PE Control Systems Exam Prep
Start your PE Control Systems exam prep with this original, unofficial problem, written to the new NCEES specification for the April 14, 2027 exam. All 36 problems, worked solutions, a study map, and a dated plan to exam day are free on this page.
Problem 1 · Flow from a DP reading
MS-01 · Measurement, 1H · Select one
A differential-pressure (DP) flow transmitter is ranged 0–100 inH2O for 0–400 gpm. No square-root extraction is applied in the transmitter. The transmitter reads 25 inH2O. What is the flow?
Show answer and solution
Answer and solution
Answer: B. 200 gpm
Across a DP flow element, differential pressure rises with the square of flow, so flow goes with the square root of DP. 25 inH2O is 25% of the DP span, and √0.25 = 0.50, so flow = 0.50 × 400 = 200 gpm.
- A: Treats DP and flow as linear (25% of 400).
- C: Uses √0.5, the answer for a 50% DP reading.
- D: Squares the fraction instead of taking the root.
Try a change: At 1% of DP span, flow is already 10% of maximum. That's why small DP errors near the bottom of the range produce large flow errors.
Source: Control Engineering, “Square root scaling for differential pressure flowmeters” (Hardy, 2017). Topic: NCEES PE Control Systems specification, April 2027, 1H.
By Castleport Test Prep Editorial Team · Exam facts last verified October 11, 2026
What changed for 2027 · Study map · All 36 problems · Plan to April 14
The exam at a glance
- One date: Wednesday, April 14, 2027, at Pearson VUE test centers. Control Systems is offered once a year. NCEES lets you reserve a seat up to a year ahead once you're registered and approved by your licensing board, and seats are first come, first served.
- 85 questions with 8.5 hours of exam time inside a 9.5-hour appointment (2-minute nondisclosure agreement, 8-minute tutorial, 50-minute scheduled break).
- $400 exam fee paid to NCEES. Your licensing board may charge its own application fee.
- References on screen: the NCEES PE Control Systems Reference Handbook plus two standards: ANSI/ISA-5.1-2024 and ANSI/ISA/IEC 61511 Part 1 (2018). You can't bring your own copies.
- Units: both SI and U.S. customary.
- Question types: multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank), all scored right or wrong.
- Results: pass or fail, typically 7–10 days after the exam.
- Pass rates: 61% of 250 first-time takers and 45% of 83 repeat takers, in the Control Systems row NCEES labels "July 2026."
- Not the same exam as PE Electrical and Computer: Electronics, Controls, and Communications. That's a separate NCEES exam (also on April 14, 2027) with its own specification and no supplied design standards.
Sources: NCEES PE Control Systems exam page; PE Control Systems specification, effective April 2027, pp. 1 and 4; NCEES Examinee Guide, May 2026, pp. 3, 5, 11, 14, and 16.
What changed for the April 2027 exam
The question count, timing, and the five knowledge areas stayed. What changed is inside them: NCEES narrowed most of the question ranges, renamed "Safety Systems" to Safety Instrumented Systems, rewrote the subtopics as specific tasks, and moved ISA-5.1 from the 2009 edition to 2024. NCEES confirmed the change in a July 2026 notice to licensing boards (NCEES memo, July 20, 2026).
Question ranges, then and now
| Knowledge area | Through 2026 (topics since October 2019) | April 2027 specification |
|---|---|---|
| 1. Measurement | 17–27 | 16–24 |
| 2. Control Systems | 17–27 | 17–26 |
| 3. Final Control Elements | 14–23 | 14–21 |
| 4. Signals, Transmission, and Networking | 11–18 | 11–17 |
| 5. Safety Systems → Safety Instrumented Systems | 11–19 | 12–18 |
Sources: specification effective April 2027, pp. 1–3; previous CBT specification, effective October 2022, pp. 1–5.
Tasks the 2027 outline now names outright include sensor testing and commissioning (loop tests, square-root extraction, range verification), final-element testing and commissioning, control-system cybersecurity as its own task, network addressing, and the safety lifecycle phase by phase, from hazard and risk assessment through modification and decommissioning.
The two standards supplied on exam day
| Standard | Before April 2027 | April 2027 |
|---|---|---|
| ANSI/ISA-5.1, Instrumentation and Control – Symbols and Identification | 2009 | 2024 |
| ISA/IEC 61511 Part 1, Functional Safety – Safety Instrumented Systems for the Process Industry Sector | 2018 | 2018 (listed as ANSI/ISA/IEC 61511 Part 1) |
NCEES scores standards-based answers against the edition on its list and says solutions based on other standards won't receive credit. It doesn't sell or lend the standards for study; they're available from the publisher or a bookseller (April 2027 specification, standards page). If your symbology study is based on the 2009 edition, redo it in the 2024 edition.
Studying from material written to the older specification?
Most of the engineering still transfers: DP flow, valve sizing, PID behavior, and SIL math didn't change because a specification did. What needs attention:
- Recheck symbols and identification in ISA-5.1-2024.
- Add the named tasks listed above, especially commissioning, cybersecurity, and network addressing.
- Reorganize safety study around the lifecycle, not just SIL calculations.
- Cross-check every topic list against the April 2027 NCEES specification. Older prep material may still teach useful engineering, but its organization and standard editions may no longer match the exam you will take.
What to study first: your 2027 study map
The specification says the exam uses "a variety of approaches and methodologies, including design, analysis, and application." So each area below asks for two kinds of skill: numbers you must calculate, and choices you must make (select a device, read a drawing, diagnose a fault, place a step in the lifecycle). Drill both.
| Area and NCEES range | Calculate | Select, interpret, diagnose | Reference to know | Move on when you can… | Practice |
|---|---|---|---|---|---|
| 1. Measurement (16–24) | DP flow and square root, level with suppression or elevation, standard-to-actual flow, thermowell wake frequency, accuracy | Pick sensors for service and materials; analyzers and sampling; fire and gas detectors; machinery monitoring; impulse lines and installation; loop tests and calibration | Handbook | choose a defensible sensor for a service and do the common flow and level conversions without notes | Problems 1–8 |
| 2. Control Systems (17–26) | Loop tuning from a process model, panel heat load and power | Read P&IDs, loop and logic diagrams, cause-and-effect matrices; explain process response; read ladder, function block, SFC, and structured text; alarm management; architecture; FAT, SAT, MOC; cybersecurity | Handbook; ISA-5.1-2024 for symbols | read a drawing or logic block and predict what the plant will do | Problems 9–16 |
| 3. Final Control Elements (14–21) | Valve Cv, split range, pressure drop, relief device sizing, pump speed and power | Valve type, trim, materials; cavitation and flashing; actuator failure modes; accessories; relief device types; motor controls, solenoids, relays, regulators | Handbook | size a valve, name its fail position and say why | Problems 17–23 |
| 4. Signals, Transmission, and Networking (11–17) | 4–20 mA scaling, loop power budgets, A/D resolution, voltage drop, IS entity checks | Signal types and conversion; hazardous-area protection; grounding and EMI; network media, protocols, and addressing | Handbook | trace a signal from field device to controller and check every voltage and limit along the way | Problems 24–29 |
| 5. Safety Instrumented Systems (12–18) | LOPA, PFDavg, voting, proof-test interval effects | Lifecycle phases; SRS content; device selection; SAT; proof testing, bypasses, MOC | 61511 Part 1 (2018) | place a scenario in the lifecycle and back a SIL claim with numbers | Problems 30–36 |
The ranges are NCEES's (April 2027 specification, pp. 1–3). The skill columns and checkpoints are our editorial summary of its subtopics, not an official weighting. NCEES also says its listed examples aren't exclusive or exhaustive. Spend the most time where an area is both large and weak for you. If your job is all DCS configuration, Final Control Elements and SIS may need real hours, while Control Systems needs only a review.
