BRAYTON AND JET PROPULSION CYCLES
Brayton Cycle is the ideal cycle for gas turbine engines. Electric power generation and
aircraft propulsion are major applications for gas-turbine engines.
4 PROCESSES:
1-2 Isentropic compression
2-3 Constant pressure heat addition
3-4 Isentropic expansion
4-1 Constant pressure heat rejection
Performing energy balance, we get:
qm  C p T3  T2 
qout  C p T4  T1 
 th 
wnet
q
 1  out
qm
qm
 T4

T1 
 1
T
T  T 
th  1  4 1  1   1 
T
T3  T2 
T2  3  1
 T2
(1)

Note:
P2 = P3
P1 = P4
Also:
T2  P2 
 
T1  P1 

k 1
k
P 
 3
 P4 
k 1
k

T3
T4
T4 T3

T1 T2
Substitute in (1), we get:
 th  1 
T1
1
1
T2
T2
T1
th  1 
1
or
k 1 k
p
r
(2)
2
where the pressure ratio, rp  P2 P .
1
DEVELOPMENT OF GAS TURBINES
► Early (1950) gas turbines had simple-cycle efficiencies of about 17% because of low
compressor and turbine efficiencies and low inlet temperature of the turbine.
► Effort to improve cycle efficiency concentrated in three areas:
(1)
Increasing the turbine inlet temperature increased from about 500°C to
1425°C presently.
(2)
Increasing the efficiencies of compressors and turbines. Efficiencies improved
due to designing the components aerodynamically with minimum losses.
(3)
Adding modifications to the basic cycle, such as regeneration, intercooling,
and reheating.
ADVANTAGES OF PRESENT GAS TURBINES
1.
2.
3.
4.
5.
High efficiencies
Lower capital cost
Shorter installation time
Better emission characteristics
Being used for base-load as well as peak load
Capacities and Efficiencies Range.
Inlet
Pressure Capacity
Efficiency
Temperature
Ratio
(MW)
540
6.5
2
26
↓
↓
↓
↓
1425
135.5
282
39.5
ACTUAL GAS TURBINE CYCLES
Actual gas turbine cycles differ from Brayton cycle. In the actual cases:
(1)
Pressure drops during heat addition and heat rejection processes.
(2)
The expansion and compression processes in the turbine and compressor,
respectively, are not isentropic.
The actual processes in the turbine and compressor can be accounted for by the isentropic
efficiencies:
3
c 
ws
wa
(3)
t 
wa
ws
(4)
BRAYTON CYCLE WITH REGENERATION
■
Heating the high-pressure air leaving the compressor by the hot exhaust gases in a
counterflow heat exchanger is known as regeneration (see Figure 8.38).
■
The thermal efficiency of the Brayton cycle increases as a result of decrease in the
heat input (thermo-fuel) for the same net power output.
■
Regeneration is used only when the compressor exit temperature is less than the
turbine exit temperature.
■
Referring to the T-s diagram, the regenerator effectiveness is given as:

qreg, act
qreg, max

h5  h2
h4  h2
(5)
Considering the cold air-standard assumptions, equation (5) reduces to:

■
T5  T2
T4  T2
(6)
Assuming cold air-standard assumptions, show that the thermal efficiency of an ideal
Brayton cycle with regeneration is given as:
k 1
 T1 
k
  rp 
 T3 
th  1  
(7)
■
Comment on the effect of temperature and pressure ratios on the thermal efficiencies.
■
See Example 8.7.
BRAYTON CYCLE WITH INTERCOOLING, REHEATING,
AND REGENERATION
●
Using multistage compression with intercooling reduces the total work of the
compressor operating between two pressures.
●
Similarly using multistage expansion with reheating increases the workout of a
turbine operating between two pressures.
4
●
Even-though intercooling and reheating improves the back work ratio of a gas turbine
cycle, but it does guarantee an improvement in the thermal efficiency (why?).
●
Intercooling and reheating have to be used in conjunction with regeneration for the
thermal efficiency to improve.
●
The best performance is achieved when equal pressure ratios are maintained across
each stage. For example (considering Figure 8.44) when
P
P
P2 P4

and 6  8 .
P1 P3
P7 P9
●
See Example 8.8.
IDEAL JET-PROPULSION CYCLES
► Aircrafts are powered by gas turbines that operate on jet-propulsion open cycles.
► In jet-propulsion cycles the gases are expanded in the turbine such that the power
produced is just sufficient to drive the compressor and the auxiliary equipment.
► The thrust to propel the aircraft is provided by a nozzle in which high-pressure gases
exiting the turbine do expand.
► A schematic of turbojet engine is shown in Figure 8.48.
► Aircrafts are propelled by either slightly accelerating a large mass of fluid (propellerdriven engine) or greatly accelerating a small mass of fluid (turbojet engine) in the
opposite direction to motion.
► Gases leave the aircraft at high-velocity after expanding in the nozzle to atmospheric
pressure.
► Processes in the diffuser, compressor, turbine, and nozzle are assumed to be isentropic
in the ideal turbojet cycle as shown by the T-s diagram (Figure 8.48).
► The net thrust developed by the engine is given by:
F  m Vexit  Vinlet 
(N)
where:
Vexit is the exit velocity of exhaust gases relative to aircraft.
Vinlet is the air inlet velocity relative to aircraft.
► The propulsive power, Wp is given as:
Wp  Thrust * Vaircraft
(kW).
5
► The propulsive efficiency, ηp is given by:
p 
Wp
Qm
.
where Q is the thermal energy of the fuel. See Example 8.9.
PROBLEM
A gas turbine at Dammam Electrical Power Station takes in 108,000 kg/h of filtered outside
air at 27°C and compresses it to 6.516 atmospheres. The combustion of gas adds 30 MW of
heat to the air. If the turbine exhausts to atmospheric pressure and both the compressor and
turbine are 75% efficient:
(a)
(b)
Draw the T-s diagram taking the inlet to the compressor as State 1.
Determine the net power output.
SOLUTION
a)
T-s diagram.
T
s
b)
Determination of the net power output:
Given: T1  27  273  300 K, rp  6
Wnet  T  c
c 
h2 s  h1
c
 h2 a  h1
h  300.19 kJ kg
T1  300 K at T1  1
 Pr1  1.386
6
Pr2 P2

 6.516
Pr1 P1
 Pr2  Pr1 * 6.516  1.386  6.516  9.031
Hence


T2 s  510 K
h2 s  513.32 kJ kg
513.32  300.19
 284.17 kJ kg
0.75
h2 a  h1   c  300.19  284.17  584.36 kJ kg
 c 
Q
30  103
 h2 a 
 584.36
m
30.50
 h3  1000  584.36  1584.36 kJ kg
qin  h3  h2 a  h3  qin  h2 a 
 T3  1457.3 K, and Pr3  533
Pr4 P4
 1 

 Pr4  
 Pr3
Pr3 P3
 6.516 
533
Pr4 
 81.8
6.516
Hence, h4s = 954.47 kJ/kg
t  t  h3  h4 s 
t  0.75(1584.36  954.47)  472.42 kJ kg
 net  t  c  472.42  284.17  188.25 kJ kg