Thermodynamics: help regarding the coefficient of heat transfer

In summary, the conversation revolves around solving a problem involving the calculation of heat transfer through a concrete wall in a building. The equation ## \dot{Q} = \frac{k}{s} A \Delta T## is used, but there is also a need to calculate the thermal resistance between the wall and the air at the inside and the outside to obtain the heat transition coefficient. The conversation also includes suggestions and corrections for the calculations.
  • #1
Hi!

Basically I have some tasks in thermodynamics calculating the heat transfer through a wall.

I have been attempting to solve the following problem:

“Calculate the heat transfer through the exterior wall in a building, where the temperature inside the building is 20 °C and the temperature outside the building is 5 °C. The wall is concrete, and is 4 m high and 8 m long, and have a thickness of 200 mm. The thermal conductivity for the concrete wall is 0,40 W/(mK), the coefficient of heat transfer at the wall inside is 1,2 W/(m2K) and the coefficient of heat transfer at the wall outside is 1,9 W/(m2K).”

I have attempted the following:

q = (k / s) A dT

= U A dT

= [(0,40 W/(mK) / (0.2 m)] [(4 m) (8 m)] [(20 °C) - (5 °C)] I'm not sure where to begin and what to add into the calculations. The two coefficient of heat transfer confuses me. How do I solve this?

Help is appreciated, thanks.
 
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  • #2
Hi,

first of all, please use the homework template and do not delete or change it in your posts - thanks.

However, your first equation ## \dot{Q} = \frac{k}{s} A \Delta T## basically is correct, but only describes the heat transfer due to conduction in the wall. You also have to calculate the thermal resistance between the wall and the air at the inside and the outside to obtain the heat transition coefficient. Like in electrical engineering you can think of the three thermal resistances (air/wall - wall - wall/air) as a series of resistances. Do you know how to calculate the total resistance consisting of several resistances in series?
 
  • #3
Hi

Thanks for the reply. Yes I do. But how to apply two coefficient of heat transfer still confuses me. I am not sure how to do it.
 
  • #4
Ian von Hegner said:
Hi

Thanks for the reply. Yes I do. But how to apply two coefficient of heat transfer still confuses me. I am not sure how to do it.

The approach is like this:

1) Formulate the 3 thermal resistances (the inverse value of the conductances)
2) Sum them up to obtain the total resistance (which is the inverse value of the total conductance or the heat transition coefficient respectively)
3) Plug the total confuctance into your formula

So try to find the expressions for the three resistances ##R_1## - ##R_3##. Any suggestions/ideas?
 
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  • #5
Hi!

Thanks for the reply.

Since I shall find the heat transfer through the wall, and have two coefficient of heat transfers, then this might be the solution:

Q = (TA – TB) / (1/h1A) + (Δx/kA) + (1/h2A)

= (20 °C – 5 °C) / ((1/1,2 W/m2K x 32 m2) + (02 m/0,40 W/mK x 32 m2) + (1/1,9 W/m2K x 32 m2))

= (15 °C) / ((0.023 W/m2K) + (0.016 W/mK) + (0.016 W/m2K))

= (15 °C) / ((0.023 W/m2K) + (0.016 W/mK) + (0.016 W/m2K))

= 15 °C / 0.055 W/m2K

= 15 °C / 0.055 W/m2K

= 272, 7.
 
  • #6
Your calculation is almost correct, I obtain a slightly different value ##\dot{Q}=258.11\;W##. First of all, I apparently used more significant digits. Secondly your value for ##\frac{1}{h_{in}}## seems to be wrong. Please try to correct it and see if we get the same result.

Two more comments:

1) It would be easier to keep the area of the wall above the fraction bar:

$$\dot{Q}=\frac{A\Delta T}{\frac{1}{h_{in}}+\frac{s}{k}+\frac{1}{h_{out}}}$$

2) Writing the units/dimensions of the different values is a good idea, but you also have to check them for conistency. Unfortuanately your units are wrong starting from the third line downwards
 
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  • #7
HI!

I got the same results now, using both approches. Thank you for the help.
 
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  • #8
You're welcome
 

1. What is the coefficient of heat transfer?

The coefficient of heat transfer is a measure of how easily heat can transfer from one object to another. It is a value that represents the rate of heat flow per unit area per unit temperature difference. It is commonly denoted as 'k' and has units of watts per meter per kelvin.

2. How is the coefficient of heat transfer calculated?

The coefficient of heat transfer can be calculated using the formula k = (q / AΔT), where q is the amount of heat transferred, A is the surface area, and ΔT is the temperature difference between the two objects. The value of k can also be experimentally determined using various methods and equipment.

3. What factors affect the coefficient of heat transfer?

The coefficient of heat transfer is affected by several factors, including the thermal conductivity of the materials in contact, the surface area, the temperature difference, and the presence of any insulating materials. It can also be influenced by external factors such as air flow or pressure differences.

4. Why is the coefficient of heat transfer important in thermodynamics?

The coefficient of heat transfer is an essential concept in thermodynamics because it helps us understand how heat is transferred between objects. It is used in calculations for various thermodynamic processes and is crucial in designing and optimizing heat transfer systems, such as heat exchangers, engines, and refrigeration systems.

5. How can the coefficient of heat transfer be improved?

To improve the coefficient of heat transfer, one can increase the thermal conductivity of the materials in contact, increase the surface area, or decrease the temperature difference between the objects. The use of insulating materials can also help reduce heat loss and improve the overall coefficient of heat transfer in a system.

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