For a Kaplan (axial flow) turbine, the outlet blade velocity diagram at a section is shown in figure.

The diameter at this section is 3 m. The hub and tip diameters of the blade are 2 m and 4 m, respectively. The water volume flow rate is 100 m³/s. The rotational speed of the turbine is 300 rpm. The blade outlet angle β is _________ degrees (round off to one decimal place).

The correct answer is 12.7

1. Identify the given parameters from the problem description and the provided diagram:
- Hub diameter of the blade: $D_h=2\text{m}$
- Tip/outer diameter of the blade: $D_o=4\text{m}$
- Section diameter where the outlet velocity diagram is drawn: $D=3\text{m}$
- Water volume flow rate: $Q=100{\text{m}}^3\text{/s}$
- Rotational speed of the turbine: $N=300\text{rpm}$
- From the outlet blade velocity diagram shown in the image, we observe a right-angled triangle where the flow velocity $C_f$ is perpendicular to the blade velocity $C_b$, and the relative velocity $C_r$ is the hypotenuse. The blade outlet angle $\beta$ is the angle between $C_r$ and $C_b$.

2. Calculate the blade velocity ($C_b$) at the given section diameter of 3 m:
The blade linear velocity $C_b$ is given by the formula:
$$C_b=\frac{\pi DN}{60}$$
Substituting the given values:
$$C_b=\frac{\pi \times 3\times 300}{60}=15\pi \approx 47.124\text{m/s}$$

3. Calculate the flow velocity ($C_f$):
For a Kaplan turbine, the flow is axial, and the volume flow rate $Q$ is related to the flow velocity $C_f$ by the area of the flow passage:
$$Q=\frac{\pi }{4}\left(D_o^2-D_h^2\right)C_f$$
Substituting the values of $Q$, $D_o$, and $D_h$:
$$100=\frac{\pi }{4}\left(4^2-2^2\right)C_f$$
Simplifying the expression:
$$100=\frac{\pi }{4}\left(16-4\right)C_f=3\pi C_f$$
Solving for $C_f$:
$$C_f=\frac{100}{3\pi }\approx 10.610\text{m/s}$$

4. Calculate the blade outlet angle ($\beta$):
From the right-angled velocity triangle at the outlet:
$$tan\beta =\frac{C_f}{C_b}$$
Substituting the calculated values of $C_f$ and $C_b$:
$$tan\beta =\frac{10.610}{47.124}\approx 0.22516$$
Taking the arctangent to find $\beta$:
$$\beta ={tan}^{-1}\left(0.22516\right)\approx 12.69\text{°}$$
Rounding off to one decimal place, we get:
$$\beta \approx 12.7\text{degrees}$$