Compression of High-Speed Air in an Aircraft An aircraft is flying at a cruising speed of 250 m/s at an altitude of 5000 m where the atmospheric pressure is 54.05 kPa and the ambient air temperature is 255.7 K. The ambient air is first decelerated in a diffuser before it enters the compressor

Question

Compression of High-Speed Air in an Aircraft

An aircraft is flying at a cruising speed of 250 m/s at an altitude of 5000 m where the atmospheric pressure is 54.05 kPa and the ambient air temperature is 255.7 K. The ambient air is first decelerated in a diffuser before it enters the compressor (Fig. 12–5). Approximating both the diffuser and the compressor to be isentropic, determine (a) the stagnation pressure at the compressor inlet and (b) the required compressor work per unit mass if the stagnation pressure ratio of the compressor is 8.

Accepted Answer

SOLUTION High-speed air enters the diffuser and the compressor of an aircraft. The stagnation pressure of the air and the compressor work input are to be determined.

Assumptions 1 Both the diffuser and the compressor are isentropic. 2 Air is an ideal gas with constant specific heats at room temperature.

Properties The constant-pressure specific heat [latex]c_{p}[/latex] and the specific heat ratio k of air at room temperature are

[latex]c_{p}=1.005 kJ / kg \cdot K \quad ext { and } \quad k=1.4[/latex]

Analysis (a) Under isentropic conditions, the stagnation pressure at the compressor inlet (diffuser exit) can be determined from Eq. 12–5. However, first we need to find the stagnation temperature [latex]T_{01}[/latex] at the compressor inlet. Under the stated assumptions, [latex]T_{01}[/latex] is determined from Eq. 12–4 to be

[latex]\frac{P_{0}}{P}=\left(\frac{T_{0}}{T} ight)^{k /(k-1)}[/latex] (Eq. 12-5)

[latex]T_{0}=T+\frac{V^{2}}{2 c_{p}}[/latex] (Eq. 12-4)

[latex]\begin{aligned}T_{01} &=T_{1}+\frac{V_{1}^{2}}{2 c_{p}}=255.7 K +\frac{(250 m / s )^{2}}{(2)(1.005 kJ / kg \cdot K )}\left(\frac{1 kJ / kg }{1000 m ^{2} / s ^{2}} ight) \&=286.8 K\end{aligned}[/latex]

Then from Eq. 12–5,

[latex]\begin{aligned}P_{01} &=P_{1}\left(\frac{T_{01}}{T_{1}} ight)^{k(k-1)}=(54.05 kPa )\left(\frac{286.8 K }{255.7 K } ight)^{1.4 /(1.4-1)} \&=80.77 kPa\end{aligned}[/latex]

That is, the temperature of air would increase by 31.1°C and the pressure by 26.72 kPa as air is decelerated from 250 m/s to zero velocity. These increases in the temperature and pressure of air are due to the conversion of the kinetic energy into enthalpy.

(b) To determine the compressor work, we need to know the stagnation temperature of air at the compressor exit [latex]T_{02}[/latex]. The stagnation pressure ratio across the compressor [latex]P_{02} / P_{01}[/latex] is specified to be 8. Since the compression process is approximated as isentropic, [latex]T_{02}[/latex] can be determined from the ideal-gas isentropic relation (Eq. 12–5):

[latex]T_{02}=T_{01}\left(\frac{P_{02}}{P_{01}} ight)^{(k-1) / k}=(286.8 K )(8)^{(1.4-1) / 1.4}=519.5 K[/latex]

Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from Eq. 12–8:

[latex]\left(q_{ in }-q_{ out } ight)+\left(w_{ in }-w_{ out } ight)=c_{p}\left(T_{02}-T_{01} ight)+g\left(z_{2}-z_{1} ight)[/latex] (Eq. 12-8)

[latex]\begin{aligned}w_{ ext {in }} &=c_{p}\left(T_{02}-T_{01} ight) \&=(1.005 kJ / kg \cdot K )(519.5 K -286.8 K ) \&=233.9 kJ / kg\end{aligned}[/latex]

Thus the work supplied to the compressor is 233.9 kJ/kg.

Discussion Notice that using stagnation properties automatically accounts for any changes in the kinetic energy of a fluid stream.

Question 12.1: Compression of High-Speed Air in an Aircraft An aircraft is ...

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