SME
Propulsion
Gas turbine cycles
Use this for “An ideal Brayton cycle takes in air at 300 K, has a pressure ratio of 10 and a turbine inlet at 1400 K. Find its net work and efficiency.”
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 = 48.21%
This is an example. Type your own values to start a new problem.
T₁
Compressor inlet temperature
example · K
rₚ
Pressure ratio
example
T₃
Turbine inlet temperature
example · K
T₂
Compressor exit temperature
calculated · K
T₄
Turbine exit temperature
calculated · K
wc
Compressor work
calculated · kJ/kg
wₜ
Turbine work
calculated · kJ/kg
wₙₑₜ
Net work
calculated · kJ/kg
η
Thermal efficiency
calculated · %
Formulas
T₂
=
T₁
×
rₚ
0.4 ÷ 1.4
579.2
=
300 ×
10
0.4 ÷ 1.4
T₄
=
T₃
rₚ
0.4 ÷ 1.4
725.1
=
1,400
10
0.4 ÷ 1.4
wc
=
1.005
×
(
T₂
−
T₁
)
280.6 = 1.005 × (579.2 − 300)
wₜ
=
1.005
×
(
T₃
−
T₄
)
678.2 = 1.005 × (1,400 − 725.1)
wₙₑₜ = wₜ − wc
397.6 = 678.2 − 280.6
η
=
100
×
(1
−
1
rₚ
0.4 ÷ 1.4
)
48.21
=
100
×
(1
−
1
10
0.4 ÷ 1.4
)
Assumptions
•
Air-standard cycle: air alone flows round it, with constant cₚ = 1.005 kJ/(kg·K) and γ = 1.4.
•
The compressor and turbine are isentropic (adiabatic and reversible).
•
Heat is added and rejected at constant pressure, so both run over the same rₚ.
Step-by-step
WE KNOW
T₁ = 300 K, rₚ = 10, 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
γ − 1
γ
,
then
multiply
by
T₁
.
T₂
=
300 ×
10
0.4 ÷ 1.4
T₂
=
300 ×
10
0.2857
T₂
=
300
×
1.931
T₂
=
579.2
K
Step 2 · Find turbine exit temperature (T₄)
Use T₄ = T₃ ÷ rₚ0.4 ÷ 1.4
The
turbine
expands
isentropically
over
the
same
ratio:
divide
T₃
by
rₚ
0.4 ÷ 1.4
.
T₄
=
1,400
10
0.4 ÷ 1.4
T₄
=
1,400
10
0.2857
T₄
=
1,400
1.931
T₄
=
725.13
K
Step 3 · Find compressor work (wc)
Use wc = 1.005 × (T₂ − T₁)
The compressor’s work is cₚ times the temperature rise it gives the air.
wc = 1.005 × (579.2 − 300)
wc = 1.005 × 279.2
wc = 280.6 kJ/kg
Step 4 · Find turbine work (wₜ)
Use wₜ = 1.005 × (T₃ − T₄)
The turbine’s work is cₚ times the temperature drop across it.
wₜ = 1.005 × (1,400 − 725.13)
wₜ = 1.005 × 674.87
wₜ = 678.2 kJ/kg
Step 5 · Find net work (wₙₑₜ)
Use wₙₑₜ = wₜ − wc
The turbine drives the compressor; what is left over is the net work.
wₙₑₜ = 678.2 − 280.6
wₙₑₜ = 397.6 kJ/kg
Step 6 · Find thermal efficiency (η)
Use η = 100 × (1 − 1 ÷ rₚ0.4 ÷ 1.4)
The ideal Brayton efficiency depends on the pressure ratio alone; times 100 for a percent.
η
=
100
×
(1
−
1
10
0.4 ÷ 1.4
)
η
=
100
×
(1
−
1
10
0.2857
)
η
=
100
×
(1
−
1
1.931
)
η
=
100
×
(1
−
0.5179)
η
=
100
×
0.4821
η
=
48.21%
Check
Put the numbers back in. Both sides should match.
579.2
=
300 ×
10
0.4 ÷ 1.4
✓
725.13
=
1,400
10
0.4 ÷ 1.4
✓
280.6 = 1.005 × (579.2 − 300) ✓
678.2 = 1.005 × (1,400 − 725.13) ✓
397.6 = 678.2 − 280.6 ✓
48.21
=
100
×
(1
−
1
10
0.4 ÷ 1.4
)
✓
Topic check
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