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Laboratory Exercise No. 6 Chemical Reactors: Plug Flow (Isothermal and Non-isothermal), Batch and Continuous Stirred Tank 1. Objective: The activity aims to develop equations and solve problems involving plug flow, batch and continuous stirred tank chemical reactors. 2. Intended Learning Outcomes (ILOs): The students shall be able to: 2.1 develop equations for plug flow, batch and continuous stirred tank chemical reactors 2.2 solve plug flow, batch and continuous stirred tank chemical reactors problems using matlab. 3. Discussion: Ethane flowing through a cylindrical tube can be heated to the point that it reacts into hydrogen and ethylene, the precursor to polyethylene. Many reactors, though, require a catalyst to speed up the reaction. The water gas shift reactor makes hydrogen using a cylindrical tube that is filled with catalyst and past which the gas flows. One feature that sets chemical engineers apart from other engineering disciplines is their ability to handle chemical reactions, the process of converting one or more chemicals into other chemicals that are more valuable. Thus, chemical engineers need to understand chemical reactions and be able to model equipment in which the chemical reactions take place, that is, chemical reactors. Sometimes reactors are as simple as a cylindrical tube, with reactants fed in one end and products (and unreacted reactants) coming out of the other end. Reactants may be placed in a vessel and stirred to create new chemicals. Real-life complexities challenge engineers as they design chemical reactors. In some reactors, the reaction rate is slow. When this happens, a bigger reactor is needed to achieve a specified production rate, and it may be too big to be economical. Then you look for a catalyst, which speeds up the reaction but remains unchanged after the reaction. The tube is packed with catalyst and reactants flow around the catalyst. Often the reaction is exothermic and the reactor becomes hot. The energy must be removed somehow, and heat transfer must be included in the calculations. Sometimes the reaction occurs inside a porous catalyst, perhaps an alumina pellet (Al2O3) with catalyst coating the internal surface. Then it may be necessary to include heat and mass transfer effects inside the pellet. When it is necessary to include these effects – slow reaction rates, catalysts, heat transfer, and mass transfer – it can make an engineering problem extremely difficult to solve. Numerical methods are a must, but even numerical methods may stumble at times. This laboratory exercise considers only relatively simple chemical reactors, but to work with these one must learn to solve ordinary differential equations as initial value problems. MATHEMATICAL FORMULATION OF REACTOR PROBLEMS A differential equation for a function that depends on only one variable is called an ordinary differential equation. The independent variable is frequently time, t, but for reactors it can also be length down a plug flow reactor. An example of an ordinary differential equation is
1
The general solution to the differential equation includes many possibilities; the engineer needs to provide initial conditions to specify which solution is desired. If all conditions are available at one point [as in Eq. (1)], then the problem is an initial value problem and can be integrated from that point on. If some of the conditions are available at one point and others at another point, then the ordinary differential equation becomes a two-point boundary value problem . Initial value problems as ordinary differential equations arise in the control of lumped parameter models, transient models of stirred tank reactors, and generally in models where there is no diffusion of the unknowns. Plug Flow Reactor and Batch Reactor Consider a plug flow reactor – a cylindrical tube with reactants flowing in one end and reacting as they flow to the outlet. A mole balance is made here for the case when the operation is steady. Let F be the molar flow rate of a chemical, which changes dF in a small part of the tube; take the volume of that small part to be dV. The rate of reaction is expressed as the moles produced per unit of time per unit of volume. If a chemical specie reacts, then the r for that specie is negative. Thus, the overall equation for a small section of the tube is dF = rdV
2
The molar flow rate can be related to the velocity in the tube. F = u Ac
3
In many cases the velocity and cross-sectional area of the tube are constant. Then you can write dF = uA dc , dV = Adz
4
If you put these into Eq. (2) and divide by dz you get u dc/dz = r , plug flow reactor
5
Normally r is a function of c, and this is an ordinary differential equation for c(z). In addition, you need to know the initial value of the concentration, that is, c(0) = co 6 If the reactor contains a catalyst, then the reaction rate is expressed as the molar rate of change per mass of catalyst, r ‘. One must then multiply by the mass of catalyst per unit volume, but otherwise the situation is the same.
dF = ρBr’ dV and u dc/dz = ρBr’, catalytic flow reactor
7
In a batch reactor, let N be the moles of a chemical; the molar change of a chemical is the reaction rate (molar change per unit time per unit volume) times the volume and the time. dN = rV dt
8
This can be put into a differential equation, too: dN/dt = rV
9
If the volume of the batch reactor is constant, then c = N/V
10
dc/dt = r
11
and the equation becomes
This equation is similar to the equation for a plug flow reactor, with the time t taking the place of the variable z/u. One can also introduce catalyst into the batch reactor, in which case the reaction rate r is replaced by ρB r’. In addition, you need an initial condition to go with Eq. (11), c(t =0) = c0. The structure of the problem is the same as that for a plug flow reactor, and if one can solve one, one can solve the other. So, there is some economy of effort. Many chemicals are involved, and you need a reaction rate for each reaction occurring. Thus, use {cj} f or all the chemical concentrations and {ri] f or all the reactions. If the reaction is aA + bB
cC + dD
12
then the rate of reaction of B is b/a times the rate of reaction of A, and the rates of reaction of C and D are –c/a and –d/a times the rate of reaction of A. The important point is that the rate of reaction is an algebraic function of the concentrations, temperature, and pressure. Each reaction rate needs to be known specifically for the reaction in question. Continuous Stirred Tank Reactor Another type of reactor is known as a continuous, stirred-tank reactor (CSTR). This reactor is a vessel with an inlet stream and an outlet stream, and the contents are constantly being stirred. Engineers often assume that this reactor is so well mixed that the concentration is the same everywhere. The mole balance on this reactor is Fout – Fin = rV
13
where V is the volume of the reactor. The molar flow rates are related to the volumetric flow rates, Q, as follows: F = Qc
14
When the volumetric flow rates are constant, Eq. (13) becomes Q(cout – cin ) = rV, continoud stirred tank
15
In a well-mixed reactor (CSTR), the concentration going out of the reactor is the same as the concentration in the reactor. Since the rate expression is a function of c, it is a function of cout. Then Eq. (15) becomes an algebraic equation in one variable (here) for cout. 4. Resources: Matlab 5. Procedure: 1. To solve the single differential equation: dy/dt = -10y , y(0) = 1 from t = 0 to t = 1. a. Construct a function file that contains: % filename rhs.m function yprime=rhs(t,y) % compute the rhs of eqns.for any % input values t, y yprime=-10*y; b.
