3.1 Brayton Cycle
Used for gas turbines which usually operate on an open cycle.
Fuel Combustion chamber
3
2
Compressor Turbine w net
1 Fresh air Exhaust gases 4
OPEN-CYLCE GT ENGINE (Real GT engine)
The cycle above can be modeled as a closed cycle, by utilizing the air-standard assumptions .
The combustion process is replaced by a constant-pressure heat addition process from an external source. The exhaust process is replaced by a constant-pressure heat rejection process to ambient air. This ideal closed-loop cycle is known as the Brayton cycle; made up of 4 internally reversible processes:
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
q in
Heat exchanger
3
2
Compressor Turbine w net
1 Heat exchanger 4
q out
Closed-cycle gas-turbine engine operating on Brayton Cycle
T 3 P 2 3
q in
P=c
s=C s=C
2 4
q out
1 4
1 P=c
s v
1

2
All four processes of an ideal Brayton cycle are steady-flow processes. Neglecting changes in kinetic and potential energies, we can write the following steady flow energy equations: q in q out

 h
3 h
4

 h
2 h
1

 c c p p


T
3
T
4


T
2
T
1


Under cold-air-standard assumptions (ie constant c p
at 25 o
), the thermal efficiency becomes
 th , Brayton
 w net q in
= 1
 q in q out

1
 c p c p


T
4
T
3

T
1

T
2


= 1

T
1 


T
4
T
2 


T
3
T
1
T
2

1




1



Processes 1-2 and 3-4 are isentropic (P
2
= P
3
and P
4
= P
1
). Thus
T
2
T
1




P
2
P
1


 k

1 k




P
P
4
3


 k

1 k

T
T
4
3
When substituted into the equation for thermal efficiency gives,
 th , Brayton

1

1 r p
( k

1 ) / k where r p
is the pressure ratio, P
2
/ P
1
and k is the specific heat ratio (i.e c p
/c v
).
The thermal efficiency increases with both r p
and k , which also occurs with real gas turbines.
For fixed values of T min
= T
1
and T max
= T
3
, the net work output of Brayton cycle first increases with pressure ratio, then reaches a maximum at r p
=



T max
T min



2 k
 
and finally decreases. A low work output per cycle means a larger mass flow rate is needed to maintain the same power output (ie bigger plant) which is not economical. In common gas turbines, pressure ratio ranges from 11 to 16, where a compromise is made between high r p and reasonable net work output.
Two major application of gas-turbine engines are : aircraft propulsion and electric power generation. The ratio of compressor work input to turbine work output is known as back work ratio, which is high for gas-turbine power plants (usually more than 50%), whereas for steam power plant, the ratio (pump work input divided by turbine work output) is only a few percent. Therefore, for similar net power output, turbines in gas-turbine plants are larger than those used in steam power plants.
The gas-turbine cycle thermal efficiency improvement covers three areas:
(1) Increasing the turbine inlet (firing) temperatures.
Today the inlet temperature can reach 1425 o
C made possible by new materials and innovative cooling techniques such as ceramic coating of turbine blades and blade cooling by discharged air from the compressor.
(2) Increasing the efficiencies of turbomachinery components.
Made possible by computer-aided design which results in minimal energy losses in compressors and turbines.
(3) Adding modifications to the basic cycle.
By incorporating intercooling, regeneration (or recuperation), and reheating. These come with added initial and operation costs which can be justified when decrease in fuel costs offsets the increase in other costs.

3
Example 1
A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilizing airstandard assumptions, determine (a) the gas temperature at the exits of the compressor and turbine, (b) the back work ratio, and (c) the thermal efficiency.
. 1300 K
T 3
q in
P=c
r p
= P
2
/P
1
= P
3
/P
4
= 8
2 4
q out
300K
P=c
1
s
(a)

Process 1–2 is isentropic compression of ideal gas,
T
1
= 300 K
Table A-17
h
1
= 300.19 kJ/kg, P r1
= 1.386
P
2
P
1

P r
P r 1
2 = 8 P r 2

P r 1


P
2
P
1

 = 1.386 (8) = 11.09
Table A-17
T
2
= 540 K, h
2
= 544.35

Process 3– 4 is isentropic expansion of ideal gas,
T
3
= 1300 K
Table A-17
h
3
= 1395.97 kJ/kg, P r3
= 330.9
P
4
P
3

