Pelton Wheel Scaling — Unit Discharge Calculation A Pelton wheel develops 1750 kW under a head H = 100 m while running at N = 200 rpm and discharging Q = 2500 litres per second. Compute the unit discharge of the wheel (i.e., discharge at 1 m head, Q_unit).

Introduction:
Unit quantities allow comparison of turbines independent of the operating head by normalizing speed, discharge, and power to a 1 m head. This problem asks for unit discharge, a commonly used parameter for model–prototype scaling and performance catalogs.

Given Data / Assumptions:

• Pelton wheel (impulse turbine), head H = 100 m.
• Actual discharge Q = 2500 L/s = 2.5 m^3/s.
• Standard definition Q_unit = Q / sqrt(H).

Concept / Approach:
Unit discharge is defined as the discharge at unit head (1 m) for dynamic similarity. Since discharge scales with sqrt(H) for dynamically similar operation of impulse turbines, we divide the actual discharge by sqrt(H).

Step-by-Step Solution:
Convert units: Q = 2500 L/s = 2.5 m^3/s.Compute sqrt(H) = sqrt(100) = 10.Q_unit = Q / sqrt(H) = 2.5 / 10 = 0.25 m^3/s.Therefore, the unit discharge is 0.25 m^3/s.

Verification / Alternative check:
Dimensional analysis yields Q ∝ D^2 * V and V ∝ sqrt(g * H); with D scaled to preserve similarity at unit head, Q scales with sqrt(H), confirming the formula.

Why Other Options Are Wrong:

• 0.5 m^3/s: Implies division by sqrt(25) or an arithmetic slip.
• 1.5 m^3/s and 2.5 m^3/s: Do not apply the head normalization; 2.5 m^3/s is the given actual discharge.

Common Pitfalls:
Forgetting to convert litres to cubic metres and neglecting the square-root head scaling.

Final Answer:
0.25 m^3/s