Annular flow: What is the ratio of the equivalent diameter used for pressure-drop calculations to the equivalent diameter used for heat-transfer calculations for a concentric annulus when both inner and outer walls are heat-transfer surfaces?

Difficulty: Easy

Correct Answer: 1 (they are equal)

Explanation:


Introduction / Context:
When analyzing flow in non-circular passages such as annuli, engineers often use an “equivalent” or “hydraulic” diameter. For design, two related but sometimes different diameters appear: one applied in pressure-drop correlations and the other in convective heat-transfer correlations. Knowing when they coincide simplifies calculations and reduces confusion.


Given Data / Assumptions:

  • Concentric circular annulus.
  • Both the inner and outer walls participate in heat transfer.
  • Single-phase, fully developed flow regime assumptions are acceptable.


Concept / Approach:
The general definition for equivalent diameter is Deq = 4 * (flow area) / (wetted perimeter). For an annulus with both walls “active,” the wetted perimeter includes both inner and outer circumferences, and the flow area is the ring area. Under these conditions, the equivalent diameter for pressure drop and for heat transfer are numerically the same and reduce to the simple difference of diameters: Deq = Do – Di.


Step-by-Step Solution:

Write flow area A = (π/4) * (Do^2 – Di^2).Write wetted perimeter P = π * (Do + Di) when both walls are “wetted.”Compute Deq = 4A/P = (Do^2 – Di^2) / (Do + Di) = Do – Di.Hence Deq for pressure drop equals Deq for heat transfer → ratio = 1.


Verification / Alternative check:
If only one wall were heated (say, the inner), the heat-transfer equivalent diameter would use only that wall’s perimeter, and the equality would not hold. The present question specifies both walls are active, making the ratio unity.


Why Other Options Are Wrong:

  • Greater than 1 / less than 1: arise only when heated perimeter differs from wetted perimeter.
  • “Depends only on which wall is heated”: not correct under the stated condition (both walls heated).
  • 0.5: no basis for this constant in standard derivations.


Common Pitfalls:
Using Deq = Do – Di without checking which surfaces exchange heat; forgetting to adjust the heated perimeter when only one wall is active can lead to significant Nu and h errors.


Final Answer:
1 (they are equal)

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