How to study with the handbook and the two standards
The exam is closed book with electronic references. You'll have the PE Control Systems Reference Handbook and the two listed standards as searchable PDFs; those are the only reference materials you may use (Examinee Guide, p. 10). Download the current handbook free from your MyNCEES account and study from it, not from a summary.
Run every session through four steps:
- Learn. Review the concept and what each equation assumes.
- Find. Locate the relation in the handbook, or the requirement in the standard. Write down the search words that worked.
- Solve. Work fresh problems before you look at any solution.
- Log. Record every miss and every lucky guess by cause (see the error table).
On exam day you search with a box on the left side of the reference; Ctrl+F doesn't work (Examinee Guide, p. 10). NCEES posts a video on searching the on-screen references.
A lookup drill. Pick 10 prompts from the week's area. Give yourself 60 to 90 seconds each to find the governing equation, table, or requirement. Search with one distinctive term, not the whole question. Read the variable definitions and units before you use anything. Log a failed lookup separately from a concept error. (The 60–90 seconds is our drill target, not an NCEES rule.)
What each reference is for:
| Reference | Use it for | Practice finding |
|---|---|---|
| PE Control Systems Reference Handbook | Equations and general technical data | Check whether it covers each calculation in the study map, such as DP flow, valve sizing, and PFD, and note the search term that finds it |
| ANSI/ISA-5.1-2024 | Instrument symbols and identification letters | How functions, signal lines, and devices are shown and tagged |
| ANSI/ISA/IEC 61511 Part 1 (2018) | The SIS lifecycle and its requirements | Where the standard covers hazard and risk assessment, the SRS, design, validation, operation and maintenance, bypasses, and modification |
36 original practice problems with worked solutions
These are original Castleport problems written to the April 2027 PE Control Systems topic list. They aren't NCEES questions, they aren't a full-length exam, and they don't cover every subtopic. Every value you need is given in the problem, so you can solve each one without the standards.
The set has 28 single-answer multiple-choice problems, 5 numeric-entry problems, 2 select-all-that-apply problems, and 1 ordering problem. NCEES also uses point-and-click and drag-and-drop items, which this page doesn't reproduce. On the real exam every question, including multiple-correct items, is scored right or wrong with no partial credit (Examinee Guide, p. 11). We score our select-all and ordering problems the same way.
Work each problem before you open the solution. When you miss one, write down the first step that went wrong.
Measurement (Problems 1–8)
What this area tests most often:
- DP flow is a square-root relationship. Mix up DP percent and flow percent and every downstream number is wrong.
- Hydrostatic level is height × specific gravity, and wherever the transmitter sits relative to the tap shifts the calibrated range.
- Convert to absolute pressure and temperature before any gas-flow correction.
- Installation is fair game: impulse-line orientation, block valves, and where a sensor actually sees the process.
Problem 1 is at the top of this page.
Problem 2 · Level range with zero suppression
MS-02 · Measurement, 1I · Select one
An open tank holds a liquid with specific gravity 0.90. The level range is 0–10 ft above the lower tap. The DP transmitter is mounted 3 ft below the lower tap; its low side is vented and the high-side line is full of process liquid. What calibrated range should the transmitter have, in inH2O?
Show answer and solution
Answer and solution
Answer: B. 32.4 to 140.4 inH2O
Each foot of this liquid is 12 × 0.90 = 10.8 inH2O. At zero level the transmitter still sees the 3 ft of liquid between the tap and the transmitter: 3 × 10.8 = 32.4 inH2O. At full level it sees 13 ft: 13 × 10.8 = 140.4 inH2O. A transmitter mounted below the zero reference needs zero suppression.
- A: Ignores the 3 ft of liquid below the tap.
- C: Treats the offset as zero elevation (wrong sign).
- D: Ignores specific gravity.
Source: Automation.com/ISA, “Level reading with a differential pressure cell” (2017); Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 1I.
Problem 3 · Standard to actual gas flow
MS-03 · Measurement, 1J · Enter a number
An ideal gas (Z = 1) flows at 1,000 SCFM, with standard conditions of 14.696 psia and 60 °F. Line conditions are 100 psig and 140 °F. Enter the actual volumetric flow in ACFM.
Show answer and solution
Answer and solution
Answer: 148 ACFM (accepted ±2 ACFM).
Convert to absolute units first: 100 psig = 114.696 psia; 60 °F = 519.67 °R; 140 °F = 599.67 °R. ACFM = 1,000 × (14.696 ÷ 114.696) × (599.67 ÷ 519.67) = 1,000 × 0.1281 × 1.1539 ≈ 148 ACFM.
About 128 ACFM means the temperature correction was skipped. About 170 ACFM used gauge pressure (100) instead of absolute (114.696). About 111 ACFM inverted the temperature ratio.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 1J.
Problem 4 · Thermowell wake frequency
MS-04 · Measurement, 1K · Select one
Water flows at 20 ft/s past a thermowell with a 0.75 in tip diameter. Using a Strouhal number of 0.22 (given) and f = St × V ÷ d, what is the vortex-shedding (wake) frequency?
Show answer and solution
Answer and solution
Answer: A. 70 Hz
Convert the diameter: 0.75 in = 0.0625 ft. f = 0.22 × 20 ÷ 0.0625 = 70.4 Hz. The design check compares this frequency with the thermowell's natural frequency; vortex shedding near the natural frequency can drive damaging resonance.
- B: Leaves the diameter in inches with velocity in ft/s.
- C: Inverts the Strouhal relation (V ÷ (St × d)).
- D: Is the ratio of 70.4 Hz to a 150 Hz natural frequency, not a frequency.
Try a change: Double the velocity and the wake frequency doubles too. That's why the highest expected velocity is the governing case.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 1K.
Problem 5 · Accuracy: % of span vs. % of reading
MS-05 · Measurement, 1E · Select one
A pressure transmitter is calibrated 0–500 psig with an accuracy of ±0.25% of calibrated span. At a reading of 100 psig, the possible error expressed as a percentage of the reading is closest to:
Show answer and solution
Answer and solution
Answer: B. ±1.25%
±0.25% of a 500 psi span is ±1.25 psi at any point in the range. At 100 psig, 1.25 ÷ 100 = ±1.25% of reading. A percent-of-span spec gets worse, relative to the reading, the lower you operate in the range.
- A: Confuses % of span with % of reading.
- C: Applies 0.25% to 100 psi and then divides by the span.
- D: Is a decimal slip (×10).
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 1E.
Problem 6 · pH electrode output
MS-06 · Measurement, 1B · Select one
An ideal pH glass electrode at 25 °C has a slope of 59.16 mV per pH unit. By how much does its output change when the process goes from pH 7 to pH 4?
