Question:medium

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 \(15^{\circ}\), then the inlet vane angle of the runner is ________ degree (rounded off to 1 decimal place).

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Draw the inlet velocity triangle and use the guide blade angle to split the absolute velocity into flow and whirl parts.
Updated On: Jul 27, 2026
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Correct Answer: 59.4

Solution and Explanation

Step 1: List what the inlet triangle needs.
Three things fix the triangle at inlet: the blade speed $u$, the flow speed $V_{f1}$, and the whirl speed $V_{w1}$ set by the guide vane.

Step 2: Get the blade speed from the RPM.
$u = \pi D N / 60 = \pi(1)(600)/60 = 31.42$ m/s.

Step 3: Get the whirl speed from the guide blade angle.
Since the guide angle is measured from the tangential direction, $V_{w1} = V_{f1} \cot(15^{\circ}) = 10 \times 3.732 = 37.32$ m/s.

Step 4: Build the runner triangle and solve for the vane angle.
The relative whirl left after subtracting blade speed is $37.32 - 31.42 = 5.90$ m/s, so $\tan \beta = 10/5.90 = 1.693$, giving $\beta = 59.4^{\circ}$.

Final Answer:
The runner inlet vane angle is about $59.4^{\circ}$. \[ \boxed{\beta \approx 59.4^{\circ}} \]
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