Gas turbine cycle

1
 GAS TURBINES
Introduction
Gas turbines are prime movers
producing mechanical power
from the heat generated by the
combustion of fuels.
They are used in aircraft, some
automobile units, industrial
installations and small – sized
electrical power generating
units.
A schematic diagram of a
simple gas turbine power plant
is shown below.
This is the open cycle gas
turbine plant. 2
 Working:
Air from atmosphere is
compressed adiabatically
(idealized) in a compressor
(usually rotary) i.e., Process
1–2.
This compressed air enters the
combustion chamber, where
fuel is injected and undergoes
combustion at constant
pressure in process 2–3.
The hot products of
combustion expand in the
turbine to the ambient
pressure in process 3–4 and
the used up exhaust gases are
let out into the surroundings. 3
The compressor is usually
coupled to the turbine, so that the
work input required by the
compressor comes from the
turbine.
The turbine produces more work
than what is required by the
compressor, so that there is net
work output available from the
turbine.
Since the products of combustion
cannot be re–used, real gas
turbines work essentially in open
cycles. The p–v and T–s diagrams
of such a plant are shown above.
4
The compressor is usually
coupled to the turbine, so that the
work input required by the
compressor comes from the
turbine.
The turbine produces more work
than what is required by the
compressor, so that there is net
work output available from the
turbine.
Since the products of combustion
cannot be re–used, real gas
turbines work essentially in open
cycles. The p–v and T–s diagrams
of such a plant are shown above.

5
 Closed Cycle Model

The open gas-turbine cycle can be

modelled as a closed cycle, using the
air-standard assumptions

The compression and expansion

processes remain the same, but 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 the ambient air.
6
Brayton Cycle:
This is the air–
standard cycle for
the gas turbine plant.
It consists of
two reversible
adiabatic processes
and two reversible
isobars (constant
pressure processes).
The p–v and T–s
diagrams of a
Brayton Cycle are as
shown.

7
Process 1 - 2: Reversible
adiabatic compression.
2 – 3: Reversible constant
pressure heat addition.
3 – 4: Reversible adiabatic
expansion.
4 – 1: Reversible constant
pressure heat rejection.
A schematic flow diagram of
this somewhat hypothetical
gas turbine plant is shown
below.

8
Though this plant works on a closed cycle, each of the four
devices in the plant is a steady–flow device, in the sense that
there is a continuous flow of the working fluid (air) through
each device.
Hence, the steady–flow energy equation is the basis for
analysis, and can be applied to each of the four processes.
Neglecting changes in kinetic and potential energies, the
steady flow energy equation takes the from
Q – W = ∆h = Cp.∆T (Since air is assumed to be an ideal
gas)
Process 1 – 2 is reversible adiabatic, hence Q1-2 = 0
W1-2 = - Cp.∆T = - Cp (T2 – T1): - ve, work input
Work of compression
Wc = |W1 – 2| = Cp (T2-T1)

9
Process 2–3 is a constant pressure
process

Heat added,

Process 3-4 is again reversible

adiabatic,

+ve work output.

10
Process 4-1 is also a constant
pressure process

: -ve, i.e., heat is rejected

Heat rejected,
Q2 = |Q4-1| = Cp (T4-T1)

Therefore the cycle efficiency,

11
Therefore the cycle efficiency,

For isentropic process 1-2,

& for process 3-4,

12
Since p3 = p2 and p4 = p1

Compression ratio,
13
 Therefore,

Pressure ratio,

Thus it can be seen that, for the same compression ratio,

A closed cycle turbine plant is used in a gas–cooled nuclear

reactor plant, where the source is a high temperature gas
cooled reactor supplying heat from nuclear fission directly
to the working fluid (gas/air).
14
Effect of pressure Ratio rp on simple Brayton
Cycle:-

That means, the more the pressure ratio, the more will be the efficiency.
Temperature T1 (=Tmin) is dependent on the temperature of
surroundings.
Temperature T3 (=Tmax) is limited by metallurgical considerations and
heat resistant characteristics of the turbine blade material.
For fixed values of Tmin and Tmax, the variation in net work output, heat
added and efficiency with increasing pressure ratio rp can be explained
with the help of a T-s diagram as shown.

