Refrigerator (Thermodynamics)

[roam]

Homework Statement

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The Attempt at a Solution

For the first part, if we assume that the cycle is reversible from the 2nd law we have

$\Delta S = \frac{\Delta Q}{T}+S_{gen} = 0$

And here is the Clasius Inequality

$S_{gen} \geq 0 \implies \frac{\Delta Q}{T} \leq 0$

$\frac{Q_x}{T_1}-\int^{T_2}_{T_1} \frac{\Delta Q}{T} \leq 0$

Since ΔQ = dT

$\frac{Q_x}{T_1}-\int^{T_2}_{T_1} \frac{dT}{T} = \frac{Q_x}{T_1}- \ln \frac{T_2}{T_1} \leq 0$

$T_2 \geq T_1e^{\frac{Q_x}{T_x}}$

Since we do not know the temprature values we can't make numerical calculation of the lower limit for the final temperature of y. But did I derive the correct equation?

For the second part of the equation, to find the minimum value of work that needs to be supplied to the heat pump I tried to use the 1st Law;

Q − W = U = 0 → W = mcΔT - U

How can we use this to show the above expression for W_min?

 

[roam]

We do not know if the cycle is reversible or irreversible. But I assumed that it is reversible because if it is not then more and more work is required due to the entropy generated.

 

[roam]

Okay, I figured out part (1) , but I'm still having some trouble working out the minimum work.

The entropy change of x is

$\Delta S_x = \int^{T_2}_{T_1} C_p \frac{dT}{T} = C_p \ln \frac{T_2}{T_1}$

The entropy change of y is

$\Delta S_y = \int^{T_2’}_{T_1} C_p \frac{dT}{T} = C_p \ln \frac{T_2’}{T_1}$

The total entropy change is

$\Delta S_{tot} =\Delta S_x + \Delta S_y = C_p \ln \frac{T_2}{T_1} + C_p \ln \frac{T_2’}{T_1}$

And we know from the entropy principal that ΔS ≥ 0, so

$\left( C_p \ln \frac{T_2}{T_1} + C_p \ln \frac{T_2’}{T_1} \right) \geq 0$

$C_p \ln \frac{T_2T_2’}{T_1^2} \geq 0$

The minimum value of T₂’ would be

$C_p \ln \frac{T_2T_2’}{T_1^2} = 0 = \ln 1$

$\therefore T_2’ = \frac{T_1^2}{T_2}$

Okay now the minimum work:

Heat removed from x to cool it is Q=C_p (T₁-T₂) and the heat added to y is Q+W=C_p (T₂'-T₁) so

W= C_p(T₂'-T₁)-C_p(T₁-T₂)=C_p(T'₂+T₂-2T₁)

Substituting the minimum temprature we found we end up with

$W_{min} = C_p \left( \frac{T_1^2}{T_2} +T_2-2T_1 \right) = C_p \frac{(T_1-T_2)^2}{T_2}$

But the expression I need is:

$W_{min} = \frac{mc (T_1-T_2)^2}{T_2}$

What could we do to arrive at that expression?

 

[roam]

The only way I can have:

$W_{min} = C_p \frac{(T_1-T_2)^2}{T_2} = \frac{mc (T_1-T_2)^2}{T_2}$

is if C_p is equal to mc. But how could this be?

 

[asteriatic]

I would like to provide a response to the content by first acknowledging that the problem presented is a classic example of refrigeration using the principles of thermodynamics. The goal of a refrigerator is to transfer heat from a cold reservoir (the inside of the fridge) to a hot reservoir (the outside of the fridge) in order to maintain a lower temperature inside.

In the first part of the problem, the approach taken to derive the expression for the minimum final temperature is correct. The use of the second law of thermodynamics and the Clausius inequality is appropriate and leads to the correct expression. The final result shows that the minimum temperature of the cold reservoir (y) is dependent on the initial temperature (x) and the amount of heat transferred (Qx) from the hot reservoir. This is a fundamental result in refrigeration and is known as the Carnot efficiency.

In the second part of the problem, the approach taken to find the minimum work needed to be supplied to the heat pump is not entirely correct. The first law of thermodynamics states that the change in internal energy (ΔU) is equal to the heat added to the system (Q) minus the work done by the system (W). In this case, the internal energy does not change (ΔU = 0) and therefore the work done by the system is equal to the heat added (W = Q). However, this does not necessarily give us the minimum value for work.

To find the minimum value for work, we can use the Carnot efficiency equation derived in the first part of the problem. This equation shows that the minimum work needed is dependent on the temperature difference between the hot and cold reservoirs. Therefore, to minimize the work, we need to maximize the temperature difference. This can be achieved by using a heat pump with a high temperature difference and efficient insulation to reduce heat loss.