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HANDOUT RAC 6 SEMESTER

ADAPTED FROM PRINCIPLES OF TROPICAL AIR CONDITIONING By Chris A. Ibe, Emmanuel E. Anyanwu

CHAPTER SIX 6.0 AIR CONDITIONING HEAT GAIN Air conditioning systems are required to provide the temperature and humidity levels necessary for thermal comfort in occupied spaces. This is expected to satisfy either comfort cooling or heating needs of even the warmest or coldest day in such locality. This implies that the cooling/heating system must have sufficient capacity to provide thermal comfort during the most inclement weather condition of the locality. During cooling the installed system removes as much heat from the occupied space as is generated. On the other hand a heating air conditioning system must add as much heat into the space as is lost to the environment. Quantities of this heat must be carefully determined to appropriately design air conditioning system for any application. In all cases, the indoor and outdoor design conditions must clearly be specified. While the indoor design condition depends on the activity levels of occupants and the expected use of the conditioned space, the outdoor design condition depends on the locality weather data. It is important to note that the choice of outdoor design condition is the first step in heat load calculations. It influences not only the selection of air conditioning equipment but also the indoor comfort condition. 6.1 LOAD ESTIMATING Installation of air conditioning system and equipment is usually preceded by careful evaluation of their cooling and/or heating demands. This is called load estimating. It is perhaps the most important task in the design of an air conditioning system. Unless a proper assessment is made, the system will not function efficiently, even if other aspects of the design and application are correct. Air conditioning load demand may be either cooling load or heating load depending on whether heat is removed or supplied per unit time to achieve the design condition within the space. The load may also be classified as indoor or outdoor load. Indoor load is the air conditioning cooling and heating demand of the space. When the demand is external to the conditioned space, it is outdoor load. This later load is evaluated from location specific climate data. The cooling load may be defined as the amount of heat which must be removed by the air conditioning equipment per unit time to produce and maintain the desired indoor climate of the conditioned space. On the other hand, the heat gain is the amount of heat which instantaneously enters the conditioned space. In some texts, the two terms are used interchangeably, but in actual sense, the values of the cooling load and the instantaneous heat gain usually vary slightly due to the thermal inertia or storage effect of the building structure. 6.2 CLASSIFICATION OF HEAT GAINS Heat gains may be classified under two main groups, namely:

(a) Sensible heat gain. As the name implies sensible heat gain results in an increase in the dry bulb temperature of the conditioned space. It is transmitted either by conduction or convection or radiation or simultaneously by

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combination of these heat transfer mechanisms. The sources of sensible heat gain which may be external or internal include:

(i) Transmission heat gain through building fabrics such as walls, doors, windows, internal partitions, roofs, floors, etc.

(ii) Effect of direct and scattered solar radiation through glass windows and doors.

(iii) Electric lighting of space (iv) Outside air brought into space, due to infiltration or for the purpose of

ventilation. (v) Occupancy (number of people in the space to be conditioned) (vi) Heat dissipated by miscellaneous equipment or sources such as

motors, office equipment, electric kettles, cooking appliances, etc. (vii) In the tropics, the sun is also a source of sensible heat, directly through

building windows, and by heating the surface it strikes

(b) Latent heat gain. This results in moisture addition to the conditioned space but no change in the dry bulb temperature. The sources of latent heat gain, which again may be internal or external, include: (i) Outside air brought into the space due to infiltration or for the purpose

of ventilation. (ii) Occupancy (iii) Heat dissipated by products brought into space, resulting in the

evaporation of moisture. (iv) Heat generated from other internal sources

6.3 TRANSMISSION HEAT GAIN THROUGH BUILDING FABRICS 63.1 WaIls. Heat flow through an external wall to the inside of a building is not usually a straight forward steady state heat conduction equation where by the heat flow is given by

Q= U A ( to ti) 6.1 Where

U = thermal transmittance (overall heat transfer coefficient) W/m2 K A = area of wall surface (m2) to = outside air temperature ti = inside air temperature, C

The above relationship would clearly lead to an inaccurate computation of the wall heat gain due to the following reasons:

(i) Even though the internal temperature of the space can be kept constant throughout the day, the outside air temperature varies extensively during any one day.

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(ii) Due to the thickness of the wall, the effect of solar radiation striking the outside surface is not immediately felt at the inside surface, but gets there after a few hours. For example, solar radiation that comes in contact with the external wall at say 900am may eventually get to the inside wall surface at 3.00 p.m.

The above phenomena therefore result in an unsteady heat flow situation through the building fabric. The computation of the solar heat gain is therefore rather complicated. Two main approaches have been developed for the determination of the heat gain through a wall, namely:

(i) The Equivalent Temperature Difference (ETD) method and (ii) The Chartered Institution of Building Services (CIBS) method.

Both methods make use of the sol-air temperature concept, which was first developed by Mackey and Wright (1944). 6.4 SOL-AIR TEMPERATURE Sol-air temperature may be defined as the hypothetical outside air temperature which, in the absence of all radiation effects, would give the same temperature distribution and rate of heat transfer into the external surface of the wall as exists with the combination of the actual dry bulb temperature distribution of the outdoor air and the actual solar radiation incident on the surface. Thus, although the maximum outdoor temperature for a design day may be 35C, occurring at 3.00 p.m., the peak sol-air temperature for that day may be about 50C, and may occur at 1.00 p.m. The CIBS Guide gives the following equation for the determination of sol-air temperature, t

tsa = tso + [R ( lt Iw)] 6.2 where t0 outside temperature, C

R0 = Resistance of outside wall surface (0.042 m2 K / W for walls in tropics), It = Intensity of direct and scattered solar radiation W/m2 = Emissivity of outer surface to long wave radiation (0.9 for most building materials) Iw = long wave radiation from black surface at air temperature (Iw = 0 for

vertical wall and Iw = 100 W/m2 for roof ) = absorption coefficient applying to the outer surface of wall or roof (for light

surfaces, = 0.4, and for dark surfaces = 0.9). The relationship for Sol-air temperature developed by the American Society of Heating Refrigerating and Air-conditioning Engineers (ASHRAE) is similar to the CIBS equation.

