process dynamics and control Seborg 4th edition chapter solution manual pdf

1. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/

2. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ Chapter 2 2.1 a) Overall mass balance: ρ ( ) d V = + − (1) w w w 1 2 3 dt Energy balance: d C    = − ρ ( V T T ) 3 ref − + − ( ) ( ) wC T T w C T T 1 1 2 2 ref ref (2) dt ( ) − − w C T T 3 3 ref V = Because ρ = constant and V = constant, Eq. 1 becomes: = + w w w (3) 3 1 2 b) From Eq. 2, substituting Eq. 3 ( ρ CV dt − ) d T T dT dt − 3 ref = = − + − 3 CV wC T T w C T T ρ ( ) ( ) 1 1 2 2 ref ref (4) ) ( ) ( + − w w C T T 1 2 3 ref Constants C and Tref can be cancelled: dT V ρ = + − + 3 ( ) w T w T w w T (5) 1 1 2 2 1 2 3 dt The simplified model now consists only of Eq. 5. Solution Manual for Process Dynamics and Control, 4th edition Copyright © 2016 by Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp, and Francis J. Doyle III 2-1

3. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ Degrees of freedom for the simplified model: Parameters : ρ, V Variables : w1, w2, T1, T2, T3 NE = 1 NV = 5 Thus, NF = 5 – 1 = 4 Because w1, w2, T1 and T2 are determined by upstream units, we assume they are known functions of time: w1 = w1(t) w2 =w2 (t) T1 = T1(t) T2 = T2(t) Thus, NF is reduced to 0. 2.2 Energy balance:    = − d ρ ( V T T ) ref − − − − − + ( ) ( ) ( ) C wC T T wC T T UA T T Q p p i ref p ref s a dt Simplifying dT dt dT dt = − − − + VC wC T wC T UA T T Q ρ ( ) p p i p s a = − − − + VC wC T T UA T T Q ρ ( ) ( ) p p i s a b)T increases if Ti increases and vice versa. T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if Q > UAs(T-Ta), the contents are heated, and T >Ti. 2-2

4. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.3 a) Mass Balances: dh ρ = − − (1) 1 A w w w 1 1 2 3 dt dh ρ = (2) 2 A w 2 2 dt Flow relations: Let P1be the pressure at the bottom of tank 1. Let P2be the pressure at the bottom of tank 2. Let Pabe the ambient pressure. P P w = − ρ g = − Then (3) 1 2 ( ) h h 2 1 2 R g R 2 2 c − ρ P P g = = 1 a w h (4) 3 1 R g R 3 3 c Seven parameters: ρ, A1, A2, g, gc, R2, R3 b) Five variables : h1, h2, w1, w2, w3 Four equations Thus NF = 5 – 4 = 1 1 input = w1 (specified function of time) 4 outputs = h1, h2, w2, w3 2-3

5. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.4 Assume constant liquid density, ρ . The mass balance for the tank is Ah ρ + ( ) d m g = ρ − ( ) q q i dt Because ρ, A, and mgare constant, this equation becomes dh = i− (1) A q q dt The square-root relationship for flow through the control valve is / 1   2   ρ gh   = + − (2) q C P P v g a g c From the ideal gas law, ( / H ) m M RT g = (3) P g − ( ) A h 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 Pg from Eq. 3: 1/2       ( / ) m M RT h − ρ dh dt gh g g = − + − (4) A q C P i v a ( ) A H c Because the model contains Pa, operation of the system is not independent of Pa. For an open system Pg= Pa and Eq. 2 shows that the system is independent of Pa. 2-4

6. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.5 a) For linear valve flow characteristics, P w = − − P P 1− P P P 2 f = = 1 d 2 w , , (1) w a b c R R R a b c Mass balances for the surge tanks dm dm = − = − , (2) 1 2 w w w w a b b c dt dt where m1 and m2 are the masses of gas in surge tanks 1 and 2, respectively. If the ideal gas law holds, then m m = = , (3) 2 1 P V RT P V RT 1 1 1 2 2 2 M M where M is the molecular weight of the gas T1 and T2 are the temperatures in the surge tanks. Substituting for m1 and m2 from Eq. 3 into Eq. 2, and noticing that V1, T1, V2, and T2 are constant, V M dP V M dP = − = − and (4) 2 2 1 1 w w w w a b b c RT dt RT dt 2 1 The dynamic model consists of Eqs. 1 and 4. For adiabatic operation, Eq. 3 is replaced by b) γ γ     V V         = = , a constant (5) 1 2 P P C 1 2 m m 1 2 γ γ / 1 / 1     γ γ P V P V         = = 1 C 1 2 2 or and (6) m m 2 1 C Substituting Eq. 6 into Eq. 2 gives, γ / 1   γ V dP 1     − γ γ 1 ( 1 / ) = − 1 1 P w w a b γ C dt 2-5

7. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ γ / 1   γ V dP 1     − γ γ 1 ( 2 / ) = − 2 2 P w w b c γ C dt as the new dynamic model. If the ideal gas law were not valid, one would use an appropriate equation of state instead of Eq. 3. 2.6 a) Assumptions: 1. Each compartment is perfectly mixed. 2. ρ and C are constant. 3. No heat losses to ambient. Compartment 1: Overall balance (No accumulation of mass): 0 = ρq−ρq1 thus q1 = q (1) Energy balance (No change in volume): dT dt 1 = − − − (2) V C qC T T UA T T ρ ρ ( ) ( ) 1 1 1 2 i Compartment 2: Overall balance: 0 = ρq1−ρq2 thus q2 = q1= q (3) Energy balance: dT dt 2 = − + − − − (4) V C qC T T UA T T U A T T ρ ρ ( ) ( ) ( ) 2 1 2 1 2 2 c c c Eight parameters: ρ, V1, V2, C, U, A, Uc, Ac Five variables: Ti, T1, T2, q, Tc Two equations: (2) and (4) b) 2-6

8. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ Thus NF = 5 – 2 = 3 2 outputs = T1, T2 3 inputs = Ti, Tc, q (specify as functions of t) c) Three new variables: ci, c1, c2 (concentration of species A). Two new equations: Component material balances on each compartment. c1 and c2 are new outputs. ci must be a known function of time. 2.7 As in Section 2.4.2, there are two equations for this system: 1( ρ dV dt dT dt = − ) w w i w V Q VC ρ ( ) = − + i T T i ρ Results: (a)Since w is determined by hydrostatic forces, we can substitute for this variable in terms of the tank volume as in Section 2.4.5 case 3. 1 i v w C dt A ρ         dV V = − w ρ dT dt Q VC ρ ( ) = − + i T T i V This leaves us with the following: 5 variables: , , 4 parameters: , , 2 equations The degrees of freedom are 5 2 have: 2 output variables: , V T w T Q C ρ , , C A i i , v − = . To make sure the system is specified, we 3 T V 2-7

9. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ (b)In this part, two controllers have been added to the system. Each controller provides an additional equation. Also, the flow out of the tank is now a manipulated variable being adjusted by the controller. So, we have 4 parameters: , , , sp sp C T V ρ 6 variables: , , , , , i i V T w T Q w 4 equations 2 manipulated variables: , 1 disturbance variable: Q w i iT − = . To specify the two degrees of freedom, we The degrees of freedom are 6 4 set the variables as follows: 2 output variables: , 2 manipulated variables (determined by controller equations): , 2 disturbance variables: , i T w 2 T V Q w i 2.8 Additional assumptions: (i) Density of the liquid, ρ, and density of the coolant, ρJ, are constant. (ii) Specific heat of the liquid, C, and of the coolant, CJ, are constant. Because V is constant, the mass balance for the tank is: dV ρ = − = ; thus q = qF 0 q q F dt Energy balance for tank: dT 8 . 0 VC ρ = ρ − − − (1) ( ) ( ) q C T T Kq A T T F F J J dt Energy balance for the jacket: dT 8 . 0 ρ = ρ − + − (2) J ( ) ( ) V C q C T T Kq A T T J J J J J J i J J J dt where A is the heat transfer area (in ft2)between the process liquid and the coolant. 2-8

10. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ Eqs.1 and 2 comprise the dynamic model for the system. 2.9 Assume that the feed contains only A and B, and no C. Component balances for A, B, C over the reactor give. dc dt − 1/ E RT A = − − (1) V i Ai q c qc Vk e c 1 A A dc dt − − / / E RT E RT B = − + − (2) ( ) V i Bi q c qc V k e c k e c 1 2 1 2 B A B dc dt − 2/ E RT C = − + (3) V qc Vk e c 2 C B An overall mass balance over the jacket indicates that qc = qci because the volume of coolant in jacket and the density of coolant are constant. Energy balance for the reactor: ( )    = + + d Vc M S Vc M S Vc M S T ( )( A A A B B dt ) ( + −∆ B C C C + − ) i Ai q c M S i Bi q c M S T T A A B B i − − / / E RT E RT UA T − − + −∆ (4) ( ) ( ) T H Vk e c H Vk e c 1 2 1 1 2 2 c A B where MA, MB, MC are molecular weights of A, B, and C, respectively SA, SB, SC are specific heats of A, B, and C. U is the overall heat transfer coefficient A is the surface area of heat transfer Energy balance for the jacket: dT dt c = − + − (5) S V S q T T UA T T ρ ρ ( ) ( ) j j j j j ci ci c c where: ρj, Sj are density and specific heat of the coolant. Vj is the volume of coolant in the jacket. Eqs. 1 - 5 represent the dynamic model for the system. 2-9

11. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.10 The plots should look as shown below: Notice that the functions are only good for t = 0 to t = 18, at which point the tank is completely drained. The concentration function blows up because the volume function is negative. 2-10

12. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.11 a) Note that the only conservation equation required to find h is an overall mass balance: ρ ( ) dm dt d Ah dt dh dt (1) = = ρ = + − A w w w 1 2 ρ g v ′ = C h C h Valve equation: w = (2) v g c ρ g v ′ = C C where (3) v g c Substituting the valve equation into the mass balance, 1 A ρ dh = + − ( ) w w C h (4) 1 2 v dt Steady-state model: + − w w C h 0 = (5) 1 2 v + + 2.0 1.2 2.25 3.2 1.5 kg/s m w w 1 2 = = = = 2.13 C b) v 1/2 h c) Feedforward control 2-11

13. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ Rearrange Eq. 5 to get the feedforward (FF) controller relation, 1 2 w h C w R v − = where 2 . 3 ) 5 . 1 )( 13 . 2 ( w w = − = h = 2.25 m R − w (6) 2 1 1 Note that Eq. 6, for a value of w1= 2.0, gives w2 = 3.2 –1.2 = 2.0 kg/s which is the desired value. = 1 . 1 − 2 . 3 w w If the actual FF controller follows the relation, transmitter 10% higher), turned on, w2= 3.2 –1.1 (2.0) = 3.2 – 2.2 = 1.0 kg/s (instead of the correct value, 1.2 kg/s) Then 0 . 2 13 . 2 = = h h Cv 3 = = h (flow 2 1 w will change as soon as the FF controller is 2 0 . 1 + and h= 1.983 m (instead of 2.25 m) or . 1 408 . 2 13 Error in desired level = 2.25 1.983 − × = 100% 11.9% 2.25 2-12

14. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ The sensitivity does not look too bad in the sense that a 10% error in flow measurement gives ~12% error in desired level. Before making this conclusion, however, one should check how well the operating FF controller works for a change in w1 (e.g., ∆w1 = 0.4 kg/s). 2.12 a) Model of tank (normal operation): dh dt π = ρ = + − (Below the leak point) A w w w 1 2 3 2 (2) 4 2 = π = 3.14 m A dh = + − = (800)(3.14) 120 100 200 20 dt 20 dh dt = = 0.007962 m/min (800)(3.14) Time to reach leak point (h = 1 m) = 125.6 min. , , w w w b) Model of tank with leak and constant: 1 2 3 dh dt − − = 20 − 20 , h≥ 1 = − d = − 1 h A q h ρ 20 20 ρ(0.025) 1 4 To check for overflow, one can simply find the level hm at which dh/dt = 0. That is the maximum value of level when no overflow occurs. 0 = 20 − 20 1 − m h or hm= 2 m Thus, overflow does not occur for a leak occurring because hm< 2.25 m. 2-13

15. https://gioumeh.com/product/process-dynamics-and-control-solution/https://gioumeh.com/product/process-dynamics-and-control-solution/ 2.13 Model of process Overall material balance: dh + − ρ = + − w w C h = (1) AT w w w 1 2 v 1 2 3 dt Component: ( 3) d hx ρ = + − AT w x w x w x 1 1 2 2 3 3 dt dx dh ρ + ρ = + − 3 A h A x w x w x w x 3 1 1 2 2 3 3 T T dt dt Substituting for dh/dt (Eq. 1) dx ρ + + − = + − 3 ( ) AT h x w w w w x w x w x 3 1 2 3 1 1 2 2 3 3 dt dx ρ = − + − 3 ( ) ( ) AT h w x x w x x (2) 1 1 3 2 2 3 dt dx 1 A [ ] ) 3 = − + − 3 ( ) ( w x x w x x or (3) 1 1 3 2 2 ρ dt h T a) At initial steady state , = + = + = 120 100 220 Kg/min w w w 3 1 2 220 = 166 3 . Cv = . 1 75 b) If x1 is suddenly changed from 0.5 to 0.6 without changing flowrates, then level remains constant and Eq.3 can be solved analytically or numerically to find the time to achieve 99% of the x3response. From the material balance, the final value of x3= 0.555. Then, dx dt 1 [ ] = − ) 100(0.5 + − 120(0.6 ) 3 x x 3 3 π (800)(1.75) 2-14