Chapter 3

Process Process Modelli Modelling, ng, Simula Simulatio tion n and Contro Controll for Chemic Chemical al Engi Enginee neeri ring ng.. Solv Solved ed probl problem ems. s. Chap Chapte ter r 3: Exam Exampl ples es of  mathematical models of chemical engineering systems. This document contains my own solutions to the problems proposed at the end of each chapter of the book ”Process Modelling, Simulation and Control for Chemica Chemicall Engine Engineers ers”” Second Second Edition Edition,, by William William L. Luyben. Luyben. At such, such, I can’t guarantee guarantee that the proposed proposed solutions solutions are free from errors. errors. Think about them as a starting point for developing or as a means of checking your own soluti solutions ons.. Any Any com commen ments ts or correc correctio tions ns will be apprec appreciat iated. ed. Contac Contactt me at [email protected]

Problem 1 A fluid of constant density ρ density  ρ is  is pumped into a cone-shaped tank of total volume 2 H πR /3 (Figure ??). The flow out of the bottom bottom of the tank tank is proportio proportional nal to the square root of the height h   of liquid in the tank. Derive Derive the equations equations describing the system.

Figure 1: Cone-shaped tank. Solution

The volumetric balance (constant density) for the fluid inside the tank is then: √  dV  =  F 0 − K  h dt Calling r Calling  r  to the radius of the fluid cone inside the tank, and if  θ if  θ  is the angle formed by the cone-shaped tank and the horizontal, we have: tgθ = tgθ  =

H  h = R/2 R/2 r/2 r/2

The volume of fluid inside the tank is: hπr 2 π V  = = 3 3 1

2

R H 

h3

The last equation, together with the volumetric balance expression, allow to solve for h for  h  and  V  as  V   as a function of time.

Problem 2 A perfect gas with molecular weight M   M   flows at a mass flow rate W 0   into a cylinder through a restriction. The flow rate is proportional to the square root of the pressure drop over the restriction: W 0  = K   =  K 0

 P  − P  0

where P  where P  is  is the pressure in the cylinder and P  and  P 0  is the constant upstream pressure. The system is isothermal. Inside the cylinder, a piston is forced to the right as the pressure P  builds builds up. A spring spring resist resist the move movemen mentt of the piston with a force that is proportional to the axial displacement  x  of the piston. F s  =  K s x The piston is initially at x  = 0 when the pressure in the cylinder is zero. The cross-sectional area of the cylinder is  A.  A . Assume Assume the piston has negligible negligible mass and friction. 1. Derive Derive the equations equations describing the system. system. 2. What will the steady state piston displaceme displacement nt be?

Figure 2: Piston + cylinder. Solution

Because the mass and friction of the cylinder are negligible, we can assume that forces at each side of him are always balanced. The pressure at the spring side of the cylinder, according to Figure  ??, is atmospheric, so the initial pressure must be atmospheric (the barometric value is 0). A force balance for the piston gives: P atm x/A =  = P   P  atm  + K s x/A The volume variation of the cylinder gives: dx W 0 A = = dt ρ

 RT  √ P  − P   RT   P  − P  0

M 

P 

=

2

0

M 

atm atm

− K sx/A

P atm atm  + K s x/A

dx = dt

 RT   P  − P 

atm

0

M 

− K sx/A

AP atm + K s x

with initial condition x t=0  = 0. The steady state piston displacement will be reached when the volume no longer changes: P 0 − P atm − x =

K s x =0 A

A(P 0 − P atm ) K s

Problem 3 A perfectly mixed, isothermal CSTR has an outlet weir. The flow rate over the weir is proportional to the height of liquid over the weir ( how ) to the 1.5 power. The weir height is hw . The cross sectional area of the tank is A. Assume constant density. A first order reaction takes place in the tank: k1

A −→ B Derive the equations describing the system.

