SESSION 22
SAMPLE PROBLEM
The following data were obtained from space cooling load survey of a centralized type
Air-Conditioning (AC) system of a Conference Hall:
Sensible Heat Load
Occupants -------------------- 46.6 kW
Lightings --------------------3.0 kW
Air Infiltration ---------------5.5 kW
Heat Transmission ----------- 7.5 kW
Internal Miscellany Loads -- 1.5 kW
Latent Heat Load
Occupants --------------------Air Infiltration -----------------Internal Miscellany Loads ---
5.4 kW
4.3 kW
2.7 kW
Design inside condition is set at 26 C dry bulb temperature and 50 percent relative humidity and
cold air is supplied at 16 OC. The minimum ventilation requirement is 35.5 cubic meter per min
second based on outside air condition of 32 OC dry bulb and 70 % relative humidity.
Recommend an AC system suitable to the set-up above.
a) fan capacity needed that supplies air to the hall in liters/s
b) what is cooling coil probable Bypass Factor ( BF )
c) refrigeration capacity of the AHU in tons
from Psychrometric chart,
@ 26 C dry bulb & 50% RH
hR = 52.5 - 0.2
= 52.3 kJ/kg da
from Law of Conservation of Mass & Energy
with ΔKE & ΔPE nil, applied to the ROOM
qs + qL = ma ( hR – hs )
ma = ( 64.1 + 12.4 )
( 52.3 – 40.8 )
= 6.68 kg da/s
@ 32 C dry bulb & 70% RH
ho = 87.4 – 0.2
= 87.2 kJ/kg da
mo = 35.5 m3/min (1 min/60s)
ѵo m3/kg da
= 35.5 (1/60)
RSHF = qs /(qs + qL)
0.895
= 64.1/(64.1+12.4 )
= 0.661 kg da/s
= 0.84
since ma = mo + mR
then mR = 6.68 – 0.66
@ Room process (@ RSHF)
mR = 6.02 kg da
and t = 16 C, locate pt. s
read, hs = 40.8 kJ/kg da
from Law of Conservation of Mass & Energy
with ΔKE & ΔPE nil, applied @ mixing point
mo to + mR tR = ma tM
0.661 (32) + 6.02 (26) = 6.68 tM
tM = 26.59 C
read hM = 56.9 - 0.21 = 56.69 kJ/kg da
ѵo = 0.895 m3/kg da
from pt. M , straight line is projected to pt. S .
This is the cooling coil process curve, if extended up to
saturation curve, the intersection is the apparatus
dewpoint temperature, tADP = 12 C
a) Qa, liters/hr of supply air to the room
Qa = ma kg da/s ( ѵs m3/kg da ) ( 1000 liter/m3 ) ( 60 s/min )
= 6.68 ( 0.895 ) ( 1000 )
Qa = 5979 liters/s
b) BF = ts – tADP
tM – tADP
= (16 – 12) / (26.59 – 12)
BF = 0.27
c) from Law of Conservation of Mass & Energy
with ΔKE & ΔPE nil, applied to the COOLING APPARATUS
qREF = ma ( hM – hs )
= 6.68 (56.69 - 40.8 )
= 106.4 kW
OR
must be
EQUAL
from Law of Conservation of Mass & Energy
with ΔKE & ΔPE nil, applied to the AIR-CONDITIONING SYSTEM
qREF =
Refrigeration
Load
(Grand Load)
( qs + qL
) + mo ( ho – hR )
Room Load
Outside Air Load
d
qREF = 64.1 + 12.4 + 0.661(87.2 - 52.3)
= 99.57 kW
16 C
tADP = 12 C
qs
qL
A PE workout gym is to be airconditioned at 22 C dry bulb temperature & 60 % relative humidity.
The AC system is designed with mass ratio of outside air to recirculated room air is 1:3. Outside
air is at 32 C dry bulb temperature and 70 % relative humidity.
The cooling apparatus has following specifications:
 DX type cooling coil
 2-row, R-134a refrigerant cooled
 8 OC dewpoint,
 Bypass Factor (BF) of 0.20
Fan heating, duct heat gain and other miscellaneous heat gain are considered nil.
The estimated space cooling loads are:
 Sensible Heat load --------- 75 kW
 Latent Heat load --------- 50 kW
Determine:
a) show the schematic diagram of AC system as described above and plot the
processes on the psychrometric chart completely labeled
b) temperature of cooled air supplied to the gym
c) heating requirement, kW
d) refrigeration load, tons