plot(t,y)
Construct a script file that contains: % filename yourSurname_le10_p1b.m y0=1; % set the initial condition to 1 tspan=[0 1]; % set the initial time to zero and the final % time to 1 % call the routine ode45; use the m-file called % rhs, % start with y=y0 at tspan(1) % and integrate until tspan(2) [t,y]=ode45(@rhs,tspan,y0) % plot the solution
c. Run the program. Show the results. 2. Suppose you take a reactor in which components A and C are fed in equimolar amounts, and the following reaction takes place: 2A B 1 You assume that the reaction takes place in the liquid phase and that the volumetric flow
rate remains constant even when reaction occurs. The equations are Eq. (5) for each species. u dCj/dz = rj , j = 1…3 2 where Cj is the molar concentration of the jth species, u is the velocity, z is the distance down the tube from the inlet, and rj is the rate of reaction of the jth species, in mol/vol time. Here the rate of reaction is taken as second order: rate of formation of B = kCA2 3 where the units of k are vol/mol time. The equations for all three species are then 4
At the inlet you take CA(0) = 2 kmol/m3, CB(0) = 0, CC(0) = 2 kmol/m3 5 and we take u = 0.5 m/s, k = 0.3 m 3 /kmol s, and the total reactor length as z = 2.4 m. a. Create a function file that contains: rate1.m % % This function gives the right-hand side for a simple % reactor problem which is isothermal. function yprime=rate1(VR,y) % y(1) is CA, y(2) is CB, y(3) is CC % k=0.3 and u=0.5 CA=y(1); rate=0.3*CA*CA; yprime(1)=-2.*rate/0.5; yprime(2) =+rate/0.5; yprime(3)=0.; yprime=yprime’; b. Create an m-file that contains: % yourSurname_le10_p2b,m % This is the driver program to solve the simple flow % reactor example. % set the initial conditions y0=[2 0 2] % set the total volume of the reactor zspan=[0 2.4] % call the ode solver [z y]=ode45(@rate1, zspan , y0) % plot the result plot(z,y)
xlabel(‘length (m)’) ylabel(‘concentrations (kgmol/m^3)’) legend(‘A’, ‘B’, ‘C’) c. Run the program. Show the results. 3. Most reactions have heat effects, and this means that reactor models must be able to model the heat transfer as well as the mass balances. The following example models a simple reactor oxidizing SO2 to form SO3 . After some manipulation, the equations are 1 where the reaction rate is
with the parameters: Tsurr = 673.2, T(0) = 673.2, X(0) = 1. The variable X is the concentration of SO2 divided by the inlet concentration, 1 - X is the fractional conversion, and T is the temperature in K. The first equation is the mole balance on SO2, and the second is the energy balance. The first term on the right-hand side of Eq. (1) represents cooling at the wall; the second term there is the heat of reaction. These equations are appropriate when radial and axial dispersion are not important. Axial and radial dispersion are actually important in this case, as discussed by Young and Finlayson (1973), but the simpler case is a good vehicle for seeing how to include temperature effects in the problem. The reaction rate equation shows that equilibrium can be reached (then R’ = 0), and the equilibrium constant depends on the temperature. Thus, the conversion depends intimately on the temperature. a. Create a function file that contains: % rateSO2.m for SO2 reaction function yprime=rateSO2(z,y) % X is concentration SO2 divided by the inlet concentration X=y(1); % T is temperature in degrees K T=y(2); k1=exp(-14.96+11070/T); k2=exp(-1.331+2331/T); Keq=exp(-11.02+11570/T); term1=X*sqrt(1-0.167*(1-X)); term2=2.2*(1-X)/Keq; denom=(k1+k2*(1-X))^2; rate=(term1-term2)/denom; yprime(1)=-50*rate; yprime(2)=-4.1*(T-673.2)+1.02e4*rate;
yprime=yprime’; b. Create an m-file that contains: % yourSurname_le10_p3b.m % set the dimensionless initial conditions y0=[1 673.2] % set the integration range zspan=[0 1] % call the solver [z y]=ode45(@rateSO2,zspan,y0) % plot the result plot(z,y(:,1),‘r-’) xlabel(‘dimensionless axial position’) legend(‘fraction converted’) pause plot(z,y(:,2),‘g-’) xlabel(‘dimensionless axial position’) legend(‘temperature (K)’) c. Run the program. Show the results. 4. The situations in Procedures 2 and 3 are for liquid streams or dilute gaseous streams where the density does not change drastically. Sometimes the density does change, though, and such problems are not much harder to solve numerically than their constant-density counterparts. The governing equation is Eq. (2), here expressed as a differential equation:
where Fj is the molar flow rate of the jth species, VR is the volume of reactor (from the inlet), and rj is the rate of reaction of the jth species. Let us take the same reaction, Procedure 2 Equation 1 and reaction rate, Procedure 2 Equation 3. Putting these together gives the differential equations to be solved:
The equations are differential equations for the molar flow rates, FA, FB, and FC, but the rates of reaction are expressed as concentration, CA. It is possible to derive the concentration of each chemical from the molar flow rates. In a gas phase, the total concentration is governed by the perfect gas law, although more complicated equations of state are possible:
Thus, at any point in the reactor, if you know the pressure and temperature, you can find CT. The pressure is determined by fluid mechanics and the pressure drop, and the temperature is determined by the energy equation. Then the concentration of a chemical is the mole fraction times the total concentration; for example
And the mole fraction is
If the temperature and pressure are constant, the total concentration is the total concentration at the inlet, which is taken here to be 5 kmol/m3. Thus, the flow rate is
The problem is summarized here as
Write a function file that calculate the right-hand side, given the input, which are the flow rates of all species. The variables are: y(1)=FA , y(2)=FB , y(3)=FC The function will have available to it the accumulated reactor volume VR, and the local value of y(j), j=1,..., 3 at the same reactor volume. To calculate the concentrations at that reactor volume use:
