Question

A jet of water having a velocity of 20 m/s strikes a series of plates fixed radially on a wheel revolving in the same direction as the jet at 15 m/s. What is the percentage efficiency of the plates? (round off to one decimal place)

• A. 37.5
• B. 66.7
• C. 50.0
• D. 88.9

Hint:

The maximum efficiency of jet plates occurs when the plate velocity is half the jet velocity (\(u = V/2\)).

Answer 1

The Correct Option is A

Step 1: Given data.
Jet velocity, \( V = 20 \, \text{m/s} \)
Plate velocity, \( u = 15 \, \text{m/s} \)

Step 2: Work done per unit weight of water.
The work done per unit weight is proportional to the product of plate velocity and the difference between jet velocity and plate velocity: \[ W \propto u (V - u) \]

Step 3: Efficiency of the plates.
Efficiency is defined as the ratio of work done to the kinetic energy of the jet: \[ \eta = \frac{2u (V - u)}{V^2} \]

Step 4: Substitution.
\[ \eta = \frac{2 \times 15 (20 - 15)}{20^2} = \frac{2 \times 15 \times 5}{400} = \frac{150}{400} = 0.375 \]

Step 5: Convert to percentage.
\[ \eta \times 100 = 37.5 % \] \[ \boxed{ \text{Efficiency of the plates = 37.5%} } \]