Internal flow – head loss dependence in laminar regime For fully developed laminar flow in a circular pipe, the loss of pressure head is proportional to which power of the mean velocity?

Mechanical Engineering Hydraulics and Fluid Mechanics Difficulty: Easy
Choose an option
  • A
    velocity (first power)
  • B
    velocity squared
  • C
    velocity cubed
  • D
    velocity to the fourth power

Answer

Correct Answer: velocity (first power)

Explanation

Introduction / Context:Understanding how head loss scales with velocity helps select pumps and size pipes. The relationship differs markedly between laminar and turbulent regimes. This question focuses on the laminar case.

Given Data / Assumptions:

  • Steady, incompressible, fully developed, laminar flow in a straight circular pipe.
  • Newtonian fluid with constant viscosity μ.
  • No entrance or exit losses considered for the proportionality discussion.

Concept / Approach:Hagen–Poiseuille law: Δp = 32 μ L V / D^2 for laminar pipe flow. Converting to head loss h_f = Δp/(ρ g) shows h_f ∝ V. Thus, in laminar flow, head loss varies linearly with mean velocity. In contrast, turbulent empirical correlations (e.g., Darcy–Weisbach with f depending on Re and roughness) often show approximately V^2 dependence over ranges.

Step-by-Step Solution:

Start: Δp = 32 μ L V / D^2.Divide by ρ g: h_f = Δp/(ρ g) = (32 μ L /(ρ g D^2)) * V.Therefore, h_f ∝ V (first power).

Verification / Alternative check:Darcy–Weisbach with laminar friction factor f = 64/Re yields h_f = f (L/D) (V^2/(2g)) = (64/Re)(L/D)(V^2/(2g)) = (64 μ /(ρ V D))(L/D)(V^2/(2g)) ∝ V, confirming linear dependence.

Why Other Options Are Wrong:

  • V^2, V^3, V^4 correspond to turbulent or nonphysical scalings for this regime.

Common Pitfalls:Memorizing V^2 from turbulent flow and applying it universally. Always check Reynolds number to identify regime.

Final Answer:velocity (first power)

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