Numerical designing (Mathematical modelling) of an air conditioner ( by two methods)

Human do not feel comfort at  high temperatures or extremely low temperatures. The situation of body when it feel comfort with the surrounding conditions like temperature, RH%, air flow etc is called the thermal comfort. This comfort situation is at 27 DBT and 19 WBT.
 To provide this thermal comfort condition, Air conditioners are used which supply the air at required temperatures.

The schematic of an air conditioner is shown as below:

Consider a 1 TR air conditioner with ambient 35 DBT and 24 WBT and indoor design conditions are 27 DBT and 19 WBT.
So the inlet air to evaporator is at 27 DBT and 19 WBT.

The design conditions are as per below:

Outdoor conditions: 35 DBT, 24 WBT
Indoor conditions: 27 DBT and 19 WBT
Evaporator temperature in: 12 deg C
Evaporator temperature out: 15 deg C
Condensing temperature: 54 deg C
Liquid temperature: 46 deg C (ASHRAE conditions)

Now main thing to calculate is required mass flow rate of air (CFM) and air out condition from evaporator as per below steps. (Two methods)

First method: By fixing mass flow rate of air

Step 1: Find the condition of air at designed indoor conditions (27 DBT and 19 WBT) using psychrometric chart

H_indoor= 53.78 kJ/kg   (Enthalpy)
RH % = 46.9
Absolute humidity=10.441 g/kg of dry air
density = 1.169 kg/m3

Step 2: Decide the CFM generally it is taken as 400.Converting this in to m3/s .
1 CFM = 0.000472 m3/s
So 400 CFM = 0.1888 m3/s = volume flow rate of air

in terms of kg/s = volume flow rate of air*density of air

so mass flow rate of air in kg/s = 1.169*0.1888 = 0.2207072 kg/s

Step 3:  Divide the cooling capacity (1 TR = 3500W) by mass flow rate, so that we can calculate the change in enthalpy.

change in enthalpy of air = 3500/mass flow rate of air in kg/s
                                        = 3500/0.2207072 = 15858 J/kg = 15.858 kJ/kg

Step 4: So from step 3, the enthalpy of air leaving the evaporator can be calculated as below:

Enthalpy of air leaving the evaporator = enthalpy of air entering the evaporator (27 DBT and 19 WBT) - change in entahlpy

H_out = H_in - Delta H

H_out = 53.78 - 15.858 = 37.922 kJ/kg

 Step 5: Using the psychrometric chart, Denote the inlet designed condition of air (27 DBT, 19 WBT) and the evaporator temperature on saturate line (100 % RH) and find the enthalpy of outlet air calculated above on that line by drawing a line from enthalpy of 37.922 kJ/kg parallel to enthalpy lines cut the line at C as below fig. Point C is the outlet condition of air. (psychrometric chart is copied from internet)

The point C lies on the line, and can find the temperatures and following condition of outgoing air is obtained
DBT = 14.7 deg C
WBT = 13.51 deg C
Enthalpy = 37.922 (calculated in step 4)
RH = 87.9%
Absolute humidity = 9.154 g/kg of dry air

From the above data, we can calculate the amount of water condensed per second or per hour as per below:

Amount of water reduced in outgoing air = Absolute humidity of air before evaporator - absolute humidity of air after evaporator

Delta humidity = 10.441 - 9.154 = 1.287 g/kg of dry air

Amount of water condensed per sec = mass flow rate of air*delta humidity

amount of water condensed per sec = 0.22070702*1.287 = 0.28405 g/sec

amount of water condensed per hour = 0.28405*3600/1000 = 1.023 kg/hour

Amount of water condensed per day = 1.023*24 = 24.552 kg/day.

Step 6: Division of the cooling capacity in sensible and latent heat load.

Latent heat of water = 2230 kJ/kg

Latent heat load = amount of water condensed per sec (calculated in step 5)*Latent heat of water

Latent heat load = 0.28405*2230 = 633.43 W

Sensible heat load = mass flow rate of air*specific heat of air*difference in temperature of air in and out of evaporator

Sensible heat load = 0.22070702*(27-14.7)*1.05*1000 = 2850.435W


Second method: (By fixing the out air condition from evaporator)

Step 1: Find the condition of air at designed indoor conditions (27 DBT and 19 WBT) using psychrometric chart

H_indoor= 53.78 kJ/kg   (Enthalpy)
RH % = 46.9
Absolute humidity=10.441 g/kg of dry air
density = 1.169 kg/m3

Step 2: Decide the temperature of air leaving the evaporator let it be 15 DBT and 13.5 WBT.

So find the condition of air at 15 DBT and 13.5 WBT. 

Enthalpy = 37.89 kJ/kg

RH % = 85

Absolute humidity = 9.021 g/kg of dry air.

Step 3: Now from pointing the two conditions of air (Entering and leaving the evaporator) on psychrometric chart, we can find out the evaporator temperature by extending the line joining the air entering and leaving the evaporator till saturation line, we can calculate the evaporator temperature as below:

By extending the line AC to Saturated line on psychrometric chart which meet it at point B. Point B is the temperature where evaporator will balance, which is in this example is 12.8 deg C. 

Step 4: Calculate the difference between enthalpies of air leaving and entering the evaporator.

Delta H = Enthalpy of air entering - enthalpy of air leaving

Delta H = 53.78 - 37.89 = 15.89 kJ/kg

Now divide the cooling capacity(1TR =3500W) to this enthalpy difference to calculate the mass flow rate of air.

mass flow rate of air = cooling capacity/(1000*Delta H)

m_air = 3500/(1000*15.89) = 0.2202643 kg/s

volume flow rate of air = m_air/density of air

Vol_air = 0.2202643/1.169 = 0.18842 m3/s

CFM = Vol_air/0.000472 

CFM = 0.18842/0.000472 = 399.2

Step 5: Amount of water condensed can be calculated as per first method.

Step 6: The sensible and latent heat can be calculated as per first method.