An operator is the attendant for an entry into a drained deaerator storage tank. One entrant stops answering and appears to have collapsed. What does the attendant do?

Raise the alarm and call the rescue team, staying outside.

Test the tank atmosphere from the manway, then call for rescue.

Ventilate the tank for ten minutes, then go in alone.

Enter at once and drag the entrant to the manway.

A butt weld in a header passes visual inspection: good profile, no undercut, and no cracks or porosity at the surface. Which defect could it still contain?

an unfilled crater at the end of the bead

overlap of weld metal at one toe

spatter on the plate beside the weld

During an outage, a horizontal lift check valve in the discharge line of a condensate transfer pump was replaced. The new valve went in with the arrow cast on its body pointing toward the pump. What happens when the pump is started?

Pump pressure seats the disc, so no condensate is delivered.

Condensate flows normally, as a lift check opens either way.

Condensate flows, but the valve will not close if the pump stops.

Condensate flows, but with a much larger pressure drop across it.

In a three-wire start–stop circuit with a seal-in contact, an operator holds the Start button in and presses the Stop button at the same time. What does the motor do?

Stops, because Stop opens the path to the coil

Keeps running, because Start bypasses the Stop button

Keeps running, held in by its seal-in contact

Runs while Start is held, then stops

Which of these boilers must be built as a power boiler to ASME Section I? Pressures are gauge.

A 600 kW hot-water boiler at 1 170 kPa and 124 °C

A 4 500 kW steam boiler at 100 kPa

A 2 000 kW hot-water boiler at 1 050 kPa and 118 °C

A 3 000 kW hot-water boiler at 1 100 kPa and 121 °C

A pulp mill's power boiler has metered combustion control with cross-limiting: the fuel flow and the air flow are each measured and controlled. When steam demand falls, how does the control change the two flows?

The fuel flow is cut first and the air flow follows it

The air flow is cut first and the fuel flow follows it

Fuel and air are cut together through a fixed linkage

The fuel is cut, and the O₂ trim alone brings the air down

10 Free 4th Class Power Engineering Exam Questions (4A) + PDF

Ten questions for paper A (4A) of the 4th Class Power Engineer exam, in the format of SOPEEC's sample questions: four options, one correct answer. Paper A covers the theory, codes, safety and the boiler itself: mechanics, thermodynamics and steam, safety, materials, piping, electricity and controls; paper B covers operating the plant. The questions are free to use, print and share. There is nothing to sign up for on this page.

01How to use these

The questions come from the TicketPrep 4A bank and follow the units of the SOPEEC Fourth Class syllabus in order: A (Elementary Mechanics and Dynamics), B (Elementary Chemistry and Thermodynamics), D (Power Plant/Heating Plant Safety), E (Environment), F (Material and Welding), G (Piping and Valves), H (Electricity), I (Energy Plant Instrumentation and Controls), K (Boilers), L (Boiler Systems). None of SOPEEC, ABSA, Technical Safety BC, TSASK or TSSA publishes how many exam questions each unit carries, so the ten are spread by TicketPrep's practice-bank split. That split gives one question each to ten of the twelve units; legislation, codes and standards (C) and plant communication (J) are in the full bank. Four of the ten are calculations. Answer each one before opening the explanation. Every explanation covers all four options, because knowing why a wrong answer is wrong is most of the skill.

Pace. The syllabus sets paper A at 100 multiple-choice questions in 3 hours, about 1 min 48 s per question. Try these at that pace.
Data. At the sitting you are normally given the handbook of formulae and physical constants and the steam tables. The questions here put the values they need in the question, so you can work them without either.
Print. The printed version shows the questions and the answer key without the explanations, so it works as a handout. The explanations are on the web page. Download PDF
Mark. Each paper is passed on its own at 65 percent; the two papers are not averaged. Seven of ten here clears that bar.

