In a Brayton cycle with regeneration: 1) Without regeneration, the cycle efficiency is 19.9%. 2) With a regenerator that is 75% effective, the cycle efficiency increases to 28.37%. 3) This is a 42.56% increase in efficiency due to regeneration. For an ideal Brayton cycle with an air inlet of 1 atm and 300K compressed to 6 atm and a maximum temperature of 1100K: 1) The thermal efficiency of the cycle is 40.1% 2) The work ratio is 0.545 3) The power output is 40.1 MW 4) The exergy flow rate of the exhaust gas is 20.

1. Brayton Cycle
Guided by;
Prof. Sankalp Kulkarni
G.H Patel College of Engineering and
Technology
By:
150110119126 ( Sharvil )
150110119127 ( Vishvak )
150110119128 ( Festus )
( 2C19 )Mechanical Engineering

2. • In a gas turbine plant, working on the Brayton cycle with a re-
generator of 75% effectiveness, the air at inlet to the compressor is at
0.1 MPa, 30°C, the pressure ratio is 6, and the maximum cycle
temperature is 900°C. If the turbine and compressor have each an
efficiency of 80%, find the percentage increase in the cycle efficiency
due to regeneration.
Solution: p₁ = 0.1 MPa
T₁ = 303 K
T₃ = 1173 K
rp = 6, ƞT = ƞC = 0.8
T
s
2s 2
6
3
4
4s
5
1

3. Without a regenerator
T₂s
T₁
= (p₂
p₂
) = T₃
T₄s
= (6) = 1.668
T₂s = 303*1.668 = 505 K
T₄s = 1173
1.668
= 705 K
T₂ - T₁ = T₂s− T₁
ƞC
= 505−303
0.8
= 252 K
T₃ - T₄ = ƞT (T₃ - T₄s) = 0.8 (1173 – 705) = 375 K
∴ WT = h₃ - h₄ = cp (T₃ - T₄) = 1.005 * 375 = 376.88 𝑘𝐽 𝑘𝑔
∴ Wc = h₂ - h₁ = cp (T₂ - T₁) = 1.005 * 252 = 253.26 𝑘𝐽 𝑘𝑔
T₂ = 252 + 303 = 555 K
Q₂ = h₃ - h₂ = cp (T₃ - T₂) = 1.005 (1173 – 555) = 621.09 𝑘𝐽 𝑘𝑔
∴ ƞT = WT−Wc
Q₂
=376.88−253.26
621.09
= 0.199 or 19.9%
(Ƴ − 1) Ƴ
(0.4) 1.4

4. With regenerator
T₄ = T₃ - 375 = 1173 – 375 = 798 K
Regenerator effectiveness = T₆− T₂
T₄− T₂
= 0.75
T₆ - 555 = 0.75 (798 – 555)
∴ T₆ = 737.7 K
∴ Q₁ = h₃ - h₆ = cp (T₃ - T₆)
= 1.005(1173- 737.3) = 437.88 𝑘𝐽 𝑘𝑔
Wnet remains the same.
ƞ = Wnet
Q₁
= 123
437.9
= 0.2837 or 28.37%
∴ Percentage increase due to regeneration
= 0.2837−0.199
0.199
= 0.4256 or 42.56%

5. • In an ideal Brayton cycle, air form the atmosphere at 1 atm, 300K is
compressed to 6 atm and the maximum cycle temperature is limited to
1100 Kby using a large air-fuel ratio. If the heat supply is 100 MW, find
(a) the thermal efficiency of the cycle, (b) work ratio, (c) power output,
(d) exergy flow rate of the exhaust gas leaving the turbine.
Solution:
the cycle efficiency,
ƞcycle = 1 - 1
𝑟
= 1 - 1
6
= 0.401 or 40.1 %
𝑇₂ 𝑇₁ = (rp ) = 1.67
T₂ = 501 K
𝑇₃ 𝑇₄ = 1.67, T₄ =1100 1.67 = 658.7 K
(Ƴ − 1) Ƴ
(Ƴ − 1) Ƴ (0.4 1.4)

6. WC = 1.005 ( 501 – 300 ) = 202 𝑘𝐽 𝑘𝑔
WT = 1.005 ( 1100 – 658.7 ) = 443.5 𝑘𝐽 𝑘𝑔
- Work ratio = WT−Wc
WT
= 241.5
443.5
0.545
- Power output = 100 * 0.401 = 40.1 MW
- Q₁ = ṁ CP (T₃ - T₂) = 100000 kW
- ṁ = 166.1 kg s
- Exergy flow rate of the exhaust gas stream
= ṁ CP T₀( T₄
T₀
-1 - lnT₄
T₀
)
= 166.1 * 1.005 * 300 ( 658
300
-1 - ln658.7
300
)
= 20.53 MW

7. Thank You