Sensible Load Latent Load Temperature Humidity Heat Transfer This paper focus on cooling load calculation of the meeting hall [4m*15m*7m] in the location of 8.55 north latitude, East longitude 39.27 and Altitude 1726 m elevation above sea level. The total building cooling load consists of inside design condition of building, outside design condition of building, consider building mater and wall facing to sun and etc.by categorized in to sensible and latent heat gain from ventilation, infiltration and occupants. From different Room heat gain component, the total heat load 21,301.66 w.
## 1. INTRODUCTIONDesign information to calculate the space cooling load, detailed building information, location, site and weather data, internal design information and operating schedules are required. Information regarding the outdoor design conditions and desired indoor conditions are the starting point for the load calculation and is discussed below. · Outdoor Design Weather Conditions · Building Characteristics · Operating Schedules · Indoor Design Conditions and Thermal Comfort · Indoor Air Quality and Outdoor Air Requirements Components of
cooling load, the total building cooling load consists of heat
transferred through the building envelope walls Roof doors etc. Heat transfer
by fenestration through windows and skylights. All of the above loads are
sensible loads. Heat transfer through infiltration, consists of sensible and
latent loads. Heat transfer through ventilation is not a load on the building
but it’s a load on the system. It also consists of sensible and latent loads is
called as external load. Sensible and latent loads due to occupants, products
and processes and appliances. Sensible heats due to lighting, and other
equipment are called as internal loads. The percentage of external versus
internal load varies with building type, site climate, and building design. The
total cooling load on any building consists of both sensible as well as latent
load components. The sensible load affects the dry bulb temperature, while the
latent load affects the moisture content of the conditioned space. Buildings
may be classified as externally loaded and internally loaded. In externally
loaded buildings the cooling load on the building is mainly due to heat
transfer between the surroundings and the internal conditioned space. In
internally loaded buildings the cooling load is mainly due to internal heat
generating sources such as occupants, lights or appliances. In general, the
system design strategy for an externally loaded building should be different
from an internally loaded building. Hence, prior knowledge of whether the
building is externally loaded or internally loaded is essential for effective
system design. Therefore, most of the designers
do not use these methods. They prefer a more compact and easy method for
calculating the cooling load of a building. A more basic version for
calculating a cooling load using the transfer function method is to use the one
step procedure, which was first presented in the ASHRAE Handbook of
Fundamentals in the year 2005. This method is called the cooling load
temperature differences (CLTD) method. In this method, hand calculation is used
to calculate cooling load. The general step by step
procedures for calculating the total heat load are as follows: ·
Select inside design
condition (Temperature, relative humidity). ·
Select outside design
condition (Temperature, relative humidity). ·
Determine the overall
heat transfer coefficient ·
Calculate area of
wall, ceiling, floor, door, windows. ·
Calculate heat gain
from transmission. ·
Calculate solar heat
gain ·
Calculate sensible and
latent heat gain from ventilation, infiltration and occupants. ·
Calculate lighting
heat gain ·
Calculate equipment
heat gain ·
Calculate total heat
gain ## 2. METHODS: COOLING LOAD CALCULATIONGiven data is obtained from the
metrology office of Eastern Ethiopia ·
situated at 8.55 north
latitude ·
East longitude 39.27 ·
Altitude 1726 m
elevation above sea level ·
Outside dry bulb
temperature 33.5 ·
Outside wet bulb
temperature 20.4 The design is done to cool a hall
in Adama city. As we all notice that the geographical
location of Adama is on the north and east
hemisphere. Our load estimation based on 15m*4m*7m meeting hall with 80
occupants. There are two unconditioned space on the adjacent two side (south
and west). ## 3. DOMINATIONS OF THE HALL·
Height = 4m ·
Length = 15m ·
Width = 7m Inside design condition for
comfort this is related to the amount of heat being removed from a room, where
a higher number means more cooling power is needed to cool the room. It
determines base on the orientation of the room material characteristic of the
wall and the window, occupants, light and infiltration.
