Solution to Question 4: Heat Transfer in a Com-

posite Furnace Wall

1. Problem Statement Analysis
The problem asks us to determine the optimal thickness of each layer in a three-
layer composite furnace wall to achieve a minimum heat loss, given a total
thickness constraint and a target heat loss rate. The furnace wall consists of:
1. Refractory brick: This is the innermost layer, exposed to the highest
temperature.
2. Insulating brick: This layer is designed to reduce heat transfer.
3. Steel plate: This is the outermost layer, providing mechanical protection.
We are provided with the following data:
• Inner furnace wall temperature (T_inner) = 1371°C
• Outer furnace wall temperature (T_outer) = 37.78°C
• Total load thickness (L_total) = 1577.95 (units are not explicitly stated,
but typically this would be in mm or cm for furnace walls, or perhaps a to-
tal thermal resistance value if it’s a ‘load’ in terms of resistance. Given the
context of calculating thickness, it’s likely a total thickness in some length
unit. Assuming mm for now, as it’s a common unit for such applications,
and will verify if it makes sense with the final calculations).
• Target heat loss (Q/A) = 15772.95 W/m²
• The layers are assumed to be in good thermal contact, implying negligible
contact resistance.

Material Properties:
Refractory brick: * Maximum service temperature = 1426.67°C * Thermal
Conductivity (k) at 37.78°C = 10.22 W/mK * Thermal Conductivity (k) at
1371.11°C = 20.44 W/mK
Insulating brick: * Maximum service temperature = 1093.33°C * Thermal
Conductivity (k) at 37.78°C = 5.11 W/mK * Thermal Conductivity (k) at
1371.11°C = 10.22 W/mK
Steel Plate: * Thickness (L_steel) = 0.635 cm = 0.00635 m * Thermal Con-
ductivity (k) at 37.78°C = 13.48.70 W/mK (This value seems unusual, possibly
a typo. 1348.70 W/mK is very high for steel. Typical thermal conductivity
for steel is around 15-50 W/mK. I will proceed with the given value but note
this discrepancy. If it’s 13.4870, that’s more reasonable. Given the format, it’s
likely 1348.70. I will use 1348.70 and proceed, assuming it’s a specific alloy or
condition.)

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2. Theoretical Background
Heat transfer through a composite wall under steady-state conditions and one-
dimensional heat flow can be analyzed using the concept of thermal resistance.
The rate of heat transfer (Q) through a plane wall is given by Fourier’s Law of
Heat Conduction:
𝑄 = −𝑘𝐴 𝑑𝑇
𝑑𝑥

For a composite wall, the heat transfer rate per unit area (Q/A) is constant
through each layer and can be expressed as:
𝑄 𝑇𝑖𝑛𝑛𝑒𝑟 −𝑇𝑜𝑢𝑡𝑒𝑟
𝐴 = 𝑅𝑡𝑜𝑡𝑎𝑙

Where 𝑅𝑡𝑜𝑡𝑎𝑙 is the total thermal resistance per unit area of the composite wall.
For layers in series, the total thermal resistance is the sum of individual thermal
resistances:
𝑅𝑡𝑜𝑡𝑎𝑙 = 𝑅1 + 𝑅2 + 𝑅3
For each layer, the thermal resistance per unit area (𝑅𝑖 ) is given by:
𝐿𝑖
𝑅𝑖 = 𝑘𝑖

Where: * 𝐿𝑖 is the thickness of layer i * 𝑘𝑖 is the thermal conductivity of layer i

Therefore, the heat transfer rate per unit area can be written as:
𝑄 𝑇𝑖𝑛𝑛𝑒𝑟 −𝑇𝑜𝑢𝑡𝑒𝑟
𝐴 = 𝐿𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑜𝑟𝑦 𝐿𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑛𝑔 𝐿
𝑘𝑟𝑒𝑓𝑟𝑎𝑐𝑡𝑜𝑟𝑦 + 𝑘𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑛𝑔 + 𝑘 𝑠𝑡𝑒𝑒𝑙
𝑠𝑡𝑒𝑒𝑙

Temperature-Dependent Thermal Conductivity

For materials where thermal conductivity varies significantly with temperature,
an average thermal conductivity (𝑘𝑎𝑣𝑔 ) is used. This is typically calculated as
the arithmetic mean of the thermal conductivities at the inner and outer surface
temperatures of the layer:
𝑘1 +𝑘2
𝑘𝑎𝑣𝑔 = 2

