Free PE Control Systems Practice Test
Work through 85 original, unofficial questions mapped to the NCEES PE Control Systems specifications effective beginning April 2027, across all five domains. Each answer sits under its question with a worked explanation and a source you can check.
Practice test
Measurement (Questions 1–19)
Question 1
M01 · 1A General measurement sensors · Multiple choice
A plant needs to meter diesel fuel in a 3-in. line. Which flowmeter technology cannot measure this fluid?
Show explanation
Answer and explanation
Answer: A. Magnetic flowmeter
A magnetic flowmeter makes its signal from a conducting liquid moving through a magnetic field, so the liquid must conduct electricity reasonably well. Oils such as diesel have negligible conductivity, so no usable signal forms.
Why not the others: Coriolis meters sense the effect of mass flow on vibrating tubes, turbine meters count rotor speed, and transit-time ultrasonic meters time sound pulses. None of them needs a conductive fluid.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Velocity-based flowmeters, Magnetic flowmeters
Question 2
M02 · 1A General measurement sensors · Multiple choice
Which flowmeter type measures mass flow directly and can also report fluid density?
Show explanation
Answer and explanation
Answer: D. Coriolis meter
A Coriolis meter is a true mass flow instrument. The same vibrating tubes give density from their resonant frequency, so one meter reports mass flow, density, and temperature.
Why not the others: Vortex and magnetic meters measure velocity, which gives volume flow. An orifice infers flow from differential pressure and needs density from somewhere else to give mass flow.
Source: Kuphaldt, Lessons in Industrial Instrumentation: True mass flowmeters, Coriolis flowmeters; density measurement
Question 3
M03 · 1A General measurement sensors · Multiple choice
Temperature must be measured in a furnace convection section at about 1,800 °F (982 °C). Which primary element fits best?
Show explanation
Answer and explanation
Answer: D. Type N or Type K thermocouple in a protective sheath and thermowell
Base-metal thermocouples cover this range: Type K to about 1,200 °C and Type N to about 1,260 °C in Emerson's sensor tables. The industrial platinum RTD standard, IEC 60751, stops at 850 °C (1,562 °F), and typical Pt100 sensors are rated lower still.
Why not the others: A Pt100 is out of range here. Thermistors are low-temperature devices. A bimetal dial gives only a local reading with no transmitted signal.
Source: Emerson, About Rosemount temperature sensors, sensor type ranges; Reliable Plant, IEC 60751 standard summary, RTD temperature range
Question 4
M04 · 1B Analytical instruments and sampling · Numeric entry
An ideal pH glass electrode has a slope of 59.16 mV per pH unit at 25 °C, and the slope is proportional to absolute temperature. What is the slope at 50 °C? Enter mV/pH to two decimals.
Show explanation
Answer and explanation
Answer: 64.12 mV/pH (accepted range 64 to 64.2)
Convert both temperatures to kelvin and scale: 59.16 × (323.15 ÷ 298.15) = 64.12 mV/pH. This is why pH transmitters need temperature compensation: the same millivolt reading means a different pH at a different temperature.
Common mistake: Scaling by the Celsius ratio (50 ÷ 25) gives 118 mV/pH, which is far too large. Temperature ratios in a proportional relationship always use absolute temperature.
Source: Worked calculation from the data given in the question
Question 5
M05 · 1B Analytical instruments and sampling · Numeric entry
An analyzer sample travels through 100 ft of tubing with a 0.18-in. inside diameter at 0.5 gpm. Ignoring fittings and analyzer volume, what is the transport lag in seconds? (1 ft³ = 7.4805 gal)
Show explanation
Answer and explanation
Answer: 15.9 s (accepted range 15 to 17)
Area = π/4 × (0.18/12)² = 1.767 × 10⁻⁴ ft². Volume = 1.767 × 10⁻⁴ × 100 = 0.01767 ft³ = 0.1322 gal. Lag = 0.1322 ÷ 0.5 gpm = 0.264 min = 15.9 s. Transport lines add this delay to every analyzer reading; a fast loop or a higher sample flow cuts it.
Common mistake: A common slip is using the diameter in inches without converting to feet, which inflates the volume by a factor of 144.
Source: Worked calculation from the data given in the question; Swagelok, Understanding time delay in analyzer sampling systems, transport line delay; fast loops
Question 6
M06 · 1C Fire and gas detectors · Multiple choice
Which gas-detection technology cannot detect a hydrogen leak?
Show explanation
Answer and explanation
Answer: C. Infrared absorption point detector
Hydrogen does not absorb infrared radiation, so an IR gas detector is blind to it. Catalytic bead sensors are a standard choice for hydrogen at lower-flammable-limit levels.
Why not the others: Electrochemical cells are made for specific gases, including hydrogen. Ultrasonic detectors hear the hiss of a pressurized leak no matter which gas is escaping.
Source: Det-Tronics, Hydrogen fire safety white paper 74-1019 (2020), p. 2 gas detection; p. 3 flame detection
Question 7
M07 · 1C Fire and gas detectors · Multiple choice
UV-only flame detectors in a process unit keep alarming during maintenance turnarounds. What is the most likely cause, and the best fix?
Show explanation
Answer and explanation
Answer: A. Arc welding UV; use UV/IR or multispectrum IR detectors
Arcs, sparks, and welding emit UV that a UV-only detector can mistake for flame. UV/IR detectors need both signatures, and multispectrum IR detectors don't false alarm on welding, so both reject this source much better.
Why not the others: Smoke detection suits enclosed rooms, not open process areas. Raising sensitivity makes false alarms worse. A second UV detector sees the same welding arc, so 1oo2 voting doesn't help.
Source: Det-Tronics, Hydrogen fire safety white paper 74-1019 (2020), p. 2 gas detection; p. 3 flame detection
Question 8
M08 · 1D Machinery monitoring · Multiple correct
A large centrifugal compressor runs on hydrodynamic (fluid-film) bearings. Which TWO measurements are standard for protecting its shaft?
Show explanation
Answer and explanation
Answer: B. X–Y non-contacting proximity probes at each radial bearing and D. Axial proximity probes watching the thrust collar or shaft end
On fluid-film machines, the oil film adds a lot of damping, so shaft motion is not fully seen at the casing. Proximity probes measure the shaft directly relative to the bearing, and axial probes track thrust-collar position.
Why not the others: Casing accelerometers and baseplate velocity sensors under-report shaft vibration on these machines, so neither can be the main protection.
Source: Metrix, Advantages of proximity vibration monitoring for critical machinery, fluid-film bearings; XY and thrust probes; Baker Hughes Orbit, Shaft position changes (2015), thrust position measurement
Question 9
M09 · 1E Sensor characteristics · Multiple choice
A gauge checked against a 50.0 psig reference reads 49.2, 49.3, 49.2, and 49.3 psig. How would you describe it, and what fixes it?
Show explanation
Answer and explanation
Answer: A. Precise but not accurate; calibrate it against appropriate reference points
The readings agree with each other within 0.1 psi, so the gauge is precise (repeatable). They average 49.25 psig against a 50.0 psig reference, so the check shows a systematic error at this point. Calibrate the gauge against appropriate reference points to determine and correct the actual zero/span adjustment needed.
Why not the others: The readings are not accurate at this test point, so B and C are wrong. The tight grouping shows useful repeatability, so replacement is not justified from this check alone.
Source: Control.com, What is the difference between accuracy, precision, and resolution?, dartboard analogy; Kuphaldt, Lessons in Industrial Instrumentation: Calibration errors and testing, Zero shift calibration error
Question 10
M10 · 1E Sensor characteristics · Multiple choice
An orifice DP transmitter is spanned for 100% flow. At 10% of maximum flow, the differential pressure is what percent of span?
Show explanation
Answer and explanation
Answer: B. 1%
Flow is proportional to the square root of DP, so DP is proportional to flow squared: (0.10)² = 0.01, or 1% of span. That tiny low-flow signal is why practical turndown for an orifice meter is limited.
Why not the others: 10% assumes a linear relationship. 31.6% is √0.10, the inverse mistake. 0.1% squares the wrong number.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Pressure-based flowmeters, Volumetric and mass flow calculations; square-root characterization
Question 11
M11 · 1F Sensor selection and material compatibility · Multiple choice
A DP level transmitter on a viscous polymer service keeps losing its signal because the impulse connections plug. What is the best fix?
Show explanation
Answer and explanation
Answer: D. Flush-mounted diaphragm seals with capillaries
Diaphragm seals keep the process fluid from reaching the transmitter and remove the dead legs where viscous or solids-laden fluid plugs. Viscous, plugging service is a standard reason to specify them.
Why not the others: Tracing and longer lines leave the dead leg in place. Smaller bore plugs faster. Moving the transmitter doesn't change what's in the lines.
Source: Emerson, Rosemount DP level transmitters and 1199 diaphragm seal systems (2023), p. 1 applications
Question 12
M12 · 1G Sensor installation · Multiple choice
For DP flow measurement on a gas line that may carry some condensate, which installation is preferred?
Show explanation
Answer and explanation
Answer: C. Transmitter above the taps, so liquid drains back into the pipe
Mounting above the taps keeps condensable liquid from collecting in the impulse piping. Trapped liquid would add a false head to the DP.
