For incompressible flow in a circular pipe at fixed capacity (constant volumetric flow rate), how does required head (frictional head loss) vary with pipe diameter D, assuming similar roughness and length?

Introduction / Context:Sizing process lines requires understanding how frictional head loss depends on diameter at a set flow rate. This relation drives energy consumption and pumping cost and explains why larger diameters can dramatically reduce head and power for the same capacity.

Given Data / Assumptions:

• Single-phase, incompressible liquid; steady flow.
• Circular pipe of length L and diameter D; similar roughness.
• Constant volumetric flow rate Q.

Concept / Approach:Use the Darcy–Weisbach equation: h_f = f * (L/D) * (V^2 / (2g)). At fixed Q, velocity V = Q / A and A ∝ D^2, so V ∝ 1/D^2 and V^2 ∝ 1/D^4. Substituting gives h_f ∝ (L/D) * (1/D^4) = L * 1/D^5 (neglecting the weak dependence of f on Reynolds number for the main trend). Thus, at constant capacity, frictional head varies approximately as 1/D^5—an extremely strong sensitivity.

Step-by-Step Solution:Start: h_f = f * (L/D) * (V^2 / (2g)).With Q fixed, V = 4Q / (π D^2) → V^2 ∝ 1/D^4.Therefore h_f ∝ (1/D) * (1/D^4) = 1/D^5 (dominant dependence).

Verification / Alternative check:Engineering nomographs and pump system curves show steep head reduction with increased diameter at constant Q, consistent with ≈ D^−5 scaling in turbulent regimes when friction factor changes modestly.

Why Other Options Are Wrong:(a) and (b) underestimate the dependence; (d) inverts the trend; (e) contradicts fundamental hydraulics.

Common Pitfalls:Forgetting that friction factor f can change with Re; ignoring minor losses (they follow similar velocity-squared trends); applying the result to compressible gas without correction.

Final Answer:Proportional to 1/D^5