Figure 3: CSTR. Solution

The total continuity equation gives (assuming constant density of the process fluid): dhow A = F 0 − K F (how )3/2 dt The A component continuity equation gives: A

d(how C A ) = F 0 C A0 − K F (how )3/2 C A − k1 Ahow C A dt

3

Problem 4 In order to ensure an adequate supply for the upcoming set-to with the Hatfields, Grandpa McCoy has begun to process a new batch of his famous Liquid Lightning moonshine. He begins by pumping the mash at constant rate  F 0  into an empty tank. In this tank the ethanol undergoes a first-order reaction to form a product that is the source of the high potency of McCoy’s Liquid Lightning. Assuming that concentration of ethanol in the feed,  C 0 , is constant an that the operation is isothermal, derive the equations that describe how the concentration C  of ethanol in the tank and the volume  V  of liquid in the tank vary with time. Assume perfect mixing and constant density. Solution

The total continuity equation gives: dV  = F 0 dt The ethanol continuity equation gives: d(V C ) = F 0 C 0 − kV C  dt With initial conditions: V t=0  = 0 and C t=0  =  C 0 .

Problem 5 A rotating metal drum heat exchanger is half submerged in a cool stream, with its other half in a hot stream. The drum rotates at a constant angular velocity ω  (radians per minute). Assume T H  and T C  are constant along their respective sections of the circumference. The drum length is  L, thickness d, and radius R. Heat transfer coefficients in the heating and cooling zones are constant ( U H  and U C ). Heat capacity C P   and density of the metal drum are constant. Neglect radial temperature gradients and assume steady state operation. 1. Write the equations describing the system. 2. What are the appropiate boundary conditions? Solution

Assumptions: 1. The entalphy of the heat exchanger material can be represented as h = C  p T . 2. The energy transport by conduction in the θ   direction is negligible. 3. The thickness is much smaller than the exchanger radius. For a time interval ∆t, the energy entering and leaving a volume of size d∆θRL is: 4

Figure 4: Rotating heat exchanger.

•   Entering: ∆tωRdLρC  pT θ •   Leaving: ∆tωRdLρC  p T θ  + dT  dθ ∆θ



 + 2∆tL∆θRU  (T  − T  ) j

j

With U j and T j  corresponding to the heat transfer coefficient and temperature at the cooling or heating zone. An energy balance in the cooling zone, gives, after dividing by ∆ t∆θRL: dT   wdC  p ρ = dθ

−2U C (T  − T C )

dT   − U C (T  − T C ) = dθ wdC  p ρ The analogous energy balance for the heating zone gives: dT   − U H (T  − T H ) = dθ wdC  p ρ The energy balance for the cooling zone applies in the range θ : [π/2, π], whereas the energy balance for the heating zone applies in the range  θ : [0, π/2]. Solution of the differential equations gives two integration constants, which are determined applying the following boundary conditions: T h (0) = T c (2π) T h (π/2) = T c (π/2)

Problem 6 Consider the system that has two stirred chemical reactors separated by a plugflow dead time of  D seconds. Assume constant holdups (V 1 and V 2 ), constant throughput (F ), constant density, isothermal operation at temperatures  T 1  and T 2 , and first order kinetics with simultaneous reactions: k1

A −→ B 5

k2

A −→ C No reactions occur in the plug-flow section. Write the equations describing the system.

Figure 5: CSTRs separated by a dead time. Solution

The species balances for the first reactor are:  dC A,1,t V  = F C A0 − F C A,1,t − k1,T  V C A,1,t − k2,T  V C A,1,t dt  dC B,1,t V  = −F C B,1,t + k1,T  V C A,1,t dt  dC C,1,t V  = −F C C,1,t  + k2,T  V C A,1,t dt The species balances for the second reactor are equivalent, but using a time variable t equal to t + D: 1

1

1

1

∗

 dC A,2,t V  = F C A,1,t dt ∗

∗

∗

 − F C A, ,t  − k ,T  V C A, ,t  − k ,T  V C A, ,t 2

 dC B,2,t V  = F C B,1,t dt  dC C,2,t V  = F C C,1,t dt ∗

∗

∗

∗

∗

∗

∗

1

2

2

∗

2

2

2

∗

 − F C B, ,t  + k ,T  V C A, ,t 2

∗

1

2

2

 − F C C, ,t  + k ,T  V C A, ,t 2

∗

2

2

2

∗

∗

Problem 7 Consider the isothermal hydraulic system sketched below. A slightly compressible polymer liquid is pumped by a constant-speed, positive displacement pump so that the mass flow rate  W 1  is constant. Liquid density is given by: ρ = ρ0  + β (P  − P 0 ) where ρ 0 , β , and P 0  are constants,  ρ  is the density, and  P   is the pressure. Liquid is pumped through three resistances where the pressure drop is proportional to the square of the mass flow: ∆ P  = RW 2 . A surge tank of volume V   is located between  R 1 and R 2  and is liquid full. The pressure downstream of  R3  is atmospheric.

