.
When a fluid flows through a duct (tube) of varying cross-sectional area the kinetic
energy and the pressure change. For reversible and adiabatic processes the entropy remains
constant (isentropic process). This fact is used to calculate the enthalpy in state 2 if the
pressure is known. Since s2 = s1, it holds
)s,p(h)s,p(hh 12222
For the steady state flow m through a tube of constant cross-sectional area A the change in the kinetic energy is given according to the continuity equation (6.21) as
21222
2
1
2
22
1)(
2
1
A
mcc
(6.28)
6.2.2 Isentropic flow of an ideal gas
We now derive some relations for the steady flow of gases by taking the model of
ideal gas. Neglecting the change in potential energy it follows from equation (6.24) that
10
)(2 212
12 hhcc (6.29)
If we assume the specific heat to be independent of temperature then we can write for
specific enthalpy
)( 2121 TTchhig
p (6.30)
It follows from (6.29) and (6.30) that
)(2 212
12 TTcccig
p (6.31)
Putting the value of T2 from equation (6.16) and 1
)/(
MRcigp in equation (6.31) we
get the relation between the velocities at the state point 2 and state point 1 as
])(1[1
2
1
1
21
2
12
p
pT
M
Rcc (6.32)
where is the isentropic coefficient (exponent).
The velocity at state point 2 thus depends on the temperature and velocity in state 1
and on the pressure ratio (1
2
p
p). If the pressure remains constant then there is no change in
velocity during the flow. If the pressure decreases during the flow the velocity increases and
reaches a maximum when 1
2
p
p=0. If the pressure increases during the flow the velocity
decreases at the state point 2 and attains the minimum value of zero (c2=0) at some maximum
pressure pmax.
Equation (6.32) may be seen as the equation for the mass and energy balance for ideal gas
during isentropic flow.
If the internal area of the duct (tube or channel) remains constant then we also [see
equation (6.22)] have
1
2
2
1
1
2
c
c (6.33)
and hence using (6.17) we obtain the relation:
/1
2
1
1
2 )(p
p
c
c (6.34)
11
Equation (6.34) is a special case of equation (6.32). These equations are consistent only under
the conditions p2=p1 and c2=c1. Therefore the isentropic flow of ideal gas in a duct of constant
area is isobaric and without any change in kinetic energy.
6.2.3 Mass flow through a nozzle
We now consider the flow of an ideal gas through a nozzle (converging nozzle). For
simplification we consider the continuous expansion from initial state where the velocity is
zero (c1=0). Using ideal gas law and equation (6.16) one can write the specific density in state
2 as
1
1
21
2
2
22
)()(
p
pT
M
R
p
TM
R
p
The mass flow rate may now be calculated using equation (6.32)
21/1
1
2
1
1
1
2222 )(
)(
1])(1[
1
2Ap
p
p
TM
Rp
pAcm
or 2/1
1
2
1
1
1
1
2 )(])(1[1
2A
p
pp
p
pm
(6.35)
or 1
12
pAm (6.36)
where
/1
1
2
1
1
2 )(])(1[1
2
p
p
p
p
(6.37)
is a function of the pressure ratio (1
2
p
p). Hence the mass flow rate depends on the pressure
ratio.
For 1
2
p
p = 0 = 0
and for 1
2
p
p = 1 = 0
For a certain 1
2
p
pratio we have a maximum value of and of mass flow rate. The pressure
ratio for the maximum m may be determined by differentiating (6.37) with respect to (1
2
p
p)
and equating to zero.
12
0
)(1
2
p
pd
d
The result is that m is a maximum when
1*
1
2 )1
2()(
p
p (6.38)
This pressure ratio corresponding to maximum flow is called the critical pressure ratio.
6.3 Model Processes for Conversion of Heat into Work
With the growth of civilization and the consequent improvement in the living
standards of mankind, the requirement of power has steadily increased. To meet this ever
increasing requirement of power, a large number of power-producing devices have been
developed. These devices convert the internal energy of fuels into work. It is convenient to
analyze the performance of these devices by means of idealized cycles which are theoretical
approximations of the real cycles.
6.3.1 The Carnot Cycle
First we focus our attention on cycles where all the processes are reversible. Such
cycles, called reversible cycles play an important role in thermodynamics. Carnot cycle is one
of these reversible cycles. It involves
- a system e.g. a gas (not necessarily an ideal gas) - an energy reservoir at some temperature T - an energy reservoir at some lower temperature T0 - some means of periodically insulating the system from one or both of the reservoirs - a part of the surroundings that can both take work from the system and do work on the
system.
For a stationary system, as in a piston and cylinder machine (see Figure 6.7), the cycle
consists of the following four successive processes:
(i) A reversible isothermal expansion process in which heat Q enters the system at T
reversibly from a constant temperature source at T when the cylinder cover is in
contact with the diathermic cover A.