The rates of reaction are then evaluated, and the function returns the numerical value of the right-hand side. Solve this procedure for a total volume of 1.2 m3 , k=0.3 m3/kmol s, a total concentration of 5 kmol/m3. 5. When a reaction occurs on a catalyst particle, it is necessary for the reactants to get to the catalyst. Sometimes this transfer is hindered by mass transfer, and then the concentration near the catalyst is not the same as it is in the bulk stream. This phenomena is demonstrated with Procedure 2 posed in Eqs. (4) – (5), except that now the rate of reaction is evaluated at a concentration that is near the catalyst. The concentration Cs is the concentration on the surface, expressed as the kmol of a species per volume of the catalyst. There are various ways to represent this concentration. The net effect, however, is that Procedure 2 Eq. (4) is changed to :
5.1 You will be solving for CA(z), but you cannot evaluate the rate of reaction in Eq. (5.1) because you do not knowCA,s. You need a mass balance relating the rate of mass transfer to the catalyst to the rate of reaction. One form of that is 5.2
where km is a mass transfer coefficient in units of m/s, determined from correlations derived in fluid mechanics and mass transfer courses,a is the surface area exposed per volume of the reactor (m2/m3), and ks is a rate of reaction rate constant (here m3/kmol s). Looking closely at Eqs. (5.1) –(5.2), one can see that, in order to solve the differential equations in Eq. (5.1), you must solve Eq. (5.2) at every position z. Thus, this is a problem that combines ordinary differential equations with nonlinear algebraic equations.
a. The function file that will evaluate Eq. 5. 2 contains: function y=mass_rxn(CAs,CA) global k km % mass_rxn.m % this m-file evaluates the mass transfer equations % km in the program includes the a % the first entry is the guess of CAs and % the second entry is the fluid concentration y=km*(CA-CAs)-k*CAs*CAs; b. To evaluate the right-hand side of Eq. 5.1 , the m-file contains the following: % yourSurname_le10_p5b.m % This function gives the right-hand side for a % reactor problem which is isothermal but has % mass transfer effects function yprime=rate1_mass(VR,y) global k vel km % y(1) is CA, y(2) is CB, y(3) is CC CA=y(1); % solve the mass transfer problem % the first entry is the guess of CAs and % the second entry is the fluid concentration OPTIONS=[]
CAguess=CA; CAs=fzero(@mass_rxn,CAguess,OPTIONS,CA); rate=k*CAs*CAs; yprime(1)=-2.*rate/vel; yprime(2)=+rate/vel; yprime(3)=0.; yprime=yprime’; c. Using the value of km = 0.2, the m-file that will solve the problem contains: % yourSurname_le10_p5c.m global k vel km k=0.3; vel=0.5; km=0.2; % set the initial conditions y0=[2 0 2] % set the total volume of the reactor zspan=[0 2.4] % call the ode solver [z y]=ode45(@rate1_mass,zspan,y0) % plot the result plot(z,y(:,1),‘*-’,z,y(:,2),‘+-’,z,y(:,3),‘x-’) % or plot(z,y) xlabel(‘length (m)’) ylabel(‘concentrations (kgmol/m^3)’) legend(‘A’, ‘B’, ‘C’) d. Run the m-file. Show the results. 6. This procedure considers a CSTR in which a first-order reaction occurs, but the temperature also changes owing to the heat of reaction. The equations to be solved are: 6.1
The left-hand sides are the flow rate times a concentration or temperature difference between the input and output, divided by the volume. The equations have been normalized by the inlet concentration and temperature. The right-hand sides are the rate of reaction and the rate of energy generation due to reaction, respectively. The case described by Eq. (6.1) is for an adiabatic reactor. When the reactor is adiabatic, the equations can be combined by multiplying the first equation by β and adding it to the second equation; then the right-hand side vanishes. 6.2
This equation can be solved for T: 6.3
Now the mass balance can be considered a single equation in one unknown (c) 6.4
a. Construct a function file representing Eq. 6.4 that contains the following: % rate_T function fn=rate_T(c) global beta gamma flowvol T=1+beta*(1-c) rate=c*exp(gamma*(1-1/T)) fn=flowvol*(1-c)-rate b. To solve the problem, type the following commands into the command window: >> fzero(@rate_T,0.5) c. Show the results 7. Reactors do not always run at steady state. In fact, many pharmaceuticals are made in a batch mode. Such problems are easily solved using the same techniques presented above because the plug flow reactor equations are identical to the batch reactor equations.Even CSTRs can be run in a transient mode, and it may be necessary to model a time dependent CSTR to study the stability of steady solutions. When there is more than one solution, one or more of them will be unstable. Thus, this section considers a time dependent CSTR as described by Eq. (7.1): 7.1
The variables are V= reactor volume, Q = volumetric flow rate C = concentation, t = time k0 = reaction rate constant,E = activation energy Φ = void fraction, ρ = density, = heat capacity per mass subscript f for fluid, s for solid - ΔHrxn = heat of reaction The nondimensional form of these equations is
7.2
The parameter Le is a Lewis number, and it includes the heat capacity of the system. The Da is a Damkohler number and includes the rate of reaction. The parameters are taken as 7.3
a. A function file that computes the right hand side contains : % rate_limit.m % solve for particle with limit cycle function yprime=rate_limit(t,y) global beta gamma Damk Lewis Tin c=y(1); T=y(2); rate=c*Damk*exp(gamma*(1-1/T)); yprime(1)=1 - c- rate; yprime(2)=Tin -T+ beta*rate; yprime(2)=yprime(2)/Lewis; ydot=ydot’; b. An m-file contains the following: % yourSurname_le10_p7b.m global beta gamma Damk Lewis Tin beta=0.15 gamma=30 Lewis=1080 Damk=0.115 Tin=1 [t y]=ode45(‘rate_limit’,[0 2],[0.7 1.0]) plot(t,y(:,1),‘*-k’,t,y(:,2),‘o-k’) xlabel(‘time’) legend(‘concentration’,‘temperature’) c. Run the m-file. Show the results. 8. Solve for the concentration distribution in a plug flow reactor governed by the following equations:
8.1
8.2
9. Solve for the concentration distribution in a plug flow reactor with the following reaction rate expression and parameters: 9.1
10. P.142
Course: CHE 508 Group No.: Group : AQUINO, AIRAH DS.
