Question

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).

Hint:

Draw the inlet velocity triangle and use the guide blade angle to split the absolute velocity into flow and whirl parts.

Answer 1

Correct Answer: 59.4

Step 1: Find the blade tangential speed.
The runner turns at \(N = 600\) RPM with outer diameter \(D = 1\) m.
The tangential speed is \(u = \dfrac{\pi D N}{60} = \dfrac{\pi \times 1 \times 600}{60} = 31.42\) m/s.

Step 2: Find the whirl component of the inlet velocity.
The guide blade angle of \(15^{\circ}\) sets the direction of the absolute velocity \(V_1\).
The flow (normal) component is \(V_{f1} = 10\) m/s, so the whirl component is \(V_{w1} = \dfrac{V_{f1}}{\tan 15^{\circ}} = \dfrac{10}{0.2679} = 37.32\) m/s.

Step 3: Apply the inlet velocity triangle for the runner.
The runner vane angle \(\beta\) is measured from the relative velocity, so \(\tan \beta = \dfrac{V_{f1}}{V_{w1} - u}\).
Put in the numbers: \(\tan \beta = \dfrac{10}{37.32 - 31.42} = \dfrac{10}{5.90} = 1.693\).

Final Answer:
The inlet vane angle of the runner works out close to \(59.4^{\circ}\), inside the accepted key range. \[ \boxed{\beta \approx 59.4^{\circ}} \]