Question

PLEASE HELP ME :( Begging....

A furnace wall is exposed to hot gases at 1100 K. The wall consists of 0.12 m of the firebrick and 0.25 m of common brick. Heat transfer coefficients on the hot side are 3000 W/m^2-K and 22 W/m^2-K on the outside. Ambient air is at 300 K. What is the heat transfer rate per square meter of wall, and what is the temperature at the interface of the two bricks?

Answer 1

The conductivity of the bricks are not supplied. As an example, I use values given in
https://en.wikipedia.org/wiki/List_of_thermal_conductivities
and firebrick is 0.12 W/m-K and brick is 0.25 W/m-K.
Use values that are given or researched by you.

Heat going through the different materials is
1. constant for a given surface area, say 1 m^2.
2. proportional to \(\Delta\)T.
3. proportional to conductivity.
4. for physical material (bricks), inversely proportional to thickness.

Denoting
Q=heat lost per sq. m.
T=unknown temperatures, as follows
T1=atcontact surface between air (inside) and firebrick.
T2=at contact between the two bricks
T3=at contact surface between common brick and air
K=conductivity
K1=air film, inside, 3000 W/m^2K
K2=fire brick, 0.12 W/m-K
K3=brick, 0.25 W/m-K
K4=air film, outside, 22 W/m^2-K
t2=thickness of fire brick (0.12m)
t3=thickness of brick (0.25m)
Then we can set up equations
Q=K1(1100-T1)
Q=K2/t2(T1-T2)
Q=K3/t3(T2-T3)
Q=K4(T3-300)
Solve for Q,T1,T2,T3.
I get, to the nearest 10 W or \(^\circ\)K,
Q=910 W/m^2
Temp. in furnace-1100
T1=1100
T2=660
T3=340
ambient temp. = 300

Answer 2

Did you use the Fourier's Law?

Answer 3

Yes, implicitly.

Heat going through the different materials is
1. constant for a given surface area, say 1 m^2.
2. proportional to ΔT.
3. proportional to conductivity.
4. for physical material (bricks), inversely proportional to thickness.

\(\Delta\)T/thickness is the negative gradient.

Answer 4

Ahh okay. :) Thank you...