Show answer and solution
Answer and solution
Answer: B. 177 mV
Three pH units × 59.16 mV = 177.5 mV. The slope comes from the Nernst equation and changes with absolute temperature (about 55 mV/pH at 5 °C and 60 mV/pH at 30 °C), which is why pH analyzers need temperature compensation.
- A: One pH unit only.
- C: Four pH units.
- D: Multiplies 7 × 59.16, as if pH 7 were not the zero-mV point.
Source: Process Instruments, Technical Note 147, “The Effect of Temperature on pH”. Topic: NCEES PE Control Systems specification, April 2027, 1B.
Problem 7 · Impulse lines for gas service
MS-07 · Measurement, 1G · Select one
You are installing a DP transmitter for a dry gas flow measurement on a horizontal line. Where should the transmitter normally go relative to the process taps?
Show answer and solution
Answer and solution
Answer: A. Beside or above the taps, so any condensed liquid drains back to the line
For gas, you want the impulse lines to stay gas-filled. Mounting beside or above the taps lets liquid drain back into the pipe instead of collecting in the lines and adding a false head. Liquid and steam service are the opposite case: beside or below, so gas drains back up and the lines stay full of liquid.
- B: Is the liquid/steam arrangement.
- C: Horizontal lines can trap liquid or gas pockets.
- D: A manifold isolates and equalizes the transmitter; it doesn't prevent trapped liquid.
Source: Emerson, Rosemount 3051S Quick Start Guide, §1.1–1.3, pp. 5–7. Topic: NCEES PE Control Systems specification, April 2027, 1G.
Problem 8 · Measuring shaft vibration
MS-08 · Measurement, 1D · Select one
A large compressor runs on fluid-film (sleeve) bearings. You need to measure radial shaft vibration relative to the bearing. Which sensor is the best fit?
Show answer and solution
Answer and solution
Answer: A. Non-contacting eddy-current proximity probes mounted at the bearing (X and Y)
Proximity probes see the gap between probe and shaft, so they measure shaft motion relative to the bearing directly. With fluid-film bearings, shaft vibration may not show up clearly on casing-mounted sensors. Bently Nevada's ORBIT describes a gearbox where shaft probes showed high vibration while casing transducers read normal.
- B: Measures casing (absolute) vibration, not shaft-relative motion.
- C: Is even further from the shaft.
- D: Measures bearing temperature, a separate protection parameter.
Source: Baker Hughes Bently Nevada, ORBIT Vol. 30 No. 1 (2010), “Are XY Radial Proximity Probes Really Necessary?” pp. 15–16. Topic: NCEES PE Control Systems specification, April 2027, 1D.
Control Systems (Problems 9–16)
What this area tests most often:
- Drawings are data. P&IDs, loop diagrams, logic diagrams, and cause-and-effect matrices answer questions if you read them carefully.
- Know what each control strategy fixes: cascade for fast inner disturbances, feedforward for measured ones, split range for two actuators.
- Logic questions reward tracing, scan by scan, what the code or ladder actually does.
- Cybersecurity and commissioning are named tasks in the 2027 outline.
Problem 9 · Reading a cause-and-effect matrix
CS-01 · Control Systems, 2A · Select one
A cause-and-effect matrix for a reactor shows an X where a cause triggers an effect:
| Cause | Close XV-201 (feed) | Close XV-202 (steam) | Stop P-301 (transfer pump) | Open XV-203 (vent) |
|---|---|---|---|---|
| PSHH-101 reactor pressure high-high | X | X | X | |
| LSLL-102 feed drum level low-low | X | X | ||
| TSHH-103 reactor temperature high-high | X | X |
PSHH-101 trips. Which actions should occur?
Show answer and solution
Answer and solution
Answer: A. Close XV-201, close XV-202, and open XV-203
Read across the PSHH-101 row: Xs under close XV-201, close XV-202, and open XV-203. P-301 stops only on LSLL-102.
- B: Is the LSLL-102 row.
- C: Adds the pump stop, which isn't in the PSHH-101 row.
- D: Misses two marked effects.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 2A.
Problem 10 · IMC (lambda) PI tuning
CS-02 · Control Systems, 2B · Select one
A bump test gives a first-order-plus-dead-time model: process gain Kp = 2.0 %/%, time constant τp = 60 s, dead time θp = 10 s. You choose a closed-loop time constant τc = 50 s. Using the IMC PI correlation Kc = τp ÷ [Kp(τc + θp)] and Ti = τp, what are the settings?
Show answer and solution
Answer and solution
Answer: A. Kc = 0.50, Ti = 60 s
Kc = 60 ÷ [2.0 × (50 + 10)] = 60 ÷ 120 = 0.50. Ti = τp = 60 s. A smaller τc makes the loop faster and more aggressive; a larger τc makes it slower and more robust.
- B: Leaves dead time out of the denominator.
- C: Leaves τc out and divides by dead time alone.
- D: Sets integral time equal to dead time.
Try a change: Raise τc to 110 s and Kc drops to 0.25 — half as aggressive, with Ti unchanged.
Source: APMonitor (BYU), PI control tuning, IMC correlation; Dataforth AN124, “Tuning Control Loops with the IMC Tuning Method”. Topic: NCEES PE Control Systems specification, April 2027, 2B.
Problem 11 · Tuning a cascade loop
CS-03 · Control Systems, 2B · Select one
A reactor temperature controller (primary) sets the setpoint of a jacket temperature controller (secondary). How should you tune the pair?
Show answer and solution
Answer and solution
Answer: A. Tune the secondary (inner) loop first with the primary in manual, then tune the primary with the secondary in cascade
The inner loop has to work on its own before the outer loop can rely on it. Put the primary in manual, tune the secondary, put it in automatic, then tune the primary with the secondary in cascade. Cascade pays off when the inner loop responds faster than the outer loop.
- B: The primary's response depends on how the secondary is tuned.
- C: The loops have different dynamics.
- D: Gain doesn't decide the tuning order.
Source: Control Guru, “An Implementation Recipe for Cascade Control”. Topic: NCEES PE Control Systems specification, April 2027, 2B.
Problem 12 · What feedforward can and can't do
CS-04 · Control Systems, 2B · Select one
You add feedforward from inlet flow to a heat exchanger's outlet-temperature loop. Which statement is correct?
Show answer and solution
Answer and solution
Answer: A. Feedforward acts only on the measured disturbance, so you keep feedback to correct model error and unmeasured disturbances
Feedforward predicts the effect of the disturbance you measure and acts before the temperature moves. Its model is never perfect and it can't see other disturbances, so feedback trim stays in the loop.
- B: Without feedback, model error leaves a steady offset.
- C: Dead time is a property of the process, not the controller.
- D: Feedforward is built on a model of how the disturbance affects the process.
Source: Control Guru, “The Feed Forward Controller”. Topic: NCEES PE Control Systems specification, April 2027, 2B.
Problem 13 · Reading structured text
CS-05 · Control Systems, 2D · Select one
A PLC runs this structured text every scan:
IF LT101 > 80.0 THEN
XV102 := FALSE;
ELSIF LT101 < 20.0 THEN
XV102 := TRUE;
END_IF;The level falls to 15%, then rises to 50%. What is XV102 at 50%?