15
For low pressure ratio, the net work output is small and
the efficiency is also small (Cycle 1 – 2 – 3 - 4).
In the limit, as rp tends 1, efficiency tends to zero (net
work output is zero, but heat added is not zero).
As the pressure ratio increases, the work output
increases and so does the efficiency.
However, there is an upper limit for rp when the
compression ends at Tmax.
As rp approaches this upper limit (rp)max, both net work
output and heat added approach zero values.
However, it can be seen that the mean temperature
heat addition Tadd approaches Tmax, while the mean
temperature of heat rejection approaches Tmin, as rp
comes close to (rp)max.
16
Hence cycle efficiency, given
by
approaches the Carnot
efficiency i.e.,

rp - (rp)max When the

compression ends at Tmax
i.e., when state point 2 is at
Tmax.

When rp=rpmax,
17
The variation of net work output Wnet with pressure
ratio rp is shown below.

As rp increases from 1 to (rp)max, Wnet increases from

zero, reaches a maximum at an optimum value of rp
i.e., (rp)opt and with further increase in rp, it reduces
and becomes zero when rp = rpmax 18
Pressure Ratio for maximum net work output:-
Wnet= Cp[(T3 - T4) - (T2 - T1)]

T3 = Tmax & T1= Tmin

19
Condition for maximum Wnet is

i.e.,

It can be seen that,

20
Maximum net work output

Corresponding to rp = (rp)opt i.e., when Wnet is maximum, cycle

efficiency is

21
Effect of pressure Ratio rp on simple Brayton Cycle

The thermal efficiency of an ideal

Brayton cycle depends on the
pressure ratio, rp of the gas
turbine and the specific heat
ratio, γ of the working fluid.
The thermal efficiency increases
with both of these parameters,
which is also the case for actual
gas turbines.
A plot of thermal efficiency versus
the pressure ratio is shown in, for Thermal efficiency of the ideal
the case of γ =1.4. Brayton cycle as a function of
Pressure ratio

22
Effect of irreversibility’s in turbine/compressor:
In the ideal Brayton cycle,
compression and expansion of
air are assumed to be reversible
and adiabatic.
In reality, however,
irreversibility’s do exist in the
machine operations, even
though they may be adiabatic.
Hence the compression and
expansion processes are not
really constant entropy
processes.
Entropy tends to be increase (as
per the principle of increase of
entropy). 23
Effect of irreversibility’s in turbine/compressor:
The T–s diagram of a Brayton
cycle subject to irreversibility’s will
be as shown.
Irreversibility’s result in a reduction
in turbine output by (h4-h4S) and in
an increase in the compressor
input by (h2 – h2S).
Hence the output reduces by the
amount (h4–h4S )+ (h2–h2s).
Though heat input is also reduced
by (h2-h2s), the cycle efficiency is
less than that of an ideal cycle.
The extent of losses due to
irreversibility’s can be expressed
in terms of the turbine and
24
compressor efficiencies.
Turbine efficiency,

Compressor efficiency,

25
 Back Work Ratio

The fraction of the turbine work used to drive the compressor is

called the back work ratio.
BWR is defined as the ratio of compressor work to the
turbine work

The BWR in gas turbine power plant is very high, normally one-
half of turbine work output is used to drive the compressor
Back Work Ratio for gas power cycles: 40-80%
 Work Ratio

The fraction of the turbine

work that becomes the net
work is called the work
ratio.

Work Ratio is defined as the ratio of net work to the

turbine work
Methods of Improving the thermal efficiency of the open gas turbine cycle

Although the performance of the gas turbine is not especially attractive compared
with the efficiencies possible in Diesel and Petrol engine power plants, a simple
gas turbine has advantages in weight, size and vibration compared to the engine
and in size and cost compared to small steam plant. It is also superior to both in
quantity of water used, for the simple gas turbine plant uses almost no cooling
water.

Even if the components used in the gas turbine power plants are improved in
design, the efficiency and the specific output of the simple gas turbine cycle are
quite low. The efficiency handicap is surmountable, at the expense of adding
complexity to the gas turbine plant.

