Gas Power Cycles(1)

Chapter 8-1

Chapter 8: Gas Power Cycles

Our study of gas power cycles will involve the study of those heat engines inwhich the working fluid remains in the gaseous state throughout the cycle.We often study the ideal cycle in which internal irreversibilities andcomplexities (the actual intake of air and fuel, the actual combustion process,and the exhaust of products of combustion among others) are removed.

We will be concerned with how the major parameters of the cycle affect theperformance of heat engines. The performance is often measured in terms ofthe cycle efficiency.

th netin

WQ

=Carnot Cycle

The Carnot cycle was introduced in Chapter 5 as the most efficient heatengine that can operate between two fixed temperatures TH and TL. TheCarnot cycle is described by the following four processes.

Chapter 8-2

Carnot Cycle

Process Description 1-2 Isothermal heat addition 2-3 Isentropic expansion 3-4 Isothermal heat rejection 4-1 Isentropic compression

Note the processes on both the P-v and T-s diagrams. The areas under theprocess curves on the P-v diagram represent the work done for closedsystems. The net cycle work done is the area enclosed by the cycle on theP-v diagram. The areas under the process curves on the T-s diagramrepresent the heat transfer for the processes. The net heat added to the cycleis the area that is enclosed by the cycle on the T-s diagram. For a cycle weknow Wnet = Qnet; therefore, the areas enclosed on the P-v and T-s diagramsare equal.

Chapter 8-3

th Carnot LH

TT,

= 1

We often use the Carnot efficiency as a means to think about ways toimprove the cycle efficiency of other cycles. One of the observations aboutthe efficiency of both ideal and actual cycles comes from the Carnotefficiency: Thermal efficiency increases with an increase in the averagetemperature at which heat is supplied to the system or with a decrease in theaverage temperature at which heat is rejected from the system.

Air-Standard Assumptions

In our study of gas power cycles, we assume that the working fluid is air, andthe air undergoes a thermodynamic cycle even though the working fluid inthe actual power system does not undergo a cycle.

To simplify the analysis, we approximate the cycles with the followingassumptions:

The air continuously circulates in a closed loop and always behaves as anideal gas.

All the processes that make up the cycle are internally reversible. The combustion process is replaced by a heat-addition process from an

external source.

A heat rejection process that restores the working fluid to its initial statereplaces the exhaust process.

The cold-air-standard assumptions apply when the working fluid is airand has constant specific heat evaluated at room temperature (25oC or77oF).

Chapter 8-4

Terminology for Reciprocating Devices

The following is some terminology we need to understand for reciprocatingenginestypically piston-cylinder devices. Lets look at the followingfigures for the definitions of top dead center (TDC), bottom dead center(BDC), stroke, bore, intake valve, exhaust valve, clearance volume,displacement volume, compression ratio, and mean effective pressure.

The compression ratio r of an engine is the ratio of the maximum volume tothe minimum volume formed in the cylinder.

Chapter 8-5

r VV

VV

BDC

TDC

= =maxmin

The mean effective pressure (MEP) is a fictitious pressure that, if it operatedon the piston during the entire power stroke, would produce the same amountof net work as that produced during the actual cycle.

MEP WV V

wv v

net net= = max min max min

Chapter 8-6

Otto Cycle: The Ideal Cycle for Spark-Ignition Engines

Consider the automotive spark-ignition power cycle.

ProcessesIntake stroke Compression stroke Power (expansion) stroke Exhaust stroke

Often the ignition and combustion process begins before the completion ofthe compression stroke. The number of crank angle degrees before thepiston reaches TDC on the number one piston at which the spark occurs iscalled the engine timing. What are the compression ratio and timing of yourengine in your car, truck, or motorcycle?

aa

bb

c

d

c

d

a

Chapter 8-7

The air-standard Otto cycle is the ideal cycle that approximates the spark-ignition combustion engine.

