Process Modelling, Simulation and Control for ChemicalEngineering. Solved problems. Chapter 3: Examples 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 Chemical Engineers” Second Edition, by William L. Luyben. At 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] Problem 1 A fluid of constant density ρ is pumped into a cone-shaped tank of total volume HπR2 /3 (Figure ??). 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 then: √ dV = F0 − K h dt Calling r to the radius of the fluid cone inside the tank, and if θ is the angle formed by the cone-shaped tank and the horizontal, we have: tgθ = H h = R/2 r/2 The volume of fluid inside the tank is: hπr2 π V = = 3 3 1 R H 2 h3 The last equation, together with the volumetric balance expression, 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 W 0 = K 0 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 resist 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. 1. Derive the equations describing the system. 2. 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 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: Patm + Ks x/A = P The volume variation of the cylinder gives: √ p P0 − Patm − Ks x/A W0 RT P0 − P RT dx = = = A dt ρ M P M Patm + Ks x/A 2 Assume constant density. Solution The total continuity equation gives (assuming constant density of the process fluid): dhow A = F0 − KF (how )3/2 dt The A component continuity equation gives: A d(how CA ) = F0 CA0 − KF (how )3/2 CA − k1 Ahow CA dt 3 .5 power. The steady state piston displacement will be reached when the volume no longer changes: P0 − Patm − x= Ks x =0 A A(P0 − Patm ) Ks Problem 3 A perfectly mixed. The weir height is hw . A first order reaction takes place in the tank: k 1 A −→ B Derive the equations describing the system.dx = dt RT M p P0 − Patm − Ks x/A APatm + Ks x with initial condition xt=0 = 0. The flow rate over the weir is proportional to the height of liquid over the weir (how ) to the 1. isothermal CSTR has an outlet weir. The cross sectional area of the tank is A. Figure 3: CSTR. The energy transport by conduction in the θ direction is negligible. 2. 3. is constant an that the operation is isothermal. Problem 5 A rotating metal drum heat exchanger is half submerged in a cool stream. The entalphy of the heat exchanger material can be represented as h = Cp T . the energy entering and leaving a volume of size d∆θRL is: 4 . For a time interval ∆t. 1.Problem 4 In order to ensure an adequate supply for the upcoming set-to with the Hatfields. What are the appropiate boundary conditions? Solution Assumptions: 1. The drum rotates at a constant angular velocity ω (radians per minute). Assume TH and TC are constant along their respective sections of the circumference. The thickness is much smaller than the exchanger radius. Neglect radial temperature gradients and assume steady state operation. Solution The total continuity equation gives: dV = F0 dt The ethanol continuity equation gives: d(V C) = F0 C0 − kV C dt With initial conditions: Vt=0 = 0 and Ct=0 = C0 . Assume perfect mixing and constant density. 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. Grandpa McCoy has begun to process a new batch of his famous Liquid Lightning moonshine. 2. C0 . He begins by pumping the mash at constant rate F0 into an empty tank. Heat capacity CP and density of the metal drum are constant. with its other half in a hot stream. 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. The drum length is L. thickness d. Heat transfer coefficients in the heating and cooling zones are constant (UH and UC ). Assuming that concentration of ethanol in the feed. Write the equations describing the system. and radius R. gives. An energy balance in the cooling zone. whereas the energy balance for the heating zone applies in the range θ : [0. isothermal operation at temperatures T1 and T2 . π]. • Entering: ∆tωRdLρCp Tθ • Leaving: ∆tωRdLρCp Tθ + dT dθ ∆θ + 2∆tL∆θRUj (T − Tj ) With Uj and Tj corresponding to the heat transfer coefficient and temperature at the cooling or heating zone. after dividing by ∆t∆θRL: dT wdCp ρ = −2UC (T − TC ) dθ dT −UC (T − TC ) = dθ wdCp ρ The analogous energy balance for the heating zone gives: dT −UH (T − TH ) = dθ wdCp ρ The energy balance for the cooling zone applies in the range θ : [π/2. constant density. Solution of the differential equations gives two integration constants. π/2]. constant throughput (F ). which are determined applying the following boundary conditions: Th (0) = Tc (2π) Th (π/2) = Tc (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 (V1 and V2 ).Figure 4: Rotating heat exchanger. and first order kinetics with simultaneous reactions: k 1 A −→ B 5 . t∗ + k2.1.T1 V CA.1.t∗ dt∗ dCC.k 2 C A −→ No reactions occur in the plug-flow section.t V = −F CB.2.2.t∗ − F CC.T2 V CA.1.2. 