6

1. Derive the differential equation that gives the pressure P  in the tank as a function of time and  W 1 . 2. Fin the steady state value of tank pressure P .

Figure 6: Hydraulic system. Solution

A mass balance for the tank gives:  dρ V  = W 1 − W 2 dt The balance can be re-stated in terms of the pressure, using the expression for the liquid density and the expression for the pressure drop: dP  V β  = W 1 − dt

 

P  R2  + R3

The steady state value for the pressure is  P  = W 12 (R2  + R3 ).

Problem 8 Develop the equations describing an ”inverted” batch distillation column. This system has a large reflux drum into which the feed is charged. This material is fed to the top of the distillation column (which acts like a stripper). Vapor is generated in a reboiler at the base. Heavy material is withdrawn from the bottom of the column. Derive a mathematical model of this batch distillation system for the case where the tray holdups cannot be neglected. Solution

The inverted batch distillation column is shown in Figure assumptions are made in order to develop the model:

??.

The following

•  Vapor hold up is negligible. •  Trays are ideal. •  The column operates with a binary mixture. •  Tray contents, reflux drum and reboiler contents are perfectly mixed. The mole conservation equations (both total and by component) for the Nth tray are:

7

Figure 7: Inverted batch distillation column.

d(M NT  xNT  ) = Rx D − LNT  xNT  + V NT  dt

1

−

yNT 

1

−

− V NT  yNT 

dM NT  = R − LNT  + V NT  1 − V NT  dt The energy conservation conservation equation for the Nth tray is: −

dM NT  hNT  = Rh D − LNT  hNT  + V NT  1 H NT  1 − V NT  H NT  dt The mole conservation equations (both total and by component) for the reflux drum are: −

−

d(M D xD ) = V NT  yNT  − RxD dt dM D = V NT  − R dt The energy conservation equation for the reflux drum is: d(M D hD ) = V NT  H NT  − RhD  + QD dt The mole conservation equations (both total and by component) for the reboiler are: 8

d(M R xR ) = L 1 x1 − V R yR − P xR dt dM R = L 1 − V R − P  dt The energy conservation equation for the reboiler is: d(M R hR ) = L1 h1 − V R H R − P hR  + QR dt The mole conservation equations (both total and by component) for the  j th tray are: d(M j xj ) = Lj +1 xj +1 − Lj xj  + V j dt

1

−

yj

1

−

− V j yj

dM j = L j +1 − Lj  + V j 1 − V j dt Finally, the energy conservation equation for the  j th tray is: −

d(M j hj ) = L j +1 hj +1 − Lj hj  + V j 1 H j 1 − V j H j dt Where M j  is the holdup of liquid in the jth tray, reboiler (R) or reflux drum (D) respectively, h  is the entalphy of the liquid phase (in a molar basis), H  is the entalphy of the gas phase (in a molar basis), P   is the product flow and V 0  = V R . Also, in each tray, an equilibrium relation holds: −

−

yj = f (xj , P , T ) 

Problem 9 An ice cube is dropped into a hot, perfectly mixed, insulated cup of coffe. Develop the equations describing the dynamics of the system. List all the assumptions and define all the terms. Solution

Besides the conditions indicated in the problem statement, the following assumptions are made: 1. There are no internal temperature gradients in the ice cube, it remains at fusion temperature. 2. Entalphy of the liquid phase can be expressed as h = Cpw T . 3. Entalphy of the solid phase can be expressed as h = Cpw T  − λ. 4. The physical properties of the coffe are the same as those of water.

