A single jet Pelton wheel operates at 300 rpm. The mean diameter of the wheel is 2 m. Operating head and dimensions of jet are such that water comes out of the jet with a velocity of 40 m/s and flow rate of 5 m³ /s. The jet is deflected by the bucket at an angle of 165°. Neglecting all losses, the power developed by the Pelton wheel is ______________ MW (round off to two decimal places).

The correct answer is 2.65.

Here is the detailed step-by-step explanation of the solution:

1. Identify the Given Parameters:
We are given the following values for the single-jet Pelton wheel:
- Rotational speed of the wheel, N = 300 rpm
- Mean diameter of the wheel, D = 2 m
- Velocity of the water jet, V₁ = 40 m/s
- Volumetric flow rate of water, Q = 5 m³/s
- Deflection angle of the jet, δ = 165°
- Density of water, ρ = 1000 kg/m³ (or 10³ kg/m³)
- Friction and mechanical losses are neglected (losses = 0)

2. Calculate the Blade/Bucket Velocity (u):
The linear velocity of the bucket (u) at the mean pitch circle is given by the formula:

$$u=\frac{\pi \times D\times N}{60}$$
Substituting the given values:

$$u=\frac{\pi \times 2\times 300}{60}=10\pi \approx 31.4159\text{m/s}$$

3. Determine the Bucket Exit Angle (ϕ):
The relation between the angle of deflection (δ) and the bucket exit angle (ϕ) is:

$$\delta ={180}^{\circ }-\varphi$$
Therefore:

$$\varphi ={180}^{\circ }-{165}^{\circ }={15}^{\circ }$$

4. Calculate the Power Developed by the Wheel:
Neglecting all losses, the power developed by a Pelton wheel is given by the momentum equation:

$$P=\rho \times Q\times (V_1-u)\times [1+\cos (\varphi )]\times u$$
Substituting the calculated values into the formula:

$$P={10}^3\times 5\times (40-31.4159)\times [1+\cos ({15}^{\circ })]\times 31.4159$$
Let us evaluate each component:
- $V_1-u=40-31.4159=8.5841\text{m/s}$
- $\cos ({15}^{\circ })\approx 0.9659$
- $1+\cos ({15}^{\circ })\approx 1.9659$

Now multiply these factors:

$$P=5000\times 8.5841\times 1.9659\times 31.4159$$
$$P\approx 5000\times 530.468\approx 2,652,340\text{W}$$
Convert the power to Megawatts (MW):

$$P=\frac{2,652,340}{{10}^6}\approx 2.65\text{MW}$$

Rounding off to two decimal places, the power developed by the Pelton wheel is 2.65 MW.