1) Geographical location for a meeting hall in Eastern Ethiopia. 2) Estimation external load
R1 = 0.01 R2 = 0.18 R3 = 0.01 [Source: ASHREA hand book 1981
fundamental chapter 23 table 3A] Therefore, the total resistance of
wall is Rt = 0.01 + 0.18 + 0.01 = 0.2 Therefore, the heat transfer through
the wall due to conduction is Q= U*A*∆T, from a Measurement taken from the
hall; the area of the wall can be calculated as /east face/ A = 15*6-2(4*1.5) - 3 =75 Q= 5 *75* (33.5-20) = Conduction heat transfer through
the wall facing the north Q = UA∆T Where area of wall is A = L*H=
(15*6)-2(3*1.5) -(1.5*2) =78 the transfer though
the wall in Q = 5*78*(33.5-20) = ## 4. WEST AND SOUTH FACESSolar
heat gains due to conduction through the wall/west and south faces/ since there is a direct solar radiation on this side of
the wall the cooling load gain through this wall is considered differently from
the other walls. Heat loss or heat gain
Just as the human body heat exchange processed with the environment, the building
can be similarly considered as a defined unit and its heat exchange process
with outdoor environment can be examined heat energy tends to distribute itself
evenly until a perfectly diffused uniform thermal field to flow by any of these
three forms is determined by the temperature difference between the two zones
or area considered. The greater the temperature
difference, the faster the heat rate flow The cooling load due to conduction
through this wall is given by Q=UA CLTDs The material of the wall is the
same as that for the other walls therefore U =5 . The area of the wall is calculated
in west face to be = 75m The corrected cooling load
temperature difference is obtained from CLTDs :( CLTDS +Lm) *k + (25.5-Tr) + (To-33.5)
CLTD-cooling load temperature
difference and is obtained from table Lm - latitude and month correction k - Color adjustment factor From the table, for heavy weight
concrete wall + finish of 12m the wall construction is group A [ASHREA HAND
BOOK 1981 FUNDAMENTAL CHAPTER 26 For group A wall, we have the CLTD
from the table is CLTD =12 (using solar time of 9 hours)
## 5. LATITUDE AND MONTH CORRECTION (LM)For the month of March and
latitude of 8.33, we have interpolation 8 0.5 8.33 Lm
12 0.5
Therefore, by interpolation Lm = -0.5 The color adjustment factor k=0.5 (permanent light-colored
wall) Therefore ULTDs = [12+ (-0.5)] *0.5+
(25.5-28) + (24-33.5) =6.25 Again, for the south face Q Q=U*A* CLTDc
=5*78*6.25 ## 6. Conduction heat transfer through the doorMaterial of the door, wood For wood type material (1.5 inch)
from table we have resistance of the door is R= 0.33 [ASHREA HAND BOOK 1981
FUNDAMENTAL CHAPTER 23 table 3A] The overall coefficient of heat
transfer, U is U=1/R =3.03 Qd=U*A*∆T Where area of the door A=3
## 7. HEAT TRANSFER THROUGH THE ROOF AND FLOORMaterial selection wood shingles,
plain and plastic film faced /roof/ [ASHREA HAND BOOK 1981 FUNDAMENTAL
CHAPTER 23 table The coefficient of heat transfer U
is U = 1/R R=0.17 U = 5.88
Where A is area of the floor A
=15*7=105 = 5.88 * 105 * (33.5-20) =8334.9w Similarly, the conduction heat
transfer through the floor Qf= U*A* ∆T But the floor for the hall
material for floor -carpet and fibrous pad R = 0.37 U = 2.7 The conduction heat transfer
through the floor is Qf= UA∆T where A- is area
of the floor Qf = 2.70 *105 * (33.5-20) = ## 8. SOLAR HEAT GAIN THROUGH THE GLASSSolar heat gain or passive solar
gain refers to the increase in temperature in a space object or structure that
results from solar radiation. the amount of solar gain increase with strength
of the sun radiation and with the ability of any intervening material to
transmit or resists of radiation. Then from table, we have for west facing
glass For 8 and month of March, SHGF=
769 For 12 and month of March, SHGF=
757 For 8.55° and month of March, SHGF
[source: ASHREA HAND BOOK 1981
