Elementary Principles 317

# Elementary Principles 317 - 8.61(cont'd N 2 510 K H N...

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8- 38 8.61 (cont’d) N 510 K K) - K) = C) - C) [6.24 - (-1.05)] kJ / mol = 7.286 kJ / mol = 7286 J / mol 2 oo Table B.8 bg : ± ( ± ( ± ( ± ( HH H H NN N N 22 2 2 510 262 237 11 = B A( v , 262 K ): ±± HCT H K Cd T pl b v pv T b =− + + z 262 359 262 b g Δ Part (a) results for T b C pl C pv H v ,, , ± Δ ± .. H T + + + L N M O Q P = 24 354 262 31300 4 245 0 05052 2 31686 2 354 262 Jmo l A( v , 510 K ): H K T pl b v pv T b + + = z 262 354 37575 510 b g Δ l Energy balance: ²²±± . Q H nH ii == = = ∑∑ Δ out in J s 1 kW cooling kJ s kW 1060 10 106 3 8.62 a. Basis: 50 kg wet steaks/min D.M. = dry meat 0.28 D.M. 50 kg/min @ –26°C (kg H O( )/min) m 1 0.72 H O( ) 60°C Q (kW) s 2 v 2 (96% of H O in feed) 2 (kg D.M./min) m 2 (kg H O( )/min) m 3 l 2 50°C 96% vaporization: ² . m i n mv 1 0 96 0 72 50 34 56 = kg min kg H O 2 b g ² . m i n ml 3 0 04 072 50 144 = kg min kg H O 2 b g Dry meat balance: ² m 2 028 50 140 b g
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## This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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