Process Description 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection

The P-v and T-s diagrams are

s

T

A ir O tto C yc le T -s D ia g ra m

1

2

3

4

v = c o n s t

P 1

P 3

v

P

T 1

T 3

A ir O tto C yc le P -v D ia g ram

s = c o n s t

1

2

3

4

Chapter 8-8

Thermal Efficiency of the Otto cycle:

th netin

net

in

in out

in

out

in

WQ

QQ

Q QQ

QQ

= = = = 1Now to find Qin and Qout. Apply first law closed system to process 2-3, V = constant.

Q W U

W W W P dV

net net

net other b

, ,

, , ,

23 23 23

23 23 23 2

30 0

== + = + =z

Thus, for constant specific heats,

Q UQ Q mC T T

net

net in v

,

, ( )23 23

23 3 2

== =

Apply first law closed system to process 4-1, V = constant.

Q W U

W W W P dV

net net

net other b

, ,

, , ,

41 41 41

41 41 41 4

10 0

== + = + =z

Thus, for constant specific heats,

Q UQ Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )( ) ( )

41 41

41 1 4

1 4 4 1

== = = =

Chapter 8-9

The thermal efficiency becomes

th Otto outin

v

v

QQmC T TmC T T

,

( )( )

=

=

1

1 4 13 2

th Otto T TT TT T TT T T

,( )( )

( / )( / )

= =

1

1 11

4 1

3 2

1 4 1

2 3 2

Recall processes 1-2 and 3-4 are isentropic, so

TT

VV

and TT

VV

k k

2

1

1

2

1

3

4

4

3

1

= FHGIKJ =

FHGIKJ

Since V3 = V2 and V4 = V1, we see that

Chapter 8-10

TT

TT

orTT

TT

2

1

3

4

4

1

3

2

=

=

The Otto cycle efficiency becomes

th Otto TT, = 11

2

Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic,

TT

VV

TT

VV r

k

k k

2

1

1

2

1

1

2

2

1

1 11

= FHGIKJ

= FHGIKJ =FHGIKJ

where the compression ratio is r = V1/V2 and

th Otto kr, = 11

1

We see that increasing the compression ratio increases the thermalefficiency. However, there is a limit on r depending upon the fuel. Fuelsunder high temperature resulting from high compression ratios willprematurely ignite, causing knock.

Chapter 8-11

Example 8-1

An Otto cycle having a compression ratio of 9:1 uses air as the workingfluid. Initially P1 = 95 kPa, T1 = 17oC, and V1 = 3.8 liters. During the heataddition process, 7.5 kJ of heat are added. Determine all T's, P's, th, theback work ratio, and the mean effective pressure.

Process Diagrams: Review the P-v and T-s diagrams given above for theOtto cycle.

Assume constant specific heats with C v = 0.718 kJ/kg K, k = 1.4. (Use the300 K data from Table A-2)

Process 1-2 is isentropic; therefore, recalling that r = V1/V2 = 9,

T T VV

T r

KK

kk

2 11

2

1

11

1 4 117 273 9698 4

= FHGIKJ =

= +=

b g

b g( ).

.

Chapter 8-12

P P VV

P r

kPakPa

kk

2 11

21

1 495 92059

= FHGIKJ =

==

b g

b g .

The first law closed system for process 2-3 was shown to reduce to (yourhomework solutions must be complete; that is, develop your equations fromthe application of the first law for each process as we did in obtaining theOtto cycle efficiency equation)

Q mC T Tin v= ( )3 2Let qin = Qin / m and m = V1 /v1

v RTP

kJkg K

K

kPam kPa

kJmkg

11

1

3

3

0 287 290

95

0 875

=

=

=

. ( )

.

Chapter 8-13

q Qm

Q vV

kJ

mkgm

kJkg

inin

in= =

= =

1

13

3 37 50 875

38 10

1727

..

.

Then,

T T qC

K

kJkgkJ

kg KK

in

v3 2

698 41727

0 718

31037

= +

= +

=

..

.

Using the combined gas law

P P TT

MPa3 2 32

9 15= = .

Process 3-4 is isentropic; therefore,

Chapter 8-14

T T VV

Tr

K

K

k k

4 33

4

1

3

1

1 4 1

1

31037 19

12888

= FHGIKJ =

FHGIKJ

= FHGIKJ

=

( . )

.

