Maple Solutions to the Chemical Engineering Problem Set

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Reversible, Exothermic, Gas Phase reaction in a Catalytic Reactor The component material balances for a tubular reactor may be written as follows > restart: > CMB := Diff(X,W)=-r[A]/F[A,0]: CMB; The reaction rate is given by > rateeqn:=r[A]=-k*(C[A]^2-C[C]/K[C]): rateeqn; ∂ X = − ∂W r = −k A with the reaction rate coefficient calculated from > keqn:=k=k[ref]*exp(E[A]/R*(1/T[ref]-1/T)): keqn; r A F A, 0 ⎛ ⎞ ⎜ 2 ⎟ ⎜C − ⎟ ⎜ A ⎟ ⎝ ⎠ C C K C ⎛ ⎛ 1 1 ⎞ ⎞ ⎜ E ⎜ − ⎟ ⎟ A ⎟ ⎜ ⎜ T T ⎟ ⎟ ⎜ ⎝ ref ⎠ ⎟ ⎜ ⎟ ⎝ R ⎠ k = k e ref The reaction equilibrium coefficient is given by > Keqn:=K[C]=K[C,ref]*exp(Delta(H[R])/R*(1/T[ref]-1/T)): Keqn; ⎛ ⎛ 1 1 ⎞ ⎜ ΔH ( ) ⎜ − ⎟ R ⎜ ⎜ T T ⎟ ⎜ ⎝ ref ⎠ ⎜ ⎝ R K = K e C C, ref The concentrations of A and C are related to conversion by > CAeqn:=C[A]=C[A,0]*(1-X)/(1+epsilon*X)*y*T[0]/T: CAeqn; C ( 1 − X) yT A, 0 0 C = A ( 1 + εX) T > CCeqn:=C[C]=-epsilon*C[A,0]*X/(1+epsilon*X)*y*T[0]/T: CCeqn; εC XyT A, 0 0 C = − C ( 1 + εX) T The energy balance is > EB := Diff(T,W)=(U[a]*(T[a]-T)+r[A]*Delta(H[R]))/(F[A,0]*C[P,A]): EB; ∂ = ∂W T U ( T − T) + r ΔH ( ) a a A R F C A, 0 P, A The pressure drop is given by > dpeqn:=Diff(y,W)=-alpha*(1+epsilon*X)/2/y*T/T[0]: dpeqn; Page 38 ⎞ ⎟ ⎠
∂ 1 α( 1 + εX) T y = − ∂W 2 yT 0 P where y = is the pressure ratio. P 0 The parameters in the model are > params := {C[P,A]=40,C[P,C]=80,Delta(H[R])=-40000,E[A]=41800,k[ref]=0.5,K[C,ref]=25000,R=8.314,U[a]=0.8,T[a]=500,alpha =0.015,T[ref]=450,epsilon=-0.5}; params := { C = 40 , C = 80 , ΔH ( ) = -40000 , E = 41800 , k = .5 , K = 25000 , R = 8.314 , U = .8 , P, A P, C R A ref C, ref a T = 500 , α = .015 , T = 450 , ε = -.5 } a ref Values at the start of the reactor include > startvalues := {F[A,0]=5.0,P[0]=10,C[A,0]=0.271,T[0]=450,y[0]=1,C[C,0]=0}: startvalues; { F = 5.0 , P = 10 , C = .271 , T = 450 , y = 1 , C = 0 } A, 0 0 A, 0 0 0 C, 0 The initial condition is given as a list of equations. > Start := [C[A]=C[A,0],C[C]=0]: Start; [ W= 0 , X = 0 , T= T , y = y , C = C , C = 0 ] 0 0 A A, 0 C We are going to use the BESIRK code to solve this particular problem. BESIRK is an implementation of a semi-implicit Runge-Kutta method (see Schwalbe et al, 1996) and is much faster than the methods built