Question

An ideal gas refrigeration cycle using air as the working fluid is to maintain a refrigerated space at \(-23^{\circ} \mathrm{C}\) while rejecting heat to the surrounding medium at \(27^{\circ} \mathrm{C}\). If the pressure ratio of the compressor is \(3,\) determine \((a)\) the maximum and minimum temperatures in the cycle, \((b)\) the coefficient of performance, and ( \(c\) ) the rate of refrigeration for a mass flow rate of \(0.08 \mathrm{kg} / \mathrm{s}\).

Short answer

Based on the ideal gas refrigeration cycle with air as the working fluid and a pressure ratio of 3, the maximum temperature in the cycle is approximately 350.23 K and the minimum temperature is around 214.25 K. The coefficient of performance (COP) for this cycle is around 5.82, and the rate of refrigeration needed to maintain a refrigerated space at -23°C is about 2.06 kW.

Step-by-step solution

1. Determine the Temperature of the Refrigerated Space

First, let's convert the required temperature of the refrigerated space, -23°C, to Kelvin: $$ T_1 = -23 + 273.15 = 250.15 \, \mathrm{K} $$

2. Determine the Temperature of the Surroundings

Now, let's convert the temperature of the surroundings, 27°C, to Kelvin: $$ T_{3}^\prime = 27 + 273.15 = 300.15 \, \mathrm{K} $$

3. Calculate the Maximum Temperature in the Cycle

We are given the pressure ratio of the compressor, \(r_p\): $$ r_p = \frac{P_2}{P_1} $$ Since it is an ideal gas refrigeration cycle, the temperature and pressure at all stages can be related using the isentropic relation: $$ \frac{T_2}{T_1} = \left( \frac{P_2}{P_1} \right)^{(k - 1) / k}$$ Where \(k\) is the specific heat ratio of the working fluid (air), approximately equal to 1.4. Now, substitute the given pressure ratio and solve for the maximum temperature, \(T_2\): $$ T_2 = T_1 \left( r_p \right)^{(k - 1) / k} = 250.15 \left( 3 \right)^{(1.4 - 1) / 1.4} \approx 350.23 \, \mathrm{K}$$

4. Calculate the Minimum Temperature in the Cycle

Using the same isentropic relation as before: $$ \frac{T_{1}^\prime}{T_{3}^\prime} = \left( \frac{P_1}{P_2} \right)^{(k - 1) / k} $$ Rearrange the equation to find the Minimum Temperature, \(T_{1}^\prime\): $$ T_{1}^\prime = T_{3}^\prime \left( \frac{1}{r_p} \right)^{(k - 1) / k} = 300.15 \left( \frac{1}{3} \right)^{(1.4 - 1) / 1.4} \approx 214.25 \, \mathrm{K} $$

5. Determine the Coefficient of Performance (COP)

The COP for an ideal gas refrigeration cycle is given by: $$ COP = \frac{1}{(r_p)^{(k - 1) / k} - 1} $$ Substitute the values and find the COP: $$ COP = \frac{1}{(3)^{(1.4 - 1) / 1.4} - 1} \approx 5.82 $$

6. Calculate the Rate of Refrigeration

The rate of refrigeration (\(Q_1\)) can be found using the mass flow rate (\(\dot{m}\)), the specific heat at constant volume (\(c_v\)), and the temperature difference between the minimum and refrigerated space temperatures: $$ Q_1 = \dot{m} c_v (T_1 - T_{1}^\prime) $$ The specific heat at constant volume for air is approximately \(c_v = 0.718 \, \mathrm{kJ/kg \cdot K}\). Substitute the given mass flow rate and temperature values to find the rate of refrigeration: $$ Q_1 = 0.08 \, \mathrm{kg/s} \times 0.718 \, \mathrm{kJ/kg \cdot K} \times (250.15 \, \mathrm{K} - 214.25 \, \mathrm{K}) \approx 2.06 \, \mathrm{kW} $$ Now, we have found: (a) Maximum Temperature = 350.23 K, Minimum Temperature = 214.25 K (b) COP = 5.82 (c) Rate of Refrigeration = 2.06 kW