How to determine cooling load of a chilled water cooled AHU?

[Rohitha]

Hi All,

If no heat recovery is available in the AHU, and if the AHU is supplied with centralized chilled water line for cooling and if the air flow from the AHU is known, the supply and the exhaust air temperatures are known, appreciate indicate procedure to calculate the cooling load of the chilled water cooled AHU.

Thanks.

[moideen]

Hi All,

If no heat recovery is available in the AHU, and if the AHU is supplied with centralized chilled water line for cooling and if the air flow from the AHU is known, the supply and the exhaust air temperatures are known, appreciate indicate procedure to calculate the cooling load of the chilled water cooled AHU.

Thanks.\

4.5*CFM*ENTHALPY DEFFERENCE
for example:

Air quantity =370CFM
Temperature of air entering the coil DB = 78F
" WB = 67F
Temperature of air leaving the coil DB = 59F
" WB = 57F
Enthalpy of air at 67F WB = 31.62 BTU/lb
Enthalpy of air at 57F WB = 24.48 BTU/lb
enthalpy deference = 7.14BTU/lb
capacity = 4.5*370*7.14= 11888 BTU/lb

[Rohitha]

Thanks and apprecite your reply
Why do we mutiply by 4.5 ? and how the units of capacity become btu / lb , with air flow in CFM pls?

[dougheret0]

Thanks and apprecite your reply
Why do we mutiply by 4.5 ? and how the units of capacity become btu / lb , with air flow in CFM pls?

∆Qs = ∆tdb * Cc * rho * cp * min/hr
∆Ql = ∆W * Cc * rho * QLV * min/hr
∆Qt = ∆h * Cc * rho * min/hr = ∆Qs + ∆Ql

where (IP units)
∆Qs is the change in sensible heat of the flowing air (Btu/hr)
∆Ql is the change in latent heat of the flowing air (Btu/hr)
∆Qt is the change in total heat of the flowing air (Btu/hr)
Cc is the air flow rate (ft3/min)
W is the humidity ratio, lbs H20/lb dry air
h is the enthalpy of the flowing air, btu/lb
tdb is the dry bulb temperature of the flowing air, deg F
rho is the average air density of the flowing air (lb/ft3)
cp is the specific heat of air (Btu/lb/˚F)
QLV is the latent heat of vaporization of water (Btu/lb H2O)

for air at standard conditions: rho = .075, cp = .24, QLV = 1076
∆Qs = ∆tdb * Cc * 1.08
∆Ql = ∆W * Cc * 4840
∆Qt = ∆h * Cc *4.5
Note that also, Qs + Ql = Qt
The above from my PDH course
HVAC Psychrometric Analysis to Avoid Moisture Problems, PDH course M409, www.pdhonline.org