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Process Modelling, Simulation and Control for Chemical Engineers. Solved problems. Chapter 3: Examples of mathematical models of chemical engineering systems. Version 2. 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 Chemical Engineers”, Second Edition, by William L. Luyben. As such, I can’t guarantee that the proposed solutions are free from errors. Think about them as a starting point for developing or as a means of checking your own solutions. Any comments or corrections will be appreciated. Contact me at [email protected]

Changes from Version 1 *Figure references were corrected. *Equation that describes the system now contains K0 , problem 2. *A component continuity equation, problem 3. *Energy balance, domain and boundary condition specifications, problem 5. *Species mass balances, problem 6. *Equations for reflux drum, problem 8. *Energy balances, problem 15. *Energy balance for condenser and reboiler, problem 17.

Problem 1 A fluid of constant density ρ is pumped into a cone shaped tank of total volume HπR2 /3 (Figure 1). The flow out of the bottom of the tank is proportional to the square root of the height h of liquid in the tank. Derive the equations describing the system.

Figure 1: Cone shaped tank.

Solution The volumetric balance (constant density) for the fluid inside the tank is: √ dV = F0 − K h dt

(1)

Denoting by r the radius of the cone of fluid inside the tank, and if θ is the angle formed by the cone shaped tank and the horizontal, we have: tan θ =

H h = R r

(2)

hπr2 3

(3)

The volume of fluid inside the tank is: V =

1

Replacing equation (2) into equation (3) we get: π V = 3



R H

2

h3

(4)

Equation (4) together with equation (1) allow to solve for h and V as a function of time.

Problem 2 A perfect gas with molecular weight M flows at a mass flow rate W0 into a cylinder through a restriction. The flow rate is proportional to the square root of the pressure drop over the restriction: p W0 = K0 P0 − P where P is the pressure in the cylinder and P0 is the constant upstream pressure. The system is isothermal. Inside the cylinder, a piston is forced to the right as the pressure P builds up. A spring resists the movement of the piston with a force that is proportional to the axial displacement x of the piston. Fs = Ks 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. Assume the piston has negligible mass and friction. (a) Derive the equations describing the system. (b) What will the steady state piston displacement be?

Figure 2: Piston + cylinder.

Solution Because the mass and friction of the cylinder are negligible, we can assume that forces at side of him are always balanced. The pressure at the spring side of the cylinder, according to Figure 2, is atmospheric, so the initial pressure must be atmospheric (the barometric value is 0). A force balance for the piston gives: Patm + Ks x/A = P The volume variation of the cylinder is:  √ dx W0 K0 RT P0 − P = = dt ρ M P  p P0 − Patm − Ks x/A dx K0 RT A = dt M Patm + Ks x/A  p P0 − Patm − Ks x/A K0 RT dx = dt M APatm + Ks x A

With initial condition x(t=0) = 0. 2

The steady state piston displacement will be reached when the volume no longer changes: Ks x =0 A A(P0 − Patm ) x= Ks

P0 − Patm −

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: k A− →B Derive the equations describing the system.

Figure 3: CSTR.

Solution The total continuity equation (assuming constant density of the process fluid) is: A

dhow = F0 − KF (how )3/2 dt

The A component continuity equation is: A

d((hw + how )CA ) = F0 CA0 − KF (how )3/2 CA − kA(hw + how )CA dt

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 a constant rate F0 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 the concentration of ethanol in the feed, C0 , is constant and 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 is: dV = F0 dt 3

The ethanol continuity equation is: d(V C) = F0 C0 − kV C dt With initial conditions V(t=0) = 0 and C(t=0) = C0 .

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 w (radians per minute). Assume Th and Tc are constant along their respective sections of the circumference. The drum lenght is L, thickness d, and radius R. Heat transfer coefficients in the heating and cooling zones are constant(Uh and Uc ). Heat capacity Cp and density of the metal drum are constant. Neglect radial temperature gradients and assume steady state operation. (a) Write the equations describing the system. (b) What are the appropiate boundary conditions?

Figure 4: Rotating heat exchanger.

