Question

An air conditioner with refrigerant-134a as the working fluid is used to keep a room at \(26^{\circ} \mathrm{C}\) by rejecting the waste heat to the outside air at \(34^{\circ} \mathrm{C}\). The room is gaining heat through the walls and the windows at a rate of \(250 \mathrm{kJ} / \mathrm{min}\) while the heat generated by the computer, TV, and lights amounts to \(900 \mathrm{W}\). An unknown amount of heat is also generated by the people in the room. The condenser and evaporator pressures are 1200 and \(500 \mathrm{kPa}\), respectively. The refrigerant is saturated liquid at the condenser exit and saturated vapor at the compressor inlet. If the refrigerant enters the compressor at a rate of \(100 \mathrm{L} / \mathrm{min}\) and the isentropic efficiency of the compressor is 75 percent, determine (a) the temperature of the refrigerant at the compressor exit, (b) the rate of heat generation by the people in the room, (c) the COP of the air conditioner, and (d) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions.

Short answer

(a) To find the temperature of the refrigerant at the compressor exit, we first determine the enthalpy and entropy of the refrigerant at the compressor inlet and the isentropic enthalpy at the compressor exit using the given conditions. Then, we use the compressor isentropic efficiency to find the actual enthalpy at state 2 and determine the temperature of the refrigerant at state 2 using the enthalpy and condenser pressure. (b) To find the rate of heat generation by the people in the room, we first calculate the heat absorbed by the evaporator (Q_in) and the work done by the compressor (W_comp). Then, we compute the rate of heat rejection to the outdoor air (Q_out) and determine the rate of heat generation by the people in the room using the given heat gain from walls and windows, the heat generated by appliances, and the computed rate of heat rejection to the outdoor air. (c) To find the Coefficient of Performance (COP) of the air conditioner, we use the following formula: $$COP = \frac{Q_{in}}{W_{comp}}$$ Using the values of Q_in and W_comp calculated in part (b), we can determine the COP. (d) To find the minimum volume flow rate of the refrigerant at the compressor inlet, we need to maintain the same COP and work done by the compressor while minimizing the heat absorbed by the evaporator (Q_in). We first find the departure state at the evaporator exit and calculate the new mass flow rate required to maintain the same work done by the compressor. Then, we determine the specific volume at the compressor inlet and calculate the minimum volume flow rate of the refrigerant at the compressor inlet.

Step-by-step solution

(a) Temperature of the refrigerant at the compressor exit

To determine the temperature of the refrigerant at the compressor exit, we first need to find the enthalpy and entropy of the refrigerant at the compressor inlet (state 1) and the isentropic enthalpy at the compressor exit (state 2s) using the given conditions. 1. Locate the state 1 properties using the saturated vapor condition at the evaporator pressure (500 kPa): $$T_1 = T_{sat} = -6^\circ C$$ $$h_1 = h_{g}(500 \thinspace kPa)$$ $$s_1 = s_{g}(500 \thinspace kPa)$$ 2. Determine the isentropic enthalpy at the compressor exit (state 2s) using the condenser pressure (1200 kPa) and the isentropic condition: $$s_{2s} = s_1$$ $$h_{2s} = h(s_{2s}, 1200 \thinspace kPa)$$ 3. Use the compressor isentropic efficiency to find the actual enthalpy at state 2 (compressor exit): $$\eta_{isentropic} = \frac{h_{2s} - h_1}{h_2 - h_1}$$ $$h_2 = h_1 + \frac{h_{2s} - h_1}{\eta_{isentropic}}$$ 4. Finally, determine the temperature of the refrigerant at state 2 using the enthalpy and condenser pressure: $$T_2 = T(h_2, 1200 \thinspace kPa)$$

(b) Rate of heat generation by the people in the room

First, we need to calculate the heat absorbed by the evaporator (Q_in) and the work done by the compressor (W_comp). Then, we can compute the rate of heat generation by the people in the room. 1. Determine the mass flow rate of the refrigerant using the given volumetric flow rate and specific volume at the compressor inlet (state 1): $$\dot{m} = \frac{100 \thinspace L/min}{v_1}$$ 2. Calculate the heat absorbed by the evaporator (Q_in) using the mass flow rate and the enthalpy change: $$Q_{in} = \dot{m}(h_1 - h_4)$$ 3. Calculate the work done by the compressor (W_comp) using the mass flow rate and enthalpy change: $$W_{comp} = \dot{m}(h_2 - h_1)$$ 4. Determine the rate of heat rejection to the outdoor air (Q_out) using the work done by the compressor and the heat absorbed by the evaporator: $$Q_{out} = Q_{in} - W_{comp}$$ 5. Compute the rate of heat generation by the people in the room using the given heat gain from walls and windows, the heat generated by appliances, and the computed rate of heat rejection to the outdoor air: $$Q_{people} = Q_{out} - Q_{walls\thinspace and\thinspace windows} - Q_{appliances}$$