The principle Methods:

 Regeneration
 Intercooling
 Reheating

28
 Methods of Improving the thermal efficiency of the open gas turbine cycle
Thermal efficiency of the Brayton cycle :

The principle Methods:

 Regeneration
 Intercooling
 Reheating
(a) QA the heat added or supplied can be decreased by the process called
regenerative heating of gas before entering the combustion chamber.
(b) Compressor work Wc can be decreased when multistage compressors with
intercooling are used.
(c) Turbine work-output Wt can be increased by using a multistage turbine with
reheating of the gas in between the two stages. 29
Methods of Improving the thermal efficiency of the open gas turbine cycle

Regeneration:
The efficiency of the Brayton cycle can be increased by
utilizing part of the energy of exhaust air from the
turbine to preheat the air leaving the compressor, in a
heat exchanger called regenerator.
This reduces the amount of heat supplied Q1 from an
external source, and also the amount of heat rejected
Q2 to an external sink, by an equal amount.
Since Wnet = Q1 - Q2 and both Q1 and Q2 reduce by
equal amounts, there will be no change in the work
output of the cycle.

30
Regeneration:

31
Regeneration:
Heat added Q1 = h3 –h2’ = Cp (T3 – T2’)
Heat rejected Q2 = h4’ – h1 = Cp (T4’ – T1)
Turbine output WT = h3 – h4 = Cp (T3 – T4)
Compressor input WC = h2 – h1 = Cp (T2 – T1)

Regeneration can be used only if the temperature of air

leaving the turbine at 4 is greater than that of air leaving
the compressor at 2.
In the regenerator, heat is transferred from air leaving the
turbine to air leaving the compressor, thereby raising the
temperature of the latter.
The maximum temperature to which compressed air at 2
can be heated is equal to the temperature of turbine
exhaust at 4. 32
Regeneration:
This, however, is
possible only in an
ideal regenerator.
In reality, T2’<T4.
The ratio of the
actual temperature
rise of compressed
air to the maximum
possible rise is called
effectiveness of the
regenerator.

33
Regeneration:

With a regenerator, since Wnet

remains unchanged, but Q1
reduces, efficiency
η = Wnet/Q1 increases.
This is also evident from the
fact that the mean temperature
of heat addition increases and
the mean temperature of heat
rejection reduces with the use
of the regenerator, and
efficiency is also given by

34
 With regenerator,

In the regenerator,
Heat lost by hot air = Heat gained by cold air
i.e.,

With an ideal regenerator,

35
therefore,

36
 1-

For a fixed ratio , the cycle efficiency decreases

with increasing pressure ratio.

In practice, a regenerator is expensive, heavy
and bulky and causes pressure losses, which may
even decrease the cycle efficiency, instead of
increasing it.

37
Multistage compression with inter cooling

38
 In this arrangement,
compression of air is carried
out in two or more stages with
cooling of the air in between
the stages.
The cooling takes place in a
heat exchanger using some
external cooling medium
(water, air etc).
Shown above is a schematic
flow diagram of a gas turbine
plant with two-stage
compression with inter
cooling.
39
1-2: first stage compression
(isentropic)
2-3: inter cooling
(heat rejection at
constant pressure)
3-4: second stage
compression (isentropic)
4-5: constant pressure heat
addition
5-6: isentropic expansion
6-1: constant pressure heat
rejection.
40
Air, after the first stage compression is cooled before it
enters the second stage compressor.
If air is cooled to a temperature equal to the initial
temperature (i.e., if T3=T1), inter cooling is said to be
perfect.
In practice, usually T3 is greater than T1.
Multistage compressor with inter cooling actually
decreases the cycle efficiency.
This is because the average temperature of heat addition
Tadd is less for this cycle 1-2-3-4-5-6 as compared to the
simple Brayton cycle 1-2’-5-6 with the initial state 1. (refer
fig).
Average temperature of heat rejection Trej also reduces,
but only marginally.
41
Hence efficiency is less for the modified cycle.
However, if a regenerator is also used the heat
added at lower temperature range (4 to 4’) comes
from exhaust gases from the turbine.
So there may be an increase in efficiency
(compared to a simple Brayton cycle) when multi–
stage compression with inter cooling is used in
conjunction with a regenerator.
For a gas turbine plant using 2–stage compression
without a generator,
Q1 = h5 - h4 = Cp(T5 - T4)
WT = h5 - h6 = Cp(T5-T6)
WC = (h2 - h1) + (h4 - h3)
= Cp [(T2 - T1) + (T4 - T3)] 42
WC = (h2 - h1) + (h4 - h3) = Cp [(T2 - T1) + (T4 - T3)]

