# Fridge warms up a room

1. Sep 13, 2014

### franceboy

Hi there, i have some problems with the following homework. It would be nice, if you help me :)
1. The problem statement, all variables and given/known data

A well isolated house is located in a region where the sun does not shine and the temperature is constantly about 5°C outside. The house is empty except a full fridge with the constant temperature of -18°C. After a while the house has stabile temperature of 6.5°C. The fridge shall work like an ideal Carnot cooling-machine.
If you turn off the fridge and put it outside, it warms up. After 15 minutes it has a temperature of -12.4 °C and after 30 minutes -8.1°C. The heat capacity of the fridge is C= 80 J/K.

a) Determine the stabile temperature of house if you would have two fridge like in the text above.
b) Calculate the elictrical power in both cases.
c) Determine the maximal temperature of the house, if you would use a bigger fridge with same temperature, kind of working and isolation.

2. Relevant equations
3. The attempt at a solution

Okay, my idea was that the fridge is heated by the room like it is heated ouside (proportional to the difference of temperature PH->F= k*(TH-TF). Since the fridges temperature keeps equal it is cooled all the time, while heating the room.
However, now i was confused: the text says that the house is well-isolated (I thought this means no energy transfer) but then the temperature would increase without stopping because the fridge heats more than it cool (second law). This means the house must be cooled by the outer world like it is heated by the fridge. Is that right?

2. Sep 13, 2014

### franceboy

formulas

My idea was:
PH->F= Pcool
PF->H= g*Pcool and g can be calculated with the formula TF\(TH-TF)= Pcool\(PF->H-Pcool)
The last equation shall be PF->H-PH->F=PH->W
with H=House, F=Fridge, W=World.
This is the case for one fridge.
If you ust to two identical fridges PH->F will be doubled and then you can solve a system of equations if we guess H->W is proportional to the difference of temperature.

b) second law: Wel=PF->H-Pcool in both cases

c) no idea

I hope you understand my bad english and can help me :)

3. Sep 13, 2014

### Staff: Mentor

Hi franceboy,

Perhaps you can begin by determining the thermal resistances (or conductances)? What information was given that would let you find the fridge's thermal resistance? How about the house's?

4. Sep 14, 2014

### franceboy

Thank you gneill
We assume that thermal conduction is the unique kind of energy transfer.
When the fridge has been outside, the process can be described by TF(t)=a*(TW-TF(t)). The solution of this equation is TF(t)=TW+(TF(0)-TW)*e-a*t.
Using the given information,we determine a:a≈0.02/min.
Secondly, we assume that this process is the same as in the house. As a result we get the following equation: PH->F=a*C*(TH-TF=Pcool.
Since the fridge is a Carnot-machine, we can determine PF->H=((TH-TF)/TF+1)*PH->F.
The room is heated by power of PF->H-PH->F. The room has to be cooled by this power. That means:PF->H-PH->F=PH->W= k*(TH-TW). However, I am unsure about the last equation. k would be something like 2.51 W/K. Is that right?

5. Sep 14, 2014

### franceboy

Correction: k=2.51 J/(min*K)

6. Sep 14, 2014

### Staff: Mentor

Hmm. If we define the thermal conductance as the heat flow in watts divided by the temperature difference, then my tinkering with the numbers would indicate it should end up somewhere in the ballpark of 0.4 W/K, or ~ 24 J/(min*K). Its reciprocal, thermal resistance, would be around 2.4 K/W (values approximate!).

Of course, the values depend upon what you obtained for the rate of heat loss (equal to that dumped into the room by the fridge's work).

7. Sep 14, 2014

### franceboy

How did you calculate this value?
And what's about the conductance of the house?
And how do you determine the temperature in a)?
If my calculation is wrong i do not have any ideas how to solve this exercise...

8. Sep 14, 2014

### Staff: Mentor

I presume you're referring to the thermal resistance associated with the fridge? I did as you did, fit an exponential decay to the given data. The time constant in seconds is the thermal resistance multiplied by the thermal capacitance (just like in electrical circuits!).

You're given an equilibrium scenario. If you've calculated the heat generated by the heat pump in the fridge then you know the outgoing heat flow and the temperature difference, right?

I haven't looked at that. You'll have to provide an attempt first.

Can you show your calculation in detail? Formula then numbers.

9. Sep 15, 2014

### franceboy

my calculation

1. The fridge
Since the fridge was turned off and the sun does not shine, we can use Newtons law:
TF(t)=a*(TW-TF(t))
The solution of this differential equation is:
TF(t)=TW+(TF(0)-TW)*e-a*t
Then we have two eqations:
TF(15)=-12.4°C and TF(30)=-8.1°C if we measure t in min. Besides,
TW=5°C and TG(0)=-18°C
I solved both equations with a GTR and both times the solution is a=0.019/min≈0.02/min

Now we determin PW->F(t)=Q(t)=(C*(TF(t)-TF(0)))`=C*a*(TW-TF(t)).

2.The house
We assume that PH->F=a*C*(TH-TF). This must be the cooling power of the fridge because the temperature keeps constant. Since the fridge shall be a Carnot-machine we can use following equation:
PF->H=((TH-TF)/TF+1)*PH->F
Then the room is heated by the power of PF->H-PH->F. The temperature TH is stabile that means the room must be cooled by this power, too.
PF->H-PH->F=PH->W=k*(TH-TW)
(TH-TF)/TF*a*C*(TH-TF)= k*(TH-TW).
Using TH=279.5K and TW=278K and C=80J/K we get k=2.51J/(K*min)

3. a)
I only replaced C by 2*C in the last equation and calculated the new TH=8.5°C. However this is a very unlogical solution because the increase of temperature would be higher...