Laboratory Exercise No.: 6 Section: CH51FC2 Date Performed: 1/18/19 Date Submitted: 1/18/19 Instructor: ENGR. CRISPULO MARANAN
6. Data and Results: Procedure 1c
Results >> Aquino_le10_p1b t= 0 0.0050 0.0100 0.0151 0.0201 0.0406 0.0612 0.0817 0.1023 ... 0.7901 0.8098 0.8311 0.8524 0.8737 0.8950 0.9213 0.9475 0.9738 1.0000 y= 1.0000 0.9510 0.9044 0.8601 0.8180 0.6659 0.5421
0.4416 0.3598 0.2957 ... 0.0004 0.0003 0.0002 0.0002 0.0002 0.0001 0.0001 0.0001 0.0001 0.0000
2c
>> Aquino_le10_p2b y0 = 2
0
0
2.4000
zspan =
2
z= 0 0.0000 0.0000 0.0001 0.0001 0.0002 0.0003 0.0004 0.0005 0.0010 … 2.0641 2.1241 2.1841 2.2441 2.3041 2.3641 2.3731 2.3820 2.3910 2.4000 y= 2.0000 1.9999 1.9998 1.9997 1.9996 1.9991 1.9986 1.9981 1.9976 1.9951 0.3359 0.3280 0.3204 0.3132 0.3063 0.2997 0.2987 0.2978 0.2968 0.2959
0 0.0001 0.0001 0.0002 0.0002 0.0005 0.0007 0.0010 0.0012 0.0025 .... 0.8320 0.8360 0.8398 0.8434 0.8468 0.8502 0.8506 0.8511 0.8516 0.8521
2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000
3c
>> Aquino_le10_p3b y0 = 1.0000 673.2000 zspan = 0 z= 0 0.0195 0.0391 0.0586 0.0782 0.1032 0.1282 0.1532 0.1782 0.2032 …
1
0.8532 0.8782 0.9032 0.9282 0.9532 0.9782 0.9836 0.9891 0.9945 1.0000 y= 1.0000 0.9431 0.8763 0.8056 0.7377 0.6628 0.6012 0.5512 0.5102 0.4759 … 0.2278 0.2247 0.2217 0.2188 0.2161 0.2134 0.2128 0.2122 0.2117 0.2111
673.2000 684.3790 696.6215 708.6819 719.2483 729.2531 735.7154 739.3146 740.8091 740.8627 688.7606 687.8445 686.9950 686.2074 685.4774 684.8008 684.6598 684.5211 684.3846 684.2505
5
>> Aquino_le10_p5c y0 = 2
0
0
2.4000
zspan =
z= 0 0.0600 0.1200 0.1800 0.2400 0.3000 0.3600 0.4200 0.4800 0.5400 …
2
1.8600 1.9200 1.9800 2.0400 2.1000 2.1600 2.2200 2.2800 2.3400 2.4000 y= 2.0000 2.0238 2.0470 2.0694 2.0911 2.1117 2.1313 2.1495 2.1664 2.1817 … 1.8000 1.7343 1.6628 1.5854 1.5020 1.4123 1.3160 1.2131 1.1032 0.9862
0 2.0000 -0.0004 2.0238 -0.0016 2.0470 -0.0039 2.0694 -0.00732.0911 -0.01192.1117 -0.01802.1313 -0.02572.1495 -0.03512.1664 -0.04642.1817 -1.1012 1.8000 -1.20201.7343 -1.30891.6628 -1.42221.5854 -1.54191.5020 -1.66841.4123 -1.80181.3160 -1.94241.2131 -2.09041.1032 -2.24600.9862
6
Function File: function fn=rate_T(c) global beta gamma flowvol T=1+beta*(1-c) rate=c*exp(gamma*(1-1/T)) fn=flowvol*(1-c)-rate yprime(1)=rate/(1-c) yprime=yprime'; >> fzero(@rate_T,0.5) T= 1.0750 rate = 4.0547 fn = []
yprime = 7
8.1094 >> Aquino_le10_p7b t= 0 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 … 1.6000 1.6500 1.7000 1.7500 1.8000 1.8500 1.9000 1.9500 2.0000 y= 0.7000 0.7107 0.7208 0.7303 0.7393 0.7479 0.7560 0.7636 0.7708 0.7777 …. 0.8638 0.8656
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.8673 0.8689 0.8704 0.8718 0.8732 0.8744 0.8757
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
8 9
7. Conclusion: I therefore conclude that MATLAB can develop equations and solve problems involving plug flow, batch and continuous stirred tank chemical reactors.
