Process Control · Study · FE Chemical · FE → PE Prep
Process Control
4% of exam
Process dynamics (first- and second-order processes, gains and time constants, stability, damping, transfer functions), control strategies (feedback, feedforward, cascade, ratio, PID tuning), and control-loop design and hardware (sensors, control valves, interlocks, and conceptual DCS and PLC programming).
3 concepts
A. Dynamics
Process Dynamics: First- and Second-Order Response
Process gain and time constant, the 63.2% first-order step, second-order damping and overshoot, transfer functions, and what makes a loop stable.
Before you can control a process you must know how it ·moves· — how fast a tank temperature climbs when you open a steam valve, whether a level loop will overshoot and ring, how much dead time a long pipe adds. The FE handbook packs this into two compact models on the Instrumentation, Measurement, and Control pages: the first-order lag K/(τs+1) and the standard second-order form. Almost every Dynamics question is a substitution into one of them, but only if you can read a transfer function, identify K
,
τ
,
ζ
, and
ωn
from it, and translate those into a time response. Points are lost when candidates confuse the gain with the time constant, forget that the step response is measured in ·deviation· from the initial value, or mix up which
ζ
band rings and which crawls. This concept builds the fluency the handbook assumes (see 'Control Systems', First- and Second-Order Control System Models).
Process gain and time constant
Two numbers characterize a self-regulating first-order process. The steady-state (process) gain K is how far the output ultimately moves per unit of sustained input change — K=Δy∞/Δu — and it carries the units of output over input (e.g. ∘C per % valve opening). The time constant τ is how ·fast· it gets there: the time to traverse 63.2% of the total change. A bigger τ means a more sluggish process. Gain sets the destination; time constant sets the pace. Keep them straight — K has process units, τ has units of time only.
K=ΔuΔy∞,τ=time to reach 63.2% of the final change
The first-order step response
A first-order system Y(s)/R(s)=K/(τs+1) driven by a step of magnitude M responds with a single rising exponential. Working in deviation variables (so the pre-step value y0=0, the handbook's convention), the output approaches its new steady state KM asymptotically. The landmark numbers are worth memorizing: at t=τ the response is 63.2% complete, at 2τ it is 86.5%, at 3τ it is 95.0%, at 4τ it is 98.2%, and at 5τ it is 99.3% — for engineering purposes a first-order process has effectively settled after about 5τ. The slope is steepest at t=0, and the initial tangent would reach the final value in exactly one τ.
y(t)=y0e−t/τ+KM(1−e−t/τ)
Transfer functions, poles, and the dc gain
A transfer function is the ratio of Laplace transforms, G(s)=Y(s)/X(s), written as a ratio of polynomials whose denominator roots are the ·poles· and numerator roots the ·zeros·. The poles govern the character of the response: a first-order pole sits at s=−1/τ, and the farther left (more negative) it lies, the faster the mode decays. The steady-state or dc gain falls straight out by the Final Value Theorem — set s→0 — provided every pole has a negative real part. For the first-order form, lims→0K/(τs+1)=K, which is exactly the process gain.
dc gain=s→0limG(s),first-order pole at s=−τ1
Second-order systems: natural frequency and damping
Add a second energy-storage element (a thermowell on a sensor, an underdamped flow loop, a controller closing a loop on a first-order plant) and you get the standard second-order model. Two parameters now set the shape: the undamped natural frequency ωn (how fast it ·would· oscillate with no damping) and the damping ratio ζ (how much that oscillation is suppressed). The damping ratio sorts behavior into three regimes: ζ>1 is overdamped (two real poles, sluggish, no overshoot), ζ=1 is critically damped (fastest non-oscillatory), and 0<ζ<1 is underdamped (complex poles, the response overshoots and rings). The chemical-process literature often rewrites the same model with τ=1/ωn as K/(τ2s2+2ζτs+1).