Why not the others: Mounting below is the liquid and steam arrangement. Below the taps, gas service would fill with condensate. Horizontal lines trap liquid.
Source: Emerson, Rosemount Annubar remote-mount transmitter installation options (2019), §3.1 liquid, §3.2 gas, §3.3 steam
Question 13
M13 · 1H Flow calculations · Numeric entry
A DP transmitter with no square-root extraction is ranged 0–100 inH₂O = 4–20 mA. Flow at 100 inH₂O is 400 gpm. What is the flow in gpm when the output is 8.0 mA?
Show explanation
Answer and explanation
Answer: 200 gpm (accepted range 199 to 201)
DP fraction = (8 − 4) ÷ 16 = 0.25. Flow = 400 × √0.25 = 200 gpm.
Common mistake: Scaling the current linearly gives 100 gpm. That only works if the transmitter or control system extracts the square root first.
Source: Worked calculation from the data given in the question; Kuphaldt, Lessons in Industrial Instrumentation: Pressure-based flowmeters, Volumetric and mass flow calculations; square-root characterization
Question 14
M14 · 1H Flow calculations · Multiple choice
A gas orifice meter was sized for 100 psia and 520 °R. It now runs at 120 psia and 520 °R with no pressure-temperature compensation. True mass flow equals the indicated value multiplied by about: (Assume ideal gas, unchanged compressibility, and negligible expansibility change.)
Show explanation
Answer and explanation
Answer: B. 1.095
Mass flow through an orifice is proportional to √(ΔP × ρ), and ideal-gas density is proportional to P/T. The density ratio is 120/100 = 1.2, so the correction is √1.2 = 1.095. The uncompensated meter reads about 9% low.
Why not the others: 1.20 forgets the square root. 0.913 inverts the ratio. 1.44 squares it.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Pressure-based flowmeters, Volumetric and mass flow calculations; square-root characterization
Question 15
M15 · 1H Flow calculations · Multiple choice
Water (ρ = 62.4 lb/ft³, μ = 6.72 × 10⁻⁴ lb/ft·s) flows at 6 ft/s in a 4-in. inside-diameter pipe. Using Re = ρVD/μ, the Reynolds number is about:
Show explanation
Answer and explanation
Answer: C. 1.9 × 10⁵
D = 4/12 = 0.333 ft. Re = 62.4 × 6 × 0.333 ÷ 6.72 × 10⁻⁴ = 1.86 × 10⁵, well into turbulent flow.
Why not the others: 1.9 × 10⁴ and 1.9 × 10⁶ are off by a factor of ten, usually from a units or decimal slip. 2.6 × 10³ would mean near-transitional flow, which doesn't match these inputs.
Source: Worked calculation from the data given in the question
Question 16
M16 · 1I Level calculations · Multiple choice
A closed vessel uses a DP level transmitter mounted at the lower tap's elevation. The high side connects to the lower tap at zero level. The low side connects through a wet reference leg filled with liquid of SG 1.0 to a tap 120 in. above the lower tap. Process SG is 0.90 and the calibrated range is 0–100 in. of level. What are the LRV and URV in inH₂O?
Show explanation
Answer and explanation
Answer: B. −120 to −30
DP = high side − low side = 0.90h − 1.0(120). At h = 0, DP = −120 inH₂O. At h = 100 in., DP = 90 − 120 = −30 inH₂O. Vapor-space pressure acts on both sides and cancels. A filled reference leg produces a negative DP that the calibration has to shift for.
Why not the others: 0 to 90 ignores the wet leg. −30 to −120 swaps LRV and URV. 120 to 30 gets the sign of DP backward.
Source: Worked calculation from the data given in the question; Kuphaldt, Lessons in Industrial Instrumentation: Hydrostatic pressure level measurement, Compensated leg systems
Question 17
M17 · 1J Unit conversion calculations · Numeric entry
A vertical, flat-bottom cylindrical tank is 10 ft in diameter and holds liquid 12 ft deep. What is the liquid volume in gallons? (1 ft³ = 7.4805 gal)
Show explanation
Answer and explanation
Answer: 7,050 gal (accepted range 7,040 to 7,060)
V = π × (5 ft)² × 12 ft = 942.5 ft³. 942.5 × 7.4805 = 7,050 gal.
Common mistake: Using the diameter instead of the radius in πr² gives four times the true volume.
Source: Worked calculation from the data given in the question
Question 18
M18 · 1K Thermowell design · Multiple choice
A wake-frequency check shows a thermowell's vortex-shedding frequency approaching its natural frequency at high line velocity. Which change is most effective?
Show explanation
Answer and explanation
Answer: C. Shorten the insertion length (a shorter, stiffer well)
Trouble starts when the shedding (wake) frequency matches the well's natural frequency. A shorter, stiffer well raises the natural frequency and moves it away from the shedding frequency. Shortening the insertion length is a standard fix, along with a larger root diameter.
Why not the others: A longer or thinner well lowers stiffness and moves it toward resonance. Changing the sensor inside doesn't change the well's mechanics.
Source: Ashcroft, Thermowell wake frequency, fixes: shorten insertion length, increase root diameter; Control & Instrumentation, Thermowell wake frequency, why wake frequency is a concern
Question 19
M19 · 1L Sensor testing and commissioning · Numeric entry
A pressure transmitter ranged 0–200 psig = 4–20 mA reads 12.32 mA with 100.0 psig applied. What is the error as a percent of span? (Enter a positive number if the output reads high.)
Show explanation
Answer and explanation
Answer: 2 % of span (accepted range 1.95 to 2.05)
At 100 psig (50% of range) the output should be 4 + 0.5 × 16 = 12.00 mA. The error is 0.32 mA ÷ 16 mA span = +2.0% of span.
Common mistake: Dividing by 20 mA instead of the 16 mA span gives 1.6%, a common mistake.
Source: Worked calculation from the data given in the question
Control Systems (Questions 20–40)
Question 20
C01 · 2A Drawings, symbols, and terminology · Multiple choice
On an instrumentation diagram, a tag marked FC denotes which function?
Show explanation
Answer and explanation
Answer: A. Flow controller
In common P&ID instrument identification, F denotes flow and C denotes controller. The U.S. Department of Energy handbook lists FC as a flow controller.
Why not the others: A recorder uses R, so a flow recorder is FR. The cited identifier table does not use FC for a converter or a frequency controller.
Source: U.S. Department of Energy, DOE Fundamentals Handbook: Engineering Symbology, Prints, and Drawings, Vol. 1, Engineering Fluid Diagrams and Prints, Table 1 Instrument Identifiers, PDF pp. 60–61
Question 21
C02 · 2A Drawings, symbols, and terminology · Matching
Match each common instrument tag to its function.
- FI — ______
- TI — ______
- PT — ______
- TR — ______
Show explanation
Answer and explanation
Answer: FI = Flow indicator; TI = Temperature indicator; PT = Pressure transmitter; TR = Temperature recorder
The first letter identifies the measured variable and the following letter identifies the instrument function: I for indicator, T for transmitter, and R for recorder. The DOE handbook gives FI, TI, PT, and TR as examples.
Source: U.S. Department of Energy, DOE Fundamentals Handbook: Engineering Symbology, Prints, and Drawings, Vol. 1, Engineering Fluid Diagrams and Prints, Table 1 Instrument Identifiers, PDF pp. 60–61
Question 22
C03 · 2A Drawings, symbols, and terminology · Multiple correct
Use this cause-and-effect matrix (X = action). LSHH-101 trips. Which TWO effects occur?
| Cause ↓ / Effect → | Close XV-201 (inlet) | Stop P-101 (feed pump) | Open XV-301 (vent to flare) |
|---|---|---|---|
| LSHH-101, V-101 level high-high | X | X | |
| PSHH-102, V-101 pressure high-high | X | X |
Show explanation
Answer and explanation
Answer: A. Close XV-201 and B. Stop P-101
Read across the LSHH-101 row: it closes XV-201 and stops P-101.
Why not the others: XV-301 belongs only to PSHH-102. Nothing in the matrix says the two causes must vote together, so LSHH-101 acts on its own.
Source: Read directly from the information given in the question
Question 23
C04 · 2B Control theory and process dynamics · Multiple choice
A proportional-only temperature loop settles 3 °F below setpoint after a load change. What eliminates this steady-state offset?
Show explanation
Answer and explanation
Answer: D. Add integral (reset) action
With proportional-only control, a load change leaves a lasting offset between PV and SP. Integral action keeps moving the output for as long as error remains, which removes the offset.
Why not the others: More gain shrinks the offset but can't eliminate it, and it pushes the loop toward oscillation. Derivative responds only to rate of change. A deadband adds error.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Proportional-only offset, offset and integral action
Question 24
C05 · 2B Control theory and process dynamics · Multiple choice
When commissioning a reactor-temperature (primary) to jacket-temperature (secondary) cascade, what is the correct tuning sequence?
Show explanation
Answer and explanation
Answer: B. Tune the secondary first, then tune the primary with the secondary in cascade
Cascade control works only when the inner (secondary) loop responds faster than the outer one, so the secondary is tuned first. Then the primary is tuned against a well-behaved inner loop.