Solution Assumptions: 1. The entalphy of the heat exchanger material can be represented as h = Cp T . 2. The energy transport by conduction in the θ direction is negligible. 3. The thickness is much smaller than the exchanger radius. In a time interval ∆t, the energy entering and leaving a volume of size d∆θRL is: Entering: ∆tωRdLρCp Tθ Leaving:   dT ∆θ + 2∆tL∆θRUj (T − Tj ) ∆tωRdLρCp Tθ + dθ With Uj and Tj 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 ωdCp ρ = −2Uc (T − Tc ) dθ dT 2Uc (T − Tc ) =− dθ ωdCp ρ 4

The analogous energy balance for the heating zone gives: dT 2Uh (T − Th ) =− dθ ωdCp ρ The energy balance for the cooling zone applies in the range θ : [π, 2π], whereas the energy balance for the heating zone applies in the range θ : [0, π]. Solution of the differential equations gives two integration constants, which are determined applying the following boundary conditions: Th (0) = Tc (2π) Th (π) = Tc (π)

Problem 6 Consider the system that has two stirred chemical reactors separated by a plug flow deadtime of D seconds. Assume constant holdups (V1 and V2 ), constant throughput (F ), constant density, isothermal operation at temperatures T1 and T2 , and first order kinetics with simultaneous reactions: k

1 B A −→

k

2 C A −→

No reactions occurs 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: dCA1,t = F CA0 − F CA1,t − k1,T1 V1 CA1,t − k2,T1 V1 CA1,t dt CB1,t = −F CB1,t + k1,T1 V1 CA1,t V1 dt CC1,t V1 = −F CC1,t + k2,T1 V1 CA1,t dt

V1

The species balances for the second reactor are: dCA2,t = F CA1,t−D − F CA2,t − k1,T2 V2 CA2,t − k2,T2 V2 CA2,t dt CB2,t = F CB1,t−D − F CB2,t + k1,T2 V2 CA2,t V2 dt CC2,t = F CC1,t−D − F CC2,t + k2,T2 V2 CA2,t V2 dt

V2

Problem 7 Consider the isothermal hydraulic system sketched in Figure 6. A slightly compressible polymer liquid is pumped by a constant speed, positive displacement pump so that the mass flow rate W1 is constant. Liquid density is given by: ρ = ρ0 + β(P − P0 ) 5

where ρ0 , β, and P0 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 R1 and R2 and is liquid full. The pressure downstream of R3 is atmospheric. (a) Derive the differential equation that gives the pressure P in the tank as a function of time and W1 . (b) Find the steadystate value of tank pressure P .

Figure 6: Hydraulic system.

Solution A mass balance for the tank gives: dρ = W1 − W2 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: r P dP = W1 − Vβ dt R2 + R3 V

The steady state value for the pressure is P = W12 (R2 + R3 ).

Problem 8 Develop the equations describing an ”inverted” batch distillation column. This system has a large reflux drum into wich 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 in 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 can’t be neglected.

Solution The inverted batch distillation column is shown in Figure 7. The following assumptions are made in order to develop the model: • Vapor hold up is negligible. • Trays are ideal. • The column operates with a binary mixture. • Tray contents, reflux drum and reboiler are perfectly mixed. The mole conservation equations (both total and by component) for the Nth tray are: d(MN T xN T ) = RxD − LN T xN T + VN T −1 yN T −1 − VN T yN T dt dMN T = R − LN T + VN T −1 − VN T dt 6

Figure 7: Inverted batch distillation column. The energy conservation equation for the Nth tray is: d(MN T hN T ) = RhD − LN T hN T + VN T −1 HN T −1 − VN T HN T dt The mole conservation equations (both total and by component) for the reflux drum are: d(MD xD ) = VN T yN T − RxD + F xF dt dMD = VN T − R + F dt The energy conservation equation for the reflux drum is: d(MD hD ) = VN T HN T − RhD + F hF + QD dt The mole conservation equations (both total and by component) for the reboiler are: d(MR xR ) = L1 x1 − VR yR − P xR dt dMR = L1 − VR − P dt The energy conservation equation for the reboiler is: d(MR hR ) = L1 h1 − VR HR − P hR + QR dt The mole conservation equations (both total and by component) for the jth tray are: d(Mj xj ) = Lj+1 xj+1 − Lj xj + Vj−1 yj−1 − Vj yj dt dMj = Lj+1 − Lj + Vj−1 − Vj dt 7

