Exercise set

Chemical Process Control Loop Dynamics and Valve Authority Exercises

Worked chemical process control exercises for residence time, loop response, feedforward, ratio control, windup, phase margin and valve authority.

Branch: Chemical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on chemical process control as a dynamic and actuator-limited engineering problem: residence time, first-order response, controller output, feedforward, ratio control, valve flow, valve authority, saturation, windup, dead time, phase margin and release evidence.

Plant analyzer validation, alarm floods, interlocks, bypass records and operating-envelope release are handled in the companion specialist exercise set.

Release Evidence Notes

Control-loop evidence should state operating point, process boundary, controller mode, tuning basis, actuator limits, valve travel, measurement range, dead time, time constant, utility condition and acceptance rule. A stable trend is weak evidence if the loop is in manual, the valve is saturated, the analyzer is stale or the plant is outside the validated operating range.

Engineering Boundary Notes

These examples are first-pass screening calculations. Real chemical process control work should also check nonlinear operating regions, safety interlocks, analyzer lag, valve stiction, actuator rate limits, utility disturbances, startup/shutdown modes, management of change and operator handover.

Common Release Mistakes

tuning around a sticky or undersized control valve;
using one operating point for a nonlinear process;
ignoring density basis in ratio and feedforward calculations;
accepting fast response without stability margin or actuator headroom;
allowing integral action to continue while the valve is saturated;
confusing loop validation with full plant operations release.

Scenario Map

Scenario	Exercises	Main calculation	Release decision
Process dynamics	1, 2, 11, 12	Residence time, first-order response, fitted time constant and dead time	Decide whether tuning basis matches the operating point.
Control action	3, 4, 5, 6, 13, 14, 15	Controller output, integral move, feedforward, ratio control, rate limits, windup and filtering	Adjust tuning, mode logic or anti-windup.
Final element	7, 8, 9, 16	Valve coefficient, authority, saturation and cascade speed	Resize valve, revise pressure drop or limit rate change.
Release evidence	10, 17, 18	Phase-margin loss, validation residual and integrated loop gate	Release only when loop and actuator evidence agree.

Validation Package Checklist

current operating point and process boundary;
controller mode, tuning basis and anti-windup logic;
measurement location, density basis, lag and calibration state;
valve sizing, travel, pressure drop and authority;
dead time, time constant, scan time and delay margin;
release action if stability, saturation or authority gates fail.

Exercise 1: Residence Time After a Rate Increase

A stirred vessel has liquid volume $V=12\ \text{m}^3$ . Flow is increased to $Q=3.0\ \text{m}^3/\text{h}$ . Compute residence time.

Solution

\tau=\dfrac{V}{Q}=\dfrac{12}{3.0}=4.0\ \text{h}

Engineering Comment

Residence time sets a lower bound on how quickly composition can respond. Tuning based on the old rate may become too slow or too aggressive after a rate change.

Plausibility Check

At three cubic meters per hour, a twelve cubic meter inventory turns over in four hours.

Exercise 2: First-Order Temperature Response

A first-order temperature loop has time constant $\tau=8\ \text{min}$ . A step disturbance has final temperature rise $12^\circ\text{C}$ . Estimate rise after $t=8\ \text{min}$ .

Solution

\Delta T(t)=12(1-e^{-t/\tau})=12(1-e^{-1})=7.59^\circ\text{C}

Engineering Comment

One time constant gives about sixty-three percent of final response. Operators should not overcorrect before the process has had time to respond.

Plausibility Check

The result is a little over half of the final twelve-degree change.

Exercise 3: Proportional Controller Output

A proportional controller has bias $45\%$ , gain $K_c=2.0$ and error $e=4^\circ\text{C}$ . Output is limited to $0$ to $100\%$ . Compute output.

Solution

m=45+2.0(4)=53\%

Engineering Comment

The output is inside range. The engineering review should still verify direct/reverse action and valve fail position.

