Thermal resistance · Fixture Course

Watch

Explainer video0:04

Deep Dive audio0:04

Summary

Thermal resistance on one page: a wall resists heat, R = L/(kA); layers add in series; the temperature drop drives the heat through the total resistance. — InfographicTap to zoom

A wall resists heat like a resistor resists current

In steady conduction through a plane wall, the heat rate is set by the temperature drop across the wall and by the wall’s thermal resistance:

\dot{Q} = \frac{\Delta T}{R}, \qquad R = \frac{L}{k A}

A thicker wall (larger $L$ ) resists more; a better conductor (larger $k$ ) or a larger face (larger $A$ ) resists less. Layers in series add their resistances, $R_\text{total} = \sum R_i$ , and the same $\dot{Q}$ passes through every layer.

الجدار يقاوم سريان الحرارة كما يقاوم المقاوم سريان التيار: فرق درجات الحرارة يدفعها، والمقاومة

R = \frac{L}{kA}

تحدّ منها. في الشكل جدار سمكه 0.2 م تهبط حرارته من 20 إلى 5 درجات في خط مستقيم.

Assumed: supplied here, not in the Materials: $x = 0$ · $L = 0.2\ \text{m}$ · $T_1 = 20$ · $T_2 = 5\ ^\circ\text{C}$

T

(°C)

x

(m)

T_1

T_2

Fig.An illustrative wall

0.2\ \text{m}

thick; the temperature falls in a straight line

Worked examples

TopicThermal resistance

Title

Heat loss through a composite wall

ExampleW01.1

Step1 / 6

Mode

Given

A wall is plaster, brick and insulation in series. Its inside face is held at $T_1$ and its outside face at $T_4$ . Find the heat loss $\dot{Q}$ through the wall.

$A = 1\ \text{m}^2$
$T_1 = 20\ ^\circ\text{C}$ (inside face)
$T_4 = -5\ ^\circ\text{C}$ (outside face)

Layer resistances,

R = \frac{L}{kA}

Layer	$L$ m	$k$ W/(m·K)
Plaster	0.02	0.5
Brick	0.20	0.8
Insulation	0.05	0.04
Total

T

(°C)

x

(m)

T_1

T_4

Fig.The wall: plaster to

0.02\ \text{m}

, brick to

0.22\ \text{m}

, insulation to

0.27\ \text{m}

Read the wall

Three layers in series between two faces at known temperatures. The table gives each layer's

L

and

k

; the

R

and

\Delta T

columns are ours to fill.

ثلاث طبقات على التوالي بين وجهين حرارتهما معلومة: الداخلي عند 20 درجة والخارجي عند −5 درجات. نحسب مقاومة كل طبقة ثم نجمع المقاومات.

TopicThermal resistance

Title

A plate cooling from both faces

ExampleW01.2

Step1 / 5

Mode

Given

A plate $40\ \text{mm}$ thick is at $200\ ^\circ\text{C}$ throughout when both its faces are suddenly held at $20\ ^\circ\text{C}$ . Its diffusivity is $\alpha = 12\ \text{mm}^2/\text{s}$ . Find the temperatures across it after $t = 30\ \text{s}$ .

$L = 40\ \text{mm}$
$T_i = 200\ ^\circ\text{C}$
$T_s = 20\ ^\circ\text{C}$ on both faces
$\alpha = 12\ \text{mm}^2/\text{s}$
$t = 30\ \text{s}$

The plate at

t = 30\ \text{s}

, face to face

$x$ mm	$T$ $^\circ\text{C}$
0
10
20
30
40

Read the problem

The plate cools from both faces alike, so it stays symmetric about its mid-plane,

x = 20\ \text{mm}

. Work in

\theta = \frac{T - T_s}{T_i - T_s}

, which falls from

1

towards

0

.

Sim

The plate cooling from both faces

Move $t$ to watch the plate cool, or $\alpha$ to make it another material. The table reads it where W01.2 does; the map shows the whole plate over time.

Assumed: supplied here, not in the Materials: $t$ tuned from $0$ to $120\ \text{s}$ in steps of $1$ · $\alpha$ tuned from $4$ to $40\ \text{mm}^2/\text{s}$ in steps of $1$

T

(°C)

x

(mm)

Fig.The temperature across the wall at the chosen time, between where it starts and where it settles, both dashed.

t

(s)30

\alpha

(mm²/s)12

$x$ (mm)	$T$ (°C)
0	20.0
10	37.6
20	44.9
30	37.6
40	20.0

$\mathrm{Fo} = \frac{\alpha t}{L^2}$ = 0.225: the mid-plane is at 44.9 °C, the quarter points at 37.6 °C

Practice

1. A furnace burns methane, $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ , and holds the inside face of its fireclay wall at $600\ \text{K}$ above the outside face. The wall is $L = 0.25\ \text{m}$ thick with $k = 1.0\ \text{W/(m·K)}$ and a face of $A = 4\ \text{m}^2$ . Find the heat loss through the wall, in kW.

2. A plaster layer $0.02\ \text{m}$ thick, with $k = 0.5\ \text{W/(m·K)}$ , covers a face of $1\ \text{m}^2$ . Find its thermal resistance.

3. The sheet's drawing of the composite wall shows the insulation layer about $0.05\ \text{m}$ thick. Explain why that thin layer sets most of the wall's heat loss.

Scaled: measured off a drawing: $L = 0.05\ \text{m}$ (insulation, off the drawing)

Your answer