# Calculate minimum work input of refrigeration cycle

• tezktenr
In summary, The temperature of a 12-oz (0.354-L) can of soft drink is reduced from 20 to 5 ºC by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at 20 ºC to the surroundings. There are no other heat transfers. The minimum theoretical work input required by the cycle, in kJ, assuming the soft drink is an incompressible liquid with the properties of liquid water, is 18.543. To solve this problem, basically what I think was to regard it as a reversible cycle to get the ideal minimum work input for this refrigeration cycle.
tezktenr

## Homework Statement

The temperature of a 12-oz (0.354-L) can of soft drink is reduced from 20 to 5 ºC by a refrigeration cycle. The cycle receives energy by heat transfer from the soft drink and discharges energy by heat transfer at 20 ºC to the surroundings. There are no other heat transfers. Determine the minimum theoretical work input required by the cycle, in kJ, assuming the soft drink is an incompressible liquid with the properties of liquid water. Ignore the aluminum can.

ΔU = Qnet - Wnet
ΔS = Qnet/T + σ

## The Attempt at a Solution

After learning general concepts of 2nd law of thermo, I recently just started to learn entropy in thermodynamics. I am so confused with this new concept and don't know what it actually means so that I am not able to apply it in the questions.

To solve this problem, basically what I think was to regard it as a reversible cycle to get the ideal minimum work input for this refrigeration cycle.

For a refrigeration cycle, β = QL / (QH - QL) = QL / Win.
Also, for an ideal reversible cycle, β = TL / (TH - TL).

I think the hot and cold reservoirs for this refrigeration cycle is
TL = 5 + 273.15 = 278.15 K
TH = 20 + 273.15 = 293.15 K

After substituting all those known temperature, I get β = 18.543.

Now, my idea is to get the value of QL. Then, I can find the minimum work input of a refrigeration cycle.
To do that, I only look at the soft drink (incompressible liquid) system.
ΔU = Qnet - Wnet.
Looking at a system that only consists of the liquid, heat (QL) was taken from it. However, the liquid is incompressible so that the volume does not change. If the volume of the liquid does not change, Wnet for this system should be zero since Wnet = ∫P*dV.
Thus,
ΔU = - QL
So, I can calculate ΔU to find QL.
ΔU = m * cp * (T2 - T1)
T2 = TH = 293.15 K;
T1 = TL = 278.15 K;
m = V/v = 0.354 * 10-3 m3/ 10-3 m3/kg
cp = 4.2.
So, what am I doing wrong?

I knew I have to use entropy but I don't see where to use it. I have the solution.
It uses the equation: ΔS = Qnet/T + σ. To compute the minimum work input, it assumes σ = 0. (That is, assuming it is a reversible process). So far, I understood.
Then, Qnet = ΔS * T + σ * T (In which T = 293.15 and I don't understand why using this temperature here either)
I understand how to calculate ΔS but the following is what I am mainly confused about.
The solution calculates the work by using: Work minimum = Qnet - ΔU.
Since ΔU is the change of internal energy of the liquid (soft drink),
why would the work calculated here be the input work of the refrigeration cycle?
I think the input work of the refrigeration cycle is not directly applied to the liquid but instead to some other components that could help remove heat from the liquid. Am I Wrong? What concepts am I misunderstanding?
Shouldn't this work here be zero because the volume of the liquid doesn't change?

You don’t have to use entropy explicitly - the Carnot efficiency is everything you need from it. You also don’t have to care about volumes. Just use the heat capacity of water (to find its mass you need the volume, okay...).

The cooling system uses some amount of work to move heat from the soft drink to the environment. At the final temperature your calculated ##\beta## tells you how these energies are related. If you use this value, how much work do you have to put in?

You can cool it with less work. When the soft drink is still warmer you can run the machine with a higher efficiency.

But I still don't understand what I am doing wrong exactly in this problem?
I think I did calculated the carnot efficiency β and used the heat capacity to calculate ΔU. Finally, I found QL. By using the β and QL, I computed the work. But it still does not match the correct answer. I don't know why.

See the second paragraph. To cool it from 20 degrees to 19, for example, you can operate the heat pump with a much higher ##\beta##. You underestimate the initial (and average) efficiency.

tezktenr said:
But I still don't understand what I am doing wrong exactly in this problem?
I think I did calculated the carnot efficiency β and used the heat capacity to calculate ΔU. Finally, I found QL. By using the β and QL, I computed the work. But it still does not match the correct answer. I don't know why.
You calculated the work required to move some amount of heat from a heat reservoir at 5 C to a heat reservoir at 20 C. The soft drink, however, isn't a heat reservoir. Its temperature changes during the process, starting at 20 C and decreasing to 5 C. No work is needed to move an infinitesimal amount of heat dQ from the drink to the surroundings when they're both at 20 C, but to move the same amount dQ when the drink is at 19 C does require some work because the drink is colder than the surroundings.

Regarding your questions about the solution, you're confusing the work done on the liquid with the work input to the refrigerator. Qnet is what you called QH (does that clear up for you why they used TH there?), and ΔU is the heat extracted from the drink just like in your calculation.

Vela. Thank you so much! I started to realize that the soft drink is not a constant temperature reservoir.
However, that arouses my another question about entropy (ds = δQ/T). I now realized that I don't really understand this equation at all.
In this questions, why do they plug in T = 293.15 K into ds = δQ/T rather than T = 278.15 K.
What is this T in this equation? (Since I always encountered those system with changing temperature).
Is the T the temperature of the reservoir or the temperature of the system?
What if there are two reservoirs? Then, which temperature should I plug in?
I am sorry if I asked stupid questions. Entropy is one of the concepts I met in physics that I don't know what it actually means.

See above, you don't have to consider entropy if you don't want to.

## 1. How is the minimum work input of a refrigeration cycle calculated?

The minimum work input of a refrigeration cycle is calculated using the equation Wmin = Qc(1-Th/Tc), where Qc is the heat removed from the refrigerated space, Th is the high temperature of the cycle, and Tc is the low temperature of the cycle.

## 2. What is the significance of calculating the minimum work input in a refrigeration cycle?

Calculating the minimum work input in a refrigeration cycle helps determine the most efficient operating conditions for the system. It also allows for comparison between different refrigeration cycles to determine which one requires the least amount of work input to achieve the desired cooling effect.

## 3. How does the coefficient of performance (COP) relate to the minimum work input of a refrigeration cycle?

The coefficient of performance (COP) is the ratio of the desired output (cooling effect) to the required input (work). The minimum work input is inversely proportional to the COP, meaning that a lower minimum work input results in a higher COP and therefore a more efficient refrigeration cycle.

## 4. What factors can affect the minimum work input of a refrigeration cycle?

The minimum work input of a refrigeration cycle can be affected by several factors, including the type and efficiency of the compressor, the properties of the refrigerant being used, the temperature difference between the high and low temperatures of the cycle, and the desired cooling effect.

## 5. Are there any limitations to using the minimum work input to determine the efficiency of a refrigeration cycle?

While calculating the minimum work input can provide valuable insights into the efficiency of a refrigeration cycle, it is important to note that it is a theoretical value and does not take into account practical limitations such as friction, heat transfer losses, and compressor inefficiencies. Therefore, it is important to also consider other factors when evaluating the efficiency of a refrigeration cycle.

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