Introduction & Context

Integral Squared Error (ISE) is a time-domain performance index that quantifies the cumulative deviation between a controlled variable and its set-point. In process engineering it is widely used to:

Tune PID controllers—minimising ISE yields fast, well-damped responses.
Compare alternate control strategies (cascade, feed-forward, model-predictive) on an objective numerical basis.
Size relief valves and study safety-system dynamics, because the integral of pressure error squared is directly related to fluid energy and mechanical stress.

The sheet below assumes a first-order pressure response to a step change in steam flow, typical of jacketed vessels, heat exchangers or tracing lines.

Methodology & Formulas

Pressure error signal
For a step of magnitude \(ΔP\) and first-order time constant \(T_i\), the instantaneous error is \[ e(t)=ΔP\,\mathrm{e}^{-t/T_i} \]
Integral Squared Error
\[ \mathrm{ISE}=\int_0^\infty e^2(t)\,\mathrm{d}t = \frac{(ΔP)^2\,T_i}{2} \] Units: \(\mathrm{Pa^2\,s}\) when \(ΔP\) is in pascals; convert to \(\mathrm{bar^2\,s}\) by dividing by \(10^{10}\).
Steam velocity
Mass flow continuity gives \[ v=\frac{\dot m}{ρ\,A} \quad\text{with}\quad A=\frac{πD^2}{4} \]
Reynolds number
\[ \mathit{Re}=\frac{ρ\,v\,D}{μ} \]

Regime / Criterion	Threshold	Implication
Correlation validity	\(\mathit{Re}\geq 10\,000\)	Below limit, friction factor correlations are unreliable.
Erosion / noise limit	\(v\leq 40\ \mathrm{m\,s^{-1}}\)	Higher velocities cause valve seat erosion and acoustic noise.
Valve sizing	\(P_\mathrm{sp}\leq 6\ \mathrm{bar(abs)}\)	Standard control valves are rated to 6 bar saturated steam.
Safety-valve lift	\(P_\mathrm{final}\leq 3.5\ \mathrm{bar(abs)}\)	Exceeding may open PSV and dump steam to flare/condenser.

ISE is the time-integral of the squared difference between set-point and measured value. Engineers care because:

It penalises even small persistent offsets, so minimising ISE yields tight steady-state control.
It is mathematically convenient—differentiable and positive-definite—making optimisation and tuning straightforward.
Lower ISE usually correlates with less off-spec product and reduced utility consumption.

Export SP and PV at the fastest scan rate available.
Compute error e(t) = SP − PV for each timestamp.
Square every error to get e²(t).
Multiply each squared error by the time-step Δt (or use trapezoidal integration for variable steps).
Sum the products; the result is ISE in engineering units squared·time.

Use ISE when you want to strongly suppress large errors; squaring magnifies big deviations.
Use IAE (Integral Absolute Error) when all errors are equally important; it is less aggressive on outliers.
Use ITSE (Integral Time-weighted Squared Error) if you want to penalise errors that linger; the time multiplier forces the controller to settle faster.

Check for increased oscillation—aggressive gains amplify squared error.
Verify that integral time was not lengthened excessively, allowing offset to accumulate.
Look for valve stiction or measurement noise; both inflate ISE rapidly once errors are squared.

Worked Example – Estimating the Integral Squared Error for a Steam-Pressure Control Loop

A small chemical plant uses a PI controller to keep the header steam pressure at 3.2 bar. After a sudden load change the pressure rises and eventually settles at 3.55 bar. The plant engineer wants to quantify how far, and for how long, the pressure deviated from the set-point by calculating the Integral Squared Error (ISE) over a 5-min evaluation window.

Knowns

Set-point pressure, \(P_{\text{sp}}\) = 3.2 bar
Final steady pressure, \(P_{\text{final}}\) = 3.55 bar
Evaluation period, \(t_{\text{max}}\) = 300 s
Data-collection interval, \(\Delta t\) = 1 s

Step-by-Step Calculation

Convert the set-point and final pressure to pascals for internal consistency: \[ P_{\text{sp}} = 3.2\ \text{bar} \times 1.0 \times 10^{5}\ \text{Pa/bar} = 320\,000\ \text{Pa} \] \[ P_{\text{final}} = 3.55\ \text{bar} \times 1.0 \times 10^{5}\ \text{Pa/bar} = 355\,000\ \text{Pa} \]
Compute the pressure error at every time step. Because the response is essentially a first-order rise, the error decays exponentially from an initial step of 35 kPa. The error at the \(k\)-th second is: \[ e(k) = (P_{\text{final}} - P_{\text{sp}})\ \text{e}^{-k/\tau} \] For this example the time constant \(\tau\) is taken as 60 s, giving a smooth curve that reaches the new steady state within the 300 s window.
Evaluate the squared error at each second and accumulate the sum: \[ \text{ISE}_{\text{Pa}^2\text{s}} = \sum_{k=1}^{300} e(k)^2\ \Delta t \] Carrying out the summation (or integrating analytically) yields: \[ \text{ISE}_{\text{Pa}^2\text{s}} \approx 7.35 \times 10^{9}\ \text{Pa}^2\text{s} \]
Convert the ISE back to bar²·s for reporting convenience: \[ \text{ISE}_{\text{bar}^2\text{s}} = \frac{7.35 \times 10^{9}\ \text{Pa}^2\text{s}}{(1.0 \times 10^{5}\ \text{Pa/bar})^2} = 0.735\ \text{bar}^2\text{s} \]

Final Answer

The Integral Squared Error for the 5-min event is 0.735 bar²·s (or 7.35 × 10⁹ Pa²·s).

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

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