Process control formulas

Measurement, dynamics and process control. 30 formulas for GATE BT Fundamentals of Biological Engineering, each with when to use it. Each is explained in the topic notes.

All 30 Process control formulas

PH ELECTRODE (NERNST)

E = E^{0} - \frac{2.303 R T}{F} pH, \frac{2.303 R T}{F} = 59.16 mV per pH at 298 K

$E^{0}$ is a constant of the electrode pair, fixed by calibration with buffers. $R = 8.314$ J mol $^{- 1}$ K $^{- 1}$ , $F = 96485$ C mol $^{- 1}$ , $T$ in K. The slope is proportional to absolute temperature.

ORIFICE OR VENTURI FLOW METER

Q = C_{d} A_{0} \frac{2 Δ P}{ρ ( 1 - β ^{4} )}, β = \frac{d _{0}}{D}

SI units: $Q$ in m $^{3}$ s $^{- 1}$ , $Δ P$ in Pa, $ρ$ (fluid density) in kg m $^{- 3}$ . $A_{0}$ and $d_{0}$ are the orifice (or venturi throat) area and diameter, $D$ is the pipe diameter and $C_{d}$ the discharge coefficient. Because $Q \propto Δ P$ , a 10 : 1 flow range needs a 100 : 1 pressure-drop range.

VALVE FLOW EQUATION

q = C_{v} f (x) \frac{Δ P _{v}}{s.g.}

US customary units: $q$ in US gal/min, $Δ P_{v}$ (pressure drop across the valve) in psi. $C_{v}$ is the valve coefficient (the flow of water in US gal/min through the fully open valve at a 1 psi drop), s.g. the specific gravity of the liquid, and $f (x)$ the inherent characteristic.

INHERENT VALVE CHARACTERISTICS

Linear: Equal percentage: Quick opening: f = x f = R^{x - 1}, \frac{1}{f} \frac{df}{d x} = ln R f = x

At constant valve pressure drop; the quick-opening form is one common idealisation. For an equal-percentage valve, equal increments of stem travel give equal fractional changes in flow. The ideal equal-percentage curve gives $f = 1/ R$ at $x = 0$ , the minimum controllable flow; a real valve departs from the curve near closure and shuts off.

LAPLACE TRANSFORMS USED IN PROCESS DYNAMICS

L {M u (t)} L {e^{- a t}} L {\frac{d y}{d t}} L {f (t - θ) u (t - θ)} = \frac{M}{s}, L {a t} = \frac{a}{s ^{2}}, L {A δ (t)} = A = \frac{1}{s + a}, L {sin ω t} = \frac{ω}{s ^{2} + ω ^{2}} = s Y (s) - y (0), L {\int_{0}^{t} y d t} = \frac{Y ( s )}{s} = e^{- θ s} F (s)

One-sided transforms of functions defined for $t \geq 0$ . $u (t)$ is the unit step and $δ (t)$ the unit impulse. $e^{- θ s}$ is a pure dead time (transport lag) of $θ$ .

FINAL AND INITIAL VALUE THEOREMS

t \to \infty lim y (t) = s \to 0 lim s Y (s), t \to 0^{+} lim y (t) = s \to \infty lim s Y (s)

The final value theorem holds only if $y (t)$ settles, i.e. every pole of $s Y (s)$ has a negative real part. Applied to an unstable or sustained-oscillation response, it returns a meaningless number.

BLOCK-DIAGRAM ALGEBRA

Series: G_{1} G_{2}, Parallel: G_{1} \pm G_{2}, Negative feedback: \frac{G}{1 + G H}, Positive feedback: \frac{G}{1 - G H}

$G$ is the forward-path and $H$ the feedback-path transfer function.

CLOSED-LOOP RESPONSE OF A FEEDBACK LOOP

Y (s) = \frac{G _{c} G _{v} G _{p}}{1 + G _{c} G _{v} G _{p} G _{m}} Y_{s p} (s) + \frac{G _{d}}{1 + G _{c} G _{v} G _{p} G _{m}} D (s)

$G_{c}$ = controller, $G_{v}$ = valve, $G_{p}$ = process, $G_{m}$ = measuring element, $G_{d}$ = disturbance path. $G_{O L} = G_{c} G_{v} G_{p} G_{m}$ is the open-loop transfer function. The equation assumes the set point and the measurement are in the same units. If the set point is entered in process units while $K_{m} \neq = 1$ , the servo numerator carries an extra factor $K_{m}$ .

FIRST-ORDER TRANSFER FUNCTION

G (s) = \frac{Y ( s )}{U ( s )} = \frac{K}{τ s + 1}

$K$ = steady-state gain; $τ$ = time constant (in time units). The pole is at $s = - 1/ τ$ .

