In a pipe system with constant roughness, if the flow rate doubles in turbulent flow, what happens to the pressure drop?

• A. It doubles
• B. It increases, often approximately with the square of the flow rate
• C. It stays the same
• D. It decreases

Answer: (B)

In turbulent flow through a pipe with constant roughness, friction losses follow the Darcy–Weisbach relation, where the pressure drop ΔP is proportional to v^2 (ΔP ∝ v^2) because ΔP = f (L/D) (ρ v^2 / 2) and f, L, D, and density ρ stay fixed. Since flow rate Q is proportional to velocity v (Q = A v), doubling the flow rate doubles the velocity, and because the pressure drop scales with velocity squared, ΔP increases by about a factor of four. So the pressure drop rises, roughly with the square of the flow rate.