The simplified model now consists only of Eq. 5. Degrees of freedom for the simplified model: 2-1

Transcription

1 . a) Overall mass balance: d( ρv ) Energy balance: = w + w w () d V T Tref C = wc ( T Tref ) + wc( T Tref ) w C T Because ρ = constant and ( Tref ) V = V = constant, Eq. becomes: () w = + () w w b) From Eq., substituting Eq. d( T T ) dt ref CV = CV = wc T Tref + wc T Tref ( w w ) C ( T Tref ) + (4) Constants C and T ref can be cancelled: dt ρ V = wt + wt ( w + w ) T (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, nd edition, Copyright 004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp. -

2 Variables : w, w, T, T, T N E = N V = 5 Thus, N F = 5 = 4 Because w, w, T and T are determined by upstream units, we assume they are known functions of time: w = w (t) w = w (t) T = T (t) T = T (t) Thus, N F is reduced to 0.. Energy balance: d V T Tref C p = wc p ( Ti Tref ) wc p ( T Tref ) UAs ( T Ta ) + Q Simplifying dt VC p = wc pti wc pt UAs T Ta + Q dt VC p = wc p Ti T UAs T Ta + Q 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.. a) Mass Balances: -

3 dh ρ A = w w w () dh ρ A = w () Flow relations: Let P be the pressure at the bottom of tank. Let P be the pressure at the bottom of tank. Let P a be the ambient pressure. P P ρg Then w = = ( h h ) () R g R c w P P ρg = (4) a = h R g cr b) Seven parameters: ρ, A, A, g, g c, R, R Five variables : h, h, w, w, w Four equations Thus N F = 5 4 = input = w (specified function of time) 4 outputs = h, h, w, w.4 Assume constant liquid density, ρ. The mass balance for the tank is d( ρah + m g ) = ρ( q i q) Because ρ, A, and m g are constant, this equation becomes -

4 dh A = q q () i The square-root relationship for flow through the control valve is / ρgh q = C + v Pg Pa () g c From the ideal gas law, P g ( mg / M ) RT = () A( H h) where T is the absolute temperature of the gas. Equation gives the unsteady-state model upon substitution of q from Eq. and of P g from Eq. : dh ( mg / M ) RT ρgh A = qi Cv + Pa A( H h) gc / (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. shows that the system is independent of P a..5 a) For linear valve flow characteristics, Pd P P P wa =, wb =, Ra Rb Mass balances for the surge tanks w c P Pf = () R c dm = w a w b, c dm = w b w () where m and m are the masses of gas in surge tanks and, respectively. If the ideal gas law holds, then -4

5 m M P V = RT, V RT m P = () M where M is the molecular weight of the gas T and T are the temperatures in the surge tanks. Substituting for m and m from Eq. into Eq., and noticing that V, T, V, and T are constant, V M dp V M dp = w a w b and = w b wc (4) RT RT The dynamic model consists of Eqs. and 4. b) For adiabatic operation, Eq. is replaced by or γ γ V V P P = C m = m, a constant (5) / γ γ PV m = and C / γ γ P V m = (6) C Substituting Eq. 6 into Eq. gives, γ γ γ V C V C γ / γ / γ ( γ ) / γ P P ( γ ) / γ dp dp = w a w b = w b w c as the new dynamic model. If the ideal gas law were not valid, one would use an appropriate equation of state instead of Eq...6 a) Assumptions:. Each compartment is perfectly mixed.. ρ and C are constant.. No heat losses to ambient. Compartment : -5

6 Overall balance (No accumulation of mass): 0 = ρq ρq thus q = q () Energy balance (No change in volume): dt V C = qc Ti T UA T T () Compartment : Overall balance: 0 = ρq ρq thus q = q = q () Energy balance: dt V C = qc T T + UA T T UcAc T Tc (4) b) Eight parameters: ρ, V, V, C, U, A, U c, A c Five variables: T i, T, T, q, T c Two equations: () and (4) Thus N F = 5 = outputs = T, T inputs = T i, T c, q (specify as functions of t) c) Three new variables: c i, c, c (concentration of species A). Two new equations: Component material balances on each compartment. c and c are new outputs. c i must be a known function of time..7 Let the volume of the top tank be γv, and assume that γ is constant. Then, an overall mass balance for either of the two tanks indicates that the flow rate of the stream from the top tank to the bottom tank is equal to q +q R. Because the two tanks are perfectly stirred, c T = c T. -6