(ii) A reversible adiabatic expansion process in which the diathermic cover A is replaced
by the adiabatic cover B, and the work WE is done by the system adiabatically and
reversibly. The temperature of the system decreases from T to T0.
(iii) A reversible isothermal compression process in which the cover B is replaced by
cover A and heat Q0 leaves the system at T0 to a constant temperature sink at T0
reversibly.
(iv) A reversible adiabatic compression process in which B again replaces A, and work
WP is done upon the system reversibly and adiabatically. The temperature rises from
T0 to T.
13
Figure 6.7 : Carnot heat engine.
Two reversible isotherms and two reversible adiabatics constitute this cycle. It is
represented in p-V and T-S coordinates in Figure 6.8. The area within the p-V diagram of
Figure 6.8 represents the net work of this cycle. The area within the S-T diagram represents
the net heat transfer. Since the system executes a cycle, the net change in its stored energy is
zero. The net effects of this cycle are:
(a) heat is removed from the energy reservoir (source) at T, (b) heat is added to the energy reservoir (sink) at T0, (c) work is produced.
Figure 6.8 : p-V and T-S diagrams for a Carnot cycle.
Application of the first law shows that net work is
0QQWw net
If the working fluid were a liquid vapor mixture of a pure substance instead of a gas, the isothermal processes 1-2 and 3-4 would also be constant pressure processes. During
Source,T
Sink,T0
Diathermic
cover A
Adiabatic cover B
Adiabatic
Adiabatic
WE
WP
2
3
1
Rev. adiabatic
Rev. adiabatic
Rev. isotherm
Rev. isotherm
V
p
4
Gas
S S
T
T
T0
1 2
3 4
14
process 1-2 evaporation would occur, and during process 3-4 condensation would take place.
If all the liquid were evaporated and the vapor became superheated during the isothermal heat
addition, only that part of process 1-2 which occurred in two phase region would be constant
pressure process. The p-V diagrams for these are shown in Figure 6.9.
Figure 6.9 : p-V diagram for a Carnot cycle (working fluid is a liquid-vapor mixture).
For a steady flow system, the Carnot cycle is represented in Figure 6.10. Here heat Q
is transferred to the system reversibly and isothermally at T in the heat exchanger (A), work
WE is done by the system reversibly and adiabatically in the turbine (B), then heat Q0 is
transferred from the system reversibly and isothermally at T0 in the heat exchanger (C), and
then work WP is done upon the system reversibly and adiabatically by the pump (or
compressor) (D). To satisfy the conditions for the Carnot cycle, there must not be any friction
or heat transfer in pipeline through which the working fluid flows. In steady flow systems like
this, the ideal reversible process can be more closely approximated by actual processes than is
the case with the closed system engine described before.
Figure 6.10 : Carnot heat engine - steady flow process.
4
1 2
3
p
V
3
2 1
4
p
V
T0
Flow
Source, T
Sink, T0
Heat exchanger C
Turbine B
WE
WP
T Flow
Flow
Q0
Q
.
.
.
. Compressor D
Heat exchanger A
Surroundings, Atmospheric temperature T0
15
The thermal efficiency of the system shown in Figure 6.10 can be expressed in terms of the
net work output compared to the heat input
Q
Ptth
(6.39)
The net work for the cycles is
0QQWWP PEt
and hence
Q
Q
Q
QQth
00
1)(
(6.40)
For the two reversible isothermal processes
2
1
12)( STSTdQrev (6.41)
is the heat added to the system and
4
334
00 )( STSTdQrev (6.42)
is the heat rejected by the system. We know that entropy taken by the system during a
reversible cyclic process must also be rejected, i.e. SSS 34
12 .
It follows then
Crev
thT
T 01 (6.43)
The thermal efficiency of such a reversible heat power cycle (Carnot cycle ) is also known as
the Carnot efficiency (factor).
It is thus seen that the thermal efficiency of a Carnot process depends only on the
temperatures at which heat is absorbed, T and that at which heat is rejected, T0. It does not
depend on the nature of the medium (fluid) or the mechanical details of the devices. It is
always less than unity. The Carnot efficiency (factor) is the upper limit for the efficiency of
any heat plant working between the same temperatures.
16
6.3.2 The Reversed Carnot Cycle
Since all the processes of the Carnot cycle are reversible, it is possible to imagine that
the processes are individually reversed and carried out in reverse order. When a reversible
process is reversed, all the energy transfers associated with the process are reversed in
direction, but remain the same in magnitude. The reversed Carnot cycle for a steady flow
system is shown in Figure 6.11.
Figure 6.11 : Reversed Carnot heat engine steady flow process.
The reversible Carnot heat engine and the reversed Carnot heat engine are represented
in block diagrams in Figure 6.12. If E is a reversible Carnot heat engine (Fig. 6.12 a) and R is
the reversed Carnot heat engine (Fig. 6.12 b) then the quantities Q, Q0 and W remain of same
magnitude in both; only their directions are reversed. The reversed heat engine R takes heat
from a low temperature body, receives an inward flow of work (net work is done on the
system), and discharges heat to a higher temperature body. The reversed heat engine is known
as heat pump or refrigerator (see chapter 4.4.3).