R(s)Y(s)=s2+2ζωns+ωn2Kωn2
Overshoot, oscillation, and settling
For an underdamped step response the damped frequency is ωd=ωn1−ζ2 — always lower than the natural frequency. The first peak arrives at tp=π/ωd, and the fractional overshoot above the final value depends ·only· on ζ, not on ωn: lower damping means taller overshoot. The handbook's 2% settling time, Ts=4/(ζωn), comes from the decay envelope e−ζωnt. Note the split of roles: ζ alone fixes the overshoot and the shape, while ωn scales the whole response in time.
%OS=100e−ζπ/1−ζ2,tp=ωn1−ζ2π,Ts=ζωn4
Dead time (transport lag)
Real processes rarely respond the instant you move a valve — fluid must travel down a pipe, a sample must reach an analyzer. That pure delay θ is dead time, and it multiplies the transfer function by e−θs. In the time domain the entire first-order curve simply starts θ seconds later. Dead time is the controllability villain: it adds phase lag without adding gain, so a loop with a large θ/τ ratio is much harder to control and must be detuned. Recognizing e−θs as a delay (not a pole) is a frequent exam discriminator.
R(s)Y(s)=τs+1Ke−θs,y(t)=KM(1−e−(t−θ)/τ) for t≥θ
Stability in one sentence
An open-loop process whose transfer function has all poles in the left half plane is stable: every mode e−t/τ or e−ζωnt decays. A pole on the imaginary axis (ζ=0) sustains oscillation; a pole in the right half plane (which never happens for these self-regulating models alone, but can appear once a controller is added) grows without bound. Stability of the ·closed· loop is the subject of the next concept, but the rule is the same currency — left-half-plane poles, bounded response.
Stable⟺Re(pi)<0 for every pole pi
Exam strategy
Read the transfer function first and tag every parameter: the constant term in the denominator must be 1 to read K directly off the numerator (divide through if it is not), then τ is the coefficient of s. For second-order, match against s2+2ζωns+ωn2 — ωn is the square root of the constant term and ζ comes from the s-coefficient =2ζωn. Memorize the percent-complete ladder (63/86.5/95/98.2/99.3 at 1–5τ) so first-order time questions need no calculator. For overshoot, remember it is a function of ζ only. Always work step responses in deviation variables, then add the initial value back at the end. Watch the units of K — they are never just a number unless input and output share units.
Key equations
First-order transfer functionR(s)Y(s)=τs+1K
K = steady-state gain (output units / input units); τ = time constant (s). Single pole at s=−1/τ.
Only the underdamped band overshoots and oscillates; critically damped is fastest with no overshoot.
Damped natural frequencyωd=ωn1−ζ2
Percent overshoot%OS=100exp(1−ζ2−ζπ)
Peak timetp=ωdπ=ωn1−ζ2π
Two-percent settling timeTs=ζωn4
Time for the envelope
DC gain (Final Value Theorem)dc gain=lims→0G(s)
Steady-state gain, valid only when all poles have negative real parts. Equals K for the first-order form.
Worked examples
First-order sensor step response
Problem. A bare thermocouple behaves as a first-order system with time constant τ=8s. It is suddenly plunged from a 25∘C bath into an 80∘C bath. (a) What does it read 12s after immersion? (b) How long until it reads 75∘C?
Solution. Treat the bath change as a step. The total change is Δy∞=KM=80−25=55∘C, with initial value
y(12)=25+55(1−e−12/8)=67.7∘C
Second-order overshoot and settling
Problem. A closed level loop is modeled as second order with K=2, ωn=4rad/s, and ζ=0.3
First-order plus dead time
Problem. A temperature process is identified as FOPDT with gain K=2∘C/%, time constant τ=25s, and dead time θ=5s
Common pitfalls
•Confusing gain and time constant: K sets how far the output moves (with process units), τ sets how fast (units of time only). They are independent.
•Reading K and τ off an un-normalized transfer function. The denominator constant term must be 1 to read τ as the s-coefficient and K off the numerator — divide through first.
