Module 2: Activity No.
Experiment No. 2: Cooling Load Calculations (Solar Heat Gain, Internal Cooling Load and
Heat Gain from Infiltrating Air)
Members:
Dela Rosa, Eid Nolyn 9/25/2024
Mabasa, Charles
Macadadaya, Abdullah
Ramos, Reiven Lee
SFC (Briefing Room)
Cooling Load Calculations
Solar Load through transparent Glasses:
Formula:
Q=SHGF x A x SC x CLF
Where:
𝐴 = 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑇𝑟𝑎𝑛𝑠𝑝𝑎𝑟𝑒𝑛𝑡 𝑆𝑢𝑟𝑓𝑎𝑐𝑒
𝑆𝐶 = 𝑆ℎ𝑎𝑑𝑖𝑛𝑔 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
𝑆𝐻𝐺𝐹 = 𝑆𝑜𝑙𝑎𝑟 𝐻𝑒𝑎𝑡 𝐺𝑎𝑖𝑛 𝐹𝑎𝑐𝑡𝑜𝑟
𝐶𝐿𝐹 = 𝐶𝑜𝑜𝑙𝑖𝑛𝑔 𝐿𝑜𝑎𝑑 𝐹𝑎𝑐𝑡𝑜𝑟
Source: Air Conditioning & Principles System (Pita)
SHGF : 40° N latitude location SHGF=224
April as the reference month
Located West side of the room
SC : Single Glass SC=0.69
Heat absorbing type of glass
¼ inches thick
Without indoor shading
CLF : Glass without interior shading with CLF =0.58
uncarpeted floors
Located West side of the room
Thick Construction Room
17:00 as the time with the highest
value
Calculations:
Glass area: AG =( 0.149 m ) ( 0.126 m ) AG =0.0617 f t
2
( )
2
2 1f t
¿ 0.0188 m 2
0.3048 m
2 2
0.0617 f t (270) AG =16.659 f t
Total glass area
𝑄 = 𝑆𝐻𝐺𝐹 𝑥 𝐴 𝑥 𝑆𝐶 𝑥 𝐶𝐿 F
Cooling load Q=2083.96W
(224)(96−16.659)(0.69)(0.58)
( )
BTU 0.293W
7112.51
hr BTU
hr
Solar Load through Opaque Materials (Wall D): (Refer to Refrigeration & Air
Conditioning by Stoecker & Jones)(For Research):
Formula :
Q=U x A x CLTD
where :
A=Area of theOpaque Surface
U =¿−all Heat Transfer Coefficient
CLTD=Cooling Load Temperature Difference
where :
CLTD=temperature¿ table
LM =Correction for latitude∧month , ¿ table
t R =room temperature ,°F
t a=average outside temperate on a design day , ° F
CLTD t o− ( )
DR
2
where :
t o=outside design dry bulbtemperature
DR=Daily temperature range ,° F
Considerations (Wall D): (source: Air Conditioning Principles and Systems, Pita, 1981)
Type E. Wall (200mm Concrete Block with interior
and exterior finish
Orientation: West
19:00 as the time with the highest value
𝐶𝐿𝑇𝐷 = 49 °𝐹 (From Table 6.2 pg. 126)
𝐿𝑀 = 0 (From Table 6.4 pg. 128) CLTDc =49.78 ° C
CLTDc
t R = 77 °C
t a=t o−( )DR
2
t o=98.6 ,° F
• DR = 16 (Manila, from Table A.9)
t a=98.6− ( 162 )
t a=90.6
CLTDc = 𝐶𝐿𝑇𝐷 + 𝐿𝑀 + (78 − t R ) + (t a −
85)
𝐶𝐿𝑇𝐷 𝐶 = 121. 6 °𝐹 = 49. 78°C
CLTDc = 49°𝐹 + 0 + (78 − 77) + (90. 6 − 85)
1
U=
U ∑R 2 K
U =1.85 m
1 W
U=
0.029+ 0.08+0.18+2 ( 0.066 )+ 0.12
A=16 m ( 3.5 m )
( )
A 2 ft
2
183.73 f t
2
A=56 m
0.3048 m
calculation :
Q=U x A x CLTD
Q= (1.85 ) ( 183.73 f t )
2
( 0.3048
ft )
2 (49.78 ° C ¿
Q=5157.29 W
Note: Wall A and C are inside the Hallway while Wall B is adjacent to an air-conditioned room.