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Examples 6.1: Determine the sol-air temperature at 4.00 p.m. on February 21st for a dark west facing wall located at latitude 5N, given that the outside temperature is 34C. Obtain the intensity of the solar radiation from Table 6.1. Solution: Sol-air TEMPERATURE tsa = tso + [Rso ( lt Iw)] from equation 6.2 The equation can be further simplified to:

tsa = tso + [Rso ( lt )] since for vertical walls, Iw = 0 Also Rso = 0.042 m2 K/W for tropical walls,

and = 0.9 for dark walls, and lt = 821 W/m2 from Table 6.1

Hence tsa = 34 + (0.042) (0.9) (821) = 34+31 = 65C Ans

It = 1.2 Iv + Id + 0.1Ih

Table 6.1 INTENSITY OF SOLAR RADIATION (It ) FOR FEBRUARY 21, 5N LATITUDE

Solar Time N NE E SE S SW W NW Hor 0700 42 348 594 348 162 42 42 42 165 0800 77 461 821 743 281 77 77 77 420 0900 110 422 800 770 356 110 110 110 650 1000 133 301 649 697 409 133 133 133 830 1100 150 150 426 558 444 162 150 150 950 1200 155 155 155 371 455 371 155 155 995 1300 150 150 150 162 444 558 426 150 950 1400 133 133 133 133 409 697 649 301 830 1500 110 110 110 110 356 770 800 422 650 1600 77 77 77 77 281 743 821 461 420 1700 42 42 42 42 162 522 594 348 165

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Table 6.2 Sol-air Temperatures, tsa for February 21, 5N Latitude

Time Air Temp oC

(for light surfaces ,/ho = 0.026)

N NE E SE S SW W NW Hor

0100 25.4 25.4 25.4 25.4 25.5 25.4 25.4 25.4 25.4 22.1 0200 24.9 24.9 24.9 24.9 24.9 24.9 24.9 24.9 24.9 21.6 0300 24.4 24.4 24,4 24.4 24.4 24.4 24.4 24.4 24.4 21.1 0400 24.1 24.1 24.1 24.1 24.1 24.1 24.1 24.1 24.1 20.8 0500 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 24.0 20.7 0600 24.2 24.2 24.2 24.2 24.2 24.2 24.2 24.2 24.2 20.9 0700 24.6 25.9 40.2 40.2 38.4 29.0 25.9 25.9 25.9 25.8 0800 25.8 27.8 37.8 47.1 45.1 33.1 27.8 27.8 27.8 33.4 0900 27.2 30.1 38.2 48.0 47.2 36.5 30.1 30.1 30.1 40.8 1000 28.8 32.3 36.6 45.7 46.9 39.4 32.3 32.3 32.3 47.1 1100 30.7 34.6 34.6 41.8 45.2 42.2 34.9 34.6 34.6 52.1 1200 32.5 36.5 36.5 36.5 42.1 44.3 42.1 36.5 36.5 55.1 1300 33.8 37.7 37.7 37.7 38.0 45.3 48.3 44.9 37.7 55.2 1400 34.7 38.2 38.2 38.2 38.2 45.3 52.8 51.6 42.5 53.0 1500 35.0 37.9 37.9 37.9 37.9 44.3 55.0 55.8 46.0 48.6 1600 34.7 36.7 36,7 36.7 36.7 42.0 54.0 56.0 46.7 42.3 1700 33.9 35.0 35.0 35.0 35.0 38.1 47.5 49.3 42.9 34.9 1800 32.7 32.7 32.7 32.7 32.7 32.7 32.7 32.7 32.7 29.0 1900 31.3 31.3 31.3 31.3 31.3 31.3 31.3 31.3 31.3 28.0 2000 29.8 29.8 29.8 29.8 29.8 29.8 29.8 29.8 29.8 26.5 2100 28.6 28.6 28.6 28.6 28.6 28.6 28.6 28.6 28.6 25.3 2200 27.5 27.5 27.5 27.5 27.5 27.5 27.5 27.5 27.5 24.2 2300 26.6 26.6 26.6 26.6 26.6 26.6 26.6 26.6 26.6 23.3 2400 26.0 26.0 26.0 26.0 26.0 26.0 26.0 26.0 26.0 22.7

Average 28.8 30.1 31.6 33.1 33.4 32.9 33.3 33.1 31.4 33.1

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6.5 EQUIVALENT TEMPERATURE DIFFERENCE METHOD In this approach, the rate of heat flow into the room through the wall or roof can be obtained by using the relationship,

Q = U. A . te 6.3 Where U is the thermal transmittance of the wall or roof (also known as overall heat transfer coefficient or U-value), (W/m2 oC). A is the area of the wall (m2). te is the equivalent temperature difference across the wall or roof (C). It is important to stress that the factor te should not be confused with te = (t0 ti) which is the difference between the outside design temperature and the inside design temperature. Equivalent temperature difference,

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te takes into account the combined effects of solar radiation, the outside temperature and the flywheel effect of heat storage in the wall or roof. For example, for a very thick wall with large heat storage capacity, the rate of heat flow into the conditioned space will be virtually constant throughout the whole day because of its thermal capacity. In such a case, the te will be equal to the mean sol-air temperature over the 24 hours less the room temperature, (tsa mean ti). Similarly, for a very thin wall with a low thermal capacity, the equivalent temperature difference at the time of maximum heat flow into the room will be approximately equal to the maximum sol-air temperature for the day minus the room temperature, (tsa max ti). An original equation developed by Mackey and Wright (1944) and later modified by Stewart (1948) gives,

te = (tb ti) 6.4 Where tb = equivalent outside temperature determined from the equation