6 .1. Liquid density is given by: ρ = ρ0 + β(P − P0 ) where ρ0 .t∗ V = F CC.t = F CA0 − F CA. and P0 are constants. Write the equations describing the system.T2 V CA.1. ρ is the density.t∗ = F CB.2.T1 V CA.1.2.t dt The species balances for the second reactor are equivalent. and P is the pressure.t∗ + k1.2.T1 V CA.t V = −F CC.1. β.t∗ − F CB. A slightly compressible polymer liquid is pumped by a constant-speed. The pressure downstream of R3 is atmospheric. Solution The species balances for the first reactor are: dCA.1.t − k1.2.t∗ − k2.1.t∗ dt∗ V Problem 7 Consider the isothermal hydraulic system sketched below. but using a time variable t∗ equal to t + D: V V dCA.t dt dCC.t∗ dt∗ dCB.2.t∗ = F CA.t∗ − k1.2. Liquid is pumped through three resistances where the pressure drop is proportional to the square of the mass flow: ∆P = RW 2 . Figure 5: CSTRs separated by a dead time.T2 V CA.T2 V CA.t∗ − F CA.1.t + k2.1.t dt dCB.t − k2.T1 V CA.1.1. A surge tank of volume V is located between R1 and R2 and is liquid full. positive displacement pump so that the mass flow rate W1 is constant.t + k1.2. Solution A mass balance for the tank gives: dρ = W1 − W2 dt The balance can be re-stated in terms of the pressure. This material is fed to the top of the distillation column (which acts like a stripper). Problem 8 Develop the equations describing an ”inverted” batch distillation column. The mole conservation equations (both total and by component) for the Nth tray are: 7 . Derive the differential equation that gives the pressure P in the tank as a function of time and W1 . Vapor is generated in a reboiler at the base. 2. This system has a large reflux drum into which the feed is charged. • Tray contents. Heavy material is withdrawn from the bottom of the column. Figure 6: Hydraulic system. using the expression for the liquid density and the expression for the pressure drop: r dP P Vβ = W1 − dt R2 + R3 V The steady state value for the pressure is P = W12 (R2 + R3 ). 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 ??. Fin the steady state value of tank pressure P . The following assumptions are made in order to develop the model: • Vapor hold up is negligible. • Trays are ideal. reflux drum and reboiler contents are perfectly mixed. • The column operates with a binary mixture.1. 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 The energy conservation conservation equation for the Nth tray is: dMN 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 dt dMD = VN T − R dt The energy conservation equation for the reflux drum is: d(MD hD ) = VN T HN T − RhD + QD dt The mole conservation equations (both total and by component) for the reboiler are: 8 .Figure 7: Inverted batch distillation column. Develop the equations describing the dynamics of the system. 4. reboiler (R) or reflux drum (D) respectively. it remains at fusion temperature. P. 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. Entalphy of the liquid phase can be expressed as h = Cpw T .d(MR xR ) = L1 x1 − VR yR − P xR dt dMR = L1 − V R − 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 Finally. T ) Problem 9 An ice cube is dropped into a hot. in each tray. h is the entalphy of the liquid phase (in a molar basis). an equilibrium relation holds: yj = f (xj . 2. H is the entalphy of the gas phase (in a molar basis). insulated cup of coffe. perfectly mixed. Entalphy of the solid phase can be expressed as h = Cpw T − λ. There are no internal temperature gradients in the ice cube. Also. the following assumptions are made: 1. List all the assumptions and define all the terms. Solution Besides the conditions indicated in the problem statement. The physical properties of the coffe are the same as those of water. 