Introduction to gas turbines

1. Applied Thermodynamics
1
Gas Turbine Cycles
N S Senanayake

2. Air standard cycles
• Air standard cycles refers to thermodynamic cycle
with certain assumptions so as to use the
principles of thermodynamics conveniently.
• Assumptions
– Air is the working fluid and behaves as a perfect gas
– Mass and composition of the working fluid will not
change in the cycle
– Processes are reversible
– Specific heat capacity of the working fluid does not
change

3. Otto cycle (air standard)
1 – 2 Adiabatic compression
2 – 3 Constant volume heat addition
3– 4 Adiabatic expansion
4– 1 Constant volume heat rejection
Spark Ignition (SI) engines are based on this cycle

4. Otto cycle …
r
V
V
V
V
n ratioCompressio
3
4
2
1

)(, 23 TTcqaddedHeat vin 
)()(, 1441 TTcTTcqrejectedHeat vvout 
)()(, 1423 TTcTTcwworkNet vvnet 
inputHeat
workNet
EfficiencyThermal th, 

5. Otto cycle…
)(
)(
1
)(
)}(){(
23
14
23
1423
TT
TT
TTc
TTTTc
v
v
th






Consider process 1 – 2
CpV 
CVpV 1
)( 
CTV 1
1
22
1
11

 
VTVT
1
1
2
1
1
2 







 

r
V
V
T
T
Consider process 3 – 4
1
44
1
33

 
VTVT
1
1
3
4
4
3 







 

r
V
V
T
T
2
3
1

6. Otto cycle…
From equations 2 and 3
4
1
3
2
4
3
1
2
T
T
T
T
T
T
T
T

4
1
3
2
11
T
T
T
T

4
14
3
23
T
TT
T
TT 


1
4
3
14
23 


 
r
T
T
TT
TT
From equation 1
1
23
14 1
1
)(
)(
1 



 

rTT
TT
th

7. Otto cycle efficiency vs. compression ratio
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 5 10 15 20 25 30 35 40
Efficiency()
compression ratio (r)
= 1.2
= 1.4
1
23
14 1
1
)(
)(
1 



 

rTT
TT
th

8. Mean effective pressure (MEP)- Otto cycle
This is the mean pressure which is developed in the cylinder.
Defined as the ratio of net work done to the displacement of
volume of the piston.

9. Mean effective pressure (MEP)- Otto cycle
ngeVolume cha
Net
MEP
work


















11
11224433

VpVpVpVp
workNet
 )  ) 11224433
1
1
VpVpVpVpworkNet 























 11
1
1
11
22
11
44
33
44
Vp
Vp
Vp
Vp
Vp
VpworkNet


10. Mean effective pressure (MEP)- Otto cycle




















 1
1
1
1
1
1
1
2
11
4
3
44
rp
p
Vp
rp
p
VpworkNet




r
V
V
p
p
VpVp 






2
1
1
2
2211


r
V
V
p
p







3
4
4
3
Similarly
 )  ) 11
1
1 1
11
1
14 

  

rVprVpworkNet
Consider process 1 – 2
 ) 14
1
1
1
1
pp
rV
workNet 







11. 11
Terminology :
Reciprocating Engine
 volume swept out by the piston
when it moves from TDC to
BDC is called the displacement
volume.
 distance from TDC to BDC
is called stroke
• The piston is said to be at the top
dead center (TDC) when it has
moved to a position where the
cylinder volume is minimum. This
volume is called a clearance
volume.

12. Spark Ignition vs Compression Ignition
Spark-ignition engines: mixture of fuel and air are
ignited by a spark plug.
Compression ignition engines: Air is compressed to
high enough pressure and temperature that
combustion occurs spontaneously when fuel is
injected.

