--- title: "Composite furnace wall heat loss" subject: "Heat Transfer" chapter: 2 key: "semester_5/heat_transfer:composite-furnace-wall-interface-temperatures" source: https://www.bctengineer.com/q/semester_5/heat_transfer/composite-furnace-wall-interface-temperatures site: BCT Engineering --- # Composite furnace wall heat loss Heat Transfer · Chapter 2: Conduction Heat Transfer - Asked in: year not given, 8 marks - Video: [Problem on heat conduction in furnace wall](https://www.youtube.com/watch?v=v5ydSfrArNo) — Ashish Chaudhari - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/heat_transfer/composite-furnace-wall-interface-temperatures - Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-2 ## Question 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. ## 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. ## Answer #### Thermal circuit ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is heat loss per m^2 found for the layered wall? A. Average of the layer resistances B. Delta T times the sum of resistances C. Delta T divided by the largest resistance only D. Overall delta T divided by the sum of all resistances Answer: D. Resistances in series add; q = (1200-30)/1.1294. (Key point: Series resistances) 2. What is the resistance of the 115 mm insulating brick? A. 0.8214 K m^2/W B. 0.1217 K m^2/W C. 0.0200 K m^2/W D. 1.2174 K m^2/W Answer: A. R = L/k = 0.115/0.14 = 0.8214. (Key point: Layer resistance) 3. How is each interface temperature found? A. Average the two neighbouring surface temperatures B. Subtract q times the resistance step by step from 1200 C C. Divide 1200 by the number of layers D. Add q times resistance to 30 C only Answer: B. Temperature drops by qR across each element. (Key point: Interface temperatures) 4. The insulating brick gives about what share of total resistance? A. 19.6% B. 6% C. 72.7% D. 100% Answer: C. 0.8214/1.1294 = 72.7%. 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