LINEARISED OUTFLOW RESISTANCE

q_{o} = C h \Rightarrow R = (\frac{d q _{o}}{d h})_{\overset{ˉ}{h}}^{- 1} = \frac{2 h ˉ}{C} = \frac{2 h ˉ}{q ˉ _{o}}

Taylor linearisation about the steady state $\overset{ˉ}{h}$ . Valid only for small deviations. Then $τ = A R$ .

STEP RESPONSE (INPUT STEP OF SIZE M)

y (t) = K M (1 - e^{- t / τ}), t_{f} = τ ln \frac{1}{1 - f}

$y$ reaches 63.2% of its final change $K M$ at $t = τ$ , 86.5% at $2 τ$ , 95.0% at $3 τ$ , 98.2% at $4 τ$ and 99.3% at $5 τ$ . The initial slope is $K M / τ$ , so the tangent at $t = 0$ meets the final value at $t = τ$ . $t_{f}$ is the time to reach a fraction $f$ of the final change (natural log).

IMPULSE RESPONSE (IMPULSE OF AREA A)

y (t) = \frac{K A}{τ} e^{- t / τ}

The output jumps to $K A / τ$ at $t = 0^{+}$ and then decays back to the old steady state.

RAMP RESPONSE (INPUT U = A T)

y (t) = K a (t - τ + τ e^{- t / τ}) \to K a (t - τ) for t ≫ τ

After the transient dies out, the output follows the ramp but lags it by a time $τ$ . For a sensor ( $K = 1$ ) the reading trails the true value by $a τ$ .

SINUSOIDAL (FREQUENCY) RESPONSE

u = A sin ω t \Rightarrow y = \hat{A} sin (ω t + ϕ), \frac{A ^}{A} = \frac{K}{1 + ω ^{2} τ ^{2}}, ϕ = - tan^{- 1} (ω τ)

Ultimate periodic response, after the transient has decayed. $ω$ in rad per unit time. This amplitude ratio includes the gain $K$ ; dividing by $K$ gives the normalised amplitude ratio. The lag is 45° at $ω τ = 1$ and approaches, but never reaches, 90°.

FIRST ORDER PLUS DEAD TIME (FOPDT)

G (s) = \frac{K e ^{- θ s}}{τ s + 1}

The standard model fitted to a process reaction curve. Dead time adds a phase lag of $ω θ$ rad (a phase angle of $- ω θ$ , with no upper limit) and leaves the amplitude ratio unchanged.

INTEGRATING (PURE CAPACITIVE) PROCESS

G (s) = \frac{H ( s )}{Q _{i} ( s )} = \frac{1}{A s}

A tank whose outflow is fixed by a pump: a step in inflow gives a ramp in level that never settles. The process is not self-regulating, and the final value theorem does not apply to it ( $s Y (s)$ has a pole at $s = 0$ ).

TWO TANKS IN SERIES

Non-interacting: Interacting: \frac{H _{2} ( s )}{Q _{i} ( s )} = \frac{R _{2}}{( τ _{1} s + 1 ) ( τ _{2} s + 1 )} \frac{H _{2} ( s )}{Q _{i} ( s )} = \frac{R _{2}}{τ _{1} τ _{2} s ^{2} + ( τ _{1} + τ _{2} + A _{1} R _{2} ) s + 1}

Linear outflow resistances, $τ_{1} = A_{1} R_{1}$ , $τ_{2} = A_{2} R_{2}$ . In the non-interacting case tank 1 drains freely into tank 2. In the interacting case the flow between the tanks is $(h_{1} - h_{2}) / R_{1}$ . Both are overdamped, and interaction adds the term $A_{1} R_{2}$ , which makes the response more sluggish. The steady-state gain is $R_{2}$ in both.

STANDARD SECOND-ORDER TRANSFER FUNCTION

G (s) = \frac{K}{τ ^{2} s ^{2} + 2 ζ τ s + 1} = \frac{K ω _{n}^{2}}{s ^{2} + 2 ζ ω _{n} s + ω _{n}^{2}}, ω_{n} = \frac{1}{τ}

$K$ = gain, $ζ$ = damping ratio, $ω_{n}$ = natural frequency (rad per unit time), and $τ = 1/ ω_{n}$ , which is conventionally called the natural period of oscillation (the undamped oscillation actually repeats every $2 π τ$ ). Put the constant term of the denominator equal to 1 before reading off $K$ , $τ$ and $ζ$ .