9 . P is the product flow and V0 = VR . 3. the total volume in each reactor is constant.w Ti + U AHT (Mi )(Ti − Tc ) dt dt Where Mi and Ti are the mass and temperature of the solid phase. V2 10 . Cp. as a function of the mass of ice. Observation of flow patterns indicates that a two-tank system with back mixing. U is the heat transfer coefficient. Solution The A species balance for reactor 1 and 2 are: V1 dCA1 = F (CA0 − CA1 ) + FR (CA2 − CA1 ) − V kCA1 dt dCA2 = F (CA1 − CA2 ) + FR (CA1 − CA2 ) − V kCA2 dt Because both F and FR are constant. and AHT (Mi ) is the heat transfer area. The mixing is not perfect. Assuming F and FR are constant. Figure 8: Two tank system with back mixing. as shown in Figure ?? below. Mc and Tc are the mass and temperature of the liquid phase.w Problem 10 An isothermal. should approximate the imperfect mixing. write the equations describing the system.An energy balance allows to calculate the rate of fussion of ice: dMi Q U AHT (Mi )(Tc − Ti ) = = dt λ λ The mass balance for the system is: dMc Mi + =0 dt dt The energy balance for the liquid phase is: Mi d(Mc Tc ) =− Cp. λ is the heat of vaporization of ice. irreversible reaction: k A− →B takes place in the liquid phase in a constant-volume reactor. • Physical properties are constant.RW As. The mass of the reactor wall and the mass of the jacket wall are also significant.RW (Tf − TRW ) dt Cp. Solution The CSTR is shown in Figure ??.f As.f Mf dTf = F Cp. Write the energy equations for the system. • Reactor fluid and cooling fluid hold up are constant. reactor wall. • The cooling fluid in the jacket is perfectly mixed. In writing the energy equations.Problem 11 The liquid in a jacketed. Cp. and jacket wall.f (Tf − Ts ) − Us.f (Tf − Ts ) dt 11 . • There are no heat losses to the environment. • There are no internal temperature gradients in the agitator. Neglect radial temperature gradients in the agitator.f (Tf.0 − Tf ) − Us. the following assumptions are made: Figure 9: Jacketed non isothermal CSTR.f As. nonisothermal CSTR is stirred by an agitator whose mass is significant compared with the reaction mass.s Ms dTs = Us. reactor wall and jacket wall. c AJW.c (Tc − TRW ) dt dTc = Fc Cp. Write the equations describing the system.f (Tf − TRW ) + URW.Cp.c ARW. 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.JW MJW Problem 12 The reaction 3A → 2B + C is carried out in an isothermal semibatch reactor.c (Tc − TJW ) dt Where f denotes properties of the fluid inside the reactor. The relative volatilities of A and C to B are αAB = 1.c (Tc −TRW )−UJW. Product B is the desired product. Figure 10: Semi batch reactor. Assume perfect gases and constant pressure.c ARW. c properties of the cooling fluid and JW properties of the jacket wall. Cp. s propiertes of the stirrer.0 −Tc )−URW. List all assumptions. Assume FV is pure C.c Mc dTRW = URW.f ARW.2 and αCB = 10. Product C is a very volatile by-product that must be vented off to prevent a pressure buildup in the reactor.c (Tc.RW MRW Cp. Solution Assumptions: 12 . The reaction is first order in CA . RW properties of the reactor wall.c AJW.c (Tc −TJW ) dt dTJW = UJW. Heat is added or removed from the regenerator at a rate Q. VG and FV . The mole balances for the liquid phase are: d(VL CA ) = −3kVL CA dt d(VL CB ) = 2kVL CA dt The mole balance for the gas phase are: FV P P dVG = kVL CA − RT dt RT The additivity of volumes permit to write an additional relation between the concentrations of A and B: 1= MB A VL CA M VA + VB CA M A CB MB ρA + VL CB ρB = = + 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 . There are two vessels as shown in Figure ??. 13 . Problem 13 Write the equations describing a simple version of the pretroleum industry’s important catalytic cracking operation. C+O→P Combustion products are vented overhead.1C Spent catalyst is circulated to the regenerator where air is added to burn off C.• The liquid phase is composed of only A and B. The perfect-gas law is obeyed in both vessels. CB . VL . and regenerated catalyst is returned to the reactor. • There is no change of volume due to mixing in the liquid phase. 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. whereas the gas phase is composed only of C. Your dynamic mathematical model should be based on the following assumptions: 1. where aj is the surface/volume ratio for the catalyst. 