Wnet = WT – WC

= Cp [(T5 - T6) – {(T 2- T1) + (T4 - T3)}]

43
Multi-Stage expansion with reheating:

44
Here expansion of working fluid (air) is carried out in 2 or
more stages with heating (called reheating) in between
stages.
The reheating is done in heat exchangers called
Reheaters.
In an idealized cycle, the air is reheated, after each stage
of expansion, to the temperature at the beginning of
expansion.
The schematic flow diagram as well as T-s diagram for a
gas turbine plant where in expansion takes place in two
turbine stages, with reheating in between, are shown.
Multi-Stage expansion with reheating, by itself, does not
lead to any improvement in cycle efficiency. In fact, it only
reduces. 45
However, this modification together with regeneration may
result in an increase in cycle efficiency.
It can be seen from the T-s diagram that the turbine exhaust
temperature is much higher when multi stage expansion with
reheating is used, as compared to a simple Brayton cycle.
This makes the use of a regenerator more effective and may
lead to a higher efficiency.
Heat added Q1 = (h3 - h2) + (h5 - h4)
= Cp(T3 - T2) + Cp(T5 - T4)
Turbine output WT = (h3 - h4) + (h5 - h6)
= Cp(T3 - T4) + Cp(T5 - T6)
Compressor input WC = h2 - h1 = Cp(T2 - T1)

46
Ideal Regenerative cycle with inter
cooling and reheat:
Considerable improvement in efficiency is possible by
incorporating all the three modifications simultaneously.
Let us consider a regenerative gas turbine cycle with two
stage compression and a single reheat.
The flow diagram and T-S diagram of such an
arrangement is shown.
Idealized Regenerative Brayton cycle with two stage
compression with inter cooling and also two stage
expansion with reheating – ideal regenerator, equal
pressure ratios for stages, no irreversibilities, perfect inter
cooling and reheating.

47
Heat added Q1 = Cp(T5 - T4’) + Cp(T7 - T6)

Turbine output WT = Cp(T5 - T6) + Cp(T7 - T8)

Compressor input WC = Cp(T2 - T1) + Cp(T4 - T3)

If perfect inter cooling, no irreversibilities, equal pressure

ratios for stages and ideal regenerator are assumed,
T1=T3, T2=T4=T8’, T5=T7 and T6=T8=T4’
48
Then, Q1 = Cp(T5 - T4’) + Cp(T7 – T6)

= Cp (T5 - T6) + Cp(T5 - T6)

=2(T5 - T6)

Q2 = Cp(T8’ - T1) + Cp(T2 - T3)

= Cp(T2 - T1) + Cp(T2 - T1)

=2 Cp(T2 - T1)

49
.

50
It can be seen from this expression that the efficiency
decreases with increasing pressure ratio rp.
51
Open Cycle Gas Turbine
Plants:
In practice, a gas turbine plant works
on an open cycle.
Air from atmosphere is first
compressed to a higher pressure in
a rotary compressor, which is usually
run by the turbine itself, before it
enters the combustion chamber.
Fuel is injected into the combustion
chamber where it undergoes
combustion.
The heat released is absorbed by
the products of combustion and the
resulting high temperature; high
pressure products expand in the 52
turbine producing work output.
The used up combustion products
(exhaust gases) are let out into the
atmosphere.
In the ideal case, compression and
expansion are assumed to be isentropic
and combustion is assumed to take place
at constant pressure.
The schematic flow diagram and p-v and
T-s diagrams of an open cycle gas turbine
plant are as shown.

53
Advantages and disadvantages of closed cycle
over open cycle
Advantages of closed cycle:
1. Higher thermal efficiency
2. Reduced size
3. No contamination
4. Improved heat transmission
5. Improved part load 
6. Lesser fluid friction
7. No loss of working medium
8. Greater output and
9. Inexpensive fuel.
54
Disadvantages of closed cycle:
1. Complexity
2. Large amount of cooling water is required. This
limits its use of stationary installation or marine
use
3. Dependent system
4. The wt of the system pre kW developed is high
comparatively,  not economical for moving
vehicles
5. Requires the use of a very large air heater.

