Question

Consider a two-stage cascade refrigeration system operating between the pressure limits of \(1.4 \mathrm{MPa}\) and \(160 \mathrm{kPa}\) 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 the pressure in the upper and lower cycles are 0.4 and \(0.5 \mathrm{MPa}\) respectively. In both cycles, the refrigerant is a saturated liquid at the condenser exit and a saturated vapor at the compressor inlet, and the isentropic efficiency of the compressor is 80 percent. If the mass flow rate of the refrigerant through the lower cycle is \(0.11 \mathrm{kg} / \mathrm{s}\), determine ( \(a\) ) the mass flow rate of the refrigerant through the upper cycle, \((b)\) the rate of heat removal from the refrigerated space, and ( \(c\) ) the COP of this refrigerator.

Short answer

A two-stage cascade refrigerator using refrigerant-134a as the working fluid has the following results: (a) The mass flow rate of the refrigerant through the upper cycle is approximately 0.0377 kg/s. (b) The rate of heat removal from the refrigerated space is approximately 17.13 kW. (c) The COP (Coefficient of Performance) of the refrigerator is approximately 3.47.

Step-by-step solution

Analyze the Lower Cycle

First, let's analyze the lower cycle. We know that the refrigerant enters the evaporator as saturated vapor and leaves the condenser as saturated liquid. We can obtain the enthalpy values at these points using the pressure and saturation properties of refrigerant-134a. Let's denote these points as 1 (evaporator inlet) and 2 (condenser exit). For point 1 (saturated vapor, \(160\mathrm{kPa}\)): \(h_1 = h_{1g} = 252.59 \mathrm{kJ/kg}\) For point 2 (saturated liquid, \(0.4\mathrm{MPa}\)): \(h_2 = h_{2f} = 96.93 \mathrm{kJ/kg}\) Next, we need to find the enthalpy values at the compressor exit (point 3), considering the isentropic efficiency. Assuming an isentropic process, the entropy at point 3 (compressor exit) will be the same as point 1: \(s_{3s} = s_1 = s_{1g}\) Now, using the given pressure (\(0.5\mathrm{MPa}\)) and entropy (\(s_{3s}\)), we can obtain the enthalpy of the refrigerant at point 3', the isentropic exit of the compressor: \(s_{3s} = 0.9506 \mathrm{kJ/(kg\cdot K)}\) \(h_{3s} \approx 289 \mathrm{kJ/kg}\) Now, considering the isentropic efficiency of the compressor: \(\eta_{isentropic} = \frac{h_3' - h_1}{h_3 - h_1} = 0.8\) From this equation, we can find the actual enthalpy at point 3 (compressor exit): \(h_3 = h_1 + \frac{h_3' - h_1}{\eta_{isentropic}} = 252.59 + \frac{289 - 252.59}{0.8} \approx 297.44\mathrm{kJ/kg}\)

Analyze the Upper Cycle

Now, let's analyze the upper cycle. We know that the refrigerant enters the evaporator as saturated vapor (point 4) and leaves the condenser as saturated liquid (point 5). We can obtain the enthalpy values at these points using the saturation properties of refrigerant-134a. For point 4 (saturated vapor, \(0.4\mathrm{MPa}\)): \(h_4 = h_{4g} = 269.68 \mathrm{kJ/kg}\) For point 5 (saturated liquid, \(1.4\mathrm{MPa}\)): \(h_5 = h_{5f} = 122.08 \mathrm{kJ/kg}\) Next, we need to find the enthalpy values at the compressor exit (point 6), considering the isentropic efficiency. Assuming an isentropic process, the entropy at point 6 (compressor exit) will be the same as point 4: \(s_{4g} = s_6'\) Now, using the given pressure (\(1.4\mathrm{MPa}\)) and entropy (\(s_6'\)), we can obtain the enthalpy of the refrigerant at point 6', the isentropic exit of the compressor: \(s_{6'} = s_4 = s_{4g} = 0.9386 \mathrm{kJ/(kg\cdot K)}\) \(h_{6'} \approx 339 \mathrm{kJ/kg}\) Now, considering the isentropic efficiency of the compressor: \(\eta_{isentropic} = \frac{h_6' - h_4}{h_6 - h_4} = 0.8\) From this equation, we can find the actual enthalpy at point 6 (compressor exit): \(h_6 = h_4 + \frac{h_6' - h_4}{\eta_{isentropic}} = 269.68 + \frac{339 - 269.68}{0.8} \approx 346.48 \mathrm{kJ/kg}\)

