SME
Propulsion · Problem type
Compressor exit temperature
Use this for “A compressor with ηc = 0.85 takes in air at 300 K and has a pressure ratio of 10. What is its exit temperature?”
Diagram
The ideal compression ends at 2s (T = 579.2 K), straight up; the real one (dashed) ends at 2, T = 628.5 K, hotter and to the right: friction makes entropy.
This is an example. Type your own values to start a new problem.
T₂s (K)
=
T₁ (K)
×
rₚ
0.4
÷
1.4
T₂ (K)
=
T₁ (K)
+
(
T₂s (K)
−
T₁ (K)
)
÷
ηc
Formulas
T₂s
=
T₁
×
rₚ
0.4 ÷ 1.4
579.2
=
300 ×
10
0.4 ÷ 1.4
T₂
=
T₁
+
T₂s −
T₁
ηc
628.5
=
300
+
579.2 − 300
0.85
Assumptions
•
Air with constant cₚ and γ = 1.4; the compressor is adiabatic and steady.
•
Isentropic efficiency ηc: the ideal temperature rise over the real one, at the same rₚ.
•
The real compression makes entropy, so T₂ lies above T₂s on the same pressure line.
Step-by-step
WE KNOW
T₁ = 300 K, rₚ = 10, ηc = 0.85
FIND
ideal exit temperature (T₂s), real exit temperature (T₂)
Step 1 · Find ideal exit temperature (T₂s)
Use T₂s = T₁ × rₚ0.4 ÷ 1.4
An
isentropic
compression:
raise
rₚ
to
γ − 1
γ
,
then
multiply
by
T₁
.
T₂s
=
300 ×
10
0.4 ÷ 1.4
T₂s
=
300 ×
10
0.2857
T₂s = 300 × 1.931
T₂s = 579.2 K
Step 2 · Find real exit temperature (T₂)
Use T₂ = T₁ + (T₂s − T₁) ÷ ηc
The
real
compressor
needs
a
bigger
rise:
divide
the
ideal
rise
by
ηc,
then
add
T₁
.
T₂
=
300
+
579.2 − 300
0.85
T₂
=
300
+
279.2
0.85
T₂
=
300
+
328.5
T₂
=
628.5
K
Check
Put the numbers back in. Both sides should match.
579.2
=
300 ×
10
0.4 ÷ 1.4
✓
628.5
=
300
+
579.2 − 300
0.85
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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