13. Air-Standard Diesel Cycle
The Air-Standard Diesel Cycle is the ideal cycle that approximates the
compression ignition engine i.e. Compression Ignition (CI) engines are based
on this cycle
Process Description
1-2 Isentropic Compression
2-3 Constant Pressure Heat Addition
3-4 Isentropic Expansion
4-1 Constant Volume Heat Rejection
P
V
T
S
2 3
4
1 1
4
2
3
P = P2 3
P1
S = S1 2 S = S3 4
P4
V =V4 1O O
T1
T4
T2
T3
y
x
V2

14. Diesel cycle
)(, 2323 TTcqaddedHeat p 
)(, 4141 TTcqrejectedHeat v 
2
1
,
V
V
rrationCompressio 
2
3
,
V
V
ratiooffCut 
23
4123
q
qq
addedheat
heatnet
th


)(
)()(
23
1423
TTc
TTcTTc
p
vp
th




15. Diesel cycle
)(
)(1
1
23
14
TT
TT
th





 ) 1
12
1
1
2
1
1
2 







 

rTTr
V
V
T
T
Process 1 - 2
Process 2 - 3
1
1323
2
3
2
3
  
 rTTTT
T
T
V
V
1
2
3

16. Diesel cycle
Process 3 - 4
11
3
2
2
1
1
3
2
2
4
1
3
4
4
3
..




























r
V
V
V
V
V
V
V
V
V
V
T
T
4 ) 







1
1
1
1
1
34 T
r
rT
r
TT 















Substituting
from eq. 3

17. Diesel cycle
Substituting for all Ts in equation 1.











 
)1(
)1(1
1
)(
)(1
1 11
1
1
1
11





 



rrTrT
TT
th








 
)1(
)1(1
1 1






r
th

18. Diesel cycle efficiency vs. compression ratio
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0 5 10 15 20 25 30
 = 4
Efficiency()
Compression ratio (r)
 =2
 = 1.4








 
)1(
)1(1
1 1






r
th

19. Gas turbines
19

20. 20
Elements of simple gas turbine Power Plant

21. 21
 The simple gas turbine power plant mainly consists of a gas
turbine coupled to a rotary type air compressor and combustion
chamber which is placed between the compressor and turbine in
the fuel circuit.
 Auxiliaries, such as cooling fan, water pumps, etc. and the
generator itself, are also driven by the turbine.
 Other auxiliaries are starting device, lubrication system, duct
system, etc.
 A modified plant may have in addition to the above, an inter-
cooler, a regenerator and a re-heater

22. 22
Flow diagram – Gas turbine power plant

23. Gas turbine cycle
• Gas-turbines usually operate on an open
cycle
• A compressor takes in fresh ambient air
(state 1), compresses it to a higher
temperature and pressure (state 2).
• Fuel and the higher pressure air from
compressor are sent to a combustion
chamber, where fuel is burned at
constant pressure. The resulting high
temperature gases are sent to a turbine
(state 3).
• The high temperature gases expand to
the ambient pressure (state 4) in
the turbine and produce power.
• The exhaust gases leave the turbine.

24. Brayton cycle
By using the air-standard
assumptions, replacing the
combustion process by a
constant pressure heat addition
process, and replacing the
exhaust discharging process by a
constant pressure heat rejection
process, the open cycle described
above can be modeled as a
closed cycle, called ideal Brayton
cycle.

25. 25
Open Cycle Gas Turbine
Air
G
Compressor
Turbine
Combustor
Fuel
Generator
50 – 70 % of turbine
power
Pressure ratio: usually about 15, but up to 40 and more
Turbine inlet temperature (TIT): 900° - 1700°C
Turbine exit temperature (TET): 400° - 600°C
Power: 100 kW – 300 MW
Exhaust

26. 26
Closed Cycle Gas Turbine
Heat Source
G
Compressor Turbine
Generator
Condensate
from Process
Steam to
Process
Heat Exchanger
2
1
3
4
Working fluid circulates in a closed circuit and does not cause corrosion or
erosion
Any fuel, nuclear or solar energy can be used

27. The ideal Brayton cycle is
made up of four internally
reversible processes.
1-2 Isentropic compression
2-3 Constant pressure heat
addition
3- 4 Isentropic expansion
4-1 Constant pressure heat
rejection
Brayton cycle
Steady Flow Energy Equation

28. Efficiency of Brayton Cycle
 )232323 TTchhqq pin 
 )  )414141 TTchhqq pout 
23
41
23
4123
1
q
q
q
qq
addedheat
worknet
th 


























1
1
11
2
3
1
4
2
1
23
14
T
T
T
T
T
T
TT
TT
th 1

29. Efficiency of Brayton Cycle
Consider process 1 – 2, Isentropic compression
C
p
T
C
p
T
C
p
T
pCpV 





 