Tables for equivalent temperature differentials determined using the above equations are published by ASHRAE Fundamentals (2005) for various construction materials, orientation and time of the day. Amore recent technique for determining the space cooling load has been published by ASHRAE known as the Transfer Function Method, this second procedure was first introduced in 1972. This method is similar to the Equivalent Temperature Difference method except that different weighting factors are used in determining the cooling load. Consequently, a new set of tables known as the Cooling Load Temperature Difference (CLTD) was developed and has been published by ASHRAE for use in calculating air conditioning load estimates. It should be pointed out that these tables provide a convenient and rapid approach for the practicing engineer who may wish to determine the heat gain for a building or space. Other advantages of this method include the fact that all wall and roof structures can be covered by a few representative classes. Also, by making some simple adjustments, other latitudes and months of the year can be accommodated in the tables. Table 6.4 shows the cooling load temperature difference (CLTD) for sunlit walls on 5N on February 21.

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6.6 THE CIBS METHOD The CIBS method of determining the heat gain through a wall or roof is based on work by Danter (1960).The heat gain through a wall or roof is seen to comprise two components the steady state and the unsteady state or instantaneous factors. As mentioned earlier, the heat flow into a space will vary considerably during any one day due to the variation of the sol-air temperature, the thermal storage characteristics of the particular building fabric and its thickness. On a first approximation, the average heat flow into a structure for any one day may be determined from the standard steady state relationship,

Qmean = U A ( tsa mean ti) 6.6 Where tsa mean = the mean sol-air temperature during the 24 hour period,

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ti = the room temperature, which is assumed to be constant The above relationship would give an acceptable value for the heat flow into a room if the wall or roof were of infinite thickness. Hence, for very thick walls, say 700mm and above, the steady state equation can be used with a reasonable measure of accuracy. On the other hand, if the wall or roof were very thin, the heat flow into the room would vary considerably during the day, and it would be necessary to determine the heat flow at any particular time of the day. The instantaneous heat flow into a room at any time, is given by

Q = U A ( tsa ti) 6.7 Where tsa = the sol-air temperature at time

This equation is suitable for use in applications where the wall or roof is very thin. In actual practice, especially in the tropics, the walls are of medium or light construction, hence the heat flow computation will lie somewhere between the two extreme conditions mentioned above. The maximum heat flow is likely to occur at some later time , from the time, when the sun first strikes the wall surface.

The relationship for the actual heat gain is thus given by,

Q(+ ) = U A ( tsa mean ti) + U A( tsa tsa mean) f (6.8) Where f = the decrement factor of the wall or roof, and is dependent on the thickness.

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The time lag, is dependent on the density of the wall or roof and its thickness. The CIBS has produced diagrams which are used to evaluate f and , (See Fig. 6.1 and 6.2).

Example 6.2: Calculate the heat flow through a south-east facing concrete wall at 4.00p.m. suntime, given the following information:

Thickness of wall 200mm Density of wall 1500kg/rn3 U-value of wall 3.4 W/m2K External colour of wall dark Inside temperature 23C Outside temperature 32C Latitude 5N Design Day February 21

Solution: From figure 6.1, for a wall density of 1500 kg/m3, the time lag , is 6 hours. Therefore, the heat entering the room at 4.00 p.m. first came in contact with the outer surface of the wall some 6 hours earlier, that is, at 10.00 a.m. From Table 6.3, the sol-air temperature at 10.00 a.m. is 65C. The mean sol-air temperature is 37.9C. From figure 6.2, assuming bare wall, the decrement factor f is 0.50.

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Applying equation 6.8, Q(+ ) = 3.4 (37.9 23) + 3.4 (65 37.9) x 0.5 = 50.66 + 46.07 = 96.73 W/m2

6.7 HEAT GAIN THROUGH ROOFS For buildings located in the tropics, the roof plays the most important role when considering such climatic factors as rain and sunshine. If a roof is inadequately designed or constructed, it could provide the greatest problem area in the building facade, particularly in terms of solar radiation heat gain in the dry season, and the ingress of rain water into the building during the wet season. There are basically two main types of roofs, namely:

(i) The homogeneous heavyweight (flat) roof, usually built of reinforced concrete with weather-proof finishing, and

(ii) The lightweight roof, usually of single or double layers (roof and ceiling) separated by an air space.

The roof construction may be either pitched or inclined, and made of such standard material as galvanised steel (popularly called zinc sheets in West Africa), aluminium and asbestos sheets, as well as cement and clay tiles. The ceiling materials may be made of fibre-board, asbestos, cardboard or cellotex sheets. Generally, the thermal performance of the roof is influenced by the thickness and density of the structure. as well as the thermal resistance of the composite material, and the external colour of the roof. For lightweight roofs, the level of ventilation of the air space greatly influences the solar heat gain through the roof. If the air space is freely ventilated to outdoor air, the roof heat gain is drastically reduced. Light-coloured external surfaces are preferred for tropical roofs, as they reduce the amount of solar radiation absorbed by the roof surface. Polished aluminium has the least solar absorptivity factor ( = 0.2), while white surfaces have = 0.3. In tropical climates, the need for regular maintenance of external surfaces of roofs to prevent them from accumulating dust cannot be over stressed. In determining the transmission of heat gains through the roof structure, the same procedure earlier adopted for walls is used, except that instead of the consideration of different orientations, only the horizontal surface is considered.