5. 3.1 a1 k1 CA dt ρc 14 .1 M1 dCB M1 = a1 k1 CA − F1 CB dt ρc dX1 M1 = w(X2 − X1 ) + 0. • Reactions occur in the surface of the catalyst. 4. The mass balance equations for reactor are: VG. both in gas phase (Cj ) and in solid phase (xj ).2. Figure 11: Reactor regenerator system. Constant pressure is mantained in both vessels. Catalyst holdups in the reactor and in the regenerator are constant. Complete mixing occurs in each vessel.1 M1 dCA = F0 CA0 − a1 k1 CA − F1 CA dt ρc VG. Solution Assumptions: • Reactions are first order with respect to reactants. Heat capacities of reactants and products are equal and constant in each vessel. the rate of reaction is proM a portional to ρjc j . Catalyst heat capacity is also constant. Solution Assumptions: • The dynamic of the fluid in the shell side is negligible.2 Ta − F2 ρ2 Cp.2 M2 dX2 = w(X1 − X2 ) − a2 k2 CO X2 dt ρc M2 dCO = Fa COa − F2 CO − a2 k2 CO X2 dt ρc VG. Column pressure can be controlled by changing the distillate (or reflux) drawnoff rate.2 ρ2 Cp. Write the equations describing the dynamics of the condenser. Problem 14 Flooded condensers and flooded reboilers are sometimes used on distillation columns.2 M2 dCP = k2 CO X2 − F2 CP dt ρc The energy balance for the regenerator is: (M2 Cp. Thus a variable amount of heat transfer area is available to condense the vapor.1 T1 + wCp.c + VG.1 ρ1 Cp.2 is the heat capacity of reactants and products at vessel 2. 15 . As shown in Figure ??.2 ) dT2 = Fa ρa Cp. • The area available for heat transfer is proportional to the volume of the tubes not occupied by the condensing fluid. covering some of the tubes. and Cp.2 T2 + wCp.c (T1 − T2 ) dt M2 − a2 k2 CO X2 λ2 + Q ρc Where Cp. • The heat transferred to the cooling water is due to heat of condensation alone.c + VG.1 ) dT1 = F0 ρ0 Cp.c (T2 − T1 ) dt M1 − a1 k1 CA λ1 ρc The mass balance equations for the regenerator are: M2 VG.1 is the heat capacity of reactants and products at vessel 1.The energy balance for reactor is: (M1 Cp. a liquid level is held in the condenser.1 T0 − F1 ρ1 Cp. The reaction is irreversible and first-order in reactant A.w (Tw.out − Tw. Process fluid from the reactor is pumped through an external heat exchanger and back into the reactor. Neglect reactor and heat exchanger metal.in ) = Fc ρλ Where VT and AT are the total volume and total heat transfer area of the shell side. The volumetric balance of fluid inside the shell and the energy balance for the cooling water are: dV VT − V AT U (Tw. k A− →B The contents of the tank are perfectly mixed. Cooling water is added to the shell side of the heat exchanger at a rate Fw as set by the temperature controller. and FC is the rate of fluid condensation. The circulation rate through the heat exchanger is constant. Derive a dynamic mathematical model of the system. also through two perfectly mixed stages. 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.Figure 12: Flooded condenser. Problem 15 When cooling jackets and internal cooling coils do not give enough heat transfer area. Solution 16 .out + Tw. V is the volume of condensed fluid inside the shell.in ) = T− −R−D dt VT ρλ 2 Fw ρw Cp. a circulating cooling system is sometimes used. Density and specific heat are independent of concentration of A and B. The volume of fluid contained in each lump. for both tubes and shell side. 3. Assumptions: 1. Figure 14: Lumps in series. For both tubes and shell side. The system can be represented as shown in Figure ??. the volume of lumps are equal to each other. 2. Mass balances for the reactor: d(V CA ) Ea = F0 CA0 − F CA − CA V K exp − dt RT 17 .Figure 13: Circulating cooling system. is constant. CA . exothermic reaction is first order in reactants A and B.w (Tw.2 ) dt Vt dT2 = FC ρCp (T1 − T2 ) − U A(T2 − Tw. and CB with time during the batch cycle. Without solving the equations.w (Tw. and reactant A is slowly added to the perfectly stirred vessel.2 − T1 ) dt Energy equations for tubes side: VS Vt dT1 = FC ρCp (T − T1 ) − U A(T1 − Tw.2 = Fw ρw Cp.1 − T2 ) dt dTw. Maximum cooling-water flow is begun. d(V CB ) Ea = −F CB + CA V K exp − dt RT Energy balance for the reactor: d(V T ) Ea = F0 T0 + FC (T2 − T ) − F