Why not the others: Tuning the primary first gives settings that change once the inner loop closes. A secondary left in manual means there is no cascade at all.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Cascade control, inner-loop speed; tune the slave first
Question 25
C06 · 2B Control theory and process dynamics · Multiple choice
A heat exchanger's outlet temperature swings with frequent, measurable changes in feed flow. What is the best addition to the control strategy?
Show explanation
Answer and explanation
Answer: C. Feedforward from feed flow, combined with feedback trim
Feedforward uses a load sensor (here, feed flow) to act before the disturbance reaches the outlet. Feedback trim corrects whatever the feedforward model misses.
Why not the others: Slower tuning makes the swings last longer. Split range shares one output between two final elements and does nothing for a disturbance. Steam-to-condensate ratio isn't a meaningful control variable here.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Feedforward control, load compensation; feedforward with trim
Question 26
C07 · 2B Control theory and process dynamics · Numeric entry
A step in controller output from 40% to 50% moves the PV from 150 °F to 170 °F. The transmitter range is 100–300 °F. A first-order-plus-dead-time fit gives τ = 4 min and θ = 1 min. Using lambda tuning for a PI controller, Kc = τ ÷ [Kp(λ + θ)] and Ti = τ, with λ = 3 min, find Kc in %/%.
Show explanation
Answer and explanation
Answer: 1 %/% (accepted range 0.98 to 1.02)
Express the process gain in percent of span: the PV change is 20 °F ÷ 200 °F = 10% of span, so Kp = 10% ÷ 10% = 1.0. Then Kc = 4 ÷ [1.0 × (3 + 1)] = 1.0, and Ti = 4 min.
Common mistake: Using engineering units (2 °F per % output) for Kp gives Kc = 0.5, which is dimensionally wrong for a controller working in percent.
Source: Worked calculation from the data given in the question; ISA InTech, Loop tuning basics: self-regulating processes (2016), lambda tuning equations
Question 27
C08 · 2B Control theory and process dynamics · Multiple choice
Which process is integrating (non-self-regulating)?
Show explanation
Answer and explanation
Answer: A. Liquid level in a tank drained at a constant rate by a positive-displacement pump
With a constant draw out of the tank, the level holds steady only when inflow exactly matches outflow. Any mismatch makes the level ramp up or down with no natural limit.
Why not the others: The other three settle at a new steady state after a change, because the process pushes back on its own. That is self-regulation.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Process characteristics, Integrating processes
Question 28
C09 · 2B Control theory and process dynamics · Numeric entry
In a ratio control scheme, the wild stream is 300 gpm and the desired controlled-to-wild ratio is 0.25. What setpoint does the controlled-flow controller receive, in gpm?
Show explanation
Answer and explanation
Answer: 75 gpm (accepted range 74.5 to 75.5)
The ratio relay multiplies the wild flow by the ratio constant: 300 × 0.25 = 75 gpm.
Common mistake: Dividing instead (300 ÷ 0.25 = 1,200 gpm) inverts the ratio.
Source: Worked calculation from the data given in the question; Kuphaldt, Lessons in Industrial Instrumentation: Ratio control, Ratio control
Question 29
C10 · 2B Control theory and process dynamics · Multiple choice
As a centrifugal compressor's operating point approaches its surge line, what does the anti-surge controller do?
Show explanation
Answer and explanation
Answer: D. Opens the recycle (anti-surge) valve to keep flow above the surge limit
Opening the recycle valve raises flow through the machine, which holds the operating point away from the surge line. Controllers typically act with a margin of about 10%.
Why not the others: Closing the recycle valve cuts flow and pushes the machine toward surge. Tripping is a last-resort protection, not the control action.
Source: Turbomachinery Magazine, Five requirements for successful surge avoidance (2019), anti-surge recycle valve
Question 30
C11 · 2B Control theory and process dynamics · Multiple choice
In cross-limited combustion control, which describes the response to changes in firing rate?
Show explanation
Answer and explanation
Answer: B. On an increase air leads fuel; on a decrease fuel leads air
High and low selects make air rise before fuel on an increase and fuel fall before air on a decrease. The burner briefly gets extra air rather than extra fuel, which avoids a fuel-rich condition.
Why not the others: Letting fuel lead on an increase, or moving both with no limits, allows fuel-rich transients. If air led on a decrease too, it would fall before fuel, which is exactly the fuel-rich case cross-limiting prevents.
Source: Control Guru, Ratio with cross-limiting override control of a combustion process, cross-limiting high and low selects
Question 31
C12 · 2C Human factors and alarm management · Multiple choice
An operator console averages 18 annunciated alarms per hour over 30 days. What is the best assessment and first action?
Show explanation
Answer and explanation
Answer: C. Above the roughly 12 per hour maximum-manageable benchmark; find the worst-offending alarms and rationalize them
ISA-18.2 guidance puts about 6 alarms per hour as very likely acceptable and about 12 per hour as the most an operator can manage. At 18, the right move is bad-actor analysis and rationalization, which fixes the alarms that cause most of the load.
Why not the others: Blanket priority changes or mass disabling skip the rationalization process and can hide alarms that matter.
Source: Tennant, Implementing Alarm Management per the ANSI/ISA-18.2 Standard, Control Engineering (2013), via Yokogawa, alarm-rate limits table
Question 32
C13 · 2C Human factors and alarm management · Multiple choice
Which design choice best fits a high-performance HMI display?
Show explanation
Answer and explanation
Answer: A. Gray backgrounds, with strong color reserved for abnormal and alarm conditions
Modern HMI practice uses gray backgrounds and objects, and uses color sparingly, only where attention is needed. Exceptions then stand out at a glance.
Why not the others: The other options spend color and visual detail on normal states, which buries the abnormal ones.
Source: Control.com, Going gray: a new HMI standard (2020), gray displays; color only where attention is needed
Question 33
C14 · 2D Logic implementation and troubleshooting · Multiple choice
In a three-wire start/stop motor control circuit, the motor runs only while START is held down. What is missing?
Show explanation
Answer and explanation
Answer: D. A seal-in (holding) auxiliary contact wired in parallel with START
The starter's auxiliary contact, wired in parallel with START, keeps the coil energized after START is released. STOP, a normally closed button, breaks that seal.
Why not the others: The STOP button and overloads are in series to stop the motor; they don't hold it on. A second START button in series would make starting harder, not hold the motor.
Source: Kuphaldt, Lessons in Industrial Instrumentation: On/off electric motor control circuits, Motor control circuit wiring
Question 34
C15 · 2D Logic implementation and troubleshooting · Multiple choice
A pump's low-suction-pressure trip must be ignored for 10 s after the run command, then armed. Which timer should arm it?
Show explanation
Answer and explanation
Answer: B. On-delay timer (TON) started by the run command
An on-delay timer turns its output on only after its input has been true for the preset time, and it resets when the input drops. That arms the trip 10 s after each start.
Why not the others: An off-delay timer times after the input goes false. A retentive timer keeps accumulated time across stops, so it could arm the trip too early on a restart. A counter counts events, not time.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Ladder diagram structure and commands, On-delay, off-delay, and retentive timers
Question 35
C16 · 2D Logic implementation and troubleshooting · Multiple choice
In a sequential function chart (SFC), the step after a transition becomes active when:
Show explanation
Answer and explanation
Answer: C. The preceding step is active and the transition condition is true
A transition fires only when its preceding step is active and its condition is true. That is what keeps a sequence moving in order.
Why not the others: The condition alone isn't enough, because the transition must be enabled by its preceding step. Scan timing and the state of unrelated steps don't control transitions.
Source: Industrial Monitor Direct, Sequential function chart PLC programming guide, State activation rules
Question 36
C17 · 2D Logic implementation and troubleshooting · Multiple choice
This structured text line scales a raw value: PV := Raw / 32000 * 100; All variables are INT. With Raw = 24000, PV comes out 0. Why, and what fixes it?
Show explanation
Answer and explanation
Answer: A. Integer division truncates 24000/32000 to 0; multiply first (with a wide enough type) or use REAL math
Integer division truncates toward zero, so 24000 / 32000 = 0, and 0 × 100 = 0. Writing Raw * 100 / 32000 with a type wide enough for the intermediate value (DINT), or converting to REAL, gives 75.
Why not the others: Overflow alone doesn't produce exactly 0 here. Scan time and sign errors don't explain a result that is always zero below full scale.
Source: Neuron Automation, DIV block (IEC 61131-3 implementation), integer division truncates toward zero
Question 37
C18 · 2E System design · Numeric entry
A controller has MTBF = 50,000 h and MTTR = 8 h. Using A = MTBF ÷ (MTBF + MTTR), what is its inherent availability to five decimal places?
Show explanation
Answer and explanation
Answer: 0.99984 (accepted range 0.99983 to 0.99985)
A = 50,000 ÷ 50,008 = 0.99984.
Common mistake: Subtracting MTTR from MTBF in the denominator gives 1.00016, which is impossible for an availability.
Source: Worked calculation from the data given in the question
Question 38
C19 · 2E System design · Numeric entry
Equipment in a control cabinet dissipates 1,200 W. What is the heat load in BTU/h? (1 W = 3.412 BTU/h)
Show explanation
Answer and explanation
Answer: 4,094 BTU/h (accepted range 4,090 to 4,100)
1,200 × 3.412 = 4,094 BTU/h. This is the number a cabinet cooler or ventilation design must handle.
