Heat Transfer and Combustion -- reheating furnace wall consists of 200 mm of firebrick

[Tiberious]

Homework Statement

A small reheating furnace wall consists of 200 mm of firebrick. The inner surface of the wall is at a temperature of 320 °C and the outside temperature is 35 °C. Calculate the rate at which heat is transferred, by conduction, through unit area of the wall. The thermal conductivity of the firebrick used can be taken as 1.8 W m–1 K–1.

If the outside surface area of the furnace is 50 m2estimate the heat losses through the furnace wall per hour.

Homework Equations

Given equation:
ϕ = (kA(T_1-T_2))/L

The Attempt at a Solution

Inputting values:
k=1.8 W m^(-1) K^(-1)
A=50m^2
T_1=320 ,T_2=35
L=200mm ⟶ 0.2m
ϕ= ((1.8)∙(50)∙(320-35))/0.2
=128 250 W
=128 250 J S^(-1)

The heat loss per hour = ϕ∙t

128 250 ∙ 3600

=461 700 000 J
= 461.7 MJ

 

[tnich]

Homework Statement

A small reheating furnace wall consists of 200 mm of firebrick. The inner surface of the wall is at a temperature of 320 °C and the outside temperature is 35 °C. Calculate the rate at which heat is transferred, by conduction, through unit area of the wall. The thermal conductivity of the firebrick used can be taken as 1.8 W m–1 K–1.

If the outside surface area of the furnace is 50 m2estimate the heat losses through the furnace wall per hour.

Homework Equations

Given equation:
ϕ = (kA(T_1-T_2))/L

The Attempt at a Solution

Inputting values:
k=1.8 W m^(-1) K^(-1)
A=50m^2
T_1=320 ,T_2=35
L=200mm ⟶ 0.2m
ϕ= ((1.8)∙(50)∙(320-35))/0.2
=128 250 W
=128 250 J S^(-1)

The heat loss per hour = ϕ∙t

128 250 ∙ 3600

=461 700 000 J
= 461.7 MJ

That all looks correct

 

[Paul george 2838]

The question mentions " through unit area of the wall" is the answer above, Q = 128250 multiplied 3600sec correct for this statement, the way I'm reading it as this is correct for 50m^2 and therefore to find unit area of the wall we use q (heat flux) thus 2565 X 3600? Or am I confusing myself here? Thanks