9

An energy balance allows to calculate the rate of fussion of ice: dM i Q U AHT  (M i )(T c − T i ) = = dt λ λ The mass balance for the system is: dM c  M i + =0 dt dt The energy balance for the liquid phase is: d(M c T c ) M i = −  C  p,w T i + U AHT  (M i )(T i − T c ) dt dt Where M i and T i  are the mass and temperature of the solid phase. M c and T c  are the mass and temperature of the liquid phase. U   is the heat transfer coefficient, λ is the heat of vaporization of ice, and  AHT  (M i ) is the heat transfer area, as a function of the mass of ice. C  p,w

Problem 10 An isothermal, irreversible reaction: k

→B A− takes place in the liquid phase in a constant-volume reactor. The mixing is not perfect. Observation of flow patterns indicates that a two-tank system with back mixing, as shown in Figure ??  below, should approximate the imperfect mixing. Assuming F  and F R  are constant, write the equations describing the system.

Figure 8: Two tank system with back mixing. Solution

The A  species balance for reactor 1 and 2 are: V 1

dC A1 = F (C A0 − C A1 ) + F R (C A2 − C A1 ) − V kC A1 dt

dC A2 = F (C A1 − C A2 ) + F R (C A1 − C A2 ) − V kC A2 dt Because both F  and F R   are constant, the total volume in each reactor is constant. V 2

10

Problem 11 The liquid in a jacketed, nonisothermal CSTR is stirred by an agitator whose mass is significant compared with the reaction mass. The mass of the reactor wall and the mass of the jacket wall are also significant. Write the energy equations for the system. Neglect radial temperature gradients in the agitator, reactor wall, and jacket wall. Solution

The CSTR is shown in Figure following assumptions are made:

??.

In writing the energy equations, the

Figure 9: Jacketed non isothermal CSTR.

•  Reactor fluid and cooling fluid hold up are constant. •   Physical properties are constant. •  The cooling fluid in the jacket is perfectly mixed. •  There are no internal temperature gradients in the agitator, reactor wall and jacket wall.

•  There are no heat losses to the environment. C  p,f M f 

dT f  = F C  p,f (T f,0 − T f ) − U s,f As,f (T f  − T s ) − U s,RW As,RW  (T f  − T RW ) dt C  p,s M s

dT s = U s,f As,f (T f  − T s ) dt 11

C  p,RW M RW 

C  p,cM c

dT RW  = U RW,f ARW,f (T f  − T RW ) + U RW,c ARW,c (T c − T RW ) dt

dT c = F c C  p,c(T c,0 −T c )−U RW,c ARW,c (T c −T RW )−U JW,c AJW,c (T c −T JW  ) dt

dT JW  = U JW,c AJW,c (T c − T JW  ) dt Where f   denotes properties of the fluid inside the reactor, s  propiertes of  the stirrer, RW  properties of the reactor wall, c properties of the cooling fluid and J W  properties of the jacket wall. C  p,JW  M JW 

Problem 12 The reaction 3A  →  2B + C is carried out in an isothermal semibatch reactor. Product B is the desired product. Product C is a very volatile by-product that must be vented off to prevent a pressure buildup in the reactor. Gaseous C is vented off through a condenser to force any A and B back into the reactor to prevent loss of reactant and product. Assume F V  is pure C. The reaction is first order in  C A . The relative volatilities of A and C to B are αAB = 1.2 and αCB   = 10. Assume perfect gases and constant pressure. Write the equations describing the system. List all assumptions.

Figure 10: Semi batch reactor. Solution

Assumptions:

12

•  The liquid phase is composed of only  A  and  B , whereas the gas phase is composed only of  C .

•  There is no change of volume due to mixing in the liquid phase. The mole balances for the liquid phase are: d(V L C A ) = −3kV L C A dt d(V L C B ) = 2kV L C A dt The mole balance for the gas phase are: P  dV G F V  P  = kV L C A − RT  dt RT  The additivity of volumes permit to write an additional relation between the concentrations of  A  and  B : M B A V L C A M  V A  + V B C A M A  C B M B ρA + V L C B ρB 1= = = + V L V L ρA ρB

Finally the sum of the volume of the gas and the liquid phase must equal the volume of the reactor: V L  + V G  = V R This five equations allows to solve the five variables: C A , C B , V L , V G and F V  .

