ch10 - CHAPTER TEN 10.1 b. Assume no combustion n 1 (mol...

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CHAPTER TEN 10.1 b . Assume no combustion (mol C H /mol) (mol gas), 4 (°C) n 1 T 1 (mol CH /mol) x 1 2 x 26 1 – x 2 x 1 (mol C H /mol) 38 (mol air), (°C) n 2 T 2 (mol C H /mol) (mol), 200°C 4 n 3 (mol CH /mol) y 1 2 y 1 – y 2 y 1 (mol air/mol) (mol C H /mol) 3 y y 3 (kJ) Q 11 5 6 4 1231212 312 variables relations degrees of freedom material balances and 1 energy balance nnnxx yy yTTQ ,,,,,,,,,, bg A feasible set of design variables : nnxxTT 121212 ,,,,, lq Calculate n from total mole balance, from component balances, 3 yy y 12 ,, and 3 Q from energy balance. An infeasible set : nnnxxT 123121 Specifying n determines n (from a total mole balance) n 2 1 and 3 c . (mol C H /mol) (mol gas), n 1 T , P y 16 1 4 1 – y 1 (mol C H /mol) (mol gas), n 2 y 1 4 1 – y 2 (kJ) Q (mol N /mol) 2 T , P 2 1 (mol N /mol) 2 n 3 (mol C H ( )/mol), 61 4 T , P 2 l 9 4 5 1231212 22 variables relations degrees of freedom 2 material, 1 energy, and 1 equilibrium: CH 4 −= nnn yyTTQP yP P T ,,,,,,,, * di A feasible set : ny T Pn ,,,, 11 3 Calculate n from total balance, from C balance, T from Raoult’s law: [ ], Q from energy balance 2 y 2 H 4 2 T 2 = 64 2 An infeasible set : nynPT 223 2 ,,, , Once y are specified, is determined from Raoult’s law P 2 and T 2 10- 1
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10.2 10 2 21 6 12341234 34 3 4 variables material balances equilibrium relations: [ degrees of freedom nnnn xx xxTP xP xP T x P x P T BC ,,,,,,,, , ,] ** bg bgb g b g bg −= = 1 a. A straightforward set : nnnxxT 13414 ,,,,, lq Calculate n from total material balance, P from sum of Raoult's laws: 2 Px pT xPT Bc =+ ∗∗ 44 1 bg b g x from Raoult's law, from B balance 3 x 2 b . An iterative set : nnnxxx 12312 3 Calculate n from total mole balance, from B balance. 4 x 4 Guess P, calculate T from Raoult's law for B, P from Raoult’s law for C , iterate until pressure checks. c . An impossible set : nnnnTP 1234 ,,,, , Once nn are specified, a total mole balance determines . n 12 ,, and 3 n 4 10.3 2BaSO s 4C s 2BaS s 4CO g 42 b g +→ + a . (kg BaSO /kg) 100 kg ore, n 0 T x b4 (kg CO ) (kg BaS) n 2 n 32 (kJ) Q 0 (K) (kg coal), T 0 (K) (kg C/kg) x c (% excess coal) P ex (kg C) n 1 (kg other solids) n 4 T f (K) 11 5 1 1 1 5 01234 0 0 variables material balances C, BaS, CO BaSO other solids energy balance reaction relation defining in terms of and degrees of freedom bc e x 24 ex b c nnnnnxxTTQP Pn x x f ,,,,,,,, ,, di + b . Design set : xxTTP f e x ,,, , 0 ns Calculate n from ; from material balances, 0 xx P e x and n 1 through n 4 Q from energy balance 10- 2
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10.3 (cont’d) c . Design set: xxTnQ Bc ,,,, 2 0 lq B Specifying x determines n impossible design set. 2 d . Design set : xxTPQ e x ,,, , 0 Calculate n from , from 2 x B n 3 x B n from and 0 xx B c P ex n from C material balance, from total material balance 1 n 4 T from energy balance (trial-and-error probably required) f 10.4 2C H OH O 2CH CHO 2H O 25 2 3 2 +→ + 2CH COH O 2CH CHOOH 32 3 T (kJ) Q 0 (mol solution), n f x ef (mol EtOH/mol) 1 – x ef (mol H O/mol) 2 P , xs (mol air), n w T 0 0.79 (mol N ) n 2 0.21 (mol O ) n air 2 T (mol EtOH), n e n ah (mol CH CHO) 3 n ea (mol CH COOH) 3 n w (mol H O) 2 n ax (mol O ) 2 n n (mol N ) 2 P xs = % excess air) ( air a .
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ch10 - CHAPTER TEN 10.1 b. Assume no combustion n 1 (mol...

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