# Homework Help: Refrigerator (Thermodynamics)

1. Mar 23, 2012

### roam

1. The problem statement, all variables and given/known data

http://img21.imageshack.us/img21/1197/questionof.jpg [Broken]

3. 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 Wmin?

Last edited by a moderator: May 5, 2017
2. Mar 24, 2012

### 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.

Last edited: Mar 24, 2012
3. Mar 24, 2012

### 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=Cp (T1-T2) and the heat added to y is Q+W=Cp (T2'-T1) so

W= Cp(T2'-T1)-Cp(T1-T2)=Cp(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?

Last edited: Mar 25, 2012
4. Mar 25, 2012

### 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 Cp is equal to mc. But how could this be?