Document jmyjRrLgEbj8nmd9o73N8rKD5

158 CHAPTER 7 1949'Guid* (2) This establishes the maximumwhole number of coil-rows that can be.used as 6 and it is now possible to determine the actual location of the exit air conditions from Equatidnt by solving for the actual value of fa -- for a 6 row coil. : ............. 10.7 X 15 X 6 0.243 X 1740 'log. 102-80 S3 U idua 2.275 . This establishes values of 9.78 for ------- = R and 2.25 for It -- (dp>. 1* -- idp (3) Next, the exit air condition at 57.3 F dry-bulb and 56 F wet-bulb as shown at B, is found by locating a point on the load ratio line at a horizontal distance of 2.25 dry. bulb degrees from the saturation curve. (4) The surface temperature may now be found from Equation 7 which may also be written as: where Bl, ~t, l- ~--r h -- <dPi tt -- <dp It =* 9.78 X 57,3 - 102 im ' 62.3 (5) The total coil load may be calculated from the enthalpy difference across the coil and the air quantity using the weight of dry air instead of the weight of the mixture. 9t = <7. (hi - ft.) = 1700 (49.24 - 23.77) = 43,200 Btu per (hr) (eq ft of face area) where <7, = weight of dry air per (hour) (square foot of coil face area). . fti = enthalpy of air vapor mixture entering coil, Btu per pound of dry air. hi = enthalpy of air vapor mixture leaving coil, Btu per pound of dry air. (6) The refrigerant temperature may be found from Equation 9 43,200 325 15 X 6 X 15 (t. - (,) 22.1 Therefore, = (52.3 - 22.1) = 30.2. Thus a coil 6 rows deep, operating at a refrigerant temperature of 30.2 F and a face velocity of 400 fpm, is required; and it will carry a total load of 43,200 Btu per (hour) (square foot of face area). The air conditions leaving the coil are too low for the con ditions of the problem and therefore it,is necessary to by-pass air at the entering condition-to obtain the desired result of 80.5 F dry-bulb and 73 F wet-bulb. Although the preceding solution is satisfactory, it may be'more desirable in some cases to use a higher refrigerant temperature and employ reheat to obtain the desired load ratio. Such a solution is shown in Fig. 3. In this case the coil load ratio line intersects the saturation curve and therefore a coil of any depth may be selected. If a coil depth of 6 rows is maintained, the exit air conditions for the coil are indi cated at point B Fig. 3 as 72.3 F dry-bulb and 70.8 F wet-bulb and the surface temper ature will be: . . t. 9.78 X 72.3-- 102 8.78 69.0 X &&&& of Air Heating and Cooling Coils 159 ]oaa vriU be: qt = 1700 (49-24 - 34.66) = 24,800 Btu per (hour) (square foot f face area) and the refrigerant temperature will be found from Equation 9: 24,800 325 (t. - t.) - 12.7 15 X 6 X Fig. 3. Pstchhombtbic Layout for Coil Selection Using Reheat (square foot of face area) but the actual effective load will be less by the amount of reheat required. Therefore, for a given load, a larger coil and more refrigerating capacity are required when reheat is used. LETTER SYMBOLS USED IN CHAPTER 7 7 fin efficiency. -. A = external area of coil, square feet per (square foot of coil face area) (row of coil depth). g --- ^ tf fdpl tdpi D -- internal diameter of tube, inches. G -- air mass velocity', pounds per (hour) (square foot of coil face area). (?* = dry air mass velocity, poundB dry air per (hour) (square foot of coil face area). \ hi = enthalpy of air-vapor mixture entering coil, Btu per pound of dry air. h* = enthalpy of air-vapor mixture leaving coil, Btu per pound of dry air. h* = film coefficient of heat transfer between air and external coil surface, Btu per (hour) (square foot external surface) (Fahrenheit degree mean temperature difference between air and coil).