Show answer and solution
Answer and solution
Answer: A. TRUE
At 15% the second condition sets XV102 TRUE. Between 20% and 80% neither branch runs, so XV102 keeps its last value. It stays TRUE until the level passes 80%. That gap is built-in hysteresis — and when someone expects the valve to close at 50%, it's the reason for the "unexpected" behavior.
- B: Would require the level to exceed 80% first.
- C: Nothing in the code changes XV102 between 20% and 80%.
- D: A missing ELSE simply leaves the variable unchanged.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 2D.
Problem 14 · A chattering alarm
CS-06 · Control Systems, 2C · Select one
A noisy level signal makes its high-level alarm activate and clear many times an hour near the alarm limit. The alarm is still needed. What is the best first fix?
Show answer and solution
Answer and solution
Answer: A. Find the cause of the noise, then add an alarm deadband (and an on-delay if the process is slow) sized to the noise
Chatter is a nuisance that trains operators to ignore alarms. Fix the root cause first, then use deadband (the alarm clears only after the signal moves back a set amount) or an on-delay (the condition must persist before alarming). Neither should be applied blindly: delays postpone real alarms, and too much filtering can hide them.
- B: Removes a needed alarm.
- C: Makes the nuisance louder.
- D: Defeats the alarm's purpose.
Source: Control Global, “How to reduce alarm chatter” (Brown, 2016). Topic: NCEES PE Control Systems specification, April 2027, 2C.
Problem 15 · Panel heat load
CS-07 · Control Systems, 2E · Enter a number
A sealed control panel contains a PLC rack (60 W dissipation), I/O modules (90 W), and a network switch (15 W). It also holds a 24 VDC power supply that delivers 200 W to loads outside the panel at 90% efficiency. Enter the heat released inside the panel, in watts.
Show answer and solution
Answer and solution
Answer: 187 W (accepted ±2 W).
Electronics: 60 + 90 + 15 = 165 W. The power supply draws 200 ÷ 0.90 = 222.2 W to deliver 200 W, so 22.2 W is lost as heat in the panel. Total ≈ 187 W, or about 639 Btu/h (× 3.412).
165 W leaves out the power-supply loss. 387 W counts the 200 W delivered to outside loads as panel heat.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 2E.
Problem 16 · Securing remote access
CS-08 · Control Systems, 2G · Select all that apply
A vendor needs periodic remote access to support the plant's control system. Which practices match NIST's OT security guidance? Select all that apply.
Show answer and solution
Answer and solution
Answer: A and B.
NIST SP 800-82r3 recommends a DMZ architecture with firewalls so traffic doesn't pass directly between corporate and OT networks, multifactor authentication for remote access to the OT system, and separate credentials for OT users.
- C: NIST calls for separate authentication mechanisms and credentials for OT users.
- D: Exposes a controller's programming interface directly, bypassing the DMZ.
Scored all-or-nothing, like NCEES multiple-correct items: you need every correct option and nothing else.
Source: NIST SP 800-82r3, Guide to OT Security (2023), Executive Summary pp. 2–4. Topic: NCEES PE Control Systems specification, April 2027, 2G.
Final Control Elements (Problems 17–23)
What this area tests most often:
- Cv sizing starts with service conditions: specific gravity, pressure drop, and whether flow is choked or flashing.
- Compare outlet pressure with vapor pressure to tell flashing from cavitation.
- Fail position is a process-safety decision, not an actuator preference.
- Relief devices, motors, solenoids, and regulators are all final elements on this exam.
Problem 17 · Liquid valve Cv
FC-01 · Final Control Elements, 3C · Select one
A control valve must pass 200 gpm of a liquid with specific gravity 0.85 at a 16 psi pressure drop. Flow is not choked and there are no reducers (piping geometry factor = 1). Using q = Cv √(Δp ÷ G), what Cv is required?
Show answer and solution
Answer and solution
Answer: A. 46.1
Cv = q √(G ÷ Δp) = 200 × √(0.85 ÷ 16) = 200 × 0.2305 = 46.1. The basic liquid equation doesn't apply to flashing or choked service, which need the full sizing method.
- B: Ignores specific gravity.
- C: Drops the square root.
- D: Inverts the ratio inside the root.
Source: Swagelok, “Valve Sizing” technical bulletin, Liquid Flow. Topic: NCEES PE Control Systems specification, April 2027, 3C.
Problem 18 · Flashing or cavitation?
FC-02 · Final Control Elements, 3E · Select one
Liquid enters a control valve at 100 psia and leaves at 30 psia. Its vapor pressure at flowing temperature is 40 psia. What happens?
Show answer and solution
Answer and solution
Answer: A. Flashing
The outlet pressure (30 psia) is below the vapor pressure (40 psia), so vapor that forms at the vena contracta stays vapor downstream. That's flashing. Cavitation is when pressure dips below vapor pressure at the vena contracta and then recovers above it, so the bubbles collapse.
- B: Requires the outlet pressure to recover above vapor pressure.
- C: The outlet is below vapor pressure, so it can't stay single-phase.
- D: The fluid enters as a liquid.
Try a change: Raise the outlet to 50 psia and the same valve could cavitate instead, if pressure at the vena contracta still falls below 40 psia.
Source: Evans & Ritter, “Flashing and Cavitation,” Valve Magazine (Summer 2015), pp. 3–4 (Emerson reprint). Topic: NCEES PE Control Systems specification, April 2027, 3E.
Problem 19 · Choosing fail positions
FC-03 · Final Control Elements, 3D · Select one
A hazard review has established that loss of heat is the safe response for a reboiler and maximum cooling is the safe response for an exothermic reactor. Instrument air fails. Which pairing of valve failure actions matches those required safe responses?
Show answer and solution
Answer and solution
Answer: A. Steam to a reboiler: fail closed (air-to-open). Cooling water to an exothermic reactor jacket: fail open (air-to-close).
A fail-safe action is the one that protects the process and equipment when actuating energy is lost. Cutting steam removes heat input. Keeping cooling water flowing keeps removing heat from an exothermic reaction.
- B: Losing cooling to an exothermic reactor can cause a runaway.
- C: Keeps heating the reboiler with no control.
- D: Is the unsafe combination on both counts.
Source: Emerson, Control Valve Handbook, 6th ed., §1.4 p. 26 and §1.5 p. 29. Topic: NCEES PE Control Systems specification, April 2027, 3D.
Problem 20 · Split-range position
FC-04 · Final Control Elements, 3C · Select one
One temperature controller drives two valves. The cooling valve is 100% open at 0% controller output and fully closed at 50%. The heating valve is closed until 50% output and fully open at 100%. At 30% output, where are the valves?
Show answer and solution
Answer and solution
Answer: A. Cooling 40% open; heating closed
The cooling valve closes linearly from 100% to 0% over 0–50% output. At 30% output it's (50 − 30) ÷ 50 = 40% open. The heating valve doesn't start opening until 50%.
- B: Is 30 ÷ 50 — the fraction closed, not open.
- C: Copies the output directly.
- D: Puts the heating range below 50%.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 3C.
Problem 21 · Equal-percentage trim
FC-05 · Final Control Elements, 3B · Select one
An inherent equal-percentage valve has a rangeability R = 50. Using the ideal relationship Q/Qmax = R^(x − 1), where x is fractional travel, what fraction of maximum flow passes at 50% travel (constant pressure drop)?
