How is heat loss per m^2 found for the layered wall?

Overall delta T divided by the sum of all resistances

Average of the layer resistances

Delta T times the sum of resistances

Delta T divided by the largest resistance only

How is each interface temperature found?

Subtract q times the resistance step by step from 1200 C

Average the two neighbouring surface temperatures

Divide 1200 by the number of layers

Add q times resistance to 30 C only

Practice · 8 marks

Composite furnace wall heat loss

Short answer

Sum series resistances: 1/hi + L/k + 1/ho = 1.1294 K·m²/W. Heat loss q = (1200−30)/1.1294 = 1036 W/m². Interface temperatures: 1179 °C, 950 °C, 99.2 °C; outer surface 99.1 °C. Insulating brick offers 72.7% of total resistance.

A furnace wall consists of three layers: 230 mm of firebrick (k = 1.04 W/m K), 115 mm of insulating brick (k = 0.14 W/m K) and a 6 mm steel casing (k = 45 W/m K). The hot gases at 1200 °C have a heat transfer coefficient of 50 W/m^2 K to the firebrick, and the casing loses heat to still air at 30 °C with a coefficient of 15 W/m^2 K. Draw the thermal circuit and find (a) the heat loss per square metre of wall, (b) the temperatures at all interfaces and the outer surface, and (c) the percentage of the total resistance offered by the insulating brick.

Thermal circuit

 Gas     Firebrick   Ins.brick   Steel     Air
1200C --/\/\/--/\/\/--/\/\/--/\/\/--/\/\/-- 30C
        1/hi    L1/k1   L2/k2   L3/k3   1/ho

Area = 1 m $^2$ . Resistances per m $^2$ (K m $^2$ /W):

Element	Formula	Value
Gas film	$1/h_i = 1/50$	0.0200
Firebrick	$0.23/1.04$	0.2212
Insulating brick	$0.115/0.14$	0.8214
Steel	$0.006/45$	0.00013
Air film	$1/h_o = 1/15$	0.0667
Total		1.1294

(a) Heat loss

q = \frac{T_{gas} - T_{air}}{\sum R} = \frac{1200-30}{1.1294} = 1036\ \text{W/m}^2

(b) Temperatures

Subtract $qR$ step by step:

\begin{aligned} T_{1} &= 1200 - 1036(0.02) = 1179.3\ ^\circ\text{C (hot face of firebrick)}\\ T_{2} &= 1179.3 - 1036(0.2212) = 950.2\ ^\circ\text{C (firebrick/insulating brick)}\\ T_{3} &= 950.2 - 1036(0.8214) = 99.2\ ^\circ\text{C (insulating brick/steel)}\\ T_{4} &= 99.2 - 1036(0.00013) = 99.1\ ^\circ\text{C (outer steel surface)} \end{aligned}

Check: $T_4 - 1036(0.0667) = 30.0\ ^\circ$ C, which equals the air temperature.

(c) Share of insulating brick

\frac{0.8214}{1.1294}\times100 = 72.7\%

Answer: $q \approx 1036\ \text{W/m}^2$ ; interface temperatures 1179 °C, 950 °C, 99.2 °C, outer surface 99.1 °C; insulating brick gives about 72.7 % of the total resistance.

✓ Answer checked by Claude Sonnet 5.5

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Composite furnace wall heat loss

Thermal circuit

(a) Heat loss

(b) Temperatures

(c) Share of insulating brick

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