Introduction to Gas Turbines
Gas turbines are internal combustion engines which convert heat
energy of a fluid (product of combustion) into mechanical energy of a
rotating shaft or axial thrust for propulsion.
Main Applications: Aircraft Engines (Jet Engines), Helicopters, Ship Engines
Power Generation
Advantages:
• Great power-to-weight ratio (5 - 10 kW/kg) compared to reciprocating engines (<1 kW/kg).
• Smaller than their reciprocating counterparts of the same power.
Disadvantages:
• Expensive
• high speeds and high operating temperatures
• designing and manufacturing gas turbines is a tough problem from both the
engineering and materials standpoint
• Tend to use more fuel when they are idling
• They prefer a constant rather than a fluctuating load.
Main components:
1. Compressor
2. Turbine
3. Combustion Chamber
4. Nozzle for producing thrust (Turbo-jet type) or power turbine (Turbo-prop type) for generating
shaft power.
Fig-1: Open cycle gas turbine
Fig-3: Jet Engine (Gas Turbine)
Fig-2: Closed cycle gas turbine
Operating Principle:
Air at atmospheric pressure enters the
compressor and exits at high pressure. The
high pressure air enters the combustion
chamber where fuel is added and ignited
by a spark plug. This high pressure, high
temperature combustion gas then expands
through the turbine which drives the
compressor and rotates another shaft of a
machine or an electric generator or allows
further expansion through a nozzle to
generate thrust for propulsion.
Page 1 of 6
Air Standard Brayton Cycle:
Processes:
1 – 2 : Isentropic Compression
Process
2 – 3 : Constant Pressure
Combustion
3 – 4 : Isentropic Expansion
Process
4 – 1 : Constant Pressure
Exhaust
Fig. 4: Air standard Brayton cycle for Gas Turgine.
For a gas turbine working on air standard
Brayton cycle,
p
p
rp = 2 = 3
p1 p4
Pressure ratio,
T2 = T1rp
Or,
Tx ⎛ px ⎞
=⎜ ⎟
Ty ⎜⎝ p y ⎟⎠
⎛ vy ⎞
=⎜ ⎟
⎝ vx ⎠
T3 ⎛ p3 ⎞
=⎜ ⎟
T4 ⎝ p4 ⎠
k −1
Work Output
Brayton cycle Efficiency =
Heat Supplied
=
Heat Supplied − Heat Rejected
Heat Supplied
i.e.,
η=
i.e.,
η=
Or,
(2)
Again, for isentropic expansion process, 3 → 4 :
For any isentropic process, x → y ,
k −1
k
k −1
k
Or,
T3 = T4 rp
k −1
k
k −1
k
= ( rp )
k −1
k
(3)
Putting (2) and (3) into (1), we get,
T4 − T1
η = 1−
Wout Qin − Qout
=
Qin
Qin
T4 rp
k −1
k
− T1rp
k −1
k
(T4 − T1 )
= 1−
rp
k −1
k
(T4 − T1 )
ma c p (T3 − T2 ) − ma c p (T4 − T1 )
ma c p (T3 − T2 )
T −T
η = 1 − 4 1 (1)
T3 − T2
Therefore,
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
= ( rp )
k −1
k
1
rp
Now for isentropic compression process, 1 → 2 :
k −1
k
η = 1−
where,
k=
k −1
k
cp
cv
and k = 1.4 for air.
So, the efficiency of the Brayton cycle increases
with increasing pressure ratio rp.
Page 2 of 6
Problem-1:
A gas turbine working on ideal Brayton cycle has the following data:
Air (cp=1005 J/kgK) enters the compressor at 0.1 MPa, 20oC at a mass flow rate of 10
kg/s. The pressure leaving the compressor is 1 MPa, and the maximum temperature in the
cycle is 1225oC. Determine (i) the pressure ratio, (ii) the compressor work (back work),
(iii) the turbine work, (iv) the net work, (v) the work ratio, (vi) the back work ratio, (vii)
the total heat input and (viii) the overall cycle efficiency.
Solution:
Given, ma = 10 kg/s, cp = 1005 J/kgK
p1 = 0.1 MPa
T1 = 20oC = (273+20)K=293K
p2 = 1 MPa
T3 = 1225oC = 273+1225)K=1498K
Therefore, the turbine work,
wt = macp(T3 – T4)
= (10 kg/s)(1005 J/kgK)(1498K – 775.9K)
= 7.26 MW Ans.