.

P P VV

Pr

MPa

kPa

k k

4 33

43

1 4

1

915 19

422

= FHGIKJ =FHGIKJ

= FHGIKJ

=.

.

Process 4-1 is constant volume. So the first law for the closed system gives,on a mass basis,

Q mC T T

q Qm

C T T

kJkg K

K

kJkg

out v

outout

v

= = =

=

=

( )

( )

. ( . )

.

4 1

4 1

0 718 1288 8 290

717 1

Chapter 8-15

For the cycle, u = 0, and the first law gives

w q q qkJkg

kJkg

net net in out= = =

=

( . )

.

1727 717 4

1009 6

The thermal efficiency is

th Otto netin

wq

kJkg

kJkg

or

,

.

. .

= =

=

1009 6

1727

0 585 58 5%

The mean effective pressure is

max min max min

1 2 1 2 1 1

3

3

(1 / ) (1 1/ )

1009.61298

10.875 (1 )9

net net

net net net

W wMEPV V v v

w w wv v v v v v r

kJm kPakg kPa

m kJkg

= = = = =

= =

Chapter 8-16

The back work ratio is (can you show that this is true?)

BWRww

uu

C T TC T T

T TT T

or

comp

v

v

= = = =

=

exp

( )( )

( )( )

. .

12

34

2 1

3 4

2 1

3 4

0 225 22 5%

Air-Standard Diesel Cycle

The air-standard Diesel cycle is the ideal cycle that approximates the Dieselcombustion engine

Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection


Chapter 8-17

Thermal efficiency of the Diesel cycle

th Diesel netin

out

in

WQ

QQ,

= = 1Now to find Qin and Qout. Apply the first law closed system to process 2-3, P = constant.

Chapter 8-18

E E EQ W U

W W W PdV

P V V

in out

net net

net other b

= =

= + = +=

z

, ,

, , ,

( )

23 23 23

23 23 23 2

3

2 3 2

0

Thus, for constant specific heats

Q U P V VQ Q mC T T mR T T

Q mC T T

net

net in v

in p

,

,

( )( ) ( )

( )

23 23 2 3 2

23 3 2 3 2

3 2

= + = = + =

Apply the first law closed system to process 4-1, V = constant (just like theOtto cycle)

E E EQ W U

W W W PdV

in out

net net

net other b

= == + = + =z

, ,

, , ,

41 41 41

41 41 41 4

10 0

Thus, for constant specific heats

Q UQ Q mC T T

Q mC T T mC T T

net

net out v

out v v

,

, ( )( ) ( )

41 41

41 1 4

1 4 4 1

== = = =


Chapter 8-19

th Diesel outin

v

p

QQmC T TmC T T

,

( )( )

=

=

1

1 4 13 2

th Diesel vp

C T TC T T

kT T TT T T

,( )( )

( / )( / )

= =

1

1 1 11

4 1

3 2

1 4 1

2 3 2

What is T3/T2 ?

PVT

PVT

P P

TT

VV

rc

3 3

3

2 2

23 2

3

2

3

2

= =

= =

where

where rc is called the cutoff ratio, defined as V3 /V2, and is a measure of theduration of the heat addition at constant pressure. Since the fuel is injecteddirectly into the cylinder, the cutoff ratio can be related to the number ofdegrees that the crank rotated during the fuel injection into the cylinder.

What is T4/T1 ?

Chapter 8-20

PVT

PVT

V V

TT

PP

4 4

4

1 1

14 1

4

1

4

1

= =

=

where


PV PV PV PVk k k k1 1 2 2 4 4 3 3= = and Since V4 = V1 and P3 = P2, we divide the second equation by the firstequation and obtain

PP

VV

rk

ck4

1

3

2

= FHGIKJ =

Therefore,

th Dieselck

c

kck

c

kT T TT T T

kTT

rr

rr

k r

,( / )( / )

( )

( )

= = =

1 1 11

1 1 11

1 1 11

1 4 1

2 3 2

1

2

1

Chapter 8-21

What happens as rc goes to 1? Sketch the P-v diagram for the Diesel cycleand show rc approaching1 in the limit.