55
56
 Problem 1
Tutorial
1. In a Gas turbine installation, the air is taken in at 1
bar and 150C and compressed to 4 bar. The
isentropic  of turbine and the compressor are 82%
and 85% respectively. Determine (i) compression
work, (ii) Turbine work, (iii) work ratio, (iv) Th. .
What would be the improvement in the th.  if a
regenerator with 75% effectiveness is incorporated
in the cycle. Assume the maximum cycle
temperature to be 825K.

Solution: P1 = 1 bar T1 = 2880K P2 = 4 bar

T3 = 8250K C = 0.85 t = 0.82
57
Case1: Without Regeneration:
Process 1-2s is isentropic i.e.,
r 1
T2 s  P2  r
  
T1  P1 

T2 s  288 4

0.4
1.4  428.14 K
T2 s  T1 428.14  288
But C  i.e.,0.85  T2  452.87 K
T2  T1 T2  288

Process 3-4s is isentropic

r 1 0.4
T4 s  P4  r
 1 1.4
i.e.,    T4 s  825   554.96 K
T3  P3  4
But T3  T4 825  T4
t  i.e., 0.82  T4  603.57 K
T3  T4 s 825  554.96
58
 WC = CP (T2 – T1)
= 1.005 (452.87 – 288) = 165.69 kJ/kg
(ii) Turbine work,
Wt = CP (T3 – T4)
= 1.005 (825 – 603.57) = 222.54 kJ/kg

(iii) Work ratio = (222.54-165.69)/222.54 = 0.255

(iv) QH = Q2-3 = CP(T3 – T2) =1.005(825-452.87)

=373.99kJ/kg
(iv) Thermal Efficiency (th),
WT  WC 56.85
th  1
  0.152
QH 373099 59
Case2: With Regeneration:
We have effectiveness,
T5  T2 T5  452.87
 i.e., 0.75 
T4  T2 603.57  452.87
T5 = 565.890K
Heat supplied,
QH1 = Q5-3 = CP(T3 – T5)
= 1.005 (825 – 565.89)
= 260.4 kJ/kg
WT  WC 56.85
th  1
  0.218
QH 260.4
Improvement in th due to
regenerator  0.218  0.152 = 0.436
0.152
60
i.e., 43.6%
 Problem 2
Tutorial
The maximum and minimum pressure and
temperatures of a gas turbine are 5 bar, 1.2
bar and 1000K and 300K respectively.
Assuming compression and expansion
processes as isentropic, determine the th
(a) when an ideal regenerator is incorporated in
the plant and (b) when the effectiveness of
the above regenerator is 75%.
Solution:
P2 = P3 = 5 bar P1 = P4 = 1.2 bar
T3 = 1000K T1 = 300K
61
Process 1-2s is isentropic i.e.,
r 1
T2 s  P2  r
  
T1  P1 
0.4
 5  1.4
T2 s  300   451.21K
 1.2 
Process 3-4s is isentropic i.e.,
r 1
T4 s  P4  r
  
T3  P3 
0.4
 1.2  1.4
T4 s  1000   664.88K
 5 
62
Ideal regenerator: i.e., T5 = T4

Heat supplied = CP (T3 – T5)

Qin = 1.005 [1000 – 664.88] = 336.79 kJ/kg

Wnet = WT – WC = CP (T3 – T4) – CP (T2 – T1)

= 1.005 [1000 – 664.88 – 451.21 + 300] = 183.91kJ/kg

Wnet 183.91
th    0.546  54.6%
Qin 336.79

63
Regenerator with  = 0.75 i.e.,
i.e., T5  T2 actual temperatur e drop
0.75  
T4  T2 ideal temperatur e drop
T5  451.21
0.75  T5  611.46 K
664.88  451.21

Heat supplied, QH = CP (T3 – T5)