Figure 6.12: Block diagrams for Carnot heat engine and reversed Carnot heat engine.
Flow
Source, T
Sink, T0
Compressor D
Heat exchanger C
Heat exchanger A
Turbine B
WE
WP
T Flow
Flow
A
B
C
D E
A
B
C
D R WP WE
Wnet =WP-WE Wnet =WE-WP
WE
Q0
Q
WP
T0 T0
T T
Q
Q0
(a) (b)
17
6.3.3 Steam Power Plant / Rankine Cycle
A steam (vapor) power cycle continuously converts heat (energy released by burning
of fuel) into work (shaft work / technical work), in which a working fluid repeatedly performs
a succession of processes. Figure 6.13 gives the schematic diagram of a simple steam power
plant working on the vapor power cycle.
Figure 6.13 : Schematic diagram of a simple steam power plant.
In the vapor cycle, the working fluid, which is water, undergoes a change of phase. For each
process in the vapor power cycle, it is possible to assume a hypothetical or ideal process
which represents the basic intended operation and involves no external effects. For the steam
boiler, this would be a reversible constant pressure heating process of water to form steam, for
the turbine the ideal process would be the reversible adiabatic expansion of steam, for the
condenser it would be a reversible constant pressure heat rejection as the steam condenses till
it becomes saturated liquid, and for the pump, the ideal process would be the adiabatic
reversible compression of this liquid up to the initial pressure. When all these four processes
are ideal, the cycle is an ideal cycle, called an ideal Clausius-Rankine cycle (or simply
Rankine cycle). As all the processes are reversible, this is a reversible cycle. In Figure 6.14,
the cycle has been plotted on the p-, T-s and h-s planes. The numbers on the plots correspond to the numbers on the flow diagram.
Turbine
Condenser
Feed pump
Boiler |(w23)t|
(w01)t
|q30|
q12
2
3
0
1
18
Figure 6.14: p-, h-s and T-s diagram for a Rankine cycle.
The cycle consists of four processes:
1-2, heat addition in a boiler at constant pressure; q12
2-3, reversible adiabatic (isentropic) expansion from boiler pressure down to
condenser pressure; work done by the turbine revtw )( 23
3-0, isothermal heat rejection in a condenser to saturated water state; q30
0-1, reversible adiabatic (isentropic) pumping from condenser pressure back to boiler
pressure; work done on the pump revtw )( 01
Applying first law on the cycle we have the net work
30120123 )()( qqwwwrev
t
rev
t
rev
t (6.44)
Since all the components involve steady flow, the work of the turbine and pump can be found
by applying the steady flow energy equation.
The lowest temperature for heat rejection is limited by the atmospheric temperature.
The heat q30 will be rejected at a temperature higher than atmospheric temperature, tatm
t0 = tatm + t
0
1 2
3
p
v
2
C
10
s
h
C 2
3 01
s
T
1
0
2
C
10
s
h
3
19
As the heat rejection takes place in wet vapor area, the pressure (after the expansion) in the
turbine is also fixed, p0 = ps (t0). For example by heat rejection at t0 = 29C the p0 can have a
value of p0=0.04 bar.
The state of steam as it enters the turbine can also take a limited value depending upon
the nature of fuel and the material used in boiler. For coal fired boilers t2 = 550 C and p2 =
190 bar. The state points 1 and 3 may be calculated from a knowledge of the state points 0
and 2 and considering isentropic change of state (as their pressure and entropy are fixed). The
cycle is thus fully defined.
The thermal efficiency for this cycle can be expressed in terms of the net work output
compared to heat input
12,
0
12
30
12
11m
rev
trev
thT
T
q
q
q
w (6.45)
where the mean thermodynamic temperature Tm,12 for heat reception during 1-2 is obtained by
combining the first and the second law for reversible process q12 = h2-h1 =Tm,12(s2-s1).
12
12
12
1212,
ss
hh
ss
qTm
(6.46)
The mean thermodynamic temperature for heat reception depends on the properties of
working medium and the process. The thermal efficiency of a Clausius-Rankine cycle is
principally lower than the thermal efficiency of a Carnot cycle operating between the same
temperatures T2 and T0 as Tm,12 < T2.
It should be noted that even when Rankine process is a reversible process and hence there is
no devaluation of energy, only revth part of the heat supplied is converted into work.
6.3.4 Gas Turbine Power Plant / Joule Cycle
A simple gas turbine power plant is shown schematically in Figure 6.15(a).
Atmospheric air is compressed to high pressure in compressor and delivered to the burner
where fuel is injected and burnt. The combustion process occurs nearly at constant pressure.