When rc > 1 for a fixed r, th Diesel th Otto, ,< . But, since r rDiesel Otto> , th Diesel th Otto, ,> .

Chapter 8-22

Brayton Cycle

The Brayton cycle is the air-standard ideal cycle approximation for the gas-turbine engine. This cycle differs from the Otto and Diesel cycles in that theprocesses making the cycle occur in open systems or control volumes.Therefore, an open system, steady-flow analysis is used to determine the heattransfer and work for the cycle.

We assume the working fluid is air and the specific heats are constant andwill consider the cold-air-standard cycle.

The closed cycle gas-turbine engine

3

Comp Turb

1

2

4

WnetWc

Qin

Brayton cycle

3

Comp Turb

1

2

4

WnetWc

Qin

Closed Brayton cycle

Qout

Chapter 8-23

Process Description 1-2 Isentropic compression (in a compressor)2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine)4-1 Constant pressure heat rejection

The T-s and P-v diagrams are

Thermal efficiency of the Brayton cycle

th Brayton netin

out

in

WQ

QQ,

= = 1

Chapter 8-24

Now to find Qin and Qout. Apply the conservation of energy to process 2-3 for P = constant (no work),steady-flow, and neglect changes in kinetic and potential energies.

E Em h Q m h

in out

in

=+ =2 2 3 3

The conservation of mass gives

m mm m m

in out== =2 3

For constant specific heats, the heat added per unit mass flow is

( ) ( )

( )

Q m h hQ mC T T

q Qm

C T T

in

in p

inin

p

= = = =

3 2

3 2

3 2

The conservation of energy for process 4-1 yields for constant specific heats(lets take a minute for you to get the following result)

( ) ( )

( )

Q m h hQ mC T T

q Qm

C T T

out

out p

outout

p

= = = =

4 1

4 1

4 1


Chapter 8-25

th Brayton outin

out

in

p

p

QQ

qq

C T TC T T

,

( )( )

= =

=

1 1

1 4 13 2

th Brayton T TT TT T TT T T

,( )( )

( / )( / )

= =

1

1 11

4 1

3 2

1 4 1

2 3 2


TT

PP

and TT

PP

k k k k

2

1

2

1

1

3

4

3

4

1

= FHGIKJ =

FHGIKJ

( )/ ( )/

Since P3 = P2 and P4 = P1, we see that

3 32 4

1 4 1 2

orT TT TT T T T

= =

The Brayton cycle efficiency becomes

Chapter 8-26

th Brayton TT, = 11

2

Is this the same as the Carnot cycle efficiency?

Since process 1-2 is isentropic,

TT

PP

r

TT r

k k

pk k

pk k

2

1

2

1

11

1

21

1

= FHGIKJ =

=

( )/( )/

( ) /

where the pressure ratio is rp = P2/P1 and

th Braytonp

k kr, ( ) /= 1 11

Chapter 8-27

Extra Assignment

Evaluate the Brayton cycle efficiency by determining the net work directlyfrom the turbine work and the compressor work. Compare your result withthe above expression. Note that this approach does not require the closedcycle assumption.

Example 8-2

The ideal air-standard Brayton cycle operates with air entering thecompressor at 95 kPa, 22oC. The pressure ratio r p is 6:1 and the air leavesthe heat addition process at 1100 K. Determine the compressor work and theturbine work per unit mass flow, the cycle efficiency, the back work ratio,and compare the compressor exit temperature to the turbine exit temperature.Assume constant properties.

Apply the conservation of energy for steady-flow and neglect changes inkinetic and potential energies to process 1-2 for the compressor. Note thatthe compressor is isentropic.

E Em h W m h

in out

comp

=+ =1 1 2 2

The conservation of mass gives

m mm m m

in out== =1 2

For constant specific heats, the compressor work per unit mass flow is

Chapter 8-28

( ) ( )

( )

W m h h

W mC T T

wW

mC T T

comp

comp p

compcomp

p

= = = =

2 1

2 1

2 1

Since the compressor is isentropic

TT

PP

r

T T r

KK

k k

pk k

pk k

2

1

2

1

11

2 11

1 4 1 1 422 273 6492 5

= FHGIKJ =

== +=

( )/( )/

( )/

( . )/ .( ) ( ).

w C T TkJ

kg KK

kJkg

comp p= =

=

( )

. ( . )

.