= 1.005 (1000 – 611.46) = 390.48kJ/kg

Wnet 183.91
th    0.471  47.1%
Qin 390.48
64
Problem 3
Tutorial
A gas turbine plant uses 500 kg of air/min, which
enters the compressor at 1 bar, 170C. The
compressor delivery pressure is 4.4 bar. The
products of combustion leaves the combustion
chamber at 6500C and is then expanded in the
turbine to 1 bar. Assuming isentropic efficiency of
compressor to be 75% and that of the turbine to be
85%, calculate (i) mass of the fuel required /min, of
the CV of fuel is 39000KJ/Kg. (ii)net power output
(iii)Overall thermal efficiency of the plant. Assume
CP=1.13KJ/Kg-K, =1.33 for both heating and
expansion.
65
 Solution:
 a  500 kg / min  8.33 kg / sec
m
P1 = 1 bar T1 = 290 K P2 = 4.4 bar T3 = 923 K
C = 0.75 t = 0.85 , Wnet = ? , th ? m f  ?
Calorific Value = 39000 kJ/kg
Process 1-2s is isentropic compression
i.e.,
   1  1 P
P1V1  P2V2 or T1V1  T2V2 or  1  C

 1 T
 P2  
 2904.4
0.4
T2 s  T1   1.4  443.02 K
 P1 
But
T2 s  T1 443.02  290
C  i.e., 0.75  T2  494.03K
T2  T1 T2  290
66
Process 3-4s is isentropic expansion i.e.,
 1 0.32
T4 s  P4  
 1  1.33
   T4 s  923   639.18K
T3  P3   4.4 
But
T3  T4
T 
T3  T4 s
923  T4
i.e., 0.85  T4  681.76 K
923  639.18

(i) m f  ?
m a CV 500 39000
 i.e., 
We have
m f C P T3  T2  m f 1.13923  494.03

  f = 6.21 kg/min
m
67
(ii) Wnet = ?
Compressor work, WC = CP (T2 – T1)
= 1.005 (494.03 – 290)
= 205.05 kJ/kg
Turbine work, WT = CP (T3 – T4)
= 1.13 (923 – 681.76)
= 272.6 kJ/kg
Wnet = WT – WC = 67.55 kJ/kg
Net work output per minute = m a m f Wnet

= (500+6.21) (67.55) = 34194.49 kJ/min

Power output = 569.91 kW

68
(iii) th = ?
Heat supplied, Qin = CP (T3 – T2)
= 1.33 (923 – 494.03)
= 570.53 kJ/kg

Wnet 67.55
th  
Qin 570.53

= 0.118 or 11.8%

69
Problem 4
Tutorial
A gas turbine cycle having 2 stage compression with
intercooling in between stages and 2 stages of
expansion with reheating in between the stages has
an overall pressure ratio of 8.
The maximum cycle temperature is 1400K and the
compressor inlet conditions are 1 bar and 270C. The
compressors have s of 80% and turbines have s of
85%.
Assuming that the air is cooled back to its original
temperature after the first stage compression and
gas is reheated back to its original temperature after
1st stage of expansion, determine (i) the net work
output (ii) the cycle th. 70
Solution: T5 = 1400K T1 = 300K, P1= 1 bar
C1= 0.8 = C2, t1 = t2 = 0.85 ,T3 = T1 ,T7 = T5
For maximum work output,
P2 P4 P5 P7 P4 P5
      8
P1 P3 P6 P8 P1 P8

 Intermedia te Pr essure,
P2  P3  P6  P7  2.83 bar
 1
For process 1-2,  P2  
T2 s  T1  
 P1  = 300 (2.83)0.286 = 403.95 K

T2 s  T1 403.95  300
But c1  0.8   T2  429.9 K
T2  T1 T2  300 71
Since T3 = T1 and P4  P2
P3 P1

We have T4s = T2s = 403.95K

Also since C1 = C2, T4 = T2 = 429.9 K

Compressor work, WC = CP (T2 – T1) + CP (T4 – T3)

= 2 CP (T2 – T1)
= 2 (1.005) (429.9 – 300)
= 261.19 kJ/kg
For process 5 – 6,
 1
T6 s  P6 
0.286

 1 
   T6 s  1400   1039.72 K
T5  P5   2.83 
72
But
T5  T6 1400  T6
t1  i.e., 0.85  T6  1093.76 K
T5  T6 s 1400  1039.72

Since T7 = T5 and P5 P7 , then T8 = T6


P6 P8

Since t1 = t2, T6 = T8 = 1093.76 K

Turbine work, Wt = CP (T5 – T6) + CP (T7 – T8)