02The questions

Calculation · Unit A · Elementary Mechanics and Dynamics

At a gas plant in the mountains the barometer reads 88.0 kPa (standard atmospheric pressure is 101.3 kPa). The gauge on an instrument-air receiver reads 300.0 kPa. What is the absolute pressure in the receiver?
1. 212.0 kPa
2. 300.0 kPa
3. 388.0 kPa
4. 401.3 kPa
Show the answer

C. Correct. Absolute = gauge + local atmospheric = 300.0 + 88.0 = 388.0 kPa. The gauge measures above the atmosphere actually present, so the plant's own barometer reading is used, not the standard 101.3 kPa.

Why not the others

A. This subtracts the barometer reading: 300.0 − 88.0. A gauge reads above the local atmosphere, so absolute = gauge + atmospheric = 300.0 + 88.0 = 388.0 kPa.

B. This takes the gauge reading as absolute. A gauge reads zero when open to the atmosphere, so the local atmosphere must be added: 300.0 + 88.0 = 388.0 kPa.

D. This adds standard atmospheric pressure, 300.0 + 101.3. The gauge reads above the local atmosphere, which is 88.0 kPa here: 300.0 + 88.0 = 388.0 kPa.
Calculation · Unit B · Elementary Chemistry and Thermodynamics

Steam sampled from a pulp-mill boiler header at 2 500 kPa absolute has a dryness fraction of 0.90. Given: at 2 500 kPa absolute, hf = 961.9 kJ/kg and hfg = 1 840.0 kJ/kg. What is the enthalpy of the wet steam?
1. 1 656.0 kJ/kg
2. 2 521.7 kJ/kg
3. 2 617.9 kJ/kg
4. 2 801.9 kJ/kg
Show the answer

C. Correct. In wet steam only the fraction x of each kilogram has evaporated, so only x of the latent heat is present: h = hf + x hfg = 961.9 + 0.90 × 1 840.0 = 961.9 + 1 656.0 = 2 617.9 kJ/kg. All of the water carries its sensible heat hf.

Why not the others

A. This is x hfg alone: 0.90 × 1 840.0 = 1 656.0 kJ/kg. All the water, evaporated or not, holds hf = 961.9 kJ/kg, so h = 961.9 + 1 656.0 = 2 617.9 kJ/kg.

B. The dryness fraction was applied to all of hg: 0.90 × (961.9 + 1 840.0) = 0.90 × 2 801.9 = 2 521.7 kJ/kg. The water that stays liquid still has its hf, so x multiplies hfg only: 2 617.9 kJ/kg.

D. This is hg = 961.9 + 1 840.0 = 2 801.9 kJ/kg, the value for dry saturated steam. With 10 % of the water not evaporated, h = hf + 0.90 × hfg = 2 617.9 kJ/kg.
Procedural · Unit D · Power Plant/Heating Plant Safety

An operator is the attendant for an entry into a drained deaerator storage tank. One entrant stops answering and appears to have collapsed. What does the attendant do?
1. Raise the alarm and call the rescue team, staying outside.
2. Test the tank atmosphere from the manway, then call for rescue.
3. Ventilate the tank for ten minutes, then go in alone.
4. Enter at once and drag the entrant to the manway.
Show the answer

A. Correct. The attendant's job is to stay at the entrance, keep track of the entrants and summon rescue at once; a lifeline lets them start a retrieval without going in. Entry is left to the trained, equipped rescue team.

Why not the others

B. Testing first delays help when minutes count. The attendant calls the rescue team at once and stays outside; the trained rescuers test and ventilate as part of their own entry.

C. Going in alone, even after ventilating, puts a second person at risk with no one outside to raise help. The attendant stays outside and calls the trained rescue team.

D. Untrained or unprotected would-be rescuers are a large share of confined-space deaths: whatever overcame the entrant can overcome the attendant too. The attendant stays outside and calls the rescue team.
Calculation · Unit E · Environment

Two identical forced-draft fans each give a sound level of 87 dBA at an operator's work station when running alone. What sound level reaches the work station with both fans running?
1. 87 dBA
2. 90 dBA
3. 93 dBA
4. 174 dBA
Show the answer

B. Correct. Sound levels add by energy, not arithmetically: L = 10 log₁₀(10^8.7 + 10^8.7) = 87 + 10 log₁₀ 2 = 87 + 3.0 = 90 dBA. Each doubling of identical sources adds 3 dB, a large change on a logarithmic scale.

