# Estimating a thermal equilibrium on a custom made fridge

RiccardoVen
Hello,
I'm trying to build a custom made fridge made by a cube by 120cm on each side.
The material used to isolate the cube will be some polystyrene panels, with thickness s=4cm.
Let's imagine to cool the dry air inside in order to reach the internal temperature of 8 degree Celsius, while the outside air
is 18 degree Celsius.
From a given time t0 we stop the cooling, leaving the internal air to reach the thermal equilibrium with the outside.
The aim is to estimate the time needed by the internal air to reach 18 degrees.
This is how I reasoned:

Q = Deltat x P, where:

Q = heat needed to heat the air from 8 to 18 degrees
P = heat power sucked into the fridge in the Deltat time

Q could be estimated like this:

Q = cp x phoair x V x DeltaT = 1005 x 1.225 x (1.2)3x(18-8) = 21273.84 Joule

Keeping K = 0.04 W/mK the polystyrene thermal conductivity, we can get the heat power using Fourier's law over the the overall
total cube area, i.e.:

P = (KxAxDeltaT) / s = (0.04x(1.44x6)/0.04)x(18-8)) = 86.4W

So the final Deltat will be:

Deltat = Q / P = 246.225s = 4.1m

Some assumptions done: cp and K have been considered constant during the whole temperature transition.
The fridge does not touch the floor, but it's kept detached from it through some small blocks or so.

Is the reasoning I'm doing correct, please?

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Homework Helper
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I don't know about your calculations, but no refrigerator increases its internal temperature in 10 degrees Celsius in 4 minutes while the cooling system is off.

Homework Helper
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Nearly correct. The 86.4 W is in fact the power needed to maintain the desired temperature difference in steady state.
The warm up time is actually a little tricky because mathematically it takes forever to actually completely warm up to the outside T. Your calculation gives you the "time constant" for the process which is exponential. I think that is the correct number (?) but it is not very relevant because anything substantial inside will change that number a lot.

RiccardoVen
I don't know about your calculations, but no refrigerator increases its internal temperature in 10 degrees Celsius in 4 minutes while the cooling system is off.
Probably a normale fridge has some better isolation than a set of polystyrene of 4 cm. We are talking here abour formulas and Physics, so please motivate tour corcerns by that. Thanks

RiccardoVen
Nearly correct. The 86.4 W is in fact the power needed to maintain the desired temperature difference in steady state.
The warm up time is actually a little tricky because mathematically it takes forever to actually completely warm up to the outside T. Your calculation gives you the "time constant" for the process which is exponential. I think that is the correct number (?) but it is not very relevant because anything substantial inside will change that number a lot.
Thanks for your replay, It makes Sense to me. I know the real temperature profile Will be exponential. Not clear when you claim "Will change that Number a lot"

RiccardoVen
Thanks for your replay, It makes Sense to me. I know the real temperature profile Will be exponential. Not clear when you claim "Will change that Number a lot"
I think It shoukd be probably the proper time to reach 62% of the desired deltat

• hutchphd
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The heat capacity of air is tiny. Probably the heat capacity of the insulation polystyrene is bigger than that of air enclosed. Unless you just want a box of cold air then the number is irrelevant.

• sophiecentaur and Bystander
RiccardoVen
Nearly correct. The 86.4 W is in fact the power needed to maintain the desired temperature difference in steady state.
The warm up time is actually a little tricky because mathematically it takes forever to actually completely warm up to the outside T. Your calculation gives you the "time constant" for the process which is exponential. I think that is the correct number (?) but it is not very relevant because anything substantial inside will change that number a lot.

OK, thanks to your suggestion I probably approximating better the problem. As you claimed, what I found is the time constant of the problem, not the real solution. That said we can claim:

T(t) = T1 + (T2-T1)(1 - e^(-t/CT))

where CT is the time constant you are talking about. Specifically, this CT is the time needed to have a variation of (1 - e^-1) (t2-t1)= 0.632DeltaT
Replacing it, we get the T at 246s, i.e.:

T(t) = 8 + 10(1 - e^-1) = 14.32 degree

Since the system will reach 18 degree in a infinite time, we could try to get the t to reach 17 degree, for instance.

17 = 8 + 10(1 - e^(-t/CT)) -> 0.9 = 1 - e^(-t/CT) -> 0.1 = e^-(t/CT) -> ln 0.1 = -t/CT

-2.302 = -t/246.225 -> t= 566.8s = 9.5m

RiccardoVen
The heat capacity of air is tiny. Probably the heat capacity of the insulation polystyrene is bigger than that of air enclosed. Unless you just want a box of cold air then the number is irrelevant.
Currently this is an "ideal" fridge, with nothing inside so, yes we are seeking here for a box of cold air inside. This number we are looking for (see my post below: 4m is the time to reach the 62.3% of deltaT, so roughly 6 degrees) is the number the cooler would restart to cool again down to 8 degrees. So it seems, if we set the threshold to 14 degrees (which is a lot) the fridge would restart every 4m, which is really a lot. Probably the real fridges are better insulated than this, using more than 4cm panels. Also, in a real situation we'd put some meat or vegetables inside, so these ones would increase the heat latency and the overall constant time would increase significantly.

