Working of an Air Condition unit

Air conditioning unit, or air-con, AC, A/C, is the process where heat and/or moisture is removed from a building or interior space to improve the comfort of the occupants. 

It applies to both residential and commercial spaces. It is used to improve the comfort of both humans and animals. 

Sometimes, these systems are also used to dehumidify and cool spaces occupied by heat-producing devices such as computer servers as well in storage rooms with delicates stored items such as artworks. 

Air-conditioning systems are equipped with a fan(s) that are used in the distribution of conditioned air to areas of interest to improve the comfort of the occupants. 

The cooling and/or heating is achieved through the refrigeration cycle. 

However, sometimes free cooling or evaporation is also used. In some ACs, desiccants are used to remove and/or add moisture from the air.

The current experiment involves an analysis of an air-conditioning system by use of a psychrometric chart as well as a practical measurement of various components. The two sets of results are then compared and conclusions made. 

Also covered is the research and review on air-conditioning plants as far as environmental issues, energy issues, health issues, and policies are concerned.

Parts of Air conditioning Unit

A standard air conditioning unit consists of four different and basic components that work together to provide air which is conditioned at the required temperature and humidity. Basic components of an air conditioning unit are

1. Compressor

2. Condenser

3. Expansion valve

4. Evaporator

Working of Air Condition Unit

The process of the air conditioning unit starts from the compressor stage where refrigerant is compressed to the required pressure. 

Compressor on the air condition unit work as a container for the refrigerant and also as the main power consuming and producing unit of the AC system. 

Compressing the refrigerant increases its pressure and temperature where pressure is required to move the refrigerant throughout the air conditioning system. 

The increase in temperature of the refrigerant is unwanted and it is reduced in the condenser stage of the air conditioning system. 

Condenser condenses the refrigerant into the liquid form and reduces its temperature by reducing the temperature of the refrigerant. 

Condenser is a heat exchanger which removes heat from the hot fluid by passing it to the much lower temperature outside air. 

After condenser the refrigerant is passed to the expansion valve which is designed to reduce the pressure of the refrigerant. Expansion valve reduce the pressure of the refrigerant by allowing it to expand suddenly. 

This reduces the temperature and pressure of the refrigerant and thus enables the refrigerant to work as a low temperature fluid. 

In last stage of the air conditioning system the refrigerant enters the evaporator which is also a heat exchanger and it is the only indoor unit of the air conditioning system. 

Evaporator works to reduce the temperature of the indoor air by exchanging the indoor air heat with the much lower temperature refrigerant. 

Objectives

The overall objective of the experiment was to familiarize oneself with energy technology concepts and principles while solving a practical problem. 

This was to be accomplished through practical measurement of data, analyze the data, and appraise the results while presenting the outcome of the experiment in a report format.

Procedure of an Air Condition unit

The procedure for the current experiment was to fill all the relevant sections in the datasheet provided. This was accomplished by reading all the relevant values from the measuring instruments three times and while entering the data in the datasheet. For calculations, the average values were utilized.

Results and Calculations Air Condition unit

The table below shows the data collected during the experiment.

Air condition unit experimental data

Dry air flow rate

The air that flow into the air conditioning system for conditioning is added into the point D of the system. The mass flow rate of the dry air that moves into the system can be calculated from following formula. 

m ̇_a=0.05717* √(Z/v_D )

In the above formula v_Dis the specific volume of the air based on its dry bulb temperature and wet bulb temperature and z is the pressure differential in the water column due to the flow of air over it and it is measured in mmH2O. 

To measure the specific volume to calculated the mass flow of the dry air moving into the air conditioning system the dry bulb temperature and wet bulb temperature of air moving into the system are marked on psychometric chart the point at which their respective line cut each other gives the value of specific volume. 

m ̇_a  = 0.0517 * √(4.6/0.895)

m ̇_a  = 0.1172  kg_(dry_air)/S

Part-2: Water added to the system

The water that flow into the air conditioning system for conditioning the air as required is added into the system between point A and B of the system. 

The mass flow rate of the wet air that moves into the system can be calculated from following formula. 

m ̇_(water_AB )  = m ̇_a  * (ω_B  -〖 ω〗_A )

In the above formula ω_Bis the water content in air at point B, based on the air dry bulb temperature and wet bulb temperature of air at the point B. 

Similarly ω_Ais the water content in air at point A, based on the air dry bulb temperature and wet bulb temperature of air at the point B. m ̇_a is the mass flow rate of the dry air moving in the system from point A to the point B. 

To measure the water content in the air added at the point B, the dry bulb temperature and wet bulb temperature of air at the point B in the system are marked on psychometric chart, the point at which their respective line cut each other gives the value of water content in the air at point B. 

similarly when the dry bulb temperature and wet bulb temperature of air at the point A in the system are marked on psychometric chart, the point at which their respective line cut each other gives the value of water content in the air at point A.

m ̇_(water_AB )  = 0.1172 * (0.020 - 0.0098)

m ̇_(water_AB)  = 0.0012  kg/s

Part-3: Energy balance between sections A and B

Heat and Work transfer rate

Two different boilers one with power of 1 KW and one with power of 2 KW generate heat for this system. 

