Process Control practice questions

Answer: B) Pass the higher or lower of two controller outputs to a single valve

Override (selective) control uses a high or low selector to decide which of two controller outputs reaches a shared final control element. This lets a constraint controller take over from the normal controller when a limit is close.

• AIncorrect. That describes split-range control, where a single controller output moves one valve over part of its signal range and a second valve over the rest.
• BCorrect. A high or low selector picks one of two controller outputs to drive the shared final element. Normally the main controller is in charge, and the override controller takes over when a constraint (such as a pressure or level limit) is about to be broken.
• CIncorrect. That describes feedforward plus feedback control, where a measured disturbance changes the manipulated variable before the controlled variable is affected. It works by adding signals, not by choosing between them.
• DIncorrect. That describes ratio control, where the setpoint of the controlled flow is the measured wild flow times a ratio. No choice between controller outputs is involved.

Answer: A) Responds to the rate of change of error and can amplify noise

• ACorrect. It's often filtered or omitted on noisy loops.
• BIncorrect. That's integral action.
• CIncorrect. That's proportional action.
• DIncorrect. It can add damping.

Answer: A) More oscillation and smaller stability margins, leading eventually to an unstable loop

Too much feedback gain reduces stability margins, so the loop becomes oscillatory and, past the ultimate gain, unstable. Proportional gain never removes offset and never changes the process dynamics.

• ACorrect. Higher loop gain lowers the gain and phase margins. The response becomes more underdamped and oscillatory, and once the gain passes the ultimate gain the loop goes unstable.
• BIncorrect. A sluggish, overdamped response with large offset is what you get from too little proportional gain, not too much.
• CIncorrect. Raising proportional gain shrinks offset but does not remove it; only integral action does that. Higher gain also lowers damping.
• DIncorrect. Controller gain does not change the process's own dynamics (dead time, time constant). It only changes how hard the controller reacts to error.

Answer: A) 6 units

• ACorrect. Final change Δy = KΔu = 3 × 2.
• BIncorrect. Multiply, don't divide.
• CIncorrect. Multiply, don't add.
• DIncorrect. Multiply K by the step size.

Answer: C) A steady-state offset

• AIncorrect. Offset is the typical result.
• BIncorrect. P-only leaves offset.
• CCorrect. Integral action removes it.
• DIncorrect. The loop stays closed.

Answer: A) The final control element is stuck at a limit while the error persists, so the integral term keeps accumulating

Reset windup happens when the actuator saturates but the error continues, so the integral term builds up a large value. That stored value causes overshoot after the error reverses, and anti-windup logic prevents it.

• ACorrect. While the actuator is saturated, the integral term keeps growing even though the output can't move further. Once the error reverses, it takes time to unwind, which causes large overshoot. Anti-windup schemes clamp or stop the integration during saturation.
• BIncorrect. This is noise sensitivity from derivative action. It's usually handled by filtering the derivative, not by anti-windup.
• CIncorrect. This is closed-loop instability from too much gain. It isn't caused by integral accumulation during saturation.
• DIncorrect. Offset is a feature of proportional-only control. Adding integral action removes it, so this is unrelated to windup.

Answer: D) An integrating process

• AIncorrect. Level changes over time.
• BIncorrect. Outflow doesn't respond to level.
• CIncorrect. Level responds immediately.
• DCorrect. Any flow imbalance changes level continuously.

Answer: A) The primary controller's output sets the secondary controller's setpoint

• ACorrect. The inner loop rejects disturbances quickly.
• BIncorrect. They're nested.
• CIncorrect. It's nested feedback.
• DIncorrect. The inner loop should be faster.

Answer: B) It measures a disturbance and adjusts the manipulated variable before the output deviates

Feedforward control measures a disturbance and acts on the manipulated variable before the controlled variable changes. Because its model is never perfect, it is normally combined with feedback.

• AIncorrect. This describes feedback control, which acts on the error between the controlled variable and its set point.
• BCorrect. Feedforward measures a disturbance directly and uses a process model to compensate before the disturbance affects the controlled variable. In practice it is usually combined with feedback trim.
• CIncorrect. Feedforward depends on a model that relates the disturbance and the manipulated variable to the output. Acting on the error signal is what feedback does.
• DIncorrect. Model errors and unmeasured disturbances leave residual error. That is why feedback trim is added to remove the remaining offset.

Answer: A) Fail open (air-to-close), so cooling water keeps flowing if air or signal is lost

Pick the failure position from the hazard. For an exothermic reactor, losing cooling is the dangerous case, so the cooling-water valve should fail open (air-to-close).

• ACorrect. If air or signal is lost, the valve goes to full cooling. That removes heat at the highest rate and works against a temperature runaway, so it is the safe position.
• BIncorrect. Cutting off cooling while the exothermic reaction may still be generating heat can lead to a thermal runaway. Saving utility water does not justify that risk.
• CIncorrect. Fail-last-position (fail-freeze) does not keep the temperature steady. Once control is lost, the valve can no longer respond if heat generation rises, so it does not give a defined safe state for this hazard.
• DIncorrect. Standard pneumatic actuators fail to an end position (open or closed), not to mid-travel. Partial cooling also may not be enough to stop a runaway.