2 1

1005 492 5 295

19815

The conservation of energy for the turbine, process 3-4, yields for constantspecific heats (lets take a minute for you to get the following result)

( ) ( )

( )

W m h hW mC T T

w Wm

C T T

turb

turb p

turbturb

p

= = = =

3 4

3 4

3 4

Chapter 8-29

Since process 3-4 is isentropic

TT

PP

k k

4

3

4

3

1

= FHGIKJ

( )/

Since P3 = P2 and P4 = P1, we see that

TT r

T Tr

K

K

p

k k

p

k k

4

3

1

4 3

1

1 4 1 1 4

1

1

1100 16

659 1

= FHGIKJ

= FHGIKJ

= FHGIKJ

=

( )/

( )/

( . ) / .

.

w C T T kJkg K

K

kJkg

turb p= =

=

( ) . ( . )

.

3 4 1005 1100 659 1

442 5

We have already shown the heat supplied to the cycle per unit mass flow inprocess 2-3 is

Chapter 8-30

( ) . ( . )

.

m m mm h Q m h

q Qm

h h

C T T kJkg K

K

kJkg

in

inin

p

2 3

2 2 3 3

3 2

3 2 1005 1100 492 5

609 6

= =+ =

= =

= =

=

The net work done by the cycle is

w w wkJkg

kJkg

net turb comp= =

=

( . . )

.

442 5 19815

244 3

The cycle efficiency becomes

th Brayton netin

wq

kJkgkJkg

or

,

.

..

=

= =244 3

609 60 40 40%

The back work ratio is defined as

Chapter 8-31

BWR ww

wwkJkgkJkg

in

out

comp

turb

= =

= =19815

442 50 448

.

..

Note that T4 = 659.1 K > T2 = 492.5 K, or the turbine outlet temperature isgreater than the compressor exit temperature. Can this result be used toimprove the cycle efficiency?

What happens to th, win /wout, and wnet as the pressure ratio r p is increased?

Let's take a closer look at the effect of the pressure ratio on the net workdone.

Chapter 8-32

w w wC T T C T TC T T T C T T T

C Tr

C T r

net turb comp

p p

p p

pp

k k p pk k

= = = =

( ) ( )( / ) ( / )

( ) ( )( )/( ) /

3 4 2 1

3 4 3 1 2 1

3 1 11

1 1

1 1 1

Note that the net work is zero when

r and r TTp p

k k

= = FHGIKJ

1 3

1

1( )/

For fixed T3 and T1, the pressure ratio that makes the work a maximum isobtained from:

dwdr

net

p

= 0

This is easier to do if we let X = r p(k-1)/k

w C T XC T Xnet p p= 3 11 1 1( ) ( )

dwdX

C T X C Tnet p p= =3 2 10 1 1 0 0[ ( ) ] [ ]Solving for X

Chapter 8-33

X TT

rpk k2 3

1

2 1= = d i ( )/

Then, the r p that makes the work a maximum for the constant property caseand fixed T3 and T1 is

r TTp

k k

, max

/[ ( )]

work = FHGIKJ

3

1

2 1

For the ideal Brayton cycle, show that the following results are true. When rp = rp, max work, T4 = T2 When rp < rp, max work, T4 > T2 When rp > rp, max work, T4 < T2The following is a plot of net work per unit mass and the efficiency for theabove example as a function of the pressure ratio.

0 2 4 6 8 10 12 14 16 18 20 22120

140

160

180

200

220

240

260

280

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

Pratio

wnet kJ/kg

th,BraytonT1 = 22C

P1 = 95 kPa

T3 = 1100 K

t = c = 100%

rp,m ax

Chapter 8-34

Regenerative Brayton Cycle

For the Brayton cycle, the turbine exhaust temperature is greater than thecompressor exit temperature. Therefore, a heat exchanger can be placedbetween the hot gases leaving the turbine and the cooler gases leaving thecompressor. This heat exchanger is called a regenerator or recuperator. Thesketch of the regenerative Brayton cycle is shown below.