P r
P r
4
3
P r 4

P r 3


P
4
P
3

 = 330.9 / 8 = 41.36
Table A-17
T
4
= 770 K, h
4
= 789.37
(b) w comp,in
= h
2
– h
1
= 544.35 – 300.19 = 244.26 kJ/kg
w turb,out
= h
3
– h
4
= 1395.97 – 789.37 = 606.6 kJ/kg r bw
 w comp , in w turb , out

244 .
16
606 .
6
= 0.403 (ie 40.3 % of turbine work is used to drive compressor)
(c) q in
= h
3
– h
2
= 1395.97 – 544.35 = 851.62
w net
= w out
– w in
= 606.6 – 244.16 = 362.4
 th
 w net q in
362 .
4
=
851 .
62
0.426 @ 42.6%
Note: Under cold-air-standard assumptions, the thermal efficiency is,
 th , Brayton

1

1 r p
( k

1 ) / k
= 1

1
1 .
4

1
= 0.448 @ 44.8%.
8 1 .
4
which is comparable to the more accurate value of 42.6%.

4
Deviation of Actual Gas-Turbine Cycles from Ideal Cycles
The actual cycles differ from the ideal ones on accounts of:

Actual heat-addition and heat-rejection processes involve some pressure drops, whereas in ideal cycles there are none

Actual compressor work is more, and actual turbine work is less compared to ideal
Brayton cycle due to irreversibilities. It is convenient to define isentropic efficiencies for compressor and turbine as follows:
T 3

C
 w s w a
 h
2 h
2 s a

 h
1 h
1
. 2a

T
 w w s a  h
3 h
3

 h
4 h
4 a s
2s 4s
.
.
4a
. 1
.
s
Example 2
Assuming a compressor efficiency of 80% and a turbine efficiency of 85%, determine (a) the back work ratio, (b) the thermal efficiency, and (c) the turbine exit temperature of the cycle in
Example 1.
. 1300 K
T 3
q in
P=c
r p
= P
2
/P
1
= P
3
/P
4
= 8
2a
2s 4s 4a
q out
300K
P=c
1
s
For compressor,

C
 w s w a w a
 w comp , in
 w s

C

244 .
16
0 .
80
= 305.20
For turbine,

T
 w a w s w comp , in
Therefore, r bw
 w turb , out
 w a
 w turb , out
 
T w s
305 .
20
515 .
61
= 0.592 @ 59.2%
= 0.85(606.6) = 515.61
The compressor consumes 59.2% of the turbine work output (c.f 40.3% in Ex.1). The increase is caused by irreversibilities within the compressor and turbine.
(b) w comp,in
= h
2a
– h
1
h
2a
= h
1
+ w comp,in
= 300.19 + 305.20 = 605.39
kJ/kg
From Table A-17, T
2a
= 598 K.
q in
= h
3
– h
2a
= 1395.97 – 605.39 = 790.58
w net
= w out
– w in
= 515.61 – 305.20 = 210.41

5
 th
 w net
210 .
41
=
790 .
58
= 0.266 @ 26.6% q in
Compare this with 42.6% in Ex. 1. The drop in thermal efficiency is caused by irreversibilites within the turbine and compressor.
(c) Energy balance for actual turbine gives
w turb,out
= h
3
– h
4a
h
4a
= h
3
– w turb,out
= 1395.97 – 515.61 = 880.36 kJ/kg
Table A-17 gives, T
4a
= 853 K
NOTE: Turbine exit temperature is much higher than that at compressor exit (T
2a
= 598K), which suggests using regeneration to reduce fuel consumption (and cost).
3.2 BRAYTON CYCLE WITH REGENERATION
When the exhaust gas temperature leaving the turbine is much higher than the temperature of air leaving the compressor, the exhaust gas can be used to heat the air leaving the compressor.
The process takes place in a counter-flow heat exchanger known as a regenerator or a recuperator. This exercise reduces the heat input (and fuel) for the same net work output, for which the thermal efficiency increases.
3
T
q in
5’ 4
q regen
5
q saved
= q regen
6
2
q out
1
s
regenerator
6
heat
combustion 4
1 chamber
2 5 3
C T w net

Assuming the regenerator to be adiabatic, and neglecting changes in kinetic and potential energies, the actual and maximum heat transfer from the exhaust gases to the air are
q regen,act
= h
5
– h
2
q regen,max
= h
5’
– h
2
= h
4
– h
2
The regenerator effectiveness is defined as,
  q regen , act q regen , max
Using the cold-air-standard assumptions, it reduces to:
 
 h
5
T
5 h
4

T
4



T
2
T
2 h
2 h
2
A higher effectiveness saves a greater amount of fuel (due to higher pre-combustion temperature), but it requires a larger regenerator which is costly. In practice, the effectiveness is below 0.85.
Under cold-air-standard assumptions, the thermal efficiency of ideal Brayton cycle with regeneration is
 th , regen

