Heat load calculation is done to calculate the load of heat for a building or determine the capacity of an air conditioner for a building. Most reliable format is carrier's Form-E20 and one should have Carrier handbook for different temperature differences, building material properties, peak load for different exposures on walls etc. But before performing these calculations building survey is done.
Building survey:- Building survey includes following points
Building survey:- Building survey includes following points
- Building application (Hotel, Shopping Mall, School etc).
- Angle of orientation of building.
- Latitude (deg N) and Altitude (Height from sea level) of building.
- Daily Range - Difference of maximum and minimum dry bulb temperature of hottest day in summer.
- Building exposure.
- Area and Volume of the space to be air conditioned.
- Exposed wall areas according to the exposures (East wall, south wall etc).
- Glass area, Partition wall areas, skylight area etc.
- Thickness of walls and material types.
- Lights, equipments and appliances details.
Sources of Heat :- There are two types of heat sources one is sensible and other is latent heat.
Sensible heat sources -
- Walls, roof and windows exposed to sun.
- Human occupancy.
- Electrical lights, equipments and appliances.
- Glass
- Outdoor air for ventilation or fresh air.
- Partition walls.
Latent heat sources -
- Human occupancy.
- Appliances like coffee machine, laundary equipments or any other appliance that creates steam.
- Outdoor air for ventilation or fresh air.
- Swimming pool, water fountains etc.
HLC Example:- Following is an example to easily understand the heat load calculations.
- A small office dimensioning (L-9.84 X W-8.2 X H-9) feet.
- Area (Convert mm to feet; 3000mm X 2500mm)(9.84' X 8.2 ft = 80.6 sq.ft)
- Daily Range - 42-28 = 14 deg.cel (consider June 23)
- Glass area = (Area of W1+W2) = (2 X 4 X 2) + (3 X 4) = 28 sq. ft.
- Walls area = North wall - (9.84 X 9 )- (3 X 4) = 76.56 sq. ft. ; South wall- ( 9.84 X 9 )-(3 X 7)= 57.56; East wall- (8.2 X 9) - (2 X 4) = 65.8 sq. ft. ; West wall - ( 8.2 X 9) - (2 X 4) = 65.8 sq. ft.
- Roof area - 9.84 X 8.2 = 80.6 sq.ft. and thickness 4 inch.
- Occupancy leveL - 3 persons (generally considered 75sq.ft/person if not confirmed).
- Room's actual Dry Bulb Temperature (DBT)- 105 deg. F and Relative Humidity- 60 % Humidity ratio-208 (From psychrometric chart).
- Recommended design conditions are - DBT - 75 deg. F; Relative humidity - 50 %; Humidity Ratio 65 (From psychrometric chart).
- ΔT = 105-75 = 30 deg F; Difference between Humidity Ratio ΔW= 208-65= 143.
For heat load calculations the following equation is used :-
Q = U X A X ΔT
Where,
Q = Amount of heat in BTU/Hr
U = Heat transfer co-efficient (For different building materials U value will be different)
A = Area in sq.ft
ΔT = Temperature difference
Before starting the calculation we need to get 'U' values for each building materials bricks,stone and glass according to their resistance (R).
For ordinary glass we need Glass Factor and light transmission properties are :-
- Absorptivity - 6%
- Transmitivity - 86%
- Reflectivity - 8%
But only 40 % of heat absorbed, can transmitte through ordinary glass and 60 % will be reflected. So heat gain by glass will be
Heat gain or Glass factor = 0.86 + 0.4*0.06
= 0.86 + 0.02 = 0.88 BTU/Hr/sq.ft/deg.F
Shading factor for glass will be = heat gain by glass / heat gain by ordinary glass
= 0.88/0.88 = 1
Heat gain by glass may vary according to glass properties. In this example ordinary glass has been considered so the values are matching to 0.88.
Similarly 'U' value for Wall will be calculated
'U' = 1/R
Total Resistance for 9" wall
= R(still air) + R (1/2"+1/2" plaster) + R (9" brick) + R (outside air)
= 0.68+(0.1+0.1)+1.6+0.25
= 2.73
Resistance for all building materials can be obtained from carrier handbook but resistance for still air and outside air can be considered same as taken in above example.
