1. Theoretical Power and Efficiency: The work done per second (Power) by the jet on the buckets is given by the formula:
$$\text{Power} = \rho \cdot A \cdot V \cdot (V - u) \cdot (1 + \cos\phi) \cdot u$$
Where $\phi$ is the angle of the bucket at the outlet.
2. Condition for Maximum Efficiency: To find the maximum power, we differentiate the power expression with respect to $u$ and set it to zero ($dP/du = 0$). This leads to the fundamental relationship:
$$V - 2u = 0 \implies u = \frac{V}{2}$$
3. Physical Interpretation: If the bucket speed ($u$) is
half of the jet speed ($V$):
• The relative velocity of the water entering the bucket is $V - u = V/2$.
• The water leaves the bucket with a velocity relative to the bucket of $V/2$.
• Ideally, the absolute velocity of the water leaving the bucket becomes zero, meaning all of the jet's kinetic energy has been transferred to the turbine.
While real-world friction and bucket design mean the actual optimal ratio is slightly less (around 0.46 to 0.48), theoretically, the bucket speed must be exactly half of the jet speed for maximum power.