Mathematical modelling of an air conditioner using evaporative cooling (without compressor)(COP =10.99)

Air conditioners are used to provide thermal comforts in extremly hot conditions. But these are very high power consuming machines. An attempt is made to make an air conditioner using the both form of evaporative cooling.
Evaporative cooling is the most powerful and cheapest cooling technology. When air blow over the water, it gets cool. Water evaporated with blowing air, utilize latent heat from water itself thus reducing overall temperature of air and water.

For details in evaporative cooling, Please click on below link:

http://athermocreation.blogspot.in/2017/10/evaporative-cooling.html


As we know to increase the temperature of one liter of water by one degree, Sensible Heat required is 4.187 kJ.
But to evaporate one kg of water, latent heat required is 2250 kJ which is almost 550 times the sensible heat.

So to lower the temperature of water only a small portion of water evaporation required.

Evaporative technology is of two types:

1. Direct Evaproative technology
2. Indirect Evaporative technology

1. Direct Evaproative technology:

This technology used mostly in hot and dry areas. The hot and dry air is blown over wet evaporative pad which is at the wet bulb temperature of air. The water get evaproated and blown with air by taking heat from remaining water on pad and cool itself and air to almost WBT of air and increase the humidity in the air.

2. Indirect Evaproative cooling technology:

It is the advancement of direct type evaporative cooling technology. The humid and cold air coming out from the wet evaporative pad is made to exchange heat with the ambient air(DBT 35 and WBT 24) using a cross flow heat exchanger. Benefit of this type of evaporative cooling is air coming from indirect evaporative cooling pad is  having less humidity(with in comfort zone).

schematic of this is shown in below fig 1.

Consider ambient hot and dry air at condition A, is made to pass through an evaporative pad with the help of fan with 400 CFM.
This hot and dry air come in contact with the evaporative pad and absorb water by taking latent heat of water in pad thus cooling itself and water as well and reached in high humid state B.

Then this cold and high humid air is made to exchange heat with the ambient air at condition A with the help of a cross flow heat exchanger and reached again in hot and dry condition D.

The air stream at condition C is directed towards the air conditioning area.

All the above processes are shown on psychrometric chart as below:

Assumptions:
1. there is no loss if heat to surrounding.
2. Direct evaporative process is an isenthalpic process.
3. Heat transfer with in the cross flow heat exchanger without any loss.
4. There is no pressure drop of air in evaporative pad as well as in cross flow heat exchanger.


Calculations with 400 CFM are as per below:

At Condition A:

Enthalpy of air: 71.66 kJ/kg
DBT of air:        35 deg C
WBT of air:       24 deg C
Density of air:    1.136 kg/m3
Specific heat of air: 1.032 kJ/kgK
Humidity of air: 14.216 g/kg

Volume flow rate of air = CFM*0.000472 = 400*0.000472 = 0.1888 m3/s

Mass flow rate of air = Volume Flow rate*Density of air= 0.1888*1.136 = 0.21447 kg/s

At condition B:

Enthalpy of air: 71.66 kJ/kg
DBT of air:        25 deg C
WBT of air:       24 deg C
Humidity of air: 18.438 g/kg

Heat transfer from A to B (Q1): mass flow rate of air*specific heat of air*(Delta of DBT)*1000

Q1 = 0.21447*1.032*(35-25)*1000 = 2213 W (sensible only)

At condition C:

Enthalpy of air: 61.41 kJ/kg
DBT of air:        25 deg C
WBT of air:       21.2 deg C

Heat transfer from A to C (Q2): mass flow rate of air*(delta enthalpy)*1000 W

Q2 = 0.21447*(71.66-61.41)*1000 = 2198.32 W (sensible only)

At condition D:

Enthalpy of air: 81.87 kJ/kg
DBT of air:         35 deg C
WBT of air:        26.4

Heat Transfer from B to D (Q3): mass flow rate*(delta enthalpy)*1000 W

Q3 = 0.21447*(81.87-71.66)*1000 = 2189.74 W

Finally air at condition C will enter in to the air conditioning area and air at condition D will go out.

COP:

Power consumption of direc Evaporative fan: 100 W
Power consumption of indirect evaporative fan: 100 W

Total power consumption: 200 W
Total cooling capacity: 2198.32 W

COP = total cooling capacity/total power consumption

COP= 2198.32/200 = 10.99 W/W

Total water consumption: mass flow rate of air*(delta humidity from A to B)*3600/1000 Liters/hr

Total water consumption: 0.21447*(18.438-14.216)*3600/1000 = 3.26 liters/hr

Conclusion:

Direct Evaporative cooling effective only in extremly hot and dry areas like deserts where humidity is very low and it is less effective in humid condition. So to provide comfort in humid areas and in dry areas as well, Indirect evaproative cooling is better with low humidity and temperature.