Split_00056 - 8.78 a. U V = 4 C W bh g Tdb = 45 C Tdew...

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8-56 8.78 a. T T h TV a wb db dew point Fig. 8.4-1 in 2 3 C C kg H O kg D.A. C, m kg D.A. U V W = = 45 4 00050 20 4 0 908 bg . . ± . TT h wb as a == ° = 20 4 0 0151 .. C, saturated kg H O kg D.A. out 2 b. Basis: 1 kg entering sugar (S) solution m 1 (kg D.A.) m 1 (kg D.A.) 0.0050 kg H 2 O/kg DA 0.0151 kg H 2 O(v)/kg 1 kg m 2 (kg) 0.05 kg S/kg 0.20 kg S/kg 0.95 kg H 2 O/kg 0.80 kg H 2 O/kg Sugar balance: 005 1 020 025 22 . b g =⇒ = mm kg Water balance: 11 0 0050 1 0 95 0 0151 0 25 080 b gb g b gbg b gb g b g . . . += + m V 1 74 67 = kg dry air 74 kg dry air 0.908 m 1 kg D.A. m 3 3 8.79 aa 3 1 lb D.A. m h (lb H O) m2 T d = 20°F h r = 70% Coil bank 1 lb D.A. m h (lb H O) T d = 75°F Spray chamber H O 2 1 lb D.A. m h (lb H O) Coil bank 1 lb D.A. m h a (lb H O) T d = 70°F h r = 35% A B C D 3 a 2 1 Inlet air (A): T h h V db r a = U V W 20 70% 0 0017 12 2 1 F lb H O lb D.A. ft lb D.A. Fig. 8.4-2 m 3 m . ± . Outlet air (D): T h h r a db Fig. 8.4-2 m F 0.0054 lb H O lb D.A.
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