We define the regenerator effectiveness regen as the ratio of the heattransferred to the compressor gases in the regenerator to the maximumpossible heat transfer to the compressor gases.

Regenerator

3

6

5

Comp Turb

1

24

WnetWc

Qin

Regenerative Brayton cycle

Chapter 8-35

q h hq h h h h

qq

h hh h

regen act

regen

regenregen act

regen

,

, max '

,

, max

= = = = =

5 2

5 2 4 2

5 2

4 2

For ideal gases using the cold-air-standard assumption with constant specificheats, the regenerator effectiveness becomes

5 2

4 2regen

T TT T

Using the closed cycle analysis and treating the heat addition and heatrejection as steady-flow processes, the regenerative cycle thermal efficiencyis

th regen outin

qqh hh h

, =

=

1

1 6 13 5

Notice that the heat transfer occurring within the regenerator is not includedin the efficiency calculation because this energy is not a heat transfer acrossthe cycle boundary.

Assuming an ideal regenerator regen = 1 and constant specific heats, thethermal efficiency becomes (take the time to show this on your own)

Chapter 8-36

th regenk k

pk k

TT

PP

TT

r

,

( ) /

( ) /( )

= FHGIKJ

=

1

1

1

3

2

1

1

1

3

1

When does the efficiency of the air-standard Brayton cycle equal theefficiency of the air-standard regenerative Brayton cycle? If we set th, Brayton= th, regen then

th Brayton th regen

pp

k k

p

k k

rTT

r

r TT

k k

, ,

( ) /

/[ ( )]

( )( )( )/

= =

= FHGIKJ

1 1 11 13

1

3

1

2 1

Recall that this is the pressure ratio that maximizes the net work for thesimple Brayton cycle and makes T4 = T2. What happens if the regenerativeBrayton cycle operates at a pressure ratio larger than this value?

For fixed T3 and T1, pressure ratios greater than this value cause T4 to be lessthan T2, and the regenerator is not effective.

What happens to the net work when a regenerator is added?

What happens to the heat supplied when a regenerator is added?

Chapter 8-37

The following shows a plot of the regenerative Brayton cycle efficiency as afunction of the pressure ratio and minimum to maximum temperature ratio,T1/T3.

Example 8-3: Regenerative Brayton Cycle

Air enters the compressor of a regenerative gas-turbine engine at 100 kPaand 300 K and is compressed to 800 kPa. The regenerator has aneffectiveness of 65 percent, and the air enters the turbine at 1200 K. For acompressor efficiency of 75 percent and a turbine efficiency of 86 percent,determine(a) The heat transfer in the regenerator.(b) The back work ratio.(c) The cycle thermal efficiency.

Compare the results for the above cycle with the ones listed below that havethe same common data as required.(a) The actual cycle with no regeneration, = 0.(b) The actual cycle with ideal regeneration, = 1.0.(b) The ideal cycle with regeneration, = 0.65.(d) The ideal cycle with no regeneration, = 0.(e) The ideal cycle with ideal regeneration, = 1.0.We assume air is an ideal gas with constant specific heats, that is, we use thecold-air-standard assumption.

Chapter 8-38

The cycle schematic is the same as above and the T-s diagram showing theeffects of compressor and turbine efficiencies is below.

