Heat Transfer and Combustion -- A furnace wall consists of three layers of material

[Tiberious]

Homework Statement

A furnace wall consists of three layers of material as shown below.

​

The thermal conductivities are:

Firebrick = 1.15 W m–1 K–1

Insulating brick = 0.17 W m–1 K–1

Ordinary brick = 0.62 W m–1 K–1

Calculate:

1. (i) the thermal resistance of each layer
2. (ii) the heat loss per unit area
3. (iii) the temperature at the two interfaces.

Homework Equations

The Thermal Resistance of each layer.

Given equation:
R_a= L/k

The Heat Loss per unit area.

Given equation:

q= ((T_1-T_2))/R_a

The Attempt at a Solution

The Thermal Resistance of each layer.

Given equation:
R_a= L/k

Thermal Resistance of Firebrick:
L=220mm=2.2〖0m〗^(-3)
k=1.15 W m-1 K-1

R_a= (2.〖20〗^(-3))/(1.15 )=1.913 m^2 K W^(-1)
Thermal Resistance of the Insulating Brick:
L=110mm=1.1〖0m〗^(-3)
k=0.17 W m-1 K-1
R_a= (1.〖10〗^(-3))/(0.17 )=6.471 m^2 K W^(-1)
Thermal Resistance of the Ordinary Brick:
L=110mm=2.35^(-3)
k=0.17 W m-1 K-1

R_a= (2.〖35〗^(-3))/0.62=3.790 m^2 K W^(-1) The Heat Loss per unit area.

Given equation:

q= ((T_1-T_2))/R_a

Temperature One: 750
Temperature Two: 56

Heat Loss per unit area Firebrick:

q= ((750-56))/(1.913 m^2 K W^(-1) )

q=362.781 W m^(-2)

Heat Loss per unit area Insulating Brick:

q= ((750-56))/(6.471 m^2 K W^(-1) )

q=107.248 W m^(-2)

Heat Loss per unit area Ordinary Brick:

q= ((750-56))/(3.790 m^2 K W^(-1) )
q=183.113 W m^(-2)
The Temperature at the two interfaces.

Determine the total resistance

R_a= R_a12+ R_a23+ R_a34
Inputting our known values

R_a= 1.91 + 6.47+ 3.79 =12.174 m^2 K W^(-1)

= 12.174

Heat Loss per unit area

q= (T_1-T_4)/R_a

= ((750-56))/(12.174 )

≈57 W m^(-1)We can now determine the temperature at the interfaces.

Determining the temperature at interface one.

q_12=q

q=57 W m^(-1)

So,
q_12= (T_(1-) T_2)/R_a12

→ T^1-T^2= R_a12 q_12

= 1.913∙57

=109.041 °C

Determine T_2

T_2= T_1-109.041

=750-109.041
Temperature at interface one.

=640.959°C

Determining the temperature at interface two.

q_23=q

q=57 W m^(-1)

So,
q_23= (T_1-T_2)/R_a23

→ T^2-T^3= R_a23 q_23

=6.471 ∙57

=368.847°C

Determine T_3

T_3= T_2-368.85

= 640.959-368.847
Temperature at interface Two.

=272.112°C

Finally,

=3.790 ∙57

=216.03
Determine T_4

T_4= T_3-216.03

= 272.112-216.03

≈56°C

 

[CWatters]

It's a bit hard to read all that on my small phone but...

You appear to calculate the heat loss incorrectly.

The heat flows through all three materials one after the other. So you should calculate the total euivalent series resistance. Then temperature drop occurs across that equivalent resistance.