Modelling and Mathematical Methods in Process and Chemical Engineering

Transcription

1 Modelling and Mathematical Methods in Process and Chemical Engineering Solution Series 3 1. Population dynamics: Gendercide The system admits two steady states The Jacobi matrix is ẋ = (1 p)xy k 1 x ẏ = pxy k 2 y S1 : (, ) S2 : J = ( ) k2 p, k 1 1 p (1 p)y k1 (1 p)x py px k 2 (1) ˆ S1: J S1 = k1 k 2 The corresponding eigenvalues and eigenvectors are ( ) 1 ( λ 1 = k 1, z 1 = λ 2 = k 2, z 2 = 1 Thus, the origin is a stable node. ˆ S2: J S2 = (1 p) p k 2 pk 1 1 p ) (2) The corresponding eigenvalues and eigenvectors are λ 1,2 = ± k 1 k 2, z 1,2 = 1 p p k 2 k 1 ±1 (3) Thus, the finite steady state S2 is a saddle point. 1

2 .4.3 y x Figure 1: Trajectories of the system (1) for p =.517, k 1 =.1, and k 2 =.12. Triangles indicate initial conditions. Figure 1 shows a number of trajectories in the x-y-plane. 2. Population dynamics: Symbiosis We consider the two-dimensional nonlinear system dx dy where α 1, α 2, a and b are positive constants. = α 1 x(1 x + ay) = α 2 y(1 y + bx) (a) We determine the stability character of the steady states by linearizing the system around these steady states. The Jacobian matrix of the system is and, provided that δ = 1 ab >, S 1 = (, ), S 2 = (1, ), S 3 = (, 1) S 4 = ( 1 + a 1 ab, 1 + b ) 1 ab (b) Stability character: α J = 1 2α 1 x + α 1 ay α 2 by α 1 ax α 2 2α 2 y + α 2 bx 2

3 Thus, we have J (,) = α 1 α 2 and the eigenvalues of J (,) are its diagonal elements, i.e., α 1 and α 2. Since both are positive, S 1 is an unstable node in the phase plane. For S 2, the Jacobian is α J (1,) = 1 α 1 a α 2 (1 + b) with eigenvalues α 1 < and α 2 b >. Therefore, S 2 is a saddle point. For S 3, we get α J (,1) = 1 (1 + a) α 2 b α 2 Again, since α 2 < < α 1 a, we have the case of two real eigenvalues with opposite sign. Hence, S 3 is also a saddle point. If δ = 1 ab >, S 4 lies in the positive quadrant. Evaluating the eigenvalues of the Jacobian matrix in this particular case shows that the steady state S 4 is a stable node in the phase plane. However, the calculation is rather cumbersome and not necessary to determine the stability character. This stability character can also be deduced from the phase plane analysis methods discussed in the lecture (namely, the nullclines method). (c) When applying the nullclines method to the system considered in this exercise, the two cases 1 ab and 1 ab > have to be distinguished. In the former case, the ẋ- nullclines and the ẏ-nullclines do not intersect in the positive quadrant (i.e. S 4 does not exist), whereas in the latter case, they do (and S 4 exists). The first case ab 1 is shown in Fig. 2 for the arbitrarily chosen values of a = b = 1.5 and α 1 = α 2 = 1. It is easily seen that both populations become unbounded for the majority of initial conditions. 3

4 Figure 2: ab 1: unbounded growth In the second case (ab < 1), the phase plane behaviour illustrated in Fig. 3 corresponds to the arbitrary choice of parameters α 1 = α 2 = 1 and a = b =.5. The majority of trajectories is attracted to the steady state S 4 in the positive quadrant. We also see that the steady state value of each population in S 4 is bigger than its corresponding value in isolation (S 2 and S 3 ). This observation is consistent with the symbiotic nature of the interaction between the two species. 4

5 Figure 3: ab < 1: symbiotic coexistence 3. Non-adiabatic CSTR (i) The mass balance for a general CSTR is given by V dc = Q(C in C) V r(c, T ) (4) while the associated energy balance is V C p ρ dt = QC p ρ(t in T ) UA(T T c ) HV r(c, T ) (5) Rearranging both equations yields and V Q d T T in V Q d C C in = (1 T T in ) Using the following dimensionless parameters = (1 C C in ) V QC in r(c, T ) (6) UA QC p ρ ( T T c ) T in T in HV C in QC p ρt in C in r(c, T ) (7) τ = t Q V γ = C C in ϑ = T T in ψ = V r(c in, T in ) QC in φ(γ, ϑ) = r(c, T ) r(c in, T in ) α = UA C p ρ Q β = HC in C p ρt in 5

6 yields the dimensionless forms of the mass and energy balance dγ dτ = 1 γ ψφ(γ, ϑ) dϑ dτ = 1 ϑ α(ϑ ϑ c) βψφ(γ, ϑ) (ii) The conditions for the steady states are obtained by setting the right-hand sides of the differential equations to zero, i.e., = 1 γ s ψφ(γ s, ϑ s ) = 1 ϑ s α(ϑ s ϑ c ) βψφ(γ s, ϑ s ) The dimensionless concentration γ s can be expressed as function of ϑ s. However, the equation resulting from eliminating γ s from one of the above conditions still needs to be solved numerically. Notice that, since we have γ s = 1 β + αϑ c (1 + α)ϑ s β and γ s 1, we know that ϑ s is bounded as well. (iii) The linearized system is d γ dτ d ϑ dτ = γ ψ(φ γ γ + φ ϑ ϑ) = (1 + α) ϑ βψ(φ γ γ + φ ϑ ϑ) where φ = φ γ γ and φ = φ ϑ ϑ, and γ, ϑ are the perturbation variables. The Jacobian of the system is 1 ψφγ ψφ ϑ J = βψφ γ βψφ ϑ (1 + α) The trace and the determinant of this matrix determine the stability characteristicts of the linearized system. In fact, for asymptotic stability, we require tr(j) = (2 + α) ψ(βφ ϑ + φ γ ) < det(j) = (1 + α) + ψ(βφ ϑ + (1 + α)ψ γ ) > (iv) For an adiabatic reactor, there is no heat exchange with its surroundings, i.e., we have α =. In this case, the conditions for asymptotic stability are tr(j) = 2 ψ(βφ ϑ + φ γ ) < det(j) = 1 + ψ(βφ ϑ + ψ γ ) > 6

7 which can be written as ψβφ ϑ + ψφ γ > 2 ψβφ ϑ + ψφ γ > 1 Therefore, in the adiabatic case, the two conditions reduce to det(j) >. 7