Problem 13 Write the equations describing a simple version of the pretroleum industry’s important catalytic cracking operation. There are two vessels as shown in Figure ??. Component A is fed to the reactor where it reacts to form product B while depositing component C on the solid fluidized catalyst. A → B + 0.1C Spent catalyst is circulated to the regenerator where air is added to burn off C. C+O

→P

Combustion products are vented overhead, and regenerated catalyst is returned to the reactor. Heat is added or removed from the regenerator at a rate Q. Your dynamic mathematical model should be based on the following assumptions: 1. The perfect-gas law is obeyed in both vessels. 13

2. Constant pressure is mantained in both vessels. 3. Catalyst holdups in the reactor and in the regenerator are constant. 4. Heat capacities of reactants and products are equal and constant in each vessel. Catalyst heat capacity is also constant. 5. Complete mixing occurs in each vessel.

Figure 11: Reactor regenerator system. Solution

Assumptions:

•

Reactions are first order with respect to reactants, both in gas phase (C j ) and in solid phase (xj ).

•  Reactions occur in the surface of the catalyst, the rate of reaction is proM  a portional to

j

j

ρc

, where a j  is the surface/volume ratio for the catalyst.

The mass balance equations for reactor are: V G,1

dC A M 1 = F 0 C A0 − a1 k1 C A − F 1 C A dt ρc V G,1

M 1

dC B M 1 = a1 k1 C A − F 1 C B dt ρc

dX 1 M 1 = w(X 2 − X 1 ) + 0.1 a1 k1 C A dt ρc 14

The energy balance for reactor is:

(M 1 C  p,c + V G,1 ρ1 C  p,1 )

 dT 1 = F 0 ρ0 C  p,1 T 0 − F 1 ρ1 C  p,1 T 1  + wC  p,c(T 2 − T 1 ) dt 1 − M  a1 k1 C A λ1 ρc

The mass balance equations for the regenerator are: M 2 V G,2

dX 2 M 2 = w(X 1 − X 2 ) − a2 k2 C O X 2 dt ρc

dC O M 2 = F a C Oa − F 2 C O − a2 k2 C O X 2 dt ρc V G,2

dC P  M 2 = k2 C O X 2 − F 2 C P  dt ρc

The energy balance for the regenerator is:

(M 2 C  p,c + V G,2 ρ2 C  p,2 )

 dT 2 = F a ρa C  p,2 T a − F 2 ρ2 C  p,2 T 2  + wC  p,c(T 1 − T 2 ) dt 2 a2 k2 C O X 2 λ2  + Q − M  ρc

Where C  p,1  is the heat capacity of reactants and products at vessel 1, and C  p,2  is the heat capacity of reactants and products at vessel 2.

Problem 14 Flooded condensers and flooded reboilers are sometimes used on distillation columns. As shown in Figure ?? , a liquid level is held in the condenser, covering some of the tubes. Thus a variable amount of heat transfer area is available to condense the vapor. Column pressure can be controlled by changing the distillate (or reflux) drawnoff rate. Write the equations describing the dynamics of the condenser. Solution

Assumptions:

•  The dynamic of the fluid in the shell side is negligible. •  The heat transferred to the cooling water is due to heat of condensation alone.

•  The area available for heat transfer is proportional to the volume of the tubes not occupied by the condensing fluid.

15

Figure 12: Flooded condenser. The volumetric balance of fluid inside the shell and the energy balance for the cooling water are: dV  = dt

 V   − V   A U   T 

T 

V T 

ρλ

T  −

(T w,out + T w,in ) 2



−R−D

F w ρw C  p,w (T w,out − T w,in ) =  F c ρλ Where V T  and AT  are the total volume and total heat transfer area of the shell side, V  is the volume of condensed fluid inside the shell, and  F C  is the rate of fluid condensation.

Problem 15 When cooling jackets and internal cooling coils do not give enough heat transfer area, a circulating cooling system is sometimes used. Process fluid from the reactor is pumped through an external heat exchanger and back into the reactor. Cooling water is added to the shell side of the heat exchanger at a rate F w as set by the temperature controller. The circulation rate through the heat exchanger is constant. Assume that the shell side of the exchanger can be represented by two perfectly mixed lumps in series and that the process fluid flows countercurrent to the water flow, also through two perfectly mixed stages. The reaction is irreversible and first-order in reactant A. k

→B A− The contents of the tank are perfectly mixed. Neglect reactor and heat exchanger metal. Derive a dynamic mathematical model of the system. Solution

16

Figure 13: Circulating cooling system.

