First-order systems in series and interacting systems

Why it matters

Real plants are chains of capacities: tanks in series, trays in a column, a reactor followed by a cooler, a process followed by a sensor lag. Chaining first-order elements creates S-shaped, sluggish responses that look like dead time, and whether the elements interact changes how sluggish the chain is. This is the bridge from simple first-order behaviour to the second-order and dead-time models used for tuning.

Key ideas

Non-interacting systems. Two capacities are non-interacting when the output of the first drives the second but the second does not affect the first. Example: tank 1 discharges freely (over a weir or through a valve into the top of tank 2), so its outflow depends only on h₁. The overall transfer function is simply the product: G(s) = G₁(s)·G₂(s) = K₁K₂ / ((τ₁s + 1)(τ₂s + 1)).

Interacting systems. When tank 1 discharges into the bottom of tank 2 through a resistance, the flow between them is q₁ = (h₁ − h₂)/R₁; the level in tank 2 pushes back on tank 1. The transfer function is no longer a product of the separate first-order lags. For linear resistances, Q₂(s)/Q(s) = 1 / (τ₁τ₂s² + (τ₁ + τ₂ + A₁R₂)s + 1), with τ₁ = A₁R₁ and τ₂ = A₂R₂. The extra term A₁R₂ in the middle coefficient is the signature of interaction.

Consequences.

• Both kinds are second-order and overdamped; the poles are real and negative, so neither can oscillate after a step (ζ ≥ 1 for non-interacting, ζ > 1 for interacting).
• Interaction spreads the two effective time constants further apart and makes the response more sluggish than the same tanks without interaction.
• The step response of two or more lags is S-shaped: zero initial slope, an inflection point, then an exponential approach. The more lags in series, the more the response looks like a dead time followed by a first-order rise, which is why multi-capacity processes are often fitted with a first-order-plus-dead-time (FOPDT) model.
• When one time constant is much larger than the others, it dominates; the small ones behave roughly like extra dead time.
• The sum of the time constants is not a time constant of the chain. It equals the mean delay (first moment of the impulse response) of a non-interacting chain, useful for approximations only.

N equal tanks. N non-interacting stages of time constant τ give 1/(τs + 1)^N. As N grows with Nτ fixed, the response approaches a pure dead time Nτ (plug flow), linking to the tanks-in-series model in reaction engineering.

Second-element step response. For non-interacting lags with τ₁ ≠ τ₂ and a unit step, y(t)/K = 1 − (τ₁e^(−t/τ₁) − τ₂e^(−t/τ₂))/(τ₁ − τ₂).

Formulas

G(s) = K₁·K₂ / ((τ₁·s + 1)·(τ₂·s + 1))

• Non-interacting lags in series; K gains, τ time constants (s or min).

y(t) = K·M·[1 − (τ₁·e^(−t/τ₁) − τ₂·e^(−t/τ₂)) / (τ₁ − τ₂)]

• Step of size M, τ₁ ≠ τ₂, K = K₁K₂.

y(t) = K·M·[1 − (1 + t/τ)·e^(−t/τ)]

• Two equal lags (τ₁ = τ₂ = τ).

Q₂(s)/Q(s) = 1 / (τ₁·τ₂·s² + (τ₁ + τ₂ + A₁·R₂)·s + 1)

• Interacting tanks with linear resistances; τ₁ = A₁R₁, τ₂ = A₂R₂, A area (m²), R resistance (s/m²), Q volumetric flow (m³/s). Multiply by R₂ for H₂(s)/Q(s).

τ_a·τ_b = τ₁·τ₂ and τ_a + τ_b = τ₁ + τ₂ + A₁·R₂

• Effective time constants τ_a, τ_b of the interacting system: the negatives of the reciprocals of the two real poles.

Worked examples

Example 1 (standard): two non-interacting tanks. Tank 1: A₁ = 2 m², R₁ = 0.5 min/m². Tank 2: A₂ = 1 m², R₂ = 0.5 min/m². Tank 1 discharges freely into tank 2. The inlet flow to tank 1 steps up by M = 0.4 m³/min. Find the change in h₂ after 1 min and at steady state.

1. τ₁ = A₁R₁ = 1.0 min; τ₂ = A₂R₂ = 0.5 min.
2. Gain from inlet flow to h₂: K = 1·R₂ = 0.5 m per (m³/min); final change Δh₂(∞) = K·M = 0.5·0.4 = 0.20 m.
3. At t = 1 min: τ₁e^(−1/1) = 0.3679; τ₂e^(−1/0.5) = 0.5·0.1353 = 0.0677.
4. Bracket = 1 − (0.3679 − 0.0677)/(1.0 − 0.5) = 1 − 0.6004 = 0.3996.
5. Δh₂(1 min) = 0.20·0.3996 = 0.0799 m.

Δh₂(1 min) ≈ 0.080 m; Δh₂(∞) = 0.20 m

Example 2 (GATE level): effect of interaction. Two identical tanks with A = 1 m² and R = 1 min/m² (so τ₁ = τ₂ = 1 min, A₁R₂ = 1 min) are connected (a) non-interacting and (b) interacting. For a unit step in inlet flow, find the effective time constants and the fraction of the final outlet-flow change reached after 1 min.

1. (a) Non-interacting: Q₂/Q = 1/(s + 1)². Time constants 1 and 1 min. Fraction at t = 1: 1 − (1 + 1)·e^(−1) = 1 − 0.7358 = 0.264.
2. (b) Interacting: denominator τ₁τ₂s² + (τ₁ + τ₂ + A₁R₂)s + 1 = s² + 3s + 1.
3. Roots: s = (−3 ± √5)/2 = −0.382 and −2.618 min⁻¹, so τ_a = 2.618 min and τ_b = 0.382 min (check: product 1.000, sum 3.000).
4. Unit step: y(t) = 1 − (τ_a·e^(−t/τ_a) − τ_b·e^(−t/τ_b))/(τ_a − τ_b).
5. At t = 1: τ_a·e^(−0.382) = 2.618·0.6825 = 1.7868; τ_b·e^(−2.618) = 0.382·0.0729 = 0.0279; y = 1 − (1.7868 − 0.0279)/2.236 = 1 − 0.7866 = 0.213.

(a) 26.4 % and (b) 21.3 % of the final change after 1 min: interaction makes the response slower, because one effective time constant grows to 2.618 min.

Common mistakes

• Adding time constants and calling the sum "the" time constant of the chain.
• Multiplying the individual lags for an interacting system; the A₁R₂ term is then lost.
• Claiming interacting tanks can oscillate: with only resistances and capacities, the poles are always real.
• Using the general two-lag formula with τ₁ = τ₂ (division by zero); use the equal-lag form instead.
• Forgetting that the gain to a level includes R₂, while the gain to the outlet flow is 1.

For GATE CH

Expect: deriving or recognising the interacting-tank transfer function, finding the poles or effective time constants, computing the response of the second tank at a given time, comparing interacting with non-interacting arrangements, and identifying the order of a chain of capacities. Practise the quadratic-root step and the two-lag step response formula.

Quick check

1. What is the transfer function of three equal non-interacting lags of τ = 2 min?
2. Which is faster after a step: two interacting or two non-interacting tanks with the same A and R?
3. Can two non-interacting first-order lags in series give overshoot?
4. For interacting tanks with τ₁ = 2, τ₂ = 1 and A₁R₂ = 2 (min), what is the middle coefficient of the denominator?

Answers: 1. 1/(2s + 1)³. 2. The non-interacting pair. 3. No, it is overdamped (or critically damped for equal τ). 4. 2 + 1 + 2 = 5 min.