•Forgetting deviation variables: the step response formula gives the change from the initial value. Add the pre-step value back to report an absolute temperature/level.
•Treating dead time e−θs as an extra pole. It is a pure delay that shifts the whole curve by θ; it adds phase lag but no extra lag dynamics.
•Thinking overshoot depends on ωn. Percent overshoot is a function of ζ only; ωn scales the time axis (peak and settling time), not the height.
•Using ωn where ωd is required. The actual ringing frequency and peak time use ωd=ωn1−ζ2
•Assuming a process has settled at t=τ. One time constant is only 63.2% complete; full practical settling needs about 5τ (or Ts=4/ζωn
References
NCEES FE Reference Handbook — Instrumentation, Measurement, and Control (Control Systems; First- and Second-Order Control System Models)
Seborg, Edgar, Mellichamp & Doyle — Process Dynamics and Control — first- and second-order response, FOPDT identification
Coughanowr & LeBlanc — Process Systems Analysis and Control — transient response of standard transfer functions
B. Control strategies
Control Strategies, PID and Stability
Feedback, feedforward, cascade and ratio control; P/I/D action and offset; block-diagram algebra; closed-loop stability and qualitative PID tuning.
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C. Control loop design and hardware
Control Loop Hardware: Sensors, Valves and DCS/PLC
Temperature/pressure/level/flow/pH sensors and transmitters, control-valve sizing and characteristics, final control elements, interlocks, and DCS/PLC roles.
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Response to a step of magnitude M. In deviation variables y0=0. Reaches 63.2% of KM at t=τ.
.
θ = dead time (transport lag, s). The step response is delayed by θ before the exponential begins.
ωn = undamped natural frequency (rad/s); ζ = damping ratio (dimensionless); K = steady-state gain.
Equivalent form with τ=1/ωn. Common in the chemical-process literature.
Actual ringing frequency of an underdamped response (rad/s); always less than ωn.
Overshoot above the final value for a unit step; depends on ζ only, not ωn.
Time to the first (largest) peak of an underdamped step response (s).
. Find the damped frequency, the percent overshoot, the peak time, and the
2%
settling time for a step change.
Solution. Damped frequency: ωd=ωn1−ζ2=41−0.09=4(0.9539)=3.82rad/s.
Overshoot: %OS=100e−ζπ/1−ζ2=100e−(0.3)(3.1416)/0.9539=100e−0.988=37.2%.
Peak time: tp=π/ωd=3.1416/3.8158=0.823s.
Settling time: Ts=4/(ζωn)=4/(0.3×4)=3.33s.
**Answers: ωd=3.82rad/s, %OS=37.2%, tp=0.823s, Ts=3.33s.** Sanity: ζ=0.3 is lightly damped, so a large overshoot near 35–40% and several rings before settling (≈ Ts/tp≈4 oscillation half-cycles, i.e. about 2 overshoot peaks) is exactly what we expect.
%OS=100e−(0.3)π/1−0.32=37.2%
. Starting from a steady output of
50∘C
, the valve is stepped open by
Δu=10%
. (a) What is the output
20s
after the step? (b) When does the output first reach
64∘C
?
Solution. Final change =KM=(2)(10)=20∘C, so the output ultimately rises from 50 to 70∘C. Nothing happens during the dead time, so use t−θ in the exponential.
(a) At t=20s: t−θ=15s, y=50+20(1−e−15/25)=50+20(1−e−0.6)=50+20(0.4512)=59.0∘C.
(b) Set 64=50+20(1−e−(t−θ)/25): 2014=0.70=1−e−(t−5)/25, so e−(t−5)/25=0.30, giving t−5=25ln(10/3)=25(1.204)=30.1s, thus t=35.1s.
**Answers: (a) 59.0∘C, (b) 35.1s.** Sanity: at 20s only 15s of effective time has elapsed (< 1τ), so we expect just under half the change — about 45% (9.0 of 20∘C), confirmed. Ignoring the dead time would have under-predicted the time to 64∘C by exactly 5s.