The products of combustion from the burner are expanded in the turbine to atmospheric
pressure and then discharged into the atmosphere. The turbine and compressor are
mechanically coupled so that the net work equals the difference between the work done by the
turbine and the work consumed by the compressor. This gas turbine power plant may be
studied by considering a similar closed cycle gas turbine power plant shown in Figure 6.15(b).
The cycle which is also known as Brayton cycle consists of following steps:
0-1 Air is compressed reversibly and adiabatically. 1-2 Heat is added to it at constant pressure. 2-3 Air expands in the turbine reversibly and adiabatically.
3-0 Heat is rejected from air reversibly at constant pressure to bring it to the initial
state.
20
(a) (b)
Figure 6.15 : (a) Simple gas turbine power plant. (b) Schematic closed cycle gas turbine
power plant.
Figure 6.16 gives the p-, h-s and T-s diagrams for the cycle shown in Figure 6.15(b).
Figure 6.16: Joule cycle in p-, h-s and T-s diagram.
p
1 2
3 0
v
s
h 1
2
3
0 p3 = p0
p2 = p1
s
p2 = p1
T 1
2
3
0 p3= p0
(P01)t
Compressor
Turbine Heater
Cooler
|(P23)t|
1
0
3
2
p0 p0
p p
30Q
12Q
p p 2
1
0 3
|Pt|
p0patm
Air Exhaust gas
Turbine
Combutsion Chamber
Fuel
Compressor
p0
21
Expression for the thermal efficiency can be found by taking the model of ideal gas for
air. The net work may be obtained from the specific turbine and compressor work transfers:
)()()()( 01320123 hhhhwwwrev
t
rev
t
rev
t
or )()( 0132 TTcTTcwig
p
ig
p
rev
t (6.47)
The heat supplied during the cycle is
)()( 121212 TTchhqig
p (6.48)
and the cycle efficiency is therefore
)(
)]()[(
12
0132
12 TT
TTTT
q
wrevtrevth
(6.49)
We can express the temperatures in terms of the pressure ratios as 0 -->1 and 2 --> 3 are
isentropic processes.
For the above isentropic processes holds:
1
0
1
0
1 )(p
p
T
T
and
1
0
1
1
3
1
1
3
2
3
2 )p
p()
p
p()
p
p(
T
T
(since p2 = p1 and p0 = p3)
Inserting the values for T1 and T2 in Eq. (6.49) we get
11
1
revth (6.50)
where
1
0
1 )(
p
p
It is clear from equation (6.50) that the thermal efficiency of the ideal cycle is a function of
pressure ratio only. A high pressure ratio is needed to attain a high thermal efficiency. For air
as working fluid with = 1.4, the thermal efficiency is shown in Figure 6.17.
22
0,8
0,6
0,4
0,2
01 2 5 10 20 50
p/p
re
vth
0 Figure 6.17 : Thermal efficiency of Joule Processes (=1.4).
It is to be noted that like Rankine process here also only a part of the heat supplied can be
converted into work. The reason is the high temperature T3 at which heat is rejected. This
high temperature is required by the enthalpy balance around the turbine, i.e., that the entropy
remains constant during a reversible adiabatic expansion. The mean temperatures at which
heat is taken and at which heat is rejected depend on each other. For a given pressure ratio the
isobars of the process are fixed. Normally the temperature T0 is also given. The temperature
T2 will be fixed by the heat taken. The ratio of heat taken and heat given by the system and
hence the reversible thermal efficiency of a Joule process depends only on the pressure ratio.
The specific technical work is given by the area enclosed by the state points 0-1-2-3-0 in a T-s
diagram. Figure 6.18 shows the specific work as a function of pressure ratio. It is seen that in
contrast to the thermal efficiency the technical work at a fixed inlet turbine temperature
increases with increasing pressure ratio up to a maximum value and then falls.
1,5
0,5
01 2 5 10 20 50
p/p0
1,0
-w /(c T
)
tp
0
igre
v
t =1100 C2
t =800 C2
t =600 C2
Figure 6.18 : Specific work of Joule Processes (T0 = 288 K, =1.4).
With T0 at 288 K and T2 at 1000K, the optimum pressure ratio is 8.8. With this pressure ratio
the mass flow of air for a given power output is minimum, i.e. the size of the power plant is
optimum.
23
6.3.5 Spark Ignition Engine / Otto Cycle (Air standard cycle)
A combination of steadyflow components into a power plant is not the only method of obtaining mechanical power from the combustion of fuel. For small powers it is usually
preferable to employ a cycle consisting of succession of non-flow processes. A given mass of
working fluid can be taken through a series of processes in a cylinder fitted with a
reciprocating piston. Internal combustion engines in which the combustion of fuel occurs in
the engine cylinder itself are non-cyclic heat engines. Here, the working fluid, the fuel-air
mixture, undergoes permanent chemical change due to combustion, and the products of
combustion after doing work are thrown out of the engine, and a fresh charge is taken. So the
working fluid does not undergo a complete thermodynamic cycle. For modelling such engines
air standard cycle is used, in which a certain mass of air operates in a complete
thermodynamic cycle, where heat is added and rejected with external heat reservoir, and all
the processes in the cycle are reversible. Air is treated as an ideal gas.
