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23-Chem-A6 Process Dynamics and Control · December 2019

Question 1 of 8: Two Non-Interacting Tanks in Series with a Pump

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.

Problem 1: Two Non-Interacting Tanks in Series with a Pump (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. Two equal, non-interacting tanks; the pump sets the inter-tank flow:

QuantitySymbolValue
Cross-section (each tank)$A$$1$
Outlet resistance (tank 2)$R$$2$
Inlet flow (constant)$q_{in}$$10$
Inter-tank flow$q_1$pump (manipulated)
Tank-2 outflow$q_2$$h_2/R$
Initial level, tank 1$h_1(0)$$10$

Find. (a) the ODEs for $h_1(t)$ and $h_2(t)$; (b) the transfer functions relating the two levels to the inlet flow $q_{in}$.

$h_1$ Tank 1 $q_{in}=10$ pump $q_1$ $h_2$ Tank 2 $R$ $q_2=\dfrac{h_2}{R}$
Problem 1: inlet flow $q_{in}$ fills tank 1; a constant-throughput pump transfers $q_1$ to tank 2, whose outflow $q_2=h_2/R$ passes a resistance $R$. Because the pump sets $q_1$ independently of $h_1$, the two tanks are non-interacting and tank 1 acts as a pure integrator.

Approach. Write an unsteady mass balance on each tank, note that the pump makes $q_1$ an independent input (so tank 1 has no self-regulating outflow), then Laplace-transform in deviation variables to read off each transfer function.

  1. (a) Tank 1 balance. Accumulation = in − out, with $A=1$: $$\boxed{\frac{dh_1}{dt}=q_{in}-q_1.}$$ The outflow $q_1$ is fixed by the pump, not by $h_1$, so tank 1 integrates any imbalance between $q_{in}$ and $q_1$. At steady state $q_{1s}=q_{in}=10$.
  2. (a) Tank 2 balance. In flows $q_1$, out flows $q_2=h_2/R$, with $A=1$, $R=2$: $$\boxed{\frac{dh_2}{dt}=q_1-\frac{h_2}{R}=q_1-\frac{h_2}{2}.}$$ Tank 2 is self-regulating (its outflow rises with level), so $h_{2s}=R\,q_{1s}=2\times10=20$.
  3. (b) Transfer function $H_1/Q_{in}$. In deviation variables the tank-1 balance transforms to $sH_1(s)=Q_{in}(s)-Q_1(s)$. With the pump flow held at its set value ($Q_1=0$): $$\boxed{\frac{H_1(s)}{Q_{in}(s)}=\frac{1}{As}=\frac{1}{s}.}$$ A pure integrator — there is no finite time constant because the outflow does not respond to level.
  4. (b) Tank-2 dynamics. Transforming the tank-2 balance, $sH_2=Q_1-H_2/R$, so $H_2\!\left(s+\tfrac1R\right)=Q_1$ and $$\frac{H_2(s)}{Q_1(s)}=\frac{R}{ARs+1}=\frac{2}{2s+1},$$ a first-order lag with gain $R=2$ and time constant $\tau_2=AR=2$.
  5. (b) Transfer function $H_2/Q_{in}$. The inlet $q_{in}$ enters only tank 1; the pump then delivers $q_1$ to tank 2 independently of $q_{in}$. Since $q_{in}$ never appears in the tank-2 balance, $$\boxed{\frac{H_2(s)}{Q_{in}(s)}=0.}$$ The constant-speed pump completely isolates the downstream level from inlet-flow disturbances — the defining behaviour of this pumped, non-interacting pair.
QuantityResult
(a) ODE, tank 1$dh_1/dt=q_{in}-q_1$
(a) ODE, tank 2$dh_2/dt=q_1-h_2/2$
(b) $H_1/Q_{in}$$1/s$ (integrator)
(b) $H_2/Q_1$$2/(2s+1)$, $\tau_2=2$
(b) $H_2/Q_{in}$$0$ (pump isolates tank 2)
Check — reading of “transfer functions between $h$ and $q_{in}$”

The text is ambiguous because $q_1$ is a pump (an independent input), so a genuine $H_2/Q_{in}$ does not exist — it is zero. We therefore report the rigorous pair $\{H_1/Q_{in}=1/s,\ H_2/Q_1=2/(2s+1)\}$ and note $H_2/Q_{in}=0$. If the intended configuration were instead an ordinary resistance outflow $q_1=h_1/R_1$ (no pump), the tanks form the classical non-interacting cascade $H_1/Q_{in}=R_1/(AR_1s+1)$ and $H_2/Q_{in}=R_2/[(AR_1s+1)(AR_2s+1)]$; the figure and wording, however, explicitly specify a pump.

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