Common mistake: Dividing by 3.412 instead of multiplying gives 352 BTU/h, about one-twelfth of the real load.
Source: Worked calculation from the data given in the question
Question 39
C20 · 2F Testing, commissioning, and performance monitoring · Ordering
Put these project steps in their normal order.
- Site acceptance test (SAT)
- Factory acceptance test (FAT)
- Installation and loop checks
- Startup and handover
Show explanation
Answer and explanation
Answer: Factory acceptance test (FAT) → Installation and loop checks → Site acceptance test (SAT) → Startup and handover
The FAT happens at the vendor's facility with the system built but not yet shipped. The SAT happens at the plant after installation, and it comes before handover.
Common mistake: Running the SAT before installation is impossible, because there is nothing installed to test.
Source: Fabrico, Factory acceptance test vs. site acceptance test, FAT and SAT definitions
Question 40
C21 · 2G Control system cybersecurity · Multiple correct
A vendor needs remote access to a DCS engineering workstation. Which TWO measures match OT security guidance?
Show explanation
Answer and explanation
Answer: B. Use a DMZ so there is no direct traffic between the corporate and OT networks and D. Require multifactor authentication for remote access to the OT system
NIST's OT security guide recommends a DMZ architecture that blocks direct corporate-to-OT traffic, and multifactor authentication for remote access. CISA lists MFA for remote access among its primary OT mitigations.
Why not the others: Port forwarding creates exactly the direct path a DMZ exists to prevent. A shared password removes accountability for who did what.
Source: NIST SP 800-82 Rev. 3, Guide to Operational Technology Security (2023), Executive Summary; §5.2.3; §6.2.10; CISA, Primary mitigations to reduce cyber threats to operational technology, remote access MFA; segmentation
Final Control Elements (Questions 41–57)
Question 41
V01 · 3A Valve type and material selection · Multiple choice
Which construction feature does a control valve in LNG (cryogenic) service need?
Show explanation
Answer and explanation
Answer: D. An extended bonnet that keeps the stem packing near ambient temperature
The extension moves the packing away from the cold liquid. A column of warmer vapor in the extension keeps the stem seals near ambient temperature, where they still seal. That is also why manufacturers warn against insulating the extension.
Why not the others: A standard bonnet puts the packing at process temperature, where it can freeze. A valve stem still needs a stem seal of some kind.
Source: Flowserve, cryogenic extended bonnet ball valve manual, stem seals kept near ambient temperature; Emerson, Fisher EZ-C/ET-C/EWT-C cryogenic control valves instruction manual, installation notice on the extension bonnet
Question 42
V02 · 3B Valve trim · Multiple choice
At maximum flow, a control valve takes only a small fraction of the total system pressure drop. Which inherent characteristic usually fits best?
Show explanation
Answer and explanation
Answer: B. Equal-percentage
When line losses take most of the pressure drop, the valve's share shrinks as flow rises. Equal-percentage compensates and gives a more linear installed response. Linear suits a valve whose pressure drop stays nearly constant.
Why not the others: Quick-opening suits on/off service. The choice matters because installed gain decides how the loop behaves across its range.
Source: ISA Ask the Automation Pros: control valve characteristics (Mar 2025), characteristic selection
Question 43
V03 · 3B Valve trim · Multiple choice
Under ANSI/FCI 70-2, which seat-leakage class uses an air or nitrogen test and a maximum allowable leakage rate specified by port size in mL/min or bubbles per minute?
Show explanation
Answer and explanation
Answer: D. Class VI
Class VI uses air or nitrogen and limits leakage by nominal port size, with values stated in mL/min and commonly expressed as bubbles per minute. It is a very tight leakage class, but it is not literally a zero-leakage definition.
Why not the others: Classes II and IV are expressed as percentages of rated capacity. Class V uses a water test with leakage tied to port diameter and pressure differential.
Source: Emerson/Fisher, Control Valve Handbook, Section 5.2, Control Valve Seat Leakage Classifications
Question 44
V04 · 3B Valve trim · Multiple choice
Which option reduces aerodynamic valve noise at the source?
Show explanation
Answer and explanation
Answer: C. Noise-attenuating (multi-stage, multi-hole) valve trim
Noise-attenuating trims are source treatment: they reduce the noise the valve generates in the first place.
Why not the others: Insulation, thicker pipe, and silencers are path treatments. They soak up noise after the valve has already created it.
Source: Emerson, Control valve noise reduction, source vs. path treatment
Question 45
V05 · 3C Valve calculations · Numeric entry
Liquid service: Q = 200 gpm, SG = 0.81, ΔP = 16 psi. Using Cv = Q × √(SG ÷ ΔP) and ignoring piping geometry factors, what is the required Cv?
Show explanation
Answer and explanation
Answer: 45 (accepted range 44.5 to 45.5)
Cv = 200 × √(0.81 ÷ 16) = 200 × 0.225 = 45.
Common mistake: Leaving out specific gravity gives 50. Inverting the ratio inside the root gives about 889.
Source: Worked calculation from the data given in the question
Question 46
V06 · 3C Valve calculations · Numeric entry
In a split-range design, a steam valve goes linearly from 0% to 100% open as the controller output goes from 50% to 100%. How far open is the steam valve at 75% output?
Show explanation
Answer and explanation
Answer: 50 % open (accepted range 49.5 to 50.5)
The valve's share of the range is (75 − 50) ÷ (100 − 50) = 50% open.
Common mistake: Reading the controller output straight across gives 75%, which ignores the split.
Source: Worked calculation from the data given in the question
Question 47
V07 · 3D Actuators and failure modes · Multiple choice
What failure action should the fuel-gas control valve to a fired heater have?
Show explanation
Answer and explanation
Answer: A. Fail-closed (air-to-open, spring to close)
Failure mode follows process risk. Losing air or signal should cut the fuel, so the valve is air-to-open and a spring closes it. Burner shutoff valves are designed the same way, closing by spring on power loss.
Why not the others: Fail-open or fail-in-place keeps feeding fuel to a heater that may have lost its flame.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Valve failure mode, Selecting the proper failure mode; double-acting piston actuators; Kuphaldt, Lessons in Industrial Instrumentation: Safety instrumented functions and systems, SIS final control elements; burner management example
Question 48
V08 · 3D Actuators and failure modes · Multiple choice
A double-acting piston actuator with no spring loses its supply air. What is true about its failure behavior?
Show explanation
Answer and explanation
Answer: D. It has no spring-defined fail position; a defined fail-open or fail-closed action needs stored air (a volume tank) with a trip valve
With no spring, nothing drives the piston to a safe end, so it tends to fail in place. Trip-valve packages give a defined fail-up or fail-down action using a volume tank and check valve.
Why not the others: 'Always open' or 'always closed' needs a spring or stored energy. Mid-travel is not a fail mode.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Valve failure mode, Selecting the proper failure mode; double-acting piston actuators; Emerson, Fisher 377 trip valve instruction manual (2022), Table 1; volume tank for fail-up/fail-down
Question 49
V09 · 3E Fluid dynamics effects · Multiple choice
Liquid flows through a control valve with P1 = 100 psia, P2 = 8 psia, and a vapor pressure of 12 psia. What will happen?
Show explanation
Answer and explanation
Answer: B. Flashing
Downstream pressure is below the vapor pressure, so the vapor that forms stays vapor. That is flashing.
Why not the others: Cavitation is when the vena contracta pressure drops below vapor pressure and then the downstream pressure recovers above it, collapsing the bubbles. 'Choked gas flow' doesn't apply to a liquid service.
Source: Valmet, Flow control manual, ch. 3 Liquid flow, §3.3 choked flow; §3.3.1 cavitation; §3.3.3 flashing
Question 50
V10 · 3E Fluid dynamics effects · Multiple choice
Liquid flow through a valve is choked. What happens if the downstream pressure is lowered further?
Show explanation
Answer and explanation
Answer: C. Flow does not increase; size the valve using the choked pressure drop
Once flow is choked, lowering the downstream pressure no longer raises the flow rate. Sizing with the full ΔP would overstate the valve's capacity.
Why not the others: The √ΔP relationship only holds before choking. Flow doesn't stop, and lowering P2 doesn't cure the flashing or cavitation behind choked flow.
Source: Valmet, Flow control manual, ch. 3 Liquid flow, §3.3 choked flow; §3.3.1 cavitation; §3.3.3 flashing
Question 51
V11 · 3F Valve accessories · Multiple choice
A large spring-return actuator must stroke to its fail position faster on a trip. Which accessory fits best?
Show explanation
Answer and explanation
Answer: A. A quick exhaust valve that vents actuator air directly to atmosphere
A quick exhaust valve vents the actuator directly, bypassing the slow path back through the positioner or solenoid. That speeds the spring stroke.
Why not the others: A volume booster speeds strokes driven by the positioner signal, not a trip vent. Positioner gain and a regulator don't change how fast the actuator dumps air.
Source: IMI, SR quick exhaust valve, product description
Question 52
V12 · 3N Self-regulating devices · Multiple choice
As flow through a direct-operated (self-operated) pressure regulator increases, what happens to downstream pressure?
Show explanation
Answer and explanation
Answer: D. It falls below setpoint (droop)
The spring has to relax further to open the valve wider, so downstream pressure sags as flow rises. That is droop. Pilot-operated regulators have much less droop.
