Self-regulating and non-self-regulating processes

Why it matters

Before tuning a loop you must know what the process does on its own when the controller is in manual. Some processes settle at a new steady state, some drift away steadily, and a few run away faster and faster. Mistaking one kind for another leads to wrong tuning, overflowing tanks and, in reactors, real hazards.

Key ideas

Self-regulating process. After a step change in an input, the output moves to a new steady value without any controller. The process contains its own negative feedback: as the output rises, something inside the process pushes back. Examples:

• A tank draining freely through an outlet valve — a higher level gives more head, more outflow, and the level settles where outflow equals inflow.
• Outlet temperature of a heat exchanger — a hotter outlet means a smaller driving force and more heat loss, so it settles.
• Flow through a pipe after a valve opens — friction rises until a new flow is reached. Self-regulating processes are usually modelled as first order plus dead time, G(s) = K·e^(−θs)/(τs + 1). The step response has a finite steady gain K.

Non-self-regulating (integrating) process. The output keeps changing at a constant rate as long as there is any imbalance between what goes in and what comes out. Nothing inside the process pushes back. Examples:

• Level in a tank whose outflow is fixed by a pump (positive-displacement or flow-controlled) — inflow above outflow makes the level rise for ever.
• Boiler drum level — feedwater and steam flows are set independently, so an imbalance changes the inventory steadily.
• Gas pressure in a vessel where inlet and outlet flows do not depend on the vessel pressure (compressor in, controlled flow out). The model is G(s) = K_i·e^(−θs)/s. A step in input gives a ramp in output with slope K_i·Δu. There is no steady-state gain; the useful number is the integrating rate K_i (output change per unit time per unit input).

Runaway (open-loop unstable) process. The internal feedback is positive: a rise in output makes the output rise faster. The classic case is an exothermic reactor, where higher temperature speeds the reaction, releasing more heat. Model: G(s) = K/(τs − 1), with a pole in the right half plane. Such a process must be stabilised by control and protected by independent safety systems.

Degree of self-regulation. A free-draining tank has outflow q_o = h/R (linearised). A small R (big outlet) gives strong self-regulation: a small level change corrects a large imbalance. As R → ∞ the outlet stops responding to level and the tank becomes a pure integrator. So "self-regulating" and "integrating" are the two ends of the same physics.

Why the distinction matters for control.

• Testing. A self-regulating process can be stepped in manual and allowed to settle. An integrating process never settles — you measure the slope of the ramp instead, and must step back before the level hits a limit.
• Proportional-only control. With an integrating process, a P-only controller gives zero offset for set-point changes (the process itself integrates), but a sustained load change still leaves a level offset of Δd/K_c (Δd = flow disturbance, K_c in flow per unit level). With a self-regulating process, P-only leaves offset for both set-point and load changes.
• Integral action. An integrating process plus integral action gives two integrators in the loop (−180° of phase lag before anything else), so integrating loops oscillate if the reset is too fast. Use a long integral time.
• Averaging level control. Surge tanks are often deliberately loosely tuned so the level absorbs flow upsets and passes a smooth flow downstream.

Formulas

A·dh/dt = q_i − h/R → H(s)/Q_i(s) = R/(A·R·s + 1)

• A: tank cross-section (m²); h: level (m); q_i: inflow (m³/s); R: linearised outlet resistance (s/m²). Self-regulating tank: K = R (m per m³/s), τ = A·R (s).

A·dh/dt = q_i − q_o → H(s)/Q_i(s) = 1/(A·s)

• q_o: outflow fixed by a pump (m³/s). Integrating tank: K_i = 1/A (m per m³ of net inflow).

dh/dt = (q_i − q_o)/A

• Rate of level change (m/s) for a constant flow imbalance.

y(t) = y₀ + K_i·Δu·(t − θ) for t ≥ θ

• Ramp response of an integrating process to an input step Δu.

