Topic guides

Mechanical engineering / Topic guide

Conduction & convection

Separate heat flow through a wall from heat transfer into a fluid. Choose the right temperature difference, compare steady heat rates, and distinguish rate from flux.

8 min3 connected formulasWorked example

The path through this topic

  1. 01Wall conduction
  2. 02Surface convection
  3. 03Heat flux
Browse the formulas

Visual overview

One heat flow, two temperature drops

Heat passes through a plane wall from an 80 °C hot face to a 40 °C surface, then into fluid at 20 °C. The temperature plot uses cumulative thermal resistance, showing a 40 K drop through the wall and 20 K across convection.
The worked-example wall conducts 200 W and its exposed surface convects 200 W, giving a heat flux of 100 W/m². The plot is temperature against cumulative thermal resistance; it does not represent a predicted spatial temperature profile in the fluid.

Start with the question

Which formula do you need?

Identify the two endpoints for each heat-transfer mechanism. Conduction uses the wall-face temperatures; convection uses the surface and bulk-fluid temperatures. Divide a heat rate by area only when you need flux.

1

How much heat passes through the wall?

Q˙=kAΔTd\dot{Q}=\frac{kA\Delta T}{d}

Use the wall's conductivity, area, and thickness with the temperature difference between its two faces. The formula gives a heat-transfer rate in watts.

You need: Thermal conductivity k, area A, thickness d, and wall-face temperature difference ΔT.

Open formula & calculator
2

How much heat leaves the surface for the fluid?

Q˙=hA(Ts−T∞)\dot Q=hA(T_s-T_\infty)

Use the exposed surface temperature and the bulk-fluid temperature. The coefficient h summarizes this fluid-flow heat-transfer situation; it is not the wall's conductivity.

You need: Convection coefficient h, exposed area A, surface temperature Tₛ, and bulk-fluid temperature T∞.

Open formula & calculator
3

What is the heat rate per unit area?

q′′=Q˙Aq^{\prime\prime}=\frac{\dot Q}{A}

Divide the heat-transfer rate by the corresponding area. For nonuniform transfer, this is an area average. Heat flux has units of W/m² rather than W.

You need: Heat-transfer rate and the area through which that rate passes.

Open formula & calculator

Check the model first

Before you start: Temperature differences, basic algebra, and SI units.

  • Steady, one-dimensional conduction through a plane wall with constant thermal conductivity and uniform cross-sectional area.
  • No internal heat generation, contact resistance, side losses, or radiation in this example. With no energy storage, the series heat rate is the same through the wall and at the surface.
  • A convection coefficient appropriate to the actual fluid and flow, and representative surface and bulk-fluid temperatures.

Put it together

Two mechanisms. One steady heat flow.

A plane wall has area 2 m², thickness 0.10 m, and thermal conductivity 0.25 W/(m·K). Its hot face is at 80°C and its exposed face at 40°C. Fluid at 20°C contacts that face with h = 5 W/(m²·K). Calculate the two heat rates and the heat flux; check whether they are consistent with steady operation.

A=2 m2A=2\,\mathrm{m^2}d=0.10 md=0.10\,\mathrm{m}k=0.25 W/(m K)k=0.25\,\mathrm{W/(m\,K)}h=5 W/(m2 K)h=5\,\mathrm{W/(m^2\,K)}Th=80∘C, Ts=40∘C, T∞=20∘CT_h=80^\circ\mathrm C,\ T_s=40^\circ\mathrm C,\ T_\infty=20^\circ\mathrm C

  1. 01Apply the wall-face temperature difference

    For conduction, use 80 − 40 = 40 K across the wall. A Celsius temperature difference has the same numerical value as a kelvin difference.

    Q˙cond=0.25(2)(80−40)0.10\dot Q_{\mathrm{cond}}=\frac{0.25(2)(80-40)}{0.10}
    Q˙cond≈200 W\dot Q_{\mathrm{cond}}\approx 200\,\mathrm{W}
    Try your own values

  2. 02Use surface-to-fluid temperatures for convection

    The convection temperature difference is 40 − 20 = 20 K. The matching rate means that these boundary values satisfy the steady energy balance for the stated model.

    Q˙conv=5(2)(40−20)\dot Q_{\mathrm{conv}}=5(2)(40-20)
    Q˙conv≈200 W\dot Q_{\mathrm{conv}}\approx 200\,\mathrm{W}
    Try your own values

  3. 03Convert the rate into an average flux

    Divide either rate by the same 2 m² area. Choose positive heat flow from the hot face toward the cooler fluid.

    q′′=2002q^{\prime\prime}=\frac{200}{2}
    q′′≈100 W/m2q^{\prime\prime}\approx 100\,\mathrm{W/m^2}
    Try your own values

Both mechanisms transfer 200 W, so the heat flux is 100 W/m². Do not add the two rates: they describe the same heat flowing through successive parts of this steady, one-dimensional system.

Small mistakes that change the answer

Using one temperature difference everywhere

The wall sees 80 − 40 K; convection sees 40 − 20 K. The full 80 − 20 K difference spans both mechanisms and requires their combined thermal resistance.

Confusing k with h

Thermal conductivity k has units W/(m·K). The convection coefficient h has units W/(m²·K) and depends on the fluid and flow configuration.

Adding series heat rates

With no energy storage, generation, or side losses, the same heat rate passes through each series element. A mismatch means the stated steady model or boundary values need checking.

Confusing heat rate with heat flux

Watts measure total heat-transfer rate. Watts per square meter measure flux. Dividing by an unrelated surface area gives the wrong physical quantity.

What if the surface temperature is unknown?

For this one-wall, one-convection model, add the wall and convection thermal resistances. Use the hot-face-to-bulk-fluid temperature difference to get the common rate, then recover the surface temperature from either element's drop.

Rwall=dkA,Rconv=1hA,Q˙=Th−T∞Rwall+RconvR_{\mathrm{wall}}=\frac{d}{kA},\qquad R_{\mathrm{conv}}=\frac{1}{hA},\qquad \dot Q=\frac{T_h-T_\infty}{R_{\mathrm{wall}}+R_{\mathrm{conv}}}

These are whole-element resistances in K/W. Additional surfaces, contacts, radiation, or parallel heat paths must be modeled explicitly. The plotted horizontal coordinate is cumulative resistance, not distance in the fluid.

Check your understandingIf the wall thickness doubles while its two face temperatures, area, and conductivity stay fixed, what happens to its conductive heat rate?

The conductive rate halves because it is inversely proportional to thickness. If instead only the hot-face and bulk-fluid temperatures stay fixed, the surface temperature changes too: recompute the combined wall and convection resistances rather than assuming that the overall rate halves.