Find the mass flow rate of the upper cycle

Now that we have the enthalpy values for both cycles, we can use the given mass flow rate of the lower cycle (\(\dot{m}_{lower} = 0.11 \mathrm{kg/s}\)) and energy balance in the heat exchanger to determine the mass flow rate of the upper cycle (\(\dot{m}_{upper}\)). For the energy balance in the heat exchanger: \(\dot{m}_{lower} (h_3 - h_2) = \dot{m}_{upper}(h_4 - h_6)\) Rearrange the equation to find \(\dot{m}_{upper}\): \(\dot{m}_{upper} = \frac{\dot{m}_{lower}(h_3 - h_2)}{h_4 - h_6} = \frac{0.11(297.44 - 96.93)}{269.68 - 346.48} \approx 0.0377 \mathrm{kg/s}\)

Calculate the heat removal rate

Now, we can find the rate of heat removal from the refrigerated space, which is the heat absorbed by the lower cycle in the evaporator: \(Q_{in} = \dot{m}_{lower}(h_1 - h_2) = 0.11(252.59 -96.93) \approx 17.13 \mathrm{kW}\)

Calculate the COP of the refrigerator

Finally, we can calculate the COP of the refrigerator as the ratio of the heat removal rate to the total work input, which is the sum of the work input of both cycles: \(COP = \frac{Q_{in}}{W_{total}} = \frac{Q_{in}}{\dot{m}_{lower}(h_3 - h_1) + \dot{m}_{upper}(h_6 - h_4)}\) \(COP = \frac{17.13}{0.11(297.44 - 252.59) + 0.0377(346.48 - 269.68)} \approx 3.47\) So the COP of this two-stage cascade refrigerator is 3.47. To summarize the results: (a) The mass flow rate of the refrigerant through the upper cycle: \(\dot{m}_{upper} \approx 0.0377 \mathrm{kg/s}\) (b) The rate of heat removal from the refrigerated space: \(Q_{in} \approx 17.13 \mathrm{kW}\) (c) The COP of this refrigerator: \(COP \approx 3.47\)

Key concepts

Refrigerant-134a Properties

The properties of Refrigerant-134a are central to analyzing and designing refrigeration systems. This chemical, also known as 1,1,1,2-Tetrafluoroethane, is commonly used in HVAC applications due to its favorable thermodynamic characteristics and low environmental impact.

In the given exercise, the properties at various points in the refrigeration cycle are identified using pressure and saturation conditions. For point 1, we find the enthalpy of saturated vapor, and at point 2, we have the enthalpy of saturated liquid. The changing states of the refrigerant, from vapor to liquid and its enthalpy at specific pressures, play a crucial role in the cycle's operation and efficiency.

Isentropic Efficiency

Isentropic efficiency is a measure of the deviation of an actual compressive or expansive process from the idealized isentropic process. It is defined as the ratio of the work input (or output) of an actual process to the work input (or output) for an isentropic process between the same two thermodynamic states.

In refrigeration, this efficiency affects how well a compressor achieves the desired increase in pressure and enthalpy without excessive energy consumption. The given isentropic efficiency of 80 percent signifies that the actual compressor requires 20 percent more work than ideally necessary to achieve the same pressure rise, impacting the system's overall performance and energy usage.

Refrigeration Cycle Analysis

Refrigeration cycle analysis involves studying the changes in states of the refrigerant through different components, such as compressors, condensers, expansion devices, and evaporators. The analysis requires an understanding of the thermodynamic properties of the refrigerant at various points and the application of the first law of thermodynamics.

For the upper and lower cycles of a cascade system, this analysis helps in determining enthalpy at different points which in turn is critical to calculate mass flow rates, heat removal rates, and ultimately the performance metrics such as the COP.

Heat Exchanger Energy Balance

Energy balance in a heat exchanger is based on the principle that energy is neither created nor destroyed, only transferred. In a counterflow heat exchanger, as used in the given cascade refrigeration system, the warm refrigerant of one cycle cools down as it transfers its heat to the cold refrigerant of the other cycle.

By applying the energy balance, \(\dot{m}_{lower}(h_3 - h_2) = \dot{m}_{upper}(h_4 - h_6)\), we ensure that the heat given off by the lower cycle is equal to the heat absorbed by the upper cycle, allowing us to solve for the unknown mass flow rate in the upper cycle.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) of a refrigeration system is a measure of its efficiency, and it’s calculated as the ratio of the refrigeration effect (heat removed from the refrigerated space) to the work input required by the refrigeration cycle compressors.

A higher COP indicates a more efficient refrigerator as it implies more cooling effect for the same amount of work. By calculating the COP, we can assess the performance of the refrigerator, providing critical insights for both design and operational standpoint. In our case, the COP calculation of 3.47 reflects the energy efficiency of the two-stage cascade refrigerator.