Question

A commercial refrigerator with refrigerant- 134 a as the working fluid is used to keep the refrigerated space at \(-35^{\circ} \mathrm{C}\) by rejecting 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 \(50^{\circ} \mathrm{C}\) and leaves at the same pressure subcooled by \(5^{\circ} \mathrm{C}\). If the compressor consumes \(3.3 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the refrigeration load, \((c)\) the \(C O P,\) and \((d)\) the minimum power input to the compressor for the same refrigeration load.

Short answer

Answer: (a) The mass flow rate of the refrigerant: 0.0592 kg/s (b) The refrigeration load: 7.48 kW (c) The COP of the refrigerator: 2.27 (d) The minimum power input to the compressor: 2.15 kW

Step-by-step solution

Energy balance on the condenser

The energy balance on the condenser can be written as: $$Q_{cond} = m_w c_p (T_{out} - T_{in})$$ where, \(Q_{cond}\) is the heat removed by the cooling water in the condenser, \(m_w\) is the mass flow rate of the cooling water, given as 0.25 kg/s, \(c_p\) is the specific heat capacity for water, which is 4.18 kJ/kg\(^{\circ}\mathrm{C}\), \(T_{in}\) is the inlet temperature of the cooling water, given as 18\(^{\circ}\mathrm{C}\), \(T_{out}\) is the outlet temperature of the cooling water, given as 26\(^{\circ}\mathrm{C}\). Calculating \(Q_{cond}\) we have: $$Q_{cond} = 0.25 \times 4.18 \times (26-18) = 8.34 \mathrm{kW}$$

Determine the specific enthalpy values for the refrigerant

Using the given data and refrigerant-134a property tables, we can find the specific enthalpy values at different points in the cycle: 1. At the condenser inlet, \(P_1 = 1.2\ \mathrm{MPa}\) and \(T_1 = 50^{\circ}\mathrm{C}\): h1 is read to be 268 kJ/kg from the property tables. 2. At the condenser outlet, \(P_2 = 1.2\ \mathrm{MPa}\) and \(T_2 = 45^{\circ}\mathrm{C}\) (subcooled by 5\(^{\circ}\mathrm{C}\) from the inlet temperature): h2 is read to be 142 kJ/kg from the property tables.

Determine the mass flow rate of refrigerant

Using the energy balance for the refrigerant and specific enthalpy values in the condenser: $$m_R = \frac{Q_{cond}}{h_1 - h_2} $$ where, \(m_R\) is the mass flow rate of the refrigerant, \(h_1\) and \(h_2\) are specific enthalpy values of refrigerant entering and leaving the condenser, respectively. Calculating the mass flow rate of the refrigerant we have: $$m_R = \frac{8.34}{268 - 142} = 0.0592\ \mathrm{kg/s}$$ So, the mass flow rate of the refrigerant is 0.0592 kg/s.

Determine the refrigeration load

The refrigeration load, \(Q_{evap}\), can be determined using the energy balance on the evaporator and the mass flow rate of the refrigerant: $$Q_{evap} = m_R(w_{comp} - (h_2 - h_1))$$ where, \(w_{comp}\) is the power input to the compressor, given as 3.3 kW, \(m_R = 0.0592\ \textrm{kg/s}\) (from step 3). Calculating the refrigeration load, we have: $$Q_{evap} = 0.0592 (3.3 - (142 - 268)) = 7.48\ \mathrm{kW}$$ So, the refrigeration load is 7.48 kW.

Determine the COP of the refrigerator

The coefficient of performance (COP) can be calculated using the refrigeration load, \(Q_{evap}\), and the power input to the compressor, \(w_{comp}\): $$COP = \frac{Q_{evap}}{w_{comp}}$$ Calculating the COP, we have: $$COP = \frac{7.48}{3.3} = 2.27$$ So, the COP of the refrigerator is 2.27.

Determine the minimum power input to the compressor

The minimum power input to the compressor, \(w_{comp,min}\), for the same refrigeration load can be found using the Carnot COP: $$COP_{Carnot} = \frac{T_{evap}}{T_{cond} - T_{evap}}$$ where, \(T_{evap}\) is the evaporator temperature, given as \(-35^{\circ}\mathrm{C}\) (which should be converted to an absolute temperature): \(T_{evap} = -35 + 273 = 238\ \mathrm{K}\), \(T_{cond}\) is the condenser temperature, equal to the outlet cooling water temperature (given as 26\(^{\circ}\mathrm{C}\)), and should also be converted to an absolute temperature: \(T_{cond} = 26 + 273 = 299\ \mathrm{K}\). Calculating the minimum power input to the compressor, we have: $$w_{comp,min} = \frac{Q_{evap}}{COP_{Carnot}} = \frac{7.48}{\frac{238}{(299 - 238)}} = 2.15\ \mathrm{kW}$$ So, the minimum power input to the compressor for the same refrigeration load is 2.15 kW. In conclusion: \((a)\) The mass flow rate of the refrigerant: 0.0592 kg/s \((b)\) The refrigeration load: 7.48 kW \((c)\) The COP of the refrigerator: 2.27 \((d)\) The minimum power input to the compressor: 2.15 kW

Key concepts

Mass Flow Rate of Refrigerant

The mass flow rate of refrigerant is a measure of the amount of refrigerant circulating through the refrigeration cycle over a given period. It's crucial for the system's performance because it directly influences the refrigeration capacity. To calculate the mass flow rate, we must use the principle of conservation of energy in the condenser, equating the heat removed by the cooling water to the heat carried by the refrigerant.

For our commercial refrigerator scenario, using the heat transferred to the water and the specific enthalpy change of the refrigerant, we determined the mass flow rate to be 0.0592 kg/s. This value is pivotal as it allows us to calculate the refrigeration load and other key performance indicators of the refrigeration cycle.

Refrigeration Load Calculation

Refrigeration load refers to the amount of heat that needs to be extracted from the refrigerated space to maintain the desired temperature. It's a critical factor for sizing refrigeration equipment. In our problem, an energy balance on the evaporator allows us to determine the refrigeration load by considering the mass flow rate of the refrigerant and the difference in specific enthalpy from when the refrigerant enters and leaves the compressor.

With the computed mass flow rate and the given power consumption of the compressor, we concluded that the refrigeration load for our system was 7.48 kW. This figure helps us understand the cooling demand the refrigerator must meet and is instrumental in assessing the efficiency of the cooling process.

Refrigerator COP

The coefficient of performance (COP) of a refrigerator is a dimensionless number that measures the efficiency of the refrigeration cycle. COP is defined as the ratio of the refrigeration load to the power input to the compressor. The higher the COP, the more efficient the refrigerator is.

In our case, using the refrigeration load and the power input, the COP was found to be 2.27, which means for every unit of energy consumed by the compressor, the refrigerator moved 2.27 units of heat out of the refrigerated space. Understanding COP is key for selecting and comparing refrigeration systems and for optimizing energy consumption.

Minimum Compressor Power Input

The minimum power input to the compressor is the theoretical lowest amount of energy needed by the compressor to achieve the same refrigeration load under ideal conditions, following the Carnot efficiency principles. It serves as a benchmark to assess the actual performance of a real refrigeration cycle.

By calculating the Carnot COP using the temperatures of the evaporator and condenser, we then determine the minimum power input, which for the refrigerator in our exercise was 2.15 kW. This value highlights the potential for energy savings and signifies the gap between real and ideal systems, often due to irreversibilities in the actual refrigeration cycle.