(c) COP of the air conditioner

To find the Coefficient of Performance (COP) of the air conditioner, we need to use the following formula: $$COP = \frac{Q_{in}}{W_{comp}}$$ Using the values of Q_in and W_comp calculated in part (b), we can determine the COP.

(d) Minimum volume flow rate of the refrigerant at the compressor inlet

To find the minimum volume flow rate of the refrigerant at the compressor inlet, we need to use the same compressor inlet and exit conditions. Thus, the enthalpy difference will stay the same, and the work done by the compressor (W_comp) will remain constant. Therefore, to maintain the same COP, Q_in must be minimized. 1. Find the departure state at the evaporator exit (state 4) for which Q_in is minimized: $$h_4 = h_{f}(500 \thinspace kPa)$$ 2. Calculate the new mass flow rate required to maintain the same W_comp using the enthalpy difference between states 1 and 2 and the new h_4: $$\dot{m_{new}} = \frac{W_{comp}}{h_2 - h_1 - (h_1 - h_{4_{new}})}$$ 3. Determine the specific volume at the compressor inlet: $$v_{1} = v(P_1 = 500 \thinspace kPa, T_1 = -6^\circ C)$$ 4. Finally, calculate the minimum volume flow rate of the refrigerant at the compressor inlet: $$\dot{V}_{min} = \dot{m_{new}}v_1$$

Key concepts

Refrigerant Properties

Refrigerants are the working fluids used in air conditioning systems to absorb, carry, and release heat. The properties of refrigerants are crucial for the efficiency of the air conditioning process. These properties include the boiling point, pressure, enthalpy (a measure of heat content), entropy (a measure of randomness or disorder), and specific volume (the volume occupied by a unit mass of the substance).

For instance, in the given problem, refrigerant-134a is used. At the compressor inlet, it is in saturated vapor form, and at the condenser outlet, it transitions to saturated liquid. The saturated state indicates that the refrigerant is at its boiling point and ready to absorb or release heat during the phase change, which occurs without a temperature change. These phase changes are part of what make refrigerants so effective for heat transfer in cooling systems.

Isentropic Efficiency

Isentropic efficiency is a measure of the actual performance of a compressor compared to the ideal, or isentropic, performance. It is calculated as the ratio of the work output (or input for a compressor) of the actual process to the work output (or input) of an isentropic process between the same initial and final pressures.

The isentropic process is a reversible adiabatic process, where no heat is transferred in or out of the system, and the process is fully reversible. However, in reality, some amount of irreversibilities such as friction, heat loss, and pressure drop occur, which impacts the actual enthalpy change in the refrigerant. Hence, the isentropic efficiency helps us to account for these inefficiencies when calculating the actual compressor exit temperature.

Heat Transfer Rate

The heat transfer rate in an air conditioning system refers to the rate at which heat is absorbed or released by the refrigerant as it cycles through the system. Measured in units such as watts (W) or kilojoules per minute (kJ/min), the heat transfer rate is a critical factor in determining the effectiveness of the air conditioner in maintaining the desired indoor temperature.

In our example, we consider the heat absorbed by the evaporator from the room, which includes heat through the walls, windows, and heat generated by appliances and people. By quantifying the heat absorbed (Q_in) and the heat rejected to the outside air (Q_out), we can solve for unknown variables like the rate of heat generation by people in the room. Understanding heat transfer rates enables us to make crucial decisions about the capacity and efficiency of the HVAC (Heating, Ventilation, and Air Conditioning) system.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a dimensionless measure of the efficiency of a refrigeration or air conditioning system. It is defined as the ratio of the heat removed from the cold reservoir (Q_in) to the work input (W_comp) needed by the compressor. In other words, it's an indication of how effectively the air conditioning system is using energy to cool the space.

The higher the COP, the more efficient the system is. The COP provides an immediate understanding of the system's performance and is an essential factor in the design and selection of HVAC equipment. By computing the COP using the determined values of heat absorbed and work done, we gain insights into the energy efficiency of the air conditioner, which can be compared to other systems or to industry standards.