Second-Order-Reaction CSTR — Model and Transfer Function: 23-Chem-A6 Process Dynamics and Control, May 2017, Q7

Problem 7: Second-Order-Reaction CSTR — Model and Transfer Function (20%)

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 rate $r_A=k_1C_A^2$; mass flow $F$ (volumetric $q=F/\rho$); inlet $C_{A_o}$.

Find. (a) the dynamic model and steady-state $C_{As}$; (b) the transfer function $\delta C_A/\delta C_{A_o}$.

Problem 7: isothermal CSTR with an inlet disturbance $C_{A_o}$; the second-order reaction $r_A=k_1C_A^2$ provides an extra concentration-dependent removal path in addition to washout $q$.

Approach. Write a component mole balance, set the derivative to zero for $C_{As}$ (positive root of a quadratic), then linearise the non-linear rate about $C_{As}$ and transform to standard first-order form.

(a) Component balance and steady state. $$V\,\frac{dC_A}{dt}=q\,(C_{A_o}-C_A)-Vk_1C_A^{2}.$$ Setting $\dot C_A=0$: $Vk_1C_{As}^2+qC_{As}-qC_{A_o}=0$, whose physically meaningful positive root is $$\boxed{C_{As}=\dfrac{-q+\sqrt{q^{2}+4Vk_1q\,C_{A_o}}}{2Vk_1}.}$$
(b) Linearise the rate. The only non-linearity is $r_A=k_1C_A^2$; about $C_{As}$, $r_A'\approx2k_1C_{As}\,C_A'$. In deviation variables $$V\,\frac{dC_A'}{dt}=q\,C_{A_o}'-(q+2Vk_1C_{As})C_A'.$$
(b) Transfer function. Transforming and rearranging into standard first-order form, $$\boxed{\dfrac{\delta C_A}{\delta C_{A_o}}=\dfrac{K}{\tau s+1},\quad K=\dfrac{q}{q+2Vk_1C_{As}},\quad \tau=\dfrac{V}{q+2Vk_1C_{As}}.}$$ Both the gain ($K<1$) and the time constant ($\tau<V/q$) are pulled below their no-reaction values by the term $2Vk_1C_{As}$, so the reactor is self-regulating and faster than the pure residence time would suggest.

Quantity	Expression
(a) Model	$V\dot C_A=q(C_{A_o}-C_A)-Vk_1C_A^2$
(a) Steady state	$C_{As}=\dfrac{-q+\sqrt{q^2+4Vk_1qC_{A_o}}}{2Vk_1}$
(b) $K$	$q/(q+2Vk_1C_{As})<1$
(b) $\tau$	$V/(q+2Vk_1C_{As})

Concept & Theory

Topic. Non-linear CSTR modelling; linearisation about a steady state; reaction as an effective outflow path.

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

Why this works. An unreacting CSTR is a first-order lag of gain 1 and time constant $V/q$ (the residence time). A consuming reaction adds a second removal mechanism whose linearised slope $2k_1C_{As}$ appears added to $q$ in both the gain denominator and the time constant — giving a gain below 1 (some disturbance is "eaten") and a shorter $\tau$. The stronger the reaction at the operating point, the more self-regulating and faster the reactor.

Common pitfall. Taking the negative quadratic root, or linearising $C_A^2$ as $C_{As}^2$ rather than using the derivative $2k_1C_{As}$.

Source. Seborg et al., Ch. 2, 4 (dynamic models, linearisation); Fogler, Elements of Chemical Reaction Engineering, CSTR balances.

Question 7 of 8: Second-Order-Reaction CSTR — Model and Transfer Function

Problem 7: Second-Order-Reaction CSTR — Model and Transfer Function (20%)