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HEATING VENTILATING AIR CONDITIONING GUIDE 1940 gravity action). Like the roof ventilator, the stack outlet should be located, so that the wind may act upon it from any direction. With little or no wind, chimney effect depending on temperature differ ence and lower outdoor temperature will produce a removal of air from the rooms where the inlet openings are located. HEAT REMOVAL In problems of heat removal, knowing the amount of heat to be removed and having selected a desirable temperature difference, the amount of air to be passed through the building per minute to maintain this temperature difference can be determined by means of the following equation: where Q = VH c 60 (I -- to) (3) c = 0.24 = specific heat of air. V = specific volume of the air, cubic feet per pound, about 13.5. (See Chapter 1.) H = heat to be carried off, in Btu per hour. Q = air flow in cubic feet per minute. I = inside temperature, degrees Fahrenheit. to = outside temperature, degrees Fahrenheit. For disposing of odors or other air impurities, the amount of outside air to be introduced must be of such quantity to dilute the impurities to a degree that they are no longer objectionable. See Chapter 3 for the minimum of outside air necessary for ventilation. For garage ventilation, sufficient air must be admitted to dilute the carbon monoxide content of the-indoor air to 1 in 10,000 (see Garage Ventilation in this Chapter). Suggested methods for estimating the air flow due to temperature difference alone and to wind alone have already been given. It must be remembered that when both forces are acting together, even without interference, the resulting air flow is not equal to the sum of the two estimated quantities. The same openings have been assumed in both cases, and since the resistance to flow through the openings varies ap proximately with the square of the velocity*, this resistance becomes a limiting factor as the flow through the openings is increased. Recent investigations4 show that the total flow is only 10 per cent above.the flow caused by the greater force when the two forces are nearly equal, and this percentage decreases rapidly as one force increases above the other. Tests on roof ventilators indicate that this is too conservative in the direction of low total flow quantities, but there is in any case a large judgment factor involved. The wind velocity and direction, the outdoor temperature, or the indoor activities cannot be predicted with certainty, and great refinement in calculations is therefore not justified. When designing for winter conditions, an added variable is the heat lost by direct flow through walls and windows and by infiltration. *Loc. Cit. Notes 1 and 2. `This is true for turbulent flow only. It would be more correct to state that the resistance varies approxi mately with V} for high to moderate velocities, with V1-8 for moderate to low velocities, and with the first power of the velocity for very low velocities through small openings. 644 CHAPTER 37. NATURAL VENTILATION Example 1. Assume a drop forge shop, 200 ft long, 100 ft wide, and 30 ft high. The cubical content is 600,000 cu ft, and the height of the air outlet over that of the inlet is 30 ft. Oil fuel of 18,000 Btu per lb is used in this shop at the rate of 15 gal per hour (7.75 lb per gal). Temperature differences are 10 F in summer and 30 F in winter, and the wind velocity is 5 mph in summer and 8 mph in winter. What is the necessary area for the inlets and outlets, and what is the rate of air flow through the building? Solution. The system must be designed for the summer conditions as these are the more severe. The heat to be removed per hour is: H = 15 X 7.75 X 18,000 = 2,092,500 Btu. By Equation 3, the air flow required to remove this heat with a temperature difference of 10 deg is: n - VC 60 (VCI H6-0 to(t) -- 103:2~54 XXn2w6,009vX2c,510n00virr ~ 1V6,,72 cfm. This is equal to 19.6 air changes per hour. The assumption is made that the average temperature difference between indoors and outdoors is the same as the temperature rise of the air from the inlet opening to the outlet opening. Actually, the latter difference is larger and so the value of 19.6 air changes per hour is conservative as it allows for more cooling than is necessary for an average temperature difference of 10 deg. If 196,172 cfm are to be circulated by the force of the temperature difference alone, the area of opening would be, by Equation 2: ,Q 196,172 , ,,,,,, , 9.4 V H (t - <o) ~ 9.4 V 30 X 10 - sq t- If this area of openings were provided, a wind velocity of 5 mph, acting alone, would produce a flow according to Equation 1, of: Q =. EA V = 0.50 X 1,205 X 5 X 88 = 265,100 cfm. If the inlet openings do not face the wind, but are at an angle with it, about half this amount may be considered to flow. A factor of judgment must now be exercised in making the selection of the area of openings to be specified. Apparently 1205 sq ft are a very generous allowance because either a direct wind of 5 mph or an average temperature difference of 10 deg acting alone will more than suffice to carry away the heat, and when the two forces are acting together, the system may have an excess capacity of 25 per cent to 50 per cent, especially if the outlets are made up partially of roof ventilators which employ the force of the wind for producing a suction effect. On the other hand, the wind may at times come from an unfavorable direction/ or its velocity may fall below 5 mph or the building construction may not permit a full 2400 sq ft of inlet window area and an equal amount of monitor or roof ventilator outlet area. In case the two sets of openings are not equal, their effectiveness is reduced. From this example, it must be apparent that while formulas may furnish a reliable guide, the final solution of a problem of natural venti lation requires a common sense analysis of local conditions to supplement and to modify the dictates of the formulas. GENERAL RULES A few of the important requirements in addition to those already outlined are: 1. Inlet openings in the building should be well distributed, and should be located on the windward side near the bottom, while outlet openings are located on the leeward side near the top. Outside air will then be supplied to the zone to be ventilated. 645