Thus, Qroof = Ur x Ar x CLTD (6.9) Where

Ur = thermal resistance of the roof (W/m2 K) Ar = projected area of the roof (m2) CLTD = Cooling Load Temperature Difference (C)

The values of cooling load temperature difference (CLTD) may be easily obtained from standard ASHRAE Fundamentals.

Table 6.5 shows the CLTD for flat roofs based on 5 N latitude on February 21.

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6.8 UNCONDITIONED INTERIOR PARTITIONS In load calculations, there are some spaces like stores, toilets and other rooms which adjoin conditioned rooms, but are themselves not conditioned. In computing the heat gain from the unconditioned space, some temperatures lower than the outside design temperature of the conditioned space is assumed. A simple rule of thumb method is to assume a temperature that is 3C lower than the outside design temperature for normal cases. Where the adjoining unconditioned space has special heat generating equipment, for example, a kitchen or boiler room, then between 5 to 10C is added to the outdoor design temperature.

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For normal unconditioned spaces, an empirical relationship for determining the temperature of the space is given by

ts = ti + 2/3 (to t i ) (6.10) Where

ts = temperature of the unconditioned space

The heat gain, Q = UA (ts t i ) (6.11)

Example 6.3: Determine the temperature of an unconditioned store room adjoining a room conditioned at 24C, given that the outside design temperature is 32C.

Method 1: Solution, using rule of thumb method, ts = 323 = 29C. Q = A U (t0ti) (6.12)

Where A = Total area of all external glass (m2) U = thermal transmittance of glass (W/m2K)

Method 2: Using the empirical relationship, ts = ti + 2/3 (to t i ) (6.10)

= 24+2/3 (3224) = 29.3C Answer

6.9 GROUND TEMPERATURE Technical literature gives the value of ground temperature as varying between 10 and 13C. However, for load estimating considerations, it is common practice to assume no temperature differences between the conditioned space and the ground directly beneath the space.

6.10 GLAZING Glazing provides two separate sources of heat gain into a conditioned space namely:

a. Heat gain due to transmission through glass (as a result of the temperature difference between indoor and outdoor conditions), given by :

Q = A U (t0ti) (6.12) b. Solar heat gain, is a result of the combined effects of direct and scattered

radiation on the glass. In tropical climates, this constitutes one of the largest single sources of heat gain into the conditioned space.

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c. The magnitude of the solar heat gain depends on the type of glass, the shading provided (eg. internal blinds, drapery, external louvres or overhangs due to roof construction), the time of the day and the orientation of the glazing. This heat gain may be easily estimated from equation 6.13

Qg = A x MSHG x SC x SLF (6.13) Where A is the area of the glazing (m2)

MSHG is the maximum solar heat gain (W/m2) SC is the shading coefficient SLF is the storage load factor.

The maximum solar heat gain is taken for the appropriate locality latitude, month of the year, and orientation of the glazing (see Table 6.8). The shading coefficient takes into account the type of glazing and its shading device. Table 6.6 provides typical values for shading coefficient. The storage load factor is given in Table 6.7 and 6.7a, and takes into account the exposure of the glazing, time of the day arid periods of operation of the air-conditioning equipment. Minor corrections to solar heat gain include such factors as atmospheric haze, attitude. type of window frame, dewpoint temperature and the hemisphere. Type of Shading Description

Shading Factor

None

Ordinary glass (4mm) 1.00 Regular plate glass (6mm) 0.94 Regular plate glass (10mm) 0.90 Heat absorbing glass (4mm) 0.80 Heat absorbing glass (6mm) 0.70 Heat reflecting glass (gold) 0.27

Internal Ordinary glass plus light Venetian blind/curtain 0.53 Regular plate glass plus light Venetian blind/curtain 0.46

External Ordinary glass with light Venetian blind/louvred Sunbreakers with slats at 45 0.1 5 Regular plate glass with light Venetian blind/louvred sunbreakers with blades at 450 0.14

Table 6.6 SHADING COEFFICIENT FOR GLASS

6.10.1 TRANSMISSION OF SOLAR RADIATION In recent times, it has become fashionable for tropical buildings to be designed with very large glazing areas. While natural day-lighting and aesthetic values are improved, serious overheating problems are experienced in the hot seasons, thereby necessitating extra air conditioning requirements to overcome the resulting thermal deficiencies of the building envelope. When solar radiation strikes a glazed surface, part of the radiation is reflected, another part is transmitted through the glass, while a third portion is absorbed by the glass (see Fig. 6.3). The percentage of solar radiation which enters the conditioned space will depend on the type of glazing material, the incident angle of the solar ray, and the type of shading

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used (where applicable). The total solar heat gain into the internal environment will consist of the transmitted radiation, and about 40% of the heat that is absorbed by the glazing material. A 6mm single pane glass transmits about 77% of the solar radiation that comes in contact with it at an incident angle of 30, about 8% is reflected, and 15% of the radiant heat is absorbed by the glass (6% of which is converted and radiated inside, while the rest remains outside). The total heat gain to the conditioned space will therefore be about 83% of the incident radiation. For a building located at latitude 5N and at a sun time of 1600 hours (4p.m.) on February 21, the maximum solar flux is about 821 W/m2. Hence if a 6mm single pane glass is used for the window material without any shading devices, then about 680 W/m2 of radiant heat would penetrate the conditioned environment.

Fig 6.3 Solar radiation through glass panes

6.10.2 INTERNAL BLINDS In commercial buildings, the most commonly used internal shading devices are retractable venetian blinds. These apparently reduce direct solar radiation and glare. Solar radiation, however, is short wave and once it has permeated through glass into the room, remains trapped within. This is known as Greenhouse effect (see section 6.10.5) The inside temperature of the space continues to rise, thus, internal blinds tend to reduce solar heat gain. As can be seen from figure 6.3 (ii), about 49% of the solar radiation enters the conditioned space when the glazing is provided with an internal venetian blind shading.