11




1
1
2
1
2
1 1









pr
p
p
T
T
ratiopressurer
p
p
p
1
2

Consider process 3 – 4, Isentropic expansion




1
1
3
4
3
4 1









pr
p
p
T
T
2
3

30. Efficiency of Brayton Cycle
From equations 2 and 3:
Substituting equations 2, 3 and 4 in equation 1
2
3
1
4
3
4
2
1
T
T
T
T
T
T
T
T
 4




 1
2
3
2
3
1
1
1
1
1
1
1 


























pp
th
rT
T
T
T
r


 1
1
1 

p
th
r
5

31. 31
Work ratio
)(
)()(
outputWork
Net Work
34
3412
34
3412
TT
TTTT
w
ww
rw





 ) 
 )1(
3
1
1

 pw r
T
T
r

32. 32
Equation shows that the work ratio increases in direct
proportion to the ratio T3 /T1 and inversely with a power of
the pressure ratio.
On the other hand, thermal efficiency equation
shows that thermal efficiency increases with increased
pressure ratio.

33. 33
Compressor work:
w12 = - (h2 – h1 ) = -Cp(T2 – T1)
Heat supplied during the cycle:
q23 = (h3 – h2) = Cp(T3 – T2)
23
3412
inputHeat
Net Work
q
ww 

Summary of Equations



)1(
1
1










pr
Turbine work:
w34 = (h3 – h4) = Cp(T3 – T4)
34
3412
outputWork
Net Work
w
ww
rw

  ) 
 )1(
3
1
1

 pw r
T
T
r
Work ratio
Efficiency

34. 34
Improving cycle efficiency and work ratio
According to the equation (5) above, the efficiency of Brayton
cycle depends only on the pressure ratio. To maximize the
efficiency the pressure ratio has to be increased. This is done
by compressing air isentropically from 1 to 2 to the maximum
possible pressure. When pressure is increased the temperature
also rises proportionately to the order of (-1)/.
The maximum temperature (T3) is fixed by the metallurgical
properties, ability to withstand the high temperatures by the
turbine materials. In practice the minimum temperature (T1)
is limited to the atmospheric temperature.

35. 35
Isentropic compression process to
reach the maximum temperature T3
is shown by 1-2. What is shown in
the figure is the compression close
to T3. The temperature T3 is
obtained by heat addition form 2 to
3. If we reach T3 only by
compression the points 2 and 3 will
coincide. Under this condition the
compression work and turbine work
becomes equal giving zero net
work. Therefore the maximum
theoretical pressure ratio is obtained
when work ratio is zero

36. 36
1
1
3
1
2
max)(










T
T
p
p
rp
Therefore, for zero net work
 ) 
 )1(
3
1
10

 pr
T
T
It is clear from the cycle 1-2’-3’-4’of figures, the net work also becomes zero
when pressure ratio is unity i.e. p2 = p1. Here also the figures show a pressure
ratio very close to 1 for illustration. Here the maximum temperature T3 is
achieved by heat supply at the same pressure.
Therefore, the pressure ratio (rp) has to be in between (rp)max and 1.

37. 37
Pressure ratio for maximum net work
The net work out put per unit mass is given by the following equation.
)()( 3412 TTcTTcw pp 
Since
1
/)1(
/)1(
1
2
12 Tr
p
p
TT p










 3/)1(4
1
T
r
T
p
 
and
 








 

1
1
1 /)1(3
/)1(
1 

p
ppp
r
TcrTcw
0
1111
/)12(3/11 













 














 
  




p
p
p
p
p r
Tc
r
Tc
dr
dw

38. 38
)1(2/
1
3









T
T
rp
 )
)1/(
1
3
max









T
T
rp
Since
max
)( pp rr 
The maximum net work is obtained when the pressure ratio
equals the square root of maximum theoretical pressure ratio.

39. 39
Variation of efficiency and the net work out put with rp and T3
We will see how the efficiency and net work output vary with the
pressure ratio and the maximum temperature.
Let us assume the following data are available
Min. temperature, T1 = 30oC = 303K
Isentropic eff. for compressor, comp = 0.85
Isentropic eff. for the turbine, Tub = 0.90

40. 40
Isentropic efficiency
 Performance of turbines/compressors are measured by
isentropic efficiencies.
 The actual work input to the compressor is more and the
actual work output from the turbine is more because of
irreversibility.
 Isentropic efficiencies involve a comparison between the
actual performance of a device and the performance that
would be achieved under idealized circumstances for the
same inlet state and the same exit pressure.