FUNDAMENTAL CHAPTER 26 table 311] By interpolation, we have for 13.3
and month of March SHGF=763.02 Therefore, the solar heat gain
through glass is equal to Q=A*Sc*SHGF*CLF = 6 *0.37*763.020.72
=1219.6 w or 1.2196 kW Now considering the conduction
heat gain through glass the cooling load due to conduction is given by Q=UA CLTD where U= the coefficient of heat
transfer for the material of the glass A=total area of glass CLTDs = corrected cooling load
temperature difference From table Assuming solar time of
9 hours CLTD = 1 [Source: ASHREA HAND BOOK 1981
FUNDAMENTAL CHAPTER 26 table 10] The value of CLTD from a table is
calculated for an inside air temperature of 25.5 °C, Outdoor maximum
temperature of 35°C and outdoor daily range 14.6 If the conduction differs from
table explained above the CLTD should be corrected ## 9. RULE FOR CORRECTING CLTDThe rule says that the room air
temperature less than 25.5°C Add the difference between 25.5°C and room air
temperature, if greater than 25.5°C subtract the difference For this particular case, room
temperature is equal to 24°C which is less than 25.5°C Therefore, adding the
difference i.e. 25.5-25 =0.5 to
CLTD we have CLTDc= 1+0.5=1.5 Therefore the cooling load due to conduction Q=UA CLTD but A=6 , U =6.246
## 10. INFILTRATION AIROutside air may get into air
conditioning space in several way 1) As ventilation air, properly, brought in to the room 2) by normal infiltration through walls doors and windows 3) by infiltration through doors 4) By infiltration to replace air exhaust to the outside air Calculation of the quality of air infiltrated
There are 2 methods for
calculating the quality of air infiltrated ·
Air change method ·
Crack method For this particular case the air
change method is used. It is based on average number of air
change the expected in the air conditioning space Infiltration, Qi=
[H*L*W*G]/60(m/min) Where, H=height of the room L=length of the room W=width of the room G =number of air change s per
hours
H=4m W=7m L=15m From the table, the value of air
change per hour for all side exposed room is G = 1.5 air change per hours. [Source: modern refrigeration and
air conditioning for engineering Prof p.s dasae]
Door infiltration (m /min) Door opening per hour *factor from
a table]/60 For hall building door opening
/occupant /hr = 3 Flexible for swinging door the
factor read from the table is equal to 3.0 [Source: modern refrigeration and
air conditioning for engineering prof p.s dasae] Therefore for 80 occupants = 3*80
=240 door opening /hr 240*3/60 =12
Object struck by sunlight absorb
the short-wave radiation from the light and reradiate the heat at longer
infrared wave length. Certain materials and substances such as glass are more
transparent to shorter wave length than longer. When the sun shines through
such material the net result is an increase in temperature QG = radiation transmitted +
inward flow absorbed + conduction heat gain =solar heat gain + conduction heat
gain Considering first the solar heat
gain i.e. heat gain due to radiation and convection Q = A*SC *SHGF *CLF Where A-total area Sc- shading effect SHGF-maximum solar heat gain
factor CLF- cooling load factor From table, for clean glass with
an inside translucent shading SC =0.37 [Source: ASHREA HAND BOOK 1981
FUNDAMENTAL CHAPTER 27 table 35] the cooling load factor for the glass with an
interior shading and west facing fenestration CLF =0.72 [Source: ASHREA HAND BOOK 1981
FUNDAMENTAL CHAPTER 27 table 35] The maximum SHGF is obtained for
the appropriate latitude, month and surface orientation for this purpose Latitude =8.55 Month = March Then from a table,
we have for west facing glass ## 11. LOAD DUE TO OUTSIDE AIRThe load due to the outside air
will be both sensible as well as latent and can be calculated by OASH = 20.43
*(To-Ti) ω and OALH = 50* Qm
(ωo - ωi) ω Where OASH =outside air sensible
