Document xzQYjenL3RLDoxogz81pY61NG

56 CHAPTERS ' 1948 Guide are the two enthalpies calculated. Conservation of energy requires that the difference between these two enthalpies be the quantity of heat removed, or refrigeration supplied, between the two sections. Therefore, -ask = 43,072 - 34.187 = 8.885 Btu/lba The initial volume is 13.980 + 0.50 X 0.822 = 14.391 cu ft/lba. Since 20,000 cfm of air is to be processed, the total refrigeration required is -aQb = 8.885 X 20,000 4- 14.391 = 12,348 Btu per minute On the Mollier Diagram the process is represented by the horizontal line AB, Fig. 3, whose length is the quantity of refrigeration required per pound of dry air. Adiabatic Mixing of Two Air Streams A typical air conditioning process requiring special analysis is the adiabatic mixing of two air streams. Referring to Fig. 4; let mi, mj, m> denote the weights of dry air converted across sections Fi, Fj, F>, respect ively, per minute. Then nttWi, miWi, m3W3 and mjtt, mjiz will denote the weights of water and the quantities of energy similarly con verted. If the mixing is adiabatic, it must be governed by the three equations, m\ + m* -- mj m\W\ + niiWt = m%W% (8) m\h\ + mtht =* ththa Elimination of m% gives,. hi - hi IV, - W, h, - h, ~ W, - W, m, m. (9) according to which: on the Mollier Diagram the state point of the resulting mixture lies on the straight line connecting the state points of the two streams being mixed and divides the line into two segments which are in the same ratio as are the weights of dry air in the two streams. Example 8. Outside Air at 0 F and 80 per cent saturation is to be mixed adiabatically with recirculated Inside Air at 70 F and 20 per cent saturation in the ratio of one pound of dry air in the former to seven in the latter. Find the temperature and degree of saturation of the resulting mixture. Solution.. The humidity ratio W, and the enthalpy hi of the resulting mixture must satisfy Equations 9, namely, 0.003164 - W, 20.270 - hi Wi -- 0.000630 hi -- 0.668 1. 7 I Thermodynamics 57 from which: Wi = 0.002847, hi = 17.820. At the temperature of the resulting mixture, therefore, ha + 0.002847 has/W, = 17.820 At 61 F the Iefthand member of this.equation has the value 17.750; at 62 F its value is 17.991; by interpolation the temperature of the resulting mixture is 61.29 F where the humidity ratio at saturation, also by interpolation, is 0.01161; hence the degree of saturation of the resulting mixture is . p = 0.002847 4- 0.01161 = 24.52 per cent On the Mollier Diagram, Fig. 5, a straight line is drawn between point 1 (0 F, 80 per cent) and point 2 (70 F, 20 per cent); then point 3 is located on tie line one-eighth of the distance from point 2 to point 1. The temperature and degree of saturation at point 3 are read directly. Adiabatic Mixing with Injected Water Another typical air conditioning process is that of injecting water into an air stream to mix with it adiabatically. Let Wt -- Wi, denote the increase in humidity ratio of the air; this is obviously the quantity of water injected per pound of dry air; it follows that the quantity of energy injected per pound of dry air is (JF -- Wijh^, where A* denotes the specific enthalpy of the water as injected; if the process is adiabatic this produces an equal increase in the enthalpy of the air, namely, hi -- h\\ therefore, ht -- 'hi = hy,(Wt -- Wi) (10) according to which: the process of injecting water into an air stream to mix adiabatically with it is represented by a straight line on the Mollier Diagram whose direction is fixed by the specific enthalpy of the water as injected. The protractor drawn on the Mollier Diagram provided with this book provides a convenient means for determining this direction. Example 9. It is desired to increase the humidity ratio of air at 70 F dry-bulb, without changing its temperature. Under what conditions may water be injected in order to accomplish the desired result?.' Solution. At 70 F the increase of enthalpy per unit increase of humidity ratio is has/Ws = 17.27 4- 0.01582 = 1092 Btu per pound of water. This must be.the specific enthalpy of the water added if the state point of the air is to be moved along the 70 F isotherm. Saturated steam at 668 F has this specific enthalpy*. On the Mollier Diagram, Fig. 6, it is seen that the 70 F isotherm is parallel to the line on the protractor for a specific enthalpy of 1092 Btu per pound. Adiabatic Saturation Any process by which the state point of moist air is moved to the saturation curve adiabatically may properly be called adiabatic satu ration. Example 10. Liquid water chilled to 35 F is evaporated into an air stream initially at 90 F and 50 per cent saturation. How much water must be evaporated to bring the air to saturation at what temperature? Solution. The initial enthalpy of the air is 21.625 + 0.50 X 34.31 = 38.780 Btu/lba; the initial humidity ratio is 0.50 X 0.03118 = 0.01559 lbw/lba; the specific enthalpy of the chilled water is 3.06 Btu/lbwl therefore, the temperature at which the air reaches the saturation curve must be such that the enthalpy hs and humidity ratio Wt at satu ration satisfy the equation, h* - (W, - 0.01559) X 3.06 = 38.780 The solution is 75.19 F where the humidity ratio at saturation is 0.01894; consequently,