Question

Pottery kilns. Business is so good at Pottery West that the master potter plans to construct a new and larger kiln about 2 m high in which to bake his artistic creations. Estimate:

(a) The outside temperature of a vertical wall of this kiln (b) The heat loss through this wall

Data: The inside temperature of the kiln wall will be 1,150 C; room temperature is 20 C. The kiln wall will be 20 cm thick, made of high- temperature firebrick (k 1⁄4 0.1 W/m · K, ε 1⁄4 0.8).

Note: The findings of the previous problem suggest that you should not ignore the radiation from wall to room.

Answer 1

Given data

Outside Temperature T2 =?

Inside temperature T1 = 1150 + 273 = 1423 K

Room temperature T = 20 + 273 = 293 K

Thickness t = 20 cm x 1m/100cm = 0.20 m

Thermal conductivity k = 0.1 W/m-K

Emissivity ε = 0.8

Part a

Energy balance

Heat loss by conduction = Heat loss tby radiation

kA(T1 - T2) / t = Aε x (T2⁴ - T⁴)

0.1 W/m-K x (1423 - T2) K / (0.20 m) = 5.67 x 10^-8 W/m2-K4 x 0.8 x (T2⁴ - 293⁴ ) K⁴

0.5 x (1423 - T2) = 4.536 x 10^-8 x (T2⁴ - 293⁴ )

(1.1023 x 10^7) (1423 - T2) = (T2⁴ - 293⁴ )

1.568 x 10^10 - 1.1023 x 10^7 T2 - T2⁴ + 7.37 x 10^9 = 0

2.305 x 10^10 - 1.1023 x 10^7 T2 - T2⁴ = 0

T2 = 371.07 K

T2 = 98.07 °C

Part b

Heat loss = 0.1 W/m-K x (1423 - 371.07) K / (0.20 m)

= 525.97 W/m2