In a concentric (double-pipe) heat exchanger, for a fluid flowing through the annulus, what effective hydraulic diameter should be used for heat-transfer calculations (assuming both annular surfaces are heat-transferring)? D2 and D1 are the inside diameters of the outer and inner pipes respectively.

Introduction / Context:
Hydraulic diameter is used in correlations for both pressure drop and convective heat transfer. In a double-pipe exchanger annulus, choosing the proper characteristic length is essential to apply Nusselt, Reynolds, and friction factor correlations consistently.

Given Data / Assumptions:

• Concentric annulus with inner pipe ID = D1 and outer pipe ID = D2.
• Both the inner pipe outer surface and the outer pipe inner surface participate in heat transfer.
• Single-phase flow, no fins or inserts.

Concept / Approach:
The hydraulic diameter is defined as D_h = 4 * (flow area) / (wetted perimeter). For an annulus with both walls wetted and heat-transferring, this simplifies to D_h = D2 − D1. This value is valid for both flow (pressure-drop) and heat-transfer correlations when both surfaces are active.

Step-by-Step Solution:
Compute flow area: A = (π/4) * (D2^2 − D1^2).Wetted perimeter (both walls): P = π * (D2 + D1).D_h = 4A / P = 4 * [(π/4)(D2^2 − D1^2)] / [π(D2 + D1)] = (D2 − D1).Use D_h in Re, Nu, and friction correlations as usual.

Verification / Alternative check:
Textbook derivations for concentric annuli consistently reduce to D2 − D1 when both bounding surfaces contribute to wetted perimeter and heat transfer, confirming equivalence to the flow basis.

Why Other Options Are Wrong:
Less/More than D2 − D1: contradicts the 4A/P derivation.Pressure-drop only: D_h applies to both heat transfer and pressure drop here.One-wall perimeter only: applicable to cases with one heated wall; not this case.

Common Pitfalls:
Using only one wall in P when both are wetted and active.Mixing OD/ID definitions, leading to wrong D1 or D2.

Final Answer:
Equal to D2 − D1 (same as for flow area basis)