Plausibility Check

A gain of two percent output per degree gives an eight percent move from bias.

Exercise 4: Integral Move Over a Sample

A PI controller has integral gain $K_i=0.18\ \%/(\text{min}\cdot^\circ\text{C})$ , error $e=5^\circ\text{C}$ and sample time $\Delta t=2\ \text{min}$ . Compute integral output increment.

Solution

\Delta m_I=K_i e\Delta t=0.18(5)(2)=1.8\%

Engineering Comment

Small increments can accumulate into windup if the valve is saturated or the measurement is invalid.

Plausibility Check

The product of gain, error and time is under two percent.

Exercise 5: Feedforward Cooling Adjustment

A feed rate increases from $2.4$ to $3.0\ \text{kg/s}$ . Cooling duty is proportional to feed and was $1.6\ \text{MW}$ . Estimate new duty.

Solution

Q_{new}=1.6\dfrac{3.0}{2.4}=2.0\ \text{MW}

Engineering Comment

Feedforward can reduce disturbance load, but it must use the correct measured feed basis and heat of reaction or sensible heat model.

Plausibility Check

The feed rises by twenty-five percent, so duty rises by twenty-five percent.

Exercise 6: Ratio-Control Setpoint with Density Compensation

A reagent ratio requires $0.18\ \text{kg reagent/kg feed}$ . Feed flow is $8.0\ \text{m}^3/\text{h}$ with density $930\ \text{kg/m}^3$ . Reagent density is $1050\ \text{kg/m}^3$ . Compute reagent volumetric setpoint.

Solution

\dot{m}_F=8.0(930)=7440\ \text{kg/h}

\dot{m}_R=0.18(7440)=1339\ \text{kg/h}

Q_R=\dfrac{1339}{1050}=1.28\ \text{m}^3/\text{h}

Engineering Comment

Ratio stations should define mass or volume basis explicitly. Density compensation errors can create systematic off-ratio operation.

Plausibility Check

The reagent volume is much smaller than feed volume because the mass ratio is below one.

Exercise 7: Valve Flow from Cv

A liquid valve has $C_v=45$ , pressure drop $\Delta P=9\ \text{psi}$ and specific gravity $SG=1.0$ . Use $Q=C_v\sqrt{\Delta P/SG}$ in gpm.

Solution

Q=45\sqrt{9}=135\ \text{gpm}

Engineering Comment

The simplified relation supports a sizing screen. Real service should include installed pressure drop, viscosity, flashing and rangeability.

Plausibility Check

The square root of nine is three, so the flow is three times $C_v$ .

Exercise 8: Valve Authority

A control valve has pressure drop $45\ \text{kPa}$ at design flow. Total circuit pressure drop is $160\ \text{kPa}$ . Compute valve authority.

Solution

A_v=\dfrac{45}{160}=0.281

Engineering Comment

Low authority makes the installed valve gain sensitive to system changes. The loop may tune well in one condition and poorly in another.

Plausibility Check

The valve consumes a little over one quarter of the total pressure drop.

Exercise 9: Valve Saturation During Cooling Load Increase

A cooling valve is at $72\%$ open. A disturbance requires an additional $34\%$ output. What output would be requested and what is the saturation excess?

Solution

m_{req}=72+34=106\%

E=106-100=6\%

Engineering Comment

The controller asks for more cooling than the final element can deliver. This is a process capacity and anti-windup problem, not only a tuning problem.

Plausibility Check

The requested move pushes the valve slightly beyond full open.

Exercise 10: Added Delay and Phase-Margin Screen

A loop crossover frequency is $\omega_c=0.06\ \text{rad/s}$ . A new filter adds delay $\Delta t=4\ \text{s}$ . Estimate added phase lag in degrees.

Solution

\Delta \phi=\omega_c \Delta t\dfrac{180}{\pi}=0.06(4)(57.3)=13.8^\circ

Engineering Comment

Delay consumes stability margin. A filter that improves noise can make the loop less robust.

