L.D.ENGINEERING
SUBJECT:- ENGINEERING THERMODYNAMIC
TOPIC NAME:- BRAYTON CYCLE
BRNCH:- MECHANICAL B
BATCH:- 3
STUDENT NAME ENROL NO
PATEL DIPESSH 140283119018
NANE HIMANSHU 140283119015
BHAVSAR DHVANIL 140283119002
Brayton Cycle
•A gas turbine is a rotary power unit used for producing large
quantities of power in a self contained and compact unit.
•The gas turbine power plant are suitable for aircraft, marine,
military and transport application due to its smaller size and low
weight power ratio.
•These plants are free from vibration, perfect in balance, simple in
installation, operation and maintenance. These are highly reliable
•The brayton cycle is the air-standard ideal cycle approximation
for the gas-turbine engine.
• This cycle differs from the otto and diesel cycles in that the
processes making the cycle occur in open systems or control
volumes.
•We assume the working fluid is air and the specific heats are
constant and will consider the cold-air-standard cycle.
The closed cycle gas-turbine engine
The open cycle gas turbine
Process involved in a cycle :
1.Isentropic compression (1-2) :- surrounding air is compressed
isentropically pv^γ = c in the compressor from pressure p1 to p2
work of compression Wc
2.Constant pressure heat addition (2-3:)- heat is added in
combustor at constant pressure and constant temp is raised from T2
to T3 there fore pressure P3= p2
3.Isentropic expansion (3-4):-hot gases at p2, T3 expand in gas
turbine isentropically up to atmospheric pressure P1 , work
developed by turbine wt
4.Constant pressure heat rejection (4-1) :- heat is rejected by
exaust gases at constant pressure to the atmosphere.
The T-s and P-v diagrams
are
Thermal efficiency of the brayton cycle
ηth Brayton
out
in
out
in
p
p
Q
Q
q
q
C T T
C T T
,


( )
( )
= − = −
= −
−
−
1 1
1
4 1
3 2
ηth Brayton
T T
T T
T T T
T T T
,
( )
( )
( / )
( / )
= −
−
−
= −
−
−
1
1
1
1
4 1
3 2
1 4 1
2 3 2
•The steady flow energy equation can be used for cycle analysis
with changes in K.E and P.E negligible
S.F.E.E can be written as
Q-Wsf = m (∆h+ ∆ K.E + ∆ P.E)
But ∆ K.E = 0 ∆ P.E = 0 and h=Cp.T
S.F.E.E. reduce to Q-Wsf = m ∆h
1.compresser work
•Process (1-2) is irreversible adiabatic process Q1-2 = 0 and work
is nagative.
0-(-Wc) = m (h2-h1) = m Cp (T2-T1)
Wc = m Cp (T2-T1)
Heat addition process:-
Wsf = 0 therefore Q2-3 = m (h3-h2) = m Cp (T3-T2)
Turbine work:- process (3-4) is reversible adiabatic ,
hence Q=0
0-Wt = m (h4-h3)
Wt = m (h3- h4) = m Cp (T3-T4)
Shaft work Ws = turbine work (Wt) – compressor work (Wc)
= m Cp (T3-T4) - m Cp (T2-T1)
Air standard efficiency of cycle
Air standard efficiency of cycle:-
ռ = m Cp (T3-T4) - m Cp (T2-T1)
m Cp (T3-T2)
ռ= 1- (T4-T1)
(T3-T2)
(d) Efficiency in terms of pressure ratio
Let, pressure ratio Rp = P2 = P3
P1 P4
ηth Brayton
T T
T T
T T T
T T T
,
( )
( )
( / )
( / )
= −
−
−
= −
−
−
1
1
1
1
4 1
3 2
1 4 1
2 3 2
T2 _ P2 ^(γ – 1)/γ And T3 _ P3 ^(γ – 1)/γ
T1 P1 T4 P4
= (Rp) ^(γ – 1)/γ
T2 = T1 (Rp) ^(γ – 1)/γ and
T3 = T4 (Rp) ^(γ – 1)/γ
so we get ,
ռ = 1 - (T4-T1)
T4 (Rp) ^(γ – 1)/γ T1 (Rp) ^(γ – 1)/γ
ռ = 1- 1/ (Rp) ^(γ – 1)/γ
Which is the efficiency of brayton cycle in
term of pressure ratio

