Coefficient of performance of refrigerator

In summary, the conversation discusses the calculation of the coefficient of performance for a refrigerator operating on 0.850 mol of H2. The gas is assumed to be ideal and the process ab is isothermal. The correct formula for K is found to be |QC|/( |QH| - |QC| ), where QC is positive and QH is negative.
  • #1
Elias Waranoi
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Homework Statement


The pV-diagram in Fig. P20.51 (See attached file) shows the cycle for a refrigerator operating on 0.850 mol of H2. Assume that the gas can be treated as ideal. Process ab is isothermal. Find the coefficient of performance of this refrigerator.

Homework Equations


K = QC/|QH - QC|
pV = nRT
Cp = Cv + R
Isothermal expansion: Q = W = ∫p dV = nRT*ln(V2/V1)
Constant volume: Q = nCvΔT
Constant pressure: Q = nCpΔT

The Attempt at a Solution


QC = Qa-b + Qb-c = nRTa * ln(Vb/Va) + nCv(Tc - Tb) = paVa * ln(Vb/Va) + VbCv(pc - paVa/Vb)

QH = Qc-a = n(Cv + R)(Ta - Tc) = pc(Cv + R)(Va - Vc)/R

My answer K = QC/|QH - QC| = 0.462
Correct answer K = 6.23

What did I do wrong?
 

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  • #2
Elias Waranoi said:
QC = Qa-b + Qb-c = nRTa * ln(Vb/Va) + nCv(Tc - Tb) = paVa * ln(Vb/Va) + VbCv(pc - paVa/Vb)
Did you leave out R somewhere in the last term on the right side?

QH = Qc-a = n(Cv + R)(Ta - Tc) = pc(Cv + R)(Va - Vc)/R
Does this give a postive or negative value for QH? In the formula K = QC/|QH - QC|, is QH considered as positive or negative?
 
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  • #3
QC = paVa * ln(Vb/Va) + VbCv(pc - paVa/Vb)/R

QH is heat the leaves the working substance so it should be negative. Apparently K = |QC|/( |QH| - |QC| ), thanks for the help I got the correct answer now.
 

1. What is the coefficient of performance (COP) of a refrigerator?

The coefficient of performance of a refrigerator is a measure of its efficiency in removing heat from inside the refrigerator and transferring it to the outside environment. It is defined as the ratio of the heat removed from the refrigerator to the work required to remove that heat.

2. How is the COP of a refrigerator calculated?

The COP of a refrigerator is calculated by dividing the heat removed from the refrigerator (Qc) by the work required to remove that heat (W). This can be expressed as COP = Qc/W.

3. What factors affect the COP of a refrigerator?

The COP of a refrigerator can be affected by various factors such as the design and efficiency of the compressor, the type of refrigerant used, the temperature difference between the inside and outside of the refrigerator, and the insulation of the refrigerator.

4. What is a good COP for a refrigerator?

A good COP for a refrigerator would depend on the specific model and its intended use. Generally, a COP of 3 or higher is considered to be efficient, meaning that for every unit of energy used, the refrigerator can remove 3 units of heat from inside.

5. How can the COP of a refrigerator be improved?

The COP of a refrigerator can be improved by using more efficient components such as a high-efficiency compressor and better insulation. Choosing a refrigerant with a lower global warming potential can also help improve the COP. Regular maintenance and proper usage, such as keeping the door closed and not overfilling the refrigerator, can also contribute to improving its COP.

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