Answer to Question #233682 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A furnace wall consists of 200 mm layer of refractory bricks, 6 mm layer of

steel plate and a 100 mm layer of insulation bricks. The maximum temperature of the wall is

1150°C on the furnace side and the minimum temperature is 40°C on the outermost side of the

wall. An accurate energy balance over the furnace shows that the heat loss from the wall is

400 W/m2. It is known that there is a thin layer of air between the layers of refractory bricks and

steel plate. Thermal conductivities for the three layers are 1.52, 45 and 0.138 W/m°C respec￾tively. Find :

(i) To how many millimetres of insulation brick is the air layer equivalent ?

(ii) What is the temperature of the outer surface of the steel plate ?

Answer 1

Thickness of refractory bricks,

$L_A=200 \;mm = 0.2 \;m$

Thickness of steel plate,

$L_C = 6 \; mm = 0.006 \; m$

Thickness of insulation bricks,

$L_D = 100 \; mm = 0.1 \; m$

Difference of temperature between the innermost and outermost side of the wall,

∆t = 1150 – 40 = 1110°C

Thermal conductivities :

$k_A = 1.52 \;W/m°C \\ k_B = k_D = 0.138 \;W/m°C \\ k_C = 45 \;W/m°C$

Heat loss from the wall,

$q = 400\; W/m^2$

(i) The value of $x = (L_C)$ :

$Q= \frac{A∆t}{\sum \frac{L}{k}} \\ \frac{Q}{A} = q = \frac{∆t}{\sum \frac{L}{k}} \\ 400 = \frac{1110}{\frac{L_A}{k_A}+ \frac{L_B}{k_b} + \frac{L_C}{k_C} + \frac{L_D}{k_D} } \\ 400 = \frac{1110}{\frac{0.2}{1.52} +\frac{x/1000}{0.138} + \frac{0.006}{45} +\frac{0.1}{0.138} } \\ = \frac{1110}{0.1316 +0.0072x + 0.00013 + 0.7246} \\ = \frac{1110}{0.8563 + 0.0072x} \\ 0.8563 + 0.0072x = \frac{1110}{400} = 2.775 \\ x= \frac{2.775-0.8563}{0.0072}= 266.5 \;mm$

(ii) Temperature of the outer surface of the steel plate $t_{so}$ :

$q=400 = \frac{t_{so} -40}{L_D/k_D} \\ 400 = \frac{t_{so} -40}{(0.1/0.138)}= 1.38(t_{so}-40) \\ t_{so}= \frac{400}{1.38} +40 = 329.8 \;°C$