Show answer and solution
Answer and solution
Answer: A. 14.1%
50^(0.5 − 1) = 50^(−0.5) = 1 ÷ √50 = 0.141, or 14.1%. Equal-percentage trim gives the same percentage change in flow for each equal step of travel, so it's gentle at low travel and steep near the top.
- B: Is a linear characteristic.
- C: Halves the right answer.
- D: Is √0.5, a quick-opening-like guess.
Source: Emerson, Control Valve Handbook, 6th ed., §1.4 p. 26 and §1.5 p. 29; Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 3B.
Problem 22 · Relief into a shared header
FC-06 · Final Control Elements, 3H · Select one
A pressure-relief valve will discharge into a shared flare header. Other relief valves discharge into the same header, so the superimposed back pressure varies. Which valve type fits best?
Show answer and solution
Answer and solution
Answer: A. A balanced-bellows (or pilot-operated) relief valve
On a conventional valve, back pressure acts on the back of the disk and shifts its opening point. A balanced bellows keeps back pressure off the disk, so the set pressure holds even as the header pressure changes. Pilot-operated valves are a common alternative in this service.
- B: A fixed offset can't track a back pressure that varies.
- C: Orifice size doesn't fix the set-pressure shift.
- D: A rupture disk may be used in some designs, but not as a blanket answer to variable back pressure.
Source: Control Global, “Ask the experts: Relief valves and back pressure” (2022); Anderson, Greenwood & Co., “Improve safety with pilot operated relief valves,” Hydrocarbon Processing (1996), OSTI abstract. Topic: NCEES PE Control Systems specification, April 2027, 3H.
Problem 23 · VFD speed and pump power
FC-07 · Final Control Elements, 3J · Select one
A centrifugal pump draws 50 hp at full speed in a system with no static head (all friction). A variable-speed drive slows it to 80% speed. Using the affinity laws, about how much power does the pump draw?
Show answer and solution
Answer and solution
Answer: A. 25.6 hp
Power scales with speed cubed: 50 × 0.8³ = 50 × 0.512 = 25.6 hp. Flow scales with speed and head with speed squared. With no static head the system curve passes through zero, so the operating point follows the affinity laws closely. Static head reduces these savings.
- B: Uses speed squared (the head relationship).
- C: Uses a linear relationship (the flow relationship).
- D: Uses speed to the fourth power.
Source: Pumps & Systems, “Variable Speed Pump Operation Cuts Energy Costs; System Static Head Reduces Savings” (2011). Topic: NCEES PE Control Systems specification, April 2027, 3J.
Signals, Transmission, and Networking (Problems 24–29)
What this area tests most often:
- 4–20 mA scaling in both directions, with the 4 mA live zero.
- Every loop is a voltage budget: supply minus drops must cover the device's minimum.
- Intrinsic safety entity checks compare source limits with field-device limits plus cable.
- Current flows out and back, so voltage-drop problems count both conductors.
Problem 24 · Loop resistance budget
SN-01 · Signals, Transmission, and Networking, 4E · Enter a number
A two-wire transmitter is powered from 24 VDC. It needs at least 10.5 V at its terminals. The loop already has a 250 Ω input resistor and 40 Ω of total wire resistance. Designing at 20 mA, enter the maximum additional series resistance (for example, an isolator) you can add, in ohms.
Show answer and solution
Answer and solution
Answer: 385 Ω (accepted ±5 Ω).
Voltage available for the loop = 24 − 10.5 = 13.5 V. At 20 mA the total loop resistance can be 13.5 ÷ 0.020 = 675 Ω. Subtract what's already there: 675 − 250 − 40 = 385 Ω.
675 Ω is the total, not the extra. 1,200 Ω ignores the transmitter's minimum voltage (24 ÷ 0.020).
Try a change: Some designs check the budget at the transmitter's upscale alarm current, which is above 20 mA. Higher current means less room for extra resistance.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 4E.
Problem 25 · A/D resolution
SN-02 · Signals, Transmission, and Networking, 4B · Select one
A 12-bit analog input spans exactly 4–20 mA, and the transmitter is ranged 0–200 psi. What pressure change does one count represent?
Show answer and solution
Answer and solution
Answer: A. 0.049 psi
A 12-bit converter has 2¹² = 4,096 codes, or 4,095 steps from bottom to top. 200 ÷ 4,095 = 0.049 psi per count.
- B: Is an 8-bit converter (200 ÷ 255).
- C: Is a 16-bit converter.
- D: Assumes the converter spans 0–20 mA, wasting 20% of its codes.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 4B.
Problem 26 · Scaling a 4–20 mA signal
SN-03 · Signals, Transmission, and Networking, 4A · Select one
A temperature transmitter is ranged 50–250 °F. Its output reads 14.0 mA. What temperature does that represent?
Show answer and solution
Answer and solution
Answer: A. 175 °F
PV = LRV + [(mA − 4) ÷ 16] × span = 50 + (10 ÷ 16) × 200 = 50 + 125 = 175 °F. The reverse works too: mA = 4 + 16 × (PV − LRV) ÷ span. The 4 mA "live zero" lets the system tell a 0% reading from a broken loop.
- B: Forgets to add the 50 °F lower range value.
- C: Treats the signal as 0–20 mA from zero.
- D: Does both: 0–20 mA scaling plus the offset.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 4A.
Problem 27 · Intrinsic safety cable length
SN-04 · Signals, Transmission, and Networking, 4C · Enter a number
An intrinsic safety barrier has Uo = 28 V, Io = 93 mA, Co = 0.13 µF, Lo = 4.2 mH. The field transmitter has Ui = 30 V, Ii = 100 mA, Ci = 0.01 µF, Li ≈ 0. The cable is 200 pF/m and 1 µH/m. Using the entity checks, enter the maximum cable length in meters.
Show answer and solution
Answer and solution
Answer: 600 m (accepted ±5 m).
Voltage and current pass: Uo (28 V) ≤ Ui (30 V) and Io (93 mA) ≤ Ii (100 mA). Capacitance: Ci + Ccable ≤ Co, so cable capacitance ≤ 0.13 − 0.01 = 0.12 µF; 0.12 µF ÷ 200 pF/m = 600 m. Inductance would allow 4.2 mH ÷ 1 µH/m = 4,200 m, so capacitance governs.
650 m forgets the transmitter's own 0.01 µF. 4,200 m uses only the inductance limit. Real control drawings can add rules (for example, when both Ci and Li are significant), so follow the drawing for an actual installation.
Source: Pepperl+Fuchs control drawing, Entity Concept notes. Topic: NCEES PE Control Systems specification, April 2027, 4C.
Problem 28 · Subnet addressing
SN-05 · Signals, Transmission, and Networking, 4F · Select one
A control network uses 192.168.10.0/26. Which address can't be assigned to a device on that subnet?
Show answer and solution
Answer and solution
Answer: A. 192.168.10.63
A /26 leaves 6 host bits: 64 addresses, .0 to .63. The all-zeros host (.0) names the network and the all-ones host (.63) is the broadcast address, leaving .1–.62 (62 usable hosts).
- B: First usable host.
- C: A usable host.