Figure 14: Lumps in series. The system can be represented as shown in Figure Assumptions:

??.

1. The volume of fluid contained in each lump, for both tubes and shell side, is constant. 2. For both tubes and shell side, the volume of lumps are equal to each other. 3. Density and specific heat are independent of concentration of A and B. Mass balances for the reactor: d(V C A ) = F 0 C A0 − F C A − C A V K  exp dt

17



E a − RT 



d(V C B ) = −F C B  + C A V K  exp dt



E a − RT 



Energy balance for the reactor: d(V T ) = F 0 T 0  + F C (T 2 − T ) − F T  − C A V K  exp dt





E a − RT  λ

Energy equations for shell side: V S 

dT w,1 = F w ρw C  p,w (T w,0 − T w,1 ) − U A(T w,1 − T 2 ) dt

dT w,2 = F w ρw C  p,w (T w,1 − T w,2 ) − U A(T w,2 − T 1 ) dt Energy equations for tubes side: V S 

V t

dT 1 = F C ρC  p (T  − T 1 ) − U A(T 1 − T w,2 ) dt

V t

dT 2 = F C ρC  p (T 1 − T 2 ) − U A(T 2 − T w,1 ) dt

Problem 16 A semibatch reactor is run at constant temperature by varying the rate of  addition of one of the reactants, A. The irreversible, exothermic reaction is first order in reactants A and B. k → A+B − C The tank is initially filled to its 40 percent level with pure reactant B at a concentration C B0 . Maximum cooling-water flow is begun, and reactant A is slowly added to the perfectly stirred vessel. Write the equations describing the system. Without solving the equations, try to sketch the profiles of  F A , C A , and C B  with time during the batch cycle. Solution

The mass balances for each species are: d(V C A ) = F A C A0 − kC A C B V  dt d(V C B ) = −kC A C B V  dt d(V C C ) = kC A C B V  dt The energy equation is: ρC  p

d(V T ) = −kC A C B V λ + Q dt 18

Figure 15: Semibatch reactor.

Figure 16: Guessed profiles. The guessed profiles are shown in Figure ??. C A  remains relatively low whereas the cuantity of B that remains is great enought to react with the incoming A, when B is consumed A starts to grow in concentration, also, later in time the effect of dillution because of the C that is being produced is reduced. C B  diminishes at first because the product C that accumulates in the tank dilutes it, after that continues to diminish because of  the accumulation of A. F A  is low at the begining because the rate of reaction is high, after that, more A is added to compensate for the consumption of B, also the cooling system can manage a greater volume of reaction mixture because of  the lower reaction rate.

Problem 17 Develop a mathematical model for the three-column train of distillation columns shown in Figure ??. The feed to the first column is 400 kg mol/h and contains four components (1, 2, 3 and 4), each at 25 mol %. Most of the lightest component is removed in the distillate of the first column, most of the next lightest in the second column distillate and the final column separates the final two heavy components. Assume constant relative volatilities throught the system: α1 , α2 and α3 . The condensers are total condensers and the reboilers are partial.

19

Trays, column bases and reflux drums are perfectly mixed. Distillate flow rates are set by reflux drum level controllers. Reflux flows are fixed. Steam flows to the reboilers are set by temperature controllers. Assume equimolal overflow, negligible vapor holdup, and negligible condenser and reboiler dynamics. Use a linear liquid hydraulic relationship ¯ ¯ n  +  M n − M n Ln  = L β  ¯ n and M  ¯ n  are the initial steady state liquid rate and holdup and  β  is a where L constant with units of seconds.