Otto cycle is the air standard cycle for spark ignition engine used in automobiles.
Figure 6.19 shows the operations of such an engine.
Figure 6.19 : Indicator Diagram of an Otto engine.
STROKE 3Expansion
STROKE 2Compression
Exhaust valveopen
Inlet valveclosed
STROKE 4Discharge
Inlet valve openExhaust valve closed
p
patm
V
Inlet valve
Exhaust valve
Fuel-airmixture
Combustionsproducts
Ignition system
STROKE 1Sucking
Co
mb
us
tio
n
24
Stroke 1 : The inlet valve is open, the piston moves to the right, fuel-air mixture is sucked
into the cylinder at constant pressure.
Stroke 2 : Both the valves are closed, the piston compresses the mixture to the minimum
volume.
The mixture is ignited by means of a spark. On combustion the pressure and
temperature increases.
Stroke 3 : The piston is pushed to the right (work is done on the piston), and the pressure
and temperature of the gas mixture falls.
The exhaust valve opens, and the pressure drops to the initial pressure.
Stroke 4 : With the exhaust valve open, the piston moves inwards to discharge the
combustion gases from the cylinder.
This is the mechanical cycle. It is completed in four strokes of the piston. The
thermodynamic cycle corresponding to this engine is the Otto cycle which consists of two
reversible adiabatics and two reversible isochores. It is shown in Figure 6.20.
Process 1-2: Air is compressed reversibly and adiabatically.
Process 2-3: Heat is added reversibly (equivalent to combustion process in real engine) at
constant volume.
Process 3-4: Work is done by the air in expanding reversibly and adiabatically.
Process 4-1: Heat is rejected by the air reversibly at constant volume (blow-down process
in real engine).
Figure 6.20 : Otto cycle in p-V and T-S coordinates.
The intake and exhaust operations in real engines cancel each other in thermodynamic
consideration.
The net work done during a complete cycle is
4123 QQWV (6.51)
and the thermal efficiency of the cyclic process is given as
3
2
1
4
compresion
S=const
expansion
S=const.
heat
rejection
heat
addition
p
V
2
3
4
1
S
T
v = const.
v = const.
Isentropic
expansion
Isentropic
compression
25
23,
41,
23
41
23
11m
mVth
T
T
Q
Q
Q
W
(6.52)
where Tm,41 is the mean thermodynamic temperature for heat rejection and Tm,23 is the mean
thermodynamic temperature for heat reception.
We use the model of ideal gas to calculate the heat transfer.
Assuming a constant cig
for the whole temperature range, we can write:
)( 2323 TTmcQig
and )( 1441 TTmcQig
The efficiency then follows as:
23
141TT
TTth
(6.53)
Using the relations between the properties for isentropic compression and expansion
processes for ideal gas have
)1()1(
3
4
4
3)1(
2
1
1
2 )()(
T
T
T
T (6.54)
where 2
1
is the compression ratio. The compression ratio is thus fixed by the geometry
of the engine.
By algebraic manipulation of equation (6.54) follows that
)( 14)1(
1
)1(
4
)1(
23 TTTTTT (6.55)
so that ideal cycle efficiency according to equation (6.53) is
3
43
)1(
14
)1(
14 11)(
)(1
T
TT
TT
TTrevth
(6.56)
It is to be remarked at this point that the temperature T4 cannot fall below a certain minimum
value because of the restriction of the isentropic (s=const.) condition and thus limits the ideal
thermal efficiency of an Otto cycle.
It may be seen from equation (6.56) that the efficiency of the air standard Otto cycle is
a function of the compression ratio only. The higher the compression ratio, the higher the
efficiency. Practically it is limited by the ignition temperature of the fuel-air mixture. As the
value of ignition temperature is exceeded the combustion starts during the compression. The
compression ratio cannot, however, be increased beyond a certain limit, because of a noisy
and destructive combustion phenomenon, called detonation. It also depends upon the fuel, the
engine design, and the operation conditions.
26
Normally the values range between 5 and 8 (with revth =0.47 and 0.56 respectively).
For air ( =1.4) we have the following values for the efficiency:
3 4.5 5.5 7.5
1
2
p
p
4.7 8.2 10.9 16.8
rev
th 0.356 0.452 0.494 0.553
6.3.6 Compression Ignition Engine / Diesel Cycle
The limitation on compression ratio in the spark ignition engine can be overcome by
compressing air alone, instead of the fuel-air mixture, and then injecting the fuel into the
cylinder in spray form when combustion is desired. The temperature of air after compression
must be high enough so that the fuel sprayed into the hot air burns spontaneously. The engine
operating in this way is called a compression ignition engine. Diesel engine is an example of a
compression ignition engine. Its indicator diagram is shown in Figure 6.21.