1




T
T
3
1


 r p k

1 k
Regeneration is most effective at lower pressure ratios and low T
1
/T
3
ratios (see Fig. 9-40
Cengel).
Example 3
The gas-turbine in Example 2 has a regenerator with an effectiveness of 80 percent.
Determine its thermal efficiency.
  h
5 h
4 a

 h
2 a h
2 a h
5
 h
2 a
 
 h
4 a
 h
2 a

= 605.39 + 0.80(880.36 – 605.39 = 825.37
Thus, q in
= h
3
– h
5
= 1395.97 = 825.37 = 570.60
 th
 w net q in
=
210 .
41
570 .
60
= 0.360 @ 36.9 %
Note that net work output is not affected. The regenerator only reduces q in
. Therefore thermal efficiency increases.
3
T
q in
4a
4s
q regen
5
q saved
= q regen
2a 6
2s
q out
1
s
6

7
3.3 BRAYTON CYCLE WITH INTERCOOLING, REHEATING &
REGENERATION
Steady-flow compression or expansion work is proportional to the specific volume of the fluid. Recall that steady-flow work is given by w
  vdP
To lower compressor power (input), the specific volume of the working fluid should be as low as possible. To increase turbine work (output), the specific volume of the working fluid should be as high as possible. These are accomplished as follows:

By intercooling between compressor stages, the compressor work is reduced.

By reheating between turbine stages, turbine work is increased
Example 4
An ideal gas-turbine cycle with two stages of compression and two stages of expansion has an overall pressure ratio of 8. Air enters each stage of the compressor at 300 K and each stage of the turbine at 1300 K. Determine the back work ratio and the thermal efficiency, assuming (a) no regenerators and (b) and ideal regenerator with 100 percent effectiveness. Compare the results with those in Example 3.
Regenerator
10
exhaust gases
air Combustion Reheater
5 chamber
1 4 6 7 8 9
intercooler w net
2 3
Compressor I Compressor II Turbine I Tur bine II
T 6 8
q in
9
5 7
q regen
q regen
= q saved
4 2
10
q out
3 1
s

8
For best performance, it can be shown that (see Chap.7 Cengel) equal pressure ratios are maintained across each compressor and turbine stage, to minimise total compressor work required and maximise total turbine work output. Thus
P
2
P
1

P
4 
8 = 2.83 and
P
6
P
7

P
8
P
9

8 = 2.83
P
3
Assuming each compressor stage has the same isentropic efficiency (100% in this case), the temperature and enthalpy of the air at each compressor exit will be the same. A similar argument is used for the turbines.
At inlets: T
1
= T
3
, h
1
= h
3
and T
6
= T
8
, h
6
= h
8
At exits: T
2
= T
4
, h
2
= h
4
and T
7
= T
9
, h
7
= h
9
Thus, w comp-I
= w comp-II and w turbine-I
= w turbine-II
(a)
T
1
= 300 K
Table A17
h
1
= 300.19 kJ/kg, P r1
= 1.386
Process 1–2 is isentropic, so we can write
P
2
P
1

P r
P r
2
1
P r 2

P r 1
P
2
P
1

( 1 .
386 ) 8 = 3.92
TableA17
T
2
= 403.3 K, h
2
= 404.31 kJ/kg
T
6
= 1300 K
TableA17
h
6
= 1395.97 kJ/kg, P r6
= 330.9
Process 6–7 is isentropic, so we can write
P
7
P
6

P r
P r
7
6
P r 7
Thus, r bw
 w comp , in w turb , out

P r 6
P
7
P
6

( 330 .
9 )
w comp,in
= 2 w comp,in,I
= 2 (h
2
– h
1
) = 2 (404.31 – 300.19) = 208.24 kJ/kg
w turb,out
= 2 w turb,out,I
= 2 (h
6
– h
7
) = 2 (1395.97 – 1053.33) = 685.28 kJ/kg
w net
= w turb,out
– w comp,in
= 685.28 – 208.24 = 477.04 kJ/kg
q in
= q primary
+ q reheat
= (h
6
– h
4
) + (h
8
– h
7
)
= (1395.97 – 404.31) + (1395.97 – 1053.33) = 1334.30 kJ/kg

208 .
24
685 .
28
1
8
= 117.0
T A17
T
7
= 1006.4 K, h
7
= 1053.33 kJ/kg
= 0.304 @ 30.4 %
 th
 w net q in

477 .
04
1334 .
30
= 0.358 @ 35.8 %
Discussion: Compared with Ex. 3 (single-stage compression and expansion), shows that multistage compression with intercooling and multistage expansion with reheating improves the backwork ratio (drops from 40.3 to 30.4%) but hurt thermal efficiency (drops from 42.6 to
35.8%). Conclusion: Intercooling and reheating must be accompanied by regeneration (to improve thermal efficiency, as part (b) will show).