Finally, the energy conservation equation for the jth tray is: d(Mj hj ) = Lj+1 hj+1 − Lj hj + Vj−1 Hj−1 − Vj Hj dt Where Mj 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 bassis), H is the entalphy of the gas phase (in a molar basis), P is the product flow, and F is the feed. 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 assumptions and define all 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. An energy balance allows to calculate the rate of fusion of ice: Q U AHT (Mi )(Tc − Ti ) dMi = = dt λ λ The mass balance for the system is: dMc dMi + =0 dt dt The energy balance for the liquid phase is: dMi d(Mc Tc ) =− Cp,w Ti + U AHT (Mi )(Ti − Tc ) dt dt Where Mi and Ti are the mass and temperature of the solid phase. Mc and Tc are the mass and temperature of the liquid phase. U is the heat transfer coefficient, λ is the heat of fusion of ice, and AHT (Mi ) is the heat transfer area, as a function of the mass of ice. Cp,w

Problem 10 An isothermal, irreversible reaction: k

A− →B 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 8, should approximate the imperfect mixing. Assuming F and FR are constant, write the equations describing the system.

Solution The A species balance for reactor 1 and 2 are: dCA1 = F (CA0 − CA1 ) + FR (CA2 − CA1 ) − V1 kCA1 dt dCA2 V2 = F (CA1 − CA2 ) + FR (CA1 − CA2 ) − V2 kCA2 dt Because both F and FR are constant, the total volume in each reactor is constant. V1

8

Figure 8: Two tank system with back mixing.

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 9. In writing the energy equations, the following assumptions are made: • Reactor fluid and cooling fluid holdup 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.

Figure 9: Jacketed non isothermal CSTR. The energy balances are: Cp,f Mf

dTf = F Cp,f (Tf,0 − Tf ) − Us,f As,f (Tf − Ts ) − Uf,rw Af,rw (Tf − Trw ) dt

9

Cp,s Ms

Cp,rw Mrw

Cp,c Mc

dTs = Uf,s Af,s (Tf − Ts ) dt

dTrw = Uf,rw Af,rw (Tf − Trw ) − Urw,c Arw,c (Trw − Tc ) dt

dTc = Fc Cp,c (Tc,0 − Tc ) + Urw,c Arw,c (Trw − Tc ) − Uc,jw Ac,jw (Tc − Tjw ) dt

Cp,jw Mjw

dTjw = Uc,jw Ac,jw (Tc − Tjw ) dt

Where f denotes properties of the fluid inside the reactor, s properties of the stirrer, rw properties of the reactor wall, c properties of the cooling fluid and jw properties of the jacket wall.

Problem 12 The reactions 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 build up 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 Fv is pure C. The reaction is first order in CA . 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: • 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(VL CA ) = −3kVL CA dt d(VL CB ) = 2kVL CA dt 10

The mole balance for the gas phase are: P dVG FV P = kVL CA − RT dt RT The additivity of volumes permit to write and additional relation between the concentrations of A and B: 1=

MB A VL CA M VA + VB MA MB ρA + VL CB ρB = = CA + CB VL VL ρA ρB

Finally the sum of the volume of the gas and the liquid phase must equal the volume of the reactor: VL + VG = VR This five equations allows to solve the five variables: CA , CB , VL , VG and FV .

Problem 13 Write the equations describing a simple version of the petroleum industry’s important catalytic cracking operation. There are two vessels as shown in Figure 11. 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 to 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. (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.

Solution Assumptions: • Reactions are first order with respect to reactants, both in gas phase (Cj ) and in solid phase (xj ). • Reactions occur in the surface of the catalyst, the rate of reaction is proportional to Mj aj /ρj , where aj is the surface/volume ratio for the catalyst. The mass balance equations for the reactor are: dCA M1 = F0 CA0 − a1 k1 CA − F1 CA dt ρc dCB M1 VG,1 = a1 k1 CA − F1 CB dt ρc dx1 M1 M1 = w(x2 − x1 ) + 0.1 a1 k1 CA dt ρc VG,1

11

Figure 11: Reactor regenerator system. The energy balance for the reactor is: (M1 Cp,c + VG,1 ρ1 Cp,1 )

M1 dT1 = F0 ρ0 Cp,1 T0 − F1 ρ1 Cp,1 T1 + wCp,c (T2 − T1 ) − a1 k1 CA λ1 dt ρc

The mass balance equations for the regenerator are: M2 dCO = Fa COa − F2 CO − a2 k2 CO x2 dt ρc M2 dCP = k2 CO x2 − F2 CP VG,2 dt ρc M2 dx2 = w(x1 − x2 ) − a2 k2 CO x2 M2 dt ρc

VG,2

The energy balance for the regenerator is: (M2 Cp,c + VG,2 ρ2 Cp,2 )

M2 dT2 = Fa ρa Cp,2 Ta − F2 ρ2 Cp,2 T2 + wCp,c (T1 − T2 ) − a2 k2 CO x2 λ2 + Q dt ρ

Where Cp,1 is the heat capacity of reactants and products at reactor, and Cp,2 is the heat capacity of reactants and products at regenerator.