7 Component balance for chemical tracer over top tank: dct V = qcti + qrct q + qr ct () Component balance on bottom tank: or dc ( V T = q + q c q c qc R T R T T dc ( V T = q + qr ct ct () Eqs. and constitute the model relating the outflow concentration, c T, to inflow concentration, c Ti. Describing the full-scale reactor in the form of two separate tanks has introduced two new parameters into the analysis, q R and γ. Hence, these parameters will have to be obtained from physical experiments..8 Additional assumptions: (i) Density of the liquid, ρ, and density of the coolant, ρ J, are constant. (ii) Specific heat of the liquid, C, and of the coolant, C J, are constant. Because V is constant, the mass balance for the tank is: dv ρ = q F q = 0 ; thus q = q F Energy balance for tank: dt 0.8 ρ VC = qfρc( TF T ) KqJ A( T TJ ) () Energy balance for the jacket: dtj 0.8 ρ JVJC J = qjρ JC J ( Ti TJ ) + KqJ A( T TJ ) () -7

8 where A is the heat transfer area (in ft ) between the process liquid and the coolant. Eqs. and comprise the dynamic model for the system..9 Additional assumptions: i. The density ρ and the specific heat C of the process liquid are constant. ii. The temperature of steam T s is uniform over the entire heat transfer area iii. T s is a function of P s, T s = f(p s ) Mass balance for the tank: dv qf q = () Energy balance for the tank: d V ( T Tref ) C = qf C TF Tref q C T Tref + UA( T T ) s () where: T ref is a constant reference temperature A is the heat transfer area Eq. is simplified by substituting for (dv/) from Eq., and replacing T s by f(p s ), to give dt ρ VC = qfρc( TF T ) + UA[ f ( Ps ) T ] () Then, Eqs. and constitute the dynamic model for the system. -8

9 .0 Assume that the feed contains only A and B, and no C. Component balances for A, B, C over the reactor give. dca E / RT V = qicai qca Vke ca () dc V B q c qc V k e c k e c E / RT E / RT i Bi B A B = + ( ) () dcc E / RT V = qcc + Vke cb () An overall mass balance over the jacket indicates that q c = q ci because the volume of coolant in jacket and the density of coolant are constant. Energy balance for the reactor: ( + + ) d VcAM ASA VcBM BSB VcC MCSC T = ( qicai M ASA + qicbim BSB )( Ti T ) E / RT E / RT c A B + + (4) UA( T T ) ( H ) Vk e c ( H ) Vk e c where M A, M B, M C are molecular weights of A, B, and C, respectively S A, S B, S C 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: where: dtc S V = S q T T + UA T T (5) j j j j j ci ci c c ρ j, S j are density and specific heat of the coolant. V j is the volume of coolant in the jacket. Eqs. - 5 represent the dynamic model for the system. -9

10 . Model (i) : Overall mass balance (w=constant=w): d( ρv ) dh = Aρ = w + w w () A component balance: d( ρvx) = w wx d( hx) or Aρ = w wx () Note that for Stream, x = 0 (pure B). Model (ii) : Mass balance: d( V dh = Aρ = w + w w () Component balance on component A: d( ρvx) = w wx d( hx) or Aρ = w wx (4) -0

11 . a) Note that the only conservation equation required to find h is an overall mass balance: dm d( ρah) dh = = ρ A = w + w w () ρg Valve equation: w = C v h = Cv h () g where C v c ρg = C v () g Substituting the valve equation into the mass balance, dh Steady-state model : c = ( w + w Cv h) (4) ρa 0 = w + w Cv h (5) b) Cv w + w kg/s = = = =..5 / h.5 m c) Feedforward control -

12 Rearrange Eq. 5 to get the feedforward (FF) controller relation, w w = C h where h =.5 m v R w.)(.5) w =. R = ( w (6) Note that Eq. 6, for a value of w =.0, gives w =.. =.0 kg/s which is the desired value. If the actual FF controller follows the relation, w =.. w (flow transmitter 0% higher), w will change as soon as the FF controller is turned on, w =.. (.0) =.. =.0 kg/s (instead of the correct value,. kg/s) Then h =. h = C v or h = =. 408 and h =.98 m (instead of.5 m). Error in desired level = % =.9%.5 The sensitivity does not look too bad in the sense that a 0% error in flow measurement gives ~% error in desired level. Before making this -

13 conclusion, however, one should check how well the operating FF controller works for a change in w (e.g., w = 0.4 kg/s).. a) Model of tank (normal operation): dh ρ A = w + w w (Below the leak point) () A = π = π =.4 m 4 dh (800)(.4) = = 0 dh m/min = (800)(.4) = Time to reach leak point (h = m) = 5.6 min. b) Model of tank with leak and w, w, w constant: dh A = δ q4 = h = 0 0 h, h To check for overflow, one can simply find the level h m at which dh/ = 0. That is the maximum value of level when no overflow occurs. 0 = 0 0 h or h m = m m Thus, overflow does not occur for a leak occurring because h m <.5 m..4 Model of process Overall material balance: -