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6.10.3 SPECIAL GLASS Two main categories of solar control glasses are available, namely: the absorbing and the reflecting types. The tropical absorbing type glass with approximately 50% absorbing capacity, when used as glazing material, can prevent about 35% of the incident solar radiation from entering the internal space, see Fig. 6.3 (iii). Reflecting type glass material is even more effective for solar control, in that over 60% of the radiant heat of the sun is kept outside the conditioned space when used. It must be pointed out, however, that even though solar special glasses considerably reduce solar gain in buildings, their tinted nature adversely affect day lighting thus resulting in the need for a high degree of artificial lighting throughout the year. This would in turn reflect in the electric power consumption.

6.10.4 EXTERNAL SHADING External shading, clearly, provides the most effective means of reducing solar heat gain in buildings. External solar control can give as much as 90% reduction in radiation heat gain when compared with ordinary single glass, see Fig. 6.3(v). This results from the fact that the suns rays are arrested before they come in contact with the glazing, thus most of the heat is dissipated externally. External shading may be permanent or retractable. Its design, installation and operation calls into play the judgment and creativity of the architect. The effectiveness of the shading device will however depend on its ability to eliminate direct solar radiation and glare while still retaining visibility and day lighting.

6.10.5 GREENHOUSE EFFECT Glass transmits most of its radiation through short wave bands of between 0.3 and 2.8 microns. Beyond this range, the transmission through glass becomes more difficult, and is completely opaque to radiation at wavelengths of 10 microns and above. Thus, glass permits short wave direct solar radiation to penetrate into the building interior, where the heat is absorbed by the objects of the enclosure. However, the heat trapped within the space cannot be emitted outside, since such surface can only emit radiation at long wavelengths. This phenomenon is known as the Greenhouse effect. This results in the overheating of the interior of highly glazed buildings. There is thus the need to provide adequate shading devices or other means through which the effect of direct solar radiation can be reduced. Fig 6.4 illustrates the greenhouse effect.

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Fig. 6.4: Greenhouse effect

The shading coefficients listed in Table 6.6 are typical only, variations to the figures may occur due to external wind velocity, the surface coefficients of the particular glass, the cleanliness of the protection, among other factors. Table 6.7(a) illustrates the storage load factors for solar heat gain through glass for a 12 hour operation at constant space temperature, where as Table 6.7(b) shows the storage load factors for bare glass or external shade. Table 6.8 gives values of the maximum solar heat gain factor for sunlight glass for Latitude 5N.

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6.11 ELECTRIC LIGHTING, MOTORS AND OTHER APPLIANCES The heat gains resulting from these sources are generated within the conditioned space. It is important to establish whether all the heat produced is sensible, or partly sensible and partly latent. For example, the heat dissipated from electric lighting (fluorescent and tungsten) is completely sensible, while heat generated by an electric kettle will be partly sensible and partly latent. A value of between 10-20 W/m2 of floor area is allowed for fluorescent lighting where moderate illumination is required. For the same level of illumination using tungsten lighting the design value of 30 to 60 W/m2 of floor area is adopted. Tungsten lighting generates far more heat than fluorescent lighting. The heat gain resulting from electric motors and other appliances depend to a large extent on the usage of the equipment, its efficiency and whether a section or the whole unit is located within the conditioned space. Generally, the heat dissipated in the conditioned space by the equipment is equivalent to its power rating.

6.12 PEOPLE The heat generated by the occupants of a conditioned space would depend on the number of people, and the type of activity taking place in the space. As the activity becomes more energy-intensive, the amount of heat lost by the body increases. The heat gain from people will be partly sensible and partly latent. Tables are available for determining the rate of metabolism of the body for various activities, see Table 6.9

The effect of heat gain from occupants is particularly significant in such applications as restaurants, theatres, churches, banking halls, and peak periods in supermarkets. In considering heat gains in restaurants, for example, the heat gain from the food being served is also taken into account. Typically, an additional figure of about 20W per occupant is added for restaurants serving hot food with 50% of the heat gain taken as latent, and the other 50% taken as sensible heat gain. Where there is no information as to the anticipated population of the conditioned space, the following densities may be assumed, see Table 6.10.

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6.13.1 VENTILATION This is the outside air intentionally introduced into the space through the air-conditioning equipment and as a design feature to minimize odors and to maintain the freshness of the circulating air. The ventilation rate will depend on the number of occupants, and whether or not smoking is allowed. Rates will vary from a minimum of 0.0025m3/s (2.5 l/s) per occupant where no smoking is allowed to 0.020m3/s (20 l/s) per occupant where heavy smoking is anticipated. Recommended rates are 0.0035m3/s (3.5 I/s) and 0.010m3/s (10 l/s) for non-smoking and average smoking applications respectively. For some special cases, like the design of systems for operating theatres or laboratories where there is a likelihood of air contamination, 100% outside air is recommended, in which case there will be no recirculation of room air through the air conditioning equipment. Ventilation load is usually added through the equipment, thus it imposes an additional load on the unit, rather than the conditioned space, since the air would have been cooled before entering the space. Ventilation sensible heat gain can be estimated from eqn. 6.14 below

qVS = No. of occupants x ventilation rate (l/s) x (h0 hi) 6.14 specific volume of outside air

h0 = enthalpy of dry outside air (kJ/kg)

hi = enthalpy of dry inside air (kJ/kg) Ventilation latent heat gain may be estimated from eqn.6.15 below

qVL = No. of occupants x ventilation rate (l/s) x (hmo hmi) (ho hi) (6.15) specific volume of outside air

Where ho = enthalpy of dry outside air (kJ/kg) hi = enthalpy of dry inside air (kJ/kg) hmo = (actual) moisture enthalpy of outdoor air (kJ/kg) hmi = (actual) moisture enthalpy of inside air(kJ/kg).