8. Assessment (Rubric for Laboratory Performance): TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES RUBRIC FOR MODERN TOOL USAGE (Engineering Programs) Student Outcome (e): Use the techniques, skills, and modern engineering tools necessary for engineering practice in complex engineering activities. Program: Chemical Engineering Course: CHE 508 Section: CH51FC2 2nd Sem SY 2018-’19 Performance Unsatisfactor Developing Satisfactory Very Satisfactory Score Indicators y 2 3 4 1 1. Apply Fails to identify Identifies Identifies modern Recognizes the appropriate any modern modern techniques and is benefits and techniques, techniques to techniques but able to apply these constraints of skills, and perform fails to apply in performing modern engineering modern tools to discipline-spec these in discipline-specific tools and shows perform a ific engineering performing engineering task. intention to apply discipline-specif task. discipline-specifi them for engineering ic engineering c engineering practice. task. task. 2. Demonstrate Fails to apply Attempts to Shows ability to Shows ability to skills in any modern apply modern apply fundamental apply the most applying tools to solve tools but has procedures in using appropriate and different engineering difficulties to modern tools when effective modern techniques and problems. solve solving engineering tools to solve modern tools to engineering problems. engineering solve problems. problems. engineering problems. 3. Recognize the Does not Recognizes Recognizes the Recognizes the benefits and recognize the some benefits benefits and need for benefits constraints of benefits and and constraints constraints of and constraints of modern constraints of of modern modern engineering modern engineering engineering modern engineering tools and shows tools and makes tools. engineering tools. intention to apply good use of them for tools. them for engineering practice. engineering practice. Total Score Mean Score = (Total Score / 3) Percentage Rating = (Total Score / 12) x 100%
Evaluated by: ENGR. CRISPULO MARANAN Printed Name and Signature of Faculty Member
JANUARY 2019 Date
1
The general solution to the differential equation includes many possibilities; the engineer needs to provide initial conditions to specify which solution is desired. If all conditions are available at one point [as in Eq. (1)], then the problem is an initial value problem and can be integrated from that point on. If some of the conditions are available at one point and others at another point, then the ordinary differential equation becomes a two-point boundary value problem . Initial value problems as ordinary differential equations arise in the control of lumped parameter models, transient models of stirred tank reactors, and generally in models where there is no diffusion of the unknowns. Plug Flow Reactor and Batch Reactor Consider a plug flow reactor – a cylindrical tube with reactants flowing in one end and reacting as they flow to the outlet. A mole balance is made here for the case when the operation is steady. Let F be the molar flow rate of a chemical, which changes dF in a small part of the tube; take the volume of that small part to be dV. The rate of reaction is expressed as the moles produced per unit of time per unit of volume. If a chemical specie reacts, then the r for that specie is negative. Thus, the overall equation for a small section of the tube is dF = rdV
2
The molar flow rate can be related to the velocity in the tube. F = u Ac
3
In many cases the velocity and cross-sectional area of the tube are constant. Then you can write dF = uA dc , dV = Adz
4
If you put these into Eq. (2) and divide by dz you get u dc/dz = r , plug flow reactor
5
Normally r is a function of c, and this is an ordinary differential equation for c(z). In addition, you need to know the initial value of the concentration, that is, c(0) = co 6 If the reactor contains a catalyst, then the reaction rate is expressed as the molar rate of change per mass of catalyst, r ‘. One must then multiply by the mass of catalyst per unit volume, but otherwise the situation is the same.
dF = ρBr’ dV and u dc/dz = ρBr’, catalytic flow reactor
7
In a batch reactor, let N be the moles of a chemical; the molar change of a chemical is the reaction rate (molar change per unit time per unit volume) times the volume and the time. dN = rV dt
8
This can be put into a differential equation, too: dN/dt = rV
9
If the volume of the batch reactor is constant, then c = N/V
10
dc/dt = r
11
and the equation becomes
This equation is similar to the equation for a plug flow reactor, with the time t taking the place of the variable z/u. One can also introduce catalyst into the batch reactor, in which case the reaction rate r is replaced by ρB r’. In addition, you need an initial condition to go with Eq. (11), c(t =0) = c0. The structure of the problem is the same as that for a plug flow reactor, and if one can solve one, one can solve the other. So, there is some economy of effort. Many chemicals are involved, and you need a reaction rate for each reaction occurring. Thus, use {cj} f or all the chemical concentrations and {ri] f or all the reactions. If the reaction is aA + bB
cC + dD
12
then the rate of reaction of B is b/a times the rate of reaction of A, and the rates of reaction of C and D are –c/a and –d/a times the rate of reaction of A. The important point is that the rate of reaction is an algebraic function of the concentrations, temperature, and pressure. Each reaction rate needs to be known specifically for the reaction in question. Continuous Stirred Tank Reactor Another type of reactor is known as a continuous, stirred-tank reactor (CSTR). This reactor is a vessel with an inlet stream and an outlet stream, and the contents are constantly being stirred. Engineers often assume that this reactor is so well mixed that the concentration is the same everywhere. The mole balance on this reactor is Fout – Fin = rV
13
where V is the volume of the reactor. The molar flow rates are related to the volumetric flow rates, Q, as follows: F = Qc
14
When the volumetric flow rates are constant, Eq. (13) becomes Q(cout – cin ) = rV, continoud stirred tank
15
In a well-mixed reactor (CSTR), the concentration going out of the reactor is the same as the concentration in the reactor. Since the rate expression is a function of c, it is a function of cout. Then Eq. (15) becomes an algebraic equation in one variable (here) for cout. 4. Resources: Matlab 5. Procedure: 1. To solve the single differential equation: dy/dt = -10y , y(0) = 1 from t = 0 to t = 1. a. Construct a function file that contains: % filename rhs.m function yprime=rhs(t,y) % compute the rhs of eqns.for any % input values t, y yprime=-10*y; b.
