Turbine Similarity — Discharge vs Head Under dynamically similar operation, how does turbine discharge Q vary with head H?

Introduction:
Similarity relations (or model laws) connect discharge, head, and speed for turbines and pumps. Recognizing these proportionalities is essential in model testing and scaling from model to prototype.

Given Data / Assumptions:

• Geometric similarity of machines.
• Similar operating point (near best efficiency) and negligible change in efficiency.
• Same fluid properties.

Concept / Approach:
Characteristic velocity scales with the square root of head: V ∝ sqrt(2 * g * H). For a given scaled geometry, area scales with length squared, so discharge Q ∝ Area * Velocity ∝ L^2 * sqrt(H). Eliminating length under similarity leads to Q ∝ H^1/2 for a particular machine family at similar points.

Step-by-Step Solution:
Use V ∝ sqrt(H) at similar operation.Q = A * V ⇒ Q ∝ L^2 * sqrt(H).For a fixed machine, L is constant; hence Q ∝ H^1/2.

Verification / Alternative check:
Empirical performance curves plotted in unit variables show Qu = Q / H^0.5 roughly constant near best efficiency.

Why Other Options Are Wrong:
inversely proportional to H^1/2: opposite of similarity prediction.directly/inversely proportional to H^3/2: wrong power; Q scales with sqrt(H), not H^1.5.

Common Pitfalls:
Confusing turbine scaling (Q ∝ H^0.5) with power scaling (P ∝ H^1.5).

Final Answer:
directly proportional to H^1/2