= 2 CP (T5 – T6)
= 2 (1.005) (1400 – 1093.76)
= 615.54 kJ/kg

Wnet = WT – WC = 354.35 kJ/kg

73
th = ?
Heat Supplied,

Qin = CP (T5 – T4) + CP (T7 – T6)

= 1.005 (1400 – 429.9 + 1400 – 1093.76)

= 1282.72 kJ/kg

354.35
th  = 0.276 or 27.6%
1282.72
74
Problem 5
Tutorial
A two stage gas turbine cycle receives air at 100 kPa
and 150C. The lower stage has a pressure ratio of 3,
while that for the upper stage is 4 for the compressor
as well as the turbine. The temperature rise of the air
compressed in the lower stage is reduced by 80% by
intercooling. Also, a regenerator of 78%
effectiveness is used. The upper temperature limit of
the cycle is 11000C. The turbine and the compressor
s are 86%. Calculate the mass flow rate required to
produce 6000kW.

75
 Solution:
P2 P4
P1 = 1 bar T1 = 288K  3, 4
P1 P3

IC = 0.8 ε = reg = 0.78, T5 = 1373 K,

C1 = C2 = t1 = t2 = 0.86, m  ? if P = 6000 kW

Process 1-2s is isentropic compression

 1
T2 s  P2  
    T2s = 288 (3)0.286
T1  P1 
= 410.75 K
T2 s  T1 410.75  288
But  C1  i.e., 0.86  T2  430.73K
T2  T1 T2  288
Also, T2  T3 430.73  T3
 IC  i.e., 0.8  T3  316.54 K
T2  T1 430.73  288
76
Process 3-4s is 2nd stage isentropic compression
 1
T4 s  P4  
   
T3  P3 

T4s = 316.54 (4)0.286 = 470.57K

But C 2  T4 s  T3 i.e., 0.86 

470.57  316.54
T4  495.64K
T4  T3 T4  316.54

Process 5-6s is 1st stage isentropic expansion

 1
T6 s  P6  
 
 

T5  5
P
0.286
1
T6 s  1373   923.59 K
4 77
 But
T5  T6 1373  T6
t1  i.e., 0.86  T6  986.51K
T5  T6 s 1373  923.59

Process 6-7 is reheating, assume T7 = T5 = 1373K

 1

Process 7-8s is 2nd stage isentropic expansion i.e., T8 s   P8 



T7  P7 
0.286
1
T8 s  1373   1002.79 K
 3
T7  T8 1373  T8
t 2  i.e., 0.86  T8  1054.63K
But T7  T8 s 1373  1002.79

Regenerator is used to utilizes the temperature of exhaust gases

i.e.,

Tx  T4 Tx  495.64
T8  T4
i.e., 0.78  Tx = 931.65 78
K
1054.63  495.64
We have, Compressor work: WC = CP (T2 – T1) + CP (T4 – T3)
= 1.005 (430.73 – 288 + 495.64 – 316.54)
= 323.44 kJ/kg
Also, Turbine work : WT = CP (T5 – T6) + CP (T7 – T8)
= 1.005 (1373 – 986.51 + 1373 – 1054.63)
= 708.38 kJ/kg
Net work output, Wnet = WT - WC
= 384.95 kJ/kg
But, power produced, P  m  Wnet
i.e., 6000 x 1000 = 384.95 x 1000
m  = 15.59 kg/sec
We have, heat supplied, Qin = CP (T5 – Tx) + CP (T7 – T6)
= 1.005 (1373 – 931.65 + 1373 – 986.51)
Wnet
= 831.98 kJ/kg  
th  0.463 or 46.3% 79
Qin
 Problem 6 Tutorial
In a reheat gas turbine cycle, comprising one
compressor and two turbine, air is compressed
from 1 bar, 270C to 6 bar. The highest
temperature in the cycle is 9000C. The expansion
in the 1st stage turbine is such that the work
from it just equals the work required by the
compressor. Air is reheated between the two
stages of expansion to 8500C. Assume that the
isentropic s of the compressor, the 1st stage
and the 2nd stage turbines are 85% each and that
the working fluid is air and calculate the cycle .
Solution: P1 = 1 bar T1 = 300K P2 = 6 bar
T3 = 1173K WT1 = WC T5 = 1123K C = 0.85
t1 = t2 = 0.85 80
We have process 1-2 is isentropic i.e.,
 1
T2 S  P2  
  