T − CA V K exp − λ dt RT Energy equations for shell side: VS dTw.1 ) − U A(Tw. k A+B− →C The tank is initially filled to its 40 percent level with pure reactant B at a concentration CB0 . A.0 − Tw. try to sketch the profiles of FA .2 ) − U A(Tw.1 = Fw ρw Cp. Write the equations describing the system. Solution The mass balances for each species are: d(V CA ) = FA CA0 − kCA CB V dt d(V CB ) = −kCA CB V dt d(V CC ) = kCA CB V dt The energy equation is: ρCp d(V T ) = −kCA CB V λ + Q dt 18 . The irreversible.1 − Tw.1 ) dt Problem 16 A semibatch reactor is run at constant temperature by varying the rate of addition of one of the reactants. also. The feed to the first column is 400 kg mol/h and contains four components (1. 3 and 4). Figure 16: Guessed profiles. after that. 2. later in time the effect of dillution because of the C that is being produced is reduced. Problem 17 Develop a mathematical model for the three-column train of distillation columns shown in Figure ??. each at 25 mol %. 19 . also the cooling system can manage a greater volume of reaction mixture because of the lower reaction rate. CB diminishes at first because the product C that accumulates in the tank dilutes it.Figure 15: Semibatch reactor. most of the next lightest in the second column distillate and the final column separates the final two heavy components. CA 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. The guessed profiles are shown in Figure ??. Most of the lightest component is removed in the distillate of the first column. The condensers are total condensers and the reboilers are partial. after that continues to diminish because of the accumulation of A. Assume constant relative volatilities throught the system: α1 . more A is added to compensate for the consumption of B. α2 and α3 . FA is low at the begining because the rate of reaction is high. j. and negligible condenser and reboiler dynamics.j.j.i xN F.i − V1.i x2.j.i yB. Assume equimolal overflow.i − VN T.j. Steam flows to the reboilers are set by temperature controllers. component j and stage l.i yN F. Reflux flows are fixed.i + VN T −1.i dt 20 .i − L1.i = L2.i y1. the equations are analogous for every column i.i −LN F.j.i x1. Use a linear liquid hydraulic relationship ¯ ¯ n + Mn − Mn 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. Solution With negligible dynamics for condenser and reboiler.j. Figure 17: Train of distillation columns.j.i +LN F +1.i xN T. only the tray holdups must be considered.i yN T −1.i = RixR. Mass balance for feed stage: dMN F.i +VN F −1. column bases and reflux drums are perfectly mixed.i dt Mass balance for top stage: dMN T.i yN F −1. Distillate flow rates are set by reflux drum level controllers.j.j.i − LN T.j.j.j.j.j. negligible vapor holdup.i dt Mass balance for first stage: dM1.Trays.i xN F +1.i yN T.i −VN F.i = Fi zj.i + VB. i − Vl.i xl+1.N T λlv = Qc Energy balance for reboiler: VB.j.j.j. The air vent rate G is proportional to the square root of the air pressure in the box P . together with the equilibrium expressions and the relation between the hold up and the flux from the tray. Solution The mass balance for the liquid zone is: p dh = F0 − KF hgρs + P dt The mass balance for the gas zone is: A 21 .i yl−1. 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.i − Ll.i dt Energy balance for condenser: VN T.i + Vl−1.B λlv = QB This equations. Figure 18: Paper machine.i Cp.i xl.i = Ll+1.j. F is proportional to the square root of the pressure at the exit slit.Mass balance for an intermediate (l) stage: dMl. The system is isothermal. Derive a mathematical model describing the system.j.i yl. constitute the dynamic model of the distillation column train.i Cp. Feedback controllers set the inflow rates of air G0 and slurry F0 to hold P and h. A Problem 19 A wax filtration plant has six filters that operate in parallel. C= 1000[gpm](5 ∗ 60[min] + 50[min]) = 5833[gpm] 60[min] This capacity. Solution The average capacity is calculated considering that in a 1 hour period. 