plot(t,y)
Construct a script file that contains: % filename yourSurname_le10_p1b.m y0=1; % set the initial condition to 1 tspan=[0 1]; % set the initial time to zero and the final % time to 1 % call the routine ode45; use the m-file called % rhs, % start with y=y0 at tspan(1) % and integrate until tspan(2) [t,y]=ode45(@rhs,tspan,y0) % plot the solution
c. Run the program. Show the results. 2. Suppose you take a reactor in which components A and C are fed in equimolar amounts, and the following reaction takes place: 2A B 1 You assume that the reaction takes place in the liquid phase and that the volumetric flow
rate remains constant even when reaction occurs. The equations are Eq. (5) for each species. u dCj/dz = rj , j = 1…3 2 where Cj is the molar concentration of the jth species, u is the velocity, z is the distance down the tube from the inlet, and rj is the rate of reaction of the jth species, in mol/vol time. Here the rate of reaction is taken as second order: rate of formation of B = kCA2 3 where the units of k are vol/mol time. The equations for all three species are then 4
At the inlet you take CA(0) = 2 kmol/m3, CB(0) = 0, CC(0) = 2 kmol/m3 5 and we take u = 0.5 m/s, k = 0.3 m 3 /kmol s, and the total reactor length as z = 2.4 m. a. Create a function file that contains: rate1.m % % This function gives the right-hand side for a simple % reactor problem which is isothermal. function yprime=rate1(VR,y) % y(1) is CA, y(2) is CB, y(3) is CC % k=0.3 and u=0.5 CA=y(1); rate=0.3*CA*CA; yprime(1)=-2.*rate/0.5; yprime(2) =+rate/0.5; yprime(3)=0.; yprime=yprime’; b. Create an m-file that contains: % yourSurname_le10_p2b,m % This is the driver program to solve the simple flow % reactor example. % set the initial conditions y0=[2 0 2] % set the total volume of the reactor zspan=[0 2.4] % call the ode solver [z y]=ode45(@rate1, zspan , y0) % plot the result plot(z,y)
xlabel(‘length (m)’) ylabel(‘concentrations (kgmol/m^3)’) legend(‘A’, ‘B’, ‘C’) c. Run the program. Show the results. 3. Most reactions have heat effects, and this means that reactor models must be able to model the heat transfer as well as the mass balances. The following example models a simple reactor oxidizing SO2 to form SO3 . After some manipulation, the equations are 1 where the reaction rate is
with the parameters: Tsurr = 673.2, T(0) = 673.2, X(0) = 1. The variable X is the concentration of SO2 divided by the inlet concentration, 1 - X is the fractional conversion, and T is the temperature in K. The first equation is the mole balance on SO2, and the second is the energy balance. The first term on the right-hand side of Eq. (1) represents cooling at the wall; the second term there is the heat of reaction. These equations are appropriate when radial and axial dispersion are not important. Axial and radial dispersion are actually important in this case, as discussed by Young and Finlayson (1973), but the simpler case is a good vehicle for seeing how to include temperature effects in the problem. The reaction rate equation shows that equilibrium can be reached (then R’ = 0), and the equilibrium constant depends on the temperature. Thus, the conversion depends intimately on the temperature. a. Create a function file that contains: % rateSO2.m for SO2 reaction function yprime=rateSO2(z,y) % X is concentration SO2 divided by the inlet concentration X=y(1); % T is temperature in degrees K T=y(2); k1=exp(-14.96+11070/T); k2=exp(-1.331+2331/T); Keq=exp(-11.02+11570/T); term1=X*sqrt(1-0.167*(1-X)); term2=2.2*(1-X)/Keq; denom=(k1+k2*(1-X))^2; rate=(term1-term2)/denom; yprime(1)=-50*rate; yprime(2)=-4.1*(T-673.2)+1.02e4*rate;
yprime=yprime’; b. Create an m-file that contains: % yourSurname_le10_p3b.m % set the dimensionless initial conditions y0=[1 673.2] % set the integration range zspan=[0 1] % call the solver [z y]=ode45(@rateSO2,zspan,y0) % plot the result plot(z,y(:,1),‘r-’) xlabel(‘dimensionless axial position’) legend(‘fraction converted’) pause plot(z,y(:,2),‘g-’) xlabel(‘dimensionless axial position’) legend(‘temperature (K)’) c. Run the program. Show the results. 4. The situations in Procedures 2 and 3 are for liquid streams or dilute gaseous streams where the density does not change drastically. Sometimes the density does change, though, and such problems are not much harder to solve numerically than their constant-density counterparts. The governing equation is Eq. (2), here expressed as a differential equation:
where Fj is the molar flow rate of the jth species, VR is the volume of reactor (from the inlet), and rj is the rate of reaction of the jth species. Let us take the same reaction, Procedure 2 Equation 1 and reaction rate, Procedure 2 Equation 3. Putting these together gives the differential equations to be solved:
The equations are differential equations for the molar flow rates, FA, FB, and FC, but the rates of reaction are expressed as concentration, CA. It is possible to derive the concentration of each chemical from the molar flow rates. In a gas phase, the total concentration is governed by the perfect gas law, although more complicated equations of state are possible:
Thus, at any point in the reactor, if you know the pressure and temperature, you can find CT. The pressure is determined by fluid mechanics and the pressure drop, and the temperature is determined by the energy equation. Then the concentration of a chemical is the mole fraction times the total concentration; for example
And the mole fraction is
If the temperature and pressure are constant, the total concentration is the total concentration at the inlet, which is taken here to be 5 kmol/m3. Thus, the flow rate is
The problem is summarized here as
Write a function file that calculate the right-hand side, given the input, which are the flow rates of all species. The variables are: y(1)=FA , y(2)=FB , y(3)=FC The function will have available to it the accumulated reactor volume VR, and the local value of y(j), j=1,..., 3 at the same reactor volume. To calculate the concentrations at that reactor volume use:
The rates of reaction are then evaluated, and the function returns the numerical value of the right-hand side. Solve this procedure for a total volume of 1.2 m3 , k=0.3 m3/kmol s, a total concentration of 5 kmol/m3. 5. When a reaction occurs on a catalyst particle, it is necessary for the reactants to get to the catalyst. Sometimes this transfer is hindered by mass transfer, and then the concentration near the catalyst is not the same as it is in the bulk stream. This phenomena is demonstrated with Procedure 2 posed in Eqs. (4) – (5), except that now the rate of reaction is evaluated at a concentration that is near the catalyst. The concentration Cs is the concentration on the surface, expressed as the kmol of a species per volume of the catalyst. There are various ways to represent this concentration. The net effect, however, is that Procedure 2 Eq. (4) is changed to :
5.1 You will be solving for CA(z), but you cannot evaluate the rate of reaction in Eq. (5.1) because you do not knowCA,s. You need a mass balance relating the rate of mass transfer to the catalyst to the rate of reaction. One form of that is 5.2
where km is a mass transfer coefficient in units of m/s, determined from correlations derived in fluid mechanics and mass transfer courses,a is the surface area exposed per volume of the reactor (m2/m3), and ks is a rate of reaction rate constant (here m3/kmol s). Looking closely at Eqs. (5.1) –(5.2), one can see that, in order to solve the differential equations in Eq. (5.1), you must solve Eq. (5.2) at every position z. Thus, this is a problem that combines ordinary differential equations with nonlinear algebraic equations.
