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ch02 - 1234567898 1234567898 5 2.1 a Overall mass balance...

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2-1 ±²³´µ¶·¸¹¸º»µ ±²³´µ¶·¸¹¸ 2.1 a) Overall mass balance: 3 2 1 ) ( w w w dt V d - + = ρ (1) Energy balance: ( 29 3 1 1 2 2 3 3 ±² ³ ( ) ( ) ref ref ref ref d V T T C w C T T w C T T dt w C T T - = - + - - - (2) Because ρ = constant and V V = = constant, Eq. 1 becomes: 2 1 3 w w w + = (3) b) From Eq. 2, substituting Eq. 3 ( 29 ( 29 3 3 1 1 2 2 1 2 3 ( ) ± ± ² ³ ² ³ ref ref ref ref d T T dT CV CV w C T T w C T T dt dt w w C T T - = = - + - - + - (4) Constants C and T ref can be cancelled: 3 2 1 2 2 1 1 3 ) ( T w w T w T w dt dT V + - + = ρ (5) The simplified model now consists only of Eq. 5. Degrees of freedom for the simplified model: Parameters : ρ , V Solution Manual for Process Dynamics and Control, 2 nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
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2-2 Variables : w 1 , w 2 , T 1 , T 2 , T 3 N E = 1 N V = 5 Thus, N F = 5 – 1 = 4 Because w 1 , w 2 , T 1 and T 2 are determined by upstream units, we assume they are known functions of time: w 1 = w 1 (t) w 2 = w 2 (t) T 1 = T 1 (t) T 2 = T 2 (t) Thus, N F is reduced to 0. 2.2 Energy balance: ±² ³ ( ) ( ) ( ) ref p p i ref p ref s a d V T T C wC T T wC T T UA T T Q dt - = - - - - - + Simplifying ± ² ³ p p i p s a dT VC wC T wC T UA T T Q dt = - - - + ± ² ³ ² ³ p p i s a dT VC wC T T UA T T Q dt = - - - + b) T increases if T i increases and vice versa. T decreases if w increases and vice versa if ( T i – T ) < 0. In other words, if Q > UA s (T-T a ) , the contents are heated, and T > T i . 2.3 a) Mass Balances:
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2-3 3 2 1 1 1 w w w dt dh A - - = ρ (1) 2 2 2 w dt dh A = ρ (2) Flow relations: Let P 1 be the pressure at the bottom of tank 1. Let P 2 be the pressure at the bottom of tank 2. Let P a be the ambient pressure. Then ) ( 2 1 2 2 2 1 2 h h R g g R P P w c - ρ = - = (3) 1 3 3 1 3 h R g g R P P w c a ρ = - = (4) b) Seven parameters: ρ , A 1 , A 2 , g , g c , R 2 , R 3 Five variables : h 1 , h 2 , w 1 , w 2 , w 3 Four equations Thus N F = 5 – 4 = 1 1 input = w 1 (specified function of time) 4 outputs = h 1 , h 2 , w 2 , w 3 2.4 Assume constant liquid density, ρ . The mass balance for the tank is ) ( ) ( q q dt m Ah d i g - ρ = ρ Because ρ , A , and m g are constant, this equation becomes
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2-4 q q dt dh A i - = (1) The square-root relationship for flow through the control valve is 2 / 1 - ρ + = a c g v P g gh P C q (2) From the ideal gas law, ) ( ) / ( h H A RT M m P g g - = (3) where T is the absolute temperature of the gas. Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and of P g from Eq. 3: 1/2 ( / ) ( ) g i v a c m M RT dh gh A q C P dt A H h g ρ = - + - - (4) Because the model contains P a , operation of the system is not independent of P a . For an open system P g = P a and Eq. 2 shows that the system is independent of P a . 2.5 a) For linear valve flow characteristics, a d a R P P w 1 - = , b b R P P w 2 1 - = , c f c R P P w - = 2 (1) Mass balances for the surge tanks b a w w dt dm - = 1 , c b w w dt dm - = 2 (2) where m 1 and m 2 are the masses of gas in surge tanks 1 and 2, respectively.
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