5 filters operate during 60 minutes. Figure 19: Feed tank liquid level variation. gives a processing capacity for the plant of 8400000 gallons. and the remaining filter operates during 50 minutes. sketch how the liquid level in the feed tank varies over a typical three-hour period. the excess is processed over the next 50 minutes. feeding from one common feed tank. and ρs is the density of the slurry. The feed tank level variation over a three hour period is sketched in Figure ??. during the first 10 minutes of operation. Every hour. when the capacity of the plant is 6000 gpm.√ d((H − h)P ) = −KG P P + G0 P0 dt Where H is the total height of the feeder drum. The operation schedule calls for one filter to be cleaned every hour. so an excess of 8333 gallons accumulate at the feed tank. the capacity of the plant is only 5000 gpm (5 filters are operating). but the filters must be taken off-line every six hours for a cleaning procedure that takes ten minutes. over a day of activity. 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. 22 . Each filter can handle 1000 gpm when running. Derive a dynamic mathematical model of the reactor. 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). 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 reaction is exothermic. 23 . 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. Figure 20: Alkylation process. Low temperature (40 [F]) also favors the desired iC4 /C4= reaction. 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.Problem 20 Alkylation is used in many petroleum refineries to react unsaturated butylenes with isobutane to form high octane iso-octane (alkylate). The two liquid phases are completely mixed in the agitated sections. low octane polymers. k 1 iC4 + C= 4 −→ iC8 k 2 C= 4 −→ polymer Solution Assumptions: • The reactions are first order with respect to reactants. Ac T0.F0.1 dt The energy balance for stage 1 is: V1 ρM.• Only the solvent hydrocarbon is evaporated (This requires a fresh feed of hydrocarbon for compensate for the quantity that is evaporated).Ac + F0.iC4 ρiC4 Cp.HC ρHC Cp.C4= ρC4= Cp.M.C4= − k1 V1 CiC4 CC4= − k2 V1 C4= − F1 CC4= dt d(V1 CiC8 ) = k1 V1 CiC4 CC4= − F1 CiC8 dt d(V1 Cpoly ) = k2 V1 C4= − F1 Cpoly dt d(V1 CAc ) = F0. 4): d(Vj CiC4 ) = Fj−1 CiC4 − k1 Vj CiC4 CC4= − Fj CiC4 dt d(Vj CC4= ) = Fj−1 CC4= − k1 Vj CiC4 CC4= − k2 Vj C4= − Fj CC4= dt d(Vj CiC8 ) = Fj−1 CiC8 + k1 Vj CiC4 CC4= − Fj CiC8 dt d(Vj Cpoly ) = Fj−1 Cpoly + k2 Vj C4= − Fj Cpoly dt d(Vj CAc ) = Fj−1 CAc − Fj CAc dt 24 .iC4 − k1 V1 CiC4 CC4= − F1 CiC4 dt d(V1 CC4= ) = F0.C4= T0.1 Cp.HC T0.Ac − F1 CAc dt d(V1 CHC ) = F0.1 T1 The mass balances for the intermediates steps (j = 2. The mass balances for the first stage are: d(V1 CiC4 ) = F0.C4= dt + F0.HC − F1 CHC − FV.1 ρHC λlv − k1 V1 CiC4 CC4= λ1 − k2 V1 C4= λ2 − F1 ρM. 3.HC − FV.1 Cp.iC4 + F0.1 T1 = F0.Ac ρAc Cp.iC4 T0.M. M. ρM. Cp.HC : Feed of solvent hydrocarbon.j Tj = Fj−1 ρM. F0. the concentration that appears multiplying each flow Fi correspond to the concentration of the respective component in the vessel i.j : Heat capacity of the mixture at stage j.Ac : Recirculated flow of acid.M. Assume constant density and complete miscibility.j Tj − FV.j : Flow of hydrocarbon that is evaporated in each step.j Cp.j ρHC λlv Where: F0. derive a dynamical mathematical model of the system.d(Vj CHC ) = Fj−1 CHC − Fj CHC − FV. The energy balance for stage j is: Vj ρM. Problem 21 Benzene is nitrated in an isothermal CSTR in the sequential irreversible reactions: k1 Benzene + HNO −→ Nitrobenzene + H2 O k 2 Nitrobenzene + HNO −→ Dinitrobenzene + H2 O k 3 Dinitrobenzene + HNO −→ Trinitrobenzene + H2 O Assuming each reaction is linearly dependent on the concentrations of each reactant.M.j Cp. There are two feed streams. one pure benzene and one concentrated nitric acid (98wt %).C4= : Feed of unsaturated butylenes.j−1 Tj−1 − k1 Vj CiC4 CC4= λ1 dt − k2 Vj C4= λ2 − Fj ρM. F0.1 −V (k1 CHN O CB +k2 CHN O CN B +k3 CHN O CDN B )−F CHN O dt d(V CB ) = F0 − V k1 CB CHN O − F CB dt 25 .M. The total and species mass conservation equations are: dV = F0 + F1 − F dt d(V CHN O ) = F1 CHN O.j−1 Cp. Solution A sketch of the reactor is shown in Figure ??. F0.iC4 : Feed of isobutane.j : Density of the mixture at stage j.j dt In the intermediate steps mass balances before. FV. Figure 21: Benzene nitration process. d(V CN B ) = V (k1 CHN O CB − k2 CHN O CN B ) − F CN B dt d(V CDN B ) = V (k2 CHN O CN B − k3 CHN O CDN B ) − F CDN B dt d(V CT N B ) = V k3 CHN O CDN B − F CT N B dt 26 .