Figure 17: Train of distillation columns. Solution

With negligible dynamics for condenser and reboiler, only the tray holdups must be considered, the equations are analogous for every column  i, component  j and stage l. Mass balance for feed stage: dM NF,j,i = F i zj,i +LNF +1,i xNF +1,j,i −LNF,i xNF,j,i +V NF  dt

1,i

−

yNF 

1,j,i

−

−V NF,i yNF,j,i

Mass balance for top stage: dM NT,j,i = Rix R,j,i dt

− LNT,i xNT,j,i + V NT 

1,i

−

yNT 

1,j,i

−

− V NT,i yNT,j,i

Mass balance for first stage: dM 1,j,i = L 2,i x2,j,i dt

− L ,i x ,j,i  + V B,i yB,j,i − V  ,iy ,j,i 1

1

20

1

1

Mass balance for an intermediate (l) stage: dM l,j,i = Ll+1,i xl+1,j,i − Ll,i xl,j,i  + V l dt Energy balance for condenser:

1,i

−

yl

1,j,i

−

− V l,i yl,j,i

V N T,i C  p,NT  λlv = Qc Energy balance for reboiler: V B,i C  p,B λlv  =  Q B This equations, together with the equilibrium expressions and the relation between the hold up and the flux from the tray, constitute the dynamic model of the distillation column train.

Problem 18 The rate of pulp lay-down F   on a paper machine is controlled by controlling both the pressure P   and the height of slurry h  in a feeder drum with crosssectional area A. F  is proportional to the square root of the pressure at the exit slit. The air vent rate G is proportional to the square root of the air pressure in the box P . Feedback controllers set the inflow rates of air  G 0  and slurry F 0 to hold P  and h. The system is isothermal. Derive a mathematical model describing the system.

Figure 18: Paper machine. Solution

The mass balance for the liquid zone is: dh = F 0 − K F  dt The mass balance for the gas zone is: A

21

 hgρ  + P  s

√  d((H  − h)P ) = −K G P P  + G0 P 0 dt Where H  is the total height of the feeder drum, and  ρ s  is the density of the slurry. A

Problem 19 A wax filtration plant has six filters that operate in parallel, feeding from one common feed tank. Each filter can handle 1000 gpm when running, but the filters must be taken off-line every six hours for a cleaning procedure that takes ten minutes. The operation schedule calls for one filter to be cleaned every hour. How many gallons a day can the plant handle? If the flow rate into the feed tank is held constant at this average flow rate, sketch how the liquid level in the feed tank varies over a typical three-hour period. Solution

The average capacity is calculated considering that in a 1 hour period, 5 filters operate during 60 minutes, and the remaining filter operates during 50 minutes. C  =

 1000[gpm](5 ∗ 60[min] + 50[min]) = 5833[gpm] 60[min]

This capacity, over a day of activity, gives a processing capacity for the plant of 8400000 gallons. Every hour, during the first 10 minutes of operation, the capacity of the plant is only 5000 gpm (5 filters are operating), so an excess of  8333 gallons accumulate at the feed tank, the excess is processed over the next 50 minutes, when the capacity of the plant is 6000 gpm. The feed tank level variation over a three hour period is sketched in Figure ??.

Figure 19: Feed tank liquid level variation.

22

Problem 20 Alkylation is used in many petroleum refineries to react unsaturated butylenes with isobutane to form high octane iso-octane (alkylate). The reaction is carried out in a two liquid phase system: sulfuric acid / hydrocarbon. The butylene feed stream is split and fed into each of a series of perfectly mixed tanks (usually in one large vessel). This stepwise addition of butylene and the large excess of isobutane that is used both help to prevent undesirable reaction of butylene molecules with each other to form high boiling, low octane polymers. Low temperature (40 [F]) also favors the desired  iC 4 /C 4=  reaction. The reaction is exothermic. One method of heat removal that is often used is autorefrigeration: the heat of vaporization of the boiling hydrocarbon liquid soaks up the heat of reaction. The two liquid phases are completely mixed in the agitated sections, but in the last section the two phases are allowed to separate so that the acid can be recycled and the hydrocarbon phase sent off to a distillation column for separation. Derive a dynamic mathematical model of the reactor.

Figure 20: Alkylation process. iC4  + C = 4 C= 4

−k→ iC 1

8

−k→ polymer 2

Solution

Assumptions:

•  The reactions are first order with respect to reactants. 23

•

Only the solvent hydrocarbon is evaporated (This requires a fresh feed of  hydrocarbon for compensate for the quantity that is evaporated).