Similarly two identical preheaters are also provided each with power of 1 KW.

(V^2/R)_(Pre_heater)+ (V^2/R)_(Boiler_2kW)+ (V^2/R)_(Boiler_1kW)+ Fan power_(@235.7V)=  〖235〗^2/47.4  +  〖235〗^2/23.9  +  〖235〗^2/58.7  + ≈ 135 = 4551.55  W ≈ 4.55  kW

Enthalpy change rate

Due to the conditioning of the air in the air conditioning system the enthalpy of the system changes at every step and rate at which it changes is called the enthalpy rate. 

It depends on the mass flow of dry air between the point A and B and the enthalpy of the air at the point A and B and the mass flow of water added into the system between the point A and B. 

The enthalpy of air depends on the dry bulb temperature of air, so enthalpy of air at point A depends on the dry bulb temperature of point A and enthalpy of air at point B depends on the dry bulb temperature of point B. 

Putting the value of dry bulb temperature in steam table will give the value of enthalpy of the respective point.

Enthalpy change rate= m ̇_a  * (h_B  -〖 h〗_A )  - m ̇_(water_AB )  *〖 h〗_f

At DBT_A (18.93) C,

h_f  ≈ 54  kJ/kg

Therefore;

Enthalpy change rate = 0.1172 * (85.5 - 43)  - 0.0012 * (54)  ≈ 4.92  kW

Part-4: Rate of water removed

by use of the psychrometric chart

The water that flow out of the air conditioning system for conditioning the air (as required) is removed from the system between point B and C of the system. The mass flow rate of the water content out of the system can be calculated from following formula. 

m ̇_(water_BC )  = m ̇_a  * (〖 ω〗_B  - ω_C  )

In the above formula ω_Bis the water content in air at point B, based on the air dry bulb temperature and wet bulb temperature of air at the point B. 

Similarly ω_cis the water content in air at point C, based on the air dry bulb temperature and wet bulb temperature of air at the point B. m ̇_a is the mass flow rate of the dry air moving in the system from point B to the point C. 

To measure the water content in the air removed at the point C, the dry bulb temperature and wet bulb temperature of air at the point C in the system are marked on psychometric chart, the point at which their respective line cut each other gives the value of water content in the air at point C. 

similarly when the dry bulb temperature and wet bulb temperature of air at the point B in the system are marked on psychometric chart, the point at which their respective line cut each other gives the value of water content in the air at point B.

m ̇_(water_BC )  = 0.1172 * (0.020 - 0.0145)

m ̇_(water_BC )  = 0.00064  kg/s

by measurement of the condensate collected

The amount of water removed from the system can also be measured directly from the water collected in the condenser section air conditioning system. 

The mass flow of water moving out of the system depends on the water quantity collected in the condenser.

m ̇_(water_BC )  =  (164 * 〖10〗^(-6)  * 1000)/300  

m ̇_(water_BC)  = 0.00055  kg/s

Part-5: Heat input to the air using SFEE between sections C and D

Due to the conditioning of the air in the air conditioning system the heat moves into the system and rate at which it moves is called the heat addition rate. 

It depends on the mass flow of dry air between the point C and D and the enthalpy of the air at the point C and D. 

The enthalpy of air depends on the dry bulb temperature of air, so enthalpy of air at point C depends on the dry bulb temperature of point C and enthalpy of air at point D depends on the dry bulb temperature of point D.

Putting the value of dry bulb temperature in steam table will give the value of enthalpy of the respective point.

By use of the psychrometric chart

Heat Input,Q ̇  = m ̇_a  ( h_D  - h_C  )

Heat Input,Q ̇  = 0.1172 ( 71.5 - 58 )

Heat Input,Q ̇   = 1.58  kW

By heat addition

Heat Input,Q ̇=+(V^2/R)_(Boiler_2kW)=〖235〗^2/23.9≈2.31 kW

Discussion on Working of an Air Condition Unit

Part-1: Accuracy of results stating possible errors

There was a discrepancy between the computed values and the expected changes of enthalpy in the refrigeration system.

The heat added to the system was approximately 4.55 kW while the empathy change indicated that the heat received was 4.92 kW. 

Taking the heat determined from the supplied power as the base value, the discrepancy is 8.13%. 

The rate of water removed was determined to 0.00064 kg/s from the psychrometric chart while the condensate was 0.00055 kg/s from the experiment. 

This resulted in a discrepancy of 18%. Finally, the heat added to air was determined to be 1.58 kW from the psychrometric chart while the actual heat added was 2.31 kW resulting in a discrepancy of 31.6% discrepancy.

These discrepancies can be attributed to several issues as summarized below;

Approximation of various values from the psychrometric chart especially where decimal values are encountered. 

This can also emanate from the complexity of the psychrometric chart where various values have to be read/determined from the same system point.

Human errors when reading the various values from the psychrometric chart and thus errors in calculation.

Faulty measuring instruments such as manometer, timer, and thermometers during the experiment.

The experiment can be improved through;

Mastery of using the psychrometric chart in solving the refrigeration and other thermodynamic systems.

Calibration of the measuring instruments to be used in the data collection during the experiment.