Air with a static temperature of 223 K entering a gas turbine engine intake at V

Question

Air with a static temperature of 223 K entering a gas turbine engine intake at V_1 = 300 m/s accelerates to a new velocity (V_2) and decreases in pressure (P_2) at the exit plane of the intake. The pressure recovery (P_2/P_1) through the intake is 0.833. Assume that the flow through the intake is isentropic. (a) Calculate the static temperature (T_2) of air exiting the intake. (b) Find the difference in total temperature (delta T_t) across the intake. (c) Determine the velocity (V_2) of the air exiting the intake.

Explanation / Answer

Solution:

a)

Given static temperature is 223 K

P2/P1 = 0.833

We know that at constat volume P1/T1 = P2/T2

So T1/T2 = P1/P2

T1/T2 = 0.833

T2 = T1/0.833

T2 = 233/0.833

T2 = 267.7K

b) Difference in total pressure = T2 -T1 = 267.7-233 = 44.7 K

c) We know that PV = RT

P(300) = 287*223

P1 = 213.3 Pa

P2/P1 = 0.833

P2 = 0.833*213.3

P2 = 177.7 Pa

P2V2 = RT2

V2 = (283*267.7)/177.7

V2 = 426.33 m/s

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