heat OALH = outside air latent heat Qm = Volumetric flow of outside air entering the building To =outside design conduction Ti =room design conduction ωo = specific humidity outside air ωi = specific humidity inside air OASH =6902.7w OALH = 50*Qm
(ωo - ωi) ωo and ωi are read from
psychometric chart ωo using ODBT of 33.5 =0.062 kg moisture / kg dry air = 6.2grms/ kg dry air ωi = using IDBT 25°C and IRH 80% ωi =0.015kg moisture /kg dry air = 15 grms/
kg dry air Therefore, the outside air latent
heat gain OALH = 50*39.75*(6.2-15) = - 17490w
The heat gain components that
contribute to internal heat gains are · light · people (occupant) · equipment and appliance ## 12. INTERNAL HEAT GAIN DUE TO LIGHTLight generates sensible heat by
the conversion of the electric power input into heat The cooling load due to heat gain
from lighting is q = W*Ful*Fsa where, q =
heat gain w w=total light wattage Ful=light use factor Fsa =light special allowance factor The total height wattage for a
single fluorescent lamp the wattage is 40w Select 40w The most common measure of light
output or (luminous flux) is the lumens. Light source is labeled with an output
rating in lumens. T12 40-watt fluorescent lamp may
have a rating of 3050 lumens Since there are sixteen (4) lamps
in the room the total light wattage is 16*40 =640 w The light use factor, Ful = Actual wattage in use /insulated wattage =1 for residential application,
stores, etc. =0.5 for work shop So, take Ful
=1 (residential application) Light allowance factor Fsa=1.2 for tube light =1.0 for lacandensate
temp Therefore, the cooling load due to
light is q =640w *1*7.2w = 768w ## 13. INTERNAL HEAT GAINS DUE TO OCCUPANTOccupants give out both sensible
and latent heat qs = no. of occupant * sensible heat gain * CLF ql = no. of occupant * latent heat gain * CLF Assumption -the occupant in the
dormitory are seated and doing a light work (writing) -the occupant stays in hall the
for 8 hours From table sensible heat gain = 70
w Latent heat gain = 60 w CLF for
person = 0.96 [Source: modern refrigeration and
air conditioning for engineering prof p.s dasae] Therefore, the sensible heat gain is
qs = no. of occupant * sensible heat gain * CLF = 80
* 70 *0.96 = 5376 w Similarly, the latent heat gain is
ql = no. of occupant * latent heat gain * CLF =80*60* =4806 w Therefore, the total heat gains
due to occupants Qt = qs
+ ql 33 =5376 + 4800 = 10176 w is total heat gain of the room/ hall ## 14. INTERNAL HEAT GAINS DUE TO EQUIPMENT AND APPLIANCEQs = (installed wattage) * (usage
factor) *(CLF) let CLF =1.0 maximum value Computers and monitors are only
rounding the room From table 9B, for computer =20.50
w usage factor =0.87 Monitor = 575 w usage factor =
0.231 Printer = 8360 w usage factor =
0.3 Qs,c =(2050w) (0.87) *1 =
1783.5w Qs,m =(575w)(0.231) *1 = 132.8 w Qs,p = (836w)(0.3) * 1 = 250.8 w Qs = Qs,c + Qs,m + Qs,p =1783.5 + 250.8 + 132.8 = ## 15. SUMMARY
## SOURCES OF FUNDINGNone. ## CONFLICT OF INTERESTNone. ## ACKNOWLEDGMENTThey author of this paper appreciate Ethiopian Space Science and Technology Institute, Space Engineering Research and Development Directorate staff members for their valuable support and contribution. ## REFERENCES
[1]
ASHRAE
HANDBOOK OF FUNDAMENTALS IN THE YEARS 2005
[2]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 23 table 3A
[3]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 26 table 6
[4]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 26 table 7
[5]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 26 table 9
[6]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 26 table 10
[7]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 26 table 311
[8]
ASHREA
HAND BOOK 1981 FUNDAMENTAL CHAPTER 27 table 35
[9]
MODERN
REFRIGERATION AND AIR CONDITIONING FOR ENGINEERING, PROF P.S DASEA
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