UNDERDAMPED STEP-RESPONSE CHARACTERISTICS

Overshoot Decay ratio Period T Time of first peak t_{p} Rise time t_{r} = \frac{A}{B} = exp (\frac{- π ζ}{1 - ζ ^{2}}) = \frac{C}{A} = (overshoot)^{2} = exp (\frac{- 2 π ζ}{1 - ζ ^{2}}) = \frac{2 π τ}{1 - ζ ^{2}} = \frac{2 π}{ω _{d}}, ω_{d} = \frac{1 - ζ ^{2}}{τ} = \frac{π τ}{1 - ζ ^{2}} = \frac{τ ( π - ϕ )}{1 - ζ ^{2}}

$B$ = final change ( $K M$ ), $A$ = height of the first peak above the final value, $C$ = height of the second peak above it. The rise time here is the first time the response reaches its final value, with $ϕ = cos^{- 1} ζ$ in radians. Overshoot and decay ratio depend only on $ζ$ , while the period and peak times also need $τ$ . A quarter decay ratio corresponds to $ζ \approx 0.215$ and an overshoot of 0.5.

SETTLING TIME (APPROXIMATE)

t_{s} \approx \frac{3 τ}{ζ} (\pm 5%), t_{s} \approx \frac{4 τ}{ζ} (\pm 2%)

From the decay envelope $e^{- ζ t / τ}$ . An estimate for the underdamped case, not an exact result.

SECOND-ORDER FREQUENCY RESPONSE

\frac{A ^}{A} = \frac{K}{( 1 - ω ^{2} τ ^{2} ) ^{2} + ( 2 ζ ω τ ) ^{2}}, ϕ = - tan^{- 1} \frac{2 ζ ω τ}{1 - ω ^{2} τ ^{2}}

The phase runs from 0 to −180° and is −90° at $ω τ = 1$ (for $ω τ > 1$ , take the angle in the range −90° to −180°). For $ζ < 1/ 2$ the amplitude ratio has a resonant peak at $ω_{r} τ = 1 - 2 ζ^{2}$ of height $K / (2 ζ 1 - ζ^{2})$ .

BODE-PLOT ASYMPTOTES

First order: Second order: Decibels: A R_{N} \to 1 (ω τ ≪ 1), A R_{N} \approx \frac{1}{ω τ} (ω τ ≫ 1), ω_{c} = \frac{1}{τ} A R_{N} \to 1 (ω τ ≪ 1), A R_{N} \approx \frac{1}{( ω τ ) ^{2}} (ω τ ≫ 1) A R_{dB} = 20 lo g_{10} A R

$A R_{N} = \hat{A} / (K A)$ is the normalised amplitude ratio. On log–log axes the high-frequency asymptote has slope −1 for a first-order system (−20 dB per decade) and −2 for a second-order system (−40 dB per decade). The two first-order asymptotes meet at the corner frequency $ω_{c} = 1/ τ$ , where the true value is $1/ 2$ (about −3 dB) and the phase is −45°. For elements in series, amplitude ratios multiply (their dB values add) and phase angles add.

CONTROLLER LAWS (IDEAL, PARALLEL FORM)

P: PI: PID: p (t) = \overset{p}{ˉ} + K_{c} e (t), p (t) = \overset{p}{ˉ} + K_{c} [e + \frac{1}{τ _{I}} \int_{0}^{t} e d t], p (t) = \overset{p}{ˉ} + K_{c} [e + \frac{1}{τ _{I}} \int_{0}^{t} e d t + τ_{D} \frac{d e}{d t}], G_{c} G_{c} G_{c} = K_{c} = K_{c} (1 + \frac{1}{τ _{I} s}) = K_{c} (1 + \frac{1}{τ _{I} s} + τ_{D} s)

$e = y_{s p} - y_{m}$ . $\overset{p}{ˉ}$ = bias (the output when $e = 0$ ), $K_{c}$ = controller gain, $τ_{I}$ = integral (reset) time, $τ_{D}$ = derivative time; $τ_{I}$ and $τ_{D}$ are in time units. $G_{c} = P^{'} (s) / E (s)$ in deviation variables. Some controllers are set in repeats per minute $= 1/ τ_{I}$ (with $τ_{I}$ in minutes).

PROPORTIONAL BAND

P B = \frac{100}{K _{c}} %

For a dimensionless $K_{c}$ (input and output both as % of span). PB is the % change in the measured variable that moves the output through its full range. A narrow band means a high gain.

