Final answer:
To calculate the compressor power, refrigeration capacity, and coefficient of performance for an air conditioning system, one must use the principles of the vapor-compression refrigeration cycle along with thermodynamic properties of the refrigerant, which were not provided in the question.
Step-by-step explanation:
Calculating Compressor Power, Refrigeration Capacity, and Coefficient of Performance
To calculate the compressor power, refrigeration capacity, and coefficient of performance (COP), we use the principles of the vapor-compression refrigeration cycle. First, we find the isentropic work done by the compressor using the isentropic efficiency. The refrigeration capacity can be found by using the mass flow rate of the refrigerant and the enthalpy difference across the evaporator. Finally, the COP is calculated as the ratio of the refrigeration effect to the actual work input to the compressor.
Note that these calculations require the use of thermodynamic properties of Refrigerant 22 (R-22) which are not provided in the question. You'd typically use tables or software to find these properties at given states (e.g., 4 bar, 10°C; 9 bar, saturated liquid, etc.).
The compressor power (in kW) is found using the equation:
W_actual = (h2 - h1)/(η_isen)
where η_isen is the isentropic efficiency, h1 is the specific enthalpy of refrigerant entering the compressor, and h2 is the specific enthalpy of refrigerant after isentropic compression. As the actual discharge state is at the same pressure albeit with higher enthalpy due to inefficiencies, h2_actual can be found from the isentropic efficiency.
The refrigeration capacity in tons can be calculated by:
Q = (m_dot)(h1 - h4)
where m_dot is the mass flow rate and h1 and h4 are the specific enthalpies at the evaporator inlet and outlet.
To convert to tons, use the fact that 1 ton of refrigeration is equivalent to the removal of 3.517 kW of heat.
The COP is defined as:
COP = Q/W_actual
where Q is the refrigeration capacity and W_actual is the work input to the compressor.
Without the exact thermodynamic values or refrigerant properties, we cannot compute the exact figures. However, the concepts above outline how one would typically approach such problems in air conditioning systems calculations.