AIR CONDITIONING LOAD CALCULATIONS PRESENTATIONS by EVRAJU

1. PRESENTATION
E VENKATA RAJU
E VenkataRaju-MEP Projects&Facilities management Professional

2. HEATING AND COOING LOADS
To estimate the air onditioningload(Both heating and cooling
loads) the following factors are to be considered.
1.Building size, shape and orientation.
2.Materials of construction
3.Glass areas
4.People
5.Ventilation
6.Motors, lights and appliances
7.Water, gas and electrical services
8.Equipment location
9.Infiltration
10.Location of doors and windows etc..
E VenkataRaju-MEP Projects&Facilities management Professional

3. HEATING AND COOING LOADS
Sensible heat gain: When there is a direct addition of heat to the
enclosed space, a gain in the sensible heat is said to occur. This
sensible heat is to be removed during the process of summer air
conditioning. This heat gain may occur due to any one or all of
the following sources of heat transfer.
1.The heat flowing into the building by conduction through
exterior walls,floors,ceilings,doors and windows due to the
temperature difference on their two sides.
2.The heat received from solar radiation .It consists of
a) The heat transmitted directly through glass of windows,
ventilators or doors, and
b)The heat absorbed by walls and roofs exposed to solar
radiation and later on transferred to the room by conduction.
3.The heat conducted through interior partition from rooms in the
same building which are not conditioned.
4.The heat given off by lights,motors,machinery,cooking
operations,industrial process etc.
5.The heat liberated by the occupants.
E VenkataRaju-MEP Projects&Facilities management Professional

4. HEATING AND COOING LOADS
6.The heat carried by outside air which leaks in (infiltration air)
through the cracks in doors, windows, and through their frequent
openings.
7.The heat gain through the walls of ducts carrying air through
unconditioned space in the building.
8.The heat gain by the fan work.
Latent heat gain: When there is an addition of water vapour to the air of
enclosed space,a gain in latent heat is said to occur. This latent heat is
to be removed during the process of summer air conditioning. This
heat gain may occur due to any one or all of the following sources of
heat transfer.
1.The heat gain due to moisture in the outside air entering by infiltration.
2.The heat gain due to condensation of moisture from occupants.
3. 2.The heat gain due to condensation of moisture from any process such
as cooking foods which takes place within the conditioned space.
4.The heat gain due to moisture passing directly into conditioned space
through permeable walls or partitions from the outside or from
adjoining regions where the water vapour pressure is higher.
Total heat load= Sensible heat gain + Latent heat gain
E VenkataRaju-MEP Projects&Facilities
management Professional

5. HEATING AND COOING LOADS
ROOM TOTAL HEAT LOAD(RTH)=ROOM SENSIBLE
HEAT LOAD(RSH) + ROOM LATENT HEAT
LOAD(RLH)
The ratio of the RSH to the RTH(RSH/RTH) is known as the
Room sensible heat factor.
E VenkataRaju-MEP Projects&Facilities management Professional

6. HEATING AND COOING LOADS
By pass air: All the air passing over the cooling coil cannot
come in contact with the fins/tube surface of the coil due to
the gaps between the fins and tubes. It comes out of the
cooling coil at the same condition at which it entered and
so it is termed as the by-pass air. The sensible and the
latent heats in this by-pass air needs to be removed out.So
it is a part of the room load.
By pass factor: It is the ratio of the quantity of by pass air to
that of the total air passing through the coil.
New terms ,
Effective room sensible heat
Effective room latent heat
Effective room total heat
E VenkataRaju-MEP Projects&Facilities management Professional

7. HEATING AND COOING LOADS
1.SHF= (SH)/(SH+LH)=SH/TH
Where SHF-Sensible heat factor
SH-Sensible heat
LH-Latent heat
TH-Total heat
2. RSHF= (RSH)/(RSH+RLH)=(RSH/RTH)
Where RSHF-Room sensible heat factor
RSH-Room sensible heat
RLH-Room latent heat
RTH-Room total heat
3.EFSHF=(ERSH)/(ERSH+ERLH)=(ERSH)/(ERTH)
Where ERSHF-Effective room sensible heat factor
ERSH-Effective Room sensible heat
ERLH-Effective Room latent heat
ERTH-Effective Room total heat
E VenkataRaju-MEP Projects&Facilities management Professional