41. Isentropic efficiency - Turbine
The desired output from a turbine is the work output.
Hence, the definition of isentropic efficiency of a turbine is
the ratio of the actual work output of the turbine to the
work output of the turbine if the turbine undergoes an
isentropic process between the same inlet and exit
pressures.
WorkIsentropic
WorkTurbineActual
T 

42. The isentropic efficiency of turbine
can be written as
12
12
hh
hh
s
a
T



h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the
exit
12
12
TT
TT
s
a
T




43. Isentropic efficiency - compressor
The isentropic efficiency of a compressor or pump is defined as
the ratio of the work input to an isentropic process, to the
work input to the actual process between the same inlet and
exit pressures.
WorkActual
WorkCompressorIsentropic
C 

44. The isentropic efficiency of
compressor can be written as
12
12
hh
hh
a
s
C



h1 = enthalpy at the inlet
h2a = enthalpy of actual process at the exit
h2s = enthalpy of isentropic process at the
exit
12
12
TT
TT
a
s
C




45. 45
The Back Work Ratio
Therefore, the turbine used in gas-turbine power plants are larger than
those used in steam power plants of the same net power output, P.
Usually more than half of the turbine work output is used to drive the
compressor.
turbine
comp
W
W
ratioworkBack

46. 46
Deviation of Actual Gas-Turbine Cycles from
Ideal One
Pressure drop
Isentropic efficiency

47. Example 1
A four stroke SI engine has the compression ratio of 6 and swept
volume of 0.15m3. Pressure and temperature at the beginning of
compression are 98kPa and 60oC respectively. Heat supplied in
the cycle is 150kJ. cp = 1kJ/kgK, cv = 0.71kJ/kgK
Determine
(i) the pressure , volume and temperature at all main state
points
(ii) Efficiency
(iii) Mean effective pressure

48. Example 2
An ideal diesel cycle using air as working fluid has a
compression ratio of 16 and a cut off ratio of 2. The intake
conditions are 100kPa, 20oC, and 2000cm3.
Determine
(a) Temperature and pressure at the end of each process
(b) Net work output
(c) Thermal efficiency
(d) Mean effective pressure
cp = 1.0045kJ/kgK, cv 0.7175kJ/kgK

49. Example 3
In an air standard Brayton cycle the minimum and maximum
temperature are 300K and 1200K respectively. The pressure
ratio is 10.
(i) Find out temperatures after compression and expansion
(ii) Calculate the compressor and turbine work, each in kJ/kg
of air, and thermal efficiency of the cycle.

50. Example 4
A gas turbine receives air at 1bar, 300K and compresses it
adiabatically to 6.2bar. The isentropic efficiency of compressor
is 0.88. The fuel has a heating value of 44186kJ/kg and the fuel
–air ratio is 0.017kg fuel/kg of air. The turbine efficiency is 0.9.
Calculate the work of turbine and compressor per kg of air
compressed and the thermal efficiency.
For products of combustion cp = 1.147kJ/kgK, g = 1.33.
For air cp = 1.005kJ/kgK, g = 1.4.

51. 51
The ideal air-standard Brayton cycle operates with air entering the
compressor at 95 kPa, 22o
C. The pressure ratio rp is 6:1 and the air
leaves the heat addition process at 1100 K. Determine
• the compressor work
• the turbine 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.
Example 5

52. Example 6
In a gas turbine plant, working on the Brayton cycle, helium at 30° C and 22 bar
is compressed to a pressure of 64 bar and then heated to a temperature of 1200
°C. After expansion in the turbine, the gas is cooled to initial pressure and
temperature.
Assume the following:
Isentropic efficiency of the compressor – 0.85
Isentropic efficiency of the turbine – 0.8
Pressure loss in the combustion chamber – 1.2 bar
Pressure loss in the cooler – 0.5 bar
Specific heat (Cp) of the products of combustion is the same as that of helium
and it is equal to 5.1926 kJ/kg K. Ratio of specific heats of helium – 1.667
Determine the following;
• Temperature at the end of compression and expansion.
• Heat supplied, heat rejected and the net work per kg of helium.
• Thermal efficiency of the plant
• Flow rate of helium required to give an output of 12 MW.