Why not the others: A simple spring-loaded regulator works like a proportional-only controller: it holds setpoint at only one flow rate.
Source: Chemical Processing, Pressure valves: dealing with droop (2002), self-operated vs. pilot-operated droop; Emerson, Fisher Introduction to Regulators, accuracy required
Question 53
V13 · 3H Pressure relief device types · Multiple choice
A spring-loaded relief valve discharges into a flare header whose back pressure varies widely. Which valve type fits best?
Show explanation
Answer and explanation
Answer: B. Balanced-bellows valve
Superimposed back pressure acts on a conventional valve's disk and shifts the pressure at which it opens. A balanced bellows cancels most of that effect, so it suits headers with variable back pressure.
Why not the others: A conventional valve's set point drifts with back pressure. A rupture disk alone can't reclose. A weighted-lever valve doesn't address back pressure.
Source: Control Global, Ask the experts: relief valves and back pressure (2022), balanced bellows for variable back pressure
Question 54
V14 · 3I Pressure relief device calculations · Multiple correct
A relief valve set at 150 psig has a 6 psi non-recoverable inlet pressure loss (4% of set). That exceeds the common 3% design limit. Which TWO changes address it?
Show explanation
Answer and explanation
Answer: A. Use larger, shorter inlet piping and C. Use a pilot-operated relief valve with a remote-sensing pilot
Inlet loss rises with pipe length and velocity, so larger, shorter piping cuts it. A remote-sensing pilot reads pressure at the vessel instead of at the valve inlet, which is the standard remedy when inlet losses are large.
Why not the others: Lowering the set pressure changes the protection without fixing the loss. A reducer raises velocity and makes the loss worse.
Source: Dresser, F70PR pilot-operated relief valve manual, Rev M, §2.3 Installation, p. 5
Question 55
V15 · 3J Motor control · Numeric entry
A VFD slows a centrifugal pump so flow drops from 400 gpm to 300 gpm. Assume no static head and constant efficiency. What is the new shaft power as a percent of the original?
Show explanation
Answer and explanation
Answer: 42 % of original power (accepted range 41 to 43)
By the affinity laws, flow is proportional to speed and power to speed cubed. Speed ratio = 0.75, so power = 0.75³ = 0.42, or 42%.
Common mistake: Using a squared relationship gives 56%, which is the head ratio. When static head dominates, the affinity laws don't predict the savings and the real reduction is smaller.
Source: Worked calculation from the data given in the question; Engineering ToolBox, Pump affinity laws, changing wheel velocity; MechChem Africa, Pumping systems 101: advantages and pitfalls of VSDs, static head caveat
Question 56
V16 · 3J Motor control · Multiple choice
A pump needs continuous flow adjustment by speed. Which motor control should you choose?
Show explanation
Answer and explanation
Answer: C. Variable-frequency drive
A VFD controls speed throughout operation. A soft starter only manages starting and stopping, then runs the motor at full speed.
Why not the others: Across-the-line and reversing starters give full-speed on/off (and direction) control only.
Source: Control.com, The difference between variable frequency drives and soft starters, soft starter vs. VFD
Question 57
V17 · 3L Solenoid valves · Multiple choice
A solenoid valve vents an actuator to drive its valve to the safe position. For the system to fail safe on loss of power, the solenoid should normally be:
Show explanation
Answer and explanation
Answer: A. Energized, and de-energized to vent
A trip solenoid that vents when de-energized drives the valve to its safe position on a trip and on a power loss alike. That means it must be held energized in normal operation.
Why not the others: Energize-to-vent fails to act when power is lost. A latching solenoid holds its last state on power loss.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Safety instrumented functions and systems, SIS final control elements; burner management example
Signals, Transmission, and Networking (Questions 58–70)
Question 58
S01 · 4A Signal types and technologies · Multiple choice
Which statement about HART communication is correct?
Show explanation
Answer and explanation
Answer: D. It superimposes 1200/2200 Hz FSK on the 4–20 mA loop at 1200 bit/s, so analog and digital signals coexist
HART sends a 1200 Hz tone for a 1 and a 2200 Hz tone for a 0, at 1200 baud, on top of the loop current. The analog signal stays below 20 Hz, so the two don't interfere.
Why not the others: The analog current still carries the primary variable, on the same two wires. 31.25 kbit/s is not HART's rate.
Source: Texas Instruments, SLAAER4 HART-Enabled Transmitter (2025), §1.2.2 HART FSK
Question 59
S02 · 4A Signal types and technologies · Multiple choice
An analog input on a 4–20 mA loop reads 0.0 mA. What is the most likely cause?
Show explanation
Answer and explanation
Answer: B. An open circuit or loss of loop power
With a live zero, 0% of range is 4 mA, not 0 mA. A reading near zero therefore points to a failed signal, such as a cut cable or dead loop supply.
Why not the others: 0% would read 4 mA. Over-range reads above 20 mA. Calibration drift shifts readings slightly; it doesn't kill the current.
Source: Kuphaldt, Lessons in Industrial Instrumentation: 4 to 20 mA analog current signals, live zero
Question 60
S03 · 4B Signal conversion · Numeric entry
A 12-bit ADC spans 0–20 mA (0–4095 counts). It reads a 4–20 mA transmitter ranged 0–500 °F. What is the smallest temperature change the input can resolve, in °F?
Show explanation
Answer and explanation
Answer: 0.153 °F (accepted range 0.15 to 0.16)
One count = 20 ÷ 4095 = 0.00488 mA. The transmitter scale is 500 °F ÷ 16 mA = 31.25 °F per mA. Resolution = 0.00488 × 31.25 = 0.153 °F.
Common mistake: Treating the full 0–20 mA as 0–500 °F gives 0.122 °F, but the transmitter only uses 16 mA of the ADC's span.
Source: Worked calculation from the data given in the question
Question 61
S04 · 4B Signal conversion · Numeric entry
A direct-acting I/P transducer converts 4–20 mA to 3–15 psig. What is the output pressure at 12 mA?
Show explanation
Answer and explanation
Answer: 9 psig (accepted range 8.95 to 9.05)
12 mA is 50% of the 16 mA span, so the output is 3 + 0.5 × 12 = 9 psig.
Common mistake: Taking 50% of 15 psig (7.5) forgets the 3 psig live zero.
Source: Worked calculation from the data given in the question; Kuphaldt, Lessons in Industrial Instrumentation: Relating 4 to 20 mA signals to instrument variables, I/P transducer example
Question 62
S05 · 4C Hazardous area classification and installation · Numeric entry
Intrinsic safety entity check: the barrier allows Ca = 0.13 µF and La = 4.2 mH. The field transmitter has Ci = 0.01 µF and Li = 0. The cable is 60 pF/ft and 0.2 µH/ft. Voltage and current parameters are already compatible. What is the maximum cable length in feet?
Show explanation
Answer and explanation
Answer: 2,000 ft (accepted range 1,990 to 2,010)
Capacitance sets the limit: Ci + cable capacitance must not exceed Ca, so cable C ≤ 0.12 µF. 0.12 µF ÷ 60 pF/ft = 2,000 ft. Inductance check: 2,000 × 0.2 µH = 0.4 mH, which is under 4.2 mH, so it passes.
Common mistake: Ignoring the transmitter's own Ci gives about 2,167 ft, which would overload the barrier's capacitance rating.
Source: Worked calculation from the data given in the question; Pepperl+Fuchs, control drawing 116-0364, transformer isolated barriers, entity parameter rules
Question 63
S06 · 4C Hazardous area classification and installation · Multiple choice
What does Type X pressurization of an enclosure in a Class I, Division 1 location do?
Show explanation
Answer and explanation
Answer: C. Reduces Division 1 to nonhazardous inside the enclosure
Type X takes the inside of the enclosure from Division 1 to nonhazardous. Type Y reduces Division 1 to Division 2, and Type Z reduces Division 2 to nonhazardous.
Why not the others: Reducing Division 2 to nonhazardous is Type Z, and reducing Division 1 to Division 2 is Type Y. Purging is a different protection method from explosion-proof construction.
Source: Pepperl+Fuchs, Purge and pressurization systems, Type X, Y, Z
Question 64
S07 · 4D Power, grounding, and EMI · Multiple choice
To avoid ground loops, where is the shield of a twisted-pair analog signal cable normally grounded?
Show explanation
Answer and explanation
Answer: A. At one end only, usually the power supply or control-system end
Grounding the shield at one point keeps ground-potential differences from driving current through it. Transmitter manuals call for grounding it at the power-supply end only.
Why not the others: Grounding at two or more points creates exactly the ground loop you're trying to avoid. An ungrounded shield doesn't drain interference.
Source: Emerson, Rosemount power and transient protection best practices (2015), §3.1 p. 3; checklist p. 14; Emerson, Rosemount 3144P quick start guide (2024), p. 9 shield grounding
Question 65
S08 · 4D Power, grounding, and EMI · Numeric entry
A 4–20 mA loop has a 24 VDC supply, a 250 Ω sense resistor, a 340 Ω IS barrier, and 20 Ω of round-trip cable. The transmitter needs at least 12 V at its terminals. What is the terminal voltage at 20 mA?
