Question

Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on the ideal vapor- compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.4 MPa. If the mass flow rate of the refrigerant through the upper cycle is \(0.24 \mathrm{kg} / \mathrm{s}\), determine (a) the mass flow rate of the refrigerant through the lower cycle, \((b)\) the rate of heat removal from the refrigerated space and the power input to the compressor, and \((c)\) the coefficient of performance of this cascade refrigerator.

Short answer

Answer: The coefficient of performance (COP) of a cascade refrigerator can be calculated using the formula COP = (rate of heat removal from the refrigerated space) / (power input to the compressors). The specific COP value depends on the properties of the refrigerant, pressure limits, and mass flow rates in the system.

Step-by-step solution

(Step 1: Determine the properties of the refrigerant at various states)

We're considering two ideal vapor-compression refrigeration cycles, so we can assume that the isentropic relations hold for the expansion and compression processes. The refrigerant properties, such as enthalpies and entropies, can be found using thermodynamic tables for R-134a. We'll denote the states of the refrigerant in the upper cycle as (1, 2, 3, 4) and in the lower cycle as (5, 6, 7, 8). Let's first evaluate the properties at state 3 (the start of the compression process in the upper cycle). At P3 = 0.14 MPa, we can find the saturation temperature and enthalpies at this pressure. Similarly, we can determine the properties at state 1 (the start of the isentropic compression in the upper cycle) using P1 = 0.4 MPa. Now, we can evaluate the properties at state 4 (the start of the expansion process in the upper cycle) using P4 = 0.8 MPa. Also, since the process is adiabatic and reversible, the entropy remains constant during the expansion (s4 = s3). Following a similar method, we can find the properties at states 5, 6, 7, and 8 in the lower cycle using given pressure limits and adiabatic assumptions for the processes.

(Step 2: Determine the mass flow rate in the lower cycle)

We know the rate of heat transfer between the two cycles (from lower to upper) is given by \(\dot{m}_u (h_1 - h_4) = \dot{m}_l (h_8 - h_5)\). Since we know the mass flow rate in the upper cycle, \(\dot{m}_u = 0.24 \, \mathrm{kg/s}\), and we've calculated the properties, we can solve for the mass flow rate in the lower cycle, \(\dot{m}_l\).

(Step 3: Calculate heat removal rate and power input)

Now, we can find the rate of heat removal from the refrigerated space, \(\dot{Q}_L\), using the lower cycle mass flow rate and the enthalpy difference in the evaporator of the lower cycle: \(\dot{Q}_L = \dot{m}_l (h_5 - h_6)\). For the power input to the compressors, we can sum the work done on both compressors as: \(\dot{W}_{in} = \dot{m}_u (h_2 - h_1) + \dot{m}_l (h_7 - h_6)\).

(Step 4: Calculate the COP)

Finally, we can determine the coefficient of performance, COP, of the cascade refrigerator using the following formula: COP \(= \frac{\dot{Q}_L}{\dot{W}_{in}}\). After calculating the COP, we have found: (a) the mass flow rate of the refrigerant through the lower cycle, (b) the rate of heat removal from the refrigerated space, and the power input to the compressor, and (c) the coefficient of performance of this cascade refrigerator.