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

qq

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

qq

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

qq

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