Question

Water from a dam falls on a turbine 30 m below at the rate of 60 m³/min. After
leaving the turbine wheel the water has a velocity of 9 m/s. What is the maximum
efficiency of the turbine? What is the maximum power developed?

Answer 1

The maximum efficiency is

$$\eta_\text{max}=\frac{v_\text{ptactical}}{v_\text{theoretical}}=\frac{v_\text{ptactical}}{\sqrt{gH}},\\\space\\ \eta_\text{max}=\frac{9}{\sqrt{9.8\cdot30}}=0.52$$

The maximum power of the turbine is

$$P_\text{max}=QgH.$$

Convert 60 m³/min to kg/s:

$$Q=60\cdot1000/60=1000\text{ kg/s}.\\ P_\text{max}=1000\cdot9.8\cdot30=294000\text{ W}.$$