Thermodynamics heat loss fridge

  • #1
Hello

To calculate the heat loss of a fridge, do I need to take the surface area of the in- or outside of the fridge?

Heat loss formula = k * delta T * A

k = heat transmission coefficient
A = surface area
 

Answers and Replies

  • #2
Bystander
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What's the uncertainty in the value of "k?" In ΔT? In Ain? In Aout? In Agasket? What are contributions at the edges for both inside and outside cases? And, how small an uncertainty can you stand in your calculation?
 
  • #3
Let's just say there is no uncertainty, because the values are given. If you take the in- or the outside surface area, it has a large impact on the outcome. Especially with thick walls. So my question is, in this formula, should the in- or the outside surface area be used?
 
  • #4
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Let's just say there is no uncertainty, because the values are given. If you take the in- or the outside surface area, it has a large impact on the outcome. Especially with thick walls. So my question is, in this formula, should the in- or the outside surface area be used?
If you feel that it really matters that much, then you can't use the 1D heat conduction equation. A 1D analysis of the heat conduction will not not be adequate. You will need to solve the 2D heat conduction equation within the walls of the frig.

Chet
 
  • #5
russ_watters
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Since k is a rolled-up constant that includes a number of factors, such as internal and external convection and internal conduction, the answer to what k means must be provided by whomever provided k.
 
  • #6
OmCheeto
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Let's just say there is no uncertainty, because the values are given. If you take the in- or the outside surface area, it has a large impact on the outcome. Especially with thick walls. So my question is, in this formula, should the in- or the outside surface area be used?
Neither.
The corners complicate things to the point to where I don't think even the guru's of the highest levels of PF's maths forums would spend their time on such a problem.
Though, I may attempt a solution in the morning.
But don't hold your breath.

I did earlier google "heat transfer rate through an isosceles trapezoidal object", but got very bored, as, after 10 minutes of scrolling, it appeared that no one in the history of the planet has attempted to solve the problem.
 
Last edited:
  • #7
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Heat transfer problems in complicated geometries are routinely handled using finite element (numerical) analysis. Even with finite difference analysis, heat transfer in an isosceles trapezoidal object seems pretty simple if the trapazoid is mathmatically mapped onto a rectangle by transformation of the spatial independent variables.

Chet
 
  • #8
OmCheeto
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Heat transfer problems in complicated geometries are routinely handled using finite element (numerical) analysis. Even with finite difference analysis, heat transfer in an isosceles trapezoidal object seems pretty simple if the trapazoid is mathmatically mapped onto a rectangle by transformation of the spatial independent variables.

Chet
I stand corrected.
But what for you "seems pretty simple", would take me most of tomorrow to figure out.
Or were you going to use software for the analysis? I was going to do it by hand.

ps. mrquestion123, do not bother holding your breath.
 
  • #9
Thank you all for your replies,

@MR chestermiller it seems pretty complicated to do.a 2d analysis.
@Russ waters, yes maybe i should just ask the questioner. The book writes K is transport through a flat wall with a surface area, which does not answer my question yet. The previous question is to calculate the interior volume (for heat loss opening the door) and the exterior size, which why I assume they mean the outside surface area should be used.
@OmCheeto, hmm they do not expect me to calculate the corners. This should be covered with the in- or outside surface area.
 
  • #10
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Thank you all for your replies,

@MR chestermiller it seems pretty complicated to do.a 2d analysis.
@Russ waters, yes maybe i should just ask the questioner. The book writes K is transport through a flat wall with a surface area, which does not answer my question yet. The previous question is to calculate the interior volume (for heat loss opening the door) and the exterior size, which why I assume they mean the outside surface area should be used.
@OmCheeto, hmm they do not expect me to calculate the corners. This should be covered with the in- or outside surface area.
If the heat flow in the corners is negligible, then you use the inside area.

Chet
 
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