Discussion
A single-stage gas turbine operates with an axial absolute flow at the entry and exit from the stage. The absolute flow angle at the nozzle exit is 70o. The turbine stage generates a specific work of 228 kJ/kg when operating with a mean blade speed of 440 m/s. The absolute velocity at the rotor entry is
Answer: B
The absolute velocity at rotor entry is about 551.5 m/s.
With axial absolute flow leaving the stage, the swirl at rotor exit is zero, so Euler's equation gives the specific work as \(w = U\,V_{\theta 2}\), where \(V_{\theta 2}\) is the swirl at rotor entry. The 70° nozzle exit angle is measured from the axial direction, so \(V_{\theta 2} = V_{2}\sin 70^\circ\).
\[ \begin{aligned} V_{\theta 2} &= \frac{228000}{440} = 518.2\ \text{m/s} \\ V_{2} &= \frac{518.2}{\sin 70^\circ} = \frac{518.2}{0.9397} \\ &\approx 551.5\ \text{m/s} \end{aligned} \]
275.7 m/s is half of this, and 1103.0 and 1654.5 m/s are exactly two and three times it. The absolute velocity must exceed its own swirl component of 518 m/s, which already rules out 275.7, and the two larger values would need far more work than 228 kJ/kg at this blade speed.
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