Figure 6.21 : p-V Diagram of a Diesel engine.
Stroke 1: Inlet valve open, piston moves to the right. Air is sucked into the cylinder at
constant pressure.
Stroke 2: Both valves are closed, the air is compressed to the minimum volume. Fuel is
sprayed. Ignition takes place.
Stroke 3: The piston is pushed to the right and the pressure and temperature fall. The
exhaust valve opens, and the pressure drops to the initial pressure.
Stroke 4: With exhaust valve open, the piston moves inwards to discharge the
combustion gases from the cylinder.
The simplified model cycle for the thermodynamic study is shown in Figure 6.22.
STROKE 3Expansion
STROKE 2Compression
STROKE 4Discharge
p
patm
V
STROKE 1 Sucking
Combustion
27
p
V
2 3
4
1
Heating
Cooling
ExpansionCompression
s=const.s
=const.
T
S
Co
mp
ressio
n
s=
co
nst.
Exp
an
sio
n
s=
co
nst.
3
4
p=co
nst.
V=co
nst
.
2
1
Figure 6.22 : p-V Diagram of a model process for Diesel engine.
In this cycle the heat addition occurs at constant pressure. It consists of following processes:
1-2 : The air is compressed isentropically V1 V2. 2-3 : Heat q23 in added while the air expands at constant pressure to volume V3. At state 3
the heat supply is cut off. The ratio =V3/V2 is called cut-off ratio.
3-4 : The air is expanded isentropically to the original volume (V4 = V1).
4-1 : Heat q41 is rejected at constant volume until the cycle is completed.
The specific heat transfers in this cycle may be calculated by taking the model of ideal gas
and considering heat capacities to be constant in the entire temperature range.
)()}({)( 23232323232323 TTchhvvpuuwuuqig
pV (6.57)
and )( 141441 TTcuuqig (6.58)
The thermal efficiency is given by
)(
)(1
23
14
23
4123
23 TTc
TTc
q
q
wig
p
ig
Vth
or )(
123
14
TT
TTth
(6.59)
The efficiency may be expressed in terms of the compression ratio and the cut-off ratio
2
3
V
V . The temperatures in Eq.(6.59) may be eliminated as shown below.
For isentropic process 1-2, we have
1
1
2
T
T
28
and hence
1
21
1
TT (6.60)
For constant pressure process 2-3, we have
2
3
2
3
V
V
T
T
and hence T3 = T2. (6.61)
Also
11
42
23
1
4
3
3
4
VV
VV
V
V
T
T (6.62)
so that 1
24
TT (6.63)
Substituting for T1, T3 and T4 in equation (6.59) we have for the ideal efficiency
}1
1{
11
)(
)1
()(
1)1(
22
1
2
1
2
TT
TTrev
th (6.64)
It may be seen from equation (6.64) that the ideal efficiency of the Diesel cycle depends upon
cut-off ratio (and hence upon the quantity of heat added) and on the compression ratio .
Since the term in braces, { }, is always greater than unity (except when =1, i.e. when there
is no heat addition), the Diesel cycle always has a lower efficiency than the Otto cycle of the
same compression ratio. This is not a very significant result because practical engines based
upon the Diesel cycle can employ higher compression ratios than those based on Otto cycles
and thus reach a higher efficiency. For cycles with air (=1.4) having a typical compression ratio =14 we get the following values for the ideal efficiency:
1.5 2.0 2.5 3 4
rev
th 0.620 0.593 0.568 0.546 0.506
29
6.4 Heat Pump / Refrigeration Cycles
In a refrigeration cycle heat is received at a low temperature and rejected at a high
temperature, while a net amount of work is done on the fluid. Thus this is reverse of what
happens in a (heat engine) power cycle. The term heat pump is applied to a device whose
purpose is to supply heat at an elevated temperature to maintain the temperature of a body
higher than the temperature of surroundings. The term refrigerator is used for a device whose
purpose is to extract heat from a cold space to maintain its temperature lower than the
temperature of the surroundings. Both are identical in principle and it is possible to use the
same device for both heating and cooling purposes [see also Heat Pump and Refrigerator in
Section 4.7]
The coefficient of performance for heat pump, abbreviated to (COP)hp and denoted by
is given as
t
H
hpP
QCOP
)( (6.65)
where HQ is the rate of heat supply at higher temperature and Pt is the net power consumed.
Since HQ is negative we take the absolute value to make a positive quantity. It may be seen
that (COP)hp is the reciprocal of the efficiency of a power cycle.
In a refrigerator, attention is confined to the heat which is to be removed continuously
from the low temperature space. The performance parameter in a refrigerator, called the
coefficient of performance for refrigerator, abbreviated to (COP)ref, and denoted by 0 is given as
tP
Q00
(6.66)
where 0Q is the rate of heat removed from the space to be cooled.
