Topic: Fluid Mechanics and Machinery
A turbine develops 500 kW power under a head of 100 m at 200 rpm. What would be its normal speed under a head of 81 m?
- A. 100 rpm
- B. 180 rpm
- C. 1800 rpm
- D. 162 rpm
Answer: B. 180 rpm
Explanation: For a hydraulic turbine, the unit speed relation is $N \propto \sqrt{H}$. Therefore, $\frac{N_{1}}{\sqrt{H_{1}}} = \frac{N_{2}}{\sqrt{H_{2}}}$. Given $N_{1}=200~rpm$, $H_{1}=100~m$, and $H_{2}=81~m$. $\frac{200}{\sqrt{100}} = \frac{N_{2}}{\sqrt{81}} \Rightarrow \frac{200}{10} = \frac{N_{2}}{9} \Rightarrow N_{2} = 20 \times 9 = 180~rpm$.