Reference ID: MET-2D7B | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Integral control reset time, denoted TR, is the single tuning parameter that determines how aggressively the integral (reset) term of a PI or PID controller eliminates steady-state error. A short TR produces a large integral gain and therefore rapid elimination of offset, but also amplifies measurement noise and can induce oscillatory or even unstable behaviour. A long TR yields smooth output but sluggish recovery from disturbances. The calculation shown here is used in process engineering to obtain a first-pass value of TR from a user-specified required change in controller output, the current control error, the controller gain, and the execution interval of the algorithm.
Methodology & Formulas
Define the control error
\[
e = \text{SP} - \text{PV}
\]
where SP is the set-point and PV is the measured process variable (both in consistent engineering units).
Determine the minimum incremental change in the integral (reset) component, ΔmI,req, expressed in percent of the controller output range.
ε = machine epsilon (≈ 10-9) to avoid division by zero.
Impose operational limits:
Condition
Interpretation
\(T_{R,\text{calc}} < T_{R,\text{min}}\)
Noise amplification likely; clamp to TR,min = 1 s or retune.
\(T_{R,\text{calc}} > T_{R,\text{max}}\)
Sluggish recovery likely; clamp to TR,max = 3600 s or retune.
Reset time (Ti) is the minutes (or seconds) per repeat—the time needed for the integral contribution to match the proportional contribution after a step change.
From step-test data: Ti = (Δt · Kp) / Δe, where Δt is the time the PV takes to traverse the proportional band, Kp is the process gain, and Δe is the controller error in % of span.
From Ziegler–Nichols open-loop rules: Ti ≈ τ, the dominant process time constant.
From closed-loop oscillation: Ti = 0.5 · Tu, where Tu is the ultimate oscillation period.
Always match the controller’s integral setting.
If the controller uses repeats per minute (RPM), Ti in minutes = 1 / RPM.
If the controller uses seconds per repeat, Ti in seconds = 1 / RPM · 60.
Multiply minutes by 60 to get seconds; divide seconds by 60 to get minutes.
Ti sets the integral speed.
Smaller Ti (faster reset) removes offset quickly but can cause oscillations.
Larger Ti (slower reset) gives sluggish return to setpoint and may leave offset.
Rule of thumb: keep Ti between 0.3 τ and 1.0 τ for dominant-lag processes.
Yes—use the natural recovery after an upset.
Measure the time it takes PV to move 63% of the way back to setpoint; call this t63.
Ti ≈ t63 for self-regulating processes.
For integrating processes (level, pressure), Ti ≈ 2 · t63 to avoid overshoot.
Worked Example – Integral Control Reset Time Calculation
A process engineer is tuning a PI flow controller on a water-supply header. The loop must eliminate a steady-state offset of 2 L/min without causing excessive integral wind-up. The available reset-time range on the DCS is 1–3600 s. Determine the minimum reset time that will produce the required 5% increase in valve position over the next 30 s.
Knowns
Set-point (SP): 85.0 L/min
Process variable (PV): 83.0 L/min
Controller gain (Kc): 2.0 %/L·min-1
Required change in integral output (ΔmI,req): 5.0 %
Execution interval (Δt): 30.0 s
Allowable reset-time range: 1.0–3600.0 s
Step-by-step calculation
Compute the current error: \( e = SP - PV = 85.0 - 83.0 = 2.0 \) L/min
Evaluate the numerator of the reset-time equation: \( K_c \cdot e \cdot \Delta t = 2.0 \cdot 2.0 \cdot 30.0 = 120.0 \)
Identify the denominator: \( \Delta m_{I,req} = 5.0 \)
Calculate the theoretical reset time: \( T_R = \dfrac{120.0}{5.0} = 24.0 \) s
Check against limits: 24.0 s lies within 1.0–3600.0 s, so the value is admissible.
Final Answer
The integral reset time required is 24.0 s.
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