Document 7q332KKBRDKzmGwnKvBKDOwE

American Society of Heating and Ventilating Engineers Guide, 1934 SURFACE COOLING UNIT METHOD The fourth type of system employing a surface cooling unit of the extended-surface type will cool and dehumidify. A refrigerant is cir culated or expanded within the unit and air is blown over it. This system has the advantages of lower initial cost and low operating cost, and it does not require as much attention as a spray-type system'. For comfort cooling, water is usually used as the refrigerant. If a refrigerant with a temperature lower than 32 F is used, care must be exercised in the design to prevent frosting. Low temperature refrigerants often reduce the moisture content of the air lower than is usually required. Water at 45 to 50 F is quite practical, using a 10 to 15 deg rise in water temperature depending on the quantity of water used and the velocity of the water required through the tubes.' Although surface cooling units have advantages over the conventional spray-type systems, they are usually adaptable for both cooling and heat ing, and they can be used to control humidity in summer. The effective cooling accomplished by the unit is dependent upon many variable factors. The air. velocity through the unit, air temperature, moisture content, water temperature, and velocity of the water through the tubes, must all be considered in designing the unit. If any of these factors vary without a corresponding variation of the other factors, the effective cooling obtained by the unit will drop off. ------------- .Fresh Air 95" f. <db> 75" F. (whl 822 F <dt" 169 lb. per min. \S66~|_Fc_ <wb> pi 68* F (db) 58% F. twttt 52* F. (dp) Return Air 80 F (db.) 65s F <wb> 977 lb. per min. (dp) Water Out 60* F Water In 50* F 11461b. per min. Fig. 5. Diagram of Surface Cooling Method Example 4- Surface Cooling Method. (Fig. 5). Assuming the same required condi tions and load as in Example 1, a surface cooling unit of proper size and capacity may be determined by the following calculations: d Outside air...................... 169 lb per minute at 95 F (db), 75 F (wb) Recirculated air.............. 977 lb per minute at 80 F (db), 65 F (wb) Air leaving unit.............. 1146 lb per minute at 68 F (db), 59.56 F (wb), 54.17 F (dp) Water, in ................... 50 F Water, out..................... 60 F Water flowing counter to air flow. It will be noticed from Fig. 5 that the dew-point temperature of the entering mixture is lower than the leaving water temperature but higher than the entering water tempera ture. Condensation of the moisture in the entering air will therefore occur some place in the unit where the temperature of the pipe or surface is lower than 57 A F. Condensation will continue to the leaving end of the unit and will result in a dew-point temperature of the leaving air between 50 F and 54.17 F, or approximately 52 F. With this dew-point temperature and the dry-bulb temperature of the air leaving of 68 F, which has been set by pre-determined calculations, the new wet-bulb temperature will be 58M F*. '. *It can. readily be shown that if the entering water temperature were 54.17 F and the other conditions were such that-the leaving dew-point temperature were 54.17 F as desired, the refrigeration load would be the same as for the by-pass and local recirculation systems.' These^conditions, however, have not been approached in practice to date. 130 Chapter 9--Central Fan Air Conditioning (Systems The total heat to be removed by the water will be Total heat at 66)4 F (wb) = 30.77 Total heat at 58J4 F (wb) = 25.20 5.57 Btu per pound of air 1146 X 5.57 = 6383 Btu per minute Refrigeration = ~ 31.90 tons _' , . 6380 Btu per minute Quantity of water required = -----io deg X 8 33 = c gpra As the cooling efficiency is dependent partly on the velocity of water through the tubes, this must be determined from a manufacturer's catalog. For instance, one manu facturer gives the formula y _ (gallons per minute) X 1.235 36 ^ where V = velocity in feet per second. 36 = number of tubes per unit. 1.235 = a constant for the particular sized pipe used. Using Formula 3, the velocity would be, 76.6 X 1.235 36 = 2.63 fps To determine, the amount of cooling, surface required it is necessary to ascertain the mean temperature difference between the air and water, and'the coefficient of trans mission. The formula for the mean temperature difference is: (T, - T,) - (F, - T.) 7d (4) where 7d = mean temperature difference. Ti = entering dry-bulb temperature of air. Ti = leaving water temperature. T, = leaving dry-bulb temperature of air. T, -- entering water temperature. . (82)4 - 60) - (68 - 50) 7d = , f82)4~60\ = Ugel 68 -- 50 / 22)4 - 18 /22)4\ -- 20.2 F ***(-*) The coefficient of transmission is usually taken from the manufacturers data, knowing the water velocity through the unit and the air velocity over the tubes. The velocity of the water.through the tubes has been determined, and assuming a velocity of 600 fpm for the air the coefficient of transmission is (from manufacturer's curves): K = 9.7 Btu per square foot surface per mean temperature difference. " The cooling surface required is usually based upon the sensible heat load. The latent heat due to condensation is taken out at the same time the sensible heat is extracted, and no extra surface is required unless the latent heat exceeds approximately 40 per cent of the total heat. This is due to the higher coefficient of transmission factor because of the wetted surface. This factor holds fairly consistent up to a point where the latent heat 131