Summary of Results

Cycle type Actual Actual Actual Ideal Ideal Ideal

regen 0.00 0.65 1.00 0.00 0.65 1.00comp 0.75 0.75 0.75 1.00 1.00 1.00turb 0.86 0.86 0.86 1.00 1.00 1.00

qin kJ/kg 578.3 504.4 464.6 659.9 582.2 540.2

wcomp kJ/kg 326.2 326.2 326.2 244.6 244.6 244.6

wturb kJ/kg 464.6 464.6 464.6 540.2 540.2 540.2

wcomp/wturb 0.70 0.70 0.70 0.453 0.453 0.453

th 24.0% 27.5% 29.8% 44.8% 50.8% 54.7%

s

T

100 kPa

800 kPa

T-s Diagram for Gas Turbine with Regeneration

1

2s2a

5

3

4s4a

6

Chapter 8-39

Compressor analysis

The isentropic temperature at compressor exit is

TT

PP

T T PP

K kPakPa

K

s

k k

s

k k

2

1

2

1

1

2 12

1

1

1 4 1 1 4300 800100

543 4

= FHGIKJ

= FHGIKJ

= =

( )/

( )/

( . ) / .( ) .

To find the actual temperature at compressor exit, T2a, we apply thecompressor efficiency

compisen comp

act comp

s

a

s

a

acomp

s

ww

h hh h

T TT T

T T T T

K K

K

= =

= +

= + =

,

,

( )

.(543. )

.

2 1

2 1

2 1

2 1

2 1 2 11

300 10 75

4 300

624 6

Since the compressor is adiabatic and has steady-flow

2 1 2 1( )

1.005 (624.6 300) 326.2

comp a p aw h h C T TkJ kJK

kg K kg

= = = =

Chapter 8-40

Turbine analysis

The conservation of energy for the turbine, process 3-4, yields for constantspecific heats (lets take a minute for you to get the following result)

( ) ( )

( )

W m h hW mC T T

w Wm

C T T

turb a

turb p a

turbturb

p a

= = = =

3 4

3 4

3 4

Since P3 = P2 and P4 = P1, we can find the isentropic temperature at theturbine exit.

TT

PP

T T PP

K kPakPa

K

s

k k

s

k k

4

3

4

3

1

4 34

3

1

1 4 1 1 41200 100800

662 5

= FHGIKJ

= FHGIKJ

= =

( )/

( )/

( . )/ .( ) .

To find the actual temperature at turbine exit, T4a, we apply the turbineefficiency.

Chapter 8-41

turbact turb

isen turb

a

s

a

s

a turb s

a

ww

h hh h

T TT T

T T T TK KK T

= =

= = = >

,

,

( ). ( . )

.

3 4

3 4

3 4

3 4

4 3 3 4

2

1200 086 1200 662 5737 7

The turbine work becomes

w h h C T TkJ

kg KK

kJkg

turb a p a= = =

=

3 4 3 4

1005 1200 737 7

464 6

( )

. ( . )

.

The back work ratio is defined as

BWR ww

ww

kJkgkJkg

in

out

comp

turb

= =

= =326 2

464 60 70

.

..

Regenerator analysis

To find T5, we apply the regenerator effectiveness.

Chapter 8-42

regena

a a

a regen a a

T TT T

T T T TK KK

= + = + =

5 2

4 2

5 2 4 2

624 6 0 65 737 7 624 66981

( ). . ( . . ).

To find the heat transferred from the turbine exhaust gas to the compressorexit gas, apply the steady-flow conservation of energy to the compressor gasside of the regenerator.

( )

. ( . . )

.

m h Q m hm m m

qQ

mh h

C T TkJ

kg KK

kJkg

a a regen

a

regenregen

a

p a

2 2 5 5

2 5

5 2

5 2

1005 6981 624 6

73 9

+ == == = = =

=

Using qregen, we can determine the turbine exhaust gas temperature at theregenerator exit.

Chapter 8-43

4 4 6 6

4 6

4 6 4 6

6 4

( )

73.9737.7

1.005

664.2

a a regen

a

regenregen a p a

regena

p

m h Q m hm m m

Qq h h C T T

mkJ

q kgT T K kJCkg K

K

= += == = =

= =

=

Heat supplied to cycle

Apply the steady-flow conservation of energy to the heat exchanger forprocess 5-3. We obtain a result similar to that for the simple Brayton cycle.

q h h C T TkJ

kg KK

kJkg

in p= = =

=

3 5 3 5

1005 1200 6981

504 4

( )

. ( . )

.

