Determine air conditioning cooling time

[cktan22]

how can i calculate the require time for an air-conditioning unit to cool from 30 degree celsius to 22 degree celsius? others parameter can be ignore, i would like to know the formula used and the principal of calculation. thanks

 

[Andrew Mason]

how can i calculate the require time for an air-conditioning unit to cool from 30 degree celsius to 22 degree celsius? others parameter can be ignore, i would like to know the formula used and the principle of calculation. thanks

You have to find out the rate at which the air conditioner removes energy from the room and the volume of air in the room. You will also have to factor in the rate at which air in the room is exchanged. You can then work out the time. If you provide those numbers we can help you calculate the time.

AM

 

[cktan22]

room size : 3 m x 7 m, height 3 m
heat load : 10000 btu/h
cfm : 644
velocity: 6 m/s
temperature : from 30 celsius to 22 celsius
chilled water flow rate : 30.8 liter/minute
altitude : sea level

 

[sophiecentaur]

But what about the spec of the A/C unit? That's a very important factor.
An Air Con is a heat pump which will shift heat at a certain rate, depending upon the Power of the motor. They vary greatly in (what we could loosely call) efficiency.
Look up "Heating Seasonal Performance Factor" and "Coefficient of Performance".

 

[russ_watters]

Also, internal heat load and heat transfer from outside may play a significant role.

 

[Andrew Mason]

room size : 3 m x 7 m, height 3 m
heat load : 10000 btu/h
cfm : 644
velocity: 6 m/s
temperature : from 30 celsius to 22 celsius
chilled water flow rate : 30.8 liter/minute
altitude : sea level

Ok. You have a mixture of units so I will use the MKS units. 10000 BTU/h = 2.9 kW = 2900 J/sec.

I am not sure what the heat load means but since I asked for the rate at which energy is removed from the room, I will assume that this is that rate.

Volume of air in room = 63 m³;
Pressure = 1 atm = 101325 Pa;
Initial T =303K
number of moles of air in room = n = PV/RT = 2534 mol.

The heat capacity of air at constant pressure is Cp = 7R/2 = 29 J/mol K

So to reduce the temperature of that room of air by 8 degrees you have to remove:

Q = 29nΔT = 29 x 2534 x 8 = 588,000 Joules

Since the air conditioner removes 2900 Joules/sec. you would need 588000/2900 = 203 seconds or about 3.5 minutes to cool that room.

I take it that you wanted a ballpark figure. The above assumes that the air flow through the cooling coils of the air conditioner is optimal, ignores the slight increase in mass of the room air as temperature decreases at constant pressure, and ignores the need to remove heat from that extra mass.

AM

 

[russ_watters]

Heat load is the heat added to the room by electrical devices and people(etc) in it.

...I missed that it was provided in a follow-up.