Question

The hot combustion gases of a furnace are separated from the ambient air and its surroundings, which are at 25 oC, by a brick wall 0.15 m thick. The brick has a thermalconductivity of 1.2 W/m.K and a surface emissivity of 0.8. Under steady state conditions an outer surface temperature of 100 oC is measured. Free convection heat transfer to the air adjoining this surface is characterized by a convection coefficient h = 20 W/m2.K. What is the inner surface temperature of the brick?

Answer 1

T1 = 625.54 K

Step-by-step explanation:

We are given;

T_α = Tsur = 25°C = 298K

h = 20 W/m².K,

L = 0.15 m

K = 1.2 W/m.K

ε = 0.8

Ts = T2 = 100°C = 373K

T1 = ?

Assumption:

-Steady- state condition

-One- dimensional conduction

-No uniform heat generation

-Constant properties

From Energy balance equation;

E°in - E°out = 0

Thus,

q"cond – q"conv – q"rad = 0

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)

Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)

Thus;

K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0

This gives;

(8T1 - 2984) - (1500) - 520.31 = 0

8T1 = 2984 + 1500 + 520.31

8T1 = 5004.31

T1 = 5004.31/8

T1 = 625.54 K