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Alternatively, the ventilation sensible heat gain may be approximately computed using the equation,

qVS = m Cp (t0ti) kW (6.16) qVS = 1000 m Cp (t0ti) W (6.17)

Where m = the mass of outside aire required for ventilation, kg/s Cp = the humid heat (or humid air heat capacity) of the outside air, (kJ/kg K)

Similarly, the ventilation latent heat gain, qVL = mL(W0Wi)kW (6.18) qVL = 1000 m L(W0Wi)W (6.19)

Where W0 = moisture content of outside air, kg/kg of dry air Wi = moisture content of inside air, kg/kg of dry air L= the latent heat of vaporization of water at standard conditions = 2454kJ/kg.

Fig 6.5 illustrates the determination of ventilation heat gain on a psychrometric chart

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Example 6.4: A workers canteen located in Ikeja has an anticipated occupancy of 200 people during the peak period. If the inside design condition is 23C DBT, 60% relative humidity and the outside conditions are 32C DBT and 27C WBT, determine the amount of outside air required. Also calculate the sensible and latent heat gain due to ventilation air. Assume that 20% of the people are smokers. Solution: Recommended ventilation requirements:

a. 160 non-smokers @ 3.5 I/s = 560 l/s b. 40 smokers @ 10 I/s = 400 I/s c. Total ventilation= 960 l/s = 0.96m3/s d. Specific volume of outside air, (from the psychrometric chart) V0 = 0.893 m3/kg e. Specific volume of inside air, Vi = 0.853 m3/kg f. Enthalpy of dry outside air, h0 = 32.5 kJ/kg g. Enthalpy of dry inside air, hi = 23.1 kJ/kg h. Moisture (actual) enthalpy of outside air, hmo = 84.5 kJ/kg i. Moisture (actual) enthalpy of inside air, hmi = 50.8 kJ/kg j. Moisture content of outside air, W0 = 0.0206 kg/kg of dry air k. Moisture content of inside air, Wi = 0.0108 kg/kg of dry air l. Mass of ventilation air required

= 0.96m3s = 1.075kg/s 0.893m3/kg

Ventilation sensible load qVS = 1.075 (32.523.1) x 1000W = 10105 W Ventilation latent load = 1.075 [(84.550.8) (32.5 23)] 1000 W = 26,123 W Total ventilation load qvt = 10105 + 26123 W = 36,228 W Note also that the total ventilation load may be determined using the relationship,

qvt = m (hmo hmi) kW (6.20) = 1.075 (84.50 50.8) 1000 W = 36228W qVS = 1000 m Cp (t0ti) W (6.20a)

At STP, i.e 20C and 1.0135 bar, the specific heat of dry air = 1.01 kJ/kg K, the specific heat of water vapor = 1.89 kJ/kg K, taking into consideration that the outside air is humid (in this case, has a moisture content of 0.0206 kg/kg of dry air), then the specific heat of the outside air is computed, as

Cp = 1.01 + 1.89 (0.0206) kJ/kg K = 1.049 kJ/kg K Thus, qVS = 1000 x 1.075 x 1.049 (32 23) = 10,149 W Similarly, the ventilation latent load, qVL = 1000 m L (W0Wi) W

= 1000 x 1.075 x 2454 (0.0206 0.0108) W = 25,853 W The total ventilation load = (10,149 + 25,853) W = 36,002 W

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6.13.2 INFILTRATION Infiltration is the unavoidable outside air which enters the conditioned space through the opening of doors and windows; and cracks in doors, walls and windows. This is mostly common in commercial buildings such as banks, restaurants, departmental stores, etc. with heavy human traffic for transactions. The infiltration due to door opening alone could be quite significant. Infiltration results from differences in pressure between the outdoor and indoor air due mainly to:

a. Differences in their densities resulting from their temperature differences; and height, dimension and position of the occupied space.

b. Effect of wind which depends on its speed and direction Infiltration due to crack depends on its length, width and nature. The crack nature is usually a function of building construction, workmanship and age. Unlike ventilation air, infiltration imposes an additional load on the conditioned space, consequently, every effort is made to introduce enough ventilation air through the unit, so that a slight positive pressure is created within the space such that exfiltration occurs, thus reducing the incidence of infiltration. In determining the heat gain due to the ingress of outdoor air, it is usual to add the ventilation load to the infiltration load. The greater load is sometimes used alone, especially for small applications. This is because the actual load introduced into the space due to either of the two sources will tend to reduce the value of the other, since the pressure build-up in the conditioned space is normally not much higher than the ambient atmospheric pressure. In large commercial buildings like supermarkets and refrigerated warehouses, air curtains are introduced at main door entrances to create a barrier between the conditioned space and the outside environment, thus reducing air infiltration. Two methods are available for computing infiltration load, namely the air change method and the crack method.

a. Air Change Method: An air change may be defined as the number of times the air occupying the space is completely changed per unit time, usually one hour. This method assigns a number of air changes per hour due to infiltration, depending on the number of external walls adjoining the conditioned space. Thus,

The infiltration rate (m3/s) = Volume x air change rate (6.21) 3600

Values of 0.5 to 1.5 air change per hour (ACH) are assigned to conditioned spaces with exposed (external) walls. A value of 0.5 air change per hour is recommended for residential buildings, while 1.0 air change per hour is recommended for commercial buildings. For modern buildings the ACH may be as low as 0.2.

Example 6.5: If the floor area of the canteen mentioned in Example 6.4 is 250 m2, and has a floor to ceiling height of 3.5m, determine the amount of infiltration air entering the conditioned space. Also calculate the sensible and latent heat gain due to infiltration.