Now we have total Resistance = 2.73
So, 'U' value will be = 1/2.73 = 0.36 BTU/hr/sq.ft/deg. F
In same way we calculate 'U' value for ceiling slab
For 4" (100mm) slab 'U' value will be
= 0.31 (Only if slab is directly exposed to sun and without insulation)
Now we can start calculations to obtain heat load.
The heat transfer takes place in two different ways one is solar and other is transmission heat gain so we need to consider both.
A. Solar heat gain by glass
North window Q = A X Peak heat gain X GF [ Peak heat gain from carrier table 15]
= 12 X 24 X 1
= 288 BTU/hr/sq.ft/deg.F
East window Q = 8 X 164 X 1
= 1312 BTU/hr/sq.ft/deg.F
West window Q = 8 X 164 X 1
= 1312 BTU/hr/sq.ft/deg.F
Total solar heat gain by glass = 288 + 1312 + 1312 = 2912 BTU/hr/sq.ft/deg.F
B. Solar and Transmission heat gain by exposed walls :-
North wall - A X U X (ΔTe + ΔT)
- 76.56 X 0.36 X (8+30)
- 1047.34 BTU/hr/sq.ft/deg.F
East wall - 65.8 X 0.36 X (25+30)
- 1302.8 BTU/hr/sq.ft/deg.F
West wall - 65.8 X 0.36 X (28+30)
- 1373.9 BTU/hr/sq.ft/deg.F
South wall - 57.56 X 0.36 X (18 + 30)
- 994.46 BTU/hr/sq.ft/deg.F
Total = 4718.04 BTU/hr/sq.ft/deg.F
C. Transmission heat gain by all glass
Q = A X ΔT X Utg { Utg, means transmission gain From Table -33}
= 28 X 30 X 0.55
= 462 BTU/hr/sq.ft/deg.F
D. Solar and transmission heat gain by roof
Q = A X U X (ΔTe + ΔT)
= 80.6 X 0.31 X (37+30)
= 1674.06 BTU/hr/sq.ft/deg.F
E. Internal sensible heat gain
(a). Occupancy = No of persons X 245 = 3 X 245 = 735 BTU/hr/sq.ft/deg.F
(b). Lights = W X 3.41 X 1(Blast factor)
= 500W(assumed) X 3.41 X 1
= 1705 BTU/hr/sq.ft/deg.F
F. Transmission heat gain by partition walls, if any - A X U X ΔT
G. Sensible heat gain by, By-passed air
Q = CFMOA X 1.08 X ΔT X BF (By pass Factor)
CFMOA = (No of persons X 20) = 60 CFM
Q = 60 X 1.08 X 30 X 0.11 = 213.84 BTU/hr/sq.ft/deg.F
H. Latent heat gain by, By-passed air
Q = CFMOA X 0.68 X ΔW X BF
= 60 X 0.68 X 143 X 0.11= 641.78 BTU/hr/sq.ft/deg.F
I. Outside air sensible heat gain
Q = CFMOA X 1.08 X ΔT X CF (Contact Factor)
= 60 X 1.08 X 30 X 30 X 0.89
= 1730.16 BTU/hr/sq.ft/deg.F
J. Outside air Latent heat gain
Q = CFMOA X 0.68 X ΔW X CF (Contact Factor)
= 60 X 0.68 X 143 X 0.89
= 5192.61 BTU/hr/sq.ft/deg.F
K. Room Latent heat gain = No of persons X 205
= 3 X 205 = 615 BTU/hr/sq.ft/deg.F
L. Infiltration, Duct leakages or Duct heat gain + 5 % safety
= 12.5 % of sum of (A:F)
= 12.5 % of 11471.1 = 1433.88
safety 5 % = 1433.88 + 5% of 1433.88
= 1433.88+71.69 = 1505.57 BTU/hr/sq.ft/deg.F
ERTH Effective Room Total Heat
= ERSH Effective room sensible heat gain + ERLH Effective room Latent heat gain
= 9740.94 + 615
= 10355.94 BTU/hr/sq.ft/deg.F
OATH Outside air total heat
= Sum (G to J)
= 7778.39 BTU/hr/sq.ft/deg.F
GRAND TOTAL HEAT
= 18134.33 BTU/hr/sq.ft/deg.F
OR = 18134.33/12000
= 1.5 TR