Brayton cycle

  • 1.
    L.D.ENGINEERING SUBJECT:- ENGINEERING THERMODYNAMIC TOPICNAME:- BRAYTON CYCLE BRNCH:- MECHANICAL B BATCH:- 3 STUDENT NAME ENROL NO PATEL DIPESSH 140283119018 NANE HIMANSHU 140283119015 BHAVSAR DHVANIL 140283119002
  • 2.
    Brayton Cycle •A gasturbine is a rotary power unit used for producing large quantities of power in a self contained and compact unit. •The gas turbine power plant are suitable for aircraft, marine, military and transport application due to its smaller size and low weight power ratio. •These plants are free from vibration, perfect in balance, simple in installation, operation and maintenance. These are highly reliable •The brayton cycle is the air-standard ideal cycle approximation for the gas-turbine engine. • This cycle differs from the otto and diesel cycles in that the processes making the cycle occur in open systems or control volumes. •We assume the working fluid is air and the specific heats are constant and will consider the cold-air-standard cycle.
  • 3.
    The closed cyclegas-turbine engine The open cycle gas turbine
  • 4.
    Process involved ina cycle : 1.Isentropic compression (1-2) :- surrounding air is compressed isentropically pv^γ = c in the compressor from pressure p1 to p2 work of compression Wc 2.Constant pressure heat addition (2-3:)- heat is added in combustor at constant pressure and constant temp is raised from T2 to T3 there fore pressure P3= p2 3.Isentropic expansion (3-4):-hot gases at p2, T3 expand in gas turbine isentropically up to atmospheric pressure P1 , work developed by turbine wt 4.Constant pressure heat rejection (4-1) :- heat is rejected by exaust gases at constant pressure to the atmosphere.
  • 5.
    The T-s andP-v diagrams are
  • 6.
    Thermal efficiency ofthe brayton cycle ηth Brayton out in out in p p Q Q q q C T T C T T ,   ( ) ( ) = − = − = − − − 1 1 1 4 1 3 2 ηth Brayton T T T T T T T T T T , ( ) ( ) ( / ) ( / ) = − − − = − − − 1 1 1 1 4 1 3 2 1 4 1 2 3 2
  • 7.
    •The steady flowenergy equation can be used for cycle analysis with changes in K.E and P.E negligible S.F.E.E can be written as Q-Wsf = m (∆h+ ∆ K.E + ∆ P.E) But ∆ K.E = 0 ∆ P.E = 0 and h=Cp.T S.F.E.E. reduce to Q-Wsf = m ∆h 1.compresser work •Process (1-2) is irreversible adiabatic process Q1-2 = 0 and work is nagative. 0-(-Wc) = m (h2-h1) = m Cp (T2-T1) Wc = m Cp (T2-T1)
  • 8.
    Heat addition process:- Wsf= 0 therefore Q2-3 = m (h3-h2) = m Cp (T3-T2) Turbine work:- process (3-4) is reversible adiabatic , hence Q=0 0-Wt = m (h4-h3) Wt = m (h3- h4) = m Cp (T3-T4) Shaft work Ws = turbine work (Wt) – compressor work (Wc) = m Cp (T3-T4) - m Cp (T2-T1) Air standard efficiency of cycle
  • 9.
    Air standard efficiencyof cycle:- ռ = m Cp (T3-T4) - m Cp (T2-T1) m Cp (T3-T2) ռ= 1- (T4-T1) (T3-T2) (d) Efficiency in terms of pressure ratio Let, pressure ratio Rp = P2 = P3 P1 P4 ηth Brayton T T T T T T T T T T , ( ) ( ) ( / ) ( / ) = − − − = − − − 1 1 1 1 4 1 3 2 1 4 1 2 3 2
  • 10.
    T2 _ P2^(γ – 1)/γ And T3 _ P3 ^(γ – 1)/γ T1 P1 T4 P4 = (Rp) ^(γ – 1)/γ T2 = T1 (Rp) ^(γ – 1)/γ and T3 = T4 (Rp) ^(γ – 1)/γ so we get , ռ = 1 - (T4-T1) T4 (Rp) ^(γ – 1)/γ T1 (Rp) ^(γ – 1)/γ
  • 11.
    ռ = 1-1/ (Rp) ^(γ – 1)/γ Which is the efficiency of brayton cycle in term of pressure ratio