So first calculate the sensible heat gain of the room and the latent heat gain of the room.
Sensible Heat Gain q = (h2 –h1)
The heat absorbed by the air during sensible heating may be obtained from the psychometric chart by the enthalpy difference.
The heating of air, without any change in its specific humidity is known a sensible heating.
So we have to do cooling effectively to remove the heat gained.
The cooling of air without any change in its specific humidity is known as sensible cooling.
Let us consider the air to be cooled is sent through the evaporator of the temperature td3. Inlet air temperature is td1 and air leaving the cooling coil is td2, which will be more than td3.
Heat rejected by air during sensible cooling may be obtained by psychometric chart.
q = (h1- h2)
Humid specific heat = CPm
And it's value is taken as 1.ρ022 KJ/kg k
The heat rejected q = CPm (td1- td2).
For air conditioning purpose the sensible heat per minute is given as:
SH= Ma × CPm × Δ t
Ma= Mass of air flowing.
Ma = V × ρ.
V = Rate of dry air flowing in m3/ min.
ρ= density of moist air at 20o C and 50 % relative humidity = 1.2 Kg/ m3
Δ t= td1 – td2.
SH = V ×1.2 1.022 x Δt
SH = 1.2264 x V x Δt (KJ/ min)
SH = 0.02044 V x Δt (KW)
Therefore for sensible cooling, the cooling coil may have refrigerant, cooling water, or cooling gas flowing.
Sensible cooling can be done only up to dew point temperature (tdp) cooling below this will result in condensation of moisture.