Question

A refrigerator using helium gas operates on the reversed cycle shown in FIGURE. What are the refrigerator’s ($a$) coefficient of performance and ($b$) power input if it operates at $60$cycles per second?

Short answer

($a$) The coefficient of performance is stated as$K=2.27$

($b$) The power if it operates at$60$cycles per second is$P=119W$

Step-by-step solution

Number of moles from the gas law.

To calculate the coefficient of performance, we must first determine the quantity of warmth extracted from the cold reservoir and therefore the amount of labor required to get rid of it. In isochoric pressurizing, heat is removed, and therefore the gas performs positive add adiabatic expansion and negative add adiabatic compression.

The heat evacuated will be estimated as follows:

$Q_c=n{C}_v\Delta T$

The ideal gas law, which applies at $-23°\mathrm{C}$, will be wont to calculate the amount of moles.

$$\begin{gathered} n=\frac{pV}{RT} \\ =\frac{1.5\times {10}^5\times 1\times {10}^{-4}}{8.314\times 250} \end{gathered}$$

$=7.22\times {10}^{-3}$

Find the work.

The heat cooled is calculated as,

$$\begin{gathered} Q_c=0.00722\times 12.5\times 50 \\ =4.51\mathrm{J} \end{gathered}$$

The system's work are going to be evaluated.

$W=n{C}_v\Delta T$

This means we'll should work out what the temperature is at the graph's other two edges. Please detain mind that in an natural process.

$T{V}^{\gamma -1}=const$

This implies that we will are going to be ready to locate,

$T^{\text{'}}={\left(V/V^{\text{'}}\right)}^{\gamma -1}T$

$V/V\text{'}$are going to be $2.5$in our example, and $\gamma -1=0.40$. This enables us to see $361K$and $289K$temperatures. We will now locate the network as follows:$$\begin{gathered} W=0.00722\times 12.5\times \left[\right(361-250)-(289-200\left)\right] \\ =1.99\mathrm{J} \end{gathered}$$

The coefficient of performance (part a)

The coefficient of performance is calculated by dividing the number of energy a system produces by the quantity of energy you place into it.

This formula for calculating performance coefficients is applicable to all or any fields.

The coefficient of performance is going to be,

$$\begin{gathered} K=\frac{Q_c}{W} \\ =\frac{4.51}{1.99} \\ =2.27 \end{gathered}$$

Find the power (part b)

The power is,

$$\begin{gathered} P=Wf \\ =1.99\times 60 \\ =119W \end{gathered}$$