Show explanation
Answer and explanation
Answer: 11.8 V (accepted range 11.75 to 11.85)
Total external resistance = 250 + 340 + 20 = 610 Ω. Terminal voltage = 24 − 0.020 × 610 = 11.8 V. That is below 12 V, so the loop can't reach full scale. The most the loop can carry is (24 − 12) ÷ 0.020 = 600 Ω.
Common mistake: Checking at 4 mA hides the problem. Voltage drop is worst at 20 mA.
Source: Worked calculation from the data given in the question
Question 66
S09 · 4D Power, grounding, and EMI · Numeric entry
A UPS must carry a 2,000 W load at 0.8 power factor. What is the minimum VA rating?
Show explanation
Answer and explanation
Answer: 2,500 VA (accepted range 2,499 to 2,501)
Apparent power = real power ÷ power factor = 2,000 ÷ 0.8 = 2,500 VA.
Common mistake: Multiplying by the power factor gives 1,600 VA, which undersizes the UPS.
Source: Worked calculation from the data given in the question
Question 67
S10 · 4E Signal circuit design · Multiple choice
A four-wire transmitter has its own power supply and an active (current-sourcing) 4–20 mA output. What kind of analog input should it connect to?
Show explanation
Answer and explanation
Answer: D. A passive input that does not supply loop power
An active output already drives the loop current. It needs a receiver that only measures. A loop-powering input is meant for two-wire transmitters that draw their power from the loop.
Why not the others: Connecting two sources puts supplies in conflict and gives a wrong reading. A discrete input can't read an analog value.
Source: Kuphaldt, Lessons in Industrial Instrumentation: 4-wire passive versus active output transmitters, active vs. passive outputs
Question 68
S11 · 4F Communications architecture and protocols · Multiple choice
A PLC at 192.168.10.5/24 cannot reach an HMI at 192.168.11.5/24. Both are on the same switch, and there is no router. What is the cause?
Show explanation
Answer and explanation
Answer: B. They are on different subnets, and nothing routes between them
With a /24 mask, the first three octets identify the network. 192.168.10 and 192.168.11 are different networks, and devices on different networks need a router to communicate.
Why not the others: The addresses aren't duplicates. MAC addresses are learned automatically. Cable length would cause link errors on both devices, not a clean subnet failure.
Source: Kuphaldt, Lessons in Industrial Instrumentation: Internet Protocol (IP), subnet masks; University of Tartu, Network segmentation (course module), routers between networks
Question 69
S12 · 4F Communications architecture and protocols · Multiple choice
A network link must run 2,000 ft between two plant areas, through heavy EMI. The areas sit at different ground potentials. Which medium fits best?
Show explanation
Answer and explanation
Answer: C. Fiber optic cable
Fiber is immune to electromagnetic interference and electrically isolates the two areas, which removes ground-current problems. It also covers the distance in one run.
Why not the others: A twisted-pair copper Ethernet segment is limited to 100 m (328 ft). Shielding helps with EMI but doesn't fix the distance or the ground-potential difference. Coax still bonds the two grounds.
Source: PI North America, PROFINET cables: fiber optics, EMI immunity; electrical isolation; Cisco, Ethernet 100BaseTX and 10BaseT cables: guidelines and specifications, specifications table
Question 70
S13 · 4F Communications architecture and protocols · Multiple choice
What is the registered TCP port for Modbus TCP?
Show explanation
Answer and explanation
Answer: C. 502
IANA registers port 502 as mbap, the Modbus Application Protocol. A firewall rule for Modbus TCP traffic uses this port.
Why not the others: Port 80 is HTTP, 102 is ISO-TSAP, and 44818 is registered for EtherNet/IP messaging.
Source: IANA, Service name and transport protocol port number registry, port 502 (mbap)
Safety Instrumented Systems (Questions 71–85)
Question 71
F01 · 5A Safety life cycle · Ordering
Put these safety life-cycle phases in order.
- Safety requirements specification (SRS)
- Hazard and risk assessment
- Operation and maintenance
- Allocation of safety functions to protection layers
- Design and engineering
- Installation, commissioning, and validation
Show explanation
Answer and explanation
Answer: Hazard and risk assessment → Allocation of safety functions to protection layers → Safety requirements specification (SRS) → Design and engineering → Installation, commissioning, and validation → Operation and maintenance
IEC 61511-1 runs in this order: hazard and risk assessment and allocation (clauses 8–9), then the SRS (clause 10), then design through validation (clauses 11–15), then operation and maintenance (clause 16). Modification is clause 17.
Common mistake: You can't write requirements before you know what risk the function has to reduce, and you can't design before the requirements exist.
Source: Singapore Ministry of Manpower, Guidelines on Safety Instrumented Systems in Major Hazards Installations (Rev 0, 2020), §3.2 (BPCS credit), §4.5, §5.2, §6.3
Question 72
F02 · 5C Safety requirements specification · Multiple choice
Which document defines each safety instrumented function's required behavior and integrity before detailed design starts?
Show explanation
Answer and explanation
Answer: A. Safety requirements specification (SRS)
The SRS is the requirements document the SIS is designed to. It covers safe states, trip points, response times, proof-test intervals, and more.
Why not the others: A P&ID shows the instruments but not integrity requirements. The FAT procedure comes later and tests against the requirements. A cause-and-effect matrix can be part of an SRS, but it doesn't capture integrity, timing, or test requirements.
Source: ABB, Safety Requirements Specification factsheet (2012), SRS content list; Singapore Ministry of Manpower, Guidelines on Safety Instrumented Systems in Major Hazards Installations (Rev 0, 2020), §3.2 (BPCS credit), §4.5, §5.2, §6.3
Question 73
F03 · 5B Hazard and risk assessment and allocation · Multiple choice
A LOPA has an initiating event frequency of 0.1/yr. Credited independent layers: a BPCS loop with PFD = 0.1 and a relief valve with PFD = 0.01. The tolerable frequency is 5 × 10⁻⁶/yr. What SIF is needed?
Show explanation
Answer and explanation
Answer: D. SIL 1 (PFD ≤ 0.05)
Mitigated frequency without a SIF = 0.1 × 0.1 × 0.01 = 1 × 10⁻⁴/yr. Required SIF PFD = 5 × 10⁻⁶ ÷ 1 × 10⁻⁴ = 0.05. That falls in the SIL 1 band (from 0.01 up to, but not including, 0.1).
Why not the others: 'None' leaves the risk 20 times above tolerable. SIL 2 or SIL 3 would work but over-specify the function.
Source: Worked calculation from the data given in the question; Pepperl+Fuchs, Functional Safety in the Process Industry (2018), SIL/PFDavg table, hardware fault tolerance table, PFDavg formulas
Question 74
F04 · 5B Hazard and risk assessment and allocation · Multiple choice
In a LOPA, the initiating cause is failure of a BPCS level loop. An operator high-level alarm is generated by that same loop. How should the alarm be treated?
Show explanation
Answer and explanation
Answer: B. As not independent of the initiating cause, so it gets no IPL credit
LOPA requires independence between the initiating cause and each IPL. An alarm from the loop whose failure starts the event can be defeated by the same failure, so no credit is taken.
Why not the others: An alarm and operator response is at most one layer, and only when it is independent. It is not a SIL-rated function.
Source: Beedle, Calculating yourself into a corner using LOPA, IChemE Hazards 30 (2020), independence of layers, p. 5
Question 75
F05 · 5B Hazard and risk assessment and allocation · Multiple choice
A BPCS protection layer that is not designed to IEC 61511 may be credited with a risk reduction of at most:
Show explanation
Answer and explanation
Answer: A. 10
Credit for the BPCS is capped at a risk reduction of 10 unless it also meets higher IEC 61511 requirements.
Why not the others: Claiming 100 or more puts the BPCS in SIS territory without the SIS design discipline behind it.
Source: Singapore Ministry of Manpower, Guidelines on Safety Instrumented Systems in Major Hazards Installations (Rev 0, 2020), §3.2 (BPCS credit), §4.5, §5.2, §6.3
Question 76
F06 · 5C Safety requirements specification · Multiple correct
Which THREE belong in a safety requirements specification?
Show explanation
Answer and explanation
Answer: A. The safe state of the process for each SIF and B. Trip set points and the response time the SIS needs to bring the process to a safe state and C. Proof-test interval requirements
Safe state, trip points with response time, and proof-test intervals all appear in standard SRS content lists.
Why not the others: Warranty terms are a procurement matter, and BPCS tuning belongs to the control system, not the SIS requirements.
Source: ABB, Safety Requirements Specification factsheet (2012), SRS content list
Question 77
F07 · 5D SIS design and engineering · Numeric entry
A 1oo1 SIF has λDU = 2.0 × 10⁻⁶ per hour and a proof-test interval of one year (8,760 h). Using PFDavg ≈ λDU × TI ÷ 2, what is the PFDavg? Enter it as a decimal; the answer explains which SIL band it falls in.
Show explanation
Answer and explanation
Answer: 0.00876 (accepted range 0.0087 to 0.0088)
PFDavg = 2.0 × 10⁻⁶ × 8,760 ÷ 2 = 8.76 × 10⁻³ (0.00876). The risk reduction factor is about 114. That sits in the SIL 2 band, from 10⁻³ up to but not including 10⁻².
