Document 2j1mzGXZpmBjzoQQRJ7YLQ87p

>M HEATING VENTILATING AIR CONDITIONING GUIDE 1944 CHAPTER I. THERMODYNAMICS OF AIR AND WATER MIXTURE horizontal on the Mollier Chart; in other words, the humidity ratio of the Supply Air must be the same as that of Inside Air. > Good practice is to limit the temperature of the Supply Air to 105 F. The condition line crosses the 105 F isotherm at 15.6 per cent saturation. The thermodynamic proper ties of the Supply Air are: h = 33.91 Btu per pound of dry air, and W-. 0.00787 lb water per pound of dry air. Accordingly, the weight of dry air to be recirculated is 73,750 -h (33.91 - 25.38) minus 2590 = 6056 lb per hour. The dew-point temperature of the Supply Air is the same as that of Inside Air, namely, 50.8 F. A suitable conditioning process is to (see Fig. 9): (1-2) preheat the Ventilating Air to temperature t; (2-3) mix it adiabatically with Recirculated Air; (3-4) saturate the resulting mixture adiabatically with recirculated spray; (4-5) reheat to 105 F on the condition line. The temperature t must be chosen so that, after preheating, the wet-bulb of the Ventilating Air is 50.8 F. The determination of the preheating temperature t using the data of Table 6, though straightforward, is somewhat tedious. Graphical solution using the Mollier Chart is easier. The answer 37.1 F is obtained by drawing the line A + B so that the length of A is in proportion to the length of B as the weight of Recirculated Air 6056 lb is to the weight of Ventilating Air 2590 lb, and where this line crosses (1-2) temperature t results. The data needed to calculate the quantities of heat required for preheating and sub sequent reheating may now be assembled. Outside Air After Freheating After Mixing After Adiabatic Saturation After Reheating ...0................... 37.1............ .... 60.2...... ..... 50.8.......... ...105.0 ...0.67........ ....... 9.59.......... .... 20.65 ...... ......20.69........ ..... .33.91 ...0.00063...........0.00063.... ......0.00570... .......0.00787... .......0.00787 The quantity of heat required for preheating is 2590 X (9.59 -- 0.67) = 23,100 Btu per hour; and that required for reheating is 8646 X (33.91 --20.69) = 114,300 Btu per hour. A trial balance for the energy accounting may be made. The Ventilating Air brings in energy 1730 Btu per hour; the heat added by the preheating coil is 23,100 Btu per hour; the energy supplied by the spray is (6056 + 2590) X (20.69 --. 20.65) = 340 Btu per hour; the heat added by the reheating coil is, 114,300 Btu per hour; and the total is 139,470 Btu per hour. This is in substantial agreement with the stated requirements of the problem. The Ventilating Air brings in water of amount 1.63 lb per hour; the spray adds (6056 + 2590) X (0.00787 - 0.00570) = 18.76 lb per hour; and the total is 20.39 lb which is in agreement with the stated requirements. STEADY FLOW ENERGY EQUATION . It was previously stated that, in steady flow, the energy convected by the fluid at any section is the sum of (a) kinetic energy due to velocity; (b) gravitational energy due to elevation; (e) enthalpy due to the con7 dition of pressure, temperature and composition of the fluid. A more detailed discussion of item (a) is in order. Kinetic Energy There are reasons to believe that the so-called velocity pressure hv read by a Pitot tube is simply the kinetic energy per unit volume of the fluid immediately upstream from the tube, as application of Bernoulli's Equation suggests. Thus (see Equation 3, Chapter 35), V = 1097.3 J_*v where >* V = velocity, feet per minute. hv = velocity pressure, inches of water at 60 F. d = density of fluid, pounds per cubic foot. 36 (26) In the case of flow through a duct; the velocity pressure is found to vary considerably over the section and a traverse has to be made. The crosssectional area-of the duct is divided into a number of equal concentric areas, and measuring stations are-located at centroidal points in each area along two perpendicular diameters. Usually the ultimate object is to determine an average velocity V from which the weight of fluid crossing the section per unit time can be obtained on multiplying by the crosssectional area of the duct and by the density of the fluid. This is obtained by simply averaging the square roots of ail measured velocity pressures as follows: = ' (^) where V = average velocity, feet per minute. (A^)av = arithmetic average of the square roots of all measured velocity pressures, inches of water at 60 F. But the item of present importance is the average kinetic energy con vected with each pound of fluid. Consistently with' the previous discus sion, this can be shown to be KE = 0.006678 v (28) where KE -- average kinetic energy, Btu per pound. ' v = specific' volume, cubic feet per pound. (**),, = arithmetic average of the 3/2-powers of all measured velocity pressures, inches of water at 60 F. : If the velocity pressure were uniforiii over the 'section, Equations 27 and 28 could be combined td give . - KE (29) But, it is interesting to note that if the velocity varies parabolically from zero at the walls to maximum at the center as it does in the case of purely viscous flow in a circular duct, then the average kinetic energy is twice that given by Equation 29. Example SO. If 2000 cfm of air flows through ail 8 in. diameter circular duct, find the average kinetic energy per pound of air. Solution. The cross-sectional area of the duct is 0:349 sq ft; hence the average flow-velocity is 5730 fpm. If the velocity were uniform over the section, the average kinetic energy would be (5730 4- 13,430)* = 0.182 Btu per pound. But it is more likely that the actual distribution of velocity would approximate that characteristic of viscous flow; hence the average kinetic energy would be more nearly 2 X 0.182 = 0.364 Btu per pound. Gravitational Energy The potential energy due to elevation Z (feet) above any convenient datum is simply Z 4- 778.3 Btu per pound of fluid. In the case of moist air, 37