𝐻𝑒𝑛𝑐𝑒; 𝑄𝐴 = 𝑄𝐶 = 𝑄𝐵 = 0
Internal Cooling Load (People Load)
Formula:
𝑄𝑠 = 𝑁 × 𝑆𝐻𝐺 × 𝐶𝐿𝐹
N=Number of people
SHG=Sensible Geat Gain
Cooling Load Factor
𝑄𝐿 = 𝑁 × 𝐿𝐻G
N=Number of people
LHG=Latet Heat Gain
Source: Refrigeration & Air Conditioning by Stoecker & Jones, 1982)
People People count 20
𝑆𝐻𝐺 = 66 𝑊
𝐿𝐻𝐺 = 45 W
People’s Seated, very light work
Activity:
CLF: 12 hours total hours in space
CLF =0.58
1 hour after each entry into space
𝑄𝑠 = 𝑁 × 𝑆𝐻𝐺 × 𝐶𝐿 F
𝑄 = (20) (66𝑊) (0. 58)
Cooling
Load Q=765.6
Q=765.6
𝑄 = 𝑁 × 𝐿𝐻G
(20)(45W ) Q=900 W
Total 1665.6 W
Internal Cooling Load (Lighting Load)
Formula:
𝑄 = 𝑁 × 𝑊 × 𝐹𝑢 × 𝐹𝑏 × 𝐶𝐿 F
where :
N=number of lights
W =light watt rating
Fu=utilization factor
Fb=ballast factor
CLF =cooling load factor
source: Refrigeration and Air-conditioning (Jones & Stoecker, 1982)
Light: Light count N=8
Wattage: Light specification W =40 W
Fu: Assumption Fu=1
Fb: Fluorescent Light Fb=1.20
Typical value
11 hours after lights are turned
CLF: on CLF =0.94
11 hours lights are on
consider vented hanging light
Calculations:
cooling load
𝑄 = 𝑁 × 𝑊 × 𝐹𝑢 × 𝐹𝑏 × 𝐶𝐿 F
𝑄 = (8) (40𝑊) (1) (1.20) (0.94)
Q=360.96 W
Using the inside and outside conditions of the dry-bulb and relative humidity, the humidity ratio
grains H 2O grains H 2 O
is 171 , for the inside condition the humidity is 89 .
lb dry air lb dry air
Heat Gain from Infiltrating Air:
Formula:
QS =1.1 x CFM x ∆ T
where :
CFM =air infiltration flow rate
∆ T =temperature difference
Ql=0. 68 ×CFM ×(ωo−ωi)
where :
CFM =air infiltration flow rate
ωo=outside humidity ratio
ωi=inside humidity ratio
source: Refrigeration and Air-conditioning (Jones & Stoecker, 1982)
ACH: Tight construction of room ACH= 0. 5
Inside humidity ratio: Dry bulb temperature =25 ° C=77 ° F grains H 2 O
89 .
RH =0.65 lb dry air
Outside humidity Dry bulb temperature = = 25 ° C grains H 2O
171 ,
ratio: =89.6 °F lb dry air
RH =0.80
Calculations;
𝐶𝐹𝑀 = 𝐴𝐶𝐻 𝑥 V ; V =volume of room, ACH= Air change per hour
CFM = ( 0.560 )(52.48 ft )(19.68 ft )(11.48 ft)
3
ft
CFM =98.81
mi
Cooling load
QS =1.1 x CFM x ∆ T
( )
3
ft
QS =( 1.1 ) 98.81 (89.6−77)(° F )
mi
( )
btu 0.293W
QS =1369.51
hr btu
hr
QS =401.27 W
Ql=0. 68 ×CFM ×(ωo−ωi)
( )
3
ft
Ql=0. 68 98.81 (171−89)
mi
( )
btu 0.293W
Q l=5509.64
hr btu
hr
Ql=1614.33 W
Over-all data:
Internal Cooling Load Cooling Load (Watts)
Solar Load through Transparent Glasses 2083.96
Solar Load Through Opaque Materials 5157.29
People Load 1665.6
Lighting Load 360.96
Heat Gain from Infiltrating Air 2015.6
Total 11283.41