6.4.1 Vapor Compression Refrigeration Cycle
Figure 6.23 shows the schematic diagram of a vapor compression refrigeration process
and Figure 6.24 the property diagram (T-s diagram). The refrigerant (working fluid used in
refrigeration cycle) is first compressed from state 1 (saturated vapor state) to state 2 reversibly
and adiabatically (isentropically) where the specific work input is revtw )( 12 . State 2 lies on an
isobar of the condenser pressure whose corresponding condensation temperature is slightly
higher than the temperaure for heat transfer. The condensation takes place at constant pressure
reversibly in the process 2-3 where the heat rejection is q23. The refrigerant then expands
reversibly and adiabatically in process 3-4, where the work output is revtw )( 34 . State 4 is in the
wet vapor region. Finally it absorbs heat q41 from the surroundings at T0 to evaporate and
reach the state 1.
30
Figure 6.23 : Schematic diagram of a vapor compression refrigeration process.
Figure 6.24 : T-s diagram of a refrigeration cycle.
The application of the first and second laws of thermodynamics for the energy
conversions taking place during the reversible processes gives.
1121212 ),()( hsphhhwrev
t [since 1-2 is an isentropic process]
HHmH sTqhhq ,2323 [where Tm,H is the mean thermodynamic
temperature at which heat is rejected]
3343434 ),()( hsphhhwrev
t [since 3-4 is an isentropic process]
and 0004141 sTqhhq
(P12)t |(P34)t|
2
4 1
3
HQQ || 23
41Q
4 1
3
2
T
s
31
Since s0 = sH, we can write the coefficient of performance for the heat pump rev
as
)()(
)(
)( 3412
23
3412hhhh
hh
ww
qrev
t
rev
t
Hrev
or 0,
,
0 TT
T
q
Hm
Hm
H
Hrev
(6.67)
The value of rev depends on temperatures at which heat is rejected and received.
The coefficient of performance for the refrigeration process rev0 is
)()(
)(
)( 3412
41
3412
0
0hhhh
hh
ww
qrev
t
rev
t
rev
or 0,
0
0
00
TT
T
q
HmH
rev
(6.68)
In an actual vapor refrigeration cycle, an expansion machine is not used, since power recovery
is small and does not justify the cost of the expansion machine. A throttle valve or a capillary
tube is used to reduce the pressure of the fluid from that in the condenser down to that in the
evaporator. Figure 6.25 shows the flow diagram of such a cycle.
Figure 6.25 : Vapor compression refrigeration process with a throttle valve.
Since the flow through a throttle valve is irreversible, the expansion is accompanied by
an increase in entropy. Because the expansion is irreversible the whole cycle is irreversible.
As the throttling process is an adiabatic steady flow process in which no work crosses the
boundary (see Section 4.4.2.3), the net work is now equal to the compression work. The heat
transfer in the condenser is unaffected, but the heat extracted in the evaporator is reduced.
Both and 0 are, therefore, reduced.
Condenser
Evaporator
Throttle
valve
Compressor
(P12)t
3
2
1 4
41Q
23Q
32
Figure 6.26 shows the property diagrams of this cycle.
Figure 6.26 : Property diagrams of a vapor compression refrigeration cycle.
The coefficient of performance for the heat pump is then
)(
)()(
12
23
hh
hhCOP revhp
(6.69)
and the coefficient of performance for the refrigerator
)(
)()(
12
410
hh
hhCOP revref
(6.70)
4
s
h
3 p2
1
2
p1
4 1
p1 3 2
p2
s=c
p
V
h=c
h=c
p1
p2
2
1 4
T
3
s
33
6.4.2 Gas Cycle Refrigeration
Refrigeration can also be achieved by means of a gas cycle. It is then essential to use
an expansion machine (expander) because the temperature remains almost unchanged by
throttling. Cooler and heater replace the condenser and evaporator of a vapor compression
machine. The processes in cooler and heater are constant pressure processes and not constant
temperature processes. The ideal gas refrigeration cycle is the same as the reversed Joule
cycle. The flow diagram of such a cycle is shown in Figure 6.27 and the T-s diagram in
Figure 6.28. The gas (usually atmospheric air) is compressed reversibly and adiabatically
from state 1 to state 2 and then cooled up to atmospheric temperature (T3 = Tatm) in a cooler.
The gas then cools down further when it passes through a heat exchanger, up to a temperature
of T4 = T0 (T0 is the highest temperature for the heat reception). Then an adiabatic expansion
takes place in a turbine (temperature drops to T5). The turbine supplies some of the power
requirements of the compressor. As T5 is lower than T0, the heat )( 056 QQ can be absorbed
at T0 as the gas flows at constant pressure through the cooling space. The temperature of gas
rises to T6 (T6 = T0). The gas is then passed through a heat exchanger to complete the cycle
(i.e. attains the initial state 1 ).
Figure 6.27 : Schematic diagram of a reversible gas cycle refrigeration process.