14 dh ρ A T = w + w w = w + w Cv h () Component: ρ A T d ( hx ) = w x x + w x w dx ρ AT h + ρa T x dh = w x + w x w x Substituting for dh/ (Eq. ) dx ρ h + x( w + w w ) = w x A T + w x w x dx ρ A T h = w ( x x) + w ( x x) () dx = () ρa h or [ w ( x x ) + w ( x x )] a) At initial steady state, T w = w + w = = 0 Kg/min 0 C v = = b) If x is suddenly changed from 0.5 to 0.6 without changing flowrates, then level remains constant and Eq. can be solved analytically or numerically to find the time to achieve 99% of the x response. From the material balance, the final value of x = Then, dx = + (800)(.75) π [ 0(0.6 x ) 00(0.5 x )] = + (800)(.75) π = x Integrating, [(7 50) 0 x )] -4

15 x f t dx x = x o 0 Solving, where x o =0.5 and x f =0.555 (0.555)(0.0) = t = 47.4 min c) If w is changed to 00 kg/min without changing any other input variables, then x will not change and Eq. can be solved to find the time to achieve 99% of the h response. From the material balance, the final value of the tank level is h =.446 m. dh 800π Cv = + h dh = h 800π = h where h o =.75 and h f = (.446)(0.0) =.460 By using the MATLAB command ode45, t =.79 min Numerical solution of the ode is shown in Fig. S.4.8 h(m) time (min) Figure S.4. Numerical solution of the ode for part c) -5

16 d) In this case, both h and x will be changing functions of time. Therefore, both Eqs. and will have to be solved simultaneously. Since concentration does not appear in Eq., we would anticipate no effect on the h response..5 a) The dynamic model for the chemostat is given by: Cells: Product: dx V dp V = Vr g FX or dx F = r V X g () = Vr p FP or dp F = r V P p () ds Substrate: V = F( S f S) Vrg Y Y X / S P / S Vr P or ds F ( S f S) rg r () V Y = P Y X / S P / S b) At steady state, then, dx = 0 r g = DX µ X = DX µ = D (4) A simple feedback strategy can be implemented where the growth rate is controlled by manipulating the mass flow rate, F. c) Washout occurs if dx/ = 0 is negative for an extended period of time; that is, r g DX < 0 or µ < D Thus, if µ < D the cells will be washed out. d) At steady state, the dynamic model given by Eqs., and becomes: -6

17 From Eq. 5, From Eq. 7 0 = r g DX (5) 0 = r p DP (6) 0 = D( S f S) rg rp (7) Y Substituting Eq. 9 into Eq. 8, r Y X / S P / S DX = r g (8) Y X / S g = YX / S ( S f S) D + rp (9) YP / S DX Y S S D Y r X / S = X / S ( f ) + p (0) YP / S From Eq. 6 and the definition of Y P/S in (-9), r p = DP = DYP / S ( S f S) From Eq. 4 S DK = µ max S D Substituting these two equations into Eq. 0, DX = Y X / S S f DK S D D µ max -7

18 0.8 DX (g/l.h) MAXIMUM PRODUCTION 0. WASHOUT D (/h) Figure S.5. Steady-state cell production rate DX as a function of dilution rate D. From Figure S.5, washout occurs at D = 0.8 h - while the maximum production occurs at D = 0.4 h -. Notice that maximum and washout points are dangerously close to each other, so special care must be taken when increasing cell productivity by increasing the dilution rate..6 a) We can assume that ρ and h are approximately constant. The dynamic model is given by: Notice that: dm r d = = kac s () M = ρv dm dv = ρ () V = πr h dv dr dr = ( πrh) = A () -8

19 Substituting () into () and then into (), dr dr ρa = kac s ρ = kcs Integrating, Finally, r kc dr = s ro t 0 s r t) = ro t kc ( (4) ρ M = ρv = ρπhr then kcs M ( t) = ρπh ro t ρ b) The time required for the pill radius r to be reduced by 90% is given by Eq. 4: kcs 0.9roρ (0.9)(0.4)(.) 0.ro = ro t t = = = 54 min ρ kc (0.06)(0.5) Therefore, t = 54 min. s.7 For V = constant and F = 0, the simplified dynamic model is: dx = rg = µ max S K + S s X dp = rp = YP / X µ max S K + S s X ds = Y X / S r g Y P / X r P Substituting numerical values: dx = SX 0. + S -9

20 ds dp = SX ( 0.)(0.) + S SX = 0. + S By using MATLAB, this system of differential equations can be solved. The time to achieve a 90% conversion of S is t =.5 h. Figure S.7. Fed-batch bioreactor dynamic behavior. -0