8. HEATING AND COOING LOADS
Effective room total heat load: It determines the
quantity and temperature-humidity condition of the
supply-air.
Grand total heat load: It determines the capacity of
the refrigeration plant.
Apparatus dew point: It is the effective surface
temperature of the cooling coil which determines the
condition of (supply) air coming out of the coil.
To find out the air quantity required and to select the
equipment ERSHF,BF and ADP are required.
E VenkataRaju-MEP Projects&Facilities management Professional

9. HEATING AND COOLING LOADS
A. Effective room sensible heat gain(ERSH)
A.1.Solar Gain-glass
Heat gain= A X R X MF
Where A=Area of glass in sq.m/sq.ft
R = Solar gain in kcal/h/sq.m or BTU/h/ sq.ft
MF= Multiplying factor/s for the type of glass, shading etc.
A.2.Solar transmission gain through the walls and roof
The heat gain= A X U X EqTD
Where A=Area of the wall or roof in sq.m or sq.ft
U=Transmission coefficient in kcal/h/sq.m/ deg.cent
EqTD=Corrected equivalent temperature difference in
deg.cent
E VenkataRaju-MEP Projects&Facilities management Professional

10. HEATING AND COOING LOADS
A.3.Transmission gain through the walls and partitions
(In addition to the solar gain through the glass, there is a heat gain by
transmission through the glass because of the temperature difference between
the surroundings and the conditioned space)
Heat gain= A X U X TD
Where A: area of glass/partition in sq.m or sq.ft
U:Transmission coefficients for glass/partitions(from
tables) in kcal/sq.m/deg.
TD: Temperature difference between the surroundings
and the conditioned space in deg C
E VenkataRaju-MEP Projects&Facilities management Professional

11. HEATING AND COOING LOADS
A.4.Heat gain –(1)Through infiltration (2) (by
passed)Fresh Air
(1)Design tables give the infiltration air quantity for
various types of doors/ windows, for observable cracks
and for infiltration due to the opening of the doors.
This load is generally ignored as this can be merged
with the load due to fresh air intake for ventilation.
(1)The room load due to the by passed fresh
air(through the cooling)
= 1.08 cfm X BF X TD (Deg F)
E VenkataRaju-MEP Projects&Facilities management Professional

12. HEATING AND COOING LOADS
A.5.Internal heat gain
a)People: Multiply the sensible heat per person by the
number of persons to get the sensible heat gain due to
occupancy.
b) Lights :
1) Fluorescent light=total wattsx1.25x0.86 kcal/hr
= total watts x 1.25 x 3.4 btu/hr
Incandescent light: total watts x 0.86 kcal/hr
total watts x 3.4 BTU/hr
c)Fan horsepower in the draw through arrangement: HP X
641 kcal/hr or HP X2545 BTU/hr
d)Appliances:
E VenkataRaju-MEP Projects&Facilities management Professional

13. HEATING AND COOING LOADS
A.6. Safety factor:
An additional 5% on RSH is taken as a safety factor and
this also covers items such as heat gain by the supply
duct ,leaks etc
Effective room sensible heat Gain=
A.1.Solar Gain-glass+A.2.Solar transmission gain
through the walls and roof +A.3.Transmission gain
through the walls and partitions+ A.4.Heat gain –
(1)Through infiltration (2) (by passed)Fresh Air+
A.5.Internal heat gain+ A.6Safety factor
E VenkataRaju-MEP Projects&Facilities management Professional

14. HEATING AND COOING LOADS
B.Effective Room Latent Heat Gain
B.1. Room latent heat
a)Infiltration : Ignored
b)Outside air (by passed air)
The heat gain in
kcal/h= 0.706cmhXBF(wo -wi)
BTU/h=0.68XcfmXBF(wo -wi)
wo-wi:difference of moisture content at outside and inside
conditions respectively in grams of water vapour/kg dry air or
grains of water vapour/lb dry air(lb=7000Gr.)
BF=By –pass factor of the cooling coil
E VenkataRaju-MEP Projects&Facilities management Professional