9
(b) The addition of an ideal regenerator (no pressure drops, and 100% effectiveness) does not affect compressor work and turbine work. Therefore, the back work ratio and net work output of an ideal gas-turbine cycle are identical whether there’s a regenerator or not. A regenerator only reduces the heat input by preheating the air leaving the compressor, using hot exhaust gases.
In an ideal regenerator, the compressed air is heated to the turbine exit temperature T
9, before it enters the combustion chamber. Under air-standard assumptions, h
5
= h
7
= h
9
.
q in
= q primary
+ q reheat
= (h
6
– h
5
) + (h
8
– h
7
)
= (1395.97 – 1053.33) + (1395.7 – 1053.33) = 685.28
 th
 w net q in

477 .
04
685 .
28
= 0.696 @ 69.6 %
Discussion: The thermal efficiency almost doubles by adding a regenerator. As the number of compression and expansion stages is increased, the cycle approaches the Ericsson cyle which has the thermal efficiency of
 th , Ericsson
  th , Carnot

1

T
L
T
H

1

300
1300
= 0.769 @ 76.9%
In practice, it is not economical to increase the number of stages beyond two, because thermal efficiency improvement will be greatly offset by increased costs of additional stages.

10
Tutorial #3 – Gas Turbine Cycles
1. Air is used as the working fluid in a simple ideal Brayton cycle with a pressure ratio of 12, compressor inlet temperature of 300 K, and turbine inlet temperature of 1000 K. Determine the mass flow rate of air for a power output of 70 MW, if both compressor and turbine have isentropic efficiency of (a) 100 % and (b) 85 %. Assume constant specific heats at room temperature.
(a) 352 kg/s, (b) 1037 kg/s
2. A gas-turbine power plant operates on the simple Brayton cycle between the pressure limits of 100 and 1200 kPa. Air enters the compressor at 30 o C at a rate of 150 m 3 /min and leaves the turbine at
500 o C. Using variable specific heats for air and assuming a compressor isentropic efficiency of 82 percent and a turbine isentropic efficiency of 88 percent, determine (a) the net power output, (b) the back work ratio, and (c) the thermal efficiency.
(659 kW, 0.625, 0.319)
3. A Brayton cycle with regeneration using air as the working fluid has a pressure ratio of 7. The minimum and maximum temperatures in the cycle are 310 and 1150 K. Assuming an isentropic efficiency of 75 percent for the compressor and 82 percent for the turbine and an effectiveness of 65 percent for the regenerator, determine (a) the air temperature at the turbine exit, (b) the net work output, and (c) the thermal efficiency. Assume variable specific heats for air.
(783 K, 108.1 kJ/kg, 22.5 %)
4. Air enters the compressor of a regenerative gas-turbine engine at 300 K and 100 kPa, where it is compressed to 800 kPa and 580 K. The regenerator has an effectiveness of 72 percent, and air enters the turbine at 1200 K. For a turbine efficiency of 86 percent, determine (a) the amount of heat transfer in the regenerator and (b) the thermal efficiency. Assume variable specific heats for air.
(152.5 kJ/kg, 36.0 %)
5. For a specified pressure ratio, why does multistage compression with intercooling decrease the compressor work, and multistage expansion with reheating increase the turbine work ?
6. Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 9. The air enters each stage of the compressor at
300 K and each stage of the turbine at 1200 K. Accounting for the variation of specific heats with temperature, determine the minimum mass flow rate of air needed to develop a net power output of
110 MW. (250 kg/s)
7. Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion.
The pressure ratio across each stage of the compressor and turbine is 3. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Determine the back work ratio and the thermal efficiency of the cycle, assuming (a) no regenerator is used and (b) a regenerator with 75 percent effectiveness is used. Use variable specific heats.
(a) 33.5 %. 36.8 % (b) 33.5 %, 55.3 %