Why not the others

A. 87 dBA treats the second fan as adding nothing, like averaging the two levels. Two equal sources double the sound energy: 87 + 10 log₁₀ 2 = 87 + 3.0 = 90 dBA.

C. Adding 6 dB uses 20 log₁₀ 2, the rule for doubling sound pressure. Two separate fans double the sound energy, which adds 10 log₁₀ 2 = 3 dB: 90 dBA.

D. 87 + 87 = 174 adds the decibel values as ordinary numbers. Decibels are logarithmic, so two equal sources add only 3 dB: 90 dBA.
Procedural · Unit F · Material and Welding

A butt weld in a header passes visual inspection: good profile, no undercut, and no cracks or porosity at the surface. Which defect could it still contain?
1. an unfilled crater at the end of the bead
2. overlap of weld metal at one toe
3. spatter on the plate beside the weld
4. slag trapped between passes
Show the answer

D. Correct. Visual inspection sees only the surface. Slag left on one pass and buried by the next, like lack of fusion or internal porosity, is hidden inside the weld and needs a volumetric test such as radiography or ultrasonics to find it.

Why not the others

A. A crater at the end of a bead is on the surface and is a standard visual-inspection check. Slag trapped between passes is internal and can be present in a weld that passes visually.

B. Overlap is weld metal rolled out over the plate at the toe without fusing to it. It is a profile defect on the surface, and the weld was found to have a good profile. Slag trapped between passes is internal and can be present in a weld that passes visually.

C. Spatter sits on the plate surface and is plainly visible, so visual inspection would have found it. The defect that can hide below a clean surface is slag trapped between passes.
Procedural · Unit G · Piping and Valves

During an outage, a horizontal lift check valve in the discharge line of a condensate transfer pump was replaced. The new valve went in with the arrow cast on its body pointing toward the pump. What happens when the pump is started?
1. Condensate flows normally, as a lift check opens either way.
2. Condensate flows, but the valve will not close if the pump stops.
3. Condensate flows, but with a much larger pressure drop across it.
4. Pump pressure seats the disc, so no condensate is delivered.
Show the answer

D. Correct. The body arrow shows the direction of flow the valve is built to pass. Reversed, flow from the pump pushes the disc onto its seat and the valve blocks it, so nothing is delivered. Body markings such as the flow arrow, size, pressure class and material are checked before a valve is installed.

Why not the others

A. A lift check passes flow in one direction only: flow from under the seat lifts the disc, flow from above seats it. The arrow shows that direction; reversed, the valve blocks the pump's flow.

B. Reversed, the valve cannot pass pump flow at all, because pump pressure seats the disc. The arrow on the body shows the only direction in which it will open.

C. A reversed lift check does not merely add resistance; flow from the wrong side holds its disc shut. The pump's discharge is blocked and no condensate is delivered.
Calculation · Unit H · Electricity

A 230 V DC motor has an armature resistance of 0.25 Ω. At full speed its armature generates a counter-voltage of 218 V. What current flows in the armature, to three significant figures?
1. 48.0 A
2. 872 A
3. 920 A
4. 1 790 A
Show the answer

A. Correct. The turning armature acts as a generator, and its counter-voltage opposes the supply, so only the difference drives current: I = (230 V − 218 V) ÷ 0.25 Ω = 12 V ÷ 0.25 Ω = 48.0 A. If the motor slows under load, the counter-voltage falls and the current rises.

Why not the others

B. This divides the counter-voltage by the resistance: 218 V ÷ 0.25 Ω = 872 A. The current is driven by the net voltage, supply minus counter-voltage: 12 V ÷ 0.25 Ω = 48.0 A.

