unsw ceic3000 tutorial solution w2 process modeling and analysis

Unsw ceic3000 tutorial solution w2 process modeling and analysis

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School of Chemical Engineering CEIC3000 Process Modelling and Analysis Brief Solutions to Tutorial 2 1. Solution: (a) Material balance: ρ dV dt = F i ρ - (1) Energy balance: dH dt = F i ρ ¯ H i - ¯ H + Q (2) where ¯ H = T T ref c p dT is the specific enthalpy. H = V ρ ¯ H. Assume that heat capacity is constant (or calculated at an average temperature): ¯ H = c p ( T - T ref ) (3) ¯ H i = c p ( T i - T ref ) (4) Then we have: d ( V ρc p ( T - T ref )) dt = F i ρc p ( T i - T ref ) - Fρc p ( T - T ref ) + Q (5) ρc p V dT dt + ρc p ( T - T ref ) dV dt = F i ρc p ( T i - T ref ) - Fρc p ( T - T ref )+ Q (6) ρc p V dT dt + ρc p ( T - T ref ) ( F i - F ) = F i ρc p ( T i - T ref ) - Fρc p ( T - T ref )+ Q (7) dT dt = F i T i - F i T V + Q ρc p V (8) 2. Solution : (a) At steady-state, 0 = F V ( c Af - c As ) - k 1 c As - k 3 c 2 As (9) 0 = - F V c Bs + k 1 c As - k 2 c Bs (10) therefore, c Bs = 1 . 117 , [ F V ] s = 0 . 5714 1
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(b) u = F V and y = c B . Denote f 1 = F V ( c Af - c A ) - k 1 c A - k 3 c 2 A (11) f 2 = - F V c B + k 1 c A - k
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Unformatted text preview: 2 c B (12) Then the linearized model is ˙x = Ax + Bu (13) y = Cx , (14) where A = [ ∂f 1 ∂c A ∂f 1 ∂c B ∂f 2 ∂c A ∂f 2 ∂c B ] = ±-F V-k 1-2 k 3 c A k 1-F V-k 2 ²³ ³ ³ ³ c A = c As =3 c B = c Bs =0 . 5714 F V =1 . 117 (15) = ±-2 . 4015 . 833-2 . 238 ² (16) B = [ ∂f 1 ∂ ( F/V ) ∂f 2 ∂ ( F/V ) ] = ± c Af-c A-c B ²³ ³ ³ ³ c A = c As =3 c B = c Bs =0 . 5714 F V =1 . 117 (17) = ± 7-1 . 117 ² (18) C = ´ 0 1 µ (19) 2...
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This note was uploaded on 03/26/2012 for the course CHEM ENG CEIC at University of New South Wales.

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Unsw ceic3000 tutorial solution w2 process modeling and analysis

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