Process variables and process dynamics

Why it matters

Every control loop in a refinery, power plant or pharma unit exists to hold some process variable where it should be. Before you can choose a controller or tune it, you must know how fast the process reacts, how much it reacts and how long it waits before reacting. Those three numbers — gain, time constant and dead time — decide almost everything else in process control.

Key ideas

Process variables. A process variable is any measurable quantity that describes the state of a process: temperature, pressure, flow, level, composition, pH, density, conductivity. In a control loop the variables play distinct roles:

• Controlled variable (CV or PV) — the quantity we want to hold, e.g. outlet temperature of a heater.
• Manipulated variable (MV) — the quantity the controller adjusts, usually a flow through a control valve (steam flow, cooling-water flow).
• Disturbance variable (DV) — an input we do not control that upsets the CV, e.g. feed temperature or feed flow.
• Set point (SP) — the desired value of the CV. The error is e = SP − PV.

Process dynamics describes how the CV changes with time after a change in MV or DV. Industrial processes are built from two physical ingredients:

• Capacity (storage) — the ability to store mass or energy: tank area for level, heat capacity for temperature, vessel volume for gas pressure.
• Resistance — opposition to flow of mass or energy: a valve or pipe restriction, a thermal resistance.

One capacity with one resistance gives a first-order process. Its time constant is the product τ = R·C — the same idea as an RC circuit.

Steady-state (process) gain K. After the response settles, K = Δ(output)/Δ(input). K has units — for example °C per % valve opening, or m per (m³/s). It tells you how strongly the process responds, not how fast.

Time constant τ. For a first-order process, after a step input the output covers 63.2 % of its total change in one time constant, 86.5 % in 2τ, 95.0 % in 3τ, 98.2 % in 4τ and 99.3 % in 5τ. The initial slope of the response, if continued, would reach the final value in exactly τ. Large τ means a sluggish process.

Dead time θ (transport lag). The interval after an input change during which the output does not move at all. It comes from material travelling through pipes and conveyors (θ = distance/velocity), from analyser sample systems and from many small lags in series. Dead time adds phase lag without reducing gain, so it is the main thing that limits how tightly a loop can be tuned.

First-order-plus-dead-time (FOPDT) model. Most self-regulating plants can be approximated by G(s) = K·e^(−θs)/(τs + 1). The ratio θ/τ (controllability ratio) is a quick difficulty index: below about 0.3 the loop is easy; near or above 1 it is hard and benefits from dead-time compensation or feedforward.

Higher order. Two or more capacities in series that do not interact (each tank drains freely) give a product of first-order terms; interacting capacities give a response that is even slower. Multi-capacity processes show an S-shaped step response with no sharp corner; that "apparent dead time" is what the FOPDT model approximates.

Getting the model from a plant test. With the controller in manual, step the controller output by a few per cent and record the PV (a process reaction curve). Read K from the steady change, θ from the delay, and τ from the 63.2 % point — or use the two-point method below, which is less sensitive to noise.

Formulas

G(s) = K·e^(−θs) / (τs + 1)

• K: steady-state gain (output unit / input unit); τ: time constant (s); θ: dead time (s); s: Laplace variable (1/s). Applies to self-regulating processes approximated by one dominant lag.

y(t) = y₀ + K·Δu·[1 − e^(−(t − θ)/τ)] for t ≥ θ; y(t) = y₀ for t < θ

• y₀: initial steady output; Δu: size of the input step. Response of the FOPDT model to a step.

K = Δy_ss / Δu

• Δy_ss: final (steady-state) change in output; Δu: change in input.

τ = R·C (liquid tank: τ = A·R)

• A: tank cross-sectional area (m²); R: linearised outlet resistance, R = dh/dq (s/m²); τ in s. Steady gain of the level tank to inflow is K = R (m per m³/s).

θ = L / v

• L: transport distance (m); v: velocity of the material (m/s).

τ = 1.5·(t₆₃ − t₂₈), θ = t₆₃ − τ

• t₂₈, t₆₃: times (measured from the input step) at which the output reaches 28.3 % and 63.2 % of its total change. Two-point (Smith) method for fitting FOPDT, from θ + τ/3 = t₂₈ and θ + τ = t₆₃.