However, for a composite wall, the interface temperatures between layers are
unknown. To accurately determine the average thermal conductivity for each
brick layer, we would ideally need to solve for the interface temperatures. A
common iterative approach is to assume initial interface temperatures, calculate
average conductivities, then calculate heat flux and new interface temperatures,
and repeat until convergence. Alternatively, if the thermal conductivity varies
linearly with temperature, an integrated average can be used.
Given the problem statement provides thermal conductivities at the inner and
outer overall temperatures of the furnace, and not necessarily at the interface
temperatures, we will assume that the provided thermal conductivities corre-
spond to the temperatures at the boundaries of the respective layers, or that
an effective average needs to be determined based on the overall temperature

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difference. For simplicity and given the information, we will first attempt to use
an average thermal conductivity based on the overall temperature range for the
brick materials, and then refine if necessary.
Let’s re-examine the given thermal conductivity data. It provides k at 37.78°C
and 1371.11°C. These are the overall furnace temperatures. This suggests that
we should use an average thermal conductivity over the entire temperature range
for the materials, assuming a linear variation. However, this is only strictly
accurate if the entire temperature drop occurs across a single material. For a
composite wall, the temperature drop across each layer is different.
A more rigorous approach for temperature-dependent thermal conductivity in
composite walls is to use the concept of thermal conductivity integral or to
assume a linear variation of k with T, i.e., 𝑘(𝑇 ) = 𝑘0 (1 + 𝛽𝑇 ). In such cases,
the thermal resistance of a layer is given by:
𝐿𝑖
𝑅𝑖 = 𝑘̄ 𝑖

Where 𝑘̄ 𝑖 is the mean thermal conductivity for the temperature range across that
specific layer. Since the interface temperatures are unknown, we will need to
make an assumption or use an iterative method. For a first pass, let’s assume the
average thermal conductivity for each brick layer can be approximated by the
average of the given values at the two extreme temperatures of the furnace, as
a starting point. This is a simplification, but often used when detailed interface
temperatures are not readily available or for a preliminary calculation.
10.22+20.44
For Refractory brick: 𝑘𝑟𝑒𝑓,𝑎𝑣𝑔 = 2 = 15.33 W/mK
5.11+10.22
For Insulating brick: 𝑘𝑖𝑛𝑠,𝑎𝑣𝑔 = 2 = 7.665 W/mK
For Steel Plate, the thermal conductivity is given as 1348.70 W/mK at 37.78°C.
Since it’s a thin layer for mechanical protection, its temperature variation might
be small, or this value is considered representative. We will use 𝑘𝑠𝑡𝑒𝑒𝑙 =
1348.70 W/mK.

3. Formulation of the Problem

Let 𝐿1 be the thickness of the refractory brick, 𝐿2 be the thickness of the
insulating brick, and 𝐿3 be the thickness of the steel plate.
We are given:
1. Total thickness: 𝐿1 + 𝐿2 + 𝐿3 = 𝐿𝑡𝑜𝑡𝑎𝑙 (where 𝐿𝑡𝑜𝑡𝑎𝑙 is 1577.95, assuming
meters for now to be consistent with W/mK and W/m²). Since 𝐿3 =
0.00635 m, we have 𝐿1 +𝐿2 = 𝐿𝑡𝑜𝑡𝑎𝑙 −𝐿3 = 1577.95−0.00635 = 1577.94365
m. This total thickness value (1577.95 m) is extremely large for a furnace
wall. A typical furnace wall thickness would be in the range of centimeters
or a few decimeters (0.1 to 1 meter). If 1577.95 is indeed in meters, this
would be a massive structure. It’s highly probable that the unit is actually
millimeters (mm) or centimeters (cm), or that 1577.95 refers to a total

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 thermal resistance. Let’s assume it’s in mm for now, meaning 𝐿𝑡𝑜𝑡𝑎𝑙 =
1577.95×10−3 m = 1.57795 m. This is still quite thick, but more plausible
than 1.5 km. If it’s cm, then 𝐿𝑡𝑜𝑡𝑎𝑙 = 15.7795 m, which is also very large.
Given the context of
the problem being about calculating thickness, it’s most likely a length unit.
Let’s proceed with the assumption that 𝐿𝑡𝑜𝑡𝑎𝑙 = 1577.95 mm = 1.57795 m.
This is still a very thick wall, but more realistic than kilometers.
𝑄 2
2. Heat Loss Rate: 𝐴 = 15772.95 W/m
We need to find 𝐿1 and 𝐿2 such that the total heat loss is minimized, subject to
the total thickness constraint. However, the problem statement says “Calculate
the thickness of each layer if heat loss is to be minimum, total load thickness is
1577.95, and heat loss is to be 15772.95 W/m²”. This implies that the heat loss
is given as a target, not something to be minimized. The phrase “heat loss is to
be minimum” might refer to the design goal, but the numerical value provided
for heat loss overrides this for the calculation. Therefore, we will use the given
heat loss rate as a fixed value.
The equation for heat transfer through the composite wall is:
𝑄 𝑇𝑖𝑛𝑛𝑒𝑟 −𝑇𝑜𝑢𝑡𝑒𝑟
𝐴 = 𝐿1 𝐿2 𝐿3
𝑘𝑟𝑒𝑓,𝑎𝑣𝑔 + 𝑘𝑖𝑛𝑠,𝑎𝑣𝑔 + 𝑘𝑠𝑡𝑒𝑒𝑙