a. The function file that will evaluate Eq. 5. 2 contains: function y=mass_rxn(CAs,CA) global k km % mass_rxn.m % this m-file evaluates the mass transfer equations % km in the program includes the a % the first entry is the guess of CAs and % the second entry is the fluid concentration y=km*(CA-CAs)-k*CAs*CAs; b. To evaluate the right-hand side of Eq. 5.1 , the m-file contains the following: % yourSurname_le10_p5b.m % This function gives the right-hand side for a % reactor problem which is isothermal but has % mass transfer effects function yprime=rate1_mass(VR,y) global k vel km % y(1) is CA, y(2) is CB, y(3) is CC CA=y(1); % solve the mass transfer problem % the first entry is the guess of CAs and % the second entry is the fluid concentration OPTIONS=[]
CAguess=CA; CAs=fzero(@mass_rxn,CAguess,OPTIONS,CA); rate=k*CAs*CAs; yprime(1)=-2.*rate/vel; yprime(2)=+rate/vel; yprime(3)=0.; yprime=yprime’; c. Using the value of km = 0.2, the m-file that will solve the problem contains: % yourSurname_le10_p5c.m global k vel km k=0.3; vel=0.5; km=0.2; % set the initial conditions y0=[2 0 2] % set the total volume of the reactor zspan=[0 2.4] % call the ode solver [z y]=ode45(@rate1_mass,zspan,y0) % plot the result plot(z,y(:,1),‘*-’,z,y(:,2),‘+-’,z,y(:,3),‘x-’) % or plot(z,y) xlabel(‘length (m)’) ylabel(‘concentrations (kgmol/m^3)’) legend(‘A’, ‘B’, ‘C’) d. Run the m-file. Show the results. 6. This procedure considers a CSTR in which a first-order reaction occurs, but the temperature also changes owing to the heat of reaction. The equations to be solved are: 6.1
The left-hand sides are the flow rate times a concentration or temperature difference between the input and output, divided by the volume. The equations have been normalized by the inlet concentration and temperature. The right-hand sides are the rate of reaction and the rate of energy generation due to reaction, respectively. The case described by Eq. (6.1) is for an adiabatic reactor. When the reactor is adiabatic, the equations can be combined by multiplying the first equation by β and adding it to the second equation; then the right-hand side vanishes. 6.2
This equation can be solved for T: 6.3
Now the mass balance can be considered a single equation in one unknown (c) 6.4
a. Construct a function file representing Eq. 6.4 that contains the following: % rate_T function fn=rate_T(c) global beta gamma flowvol T=1+beta*(1-c) rate=c*exp(gamma*(1-1/T)) fn=flowvol*(1-c)-rate b. To solve the problem, type the following commands into the command window: >> fzero(@rate_T,0.5) c. Show the results 7. Reactors do not always run at steady state. In fact, many pharmaceuticals are made in a batch mode. Such problems are easily solved using the same techniques presented above because the plug flow reactor equations are identical to the batch reactor equations.Even CSTRs can be run in a transient mode, and it may be necessary to model a time dependent CSTR to study the stability of steady solutions. When there is more than one solution, one or more of them will be unstable. Thus, this section considers a time dependent CSTR as described by Eq. (7.1): 7.1
The variables are V= reactor volume, Q = volumetric flow rate C = concentation, t = time k0 = reaction rate constant,E = activation energy Φ = void fraction, ρ = density, = heat capacity per mass subscript f for fluid, s for solid - ΔHrxn = heat of reaction The nondimensional form of these equations is
7.2
The parameter Le is a Lewis number, and it includes the heat capacity of the system. The Da is a Damkohler number and includes the rate of reaction. The parameters are taken as 7.3
a. A function file that computes the right hand side contains : % rate_limit.m % solve for particle with limit cycle function yprime=rate_limit(t,y) global beta gamma Damk Lewis Tin c=y(1); T=y(2); rate=c*Damk*exp(gamma*(1-1/T)); yprime(1)=1 - c- rate; yprime(2)=Tin -T+ beta*rate; yprime(2)=yprime(2)/Lewis; ydot=ydot’; b. An m-file contains the following: % yourSurname_le10_p7b.m global beta gamma Damk Lewis Tin beta=0.15 gamma=30 Lewis=1080 Damk=0.115 Tin=1 [t y]=ode45(‘rate_limit’,[0 2],[0.7 1.0]) plot(t,y(:,1),‘*-k’,t,y(:,2),‘o-k’) xlabel(‘time’) legend(‘concentration’,‘temperature’) c. Run the m-file. Show the results. 8. Solve for the concentration distribution in a plug flow reactor governed by the following equations:
8.1
8.2
9. Solve for the concentration distribution in a plug flow reactor with the following reaction rate expression and parameters: 9.1
10. P.142
Course: CHE 508 Group No.: Group : AQUINO, AIRAH DS.