Problem 14 Flooded condensers and flooded reboilers are sometimes used on distillation columns. In Figure 12, 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) drawoff 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 condensed fluid. 12

Figure 12: Flooded condenser. The volumetric balance of fluid inside the shell and the energy balance for the cooling water are:     dV V T − V AT U Tw,out + Tw,in = T− −R−D dt VT ρλ 2 Fw ρw Cp,w (Tw,out − Tw,in ) = Fc ρλ Where VT 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 Fc 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 Fw 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

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

Figure 13: Circulating cooling system.

13

Solution The system can be represented as shown in Figure 14. Assumptions: • The volume of fluid contained in each lump, for both tubes and shell side, is constant. • For both tubes and shell side, the volume of lumps are equal to each other. • Density ans specific heat are independent of concentration of A and B.

Figure 14: Lumps in series. Mass balances for the reactor:   d(V CA ) Ea = F0 CA0 − F CA − CA V K exp − dt RT   d(V CB ) Ea = −F CB + CA V K exp − dt RT Energy balance for the reactor:   d(V T ) CA V K Ea = F0 T0 + Fc (T2 − T ) − F T − λ exp − dt ρCp RT Energy balances for shell side: dTw,1 UA = Fw (Tw,0 − Tw,1 ) − (Tw,1 − T2 ) dt ρw Cp,w dTw,2 UA Vs = Fw (Tw,1 − Tw,2 ) − (Tw,2 − T1 ) dt ρw Cp,w

Vs

Energy balances for tube side: dT1 UA = Fc (T − T1 ) − (T1 − Tw,2 ) dt ρCp UA dT2 = Fc (T1 − T2 ) − (T2 − Tw,1 ) Vt dt ρCp Vt

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 CB0 . 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 FA , CA , and CB with time during the batch cycle. 14

Figure 15: Semibatch reactor.

Solution The mass balances for each species are: d(V CA ) = FA CA0 − kCA CB V dt d(V CB ) = −CA CB V dt d(V CC ) = kCA CB V dt The energy balance is: ρCp

d(V T ) = −kCA CB V λ + Q dt

The guessed profiles are shown in Figure 16. CA remains relatively low whereas the quantity of B that remains is great enought to react with the incoming A, when B is consumed A starts to grow in concentration, also due to the reduced dilution effect of C that is being produced in smaller quantity. CB diminishes because of the product C that is produced dilutes it, and also because of the accumulation of A. FA is low at the beginning 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.

Figure 16: Guessed profiles.

Problem 17 Develop a mathematical model of the three column train of distillation columns sketched in Figure 17. 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 throughout the system: α1 , α2 , and 15

α3 . The condensers are total condensers and the reboilers are partial. 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: Ln = Ln +

Mn − M n β

where Ln and M n are the initial steady state liquid rate and holdup and β is a constant with unit 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: dMN F,j,i = Fi zj,i + LN F +1,i xN F +1,j,i − LN F,i xN F,j,i + VN F −1,i yN F −1,j,i − VN F,i yN F,j,i dt Mass balance for top stage: dMN T,j,i = Ri xR,j,i − LN T,i xN T,j,i + VN T −1,i yN T −1,j,i − VN T,i yM T,j,i dt Mass balance for first stage: dM1,j,i = L2,i x2,j,i − L1,i x1,j,i + VB,i yB,j,i − V1,i y1,j,i dt Mass balance for intermediate stage: dMl,j,i = Ll+1,i xl+1,j,i − Ll,i xl,j,i + Vl−1,i yl−1,j,i − Vl,i yl,j,i dt Energy balance for condenser: VN T,i λlv,N T,i = Qc,i Energy balance for reboiler: VB,i λlv,B,i = QB,i This equations, together with the equilibrium expressions and the relation between the holdup and the flux from the tray, constitute the dynamic model of the distillation column train. 16

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 cross sectional 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 . Feeback controllers set the inflow rates of air G0 and slurry F0 to hold P and h. The system is isothermal. Derive a dynamic mathematical model describing the system.

