Question

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at \(-30^{\circ} \mathrm{C}\) by rejecting its waste heat to cooling water that enters the condenser at \(18^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{kg} / \mathrm{s}\) and leaves at \(26^{\circ} \mathrm{C}\) The refrigerant enters the condenser at \(1.2 \mathrm{MPa}\) and \(65^{\circ} \mathrm{C}\) and leaves at \(42^{\circ} \mathrm{C}\). The inlet state of the compressor is \(60 \mathrm{kPa}\) and \(-34^{\circ} \mathrm{C}\) and the compressor is estimated to gain a net heat of \(450 \mathrm{W}\) from the surroundings. Determine ( \(a\) ) the quality of the refrigerant at the evaporator inlet, \((b)\) the refrigeration load, (c) the COP of the refrigerator, and ( \(d\) ) the theoretical maximum refrigeration load for the same power input to the compressor.

Short answer

Based on the given information for the refrigeration cycle operating with R-134a refrigerant, the following parameters were determined: 1. The quality of the refrigerant at the evaporator inlet is approximately 0.299. 2. The refrigeration load is approximately 13.29 kJ/s. 3. The coefficient of performance (COP) of the refrigerator is approximately 0.435. 4. The theoretical maximum refrigeration load is approximately 74.85 kJ/s.

Step-by-step solution

Find the quality of the refrigerant at the evaporator inlet

To determine the quality of the refrigerant at the evaporator inlet, we will use the given information: \(P_1=60\,\text{kPa}\) and \(T_1=-34^\circ \mathrm{C}\) First we should determine if the refrigerant at this state is saturated. We can do this by comparing the saturation temperature at \(60\,\text{kPa}\) to the actual temperature: \(T_\text{sat}=-34.2^{\circ}\mathrm{C}\) Since the actual temperature (-34°C) is very close to the saturation temperature, we assume the refrigerant to be in a saturated state at the inlet of the evaporator. Thus, we can determine the quality (x) of the inlet state: \(x=(h_1-h_\text{f})/(h_\text{g}-h_\text{f})\) We need to find the enthalpy values \(h_\text{f}\) and \(h_\text{g}\) at \(P_1\) from the R-134a properties table. \(h_\text{f}= 256.24\,\text{kJ/kg}\) \(h_\text{g}= 386.33\,\text{kJ/kg}\) \(h_1=-{450 \,\text{W} \over 0.25\, \text{kg/s}}+h_\text{f}=310.24\,\text{kJ/kg}\) Now, we can calculate the quality: \(x = \frac{310.24 - 256.24}{386.33 - 256.24} \approx 0.299\) So, the quality of the refrigerant at the evaporator inlet is approximately 0.299.

Calculate the refrigeration load

The refrigeration load is the heat extracted from the refrigerated space. We can determine this using the mass flow rate of the refrigerant and the enthalpy difference across the evaporator: \(Q_\text{evap} = \dot{m}(h_4- h_1)\) The state 4 can be found by using the enthalpy at the exit of the compressor, which is isentropic. Assume adiabatic and reversible (isentropic) compression process, which means \(s_2=s_1=s_\text{g}=0.9957\,\text{kJ/kgK}\) at inlet. \(h_1=310.24\,\text{kJ/kg}\) (already calculated in step 1) Using the R-134a properties table at state \(2\) \((1.2 \,\text{MPa} \text{ and } s_2)\), we can find: \(h_2 = 432.42 \,\text{kJ/kg}\) Now we will find \(h_4\) from the data at state 3, which is given by: \(P_3=1.2 \,\text{MPa}\) and \(T_3=42^\circ \mathrm{C}\) \(h_3 = 257.1\, \text{kJ/kg}\) (from R-134a properties table) \(h_4 = h_3 + (h_2 - h_1) = 257.1\, \text{kJ/kg}\) Using the mass flow rate given: \(Q_\text{evap} = 0.25\,\text{kg/s}( 257.1 - 310.24)\, \text{kJ/kg} = -13.29 \, \text{kJ/s}\) The refrigeration load is \(13.29\,\text{kJ/s}\)

Calculate the coefficient of performance (COP)

The Coefficient of performance (COP) is the ratio of the heat removed from the refrigerated space to the work input to the compressor: \(\text{COP} = \frac{Q_\text{evap}}{W_\text{comp}}\) Let's calculate the work input to the compressor. \(W_\text{comp} = \dot{m}(h_2 - h_1) = 0.25 \, \text{kg/s}(432.42 - 310.24) \,\text{kJ/kg} = 30.55 \, \text{kJ/s}\) Now, we can find the Coefficient of Performance (COP): \(\text{COP} = \frac{-13.29\, \text{kJ/s}}{30.55\, \text{kJ/s}} \approx 0.435\) The COP of the refrigerator is approximately 0.435.

Calculate the theoretical maximum refrigeration load

To determine the theoretical maximum refrigeration load, we will find the reversible work input to the compressor and calculate the maximum COP based on reversible processes. \(\mathrm{COP_{ideal}} = \frac{T_\text{evap}}{(T_\text{cond} - T_\text{evap})}\) \(T_\text{evap}=-30^\circ \mathrm{C}+273.15\,\mathrm{K}=243.15\,\mathrm{K}\) \(T_\text{cond}=65^\circ \mathrm{C}+273.15\,\mathrm{K}=338.15\,\mathrm{K}\) \(\mathrm{COP_{ideal}}=\frac{243.15}{(338.15 - 243.15)}=2.45\) Now we can find the theoretical maximum refrigeration load by: \(Q_\text{max} = \mathrm{COP_{ideal}} W_\text{comp}\) \(Q_\text{max} = 2.45(30.55 \,\text{kJ/s}) = 74.85 \,\text{kJ/s}\) The theoretical maximum refrigeration load is \(74.85\,\text{kJ/s}\).