in to dsolve/numeric that we used in another example. We read the code. > read `c:/maple/numerics/integ/besirk`: BESIRK needs three arguments. The first is a list of the differential (and algebraic equations): > DEqns := [CMB,EB,dpeqn]: DEqns; ⎡ r ⎤ ⎢ ∂ A ∂ ⎥ ⎢ X = − , = , ⎥ ⎢∂W F ∂W ⎥ ⎣ A, 0 ⎦ T U ( T − T) + r ΔH ( ) a a A R ∂ 1 α( 1 + εX) T y = − F C ∂W 2 yT A, 0 P, A 0 The algebraic equations are collected in a list as follows. > AEqns:=[rateeqn,keqn,Keqn,CAeqn,CCeqn]: AEqns; ⎛ ⎞ ⎜ 2 ⎟ r = −k ⎜ C − ⎟ A ⎜ A ⎟ ⎝ ⎠ C ⎡ ⎛ ⎛ 1 1 ⎞ ⎞ ⎛ ⎛ 1 1 ⎞ ⎞ ⎢ ⎜ E ⎜ − ⎟ ⎟ ⎜ ⎟ ⎜ A ⎜ T T ⎟ ⎜ ΔH ( ) ⎜ ⎟ ⎟ ⎢ − ⎟ R ⎟ ⎢ ⎜ ⎝ ref ⎠ ⎜ ⎜ T T ⎟ ⎟ ⎢ ⎟ ⎜ ⎝ ref ⎠ ⎟ ⎢ ⎜ ⎟ ⎜ ⎟ ⎢ C ⎝ R ⎠ ⎝ R ⎠ ⎢ , k = k e , K = K e , ⎢ K ref C C, ref ⎣ C ⎤ ⎥ C ( 1 − X) yT εC XyT ⎥ A, 0 0 A, 0 0 ⎥ C = , C = − ⎥ A ( 1 + εX) T C ( 1 + εX) T ⎥ ⎦ > read `c:/maple/utils/utils.mpl`: > read `c:/maple/thermo/td/tdtools.mpl`: > Subs(params,Start,startvalues,AEqns,right): AIC:=Subs(",",right); AIC := [ r = -.0367205 , k = .5 , K = 25000 , C = .2710000000 , C = 0 ] A C A C Page 39
Page 1 and 2: Maple Solutions to the Chemical Eng
Page 3 and 4: The Van der Waals Equation of State
Page 5 and 6: dimensionless form (in terms of the
Page 7 and 8: lprint(`Reduced pressure is `, Pr);
Page 9 and 10: Steady State Material Balances on a
Page 11 and 12: set of independent equations for th
Page 13 and 14: x = .54 , x = .35 , F x = F x + F x
Page 15 and 16: and a cubic is > p(3); a + a T + a
Page 17 and 18: -40 1000. 100. 10. -20 1. 20 40 60
Page 19 and 20: -40 1000. 100. 10. -20 1. 20 40 60
Page 21 and 22: ( Z, R1 ) = 0 ( D, R3) = 0 ( C, R2
Page 23 and 24: C = .9256703707 + .1402769617 I , e
Page 25 and 26: Terminal Velocity of Falling Partic
Page 27 and 28: Make a list of tank numbers > Units
Page 29 and 30: Diffusion and Reaction in a One Dim
Page 31 and 32: IC := {C[A](0)=0.2,y(0)=y0}; Alleqn
Page 33 and 34: Kvalues := [seq(RAOULT(i),i=compid)
Page 35 and 36: ead `c:/maple/numerics/newton.mpl`:
Page 37: 108 106 104 102 100 98 96 0.4 0.5 0
Page 41 and 42: ∂ A ∂ DAEqns X = − = ∂W F
Page 43 and 44: 0 -0.2 -0.4 -0.6 -0.8 5 W 10 15 20
Page 45 and 46: deproc := proc(n,t,y,dy) local q,qs
Page 47 and 48: params := { V = , , , , , , , } 400

Maple Solutions to the Chemical Engineering Problem Set

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