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. (Round off to one decimal place)

The correct answer is 59.4.

1. Given Data:
Outer diameter of the turbine runner, _D1 = 1 m
Rotational speed of the runner, N = 600 rpm
Normal component of absolute velocity (velocity of flow) at inlet, _Vf1 = 10 m/s
Guide blade angle, α = 15°

2. Step-by-Step Derivation:

First, we calculate the tangential velocity of the runner at the inlet (_u1):

$$u_1=\frac{\pi \times D_1\times N}{60}$$

Substituting the given values:

$$u_1=\frac{\pi \times 1\times 600}{60}=10\pi \approx 31.42\text{ m/s}$$

Next, we determine the whirl velocity component at the inlet (_Vw1) using the guide blade angle (α):

$$\tan \left(\alpha \right)=\frac{V_{f1}}{V_{w1}}$$

Rearranging the formula to solve for _Vw1:

$$V_{w1}=\frac{V_{f1}}{\tan \left(\alpha \right)}=\frac{10}{\tan \left({15}^{\circ }\right)}$$

Since tan(15°) ≈ 0.2679:

$$V_{w1}=\frac{10}{0.2679}\approx 37.32\text{ m/s}$$

Now, let θ be the inlet vane angle of the runner. From the inlet velocity triangle, the relationship between flow velocity, whirl velocity, runner tangential velocity, and runner inlet angle is:

$$\tan \left(\theta \right)=\frac{V_{f1}}{V_{w1}-u_1}$$

Substituting the values of _Vf1, _Vw1, and _u1:

$$\tan \left(\theta \right)=\frac{10}{37.32-31.42}=\frac{10}{5.90}\approx 1.6949$$

Solving for θ:

$$\theta ={\tan }^{-1}\left(1.6949\right)\approx {59.4}^{\circ }$$

Thus, the inlet vane angle of the runner is 59.4°.