In a single stage turbine, the hot gases come out of stator/nozzle at a speed of 500 m/s and at an angle of 70 degrees with the turbine axis as shown. The design speed of the rotor blade is 250 m/s at the mean blade radius. The rotor blade angle, \( \beta \), at the leading edge is _________ degrees (round off to one decimal place).
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Correct Answer: 51.5
Solution and Explanation
- Speed of gases from the nozzle: \( V = 500 \, \text{m/s} \).
- Angle with turbine axis: \( \theta = 70^\circ \).
- Rotor blade speed: \( U = 250 \, \text{m/s} \).
We use the relation for the angle of the rotor blade at the leading edge: \[ \tan\beta = \frac{V \sin\theta}{U}. \] Substitute the values: \[ \tan\beta = \frac{500 \times \sin(70^\circ)}{250} \approx \frac{500 \times 0.9397}{250} = \frac{469.85}{250} \approx 1.8794. \] Now, calculate \( \beta \): \[ \beta = \tan^{-1}(1.8794) \approx 62.4^\circ. \] Thus, the rotor blade angle at the leading edge is approximately \( 62.4^\circ \).
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