MechanicalThermodynamicsmediumFE ExamCourseworkPE / Advanced

Exit Temperature of an Isentropic Compressor

Air enters a compressor at $P_1 = 100\ \text{kPa}$ and $T_1 = 300\ \text{K}$ , and is compressed isentropically to $P_2 = 800\ \text{kPa}$ .

Treat air as an ideal gas with constant specific heat ratio $k = 1.4$ .

Find the exit temperature $T_2$ .

Given

Hint 1

Isentropic means constant entropy — there is a standard P-T relation for an ideal gas.

Hint 2

T₂/T₁ = (P₂/P₁)^((k−1)/k). The exponent is the part most people get wrong.

Hint 3

(k−1)/k = 0.4/1.4 ≈ 0.286, not 3.5.

Worked solution — try the problem first

For an ideal gas undergoing an isentropic process with constant $k$ :

$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}$

Exponent: $\frac{k-1}{k} = \frac{0.4}{1.4} = 0.2857$

Pressure ratio: $\frac{P_2}{P_1} = \frac{800}{100} = 8$

Because this is a ratio, the kPa units cancel — no need to convert to absolute pressure units, though both must be absolute pressures (they are).

$T_2 = 300 \times 8^{0.2857}$

$8^{0.2857} = e^{0.2857 \ln 8} = e^{0.2857 \times 2.0794} = e^{0.5941} = 1.8114$

$T_2 = 300 \times 1.8114 = 543.4\ \text{K}$

$\boxed{T_2 = 543\ \text{K} \approx 270^\circ\text{C}}$

Sanity check: compression must raise temperature, and it does — by roughly 243 K. An answer below 300 K would be non-physical.

Concepts:Isentropic processIdeal gas relationsCompressor analysis

Sign in to submit

Grading needs an account so your progress, attempts, and daily quota can be tracked. The problem statement and worked solution stay open to everyone.