(i) The pressure ratio, rp= p2/p1
= 1/0.1= 10 Ans.
(iv) The net work, Wnet= Wt – Wc
= 7.26 – 2.74 = 4.52 MW Ans.
(ii) The compressor work, wc= macp(T2 – T1)
(v) The work ratio, rw = Wnet / Wt
= 4.52 / 7.26 = 0.623 Ans.
We have,
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
k −1
k
= ( rp )
k −1
k
(vi) The back work ratio, rb = Wc / Wt
= 2.74 / 7.26 = 0.377 Ans.
(vii) The total heat input, Qin = ma c p (T3 − T2 )
Or,
1.4 −1
T2
= (10 ) 1.4 = (10)0.2857 = 1.9307
293
Or,
T2 = 565.7 K
Therefore, the compressor work (back work),
wc= macp(T2 – T1)
= (10 kg/s)(1005 J/kgK)(565.7K – 293K)
= 2.74 MW Ans.
= (10 kg/s)(1005 J/kgK)(1498K – 65.7K)
=9.37 MW Ans.
(viii) The overall cycle efficiency,
η = 1−
rp
(iii) The turbine work, wt = macp(T3 – T4)
We have,
Or,
Or,
T3 ⎛ p3 ⎞
=⎜ ⎟
T4 ⎝ p4 ⎠
k −1
k
= ( rp )
1
Or,
k −1
k
k −1
k
η = 1−
1
= 1−
(10)
1.4 −1
1.4
1
= 0.482
1.9307
i.e., η = 48.2% Ans.
1.4 −1
1498
= (10 ) 1.4 = (10)0.2857 = 1.9307
T4
T4 = 775.9 K
Or, The overall cycle efficiency,
W
4.52
η = net =
= 0.482
Qin 9.37
Page 3 of 6
i.e., η = 48.2% Ans
Optimum Pressure Ratio:
We have, for a gas turbine working on
an ideal Brayton cycle,
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
k −1
k
T ⎛p ⎞
= 3 =⎜ 3 ⎟
T4 ⎝ p4 ⎠
i.e.,
k −1
k
Or,
d (Wnet )
T
= ma c p (0 + 0 − 1 + 32 T1 ) = 0
dT2
T2
T2 = (T3 )(T1 )
This is the value of T2 for maximum Wnet.
T
T2 T3
= ,
i.e., T4 = 3 T1
T1 T4
T2
Again, The net work,
Wnet = Wt − Wc = ma c p (T3 − T2 ) − ma c p (T4 − T1 )
Or,
Now again from,
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
k −1
k
k
p ⎛ T ⎞ k −1
we get, 2 = ⎜ 2 ⎟
p1 ⎝ T1 ⎠
= ma c p (T3 − T2 − T4 + T1 ) = ma c p (T3 + T1 − T2 − T4 )
So, optimum pressure ratio,
T3
k
k
k
T1 )
T2
p2 ⎛ T2 ⎞ k −1 ⎛ (T3 )(T1 ) ⎞ k −1 ⎛ T3 ⎞ 2( k −1)
⎟ =⎜ ⎟
Here, T1 is compressor inlet temperature which ( rp )opt = p = ⎜ T ⎟ = ⎜⎜
⎟
T
1
1
⎝ 1⎠
⎝ T1 ⎠
⎝
⎠
is always atmospheric. T3 is the turbine inlet
k
temperature which is also fixed by turbine
2(
k
⎛ ⎞ −1)
materials thermal stress limit.
So,
( rp )opt = ⎜ TT3 ⎟
⎝ 1⎠
So to get the maximum net work we have to
differentiate Wnet with respect to T2.
Or, Wnet = ma c p (T3 + T1 − T2 −
Problem-2:
In a gas turbine, air (cp=1005 J/kgK) enters the compressor at 0.1 MPa, 20oC at a mass
flow rate of 10 kg/s. The pressure leaving the compressor is 1 MPa, and the maximum
temperature in the cycle is 1225oC. Determine (i) the optimum pressure ratio, (ii) the
compressor work for optimum pressure ratio, (iii) the turbine work for optimum pressure
ratio, (iv) the overall cycle efficiency for optimum pressure ratio.