10. Sep 15, 2014

### Staff: Mentor

I'm seeing something just a bit smaller than 2.51 for that value. So apparently we have slightly different values for the net heat produced by the operation of the fridge. What were your numbers for the House to Fridge and net Fridge to House?

Yeah. What you have is a new equilibrium condition to solve for. The greater temperature in the house will mean that the heat pump needs to work harder. I tend to look at these problems the way I look at an electrical circuit, with a controlled current source representing the heat input that the pump generates. For the single fridge problem where the temperatures are given, something like this:

Where $\alpha$ represents the Carnot efficiency term. IF depends on the temperature difference between the house and fridge, and so does $\alpha$.

Perhaps the diagram will give you an idea of how to treat adding another identical fridge?

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11. Sep 15, 2014

### franceboy

Yeah my values are:
PH->F≈39.2J/(K*min)
PF->H≈43J/(K*min)
Pnet≈3.8J/(K*min)

I did not really get how to apply the comparison with electrical circuits. Could you give me some more advice? Or the new equation?

b) I think b should be solved by the first law of thermodynamics: W=PF->H- PH->F

12. Sep 15, 2014

### Staff: Mentor

I think the units on your heat flows should be just J/min.

My numbers are slightly different.

I solved the exponential decay of the outdoor fridge for both given times and found the thermal resistances 40.3 K/W and 40 K/W. I figured the second one was more accurate since it was over a longer time period, so declared the fridge's thermal resistance to be 40 K/W. That translates to a conductance of 1.50 J/(K*min). I tend to use thermal resistance rather than conductance. Just my preference.

Then for the House to Fridge heat current I used $I_F = ΔT/R$ for the given stable condition, finding that $I_F = 0.613 W$ (or 36.8 J/min). Then applying the Carnot efficiency to find the new heat introduced that turned out to be 58.8 mW or 3.53 J/min. That number corresponds to your $P_{net}$ for which you found 3.8 J/min. Not sure where the difference in values comes from!

It's the 58.8 mW that must be lost from the house to the outside. So using R = ΔT/I with a heat current of 58.8 mW and the given ΔT yields a house thermal resistance of 25.5 K/W. That's a conductance of 2.35 J/(K*min).

With the two Thermal resistances in hand the the values can be penciled into the thermal "circuit" and then nodal analysis can be used to solve for temperatures. There's only one node that has a temperature that's free to vary and that's the house temperature (it's assumed that the fridge regulates its temperature and the outside temp is stated to be constant).

Changing the circuit to take into account a second identical fridge is not difficult. Keep in mind the "trick" that nodes at the same voltage (temperature) can be joined together into a single node... no "current" will flow through a wire between nodes at the same voltage (temperature). What matters is the total resistance (conductance) between the nodes...

Write the node equation for the "house node" using the usual methods. Here temperature corresponds to voltage and heat flow in watts corresponds to current.

Keep in mind that the two current sources are controlled sources. The value of $\alpha$ is given by the temperature ratio term in your Carnot efficiency expression, and the $I_F$ value is taken from the branch current: $I_F = \frac{T_H - T_F}{R_F}$ using the variables on the diagram.

13. Sep 15, 2014

### franceboy

If we add another fridge, will it be a series or parallel circuit?
If it is a series circuits, can we just add the thermal resistances using the same formulas?
Is my attempt for b) right?
It is so nice of you that you help me :)

14. Sep 15, 2014

### Staff: Mentor

You're welcome. I'm happy to help.

Well, it's another heat conduction path from the house to the fridge. So must be parallel. You can sum the thermal conductances or add the thermal resistances in parallel (just like resistors). Since both fridges have the same internal temperature they can share the one node in the circuit. The heat pump as it is will handle the total heat pumping requirement (its "current" will just be higher).

I'm not 100% certain, but I believe that since the heat pump is ideal the total electrical energy supplied must be exactly what leaves the house as heat.

No worries!

15. Sep 16, 2014

### franceboy

Hi
I solved the new (2*a instead of a) equilibrium equation for the house temperature. I got 8.5 C again. Is that right? It sounds unlogical...

16. Sep 16, 2014

### Staff: Mentor

It seems to be a reasonable value to me. You'll have to show your work so we can see if it's illogical or not.

By the way, I see a similar value when I solve it using the model presented above. I also simulated it with LTSpice and obtained the same result...

17. Sep 16, 2014

### franceboy

Okay,
PF->H-PH->F=PH->W
(TH-TF)/TF*2*a*C*(TH-TF)=k*(TH-TW)
I solved this equation with my calculator and got TH=8.5°C

However, the increase of temperature from using a fridge to two fridges is higher than from no fridge to one fridge. Therefore c) does not make sense because the temperature would increase without end using a very big fridge, wouldn't it?

18. Sep 16, 2014

### Staff: Mentor

I see what you're concerned about. But this isn't a linear situation. The Carnot efficiency changes as the temperature difference changes. The pump has to work harder as the temperature difference increases, compounding the problem.

If you write out the node equation you'll find a term where the Carnot efficiency term (which depends upon the temperature difference between the house and fridge) multiplies the temperature difference between the house and fridge. That makes the problem a quadratic.

19. Sep 16, 2014

### franceboy

Yeah but the Carnot efficiency is in my equation (TH-TF)/TF which depends upon the difference the temperature difference. So I solved the equation for different conductances in the form of n*a, but T seems increase without end... I thought the node equation in post 14 is right...

20. Sep 16, 2014

### Staff: Mentor

It looks like the model goes critical after n=4; the heat pump cannot maintain the fridge temperature as the room temperature is too high and getting higher from the efforts of the pump. A runaway situation.