Level Loop Sawtooth Caused by Valve Hysteresis
A 6 m³ surge tank on a refinery unit has its level controlled by a PI controller (Kc = 2.0, Ti = 4 min, derivative off) driving a 2-inch globe valve whose datasheet hysteresis is 2 % of stroke. In automatic, the level traces a sawtooth: it ramps slowly in one direction for about eight minutes, reverses sharply, and repeats. The controller output shows flat segments lasting several minutes, then jumps of 4–6 %. In manual, with the output held constant, level is steady to ±0.01 m. Raising Kc to 4 makes the sawtooth smaller and faster; lowering it to 1 makes it larger and slower. A tuning package recommends reducing Kc to 0.8 and increasing Ti to 8 min. The valve has no positioner, only an I/P transducer, and stem position is not fed back to the DCS. An upstream flow disturbance of ±5 % arrives roughly hourly and the loop must reject it.
1. Diagnose the oscillation and describe the test that would confirm it.
Read the worked answer
The evidence points to valve stiction or hysteresis, not to tuning. First, in manual the level is flat to ±0.01 m, so the process itself is stable and the oscillation is generated by the loop. Second, the controller output has flat segments followed by jumps: a PI controller with a persistent error should move its output continuously, so a stationary output means the valve is not responding and the integrator is winding up. Third, the level waveform is a sawtooth rather than the near-sinusoid of a loop at its stability limit. A 2 % dead band is enough: the controller ramps the output until the valve breaks away, the level overshoots, the output ramps back, and the cycle repeats. Confirming test: a valve signature. With the block valve open, step the controller output from 0 to 100 % and back in 0.5 % increments, recording stem travel or flow. A dead band near 2 % with a sharp breakaway proves the diagnosis; a clean linear response would send you back to tuning.
2. Assess the tuning package's recommendation and rank the available remedies.
Read the worked answer
The recommendation will make the problem worse. Limit-cycle amplitude from a dead band scales roughly as hysteresis divided by loop gain, so dropping Kc from 2.0 to 0.8 increases the level excursion by a factor of about 2.5, and doubling Ti slows the traversal of the dead band so each excursion lasts longer. Both effects match the operator's observation that Kc = 1 gave a larger, slower cycle, which is the opposite of what a genuinely over-tuned loop does. Ranked remedies: first, fit a positioner or smart valve with stem feedback, which cuts the dead band to well under 0.5 % and removes the flat output segments; second, if no positioner is possible, add an integral hold or output dead band that freezes the integrator while the error is inside a narrow band, accepting a small offset in exchange for no cycle; third, raise Kc moderately, which shrinks amplitude but does not eliminate the limit cycle; fourth, re-tune Ti only after the valve is fixed. Do not classify this as a tuning fault.
3. If the dead-band compensator is chosen as an interim fix, what does it cost the plant and how should it be sized?
Read the worked answer
The compensator freezes the controller output whenever the error lies inside a band, so the loop behaves open-loop within that band. The level will drift with any imbalance between inflow and outflow and will settle somewhere inside the band rather than at set point; the steady-state offset is set by the band width, not by the integral time. Size the band just above the valve's dead band expressed in level: if 2 % of stroke corresponds to roughly 0.15 m of level, a band of about 0.2 m is needed, which may be unacceptable against a tight level specification on a 6 m³ tank. The compensator also masks a degrading valve and complicates diagnosis, since the output no longer reflects the controller's true intent. Use it as a documented temporary measure with the band recorded in the loop sheet, and schedule the positioner retrofit; the hourly ±5 % flow disturbance will be rejected better once the valve responds continuously and Kc can be raised again.