Substituting the known values:

1371−37.78
15772.95 = 𝐿1 𝐿2 0.00635
15.33 + 7.665 + 1348.70

0.00635
Let 𝑅𝑠𝑡𝑒𝑒𝑙 = 1348.70 ≈ 4.708 × 10−6 m2 K/W. This is a very small resistance, as
expected for a thin steel plate with high thermal conductivity.
So, the equation becomes:
1333.22
15772.95 = 𝐿1 𝐿2 −6
15.33 + 7.665 +4.708×10

We also have the total thickness constraint (assuming 𝐿𝑡𝑜𝑡𝑎𝑙 is in meters):

𝐿1 + 𝐿2 + 𝐿3 = 1.57795 m
𝐿1 + 𝐿2 = 1.57795 − 0.00635 = 1.5716 m
Now we have a system of two equations with two unknowns (𝐿1 and 𝐿2 ).
Equation 1 (from heat transfer):
𝐿1 𝐿2 1333.22
15.33 + 7.665 + 4.708 × 10−6 = 15772.95
𝐿1 𝐿2
15.33 + 7.665 + 4.708 × 10−6 ≈ 0.084526
𝐿1 𝐿2
15.33 + 7.665 ≈ 0.084526 − 4.708 × 10−6
𝐿1 𝐿2
15.33 + 7.665 ≈ 0.0845213
Equation 2 (from total thickness):

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𝐿1 + 𝐿2 = 1.5716
From Equation 2, we can express 𝐿2 as 𝐿2 = 1.5716 − 𝐿1 . Substitute this into
the modified Equation 1:
𝐿1 1.5716−𝐿1
15.33 + 7.665 = 0.0845213
To solve for 𝐿1 , we can multiply the entire equation by the common denominator
of 15.33 and 7.665, which is 15.33:
𝐿1 + 2(1.5716 − 𝐿1 ) = 0.0845213 × 15.33
𝐿1 + 3.1432 − 2𝐿1 = 1.2960
−𝐿1 = 1.2960 − 3.1432
−𝐿1 = −1.8472
𝐿1 = 1.8472 m
Now, calculate 𝐿2 :
𝐿2 = 1.5716 − 𝐿1 = 1.5716 − 1.8472 = −0.2756 m
A negative thickness is physically impossible. This indicates that our initial
assumption about the unit of L_total (1577.95 mm) or the interpretation of
the problem statement (especially the
heat loss is to be minimum” part is the key to solving the problem, as it suggests
an optimization approach. For a given heat loss, the total thickness of the
insulating layers will be minimized if the materials are used to their fullest
potential, which is limited by their maximum service temperature.
Therefore, we will proceed with the following assumptions:
1. The problem is to find the thicknesses of the refractory and insulating
bricks that satisfy the given heat loss rate.
2. The insulating brick is used at its maximum service temperature, which is
a common practice in furnace design to optimize the use of materials. This
means the interface temperature between the refractory and insulating
bricks is equal to the maximum service temperature of the insulating brick.
3. The thermal conductivity of the materials varies linearly with tempera-
ture.
4. The value of “total load thickness = 1577.95” is either a typo or a red
herring, as it leads to a physically impossible solution when treated as a
total thickness constraint in meters or millimeters. We will calculate the
total thickness based on our derived layer thicknesses and comment on the
discrepancy.

4. Step-by-Step Calculation
We will now calculate the thickness of each layer based on the above assump-
tions.

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a. Calculate the temperature at the interface between the insulating
brick and the steel plate (T_interface2)
The heat flux (q = Q/A) is constant through all layers. For the steel plate, we
have:
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 −𝑇𝑜𝑢𝑡𝑒𝑟 𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 −𝑇𝑜𝑢𝑡𝑒𝑟
𝑞= 𝑅3 = 𝐿3 /𝑘3

We can rearrange this to solve for 𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 :

𝐿3
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 = 𝑇𝑜𝑢𝑡𝑒𝑟 + 𝑞 × 𝑘3

Substituting the known values:

0.00635
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 = 37.78 + 15772.95 × 1348.70

𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 = 37.78 + 15772.95 × 4.708 × 10−6

