ANALYTICAL METHOD TO CALCULATE ROOM COOLING LOAD

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.


INTRODUCTION
Design 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. In any building, heat is transmitted through external walls, top roof, floor of the ground floor, windows and doors. Heat transfer takes place by conduction, convection and radiation. The cooling load of the building is dependent on local climate, thermal characteristics of material and type of building. For cooling load calculation, there are many types of software like REFPROP6.EXE DATABASE, available which use the transfer functions method and heat balance method. These methods require a complex and lengthy data input.
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).
• 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).

DOMINATIONS OF THE HALL
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. Therefore, the total resistance of wall is Rt = 0.01 + 0.18 + 0.01 = 0.2 2 The coefficient of heat transfer U for the wall is U = 1/Rt = 5 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.

WEST AND SOUTH FACES
Solar 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 2 .
The area of the wall is calculated in west face to be = 75m 2 The corrected cooling load temperature difference is obtained from CLTDs :( CLTDS +Lm) *k + (25.5-Tr) + (To-33.5) Where CLTD-cooling load temperature difference and is obtained from

SOLAR HEAT GAIN THROUGH THE GLASS
Solar 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 If the conduction differs from table explained above the CLTD should be corrected

RULE FOR CORRECTING CLTD
The 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 2 , U =6.246 2 Q= 6.246*6*3.5 =131.16w

INFILTRATION AIR
Outside 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 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 Gedlu Solomon, and Yeshurun Alemayehu Adde (Kibret) 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

Internal Heat Gain
The heat gain components that contribute to internal heat gains are • light • people (occupant) • equipment and appliance

INTERNAL HEAT GAIN DUE TO LIGHT
Light 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

INTERNAL HEAT GAINS DUE TO OCCUPANT
Occupants 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

SOURCES OF FUNDING
None.