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Convective heat transfer

Combine convection on both sides of a wall with conduction through it using a thermal-resistance network.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Convective Heat Transfer. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

At a glance

  • Prerequisites: convection law, slab conduction, steady-state energy balance.
  • Outcomes: solve coupled convection-conduction-convection problems.
  • Key result: resistances in series give one heat rate through all layers.
  • Common trap: assuming wall-surface temperatures equal fluid temperatures.

Setup

Plane wall of thickness L and conductivity k separates two fluids:

  • Inside fluid: T_(∞,in), convection coefficient hᵢₙ.
  • Outside fluid: T_(∞,out), convection coefficient hₒᵤₜ.
  • Wall area A.

At steady state:

Q dot _(in,conv) = Q dot _(wall,cond) = Q dot _(out,conv) = Q dot.

Physical interpretation:

  • Convection “drops” temperature between the bulk fluid and the wall surface; conduction “drops” temperature through the solid wall.
  • The resistance network enforces a single heat-current through all series elements, and the largest resistance sets the bottleneck.

Core derivation / explanation

Thermal resistance network:

Rᵢₙ = 1/hᵢₙA, R_wall = L/kA, Rₒᵤₜ = 1/hₒᵤₜA.

Total resistance:

Rₜₒₜ = Rᵢₙ + R_wall + Rₒᵤₜ.

Heat-transfer rate:

Q dot = (T_(∞,in)-T_(∞,out))/Rₜₒₜ.

Surface temperatures follow from temperature drops:

T_(s,in) = T_(∞,in)-Q dot Rᵢₙ, T_(s,out) = T_(∞,out) + Q dot Rₒᵤₜ.

Bridge note:

  • This is the UY1 upgrade from separate conduction/convection formulas to one coupled model.
Pitfalls (convection + conduction networks)
  • Don’t assume wall-surface temperatures equal fluid temperatures; the convection resistances create boundary temperature drops. - Keep Q dot (W) vs q'' = Q dot/A (W m⁻²) consistent. Many textbook examples silently use A = 1. - If radiation is significant, add a radiative element (exact εσ A(T⁴-Tₛᵤᵣ⁴) or a linearized hᵣ) from UY1: Radiation.

Units check:

[R] = K W⁻¹ ⇒ [Q dot] = W.

Physical sanity check:

  • If k → ∞, wall resistance vanishes and convection dominates.
  • If either h is very small, that side becomes the bottleneck.

Heat loss through a window

Worked example 1

Window thermal resistance

Problem

Find the steady heat rate and heat flux through a 1.0 m² glass window with the following data. Neglect edge losses and radiation.

  • T_(∞,in) = 25°C, T_(∞,out) = -15°C.
  • L = 2.0 mm = 0.002 m.
  • k = 1.05 W m⁻¹K⁻¹ (glass).
  • hᵢₙ = 8 W m⁻²K⁻¹, hₒᵤₜ = 20 W m⁻²K⁻¹.
  • A = 1.0 m².
Show full solution

  1. Add the three resistances

    Method

    Calculate both convection resistances and the glass conduction resistance, then add them.

    Reason

    The same steady heat rate crosses each series element; their temperature drops add to the bulk-fluid temperature difference.

    Working

    Answer

    Rᵢₙ = 0.125, R_wall = 0.00190, Rₒᵤₜ = 0.050 K W⁻¹.

    Total Rₜₒₜ = 0.1769 K W⁻¹.

  2. Calculate the rate and flux

    Working

    Answer

    Q dot = (40 K)/(0.1769 K W⁻¹) ≈ 226 W, q'' = (Q dot)/A ≈ 226 W m⁻².

    Heat flows outward. The rate and flux have equal numerical values here only because A = 1.0 m². Convection accounts for most of the resistance.

Practice set

Check your understanding 1

For the window above, how much does the steady heat rate change if glass thickness doubles while all other data stay fixed?

Show hint
Double only R_wall = 0.00190 K W⁻¹.
Show answer
Rₜₒₜ becomes 0.125 + 0.00381 + 0.050 = 0.17881 K W⁻¹, so Q dot ≈ 40/0.17881 = 224 W. It falls by only about 1.1% because the convection resistances dominate.

Check your understanding 2

For the original window, what happens to heat loss if wind raises hₒᵤₜ from 20 to 40 W m⁻²K⁻¹ with other data fixed?

Show hint
Find the new outside resistance, then the total.
Show answer
Rₒᵤₜ halves to 0.025 K W⁻¹, giving Rₜₒₜ = 0.1519 K W⁻¹ and Q dot ≈ 263 W. The lower total resistance increases heat loss.

Check your understanding 3

Why are the inner and outer glass-surface temperatures different when heat flows through the window?

Show hint
Relate the temperature drop across glass to Q dot R_wall.
Show answer
The glass has finite conduction resistance, so a nonzero heat rate requires T_(s,in)-T_(s,out) = Q dot R_wall > 0. For the original window the drop is about 226(0.00190) = 0.43 K.

Summary + next steps

  • Coupled heat-transfer problems are best solved with resistance networks.
  • At steady state, one heat rate flows through all series elements.
  • Identifying the largest resistance reveals the dominant bottleneck.

Next steps:

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Syllabus and review details

No official syllabus alignment is listed for this lesson.

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