T1  P1 
0.4
6 1.4
 T2 S  300   500.5K
1

T2 S  T1 500.5  300
But  C  i.e., 0.85   T2  536K
T2  T1 T2  300
Compressor work, WC = CP (T2 – T1)
= 1.005 (536 – 300) = 237 kJ/kg

From data, WT1 = WC = 237 kJ/kg

= CP (T3 – T4) T4 = 937 kJ/kg
81
 T3  T4 1173  937
But  t1  i.e., 0.85   T4 S  895K
T3  T4 S 1173  T4 S
Process 3-4 is isentropic i.e.,

P4  T4 S   1
  
P3  T3 
1.4
 895  0.4
 P4  6   2.328 bar
 1173 
From T-S diagram, intermediate pressure, P4 = P5 = 2.328 bar
Process 5-6s is isentropic in the 2nd stage turbine
 1 0.4
T6 S  P6  
 1  1.4
i.e.,     T6 S  1123   882 K
T5  P5   2.328 
T5  T6 1123  T6
But t 2  i.e., 0.85  T6  918K
T5  T6 S 1123  882 82
WT2 = CP (T5 – T6)

= 1.005 (1123 – 918) = 206 kJ/kg

Net work output = WT – WC
= (WT1 + WT2) – WC = 206 kJ/kg
Net heat transfer or heat supplied, Qnet = Qin + Qout

= CP (T3 – T2) + CP (T5 – T4)

= 640 + 187 = 827 kJ/kg
Cycle efficiency,
Wnet 206
 cycle    25%
Qnet 827

83
 Problem 7 Tutorial
In a simple gas turbine unit, the isentropic
discharge temperature of air flowing out of
compressor is 1950C, while the actual discharge
temperature is 2400C. Conditions of air at the
beginning of compression are 1 bar and 170C. If
the air-fuel ratio is 75 and net power output
from the unit is 650kW. Compute (i) isentropic 
of the compressor and the turbine and (ii)
overall . Calorific value of the fuel used is
46110 kJ/kg and the unit consumes 312 kg/hr of
fuel. Assume for gases CP = 1.09 kJ/kg-K and 
= 1.32 and for air CP = 1.005 kJ/kg-K and  = 1.4.
84
Solution:
T2S = 195+273 = 468 K T2 = 240+273 = 513K
T1 = 290K P1=1bar A/F = 75,
Power output = Wnet = WT – WC = 650kW C = ? T = ?
cycle = ? CV = 46110 kJ/kg, CPg = 1.09 kJ/kg-k,
g = 1.30, CPa = 1.005 kJ/kg-K, a = 1.4
 f  312kg / hr  0.0867kg / s
m

We have, Compressor Efficiency,

T2 S  T1 468  290
C  i.e.,  0.79
T2  T1 513  290

 A
Also, m a   m f
F
= 75 (0.0867) = 6.503 kg/s 85
  1.4
 T2 S   1  468  0.4
Pr essure ratio  rp       5.34
 T1   290 
Applying SFEE to the constant pressure heating process 2-3,

f
m CV  m  f  CPg T3  T2 
a  m
0.0867 (46110) = (6.503 + 0.0867) 1.09 (T3 – 513)
T3 = 1069.6K
Also,
 g 1
 P4  g 1.321
T4 S  1069.65.34 
T4 S

 
 1.32
T3  P3 
T4S = 712.6K. Further,
W net  WT  W C   m  f C Pg T3  T4   m
a m  a C Pa T2  T1 

i.e., 650 = (6.503 + 0.0867) 1.09 (1069.6 – T4) – 6.503 (1.005) (513 – 290)
T4 = 776K 86
Now, Turbine Efficiency,

T3  T4 1069.6  776
T    0.822
T3  T4 S 1069.6  712.6
And, Wnet 650
 cycle    0.163
m f CV 0.086746110
Or
Wnet 650 650
cycle     0.162
Qin m a  m f CPg T3  T2  3997.9

87
Thank you

88