Document a4OyEo06XevKVJ32eDyramdDB

34 Chapter 1 1945 Guide' 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 Preheating After,., Mixing After Adiabatic Saturation After Reheating t..... ..............................0__________ .37.1________ .60.2.................50.8._______ 105.0 h...................................0.67................ 9.59.20.65...... ....... .20.69.;.33.91 W......... ................... ... 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 ' horn-; 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, iri 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; (c) enthalpy due to the con 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 h, 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 4V = 1097.3 J *v (26) where V -- velocity, feet per minute. hv = velocity pressure, inches of water at 60 F. d = density of fluid, pounds per cubic foot. 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 all 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. Thermodynamics of Air and Water Mixtures "\ 35 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 where KE = 0.006678 v (*?'),, (28) 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 uniform over the section, Equations 27 and 28 could be combined to give KE = (13,430) (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 an 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 -=- 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 + 778.3 Btu per pound of fluid. In the case of moist air, where . TM Z (1 + W) PB-------- 778T~ PE -- average potential energy, Btu per pound dry air. Z = average elevation, feet. W -- humidity ratio, pound water per pound dry air. (30) Enthalpy. No further discussion of enthalpy is required. It may be well to emphasize, however, that enthalpies have been figured on the basis of one pound of dry air. Heat and Shaft Work Between any two sections 1 and '2 in an apparatus through which steady flow occurs, there may be heat absorbed from outside, jg,, Btu per pound of dry air, and shaft work removed to outside, ik, Btu per pound ot dry air. If heat is actually rejected to outside, ig, is intrinsically negative; and if shaft work is actually put in from outside ih, is intrinsi cally negative. -