Solution

12.116
Borgnakke and Sonntag
A Brayton cycle has a compression ratio of 15:1 with a high temperature of
1600 K and an inlet state of 290 K, 100 kPa. Use cold air properties to find the
specific net work output and the second law efficiency if we neglect the “value”
of the exhaust flow.
Brayton cycle so this means:
Minimum T: T 1 = 290 K
Maximum T: T 3 = 1600 K
Pressure ratio: P 2/P 1 = 15
Solve using constant C P0
Compression in compressor: s 2 = s 1 ⇒ Implemented in Eq.8.32
k-1
T2 = T k 0.286
1(P2/P1) = 290(15) = 628.65 K
Energy input is from the combustor
T
2
1
P
3
4
P = 100 kPa
s
q H = h 3 − h 2 = C P0(T 3 − T 2) = 1.004 (1600 − 628.65) = 975.2 kJ/kg
Do the overall cycle efficiency and net work
η = W.
Q .
net
H
= wnet q
= 1 − r
H -(k-1)/k
= 1 − 15
p
-0.4/1.4 = 0.5387
w NET = η q H = 0.5387 × 975.2 = 525.34 kJ/kg
Notice the q H does not come at a single T so neglecting external irreversibility
we get
Φ qH = increase in flow exergy = ψ 3 − ψ 2 = h 3 − h 2 − Τ o (s 3 − s 2 )
= q H − Τ o C P ln(T 3/T 2) = 975.2 – 298.15 × 1.004 ln ( 1600
628.65 )
= 695.56 kJ/kg
ηII = wnet =
ψ3 − ψ2 525.34
695.56
= 0.755
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for
testing or instructional purposes only to students enrolled in courses for which this textbook has been
adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108
of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.