Laboratory Exercise No.: 6 Section: CH51FC2 Date Performed: 1/18/19 Date Submitted: 1/18/19 Instructor: ENGR. CRISPULO MARANAN
6. Data and Results: Procedure 1c
Results >> Aquino_le10_p1b t= 0 0.0050 0.0100 0.0151 0.0201 0.0406 0.0612 0.0817 0.1023 ... 0.7901 0.8098 0.8311 0.8524 0.8737 0.8950 0.9213 0.9475 0.9738 1.0000 y= 1.0000 0.9510 0.9044 0.8601 0.8180 0.6659 0.5421
0.4416 0.3598 0.2957 ... 0.0004 0.0003 0.0002 0.0002 0.0002 0.0001 0.0001 0.0001 0.0001 0.0000
2c
>> Aquino_le10_p2b y0 = 2
0
0
2.4000
zspan =
2
z= 0 0.0000 0.0000 0.0001 0.0001 0.0002 0.0003 0.0004 0.0005 0.0010 … 2.0641 2.1241 2.1841 2.2441 2.3041 2.3641 2.3731 2.3820 2.3910 2.4000 y= 2.0000 1.9999 1.9998 1.9997 1.9996 1.9991 1.9986 1.9981 1.9976 1.9951 0.3359 0.3280 0.3204 0.3132 0.3063 0.2997 0.2987 0.2978 0.2968 0.2959
0 0.0001 0.0001 0.0002 0.0002 0.0005 0.0007 0.0010 0.0012 0.0025 .... 0.8320 0.8360 0.8398 0.8434 0.8468 0.8502 0.8506 0.8511 0.8516 0.8521
2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000
3c
>> Aquino_le10_p3b y0 = 1.0000 673.2000 zspan = 0 z= 0 0.0195 0.0391 0.0586 0.0782 0.1032 0.1282 0.1532 0.1782 0.2032 …
1
0.8532 0.8782 0.9032 0.9282 0.9532 0.9782 0.9836 0.9891 0.9945 1.0000 y= 1.0000 0.9431 0.8763 0.8056 0.7377 0.6628 0.6012 0.5512 0.5102 0.4759 … 0.2278 0.2247 0.2217 0.2188 0.2161 0.2134 0.2128 0.2122 0.2117 0.2111
673.2000 684.3790 696.6215 708.6819 719.2483 729.2531 735.7154 739.3146 740.8091 740.8627 688.7606 687.8445 686.9950 686.2074 685.4774 684.8008 684.6598 684.5211 684.3846 684.2505
5
>> Aquino_le10_p5c y0 = 2
0
0
2.4000
zspan =
z= 0 0.0600 0.1200 0.1800 0.2400 0.3000 0.3600 0.4200 0.4800 0.5400 …
2
1.8600 1.9200 1.9800 2.0400 2.1000 2.1600 2.2200 2.2800 2.3400 2.4000 y= 2.0000 2.0238 2.0470 2.0694 2.0911 2.1117 2.1313 2.1495 2.1664 2.1817 … 1.8000 1.7343 1.6628 1.5854 1.5020 1.4123 1.3160 1.2131 1.1032 0.9862
0 2.0000 -0.0004 2.0238 -0.0016 2.0470 -0.0039 2.0694 -0.00732.0911 -0.01192.1117 -0.01802.1313 -0.02572.1495 -0.03512.1664 -0.04642.1817 -1.1012 1.8000 -1.20201.7343 -1.30891.6628 -1.42221.5854 -1.54191.5020 -1.66841.4123 -1.80181.3160 -1.94241.2131 -2.09041.1032 -2.24600.9862
6
Function File: function fn=rate_T(c) global beta gamma flowvol T=1+beta*(1-c) rate=c*exp(gamma*(1-1/T)) fn=flowvol*(1-c)-rate yprime(1)=rate/(1-c) yprime=yprime'; >> fzero(@rate_T,0.5) T= 1.0750 rate = 4.0547 fn = []
yprime = 7
8.1094 >> Aquino_le10_p7b t= 0 0.0500 0.1000 0.1500 0.2000 0.2500 0.3000 0.3500 0.4000 0.4500 0.5000 … 1.6000 1.6500 1.7000 1.7500 1.8000 1.8500 1.9000 1.9500 2.0000 y= 0.7000 0.7107 0.7208 0.7303 0.7393 0.7479 0.7560 0.7636 0.7708 0.7777 …. 0.8638 0.8656
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.8673 0.8689 0.8704 0.8718 0.8732 0.8744 0.8757
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
8 9
7. Conclusion: I therefore conclude that MATLAB can develop equations and solve problems involving plug flow, batch and continuous stirred tank chemical reactors.
8. Assessment (Rubric for Laboratory Performance): TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES RUBRIC FOR MODERN TOOL USAGE (Engineering Programs) Student Outcome (e): Use the techniques, skills, and modern engineering tools necessary for engineering practice in complex engineering activities. Program: Chemical Engineering Course: CHE 508 Section: CH51FC2 2nd Sem SY 2018-’19 Performance Unsatisfactor Developing Satisfactory Very Satisfactory Score Indicators y 2 3 4 1 1. Apply Fails to identify Identifies Identifies modern Recognizes the appropriate any modern modern techniques and is benefits and techniques, techniques to techniques but able to apply these constraints of skills, and perform fails to apply in performing modern engineering modern tools to discipline-spec these in discipline-specific tools and shows perform a ific engineering performing engineering task. intention to apply discipline-specif task. discipline-specifi them for engineering ic engineering c engineering practice. task. task. 2. Demonstrate Fails to apply Attempts to Shows ability to Shows ability to skills in any modern apply modern apply fundamental apply the most applying tools to solve tools but has procedures in using appropriate and different engineering difficulties to modern tools when effective modern techniques and problems. solve solving engineering tools to solve modern tools to engineering problems. engineering solve problems. problems. engineering problems. 3. Recognize the Does not Recognizes Recognizes the Recognizes the benefits and recognize the some benefits benefits and need for benefits constraints of benefits and and constraints constraints of and constraints of modern constraints of of modern modern engineering modern engineering engineering modern engineering tools and shows tools and makes tools. engineering tools. intention to apply good use of them for tools. them for engineering practice. engineering practice. Total Score Mean Score = (Total Score / 3) Percentage Rating = (Total Score / 12) x 100%
Evaluated by: ENGR. CRISPULO MARANAN Printed Name and Signature of Faculty Member
JANUARY 2019 Date
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