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Thermodynamics: help regarding the coefficient of heat transfer

  • #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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Answers and Replies

  • #2
stockzahn
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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
stockzahn
Homework Helper
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498
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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?
 
  • #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
stockzahn
Homework Helper
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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
 
  • #7
HI!

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

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