τ = (t₂ − t₁) / ln[(y_f − y₁)/(y_f − y₂)]

• y_f: final value; y₁, y₂: readings at times t₁, t₂ after the dead time has passed. Gives τ from any two points on a first-order rise.

Worked examples

Example 1 (standard) — steam heater. With the controller in manual, the steam valve output is stepped from 40 % to 50 %. The outlet temperature, initially 60 °C, starts to rise 8 s after the step and finally settles at 72 °C. It reaches 63.2 % of its total change at t = 50 s. Find K, θ, τ and the temperature at t = 100 s.

1. K = Δy/Δu = (72 − 60)/(50 − 40) = 12/10 = 1.2 °C/%.
2. θ = 8 s (time before any movement).
3. The 63.2 % point occurs at θ + τ, so τ = 50 − 8 = 42 s.
4. y(t) = y₀ + K·Δu·[1 − e^(−(t − θ)/τ)] = 60 + 12·[1 − e^(−(100 − 8)/42)].
5. (100 − 8)/42 = 2.190; e^(−2.190) = 0.1119; 12 × 0.8881 = 10.66 °C.
6. T(100 s) ≈ 70.7 °C, with K = 1.2 °C/%, θ = 8 s, τ = 42 s.

Example 2 (GATE level) — two-point fit and a tank. (a) A reaction curve reaches 28.3 % of its final change at 22 s and 63.2 % at 40 s after the step. Fit an FOPDT model. (b) A free-draining tank has A = 2 m² and a linearised outlet resistance R = 150 s/m². The inflow is stepped up by 0.01 m³/s. Find the final rise in level and the rise after 600 s.

1. (a) τ = 1.5·(t₆₃ − t₂₈) = 1.5 × (40 − 22) = 27 s.
2. θ = t₆₃ − τ = 40 − 27 = 13 s. FOPDT: τ = 27 s, θ = 13 s; θ/τ = 0.48, a moderately difficult loop.
3. (b) τ = A·R = 2 × 150 = 300 s.
4. Final rise Δh = R·Δq = 150 × 0.01 = 1.5 m.
5. At t = 600 s = 2τ: Δh = 1.5·(1 − e^(−2)) = 1.5 × 0.8647 = 1.30 m.

Common mistakes

• Measuring τ from the moment of the step instead of from the end of the dead time; the 63.2 % point is at t = θ + τ.
• Treating a first-order response as if it "reaches" its final value at a finite time and using that time in the formula — it only approaches it asymptotically.
• Using the wrong sign in the log formula: the argument (y_f − y₁)/(y_f − y₂) must be greater than 1 so τ comes out positive.
• Quoting gain without units. K = 1.2 °C/% and K = 1.2 °C/(kg/s) are very different processes.
• Confusing a large gain with a fast process. Gain is about size; τ and θ are about speed.
• Forgetting that the reaction-curve test includes the valve and transmitter, so the identified K, τ and θ belong to the whole loop seen by the controller.

For GATE IN

Expect questions that give a first-order or FOPDT transfer function and ask for the output at a given time after a step, the time to reach a certain percentage, the steady-state value from the final-value theorem, or τ for a tank or thermometer. Practise reading K, θ and τ off a reaction curve, the 63.2 % and 2τ/3τ/4τ checkpoints, and converting transport delay e^(−θs) into phase lag (−ωθ radians) for frequency-response questions.

Quick check

1. A first-order process has τ = 20 s. What percentage of the final change is complete at t = 60 s?
2. Liquid travels 9 m through a pipe at 1.5 m/s before reaching a sensor. What is the dead time?
3. A step of 5 % in controller output changes flow by 2 m³/h at steady state. What is the process gain?
4. For G(s) = 3e^(−4s)/(10s + 1), when does the output reach 63.2 % of its final change after a step?

Answers: 1. 95.0 %. 2. 6 s. 3. 0.4 (m³/h)/%. 4. At t = 14 s.