Cooler
Compressor
Heat Exchanger
Cooling Space
Turbine
|(P45)t|
1
2 3
4
5 6
(P12)t
HQQ 23
056 QQ
34
Figure 6.28 : T-s diagram of a reversible gas cycle refrigeration process.
We again use the model of ideal gas and consider the heat capacities to remain
constant. The application of the first law of thermodynamics to this steady flow process with
mass flow rate of m yields
)()()( 21212 atmig
pt TTcmhhmP
Hatmig
p
ig
p QTTcmTTcmhhmQ )()()( 2232323
)()()( 054545 TTcmhhmPig
pt
and 050565656 )()()( QTTcmTTcmhhmQig
p
ig
p
The coefficient of performance for the refrigerator is then
)()( 502
5000
TTTT
TT
P
Q
atmt
where Pt = (P12)t - (P45)t
or
1
)1(
)1(
1
1
1
5
05
2
50
20
T
TT
T
TT
TT
TT
atm
atmatm
(6.71)
Considering the isentropic changes of state T2 Tatm and T0 T5 we have:
1
05
02
p
p
T
T
T
T
atm
or
1
005
11
p
p
TT
s
T
p
p0
2
3
4
5
6
1
Tatm
T0
35
Now putting the values of temperature ratios in terms of pressure ratio we get the coefficient
of performance for refrigerator:
1
1)(
1
00
0
p
p
T
T
COP
atm
revrev
ref (6.72)
Example : In an Otto process with Vmax= 2 dm
3 and Vmin= 0.25 dm
3 the air at 20
oC and
0.1 MPa is sucked and compressed isentropically. It is then heated at constant
volume till the pressure reaches 3 MPa. Afterwards the gas expands
isentropically upto full piston stroke and attains the initial state through
isochoric cooling.
(a) Sketch the process in a p-V diagram. (b) Calculate the properties at all state points and the ideal thermal efficiency
of the process.
DATA: Molecular weight of air = 29 g/mol
for air = 1.4
Solution:
We look the process in a p-V diagram:
What we know is p1 = 0.1 MPa
V1 = Vmax = 2 dm = 2 l
T1 = 20 C =293.15 K
V2 = Vmin = 0.25 l
p3 = 3 MPa
Also that V1 = V4 and V3 = V2
3
2
p
V
4
1
V1=V4
isochoric
V3=V2
isochoric
isentropic
isentropic
36
The specific volume of air in the initial state 1 may be calculated from a knowledge of
the other state properties T1, p1 and using ideal gas equation
][101.0
]][[15.293
][29][
]][[315.8)(
6
2
1
1
1N
mK
gmolK
molNm
p
TM
R
or
kg
m
g
m 33
1 8405.00008405.0
The compression ratio may be calculated
825.0
23
3
min
max
2
1 dm
dm
V
V
V
V
kg
m312 10506.0
8
Now 1-2 is an isentropic process and so we can calculate T2 if we know T1 through the
relation
)1(
1
2
T
T or )1(12
TT
T2 = 293.15 [K] (8)0.4
= 673.48 K
][10506.0
]][[48.673
][29][
]][[315.8)(
3
2
2
2m
kgK
gmolK
molNmTMR
p
or
p2 = 1.838106
2m
N = 1.838 MPa
The temperature in state 3 may be calculated by considering the isochoric condition i.e. 3 = 2
]][[315.8][
][29]][[10506.0
][
][103
)(
3
2
633
3molNmkg
gmolKm
m
N
MR
pT
or
T3 = 1099.25 K
Process 3-4 is the isentropic expansion. The temperature T4 may be calculated using the
relation
)1(
3
4 1
T
T
37
or
KKT
T 47.4788
][25.1099)4.0()1(
3
4
Since kg
m3
14 8405.0 , we can calculate p4.
MPam
N
m
kgK
gmolK
molNmTMR
p 1632.010632.1
][8405.0
]][[47.478
][29][
]][[315.8)(
2
5
3
4
4
4
The ideal thermal efficiency may be calculated directly using the relation
565.08
11
11
)4.0()1(
revth
To calculate the heat and work transfer for isochoric process we need cig
[for isobaric
process we need cpig
].
We know that for molar heat capacities holds
Rcc igigp or Rccigig
p
and
ig
ig
p
c
c
Hence
ig
ig
c
cR
1igc
R or
)1(
Rc ig [molar]
or )1(
)(
MR
c ig for specific heat capacity
kgK
kJ
gK
Nm
gmolK
molNmc ig 7168.07168.0
4.0][29][
]][[315.8
Now q23 = cig
(T3 T2) = 0.7168(1099.25 673.48) =305.19 kJ/kg
q41 = cig
(T1 T4) = 0.7168(293.15 478.47) = -132.84 kJ/kg
Net work kg
kJqqwwwV 35.17219.30584.132)( 23413412
The ideal thermal efficiency
565.019.305
35.172
23
q
wVrevth
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