Unit speed of a Pelton wheel — compute using head scaling A Pelton wheel develops 1750 kW under a head of 100 m while running at 200 r.p.m. and discharging 2.5 m^3/s. Determine the unit speed (speed under 1 m head for a geometrically similar wheel).

Introduction:
“Unit quantities” (unit speed, unit discharge, unit power) normalize turbine performance to a head of 1 m. They are widely used to compare and scale turbines of similar geometry. For turbines, speed scales with the square root of head, enabling quick predictions at different heads.

Given Data / Assumptions:

• Pelton wheel (impulse turbine) running at N = 200 r.p.m.
• Head H = 100 m.
• Discharge Q = 2.5 m^3/s; Power P = 1750 kW (extraneous for unit speed).
• Geometric similarity and dynamic similarity for the unit conversion.

Concept / Approach:
For all hydraulic turbines, unit speed N_u (sometimes written N_1) is defined by N_u = N / sqrt(H). This follows from similarity laws: runner peripheral speed u ∝ sqrt(gH) and rotor diameter D fixed for a given machine during normalization. Therefore, to reduce the head from H to 1 m, divide the actual speed by sqrt(H).

Step-by-Step Solution:

Verification / Alternative check:
Dimensional check: rpm divided by sqrt(m) yields rpm for 1 m. Also, industry tables show Pelton unit speeds typically in the tens of rpm, which matches 20 r.p.m. for this head and speed combination.

Why Other Options Are Wrong:

• 10 r.p.m.: Corresponds to division by 20, not sqrt(100).
• 40 or 80 r.p.m.: Would require multiplying by sqrt(H) or halving incorrectly.
• 5 r.p.m.: Far too low; not consistent with scaling.

Common Pitfalls:
Using H instead of sqrt(H); confusing unit speed (normalize by head only) with specific speed (depends on both power and head: N_s ∝ Nsqrt(P)/H^(5/4)).

Final Answer: