Cooling Load Calculation
Designed for 12th April /17.00, 25.5 0C And 50% Relative Humidity
Room Dimension: 6.3*3.69*2.77 m3
In figure; W for Window & D for door
Window Dimension: 0.88*1.36 m2
Door Dimension: 1.14*2.12 m2
Open space, North Side
W
West Side
Unconditioned Room at 30 0C
Unconditioned Room at 30 0C
D
East Side
W
Partition walls (West, East & North
Wall) : with 127 mm thickness brick and
25 mm plaster on both sides.
Roof: Type 6 (No insulation) 150 mm
l.w.concrete without suspended ceiling
Floor: Type 6 (No insulation) 150 mm
l.w.concrete
South Wall: Type D with 127 mm
thickness brick and 25 mm plaster on
both sides.
Door: 2.42 m2 area ; 40 mm thickness
hardwood.
South ( Sunlit Wall )
W
W
Window: 1.2 m2 area on north & south
side.5 mm thickness glass; light
construction.
Air Exchange Rate: 1.0/hour
Lights: 4 fluorescent light of 25 watt
Occupants: 4 occupants (Approximately 15 hours) using 4 computer
(Approximately 5 hours in a day)
Heat Conduction Through Surfaces:
1) Conduction through shaded or partition surfaces
Q=U.A.TD
2) Conduction through sunlit surfaces
Q=U.A.CLTD
U→ Overall Heat Conductance
A→ Surface Area
TD→ Temperature Difference
CLTD→ Cooling Load Temperature Difference
Overall Heat Conductance, U:
ROOF:
From table -12(a)
For type 6 Code numbers of layers→ A0,E2,E3,C15,E0
RTotal=0.059+0.099+0.050+0.880+0.121 [ From table -16]
=1.209
URoof=1/RTotal=0.827
( Construction of roof and floor is same so Ufloor=Uroof=1.827)
South Wall (Sunlit) :
From table -14
For type D Code numbers of layers→ A0,A2,C4,E1,E0
RTotal=0.059+0.0 (No Insulation) +0.175+2*0.034+0.121 [ From table -16]
=0.423
USouth-wall=1/RTotal=2.364
West, East and North Walls (Partition wall) :
RTotal=Ri+R1+R2+R3+RO
=E0+ R1+R2+R3+A0
R1=R3→ Resistance for Plaster
R2→ Resistance for brick
From table -16
For Brick: C4→ K=0.727; R2=L/K=0.127/0.727=0.175
For Plaster: E1→ K=0.7277; R1=R3=L/K=0.025/0.7277=0.034
So, RTotal=0.121+0.034+0.175+0.034+0.059 [ From table -16]
=0.423
UPartition-wall=1/RTotal=2.364
Door:
R=Ri+Rd+Ro
=E0+Rd+A0
From Table -17
For Hard-Wood (Oak) ; K=0.16 → Rd=L/K=0.040/0.16=0.25
So, R=0.121+0.25+0.059
UDoor=1/R=2.33
Window:
From Table-17
Conductivity of cellular glass is 0.050w/mk
So, R=L/K=0.005/0.05=0.1
RTotal=Ri+Rw+Ro
=Eo+Rw+Ao
=0.121+0.1+0.059
=0.28
UWindow=1/RTotal=3.57
Conduction Heat Transfer Through Partition/Shaded Walls/Surfaces:
Item
Description U
A
TD
Q=A.U.TD
i)Partition Wall
East Side
2.364
17.451
4.5
185.644
ii)Partition Wall West Side
2.364
17.451
4.5
185.644
iii)Shaded Wall
North Side
2.364
5.411
7.5
95.937
iv)2Window
North Side
3.57
2.3936
7.5
64.089
v)Door
North Side
2.33
2.4168
7.5
42.234
Vi)Floor
Bottom Side 0.827
23.247
4.5
86.514
∑=660 W
CLTD For Sunlit Walls:
CLTDC=[(CLTD+LM)*K+(25.5-Ti)+(Toavg-29.4)]*f ------(i)
Here, CLTDC=Corrected Cooling Load Temperature Difference
LM=Correction for latitude and month
K=Color adjustment factor; 0.5 for light color & 1.0 for dark color.
