In turbine similarity analysis, “unit power” P_u is defined as the power developed by a turbine when operating under unit head (H = 1). Given actual power P at head H, express the unit power in terms of P and H.

Introduction:
Unit quantities (unit speed, unit discharge, unit power) allow comparison of hydraulically similar turbines at different heads. By reducing head effects via dimensional analysis, designers can scale performance between model and prototype efficiently.

Given Data / Assumptions:

• Geometric and dynamic similarity between compared operating points.
• Head H is the effective net head across the turbine.
• Define unit head condition as H = 1.

Concept / Approach:
From similarity, discharge Q ∝ H^(1/2), rotational speed N ∝ H^(1/2) (for constant specific speed form), and therefore power P ∝ ρ g Q H ∝ H^(3/2). Setting H = 1 defines unit power P_u that removes head scaling, yielding P_u = constant for dynamically similar operation. Thus, P_u relates to actual P via division by H^(3/2).

Step-by-Step Solution:
1) Use Q ∝ H^(1/2) from velocity head scaling.2) Power P = ρ * g * Q * H ⇒ P ∝ H^(1/2) * H = H^(3/2).3) Define unit power as power at H = 1: P_u = P / H^(3/2).4) Therefore, for any measured P at head H, compute unit power with the above relation.

Verification / Alternative check:
Dimensionless groups: P / (ρ g) scales like Q H, and with Q ∝ H^(1/2), the H^(3/2) law is recovered, confirming the formula.

Why Other Options Are Wrong:
P_u = P * H^(3/2): incorrectly amplifies head effect.

P_u = P / H or P * H or P / H^2: do not follow similarity exponents derived from velocity scaling.

Common Pitfalls:
Mixing unit quantities: note N_u = N / H^(1/2), Q_u = Q / H^(1/2), but P_u = P / H^(3/2).

Final Answer:
P_u = P / H^(3/2)