- D: Last usable host.
Source: IETF RFC 950, §2.1 (special addresses). Topic: NCEES PE Control Systems specification, April 2027, 4F.
Problem 29 · Voltage drop to a solenoid
SN-06 · Signals, Transmission, and Networking, 4D · Enter a number
A 24 VDC solenoid draws 0.50 A and needs at least 20.4 V at its coil. Each conductor of the field cable has 13.3 Ω per km. Enter the longest one-way cable run, in meters, that keeps the coil at or above 20.4 V.
Show answer and solution
Answer and solution
Answer: 271 m (accepted ±3 m).
Allowable drop = 24 − 20.4 = 3.6 V. At 0.50 A, total loop resistance can be 3.6 ÷ 0.50 = 7.2 Ω. Current goes out and back, so both conductors count: 7.2 ÷ (2 × 13.3 Ω/km) = 0.271 km ≈ 271 m.
About 541 m counts only one conductor. About 135 m doubles the round trip twice.
Source: Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 4D.
Safety Instrumented Systems (Problems 30–36)
What this area tests most often:
- LOPA multiplies frequencies and PFDs of independent layers, then compares the result with a tolerable frequency.
- PFDavg grows with the proof-test interval, and common cause can dominate redundant designs.
- Voting trades safety against spurious trips.
- The lifecycle has an order, and bypasses and changes are managed, not improvised.
Problem 30 · From LOPA to a SIL target
SI-01 · Safety Instrumented Systems, 5B · Select one
A LOPA scenario has an initiating event frequency of 0.2 per year. Two independent protection layers are credited: a BPCS control loop (PFD 0.1) and a relief valve (PFD 0.01). The tolerable frequency is 1 × 10⁻⁵ per year. What SIL does the new SIF need? Use the low-demand bands SIL 1: 10⁻² ≤ PFDavg < 10⁻¹; SIL 2: 10⁻³ ≤ PFDavg < 10⁻²; SIL 3: 10⁻⁴ ≤ PFDavg < 10⁻³.
Show answer and solution
Answer and solution
Answer: A. SIL 1
Mitigated frequency without the SIF = 0.2 × 0.1 × 0.01 = 2 × 10⁻⁴ per year. Required SIF PFD = 1 × 10⁻⁵ ÷ 2 × 10⁻⁴ = 0.05 (a risk reduction of 20). 0.05 falls in the SIL 1 band.
- B: Leaves out the BPCS credit (required PFD 0.005).
- C: Leaves out the relief valve credit (required PFD 5 × 10⁻⁴).
- D: 2 × 10⁻⁴ per year is still 20 times the tolerable frequency.
Source: PR electronics, “Process Industry Guide to SIL & Functional Safety” (SIL tables); Worked from first principles using only the inputs given. Topic: NCEES PE Control Systems specification, April 2027, 5B.
Problem 31 · 1oo1 PFDavg
SI-02 · Safety Instrumented Systems, 5D · Select one
A single-channel (1oo1) SIF has a dangerous undetected failure rate λDU = 2 × 10⁻⁶ per hour and is proof-tested once a year (8,760 h). Using PFDavg ≈ λDU × T ÷ 2 (no diagnostics, repair time neglected), which SIL band does its PFDavg fall in?
Show answer and solution
Answer and solution
Answer: A. About 8.8 × 10⁻³, the SIL 2 range
PFDavg ≈ 2 × 10⁻⁶ × 8,760 ÷ 2 = 8.76 × 10⁻³, inside the SIL 2 band (10⁻³ to 10⁻²). PFD alone doesn't earn a SIL claim: hardware fault tolerance and systematic capability must also meet the target.
- B: Forgets to divide by 2.
- C: Is a decimal slip.
- D: Uses the failure rate as if it were a probability.
Try a change: Stretch the test interval to 2 years and PFDavg doubles to about 1.75 × 10⁻² — now only the SIL 1 range.
Source: The 61508 Association, “SIL Calculations: Practical Guidance in the use of IEC 61508-6:2010” (2024), slide 11; PR electronics, “Process Industry Guide to SIL & Functional Safety” (SIL tables). Topic: NCEES PE Control Systems specification, April 2027, 5D.
Problem 32 · Voting trade-offs
SI-03 · Safety Instrumented Systems, 5D · Select one
Compared with a 2oo2 arrangement of the same two sensors, a 1oo2 arrangement has:
Show answer and solution
Answer and solution
Answer: A. A lower PFD but more spurious trips
In 1oo2, either sensor can trip the system. That makes a dangerous failure less likely to go unnoticed (lower PFD), but a false signal from either sensor also trips the process. 2oo2 needs both sensors to agree: fewer spurious trips, but worse PFD. 2oo3 is the usual compromise.
- B: Describes 2oo2 relative to 1oo2.
- C: Voting changes both measures.
- D: Gets both benefits, which neither arrangement offers.
Source: Kenexis, “Comparison of 2oo3 voting and 2oo2 voting”. Topic: NCEES PE Control Systems specification, April 2027, 5D.
Problem 33 · Safety lifecycle order
SI-04 · Safety Instrumented Systems, 5A · Put in order
Put these safety lifecycle activities in order, first to last.
- A. Safety requirements specification (SRS)
- B. Operation and maintenance
- C. Hazard and risk assessment
- D. Installation, commissioning, and validation
- E. Design and engineering of the SIS
- F. Allocation of safety functions to protection layers
- G. Modification and decommissioning
Show answer and solution
Answer and solution
Answer: C → F → A → E → D → B → G.
Correct order: hazard and risk assessment → allocation of safety functions to protection layers → SRS → design and engineering → installation, commissioning, and validation → operation and maintenance → modification and decommissioning. You can't specify a SIF before you know the hazard and which layer must reduce it, and you can't validate what hasn't been designed and installed.
The most common slip is writing the SRS before allocation. The SRS describes functions that allocation has already assigned to the SIS.
Scored all-or-nothing: the whole sequence must be right.
Source: UK Health and Safety Executive, “Functional safety” (safety-lifecycle sequence). Topic: NCEES PE Control Systems specification, April 2027, 5A–5G. Topic: NCEES PE Control Systems specification, April 2027, 5A.
Problem 34 · Bypassing a safety function
SI-05 · Safety Instrumented Systems, 5F · Select all that apply
Maintenance needs to bypass one transmitter in a running SIF. Which practices reflect good bypass management? Select all that apply.
Show answer and solution
Answer and solution
Answer: A and B and C.
CCPS guidance on bypassing instrumented safeguards calls for permits that manage the work and specify controls, compensating measures sufficient to cover the risk gap, and continuous communication with the control room. Where possible, it recommends timers that restore full operation automatically.
- D: Turns a temporary reduction in protection into an open-ended one.
Scored all-or-nothing, like NCEES multiple-correct items: you need every correct option and nothing else.
Source: AIChE CCPS, Bypassing or Jumpering Instrumented Safeguards: Strategies and Effective Practices. Topic: NCEES PE Control Systems specification, April 2027, 5F.
Problem 35 · De-energize to trip
SI-06 · Safety Instrumented Systems, 5D · Select one
An SIS shutdown valve uses a spring-return, fail-closed actuator with a solenoid in a de-energize-to-trip design. The 24 VDC supply to the solenoid is lost. What happens?
