SME
Engineering Thermodynamics
Power and refrigeration cycles
Use this for “An ideal Brayton cycle has a pressure ratio of 12, air at 290 K into the compressor and 1300 K into the turbine. Find the works and η.”
Diagram
Brayton cycle: compression 1→2 and expansion 3→4 are vertical (isentropic); heat enters along the high-pressure line 2→3 and leaves along the low one 4→1.
Net work wₙₑₜ = 397.6 kJ/kg
Thermal efficiency η = 0.4821
This is an example. Type your own values to start a new problem.
T₂ (K)
=
T₁ (K)
×
rₚ
0.4
÷
1.4
T₄ (K)
=
T₃ (K)
÷
rₚ
0.4
÷
1.4
η
=
1
−
1
÷
rₚ
0.4
÷
1.4
wc
Compressor work
calculated · kJ/kg
wₜ
Turbine work
calculated · kJ/kg
wₙₑₜ
Net work
calculated · kJ/kg
Formulas
T₂
=
T₁
×
rₚ
0.4 ÷ 1.4
579.2093
=
300 ×
10
0.4 ÷ 1.4
T₄
=
T₃
rₚ
0.4 ÷ 1.4
725.1265
=
1,400
10
0.4 ÷ 1.4
wc
=
1.005
×
(
T₂
−
T₁
)
280.6054 = 1.005 × (579.2093 − 300)
wₜ
=
1.005
×
(
T₃
−
T₄
)
678.2479 = 1.005 × (1,400 − 725.1265)
wₙₑₜ = wₜ − wc
+397.6425 = 678.2479 − 280.6054
η
=
1
−
1
rₚ
0.4 ÷ 1.4
0.4821
=
1
−
1
10
0.4 ÷ 1.4
Assumptions
•
Cold-air standard: cₚ = 1.005 kJ/(kg·K) and k = 1.4, both constant.
•
The compressor and turbine are isentropic; heat goes in and out at constant pressure.
•
Kinetic and potential energy changes are neglected; works are per kilogram of air.
Step-by-step
WE KNOW
rₚ = 10, T₁ = 300 K, T₃ = 1,400 K
FIND
compressor exit temperature (T₂), turbine exit temperature (T₄), compressor work (wc), turbine work (wₜ), net work (wₙₑₜ), thermal efficiency (η)
Step 1 · Find compressor exit temperature (T₂)
Use T₂ = T₁ × rₚ0.4 ÷ 1.4
The
compressor
is
isentropic:
raise
rₚ
to
k − 1
k
,
then
multiply
by
T₁
.
T₂
=
300 ×
10
0.4 ÷ 1.4
T₂
=
300 ×
10
0.2857
T₂
=
300
×
1.931
T₂
=
579.2093
K
Step 2 · Find turbine exit temperature (T₄)
Use T₄ = T₃ ÷ rₚ0.4 ÷ 1.4
The
turbine
is
isentropic
over
the
same
ratio:
divide
T₃
by
rₚ
to
the
power
k − 1
k
.
T₄
=
1,400
10
0.4 ÷ 1.4
T₄
=
1,400
10
0.2857
T₄
=
1,400
1.931
T₄
=
725.1265
K
Step 3 · Find compressor work (wc)
Use wc = 1.005 × (T₂ − T₁)
Compressor work: cₚ times the air’s temperature rise.
wc = 1.005 × (579.2093 − 300)
wc = 1.005 × 279.2093
wc = 280.6054 kJ/kg
Step 4 · Find turbine work (wₜ)
Use wₜ = 1.005 × (T₃ − T₄)
Turbine work: cₚ times the air’s temperature drop.
wₜ = 1.005 × (1,400 − 725.1265)
wₜ = 1.005 × 674.8735
wₜ = 678.2479 kJ/kg
Step 5 · Find net work (wₙₑₜ)
Use wₙₑₜ = wₜ − wc
The turbine’s work less what the compressor takes.
wₙₑₜ = 678.2479 − 280.6054
wₙₑₜ = +397.6425 kJ/kg
Step 6 · Find thermal efficiency (η)
Use η = 1 − 1 ÷ rₚ0.4 ÷ 1.4
The ideal Brayton efficiency depends on the pressure ratio alone.
η
=
1
−
1
10
0.4 ÷ 1.4
η
=
1
−
1
10
0.2857
η
=
1
−
1
1.931
η
=
1
−
0.5179
η
=
0.4821
Check
Put the numbers back in. Both sides should match.
579.2093
=
300 ×
10
0.4 ÷ 1.4
✓
725.1265
=
1,400
10
0.4 ÷ 1.4
✓
280.6054 = 1.005 × (579.2093 − 300) ✓
678.2479 = 1.005 × (1,400 − 725.1265) ✓
+397.6425 = 678.2479 − 280.6054 ✓
0.4821
=
1
−
1
10
0.4 ÷ 1.4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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