Document YGvjgmoY8O59dXneyz8M4d6vy

300 CHAPTER 13 1953 Guide cooling equipment is not sufficient to build up the required pressure to offset infiltration, the entire infiltration load'should be included in the cooling load calculations. Total Outside Air Load'. Many cooling coil manufacturers publish tables giving psychrometric data based on the average conditions of the leaving air for various coil temperatures, air velocities, and entering dry-bulb and wet-bulb conditions. When these tables are: used, it is necessary to cal culate the mixed air condition entering the coil, and determine from the tables what coil and air velocity will produce the desired leaving air con ditions as required for the space to be conditioned: When cooling, coils are listed as 80 to 95 percent efficient, the manufacturer indicates that 20 to 5 percent of the air passes through the coil without being cooled. If data of this nature are used, the uncooled portion of the air must be added to the space load before determining the effective air quantity. See later section on Apparatus Dew-Point. To determine the design cooling load caused by the introduction of out side air, the maximum rate of outside-air entry is first established. In some applications the use of special exhausters from the conditioned space may add to the outdoor-air requirements in determining the maximum.rate. Once this design quantity is established, and with the design indoor and outdoor air states known, the cooling load may be computed. There are several methods in use; the more accurate of these require rather detailed calculations. Refer to Chapter 3, and also section on Apparatus-Dew Point in this chapter. The following equations are considered to be of sufficient precision for use at usual design conditions, as their accuracy is within 1 percent. (`Sensible Load q, = Q X 60 X 0.244 X 0.075 a00923\ 0.62 / o - tl) = Q X 1.08 (to -- <i), Btu per hour Latent Load q. = Q X 60 X 0.075 X 1076 (W0 - Wi) = Q X 4840 (Wo -- WiL Btu per hour Total Load q, = q, + q. (7) (8) (9) where Q = rate of entry of outside air, cubic feet per minute, to = outdoor dry-bulb temperature, Fahrenheit, = indoor dry-bulb temperature, Fahrenheit, Wo = outdoor humidity ratio, pounds moisture per pound of dry air. Wi = indoor humidity ratio, pounds moisture per pound of dry air. 0.075 = standard air density, pounds per cubic foot. 0.244 = a constant approximating the specific heat of dry air corrected for mois ture Btu per (pound) (Fahrenheit degree). 1076 = a factor approximating the average Btu released in condensing one pound of water vapor from air. As explained later in the section on Apparatus Dew-Point, some methods of load calculations break down the ventilation air into two parts: one por tion which does not . contact the coil surfaces (i.e. bypasses the coil) in passing through the coil and thus becomes a part of the room load; and a . Cooling Load 1 301 second portion, the remainder of the air which contacts the coil surfaces and is cooled down to the apparatus dew-point. This detailed method ex plained in the literature by Ashley,30 is believed to be very easy.and ac curate to use. Similar equations, therefore, can be written: Space Sensible Ventilation Load, q,i = 0 X 1.08(f0 -- h) X b Space Latent Ventilation Load, 9,,i - Q X 4840(lVo - WO X 6 Remaining Sensible Ventilation Load, ?.x = Q X 1.08(to - <0(1 - b) Remaining Latent Ventilation Load, ?.,= X 4&40(Wo - IF0(1 - b) St = g.i - g.x gei gx . (10) . : ,(11)' (12) (13) (14) where ' b = fraction of air passing through coil which does not contact surfaces, coil by pass factor. Standard air weight (0,075) lb per cu ft) is . recommended for use in all calculations, as this is the basis for rating fans and its consistent use keeps all parts of the calculations in conformity. HOW OUTSIDE AIR LOAD AFFECTS ROOM LOAD Actually, the outdoor air used for ventilation would pass through the conditioning.equipment, and be cooled and dehumidified, to a lower tem perature and humidity ratio than room conditions before entering the room; but for heat-balance purposes the cooling load chargeable to the outdoor air is that corresponding to the difference between the outdoor and indoor air conditions. One important purpose of the cooling load estimate is to determine the conditions and quantity of air supplied to the space. All the various sen sible and latent heat loads within the space must be included. Infiltration must be included in the space load since this air enters the doors and win dows, and its heat and moisture load must be offset by the introduction .of cooler, dryer air to the space. However, since ventilation air is taken through the conditioning equipment and cooled, this portion does- not be come a part of the space load, except a small portion which passes through the coil untreated. To determine the total load on the refrigeration ma chine, the remaining ventilation air load must be included in the grand total load. Example IS: For outdoor design conditions.of 95 F dry-bulb and 75 F wet-bulb, and indoor design conditions of 80 F dry-bulb and 67 F wet-bulb, and for the supply of outdoor air at the rate of 1000 cfm and the exhaust of room air at the corresponding rate, calculate the total, sensible and latent heat gains. Solution: Substituting in Equation 7: q. = 1000 X 1.08 (95 - 80) = 16,200 Btu per hr. From psychrometric data Wo -- 0.01413, Wi = 0.01122. Substituting in Equations 8 and 9: q. = 1000 X 4840 (0.01413 - 0.01122) = 14,100 Btu per hr. gt = g. + ge = 30,300 Btu.