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Solution: Volume of space = 250 x 3.5m3 = 875 m3 Assuming 1.0 air change per hour due to infiltration, Infiltration rate (m3/s) = 875 x 1.0 = 0.243 m3/s.

3600 Mass of infiltration air = 0.243 = 0.272 kg/s

893 The sensible and latent heat loads are determined in the same way as for ventilation. Thus, infiltration sensible load, qis = 0.272 (32.5 23.1) 1000 W = 2557 W Similarly, infiltration latent load, qil = 0.272 [(84.5 50.8) (32.5 23.1)] 1000 W

= 6,610 W

Infiltration total load qit = = 2557 + 6,610 = 9,167 W Ans b. Crack Method: The crack method takes into account the effect of wind velocity

on the introduction of outside air through infiltration. Air flow due to the prevailing wind pressure has a positive effect on the windward side of the building and a negative effect on the leeward side. This results in an inflow of outside air through cracks on the building on the windward side, and exfiltration on the leeward side.

The infiltration rate using the crack method is given by: V0 = A C Pn m3/s (6.22)

Where A = effective leakage area of cracks C = flow coefficient which depends on the type of crack and the nature of flow in the

crack P = the difference between the outside and inside pressure, = (P0-Pi) n = an exponent whose value depends on the nature of the crack; and lies

between 0.4 and 1.0 i.e 0.4 n 1.0 Note that P is influenced by difference in wind pressure (Pwind), pressure difference due to stack effect (Pstack) and pressure difference due to building pressurization (Pbld).

P = (Pwind), + (Pstack) + (Pbld) (6.23) Tables are available for the computation of infiltration using this method.

6.14 SAFETY FACTOR In the computation of the space sensible heat load, a safety factor between 5 to 10% is added to the sub-total of the sensible heat load determined from known sources. This percentage takes care of the following factors:

a. Probable errors in the estimate b. Heat gain due to supply duct or piping passing through unconditioned space;

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c. The temperature rise in the supply air stream due to the heat generated by the (electric) fan motor in the air handling unit;

d. Probable air leakage from the supply air duct into unconditioned space.

6.15 SUMMARY OF LOAD ESTIMATION PROCEDURE

The steps required in the determination of the heat gain of a building for comfort air conditioning may be summarised as follows:

(i) Establish outside design conditions. (ii) Determine inside design conditions. (iii) Determine the occupancy and use of the space (iv) Calculate the heat gains through the building structure including fenestration. (v) Calculate the internal heat gains (occupancy, lighting and equipment). (vi) Determine the ventilation and infiltration loads. (vii) Estimate the total sensible and latent heat loads including a safety factor of say

10% (viii) Having determined the load estimate, study the building layout to find out the

space available and the best location for the installation of the air conditioning equipment.

(ix) Select the most economic equipment to meet your requirements.

Example 6.6: A Local Government Authority is to build a multi-purpose community hall (see Fig 6.6). The information below is contained in the design brief:

a. Desired indoor condition: 24C DBT and 55% relative humidity. b. Outside conditions: 32C DBT and 27C WBT c. Design month/day February 21 d. Latitude: 5N e. Occupancy: 600 people with an assumed smoking population of 10% f. Recommended ventilation requirements are

(1) 10 litres/sec for smokers (2) 3.5 litres/sec for non-smokers

g. Lighting requirements are assumed to be 10 watts/m2 of floor area. h. Other heat generating equipment account for 5 kW of heat dissipated within the

space to be conditioned.

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Determine the grand sensible and latent heat gains that the air conditioning plant will have to offset. Also estimate the supply air volume flow rate. The following assumptions may be made,

U-value of roof = 1.25 W/m2K U-value of wall = 1.65 W/m2K U-value of single glass = 5.7W/m2K

Solution: In tackling a load estimate of this nature, it is necessary to carry out some preliminary computation in order to determine the design time to be used in the calculation. For this problem the design month is given as February. The largest factors, apart from perhaps occupancy in the cooling load calculation, are likely to be the roof heat gain and solar heat gain for glass facing west. Usually, the highest outdoor temperatures occur between 1.00pm and 4.00pm and the greatest solar heat gains are likely to occur sometime between these hours. We may therefore need to consider the aggregate solar heat gains for the roof and west facing glass at the hours mentioned above in order to obtain the time when the peak gain will be experienced from the two sources. This time is known as the design time (also known as suntime) Using the formula for roof heat gain, QROOF we have,

QROOF = UR x AR x CLTD

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Also for solar heat gain, through west glass. from Eqn. 6.13. Qg = A x SC x MSHG X SLF

For roof: UR = 1.25 W/m2K AR = 720 m2 From Table 6.5 the following information is obtained:

Suntime (hrs) 13.00 14.00 15.00 16.00 CLTD(OC) 40 43 43 41

For glass: SC = 0.53 (from Table 6.6) MSHG = 761 W/m2 (February) (see Table 6.8)

From table (6.7) the following values for SLF are obtained. Suntime (hrs) 13.00 14.00 15.00 16.00

SLF 0.28 0.44 0.61 0.72

The summary of the estimate is shown in Table 6.11 SUNTIME

Fabric Heat Gain Area, m2 13.00 14.00 15.00 16.00 Roof U A CLTD 720 36000 W 38700 W 38700 W 36900 W West Glass A SC MSGH SLF 32 3614 W 5679 W 7873 W 9293 W TOTAL WATTS 39614 W 44379 W 46573 W 46193 W It is seen from the above computation that the maximum solar load will occur at 3pm. We can also see that the heat from the roof structure is nearly five times greater than the solar heat gain due to west glass. The entire load estimate for the building can now be determined having established the suntime as 3.00 p.m. on February 21. Heat gain due to glazing Assuming ordinary plane glass with light venetian blinds providing internal shading,

U value = 5.7 W/m2K Area of North Glass:

Windows = 2x3.5x2 = 14m2 Doors = 2x2 = 4m2 Total North Glass = 18m2 =

East Glass = 29m2 = West Glass = 32m2 = Total Glass Area = 79m2 =

Transmission heat gain through glass = U.A. (t0 ti) = 5.7 x 79 x (3224) = 3602 W.

