Engineer's Notebook §2.1

Conduction

Heat flows through a solid from hot to cold, at a rate set by the material's conductivity and the temperature gradient. This is Fourier's law:

q = −kdTdx
q
heat flux
W/m²
k
thermal conductivity
W/(m·K)
T
temperature
K
x
distance through the material
m
Φ
heat flow rate
W
A
area
m²
L
wall thickness
m
h
convection coefficient
W/(m²·K)
ℓ
pipe length
m
r
radius
m
A wall seen edge-on, hot side on the left and cold side on the right. A line shows the temperature falling straight through the wall, and an arrow shows the heat flow. k = 0.7 L = 100 mm T₁ = 20 °C T₂ = 0 °C q = 140 W/m²
Fig. 2.1-1Wall shown at a fixed width. The temperature scale fits the two faces.
dT/dx−200 K/m
q = k(T1 − T2)/L140 W/m²
Φ = qA1.4 kW

Worked example

Uses the values in Fig. 2.1-1. Change them and the working updates.

Given
  • kbrick0.7 W/(m·K)
  • Lwall thickness100 mm
  • Awall area10 m²
  • T1hot face20 °C
  • T2cold face0 °C
Find

The rate of heat flow Φ through the wall.

Solution
q = k(T1 − T2)/L = 0.7 W/(m·K) × (20 − 0) K / 0.1 m = 140 W/m² Φ = qA = 140 W/m² × 10 m² = 1,400 W
Answer
Φ = 1.4 kW

Over 24 hours that is 33.6 kWh.

Related cases

  • Layers in series

    Φ = A ΔTΣ(L/k)

  • Pipe wall

    Φ = 2πkℓ ΔTln(r2/r1)

  • Convection at a surface

    Φ = hA ΔT

Assumptions
  • Conditions are steady: temperatures no longer change with time.
  • Heat flows in one direction, straight through the wall. Edges and corners are ignored.
  • k is constant over the temperature range, and no heat is generated inside the wall.
  • The face temperatures are known. In practice the air films on each face add resistance, so use surface rather than room temperatures.
  • Conductivities are typical values. Real figures vary with density, moisture and grade.

Derivation
  1. Fourier's law: q = −k dT/dx.
  2. In steady state with no heat generated, the same flux passes every plane, so q is constant through the wall.
  3. With k constant, dT/dx is constant too: the temperature falls linearly, dT/dx = −(T1 − T2)/L.
  4. So q = k(T1 − T2)/L, and over area A, Φ = kA(T1 − T2)/L.

The minus sign in Fourier's law says heat flows down the temperature gradient, from hot to cold.