Common mistake: Forgetting the ÷2 gives 0.0175, which would fall in SIL 1.
Source: Worked calculation from the data given in the question; Pepperl+Fuchs, Functional Safety in the Process Industry (2018), SIL/PFDavg table, hardware fault tolerance table, PFDavg formulas
Question 78
F08 · 5D SIS design and engineering · Multiple choice
Under IEC 61511-1 (2016 edition, adopted by ISA in 2018), what is the minimum hardware fault tolerance for a SIL 3 SIF?
Show explanation
Answer and explanation
Answer: B. 1
SIL 3 requires HFT 1, meaning one fault can occur without losing the function (1oo2, for example). SIL 4 requires HFT 2. SIL 1, and SIL 2 in low-demand mode, require HFT 0.
Why not the others: HFT 0 is too low for SIL 3. HFT 2 is the SIL 4 requirement.
Source: Pepperl+Fuchs, Functional Safety in the Process Industry (2018), SIL/PFDavg table, hardware fault tolerance table, PFDavg formulas
Question 79
F09 · 5D SIS design and engineering · Multiple choice
Which sensor voting arrangement best balances safety with avoiding spurious trips?
Show explanation
Answer and explanation
Answer: C. 2oo3
2oo3 tolerates one dangerous failure and also one safe failure without tripping.
Why not the others: 1oo2 is safe but trips on any single spurious signal. 2oo2 rarely trips falsely, but one dangerous failure defeats it. 1oo1 tolerates no faults at all.
Source: Kenexis, Comparison of 2oo3 voting and 2oo2 voting, fault tolerance comparison
Question 80
F10 · 5D SIS design and engineering · Numeric entry
A 1oo2 SIF has λDU = 2.0 × 10⁻⁶ per hour, TI = 8,760 h, and a common-cause factor β = 0.05. Using PFDavg ≈ (λDU × TI)² ÷ 3 + β × λDU × TI ÷ 2, find the PFDavg as a decimal.
Show explanation
Answer and explanation
Answer: 0.00054 (accepted range 0.00053 to 0.00055)
λDU × TI = 0.01752. Independent term: 0.01752² ÷ 3 = 1.02 × 10⁻⁴. Common-cause term: 0.05 × 0.01752 ÷ 2 = 4.38 × 10⁻⁴. Total = 5.4 × 10⁻⁴. The common-cause term is about four times the independent term, which is why redundancy alone has limits.
Common mistake: Leaving out the common-cause term gives 1.0 × 10⁻⁴, which overstates the benefit of redundancy about fivefold.
Source: Worked calculation from the data given in the question; Pepperl+Fuchs, Functional Safety in the Process Industry (2018), SIL/PFDavg table, hardware fault tolerance table, PFDavg formulas
Question 81
F11 · 5D SIS design and engineering · Multiple choice
Why are most SIS outputs designed de-energize-to-trip?
Show explanation
Answer and explanation
Answer: A. Loss of power or a broken wire drives the process to its safe state
Removing power lets springs drive the final elements to their safe positions, so common failures like power loss or a cut wire produce a trip rather than a silent loss of protection. The process industries generally prefer this design.
Why not the others: Power use and speed aren't the reason, and BPCS output design doesn't set SIS practice.
Source: Kenexis, Safety requirements specifications: energize-to-trip requirements, de-energize-to-trip preference
Question 82
F12 · 5E Installation, commissioning, and validation · Multiple choice
What must SIS validation before startup do?
Show explanation
Answer and explanation
Answer: D. Show each SIF works end to end, sensor to final element, against the SRS
Validation confirms that the installed system meets the SRS. That means proving each complete function in the field, final elements included.
Why not the others: Testing the logic alone, or injecting signals at the marshalling cabinet, skips the field sensors and final elements. The FAT happened before installation, so it can't prove installed wiring or a valve stroking in the field.
Source: Singapore Ministry of Manpower, Guidelines on Safety Instrumented Systems in Major Hazards Installations (Rev 0, 2020), §3.2 (BPCS credit), §4.5, §5.2, §6.3
Question 83
F13 · 5F Operation and maintenance · Multiple choice
What is the main purpose of periodic proof testing of a SIF?
Show explanation
Answer and explanation
Answer: B. To find dangerous undetected failures so they can be repaired
Dangerous undetected failures build up between tests, which is why PFDavg grows with the test interval. A proof test finds the fraction of them its coverage includes.
Why not the others: Calibration may happen during a proof test, but it isn't the purpose. Proof tests don't target spurious trips, and alarm performance is a separate alarm management program.
Source: Control Global, SIS: how testing impacts reliability, proof test coverage; PFDavg
Question 84
F14 · 5F Operation and maintenance · Multiple choice
A SIF transmitter has to be bypassed for maintenance. What does good practice require?
Show explanation
Answer and explanation
Answer: C. Authorization, compensating measures, a time limit, operator indication, and a record
A bypass removes protection, so it has to be controlled. HSE expects override management covering authorization, recording, monitoring, and review. IEC 61511 requires compensating measures to keep operation safe, and time limits and operator indication are standard practice.
Why not the others: A verbal OK with no time limit leaves no control or record. Suppressing alarms, or recording the bypass only after the fact, hides an unprotected state from the people running the plant.
Source: UK HSE, COMAH technical measures: control systems, override management; Selega, Chemical Engineering Transactions vol. 48 (2016), §2.1, compensating measures during bypass
Question 85
F15 · 5G Modification and management of change · Multiple choice
An engineer wants to change a SIF trip setpoint. What does the change require?
Show explanation
Answer and explanation
Answer: A. Management of change with impact analysis, SRS update, and testing
A modification re-enters the safety life cycle at the right phase (IEC 61511-1 clause 17), under management of change. That means analyzing the impact back to the hazard assessment, updating the SRS, and re-verifying and testing the function.
Why not the others: Noting the value or getting operations approval isn't an impact analysis. A small setpoint change can still erase the margin to the process safety limit, so size doesn't exempt it from MOC.
Source: Singapore Ministry of Manpower, Guidelines on Safety Instrumented Systems in Major Hazards Installations (Rev 0, 2020), §3.2 (BPCS credit), §4.5, §5.2, §6.3
Score this set in your browser
Answer any questions above, then select Score my test. Results show correct out of attempted, by domain and in total. Partly completed matching or ordering entries count as attempted with no credit; unreadable numeric entries are listed separately.
| Domain | Correct / attempted | Items | Retry |
|---|---|---|---|
| Measurement (Questions 1–19) | — | 19 | |
| Control Systems (Questions 20–40) | — | 21 | |
| Final Control Elements (Questions 41–57) | — | 17 | |
| Signals, Transmission, and Networking (Questions 58–70) | — | 13 | |
| Safety Instrumented Systems (Questions 71–85) | — | 15 | |
| Total | — | 85 |
Marked for review: None
Assisted (answer revealed before scoring): None
Score your test
Mark each question right or wrong against this key, then total your correct answers by domain in the table below. A multiple-correct question counts only if you picked exactly the right set. A numeric answer counts if it falls inside the accepted range shown with that question.
Show the printable answer key (reveals all 85 answers)
| # | ID | Answer | Row label | # | ID | Answer |
|---|---|---|---|---|---|---|
| 1 | M01 | A | 44 | V04 | C | |
| 2 | M02 | D | 45 | V05 | 45 | |
| 3 | M03 | D | 46 | V06 | 50 % open | |
| 4 | M04 | 64.12 mV/pH | 47 | V07 | A | |
| 5 | M05 | 15.9 s | 48 | V08 | D | |
| 6 | M06 | C | 49 | V09 | B | |
| 7 | M07 | A | 50 | V10 | C | |
| 8 | M08 | B, D | 51 | V11 | A | |
| 9 | M09 | A | 52 | V12 | D | |
| 10 | M10 | B | 53 | V13 | B | |
| 11 | M11 | D | 54 | V14 | A, C | |
| 12 | M12 | C | 55 | V15 | 42 % of original power | |
| 13 | M13 | 200 gpm | 56 | V16 | C | |
| 14 | M14 | B | 57 | V17 | A | |
| 15 | M15 | C | 58 | S01 | D | |
| 16 | M16 | B | 59 | S02 | B | |
| 17 | M17 | 7,050 gal | 60 | S03 | 0.153 °F | |
| 18 | M18 | C | 61 | S04 | 9 psig | |
| 19 | M19 | 2 % of span | 62 | S05 | 2,000 ft | |
| 20 | C01 | A | 63 | S06 | C | |
| 21 | C02 | FI: Flow indicator, TI: Temperature indicator, PT: Pressure transmitter, TR: Temperature recorder | 64 | S07 | A | |
| 22 | C03 | A, B | 65 | S08 | 11.8 V | |
| 23 | C04 | D | 66 | S09 | 2,500 VA | |
| 24 | C05 | B | 67 | S10 | D | |
| 25 | C06 | C | 68 | S11 | B | |
| 26 | C07 | 1 %/% | 69 | S12 | C | |
| 27 | C08 | A | 70 | S13 | C | |
| 28 | C09 | 75 gpm | 71 | F01 | Hazard and risk assessment → Allocation of safety functions to protection layers → Safety requirements specification → Design and engineering → Installation, commissioning, and validation → Operation and maintenance | |
| 29 | C10 | D | 72 | F02 | A | |
| 30 | C11 | B | 73 | F03 | D | |
| 31 | C12 | C | 74 | F04 | B | |
| 32 | C13 | A | 75 | F05 | A | |
| 33 | C14 | D | 76 | F06 | A, B, C | |
| 34 | C15 | B | 77 | F07 | 0.00876 | |
| 35 | C16 | C | 78 | F08 | B | |
| 36 | C17 | A | 79 | F09 | C | |
| 37 | C18 | 0.99984 | 80 | F10 | 0.00054 | |
| 38 | C19 | 4,094 BTU/h | 81 | F11 | A | |
| 39 | C20 | Factory acceptance test → Installation and loop checks → Site acceptance test → Startup and handover | 82 | F12 | D | |
| 40 | C21 | B, D | 83 | F13 | B | |
| 41 | V01 | D | 84 | F14 | C | |
| 42 | V02 | B | 85 | F15 | A | |
| 43 | V03 | D |
| Domain | Questions | Items | Your correct |
|---|---|---|---|
| Measurement | 1–19 | 19 | ___ |
| Control Systems | 20–40 | 21 | ___ |
| Final Control Elements | 41–57 | 17 | ___ |
| Signals, Transmission, and Networking | 58–70 | 13 | ___ |
| Safety Instrumented Systems | 71–85 | 15 | ___ |
| Total | 1–85 | 85 | ___ |
Your percentage here tells you how you did on these 85 questions. It isn't an NCEES score and can't predict a pass. NCEES converts the number of questions you answer correctly to a scaled score and compares it with a minimum ability level set by subject-matter experts. It doesn't publish a passing score, and it doesn't pass a set percentage of examinees (NCEES exam scoring). Use your domain totals to decide where to spend your next study hours, not as a readiness verdict. With only 13 to 21 questions per domain, a difference of a question or two between domains is noise.