Solution:
Given, ma = 10 kg/s, cp = 1005 J/kgK
p1 = 0.1 MPa
T1 = 20oC = (273+20)K=293K
p2 = 1 MPa
T3 = 1225oC = 273+1225)K=1498K
(i)
(ii)
The optimum pressure ratio,
k
(r )
p opt
1.4
⎛ T ⎞ 2( k −1) ⎛ 1498 ⎞ 2(1.4 −1)
=⎜ 3 ⎟
=⎜
⎟
⎝ 293 ⎠
⎝ T1 ⎠
= ( 5.113)
1.75
=17.4 Ans.
Page 4 of 6
The compressor work, wc= macp(T2 – T1)
For optimum pressure ratio,
T2 = (T3 )(T1 ) = (1498)(293) = 662.5 K
Therefore, the compressor work (back work),
wc= macp(T2 – T1)
= (10 kg/s)(1005 J/kgK)(662.5K – 293K)
= 3.71 MW Ans.
(iii) The turbine work, wt = macp(T3 – T4)
Or,
T3 ⎛ p3 ⎞
=⎜ ⎟
T4 ⎝ p4 ⎠
k −1
k
= ( rp )
(iv) The overall cycle efficiency,
1
η = 1−
k −1
k
(rp )
1.4 −1
1498
= (17.4 ) 1.4 = (17.4)0.2857 = 2.262
T4
Or, T4 = 662.2 K
Therefore, the turbine work,
wt = macp(T3 – T4)
= (10 kg/s)(1005 J/kgK)(1498K – 662.2K)
= 8.40 MW Ans.
Or,
Or, η = 1 −
Or,
i.e.,
k −1
k
1
1.4 −1
1.4
(17.4)
1
η = 1−
= 0.557
2.262
η = 55.7% Ans.
Increasing Gas Turbine Efficiency by intercooling, reheating and regeneration:
Fig. 5: Gas Turbine with intercooling, reheating and regeneration.
Gas turbine performance can be improved by intercooling, reheating and/or regeneration.
Multistage compression (C1 and C2) with intercooling saves compressor work as well as
increases volumetric efficiency. Reheating is done by using two turbines. After partial
expansion in the high pressure turbine (T1) the gas is again heated to initial
temperature and then finally expanded in the low pressure turbine (T2) producing greater
work output. Regeneration is accomplished by utilizing some of the heat of the waste
gas leaving the turbine to increase the temperature of the air entering the combustor.
Considerable saving of energy is possible by regeneration leading to increased cycle
efficiency.
Page 5 of 6
Combined Cycle Power Plant
OCGT = Open Cycle Gas Turbine
CCGT = Combined Cycle Gas Turbine
HRSG = Heat Recovery Steam Generator
GT = Gas Turbine
ST = Steam Turbine
CC = Combustion Chamber
An Open Cycle Gas Turbine (OCGT) has a compressor, a combustor and a turbine (GT). In this
type of cycle, the input temperature to the turbine (the firing temperature), is relatively high
(900°C to 1,400°C). The output temperature of the exhaust flue gas is also high (450°C to
650°C). This is therefore high enough to provide heat for a second cycle which uses steam as the
working fluid (a Rankine cycle) which is used to drive a steam turbine (ST). Thus both GT and
ST can be coupled with generators (G) to produce electricity in a Combined Cycle Gas Turbine
(CCGT) power plant.
In a CCGT power plant, the heat of the gas turbine's exhaust is used to generate steam by passing
it through a heat recovery steam generator (HRSG) with a live steam temperature between 420°C
and 580°C. The condenser of the Rankine cycle is usually cooled by water from a lake, river, sea
or cooling towers. For large scale power generation a typical set would be a 400 MW Gas
Turbine coupled to a 200 MW Steam Turbine giving 600 MW. A big power station might
comprise between 2 and 6 such sets. The HRSG can be designed with supplementary firing of
fuel after the gas turbine in order to increase the quantity or temperature of the steam generated.
By combining both gas and steam cycles, high input temperatures and low output temperatures
can be achieved. The efficiency of the cycles add, because they are powered by the same fuel
source. If the plant produces only electricity, efficiencies of up to 60% can be achieved.
Page 6 of 6