Question 16.4: The conditions of air at the entry of an axial flow compress...
he conditions of air at the entry of an axial flow compressor stage are p_{1}=100 kN / m ^{2} and T_{1}=300 K. The air angles are \beta_{1}=51^{\circ}, \beta_{2}=10^{\circ}, \alpha_{1}=\alpha_{3}=8^{\circ} .
The mean diameter and peripheral speed are 0.5 m and 150 m/s respectively. Mass flow rate through the stage is 30 kg/s; the work done factor is 0.95 and mechanical efficiency is 90%. Assuming an isentropic stage efficiency of 85%, determine
(a) blade height at entry
(b) stage pressure ratio, and
(c) the power required to drive the stage
(for air, R = 287 J/kg K, γ = 1.4)
(a) \rho_{1}=\frac{p_{1}}{R T_{1}}=\frac{100 \times 10^{3}}{287 \times 300}=1.16 kg / m ^{3}
From Eq. (16.12),
\frac{U}{V_{f}}=\tan \alpha_{1}+\tan \beta_{1} (16.12)
Hence, V_{f}=\frac{150}{\tan 8^{\circ}+\tan 51^{\circ}}
= 109.06 m/s
\dot{m}=V_{f} \rho_{1}\left(\pi d h_{1}\right)
30=109.06 \times 1.16 \times \pi \times 0.5 h_{1}
which gives h_{1}=0.15 m
(b) From Eq. (16.19)
\Delta T_{s t}=\frac{\lambda U V_{f}}{c_{p}}\left(\tan \beta_{1}-\tan \beta_{2}\right) (16.19)
again, c_{p}=\frac{1.4}{(1.4-1)} \times 287=1005 J/kg K
Hence, \Delta T_{s t}=\frac{0.95 \times 150 \times 190.06}{1005}\left(\tan 51^{\circ}-\tan 10^{\circ}\right)
= 16.37º C
With the help of Eq. (16.20) we can write
R_{s}=\left[1+\frac{\eta_{s} \Delta T_{s t}}{T_{1 t}}\right]^{\frac{\gamma}{\gamma-1}} (16.20)
pressure ratio R_{s}=\left[1+\frac{0.85 \times 16.37}{300}\right]^{\frac{1.4}{0.4}}
= 1.17
(c) P=\frac{\dot{m} w}{\eta_{m}}=\frac{\dot{m} c_{p} \Delta T_{s t}}{\eta_{m}}
=\frac{30 \times 1005 \times 16.37}{0.9 \times 1000} kW =548.39 kW