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Solar heat gain due to glass qg = A x SC x MSGH x SLF From Tables 6.6, 6.7 and 6.8

West Glass = 32 x 0.53 x 761 x 0.61 = 7873 W East Glass = 29 x 0.53 x 761 x 0.21 = 2456 W North Glass = 18 x 0.53 x 110 x 0.98 = 1028.4 W

Total Solar heat gain due to glass = 7873 + 2456 + 1028.4 = 11,357.4 W Total heat gain due to glazing = 3602 + 11,357.4 W = 14,959.4 W

Walls: Assume solid block-wall with plaster finish (U = 1.65 W/m2K) Area of North wall = (20 x 5 less North Glass area) = (10018) = 82m2 Transmission heat gain through North wall : qi = U x A x CLTD From Table 6.4, CLTD = 6.8C

qi = (1.65 x 82 x 6.8) = 920 W Similarly, for other orientations,

West wall heat gain = 1.65 x 148 x 10.9 = 2662W East wall heat gain = 1.65 x 151 x 19.9 = 4958W

South (internal) wall adjoining unconditioned space of temperature 29C Area of wall = 96m2 South wall heat gain = 1.65 x 96 x (29 24) = 792W

Heat gain due to wooden door U = 2.16 W/m2K A = 4m2 qd = 2.16 x 4 (2924) = 43W

Total heat gain from walls = 920 + 2662 + 4958 + 792 + 43 = 9,375 W

Roof: Asbestos sheet flat roof with cardboard ceiling U = 1.25 W/m2K A = 720m2

CLTD = 43C (see Table 6.5) Roof heat gain = 38,700 W

INTERNAL HEAT GAINS Occupancy: For people, assumed seated and at rest (see Table 6.9) sensible heat from 600 occupants at 75 W/per person = 45,000 W Also from table (6.9), Latent heat gain due to people = 600 x 40 = 24,000 W

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Lighting: For lighting, assume 10 W/m2 of floor area, with unity diversity factor.

Lighting heat gain = 720 x 10 = 7,200 W

Ventilation and infiltration heat gains

For 600 people, of which 10% (60 people) are smokers, recommended ventilation requirements:

540 non-smokers @ 3.5 I/s = 1890 I/s 60 smokers @ 10 I/s = 600 I/s Total outdoor air for ventilation = 2490 l/s = 2.49 m3/s Specific volume of outside air (from the psychrometic chart) = 0.893 m3/kg Enthalpy of dry outside air, h0 = 32.5 kJ/kg Enthalpy of dry inside air, hi = 24.4 kJ/kg (h0 - hi) = 8.1 kJ/kg Mass of ventilation air required = 2.49 = 2.79 kg/s

0.893 Ventilation sensible load = 2.79(81) x 1000W = 22,599 W Actual enthalpy of outside air, hmo = 84.5 kJ/kg Actual enthalpy of inside air, hmi = 51.4 kJ/kg Difference of actual enthalpy = 33.1 kJ/kg Ventilation latent load = 2.79 (33.1-8.1)x1000 = 69,750 W Total ventilation heat gain = 92,349 W

Infiltration For infiltration, assume 0.5 air change/hour

Free volume of hall = 36 x 20 x 4.5 m3 = 3240 m3 Infiltration rate (m3/s) = 3240 x 0.5 = 0.45 m3/s

3600 (Using specific volume of outside air = 0.893 m3/kg)

Mass of infiltration air = 0.504kg/s Infiltration sensible load, as for ventilation, = 0.504 x 8.1 x 1000W = 4082 W

Infiltration latent load = 0.504 933.1 8.1)x 1000 = 12,600W Total infiltration heat gain = 16,682 W

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Table 6.12 SUMMARY OF COOLING LOAD CALCULATION Heat gain source (W) Total (W) % Sensible

Heat % Latent

Heat % Total

Heat Glazing 14,959 13.6 8.0 Walls 9,375 6.0 3.6 Roof 38,700 24.9 14.8 People (sensible) 45,000 29.0 17.2 Lighting 7,200 4.6 2.8 Ventilation (sensible) 22,599 14.5 8.6 Infiltration (Sensible) 4,082 2.6 1.6 Sub-total (sensible) 141,915 Add safety Factor, 5%, 7,095.75 4.8 2.8 Total Sensible heat 149,010.75 100.0 59.4 Latent load People 24,000 22.6 9.2 Ventilation 69,750 65.6 26.6 Infiltration 12,600 11.8 4.8 Total latent heat 106,350 100.0 40.6

Grand Total load 255,360.75 100.0 The Grand Total Load is 255.361 kW (72.96 or 73 tons of refrigeration) Required Air Quantity: From equation (8.20) Required Air quantity V = RSH .

Cp (trm ts) Where, RSH = room sensible heat (kW)

= density of supply air (kg/m3) Cp = specific heat of humid air (kJ/kg K) trm = room temperature (C) and ts = supply air temperature (C)

Assuming a permissible temperature difference of 8C, i.e. (trm ts) = 8C, or (24 ts ) = 8C, Thus ts = 16C. Cp = 1.025 kJ/kg K, and = 1.204 kg/m3

RSH = 149 kW V = 149 . = 15.092 m3/s (15,092 Iitres/s) 1.204x1.025x8

Air change rate = 15.092 x 3600 = 16.77 Air Changes per hour. Ans 3240