Figure 18: Paper machine.

Solution The mass balance for the liquid zone is: p dh = F0 − KF hgρs + P dt The mass balance for the gas zone is: √ d((H − h)P ) A = G0 P0 − KG P P dt Where H is the total height of the feeder drum, and ρs 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 clening procedure that takes ten minutes. The operating 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 19. 17

Figure 19: Feed tank liquid level variation.

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 iC4 /C4n 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. k

1 iC4 + C4= −→ iC8

k

2 C4= −→ polymer

Solution Assumptions: • The reactions are first order with respect to reactants. • Only the solvent hydrocarbon is evaporated (This requires a fresh feed of hydrocarbon to compensate for the quantity that is evaporated).

18

Figure 20: Alkylation process. The mass balances for the first stage are: d(V1 CiC4 ) dt d(V1 CC4= ) dt d(V1 CiC8 ) dt d(V1 Cpoly ) dt d(V1 CAc ) dt d(V1 CHC ) dt

= F0,iC4 − k1 V1 CiC4 CC4= − F1 CiC4 = F0,C4= − k1 V1 CiC4 CC4= − k2 V1 C4= − F1 CC4= = k1 V1 CiC4 CC4= − F1 CiC8 = k2 V1 C4= − F1 Cpoly = F0,Ac − F1 CAc = F0,HC − F1 CHC − FV,1

The energy balance for stage 1 is: d(V1 ρM,1 Cp,M,1 T1 ) = (F0 ρCp T0 )iC4 + (F0 ρCp T0 )C4= + (F ρCp T0 )Ac dt + (F0 ρCp T0 )HC − FV,1 λlv − k1 V1 CiC4 CC4= λ1 − k2 V1 C4= λ2 − F1 ρM,1 Cp,M,1 T1

19

The mass balances for the intermediates steps (j=2, 3, 4) are: d(Vj CiC4 ) dt d(Vj CC4= ) dt d(Vj CiC8 ) dt d(Vj Cpoly ) dt d(Vj CAc ) dt d(Vj CHC ) dt

= Fj−1 CiC4 − k1 Vj CiC4 CC4= − Fj CiC4 = Fj−1 CC4= − k1 Vj CiC4 CC4= − k2 Vj C4= − Fj CC4= = Fj−1 CiC8 + k1 Vj CiC4 CC4= − Fj CiC8 = Fj−1 Cpoly + k2 Vj C4= − Fj Cpoly = Fj−1 CAc − Fj CAc = Fj−1 CHC − Fj CHC − FV,j

In the intermediate steps mass balances before, the concentration that appears multiplying each flow Fj corresponds to the concentration of the respective component in vessel j. The energy balance for stage j is: d((V ρM Cp,M T )j ) = (F ρM Cp,M T )j−1 − k1 Vj CiC4 CC4= λ1 − k2 Vj C4= λ2 dt − (F ρM Cp,M T )j − FV,j λlv,j Where: F0,iC4 : Feed of isobutane. F0,C4= : Feed of unsaturated butylenes. F0,Ac : Recirculated flow of acid. F0,HC : Feed of solvent hydrocarbon. FV,j : Flow of hydrocarbon that is evaporated in stage j. ρM,j Density of the mixture at stage j. Cp,M,j : Heat capacity of the mixture at stage j.

Problem 21 Benzene is nitrated in an isothermal CSTR in the sequential reversible reactions: k

1 Benzene + HN O −→ Nitrobenzene + H2 O

k

2 Nitrobenzene + HN O −→ Dinitrobenzene + H2 O

k

3 Dinitrobenzene + HN O −→ 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 (98w%). Assume constant density and complete miscibility.

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Solution A sketch of the reactor is shown in Figure 21. The total and species mass conservation equations are: dV = F0 + F1 − F dt d(V CHN O ) = F1 CHN O,1 − F CHN O dt − V (k1 CHN O CB + k2 CHN O CN B + k3 CHN O CDN B ) d(V CB ) dt d(V CN B ) dt d(V CDN B ) dt d(V CT N B ) dt

= F0 − V k1 CB CHN O − F CB = V (k1 CHN O CB − k2 CHN O CN B ) − F CN B = V (k2 CHN O CN B − k3 CHN O CDN B ) − F CDN B = V k3 CHN O CDN B − F CT N B

Figure 21: Benzene nitration process.

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