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Determine air conditioning cooling time

  1. Aug 27, 2013 #1
    how can i calculate the require time for an air-conditioning unit to cool from 30 degree celcius to 22 degree celcius? others parameter can be ignore, i would like to know the formula used and the principal of calculation. thanks
     
  2. jcsd
  3. Aug 27, 2013 #2

    Andrew Mason

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    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
     
  4. Aug 27, 2013 #3
    room size : 3 m x 7 m, height 3 m
    heat load : 10000 btu/h
    cfm : 644
    velocity: 6 m/s
    temperature : from 30 celcius to 22 celcius
    chilled water flow rate : 30.8 liter/minute
    altitude : sea level
     
  5. Aug 27, 2013 #4

    sophiecentaur

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    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".
     
  6. Aug 27, 2013 #5

    russ_watters

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    Also, internal heat load and heat transfer from outside may play a significant role.
     
  7. Aug 27, 2013 #6

    Andrew Mason

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    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
     
  8. Aug 27, 2013 #7

    russ_watters

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    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.
     
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