Elementary Principles 118 - 5.17 Basis: Given flow rates of...

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5- 6 5.17 Basis: Given flow rates of outlet gas. Assume ideal gas behavior ± m (kg/m in ) 0.70 kg H O / kg 0.30 kg S / kg 1 2 311 m 83 C, 1 atm n (kmol / min) 0.12 kmol H O / kmol 0.88 kmol dry air / kmol 3o 3 2 /m , ± ± ± in) n (kmol air / min) V (m C, - 40 cm H O gauge 2 2 3 o 2 167 ± m (kg S / min) 4 a. 3 3 3 1 atm 311 m kmol K n 10.64 kmol min 356.2K min 0.08206 m atm == ± 2 21 1 0.12 kmol H O 18.02 kg 10.64 kmol H O balance: 0.70 m kmol kmol min m 32.9 kg min milk = ⇒= ± S olids balance 0.30 32.2 kg min m m kg S min bg b g : ±± . =⇒= 44 96 ( ) 22 Dry air balance: n 0.88 10.64 kmol min n 9.36 kmol min air =⇒ = () 3 2 2 2 3 9.36 kmol 0.08206 m atm 440K 1033 cm H O V min kmol K 1033 40 cm H O 1 atm 352 m air min = ⋅− = ± 3 3 air air 4 V (m/ s ) 352 m 1 min u (m/min)= 0.21 m/s A (m ) min 60 s (6 m) π ± b. If the velocity of the air is too high, the powdered milk would be blown out of the reactor
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