Heat Transfer: Finding temperature at the junction

In summary, we have a furnace with 0.5 m of fire brick, 0.15 m of insulating brick, and 0.25 m of ordinary building brick. The inside surface temperature is 1530K and the outside surface temperature is 525K. Using the thermal conductivities of the bricks, we can calculate the total resistance of the wall, the rate of heat flow, and the temperatures at the junctions of the bricks. The correct temperature at the (firebrick & insulating)-ordinary brick interface is 777.0231 K, according to our calculations.
  • #1
paulie
13
0

Homework Statement


A furnace is constructed with 0.5 m of fire brick, 0.15 m of insulating brick and 0.25 m of ordinary building brick. The inside surface-temperature is 1530K and the outside surface temperature is 525K. The thermal conductivities of the fire, insulating and building bricks are 1.4, 0.21, and 0.7 W/m-K, respectively. Calculate the following per square meter area:
(a) Total resistance of the wall in K/W
(b) Rate of heat flow in W
(c) The temperatures at the junctions of the bricks.

Homework Equations


Q/T = kAΔT / L

The Attempt at a Solution


(a) Total resistance of the wall in K/W: answer 1.43 K/W
(b) Rate of heat flow in W: answer 703.5 W
(c) The temperatures at the junctions of the bricks.

(Temperature drop over firebrick+insulating brick)/(Total temperature drop) = (0.3571+0.7143/1.43)
Temperature drop over firebrick+insulating brick:
(1530K-525K) ⋅ (0.3571+0.7143/1.43) = 752.9769 K
Hence the temperature at the (firebrick & insulating)-ordinary brick interface = (1530K - 752.9769 K) = 777.0231 K

The correct answer according to the answer sheet is 778 K, there might be a slight difference but I want to make sure that my computation is right?
 
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  • #3
why is the solution on letter C like that? i tried equating by (1530 - X) / (0.5 / 1.4) = (X - 525) / (0.25 / 0.7) but its not showing the correct answer...
 
  • #4
My heat flow rate equations for the three layers are as follows:
$$Q=1.4\frac{(1530-T_1)}{0.5}$$
$$Q=0.21\frac{(T_1-T_2)}{0.15}$$
$$Q=0.7\frac{(T_2-525)}{0.25}$$
What is the solution to these equations for ##T_1## and ##T_2##?
 

1. How does heat transfer occur at a junction?

Heat transfer at a junction occurs through conduction, convection, and radiation. In conduction, heat is transferred through direct contact between two objects. In convection, heat is transferred through the movement of fluids or gases. In radiation, heat is transferred through electromagnetic waves.

2. What factors affect the temperature at a junction?

The factors that affect the temperature at a junction include the thermal conductivity of the materials, the surface area of contact, the temperature difference between the two objects, and the duration of contact.

3. How is the temperature at a junction calculated?

The temperature at a junction can be calculated using the heat transfer equation, which takes into account the thermal conductivity, surface area, and temperature difference between the two objects. This equation is often represented as Q = kAΔT, where Q is the heat transfer, k is the thermal conductivity, A is the surface area, and ΔT is the temperature difference.

4. What is the difference between steady-state and transient heat transfer?

In steady-state heat transfer, the temperature at the junction remains constant over time, indicating a balance between the heat transfer from one object to another. In transient heat transfer, the temperature at the junction changes over time, indicating an unbalanced heat transfer.

5. How can the temperature at a junction be controlled?

The temperature at a junction can be controlled by adjusting the factors that affect heat transfer, such as the thermal conductivity, surface area, and temperature difference. Additionally, using insulating materials or changing the shape of the objects can also help control the temperature at a junction.

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