High-speed air enters the diffuser and the compressor of an aircraft. The stagnation pressure of 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 c_{p} and the specific heat ratio k of air at room temperature are (Table A–2a)
c_{p}=1.005 kJ / kg \cdot K \quad \text { and } \quad k=1.4
Analysis (a) Under isentropic conditions, the stagnation pressure at the compressor inlet (diffuser exit) can be determined from Eq. 17–5. However, first we need to find the stagnation temperature T_{01} at the compressor inlet. Under the stated assumptions, T_{01} can be determined from Eq. 17–4 to be
T_{0}=T+\frac{V^{2}}{2 c_{p}} (17–4)
\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}}\right) \\&=286.8 K\end{aligned}
Then from Eq. 17–5,
\frac{P_{0}}{P} =\left(\frac{T_{0}}{T}\right)^{k /(k-1)} (17–5)
\begin{aligned}P_{01} &=P_{1}\left(\frac{T_{01}}{T_{1}}\right)^{k /(k-1)}=(54.05 kPa )\left(\frac{286.8 K }{255.7 K }\right)^{1.4 /(1.4-1)} \\&=80.77 kPa\end{aligned}
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 T_{02} . The stagnation pressure ratio across the compressor P_{02} / P_{01} is specified to be 8. Since the compression process is assumed to be isentropic, T_{02} can be determined from the idealgas isentropic relation (Eq. 17–5):
T_{02}=T_{01}\left(\frac{P_{02}}{P_{01}}\right)^{(k-1) / k}=(286.8 K )(8)^{(1.4-1) / 1.4}=519.5 K
Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from Eq. 17–8:
\left(q_{\text {in }}-q_{\text {out }}\right)+\left(w_{\text {in }}-w_{\text {out }}\right)=c_{p}\left(T_{02}-T_{01}\right)+g\left(z_{2}-z_{1}\right)
\begin{aligned}w_{\text {in }} &=c_{p}\left(T_{02}-T_{01}\right) \\&=(1.005 kJ / kg \cdot K )(519.5 K -286.8 K ) \\&=233.9 kJ / kg\end{aligned}
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.