One scoring fact worth acting on: NCEES counts only correct answers and doesn't penalize wrong ones (NCEES Examinee Guide, May 2026, section 6). On exam day, never leave a question blank.
What to review next
Start with your lowest domain, rework every miss from its explanation before rereading theory, then retry those questions cold a few days later.
Measurement (Questions 1–19). If you missed the calculation questions (4, 5, 13–17, 19), drill the square-root flow relationship, wet-leg level differential pressure, and percent-of-span error until each takes under two minutes. They are quick points once automatic. If you missed the concept questions, review each sensor's operating principle (what it actually senses), installation orientation (gas meters above the taps, liquid and steam below), and how repeatability differs from accuracy.
Control Systems (Questions 20–40). Missed 20–22? Practice reading tags, symbols, and cause-and-effect matrices. ANSI/ISA-5.1-2024 is supplied in the exam, so learn to search it fast. Missed 23–30? Review loop strategy: why integral removes offset, cascade tuning order, feedforward, ratio, and integrating processes. Missed 33–36? Trace PLC logic step by step: seal-in contacts, timer types, SFC transitions, and integer math. Questions 31, 32, and 37–40 cover alarm management, HMI design, system design arithmetic, testing order, and cybersecurity.
Final Control Elements (Questions 41–57). Sizing and split-range arithmetic (45, 46) should be automatic. Review the line between flashing, cavitation, and choked flow (49, 50); fail-safe action and actuator behavior (47, 48, 57); and relief-valve selection and inlet-loss rules (53, 54).
Signals, Transmission, and Networking (Questions 58–70). Loop voltage budgets (65), intrinsic safety entity checks (62), and signal conversions (60, 61) are calculation points you can bank. Review purge types, shield grounding, active versus passive wiring, IP subnets, and media choice for the concept questions.
Safety Instrumented Systems (Questions 71–85). Know the life-cycle order cold (71). Rework the LOPA and PFDavg arithmetic (73, 77, 80) until the SIL bands are second nature. For operation and maintenance questions (83–85), practice finding bypass, proof-test, and modification requirements in the supplied IEC 61511-1 (2018), which is the edition the exam scores against.
How this test maps to the April 2027 specifications
NCEES publishes a range of questions for each domain, not a fixed count. We set this test's 85 questions near the middle of each official range. That allocation is our editorial choice, not NCEES's count for any exam form.
| Domain | Official range, April 2027 specs | Questions in this test |
|---|---|---|
| Measurement | 16–24 | 19 |
| Control Systems | 17–26 | 21 |
| Final Control Elements | 14–21 | 17 |
| Signals, Transmission, and Networking | 11–17 | 13 |
| Safety Instrumented Systems | 12–18 | 15 |
Source: NCEES PE Control Systems Exam Specifications, effective beginning April 2027.
Each question carries its spec subtopic code (for example, 1H Flow calculations), so you can see which part of the specification it practices. NCEES says CBT exams use traditional multiple-choice questions plus alternative item types that include multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank items (NCEES computer-based testing). This set uses text-friendly equivalents such as multiple-response, numeric-entry, matching, and ordering questions; it does not try to reproduce the NCEES/Pearson testing interface.
What changed from the October 2022 specifications
If your study material was written before 2027, the core engineering still applies. Check the domain names and the ISA-5.1 edition before you trust its symbol and identification answers.
| Row label | October 2022 specs | April 2027 specs |
|---|---|---|
| Questions | 85 | 85 |
| Measurement | 17–27 | 16–24 |
| Control Systems | 17–27 | 17–26 |
| Final Control Elements | 14–23 | 14–21 |
| Signals, Transmission, and Networking | 11–18 | 11–17 |
| Domain 5 | Safety Systems, 11–19 | Safety Instrumented Systems, 12–18 |
| ISA-5.1 edition supplied | 2009 | 2024 |
| IEC 61511 Part 1 edition supplied | 2018 | 2018 |
Sources: October 2022 specifications; April 2027 specifications. An NCEES memo to member boards dated July 20, 2026 confirms the exam moves to new specifications and design standards in April 2027 (memo, as posted by the Virginia board).
PE Control Systems exam facts for test day
- Date: NCEES offers the exam once a year as a single-day event. The next test date is April 14, 2027. Seats can be reserved at a Pearson VUE center up to a year ahead, after NCEES registration and board approval (NCEES Control Systems page).
- Length: 85 questions. The 9.5-hour appointment covers a 2-minute nondisclosure agreement, an 8-minute tutorial, 8.5 hours of exam time, and a 50-minute scheduled break. That works out to about 6 minutes per question (NCEES Control Systems page).
- References: The exam is closed book. NCEES supplies the electronic PE Control Systems Reference Handbook plus two design standards: ANSI/ISA-5.1-2024 and ANSI/ISA/IEC 61511 Part 1 (2018). Answers that depend on a standard are scored to those editions only. You can read the handbook in your MyNCEES account, and the standards are sold by their publisher (April 2027 specifications, design standards page).
- Calculator: You may bring one NCEES-approved calculator, and a TI-30XS is available on screen during the exam (Examinee Guide, section 3). NCEES reviews its approved model list every year and posts it on its exams page; check it before your test.
- Units: The exam uses both SI and U.S. customary units (April 2027 specifications).
- Fee: The exam fee is $400, paid to NCEES. Your state board may require its own application and fee (NCEES Control Systems page).
- Results: Results usually arrive 7–10 days after the exam, in MyNCEES (NCEES Control Systems page). NCEES diagnostic reports show a scaled score from 0 to 15 for each knowledge area (Examinee Guide, section 6).
- Not the same exam: PE Electrical and Computer: Electronics, Controls, and Communications is a separate NCEES exam with its own specifications. These questions don't target it.
How to use this test
- As a full dress rehearsal: Set aside 8.5 hours, take the scheduled break, and give each question about 6 minutes. The pacing matters as much as the content.
- One domain at a time: Take a domain in one sitting, check your answers, then take the next. This works better early in your preparation.
- With your references open: The exam is closed book, but you'll have the handbook and both standards on screen. Practice looking things up in them rather than from memory, the way the exam forces you to.
- Rework before rereading: For each miss, read the explanation, close it, and solve the question again from scratch. Then take the domain again a few days later.
Sources
Every answer explanation above links to its own technical source. Exam facts on this page come from these NCEES documents:
- NCEES, PE Control Systems exam page: date, format, fee, references, results timing, pass rates
- NCEES, PE Control Systems Exam Specifications, effective beginning April 2027: domains, ranges, design standards
- NCEES, PE Control Systems CBT Exam Specifications, effective October 1, 2022: previous ranges and editions
- NCEES, Spring 2027 exam changes memo to member boards (July 20, 2026), posted by the Virginia APELS board
- NCEES Examinee Guide, May 2026: calculators, scoring, diagnostic reports
- NCEES, Exam scoring
- NCEES, Computer-based testing: item types
Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by NCEES. All practice questions on this page are original and unofficial; they are not real, recalled, or reproduced NCEES exam questions. Exam and credential names are used only to identify the subject; trademarks belong to their respective owners.
By the Castleport Test Prep Editorial Team.
Last verified October 11, 2026: The exam date, format, fee, April 2027 specifications, design standards, calculator policy, scoring policy, and diagnostic-report scale were checked against the NCEES sources listed above. Technical sources remain cited with the individual practice explanations.