Question

can you help me how to get the h, h* and (h*-h)L (highlighted in yellow colour)?

EXAMPLE 19.3. A counterflow induced-draft cooling tower operates with inlet and exit water temperatures of 105 and 85°F when the air has dry-bulb and wet-bulb temperatures, respectively, of 90 and 76°F. The tower has 4 ft of stacked plastic fill, and the flow rates are G, = 2,000 lb/h. ft? and G: = 2,200 lb/h. ft?. (a) Determine the number of transfer units, the height of a transfer unit based on the overall gas-phase driving force, and the temperature approach. (b) If the cooling load remains the same but the air temperature drops to 70°F with a wet-bulb temperature of 60°F, predict the water temperature and the temperature approach. Solution (a) From the psychrometric chart (Fig. 19.2) the inlet humidity and humid heat are determined. 0.017 Ho = 0.017 lb water/lb air HR= 100 = 55% 21 Cs = 0.248 Btu/lb. °F Hy = 0.248(90 – 32) + 1,0756.017) = 32.7 Btu/lb 2,200(1.0)(105 – 85) = 2,000(Ha - 32.7) Ha = 54.7 Btu/lb The points Txa= 105, H = 54.7 and Txb = 85, H = 32.7 are plotted in Fig. 19.10 as the endpoints of the operating line. The number of transfer units is obtained by determining the driving force at the middle of the column and using the log mean AH to get the number of transfer units for each section. The H H* -H (H* – H) ΔΝ 85 41.5 55.5 32.7 43.7 54.7 8.8 11.8 18.3 1.08 10.2 14.8 73 0.74 Noy = 1.82 Hoy = 1.82 = 2.2 ft The temperature approach is 85 – 76 = 9°F. (b) For T, = 70°F and Tw = 60°F, Ho = 0.009 cp = 0.244 Hy = 0.244(70 – 32) + 1,075(0.009) = 18.9 Btu/lb For the same cooling load and constant water and airflows 2,200(1.0)(20) = 2,000(H. - 18.9) H.= 18.9 + 22 = 40.9

Answer 1

Ans. Given that , Intel temperature = 105 °F Erit water temperature = 85°F Gy = 8,000 1 /h-ft? Gx = 2,200 16/n-ft? Day bulb temperature - 90°f wet bulb temperature = 765 and the relative The intel humidity & 0017 Iblwater/ Ibair humidity is 0.017 x 100 0.031 = 55.1. Humid heat fg = 0.248 Btu/16f This data are obtained by Bychometric chart Hg = 0-248 x(90-32)+10753(0.017) HB = 32.7 Btu/16 9200x1x (105-85) = 2000 XC HA-32.7) Enthalat Enthalpy balance Povuation) HB = 32.7 Bulld HA = 53.7 Btu/lb Draw the operating line (HVST) as we know todo points the operating line can be constructed and the nuber d

transfer units can be found by dividing total change by log mean AH (average driving force) Here the log mean driving force in section 1 is obtained as (11.8 - 8:8)][en(11.8/818)] Foo Example DNI is obtained as (5417-32.7). 10. 2 Total change in 1108 the section divided by. avg H-H | H*_HJE doiving force | AN. . . H I 95 55.5 | 43.7 10.2 1.08 18.3 14.8 074 Noy = 1'82 Hoy * Noy = Z(Height) = 464 Hoy - 2aft temperature apparach = 85-76 = 9F

6.) If air temperature doops to 7of and wht for Then Humidity = 0.009 Humid Heat -0.0244 HB = 0.944 (70-32) +10 75(0.009) - 18.9 HA = 40.9 The opcoating like that Noy & 1.82 by to ial) The approach 75-15 = 155