P CONTROL OF A FIRST-ORDER PROCESS: GAIN, TIME CONSTANT AND OFFSET

\frac{Y}{Y _{s p}} Servo offset = \frac{K _{O L}}{1 + K _{O L}} \cdot \frac{1}{τ ^{'} s + 1}, τ^{'} = \frac{τ}{1 + K _{O L}}, K_{O L} = K_{c} K_{v} K_{p} K_{m} = \frac{M}{1 + K _{O L}}, ∣ Load offset ∣ = \frac{∣ K _{d} ∣ M}{1 + K _{O L}}

For $G_{p} = K_{p} / (τ s + 1)$ with $G_{v} = K_{v}$ and $G_{m} = K_{m}$ pure gains and $K_{O L} > 0$ (a stable loop); the servo expression takes $K_{m} = 1$ . $M$ is the step in set point (servo) or in the disturbance (load, $G_{d} = K_{d} / (τ s + 1)$ ). The closed loop stays first order, is faster by the factor $1 + K_{O L}$ , and has an offset that shrinks, but never vanishes, as $K_{c}$ increases.

P CONTROL OF A SECOND-ORDER PROCESS

τ^{'} = \frac{τ}{1 + K _{O L}}, ζ^{'} = \frac{ζ}{1 + K _{O L}}, servo offset = \frac{M}{1 + K _{O L}}

For $G_{p} = K_{p} / (τ^{2} s^{2} + 2 ζ τ s + 1)$ with pure-gain valve and sensor, $K_{m} = 1$ and $K_{O L} > 0$ . Raising $K_{c}$ makes the closed loop faster and less damped (more overshoot). With $ζ > 0$ and no dead time or extra lags, P control cannot make this loop unstable.

IDEAL AND STATIC FEED-FORWARD CONTROLLERS

G_{f} = - \frac{G _{d}}{G _{t} G _{v} G _{p}}, static: K_{f} = - \frac{K _{d}}{K _{t} K _{v} K _{p}}

$G_{t}$ = transmitter on the disturbance. Perfect compensation needs a perfect model. The feed-forward block is outside the feedback loop, so it does not change the characteristic equation or the stability.

STABILITY OF A CUBIC CHARACTERISTIC EQUATION

a_{0} s^{3} + a_{1} s^{2} + a_{2} s + a_{3} = 0 is stable ⟺ all a_{i} > 0 and a_{1} a_{2} > a_{0} a_{3}; at the limit ω_{u}^{2} = \frac{a _{2}}{a _{0}} = \frac{a _{3}}{a _{1}}

The Routh test worked out for $n = 3$ , for example three first-order lags under P control.

BODE STABILITY CRITERION, GAIN AND PHASE MARGINS

∠ G_{O L} (j ω_{co}) = - 18 0^{\circ} : stable if A R_{O L} (ω_{co}) < 1; K_{u} = \frac{1}{A R _{O L} ( ω _{co} ) _{K_{c} = 1}}, P_{u} = \frac{2 π}{ω _{co}}

Applies when the open loop is itself stable and its phase crosses −180° only once (as for lags plus dead time); the $K_{u}$ expression is for a P controller. $ω_{co}$ = crossover (phase −180°) frequency. Gain margin $GM = 1/ A R_{O L} (ω_{co})$ . Phase margin $P M = 18 0^{\circ} + ϕ$ at the frequency where $A R_{O L} = 1$ . Typical design guidelines are $GM$ of 1.7 to 2 and $P M$ of 30° to 45°. For three identical lags, each contributes −60° at crossover, so $ω_{co} τ = 3$ , each lag has $A R = 1/2$ , and $K_{u} K = 8$ ( $K$ = product of the other gains in the loop).

COHEN–COON SETTINGS (FOPDT K, Τ, Θ)

P: PI: PID: K_{c} = \frac{1}{K} \frac{τ}{θ} (1 + \frac{θ}{3 τ}) K_{c} = \frac{1}{K} \frac{τ}{θ} (0.9 + \frac{θ}{12 τ}), τ_{I} = θ \frac{30 + 3 θ / τ}{9 + 20 θ / τ} K_{c} = \frac{1}{K} \frac{τ}{θ} (\frac{4}{3} + \frac{θ}{4 τ}) τ_{I} = θ \frac{32 + 6 θ / τ}{13 + 8 θ / τ}, τ_{D} = \frac{4 θ}{11 + 2 θ / τ}

Designed, like the Ziegler–Nichols settings, for roughly a quarter decay ratio. Cohen–Coon corrects the Ziegler–Nichols reaction-curve gains with $θ / τ$ terms, so the two methods differ most when dead time is not small compared with $τ$ .

Where this was asked: 2019 (2), 2022 (1), 2023 (1), 2024 (1), 2025 (2), 2026 (2). See the 9 previous-year questions on Process control. This shows what was asked, not what will be asked.

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Process control: the notes and 9 solved previous-year questions