Show answer and solution
Answer and solution
Answer: A. The solenoid de-energizes and the valve closes to its safe position
In a de-energize-to-trip system, the de-energized state is the safe state, so losing power is treated like a trip demand. The solenoid vents the actuator and the spring drives the valve to its fail-closed position. Energize-to-trip designs, which need power to act, are the exception.
- B: Describes a fail-last-position actuator, which this isn't.
- C: Is the opposite of fail-closed.
- D: Spring-return actuators move without electrical power.
Source: Rockwell Automation, Studio 5000 Logix Designer, “De-Energize to Trip systems”; Emerson, Control Valve Handbook, 6th ed., §1.4 p. 26 and §1.5 p. 29. Topic: NCEES PE Control Systems specification, April 2027, 5D.
Problem 36 · Common cause in redundant sensors
SI-07 · Safety Instrumented Systems, 5D · Select one
A 1oo2 sensor pair has λDU = 2 × 10⁻⁶ per hour per channel, a proof-test interval T = 8,760 h, and a common-cause factor β = 5%. Using the simplified PFDavg ≈ (λDU × T)² ÷ 3 + β × λDU × T ÷ 2 (no diagnostics, repair time neglected), what is the PFDavg?
Show answer and solution
Answer and solution
Answer: A. About 5.4 × 10⁻⁴
λDU × T = 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: redundancy helps, but diversity and separation decide how much.
- B: Leaves out common cause.
- C: Uses (λDU × T)² without dividing by 3, and no common cause.
- D: Is the single-channel (1oo1) result.
Source: The 61508 Association, “SIL Calculations: Practical Guidance in the use of IEC 61508-6:2010” (2024), slide 11. Topic: NCEES PE Control Systems specification, April 2027, 5D.
How to read your results
Treat your percentage as feedback on these 36 problems, not an NCEES score or a prediction of passing. Six to eight problems per area can show which methods you can't yet explain, but they can't measure an area precisely.
NCEES counts correct answers, converts the count to a scaled score to adjust for differences in difficulty, and compares it with a standard set by subject-matter experts. It doesn't publish the passing score, so no practice percentage converts into an exam result (Examinee Guide, p. 14).
Your plan to April 14, 2027
Assumptions: you start the week of Monday, October 12, 2026, study about 8 focused hours a week (roughly 200 hours in all), and work full time. The plan takes the areas one at a time, then switches to mixed and timed practice. This is our editorial plan, not an NCEES requirement, and no number of hours guarantees a pass.
| Week of | Focus | Do this | Finish the week with |
|---|---|---|---|
| Oct 12 | Setup and baseline | Check your board's approval process and register in MyNCEES. Download the 2027 specification and the handbook. Work all 36 problems before opening any solution. | An error log and a ranked list of weak areas |
| Oct 19 | References | Arrange lawful study access to ISA-5.1-2024 and ISA/IEC 61511-1 (2018). Read each scope and table of contents. Run two lookup drills in the handbook. | You can say what each reference covers |
| Oct 26 – Nov 16 (4 weeks) | Measurement | Flow, level, pressure, temperature; analyzers and sampling; fire and gas detection; machinery monitoring; sensor selection and installation; conversions; thermowells; loop tests. Rework Problems 1–8. | DP flow and level calculations in under 6 minutes each |
| Nov 23 – Dec 14 (4 weeks) | Control Systems | Drawings and ISA-5.1 symbols; process dynamics and tuning; cascade, ratio, feedforward, split range; ladder, function block, SFC, structured text; HMI and alarms; architecture; FAT, SAT, MOC; cybersecurity. Rework Problems 9–16. | You can trace a logic block and a P&ID loop without notes |
| Dec 21 – Dec 28 | Light weeks | Two to four hours: retry earlier misses on fresh problems. | An up-to-date error log |
| Jan 4 – Jan 25 (4 weeks) | Final Control Elements | Valve types, trim, materials; sizing; cavitation, flashing, choked flow, noise; actuators and accessories; relief devices; motors and drives; solenoids, relays, regulators; commissioning. Rework Problems 17–23. | Valve sizing and fail-position choices you can justify |
| Feb 1 – Feb 15 (3 weeks) | Signals, Transmission, and Networking | Signal types and conversion; hazardous areas and intrinsic safety; grounding, EMI, UPS and fusing; loop calculations; media, protocols, and addressing. Rework Problems 24–29. | A loop budget and an entity check done from scratch |
| Feb 22 – Mar 8 (3 weeks) | Safety Instrumented Systems | Lifecycle; hazard and risk assessment and LOPA; SRS; SIL calculations and voting; SAT; proof testing, bypasses, MOC. Daily 61511-1 lookup drills. Rework Problems 30–36. | A LOPA-to-SIL calculation without notes |
| Mar 15 | Mixed practice; set up the rehearsal | Mixed sets so you choose the method yourself. Use a current 2027-aligned 85-question practice source if you have one; otherwise assemble fresh mixed problems, knowing they aren't an official exam. Block the rehearsal day. | Rehearsal day on the calendar |
| Mar 22 | Full rehearsal | 85 questions in 8.5 hours. Submit the first half before starting the second. Use only the handbook, the two standards, your approved calculator, and a whiteboard. | Results by area and error type |
| Mar 29 – Apr 5 (2 weeks) | Fix the leaks | Most of the time on the areas that cost the most questions. Check each fix on a problem you haven't seen. Confirm your appointment, ID, route, and calculator. | A short repair list built from work you actually did |
| Apr 12 | Taper | Short mixed sets early in the week. Nothing new the day before. | Rested, with an exam-day plan |
| Wed, Apr 14 | Exam day |
Timed blocks you can run with this page: Block A is the odd-numbered problems (1, 3, 5 … 35) and Block B the even-numbered ones (2, 4, 6 … 36). Each has 18 problems, so give it 18 × 6 = 108 minutes. You'll have seen them before, so treat them as pacing practice, not a fresh exam.
A typical 8-hour week
Two 90-minute sessions on concepts and handbook relations, two 60-minute problem sessions, two 30-minute lookup drills, one 90-minute mixed or timed set, and 30 minutes updating the error log. Move the blocks around work and family; keep the mix.
Starting later?
The plan compresses if you keep the order and raise the hours:
- Starting in mid-January (12 weeks, about 14 hours a week): Jan 18 setup and references · Jan 25 – Feb 1 Measurement · Feb 8 – 15 Control Systems · Feb 22 – Mar 1 Final Control Elements · Mar 8 Signals and networking · Mar 15 – 22 SIS · Mar 29 full rehearsal · Apr 5 fix the leaks and taper.
- Fewer than 8 weeks: don't try to cover everything evenly. Work all 36 problems, rank your weak areas, put most hours into the biggest weak areas, and keep daily lookup drills and at least one timed mixed set. Be honest about whether April 14 is realistic for you. Control Systems is offered once a year, so missing this date means waiting for the next annual sitting.
- Strong in one area? Keep a short review there instead of skipping it, and move the saved hours to areas you don't touch at work.