Cycle thermal efficiency

The net work done by the cycle is

(464.6 326.2) 138.4

net turb compw w wkJ kJkg kg

= = =

Chapter 8-44

The cycle efficiency becomes

th Brayton netin

wq

kJkgkJkg

or

,

.

.. .

=

= =138 4

504 40 274 27 4%

You are encouraged to complete the calculations for the other values foundin the summary table.

Chapter 8-45

Other Ways to Improve Brayton Cycle Performance

Intercooling and reheating are two important ways to improve theperformance of the Brayton cycle with regeneration.

Intercooling

When using multistage compression, cooling the working fluid between thestages will reduce the amount of compressor work required. The compressorwork is reduced because cooling the working fluid reduces the average

Chapter 8-46

specific volume of the fluid and thus reduces the amount of work on the fluidto achieve the given pressure rise.

To determine the intermediate pressure at which intercooling should takeplace to minimize the compressor work, we follow the approach shown inChapter 6.

For the adiabatic, steady-flow compression process, the work input to thecompressor per unit mass is

4 32 4

1 31 2

= = compw v dP v dP v dP v dP+ + For the isentropic compression process

0

Chapter 8-47

[ ]

2 2 1 1 4 4 3 3

2 1 4 3

1 2 1 3 4 3

( 1) /( 1) /

2 41 3

1 3

= ( ) ( )-1 -1

( ) ( )-1 -1

( / 1) ( / 1)-1

1 1-1

comp

k kk k

k kw P v Pv P v Pvk kk kRR T T T T

k kk R T T T T T T

k

k P PR T Tk P P

+

= +

= + = +

Notice that the fraction kR/(k-1) = C p.

w C T PP

T PPcomp p

k k k k

= FHGIKJ

FHG

IKJ+ FHG

IKJ

FHG

IKJ

LNMM

OQPP

12

1

1

34

3

1

1 1( )/ ( )/

Chapter 8-48

Can you obtain this relation another way? Hint: apply the first law toprocesses 1-4.

For two-stage compression, lets assume that intercooling takes place atconstant pressure and the gases can be cooled to the inlet temperature for thecompressor, such that P3 = P2 and T3 = T1.

The total work supplied to the compressor becomes

w C T PP

PP

C T PP

PP

comp p

k k k k

p

k k k k

= FHGIKJ

FHG

IKJ+ FHGIKJ

FHG

IKJ

LNMM

OQPP

= FHGIKJ +

FHGIKJ

LNMM

OQPP

12

1

1

4

2

1

12

1

1

4

2

1

1 1

2

( )/ ( )/

( )/ ( )/

To find the unknown pressure P2 that gives the minimum work input forfixed compressor inlet conditions T1, P1, and exit pressure P4, we set

dw PdP

comp ( )22

0=This yields

P P P2 1 4=or, the pressure ratios across the two compressors are equal.

PP

PP

PP

2

1

4

2

4

3

= =

Intercooling is almost always used with regeneration. During intercoolingthe compressor exit temperature is reduced; therefore, more heat must be

Chapter 8-49

supplied in the heat addition process. Regeneration can make up part of therequired heat transfer.

To supply only compressed air, using intercooling requires less work input.The next time you go to a home supply store where air compressors are sold,check the larger air compressors to see if intercooling is used. For the largerair compressors, the compressors are made of two piston-cylinder chambers.The intercooling heat exchanger may be only a pipe with a few attached finsthat connects the large piston-cylinder chamber with the smaller piston-cylinder chamber.

Extra Assignment

Obtain the expression for the compressor total work by applyingconservation of energy directly to the low- and high-pressure compressors.

Reheating

When using multistage expansion through two or more turbines, reheating between stages will increase the net work done (it also increases therequired heat input). The regenerative Brayton cycle with reheating isshown above.

The optimum intermediate pressure for reheating is the one that maximizesthe turbine work. Following the development given above for intercoolingand assuming reheating to the high-pressure turbine inlet temperature in aconstant pressure steady-flow process, we can show the optimum reheatpressure to be

P P P7 6 9=

or the pressure ratios across the two turbines are equal.

Chapter 8-50

PP

PP

PP

6

7

7

9

8

9

= =


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