f=Attic fan factor; 1 for no attic fan/duct &0.75 for +ve ventilation
From Table-9
Toutside-maximum=330C
Daily range=110C
So, Toavg=33-11/2=27.50C
CLTD For Roof:
Table 12(b) → CLTD=350C
Table-13→ LM=0.0
K=1 & f=1
So from equation (i) CLTDC=33.10C
CLTD For South Wall:
Table 15 → CLTD=150C
Table-13→ LM=-1.6
K=0.5 (For light color) & f=1
So from equation (i) CLTDC=4.80C
Conduction Heat Transfer From Sunlit Roof Or Walls:
Item
Description
U
A
CLTDC
Q=U.A.CLTDC
i)Roof
Type-6
0.827
23.247
33.1
636.356
ii)South Wall
Type-D
2.364
7.8277
4.8
88.822
iii)2 Windows
South Wall
3.57
2.3936
7
59.82
∑=785.0 W
Solar Heat Gain Through Glasses:
Q=A.SC.SHGFmax.CLF
Where,
A=Heat transfer area
SC=Shading Co-efficient
SHGFmax =Maximum solar heat gain factor
CLF=Cooling load factor
North Side Glass (2 Glasses, Shaded):
So, Q=(1.2*0.95*126*0.75)*2=215.46 Watt
[SC→Table-21, SHGFmax →Table-18(b),CLF→Table-19]
South Side Glass (2 Glasses, Sunlit):
So, Q=(1.2*0.95*237*0.43)*2=232.35 Watt
[SC→Table-21, SHGFmax →Table-18(a),CLF→Table-19]
So, Total Heat Gain Due To Glasses = (215.46+232.35)Watt=447.81Watt
Cooling Load For Air Exchange:
Sensible Heat Gain =ρcpὺ(To-Ti)
ρcp=1200 J/m3k
ὺ=6.3*3.69*2.77/3600=0.0179 m3/s
To=330C (Table-9), Ti=25.50C
So, Sensible Heat Gain =ρcpὺ(To-Ti)=1200*0.0179*(33-25.5)=161.1 Watt
Latent Heat Gain=ρhfgὺ(ω0- ωi)=3010*1000*0.0179*(0.021-0.010)=619.61 Watt
[ ρhfg =3010*1000 J/m3, Ti,dry_bulb=25.50C, RH=50%; So, From psychometric chart
ωi =0.010 kg/kg ;To_dry_bulb=330C, To_wet_bulb=270C, ω0=0.0215 kg/kg ]
Heat Gain Due To Occupants:
QS=75*4=300 Watt; QL=55*4=220 Watt [From table-28]
Heat Gain Due To Equipment:
QS=P*CLF=200*1*4=800 Watt
[ P=Wattage of each computer, Number of computer =4]
Heat Gain Due To Lights:
QS=PL*BF*D*CLF
Where, PL =Wattage of light=25 Watt*4=100 Watt [ No of light is 4]
BF=Balast factor=1 for incandescent light
=1.2 for fluorescent light
D=Diversity factor for not all lights being on=1 (Let all time on)
CFL=Cooling load factor=0.85 for light
So, QS=PL*BF*D*CLF=100*1.2*1*0.85=102 Watt
Item
QS
QL
i)Conduction heat gain through
partition/shaded walls/surfaces
ii) Conduction heat gain through sunlit
wall/roof
iii)Solar heat gain through glasses
660
iv)Heat gain due to air exchange
161.1
619.61
v)Heat gain due to occupants
300
220
vi)Heat gain due to equipments
800
vii)Heat gain due to lights
102
QTotal
785
447.81
∑All=
4095.52
Watt
So, QTotal=4095.52 Watt=4.0955 Kilowatt=4.0995/3.517 =1.164 Ton of refrigerator
So, 1.5 Ton refrigerator is required for my room.
(Room No. 5005; Dr. M. A. Rashid Hall; BUET)
Calculating a Home's Cooling Load
1. Determine the dimensions of the room or home that needs cooling. For a
room, you can do this with a measuring tape. ...
2. For a room, multiply its length by its width to get the square footage.
3. Multiply the square footage by 20. This is the measure of the BTU cooling
load of the space.
Selecting the Right Air Conditioning Unit for Your Needs
Now that you know how many BTUs are needed, it is time to select from the
myriad possibilities for air conditioning systems. Our heating and cooling service
technicians are happy to provide you with recommendations for units that we know
to be reliable and energy efficient. Another way to research is to use consumer
review publications and websites. When looking for a new air conditioning system,
be sure to choose one with the highest SEER (Seasonal Energy Efficiency Ratio)
rating that fits into your budget. The higher the SEER rating, the less electricity the
system will consume in cooling the space.
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Corporation. In order to achieve this honor, each one of our service team members
has surpassed Carrier’s intense requirements for skill, knowledge and customer
service on all types of HVAC related issues. We offer a 100 percent customer
satisfaction guarantee on all maintenance, repairs and installations of Carrier
products. Our company maintains a full supply line of Carrier factory authorized
replacement parts and new heating and cooling products.
At Berkeley Heating & Air Conditioning, we have helped home and small business
owners in the Hanahan, SC, community achieve and maintain indoor comfort for
many years. By helping you estimate the necessary cooling capacity, you can be an
informed consumer when a new air conditioner is needed. Give us a call any time
for more information.