Turn each miss into your next study task
| What went wrong | Next step | How you know it's fixed |
|---|---|---|
| Didn't know the concept | Go back to the concept and its assumptions before doing more problems | You can explain why the method applies |
| Couldn't find the relation or requirement | Repeat the lookup; record the search words that worked | The same lookup is faster next time |
| Used the wrong standard edition | Re-find it in ISA-5.1-2024 or 61511-1 (2018) | You check the edition before you start |
| Picked the wrong model or form (gauge vs. absolute, linear vs. square root, fail open vs. fail closed) | Write one sentence on which form applies and why, then rework with one input changed | You can say why the other form doesn't fit |
| Units | Redo it with units on every line | Units cancel to what the question asks for |
| Missed a given condition in the stem | Underline every given value and condition, then rework | Nothing in the stem goes unused |
| Arithmetic or sign | Redo from scratch, then check magnitude and direction | The check agrees without copying your first try |
| Ran out of time | Note where the time went; practice flagging and returning in a timed block | Your average moves toward 6 minutes |
Error log fields: problem or source · date · area and subtopic · minutes spent · your answer · correct, wrong, or lucky guess · first wrong step · error type (from the table) · where you found the fix (reference, search words) · date you'll retry on a fresh problem.
Your last week
Don't start anything new. Pick your two most repeated errors and your two slowest lookups, fix those, and run one timed block. Confirm your appointment, ID, and calculator. If you need to move your appointment, you must reschedule at least 48 hours ahead, and Pearson charges $50 (Examinee Guide, p. 6). Moving off April 14, 2027 means waiting for the next annual date.
Exam-day rules that change how you practice
- About 6 minutes per question on average. That's 510 minutes ÷ 85 questions. It's a budget, not a per-question limit.
- Two sections, one clock. After roughly half the questions, you review and submit them, and you can't go back. The sections aren't timed separately, so time you save early carries forward.
- Breaks: the optional 50-minute break comes after you submit the first section. Unused break time doesn't add exam time, unscheduled breaks come out of your exam time, and you can't leave the building during them.
- No penalty for wrong answers, so answer everything. Some questions are unscored pretest items that look like all the others.
- One approved calculator. For 2026 exams, NCEES approved the HP 33s and HP 35s, any Casio fx-115 or fx-991 model, and any TI-30X or TI-36X model. An on-screen TI-30XS is also available. NCEES hadn't posted its 2027 list when we checked, so confirm it before April.
- Scratch work: you get two reusable booklets and three markers, not paper and pencil. Work some practice problems on a whiteboard so it isn't new on exam day.
- Arrive 30 minutes early, and make sure the first and last names on your appointment confirmation match your ID.
Sources: NCEES Examinee Guide, May 2026, pp. 6 and 8–14; NCEES 2026 calculator list, p. 1.
Quick answers
Is PE Control Systems the same as the PE Electrical and Computer exam?
No. PE Control Systems is its own NCEES exam, with design standards supplied (ISA-5.1 and 61511-1). PE Electrical and Computer: Electronics, Controls, and Communications is a different exam, also on April 14, 2027, with no supplied design standards. Both are 85 questions in a 9.5-hour appointment (Examinee Guide, p. 16). If your work is power distribution, also look at PE Electrical and Computer: Power exam prep.
What's the pass rate?
In the Control Systems row NCEES labels "July 2026," 61% of 250 first-time takers passed, as did 45% of 83 repeat takers (NCEES). Those numbers describe a past group of examinees on the old specification, not your personal odds or a readiness target.
What score do I need?
NCEES doesn't publish a passing score; results are reported as pass or fail, and the passing standard varies slightly with exam difficulty (Examinee Guide, p. 14). Claims that you need a specific percentage aren't based on anything official.
I failed in 2026. How do I use my diagnostic report?
The report shows your performance in each knowledge area on a 0–15 scale, compared with the average of passing examinees (Examinee Guide, pp. 20–21). Your report uses the old area names, but the five areas map straight across (Safety Systems became Safety Instrumented Systems). Rank your weak areas, check the 2027 changes inside each one, and confirm each weakness with fresh problems.
Is it open book?
No. It's closed book with an electronic reference: the handbook and the two listed standards on screen. You can't bring your own books, notes, or copies of the standards (Examinee Guide, pp. 8 and 10). If a prep page calls it open book, it's describing the old format: the October 2011 specification described an 8-hour, 80-question open-book exam.
Do I need the FE first, or four years of experience?
Your state licensing board decides who can sit for the PE. NCEES designs the exam for engineers with at least four years of post-college experience, but whether you can test earlier is your board's rule (NCEES; Examinee Guide, p. 2). Find yours in the NCEES board directory. If you still need the FE, start with FE exam prep.
Does passing make me a licensed PE?
No. Your licensing board issues the license. Passing the exam is one of its requirements.
Do I need a prep course?
No course is required. Whatever you use, check that it matches the April 2027 specification and ISA-5.1-2024. Material built to the older outline won't reflect the 2027 changes.
How do registration, results, retakes, and accommodations work?
You register and pay in MyNCEES after checking your board's approval process. NCEES allows one attempt per testing window and no more than three in 12 months, and some boards are stricter; for Control Systems, the once-a-year date is the practical limit. Accommodations must be requested during registration (Examinee Guide, pp. 3–5). See NCEES exam registration, NCEES exam results, NCEES retake rules, and NCEES exam accommodations.
Comparing Control Systems with other PE exams? See PE exam prep.
Sources
Exam facts on this page were last verified October 11, 2026, against the NCEES sources below: the exam date, format, fee, references and editions, question ranges, calculator list, scoring and retake rules, and pass rates. We re-solved every practice problem and recomputed every calculation. The problems have not been reviewed by NCEES or by a licensed professional engineer. This guide was developed with AI-assisted research and editing; see how Castleport uses sources and AI assistance.
NCEES
- PE Control Systems exam page (date, format, fee, references, pass rates)
- PE Control Systems specification, effective April 2027
- PE Control Systems CBT specification, effective October 2022 (historical)
- NCEES memo to member boards, July 20, 2026 (copy hosted by the Virginia DPOR)
- NCEES Examinee Guide, May 2026
- NCEES computer-based testing and item types
- NCEES 2026 calculator list
Teaching sources for the practice problems
- Measurement: Control Engineering on DP square-root scaling; Automation.com/ISA on DP level and zero suppression; Process Instruments TN 147 on pH and temperature; Emerson Rosemount 3051S Quick Start Guide; Bently Nevada ORBIT, Vol. 30 No. 1
- Control: APMonitor PI tuning; Dataforth AN124 on IMC tuning; Control Guru on cascade and feedforward; Control Global on alarm chatter; NIST SP 800-82r3
- Final control elements: Swagelok valve sizing bulletin; Valve Magazine on flashing and cavitation; Emerson Control Valve Handbook, 6th ed.; Control Global on relief valves and back pressure; OSTI abstract on pilot-operated relief valves; Pumps & Systems on variable-speed pumping
- Signals and networking: Pepperl+Fuchs entity-concept control drawing; IETF RFC 950
- Safety instrumented systems: PR electronics SIL guide; 61508 Association on IEC 61508-6 simplified formulas; Kenexis on voting; UK HSE on the functional-safety lifecycle; AIChE CCPS on bypassing instrumented safeguards; Rockwell Automation on de-energize-to-trip systems
These teaching sources explain the methods. They aren't the references supplied on the exam.
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