G(s) = K/(τs − 1)

• Runaway process; output grows as e^(t/τ) after a disturbance.

offset = Δd / K_c (P-only control of an integrating level, outflow manipulated)

• Δd: sustained inflow disturbance (m³/s); K_c: controller gain (m³/s per m of level error). Offset in level (m).

Worked examples

Example 1 (standard) — free-draining tank. A tank of area A = 1.5 m² drains through a valve with linearised resistance R = 400 s/m². Inflow is 0.005 m³/s at steady state and is stepped to 0.006 m³/s. Find the initial level, the final level, τ, and the time for the level to reach 2.3 m.

1. Initial level h₀ = R·q_i = 400 × 0.005 = 2.0 m.
2. Final level h_f = 400 × 0.006 = 2.4 m — the process self-regulates.
3. τ = A·R = 1.5 × 400 = 600 s.
4. h(t) = 2.0 + 0.4·(1 − e^(−t/600)). Set h = 2.3: 0.3/0.4 = 0.75 = 1 − e^(−t/600).
5. e^(−t/600) = 0.25 → t = 600·ln 4 = 600 × 1.386.
6. t ≈ 832 s (about 13.9 min), with h₀ = 2.0 m, h_f = 2.4 m, τ = 600 s.

Example 2 (GATE level) — pumped tank. A tank of area 3 m² and height 4 m has inflow 0.020 m³/s and a pump drawing a fixed 0.020 m³/s; level is 2.0 m. The inflow suddenly rises to 0.026 m³/s. (a) Time to overflow with no control. (b) A P-only level controller now manipulates the pump flow with K_c = 0.01 (m³/s)/m. Find the steady offset after the same disturbance, and the closed-loop time constant.

1. Transfer function: H(s)/Q_i(s) = 1/(3s) — integrating.
2. dh/dt = (0.026 − 0.020)/3 = 0.002 m/s.
3. Rise needed = 4.0 − 2.0 = 2.0 m → t = 2.0/0.002 = 1000 s (≈ 16.7 min) to overflow.
4. (b) With q_o = q_o,0 + K_c·(h − h_sp): A·dh/dt = Δd − K_c·(h − h_sp). At steady state the pump must take the extra 0.006 m³/s, so K_c·(h − h_sp) = 0.006.
5. Offset = Δd/K_c = 0.006/0.01 = 0.6 m above set point.
6. Closed-loop time constant = A/K_c = 3/0.01 = 300 s — the controlled level now behaves like a self-regulating first-order process.

Common mistakes

• Calling a furnace or heat-exchanger temperature "non-self-regulating" — heat losses rise with temperature, so these settle. Integrating behaviour comes from inventories with independently fixed in and out flows.
• Trying to read a steady-state gain from an integrating process test. Report the ramp slope per unit input instead.
• Assuming P-only control on a level never has offset. It is offset-free for set-point changes but not for sustained load changes.
• Using fast integral action on level loops and then blaming the valve for the slow oscillation that follows.
• Ignoring inverse response (swell and shrink) in boiler drum level, which looks like the wrong-direction start of a ramp.

For GATE IN

Typical questions: write the differential equation of a tank and its transfer function; decide whether a given configuration is self-regulating or integrating; find the time to fill or empty a tank for a flow imbalance; find the level response of a first-order tank to a step in inflow; and compute steady-state error of P or PI control on a type-0 versus type-1 process using the final-value theorem. Practise linearising q_o = c·√h around an operating point to get R.

Quick check

1. Is the level in a tank emptied by a constant-speed positive-displacement pump self-regulating?
2. A tank of area 2 m² has a net inflow of 0.01 m³/s. How fast does the level rise?
3. What is the time constant of a free-draining tank with A = 0.5 m² and R = 120 s/m²?
4. Which process has the transfer function K/(τs − 1), and why is it dangerous?

Answers: 1. No — it is integrating. 2. 0.005 m/s (0.3 m per minute). 3. 60 s. 4. A runaway (open-loop unstable) process such as an exothermic reactor; its output grows exponentially unless controlled.