Second-Order-Reaction CSTR — Nonlinear Model, Steady State and Linearised Transfer Function: 20-Bio-A2 Process Dynamics and Control, May 2017, Q7

Problem 7: Second-Order-Reaction CSTR — Nonlinear Model, Steady State and Linearised Transfer Function (20% total)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Isothermal, constant-volume, constant-density CSTR; second-order consumption $r_A=k_1C_A^2$; constant mass flow $F$ (volumetric flow $q=F/\rho$); inlet concentration $C_{A,o}$ is the disturbance.

Find. (a) the dynamic model for $C_A(t)$ and the steady-state $C_{As}$; (b) the linearised transfer function $\delta C_A/\delta C_{A,o}$.

Problem 7: isothermal CSTR with feed disturbance $C_{A,o}$; the second-order reaction $r_A=k_1C_A^2$ is an extra concentration-dependent removal path alongside washout $q$.

Approach. Write an unsteady component mole balance, set the accumulation term to zero and solve the resulting quadratic for the positive physical root $C_{As}$, then linearise the single nonlinearity ($C_A^2$) about that steady state and Laplace-transform the resulting linear ODE in deviation variables.

(a) Component balance. With volumetric flow $q=F/\rho$ (constant, since $\rho$ is constant), a mole balance on species $A$ over the well-mixed reactor gives $$V\frac{dC_A}{dt}=q\,(C_{A,o}-C_A)-Vk_1C_A^2.$$
(a) Steady state. Setting $\dot C_A=0$: $Vk_1C_{As}^2+qC_{As}-qC_{A,o}=0$, a quadratic in $C_{As}$ whose physically meaningful (positive) root is $$\boxed{C_{As}=\frac{-q+\sqrt{q^2+4Vk_1qC_{A,o}}}{2Vk_1}.}$$
(b) Linearise. The only nonlinearity is $r_A=k_1C_A^2$; a first-order Taylor expansion about $C_{As}$ gives $r_A'\approx k_1(2C_{As})\,C_A'$, where the prime denotes a deviation variable. Substituting into the balance and subtracting the steady state, $$V\frac{dC_A'}{dt}=q\,C_{A,o}'-\big(q+2Vk_1C_{As}\big)\,C_A'.$$
(b) Transfer function. Laplace-transforming from rest and rearranging into standard first-order form, $$\boxed{\frac{\delta C_A}{\delta C_{A,o}}=\frac{K}{\tau s+1},\qquad K=\frac{q}{q+2Vk_1C_{As}}<1,\qquad \tau=\frac{V}{q+2Vk_1C_{As}}<\frac{V}{q}.}$$ Both the gain and the time constant are pulled below their no-reaction (pure washout) values $1$ and $V/q$ by the linearised reaction term $2Vk_1C_{As}$: the CSTR is more self-regulating and responds faster than a non-reacting tank of the same size.

Quantity	Expression
(a) Model	$V\dot C_A=q(C_{A,o}-C_A)-Vk_1C_A^2$
(a) Steady state	$C_{As}=\big[-q+\sqrt{q^2+4Vk_1qC_{A,o}}\big]/(2Vk_1)$
(b) Gain $K$	$q/(q+2Vk_1C_{As})<1$
(b) Time constant $\tau$	$V/(q+2Vk_1C_{As})<V/q$

Concept & Theory

Topic. Nonlinear CSTR modelling; steady-state solution of a quadratic; linearisation of a nonlinear reaction rate about an operating point.

Key relations. Component balance $V\dot C_A=q(C_{A,o}-C_A)-Vr_A$; linearised slope $\partial r_A/\partial C_A|_{C_{As}}=2k_1C_{As}$; resulting first-order lag $K/(\tau s+1)$.

Why this works. A non-reacting CSTR is a plain first-order lag of unit gain and residence time $V/q$. Adding a concentration-consuming reaction gives the system a second, state-dependent removal path whose local (linearised) strength is $2k_1C_{As}$; that term adds directly to $q$ in both the gain's denominator and the time constant, so a reaction that is faster at the operating point makes the loop both less sensitive to inlet disturbances and quicker to settle.

Common pitfall. Taking the negative (unphysical) root of the steady-state quadratic, or linearising $C_A^2$ as $C_{As}^2$ instead of using the derivative $2k_1C_{As}C_A'$.

Source. Seborg et al., Process Dynamics and Control, Ch. 2, 4 (dynamic modelling, linearisation of nonlinear ODEs).

Question 7 of 8: Second-Order-Reaction CSTR — Nonlinear Model, Steady State and Linearised Transfer Function

Problem 7: Second-Order-Reaction CSTR — Nonlinear Model, Steady State and Linearised Transfer Function (20% total)