Overall heat transfer coefficient (U): In which type of problems do we use the overall heat transfer coefficient concept?

Difficulty: Easy

Correct Answer: Combined conduction and convection (resistances in series)

Explanation:


Introduction / Context:
Real heat exchangers involve multiple layers and mechanisms: conduction through walls and fouling layers, plus convection on fluid sides. To simplify analysis, we combine these effects into a single parameter called the overall heat transfer coefficient U.



Given Data / Assumptions:

  • Steady-state heat transfer.
  • Planar or cylindrical walls separating fluids.
  • Thermal resistances can be added in series.


Concept / Approach:
The overall coefficient U encapsulates all resistances between two bulk fluids. For a flat wall, 1/U = 1/h_hot + R_cond + 1/h_cold (+ fouling terms). For a tube, appropriate logarithmic mean area is used. U enables the compact expression Q = U * A * ΔT_mean for exchanger design or rating.



Step-by-Step Solution:

Write individual resistances: R_conv,hot = 1/(h_hot * A), R_cond = L/(k * A), R_conv,cold = 1/(h_cold * A).Total resistance R_total = R_conv,hot + R_cond + R_conv,cold (+ fouling).Define U from Q = ΔT_overall / R_total = U * A * ΔT_mean, hence 1/U = (R_total / A).Apply appropriate area (inside/outside/log-mean) for cylindrical walls.


Verification / Alternative check:
Compare predictions with detailed CFD or layer-by-layer models; the series-resistance model provides good engineering estimates for design.



Why Other Options Are Wrong:
Using U for a single pure mechanism (conduction only or convection only) is unnecessary; radiation alone also does not require U unless combined with other modes in a composite resistance network.



Common Pitfalls:
Mixing areas when summing resistances in cylindrical systems; neglecting fouling, which significantly lowers U in service.



Final Answer:
Combined conduction and convection (resistances in series)

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