Plausibility Check

The product $\omega_c \Delta t$ is $0.24$ radians, which is about fourteen degrees.

Exercise 11: Time Constant from Step Test

A process reaches $63\%$ of a measured step response after $11\ \text{min}$ . Estimate time constant.

Solution

For a first-order response:

\tau\approx11\ \text{min}

Engineering Comment

The estimate should be tied to the tested operating point and input move. Nonlinear plants can have different time constants elsewhere.

Plausibility Check

The $63\%$ response point is the standard first-order time constant.

Exercise 12: Dead Time from Test Data

A flow step is made at $t=0$ . The measured temperature does not begin changing until $t=95\ \text{s}$ . Estimate dead time.

Solution

\theta=95\ \text{s}

Engineering Comment

Dead time may come from transport delay, sensor location, filtering or analyzer lag. It limits achievable control speed.

Plausibility Check

The output response starts a little over one and a half minutes after the input step.

Exercise 13: Output Rate-Limit Screen

A valve actuator can move at $12\%/\text{s}$ . A controller demands a $48\%$ move within $3\ \text{s}$ . Can the actuator complete it?

Solution

T=\dfrac{48}{12}=4\ \text{s}

It cannot complete the move in $3\ \text{s}$ .

Engineering Comment

Rate limits can create effective delay and overshoot. Tuning should match actuator speed.

Plausibility Check

At twelve percent per second, four seconds are needed for a forty-eight percent move.

Exercise 14: Integral Windup During Saturation

A saturated loop keeps error $e=6^\circ\text{C}$ for $5\ \text{min}$ . Integral gain is $0.20\ \%/(\text{min}\cdot^\circ\text{C})$ . Estimate accumulated integral demand.

Solution

I=0.20(6)(5)=6\%

Engineering Comment

The accumulated demand can delay recovery after the valve leaves saturation. Anti-windup should freeze or back-calculate integral action.

Plausibility Check

The integral term adds a little over one percent per minute for five minutes.

Exercise 15: Measurement Filter Lag

A measurement filter has time constant $12\ \text{s}$ . A loop already has process dead time $48\ \text{s}$ . Estimate total apparent delay for a conservative screen.

Solution

\theta_{app}=48+12=60\ \text{s}

Engineering Comment

This simple sum is conservative, but it reminds the reviewer that filtering can trade noise reduction for delay.

Plausibility Check

Adding a twelve-second filter to forty-eight seconds gives one minute.

Exercise 16: Cascade Speed Ratio

A cascade control design has secondary-loop time constant $20\ \text{s}$ and primary-loop time constant $180\ \text{s}$ . Compute speed ratio.

Solution

R=\dfrac{180}{20}=9

Engineering Comment

The secondary loop is about nine times faster, which supports cascade control in a first screen.

Plausibility Check

Twenty seconds fits nine times into one hundred eighty seconds.

Exercise 17: Loop Validation Residual

A validation step predicts a final temperature change of $10.0^\circ\text{C}$ . The measured final change is $9.1^\circ\text{C}$ . Compute percent residual relative to prediction.

Solution

r=\dfrac{10.0-9.1}{10.0}=9.0\%

Engineering Comment

The residual may be acceptable for tuning but weak for a tight quality or safety release. Model boundary should be stated.

Plausibility Check

The measured response is $0.9^\circ\text{C}$ lower than a ten-degree prediction.

Exercise 18: Control Loop Release Gate

A loop release requires phase-margin loss below $10^\circ$ , valve authority above $0.30$ , no saturation in the tested load step and validation residual below $10\%$ . Results are $7^\circ$ , $0.28$ , no saturation and $8\%$ . Does it release?

Solution

The valve authority gate fails:

0.28<0.30

The loop release fails.

Engineering Comment

A stable test at one condition cannot hide weak installed valve authority. The control package should address pressure-drop allocation or operating restrictions.

Plausibility Check

Only one gate fails, but the release rule treats each gate as a blocker.

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