Mentor
You omitted the thermal inertia of the 4 cm thick walls, which is much greater than the air inside. Also omitted is the convective heat transfer resistance between the outside air and the walls.

• sophiecentaur, Lnewqban and hutchphd
RiccardoVen
The heat capacity of air is tiny. Probably the heat capacity of the insulation polystyrene is bigger than that of air enclosed. Unless you just want a box of cold air then the number is irrelevant.

You omitted the thermal inertia of the 4 cm thick walls, which is much greater than the air inside. Also omitted is the convective heat transfer resistance between the outside air and the walls.
Can you help in including them in my computations, please? However even putting 3l of Waters inside increased the Tau to 28m

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What is it specifically you wish to calculate?

RiccardoVen
What is it specifically you wish to calculate?
As you can sere from my original post: The aim is to estimate the time needed by the internal air to reach 18 degrees.

Actually that time Is infinite so let's keep It real: when I stop the cooling, how Much the air inside Will take to go from 8 to 10 degrees?

As told above putting 3liters of water inside Will raise those 4m to 28m, a bit more real

Mentor
You basically have a slab of polystyrene 4 cm thick with an initial temperature profile varying linearly from 8 C on one side to 18 C on the other side. Transient heat conduction is occurring within the slab while convective heat transfer is occurring on the high temperature side from air at 18 C to the outside wall temperature (which will drop below 18 C as time progresses, and then rise again at the end). On the inside face of the slab, one can assume zero heat flux. The solution to this problem is probably in Carslaw and Jaeger, Conduction of Heat in Solids.

• Lnewqban
Homework Helper
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As told above putting 3liters of water inside Will raise those 4m to 28m, a bit more real
You can solve the slab problem as @Chestermiller suggests. That is an interesting and useful exercise, good knowledge to have, and I would recommend it.
But if your purpose is to actually design a fridge you need to worry about the stuff that drives the design.

RiccardoVen
You can solve the slab problem as @Chestermiller suggests. That is an interesting and useful exercise, good knowledge to have, and I would recommend it.
But if your purpose is to actually design a fridge you need to worry about the stuff that drives the design.
As told I've Just 3l of water inside making It a bit more real and raising the Tau to 28m. I Need an approximation, not the real temperature profile. However I Will try with your suggestions. Thanx

Homework Helper
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What does the warmup/cooldown time have to do with the design? Seems not important to me . Incidentally I run my fridge at 4C.

RiccardoVen
On the inside face of the slab, one can assume zero heat flux
Why we can assume this on the internal face? How the internal is heated to reach 18 degrees if not heat flux is there to do that?

RiccardoVen
What does the warmup/cooldown time have to do with the design? Seems not important to me . Incidentally I run my fridge at 4C.
We had in mind to estimate how much the cooling system will be stressed to be activated on a given DeltaT. So if it has to switch on every 4m is really too much and the isolation would not be enough. But putting some water inside shows the Tau is actually bigger, since the air does not contain enough "cold" to be heated for a longer time. Probably considering the missing thermal inertia and convection on the outside would make this Tau a bit more relaxed and precise.

RiccardoVen
I finally lend to some good or at least "meaningful" results with my simplified toy model.
Let's assume to put 500Kg of potatoes inside the fridge.
The cp for the potatoes is cp=4184x0.84=3514.56 J/Kg degree C
So, to cool down 500Kg of potatoes from 18 to 8 degrees we would take:

Q = 3514.56 x 500 x 10 = 17572800 J

The sucked heat power is still the same, i.e. 86.4W, so to estinguish those Q we would take:

Tau = 17572800 / 86.4 = 56.49 hours

This is the Tau to reach 62.3% of DeltaT = 10, i.e. 6.2 degrees

So the potatoes would reach roughly 14 degrees after 56.5 hours

Using the exponential law to reach 10 degrees from 8 above we would get:

10 = 8 + 10x (1 -e^(-t/Tau)), i.e.:

t = -Tau x ln 0.8 = 12,6 hours

Which is a meaningful amount, approximating the "reality" of the process, modulo thermal resistance of the slab and convection on the outside.

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Mentor
Why we can assume this on the internal face? How the internal is heated to reach 18 degrees if not heat flux is there to do that?
If it only contains air internally, that is negligible compared to the thermal inertia of the walls.

RiccardoVen
If it only contains air internally, that is negligible compared to the thermal inertia of the walls.
Actually the air is also on the external part of the slab. So, why we should take it into account on the external side and not on the internal part? Thanks

Mentor
Actually the air is also on the external part of the slab. So, why we should take it into account on the external side and not on the internal part? Thanks
We are only interested in the time it takes the slab to reach 18 C. The bulk outside are is already at 18 C, and its effect is taken into account with the outside heat transfer coefficient (which is part of the slab analysis).

RiccardoVen
We are only interested in the time it takes the slab to reach 18 C. The bulk outside are is already at 18 C, and its effect is taken into account with the outside heat transfer coefficient (which is part of the slab analysis).
OK thanks