23-Chem-A6 Process Dynamics and Control · May 2018
Question 1 of 8: Integrating Water-Storage Tank — Transfer Function and Overflow Time
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: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion, state-space models and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order tank/thermal dynamics, step/ramp 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: Integrating Water-Storage Tank — Transfer Function and Overflow Time (20%)
Given. A tank whose inlet and outlet flows are set independently (the outlet does not depend on level):
Quantity
Symbol
Value
Cross-sectional area
$A$
$100\ \mathrm{ft^2}$
Initial / final inlet flow
$q_{in}$
$5\to6\ \mathrm{ft^3/min}$
Outlet flow (held)
$q_{out}$
$5\ \mathrm{ft^3/min}$
Initial level
$h_0$
$4$ ft
Tank height (overflow)
$h_{top}$
$10$ ft
Find. (1) the transfer function $H(s)/Q_{in}(s)$; (2) the clock time at which the level reaches $10$ ft and overflows.
Problem 1: the outlet flow is fixed independently of level, so the tank is a pure integrator; a net $1\ \mathrm{ft^3/min}$ raises the level steadily from $4$ ft toward the $10$ ft rim.
Approach. Because the outlet is set by hand and does not respond to level, the mass balance has no self-regulating term — the tank integrates the flow imbalance — so the transfer function is a pure integrator and the overflow time is fill volume divided by net inflow.
Unsteady mass balance. With constant density, volume balance on the tank gives $A\dfrac{dh}{dt}=q_{in}-q_{out}$. Since $q_{out}$ is held constant, in deviation variables ($H=h-h_0$, $Q_{in}=q_{in}-5$, $Q_{out}=0$) this is $$A\frac{dH}{dt}=Q_{in}.$$
Transfer function. Laplace-transforming with $H(0)=0$: $A\,sH(s)=Q_{in}(s)$, hence $$\boxed{\frac{H(s)}{Q_{in}(s)}=\frac{1}{A\,s}=\frac{1}{100\,s}.}$$ There is no $\tau s+1$ denominator — an independent outlet removes the level-dependent feedback, leaving a non-self-regulating (integrating) process.
Net inflow after the step. At 1 PM the inlet steps to $6$ while the outlet stays at $5$, so the constant net inflow is $$q_{in}-q_{out}=6-5=1\ \mathrm{ft^3/min}.$$
Level trajectory. Integrating $A\,dh/dt=1$ from $h_0=4$: $h(t)=4+\dfrac{1}{100}\,t$. The level climbs linearly at $0.01\ \mathrm{ft/min}$.
Overflow time. The volume needed to fill from $4$ to $10$ ft is $V=(10-4)\times100=600\ \mathrm{ft^3}$; dividing by the net inflow, $$\boxed{t_{of}=\frac{V}{q_{in}-q_{out}}=\frac{600}{1}=600\ \text{min}=10\ \text{h}.}$$ Starting at 1 PM, the tank overflows $10$ hours later, at 11 PM.
