Heat Transfer: Calculating heat transfer rate?

In summary, you need to solve for the heat transfer rate for the left scenario where the bottom half of the domain is insulated on the left and right, while the top surface is held at a constant temperature. To do this, you need to solve for the thermal resistance and then use the first two rows of grid points at the top to estimate the temperature gradient variation at the top.
  • #1
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Homework Statement


Given the two domains pictured below, calculate the heat transfer rate (Q [W]) for each case.
http://img6.imagebanana.com/img/83fe4j54/Selection_001.png

The domain on the left is insulated on the left, right, and right half of the bottom side, while the domain on the right is only insulated on the left and right sides. The domain on the left has the left half of the bottom side held at constant Tc and the domain on the right has the entire bottom side held at Tc. Both domains have the top surfaces held at Th.

Homework Equations


[itex]Q=-\frac{kA}{L}\Delta{T}[/itex]

The Attempt at a Solution


I've solved for the temperature distributions numerically; however, I don't know how to calculate the heat transfer rate for the left scenario.

I believe the domain on the right would simply be:
[itex]Q=-\frac{kA}{L}(Th-Tc)[/itex]

But I am not sure how to treat the domain on the left because of the half-insulated base. Any help/hints would be appreciated. Thanks!
 
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  • #2
If you have the numerical solution, use the first two rows of grid points at the top to estimate the temperature gradient variation at the top. Multiply by the thermal conductivity k to get the heat flux variation along the top. Add up all the heat fluxes time the incremental widths along the top the get the overall heat flow rate.
 
  • #3
Chestermiller,

Thank you for your reply. The answer is so obvious one you explained it. I've calculated the two heat transfer rates and they are definitely different (good). I also checked their dependencies on grid resolution and they always converge to the same numbers (good).

Thanks very much for your help.
 
  • #4
Hi

I am trying to solve one problem:

A wall is required to be insulated by embedding wood within concrete. the total thickness of the concrete and wood are 10cm and 1 cm respectively and are fixed due to structural constraints. determine the minimum heat leak into the room if the location of the wood can be changed within the concrete. Take K for wood as 0.1w/mk and for concrete 1.0w/mkI tried solving using thermal resistance concept by equating derivative of the total thermal resistance to 0, but not able to get the answer.

Please help.
 
  • #5
onquest said:
Hi

I am trying to solve one problem:

A wall is required to be insulated by embedding wood within concrete. the total thickness of the concrete and wood are 10cm and 1 cm respectively and are fixed due to structural constraints. determine the minimum heat leak into the room if the location of the wood can be changed within the concrete. Take K for wood as 0.1w/mk and for concrete 1.0w/mk


I tried solving using thermal resistance concept by equating derivative of the total thermal resistance to 0, but not able to get the answer.

Please help.

You should start a separate post for this problem.
 

1. How is heat transfer rate calculated?

The heat transfer rate is calculated by multiplying the thermal conductivity (k) of the material, the temperature difference (ΔT) between the two objects, and the surface area (A) of the material. The formula for heat transfer rate is Q = k x ΔT x A.

2. What is thermal conductivity?

Thermal conductivity is a measure of how well a material can transfer heat. It is defined as the amount of heat (Q) transferred through a material with a thickness (L) in a given time (t), with a temperature difference (ΔT) between the two ends. The formula for thermal conductivity is k = Q x L / (A x ΔT x t).

3. How does temperature difference affect heat transfer rate?

The greater the temperature difference between two objects, the higher the heat transfer rate will be. This is because there is a larger temperature gradient that drives the heat flow from the warmer object to the cooler object.

4. What is the role of surface area in heat transfer rate?

The surface area of a material plays a significant role in determining the heat transfer rate. A larger surface area means that there is more area for heat to flow through, resulting in a higher heat transfer rate. This is why materials with high surface area, such as fins or coils, are often used to increase heat transfer in systems.

5. How can heat transfer rate be increased?

Heat transfer rate can be increased by increasing the thermal conductivity of the material, maximizing the temperature difference between objects, and maximizing the surface area for heat transfer. Other factors that can affect heat transfer rate include the type of material, the medium through which heat is transferred (conduction, convection, or radiation), and the presence of insulation or barriers that may impede heat flow.

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