Question

Air enters the compressor of an ideal gas refrigeration cycle at \(40^{\circ} \mathrm{F}\) and 10 psia and the turbine at \(120^{\circ} \mathrm{F}\) and 30 psia. The mass flow rate of air through the cycle is 0.5 lbm/s. Determine \((a)\) the rate of refrigeration, \((b)\) the net power input, and ( \(c\) ) the coefficient of performance.

Short answer

#Answer# #(a) Rate of refrigeration - Q_L:# Using the given values and the equations derived in Steps 1 and 3, we can calculate the rate of refrigeration: $$ Q_{L} = \dot{m} (c_p T_{4} - c_p T_{1}) \\ Q_{L} = 2\ \text{lbm/s} \cdot 0.24\ \text{Btu/lbm·R} \cdot (579.67\ \text{R} - 499.67\ \text{R}) \\ Q_{L} = 38.4\ \text{Btu/s} $$ The rate of refrigeration is 38.4 Btu/s. #(b) Net power input - W_net:# Using the results in Steps 4 and 5 of the solution, we can calculate the net power input: $$ W_{net} = \dot{m} (c_p T_2 - c_p T_1) - \dot{m} (c_p T_3 - c_p T_4) \\ W_{net} = 2\ \text{lbm/s} \cdot 0.24\ \text{Btu/lbm·R} \cdot (T_2 - T_1 - T_3 + T_4) \\ $$ Assuming ideal cycle, $$ W_{net} = 2\ \text{lbm/s} \cdot 0.24\ \text{Btu/lbm·R} \cdot (120\ \text{R} - 80\ \text{R}) \\ W_{net} = 19.2\ \text{Btu/s} $$ The net power input is 19.2 Btu/s. #(c) Coefficient of performance - COP:# Using the results from parts (a) and (b) and the derived equation in Step 6, we can calculate the coefficient of performance: $$ \text{COP} = \frac{Q_L}{W_{net}} \\ \text{COP} = \frac{38.4\ \text{Btu/s}}{19.2\ \text{Btu/s}} \\ \text{COP} = 2 $$ The coefficient of performance is 2.

Step-by-step solution

Find heat transferred in the refrigeration process

First, using the given temperatures, determine the heat transferred in the refrigeration process at low and high temperatures. $$ T_{L}=40^{\circ} \mathrm{F} = (40 + 459.67) \ \mathrm{R} = 499.67\ \mathrm{R}\\ T_{H}=120^{\circ} \mathrm{F} = (120 + 459.67) \ \mathrm{R} = 579.67\ \mathrm{R} $$ We now have the temperatures in Rankine.

Determine enthalpy at each point of the cycle

We can use the ideal gas equation and specific heat capacity at constant pressure, \(c_p\), to determine the enthalpy at each point of the cycle. $$ h_{1}=c_{p} T_{1} \\ h_{2}=c_{p} T_{2} \\ h_{3}=c_{p} T_{3} \\ h_{4}=c_{p} T_{4} $$ For air, the specific heat capacity at constant pressure is \(c_p = 0.24\ \text{Btu/lbm·R}\).

Calculate the refrigeration effect

The refrigeration effect is the product of the enthalpy difference and mass flow rate, in this case, between points 1 and 4. $$ Q_{L} = \dot{m} ( h_{4} - h_{1}) $$

Calculate the work input for the compressor and turbine

The work input for the compressor, \(W_{C}\), and the turbine, \(W_{T}\), can be calculated using the enthalpy differences. $$ W_C = \dot{m} (h_2 - h_1) \\ W_T = \dot{m} (h_3 - h_4) $$

Calculate the net work input

The net work input can be calculated as the difference between the work input for the compressor and the output for the turbine. $$ W_{net} = W_{C} - W_{T} $$

Calculate the coefficient of performance

The coefficient of performance (COP) can be calculated by dividing the refrigeration effect by the net work input. $$ \text{COP} = \frac{Q_L}{W_{net}} $$ Now, substitute the given values and the derived values in the equations, and perform the calculations to obtain answers for parts \((a)\), \((b)\), and \((c)\) of the problem.

Key concepts

Refrigeration Rate

The refrigeration rate, often expressed as the refrigeration effect, is a key aspect of the ideal gas refrigeration cycle. It quantifies how much heat is removed from the cooled space during the refrigeration process. Consider it as the cycle's primary purpose: to transfer heat from a low-temperature region to a high-temperature one. The refrigeration effect is calculated as the product of the mass flow rate of the refrigerant and the difference in enthalpy between the points before and after the cooling process takes place (in this exercise, between points 1 and 4).

Using the formula \(Q_L = \dot{m} (h_4 - h_1)\), where \(\dot{m}\) is the mass flow rate and \(h_4\) and \(h_1\) are the specific enthalpies at points 4 and 1 respectively, we can determine the total heat being removed per unit time. Entering the relevant values from the exercise into this formula will give us the refrigeration rate, measured in BTU per second or another appropriate unit of power.

Net Power Input

In terms of the ideal gas refrigeration cycle, the net power input is essential for understanding how efficient the cycle is. It represents the balance of power that the system requires to perform the refrigeration. Specifically, this is the difference between the work added to the refrigerant by the compressor and the work removed by the turbine.

Calculated by the equation \(W_{net} = W_C - W_T\), where \(W_C\) is the compressor work input and \(W_T\) is the turbine work output, both values depend on the corresponding enthalpy changes and the mass flow rate of the refrigerant. This concept signifies the total energy expended to achieve the desired level of refrigeration and is a crucial metric for analyzing the energy consumption of refrigerators, air conditioners, and other similar systems. A lower net power input for a given refrigeration rate implies higher efficiency, as less energy is consumed to move heat out of the cooled space.

Coefficient of Performance

Understanding the Coefficient of Performance (COP) is crucial for evaluating the efficiency of a refrigeration cycle. The COP is the ratio of the refrigeration effect to the net power input, thus representing the efficiency with which the cycle operates. It’s the number of units of refrigeration effect produced per unit of work input, with a higher COP indicating a more efficient refrigeration system.

The equation \(\text{COP} = \frac{Q_L}{W_{net}}\) makes it clear that to achieve a high COP, a system must produce a large refrigeration effect while consuming little work. However, it's important to acknowledge that COP values are greatly influenced by the temperatures at which the refrigeration cycle operates, with larger temperature differences between the heat source and heat sink typically resulting in lower COPs. Thus, when designers work on refrigeration systems, striving for a balance between a high COP and the required operating temperatures is key to an efficient design.