The mass balances for the first stage are: d(V 1 C iC  ) = F 0,iC   − k1 V 1 C iC  C C  dt 4

4

d(V 1 C C  ) = F 0,C  dt = 4

= 4

= 4

4

− k V  C iC  C C  − k 1

1

= 4

4

d(V 1 C iC  ) = k1 V 1 C iC  C C  dt 8

1

4

V 1 C 4= − F 1 C C 

= 4

− F  C iC 

= 4

4

2

− F  C iC 

1

8

d(V 1 C  poly ) = k 2 V 1 C 4= − F 1 C  poly dt d(V 1 C Ac ) = F 0,Ac − F 1 C Ac dt d(V 1 C HC ) = F 0,HC  − F 1 C HC  − F V, 1 dt The energy balance for stage 1 is: V 1 ρM,1 C  p,M,1 T 1 = F 0,iC  ρiC  C  p,iC  T 0,iC   + F 0,C  ρC  C  p,C  T 0,C  dt + F 0,Ac ρAc C  p,Ac T 0,F  ,Ac  + F 0,HC ρHC C  p,HC T 0,HC  − F V,1 ρHC λlv − k1 V 1C iC  C C  λ1 − k2 V 1 C 4= λ 2 4

4

4

= 4

4

= 4

= 4

= 4

0

4

− F  ρM, 1

1

= 4

C  p,M,1 T 1

The mass balances for the intermediates steps ( j = 2, 3, 4): d(V j C iC  ) = F j dt 4

d(V j C C  ) = F j dt = 4

1

−

1

−

C C 

d(V j C iC  ) = F j dt

= 4

4

4

= 4

− F j C iC 

4

= 4

− k V j C iC  C C  − k V j C  − F j C C 

8

1

−

C iC   − k1 V j C iC  C C  1

4

= 4

2

C iC   + k1 V j C iC  C C 

d(V j C  poly ) = F j dt

8

1

−

4

= 4

− F j C iC 

C  poly + k2 V j C 4= − F j C  poly

d(V j C Ac ) = F j dt

1

−

24

= 4

C Ac − F j C Ac

8

d(V j C HC ) = F j 1 C HC  − F j C HC  − F V,j dt In the intermediate steps mass balances before, the concentration that appears multiplying each flow F i  correspond to the concentration of the respective component in the vessel  i. The energy balance for stage j is: −

V j ρM,j C  p,M,j T j = F j 1 ρM,j 1 C  p,M,j 1 T j 1 − k1 V j C iC  C C  λ1 dt − k2 V j C 4= λ 2 − F j ρM,j C  p,M,j T j − F V,j ρHC λlv −

−

−

−

4

= 4

Where: F 0,iC  : Feed of isobutane. F 0,C  : Feed of unsaturated butylenes. F 0,Ac : Recirculated flow of acid. F 0,HC : Feed of solvent hydrocarbon. F V,j : Flow of hydrocarbon that is evaporated in each step. ρM,j : Density of the mixture at stage j. C  p,M,j : Heat capacity of the mixture at stage j. 4

= 4

Problem 21 Benzene is nitrated in an isothermal CSTR in the sequential irreversible reactions: k Benzene + HNO −→ Nitrobenzene + H2 O 1

k2

Nitrobenzene + HNO −→ Dinitrobenzene + H2 O k3

Dinitrobenzene + HNO −→ Trinitrobenzene + H2 O Assuming each reaction is linearly dependent on the concentrations of each reactant, derive a dynamical mathematical model of the system. There are two feed streams, one pure benzene and one concentrated nitric acid (98wt %). Assume constant density and complete miscibility. Solution

A sketch of the reactor is shown in Figure ??. The total and species mass conservation equations are: dV  = F 0  + F 1 − F  dt d(V C HN O ) = F 1 C H NO,1 −V (k1 C HN O C B +k2 C HN O C NB +k3 C HN O C DN B )−F C HN O dt d(V C B ) = F 0 − V k1 C B C HN O − F C B dt 25

Figure 21: Benzene nitration process.

d(V C NB ) = V (k1 C HN O C B − k2 C HN O C NB ) − F C NB dt d(V C DN B ) = V (k2 C HN O C NB dt

− k C HN O C DNB ) − F C DN B 3

d(V C T NB ) = V k3 C HN O C DNB dt

26

− F C T NB