15. HEATING AND COOING LOADS
© People(Occupancy):
Heat gain= nXwXh
Where n: no. of persons
w:moisture released/person in g/person
h:Latent heat of condensation of moisture in
kcal/g or BTU/gr.
In Kcal/ h =n X w X 0.5883
In BTU/h=nXwX0.15
E VenkataRaju-MEP Projects&Facilities
management Professional

16. HEATING AND COOING LOADS
(d) Steam: In industrial or special applications, where
steam is used, the corresponding heat gain has to be
taken into account.
(e) Appliance : If moisture generating appliances are
used in the conditioned space,
Heat gain=w X h
where w: moisture generated per hour in g or gr.
h:latent heat of moisture in kcal/g
In Kcal/h:wX0.5883(w= g.moisture generated per hour)
In BTU/h:w X 0.15 (w= Gr.moisturegenerated per hour)
E VenkataRaju-MEP Projects&Facilities management Professional

17. HEATING AND COOING LOADS
B2. Adding up the valuesof B.1 (a) to (e) and adding up 5%
as safety factor, effective room latent heat(ERLH) is
obtained.
C.Effective Room Total Heat Gain(ERTH)
It is the sum of ERSH + ERLH.
D.Outside air heat
sensible heat: in Kcal/hr=17.28Xcu.m/min(1-BF)XTD(0C)
in BTU/hr= 1.08 X (1-BF) X TD (F)
Latent heat: in kcal/hr=0.706xcmhx(1-BF)(wo-wi)
In BTU/hr: 0.68 X cfm(1-BF)(wo-wi)
E VenkataRaju-MEP Projects&Facilities management Professional

18. HEATING AND COOING LOADS
(E).Return Duct Heat Gain
The gain due to the fan horse power(blow-through
system only)
In kcal/h: 641 X BHP of fan
In BTU/h:2545 X BHP of fan
(F) Grand Total Heat (GTH)
It is sum of Effective room total heat gain +outside air
heat+ Return duct heat gain
E VenkataRaju-MEP Projects&Facilities management Professional

19. HEATING AND COOING LOADS
Determination of air quantity (cu.m/min or cfm)
In cu.m/min= (ERSH)/(17.28(trm -tadp)(1-BF))
cfm= (ERSH)/(1.08(trm -tadp)(1-BF) )
where ERSH=Effective room sensible heat
trm=Room design temp.
tadp=Apparatus dew point temp
BF = By pass factor of the coil.
E VenkataRaju-MEP Projects&Facilities management Professional

20. SERVER /DATA CENTER Heat load
Calculation
1.Room Area BTU = Length(m) X Width(m)X337
= 50 X 30 X337
=505500
2.Window Size and Position
South window BTU=window L(m)XW(m)X870
North Window BTU=window L(m) XW(m)X165
If blinds on the windows multiply by 1.5
If no windows, ignore this.
3.Occupants
Total occupants BTU= No.of occupants X 400
= 15 X 400=6000
4.Equipment
Add together all the wattages for servers,switches,Routers and multiply by 3.5
Equipment BTU= Total wattage for all equipment X 3.5
= 100000 X 3.5=350000
5.Lighting
Take the total wattage of the lighting and multiply by 4.25
Lighting BTU = Total wattage for all lighting X 4.25
=1000 X 4.25=42500
E VenkataRaju-MEP Projects&Facilities management Professional

21. SERVER /DATA CENTER Heat load
Calculation
Total Heat Load=( Room Area BTU + Windows BTU +
Total Occupants BTU + Equipment BTU + Lighting
BTU)
= 505500 + 0 + 6000 + 350000 + 42500
=904000 BTU
=(904000)/12000=75.33 TR
(1 TR=12000 BTU)
Refrigeration plant capacity= 75.33 TR
E VenkataRaju-MEP Projects&Facilities management Professional

22. THANK YOU
E. VENKATA RAJU
20+ years experience in MEP (Building services ) and
Facilities services i.e. HVAC , ELECTRICAL , FIRE FIGHTING ,
PLUMBING , SOLAR SYSTEMS ,WATER&WASTE WATER
TREATMENT SYSTEMS etc..