Difficulty: Medium
Correct Answer: Increase by 8
Explanation:
Introduction / Context:
In heat exchanger design, changing the number of tube passes affects flow distribution, velocity, and frictional losses. This question tests the understanding of how converting from a single-pass to a two-pass tube arrangement impacts tube-side pressure drop for the same overall flow rate and tube bundle length.
Given Data / Assumptions:
Concept / Approach:
For fully developed turbulent flow in a smooth tube, pressure drop per tube can be approximated as: ΔP ∝ (L/D) * (ρ * v^2 / 2). Converting to two passes: (1) Path length along a tube doubles (L → 2L), and (2) Velocity in each tube doubles because only half as many tubes are in parallel (v → 2v).
Step-by-Step Solution:
Single-pass baseline: ΔP₁ ∝ L * v^2.Two-pass length effect: L → 2L gives a factor of 2.Two-pass velocity effect: v → 2v gives v^2 → (2v)^2 = 4v^2.Combined factor: 2 (length) * 4 (velocity-squared) = 8.Therefore, ΔP₂ ≈ 8 * ΔP₁ for the same total flow rate.
Verification / Alternative check:
A quick network analogy: halving parallel paths doubles flow per path (velocity roughly doubles). Two series segments (two passes) double the frictional contribution. Multiplying effects again gives 8×, confirming the result.
Why Other Options Are Wrong:
Increase by 4: ignores the doubled path length.Increase by 2: considers only the added pass length, not velocity change.Approximately unchanged: contradicts stronger velocity and length effects.Decrease by 2: physically implausible for the stated conversion.
Common Pitfalls:
Forgetting that tubes per pass are halved in two-pass designs.Assuming friction factor changes cancel the velocity effect; they generally do not for a first estimate.Comparing total exchanger pressure drop but not normalizing per unit length consistently.
Final Answer:
Increase by 8
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