∘
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 = 37.78 + 0.07426 = 37.854 C

b. Assume the interface temperature between the refractory and

insulating brick (T_interface1)
To optimize the use of the insulating material, we assume it is operated at its
maximum service temperature:
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 = 1093.33∘ C

c. Calculate the average thermal conductivity for each layer

Since the thermal conductivity is temperature-dependent, we will first determine
the linear relationship 𝑘(𝑇 ) = 𝑎𝑇 +𝑏 for both brick materials and then calculate
the average thermal conductivity for the specific temperature range each layer
is exposed to.
For the refractory brick:
We have 𝑘(37.78) = 10.22 and 𝑘(1371.11) = 20.44. The slope a is:
20.44−10.22 10.22
𝑎1 = 1371.11−37.78 = 1333.33 = 0.007665
The intercept b is:
𝑏1 = 10.22 − 0.007665 × 37.78 = 9.9305
So, 𝑘1 (𝑇 ) = 0.007665𝑇 + 9.9305. The temperature range for the refractory
brick is from 𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 = 1093.33∘ C to 𝑇𝑖𝑛𝑛𝑒𝑟 = 1371∘ C. The average thermal
conductivity is:
𝑘1,𝑎𝑣𝑔 = 𝑘1 ( 1371+1093.33
2 ) = 𝑘1 (1232.165) = 0.007665 × 1232.165 + 9.9305 =
19.37 W/mK
For the insulating brick:
We have 𝑘(37.78) = 5.11 and 𝑘(1371.11) = 10.22. The slope a is:

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 10.22−5.11 5.11
𝑎2 = 1371.11−37.78 = 1333.33 = 0.0038325
The intercept b is:
𝑏2 = 5.11 − 0.0038325 × 37.78 = 4.965
So, 𝑘2 (𝑇 ) = 0.0038325𝑇 +4.965. The temperature range for the insulating brick
∘
is from 𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 = 37.854 C to 𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 = 1093.33∘ C. The average thermal
conductivity is:
𝑘2,𝑎𝑣𝑔 = 𝑘2 ( 1093.33+37.854
2 ) = 𝑘2 (565.592) = 0.0038325 × 565.592 + 4.965 =
7.132 W/mK

d. Calculate the thickness of the refractory brick (L1)

The heat flux through the refractory brick is:
𝑇𝑖𝑛𝑛𝑒𝑟 −𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1
𝑞= 𝐿1 /𝑘1,𝑎𝑣𝑔

Solving for 𝐿1 :
𝑘1,𝑎𝑣𝑔 ×(𝑇𝑖𝑛𝑛𝑒𝑟 −𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 ) 19.37×(1371−1093.33) 19.37×277.67
𝐿1 = 𝑞 = 15772.95 = 15772.95 = 0.3407 m

e. Calculate the thickness of the insulating brick (L2)

The heat flux through the insulating brick is:
𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 −𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2
𝑞= 𝐿2 /𝑘2,𝑎𝑣𝑔

Solving for 𝐿2 :
𝑘 ×(𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒1 −𝑇𝑖𝑛𝑡𝑒𝑟𝑓𝑎𝑐𝑒2 ) 7.132×(1093.33−37.854) 7.132×1055.476
𝐿2 = 2,𝑎𝑣𝑔 𝑞 = 15772.95 = 15772.95 =
0.4772 m

5. Results and Conclusion

The calculated thicknesses of the layers are:
• Refractory brick (L1): 0.3407 m or 34.07 cm
• Insulating brick (L2): 0.4772 m or 47.72 cm
• Steel plate (L3): 0.00635 m or 0.635 cm (given)
The total thickness of the furnace wall is:
𝐿𝑡𝑜𝑡𝑎𝑙,𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 = 𝐿1 + 𝐿2 + 𝐿3 = 0.3407 + 0.4772 + 0.00635 = 0.82425 m or
82.425 cm.
This calculated total thickness (82.4 cm) is significantly different from the given
“total load thickness” of 1577.95 (units unspecified). This confirms our suspicion
that the given value is either a typo or refers to something other than the
physical thickness of the wall. If we were to enforce the given total thickness,
the problem would not have a physically meaningful solution.

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Therefore, the most reasonable solution, based on standard engineering practice
for such problems, is to use the material properties and constraints to derive the
thicknesses, which leads to the results above. The phrase “if heat loss is to be
minimum” is interpreted as a design guideline that justifies using the insulating
material at its maximum service temperature, thus optimizing its use.
Final Answer: The required thickness of the refractory brick is 34.07 cm, and
the thickness of the insulating brick is 47.72 cm to meet the specified heat loss
rate of 15772.95 W/m², while respecting the material temperature limits.