GATE 2026 ME – Question 58
An inward flow reaction turbine, having an outer diameter of 1 m, runs at 600 RPM. The normal component of absolute velocity at the inlet is 10 m/s. If the guide blade angle is 15°, then the inlet vane angle of the runner is ________ degree (*rounded off to 1 decimal place*).
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Correct answer: 59.10 to 59.70
Explanation
The blade speed at the inlet is $U_1 = \frac{\pi \times 1 \times 600}{60} = 31.42$ m/s. The flow velocity (normal component) is $V_f = 10$ m/s and the guide blade angle is 15°, so the whirl velocity is $V_{w1} = \frac{10}{\tan 15°} = 37.32$ m/s. At the inlet, $\tan\theta = \frac{V_f}{V_{w1} - U_1} = \frac{10}{5.90} = 1.694$, so the inlet vane angle is $\theta = 59.4°$.