C. This ignores the counter-voltage: 230 V ÷ 0.25 Ω = 920 A. That is what would flow at the instant of starting, before the armature turns, which is why DC motors need starting resistance. At full speed: (230 − 218) ÷ 0.25 = 48.0 A.

D. This adds the counter-voltage to the supply: (230 + 218) ÷ 0.25 = 1 792, about 1 790 A. The counter-voltage opposes the supply, so it is subtracted: (230 − 218) ÷ 0.25 = 48.0 A.
Procedural · Unit I · Energy Plant Instrumentation and Controls

In a three-wire start–stop circuit with a seal-in contact, an operator holds the Start button in and presses the Stop button at the same time. What does the motor do?
1. Keeps running, because Start bypasses the Stop button
2. Stops, because Stop opens the path to the coil
3. Keeps running, held in by its seal-in contact
4. Runs while Start is held, then stops
Show the answer

B. Correct. The normally closed Stop button is in series ahead of the parallel Start and seal-in branch, so pressing Stop breaks the only path to the coil whatever Start is doing. Stop always takes priority in this circuit.

Why not the others

A. Start is wired in parallel with the seal-in contact, not around the Stop button. Opening Stop breaks the circuit to both, so the motor stops.

C. The seal-in contact is in the same parallel branch as Start, behind the Stop button. Opening Stop de-energizes the coil and opens the seal-in, so the motor stops.

D. Holding Start does not help while Stop is open, because Stop is in series ahead of the Start branch. The motor stops as soon as Stop is pressed.
Procedural · Unit K · Boilers

Which of these boilers must be built as a power boiler to ASME Section I? Pressures are gauge.
1. A 4 500 kW steam boiler at 100 kPa
2. A 600 kW hot-water boiler at 1 170 kPa and 124 °C
3. A 2 000 kW hot-water boiler at 1 050 kPa and 118 °C
4. A 3 000 kW hot-water boiler at 1 100 kPa and 121 °C
Show the answer

B. Correct. A hot-water boiler is a heating boiler only while it does not exceed 1 100 kPa (160 psi) and 121 °C (250 °F). This one exceeds both, so it is a power boiler built to Section I, however small its output. Classification follows pressure and temperature, not size.

Why not the others

A. Size does not decide the class; pressure and temperature do. A steam boiler is a heating boiler while its pressure does not exceed 103 kPa (15 psi), so this large boiler at 100 kPa is a Section IV heating boiler.

C. At 1 050 kPa and 118 °C this boiler is below both hot-water limits, 1 100 kPa (160 psi) and 121 °C (250 °F), so it is a Section IV heating boiler despite its size.

D. This boiler sits exactly at the hot-water limits, 1 100 kPa (160 psi) and 121 °C (250 °F). The limits read 'not exceeding', so a boiler at them is still a heating boiler; it must go above them to be a power boiler.
Procedural · Unit L · Boiler Systems

A pulp mill's power boiler has metered combustion control with cross-limiting: the fuel flow and the air flow are each measured and controlled. When steam demand falls, how does the control change the two flows?
1. The air flow is cut first and the fuel flow follows it
2. Fuel and air are cut together through a fixed linkage
3. The fuel is cut, and the O₂ trim alone brings the air down
4. The fuel flow is cut first and the air flow follows it
Show the answer

D. Correct. Cross-limiting holds the air at or above what the measured fuel flow needs. On a falling load the fuel is reduced first and the air set point follows the fuel flow down, so the furnace never has more fuel than air to burn it. On a rising load the order reverses: air first, then fuel.

Why not the others

A. Cutting the air first would leave the furnace, for a moment, with more fuel than the air can burn: CO, smoke and unburned fuel that can collect and ignite. Cross-limiting keeps the air at or above what the measured fuel flow needs, so on a falling load the fuel leads and the air follows.

B. One linkage moving the fuel valve and the air damper together is single-point positioning (jackshaft) control, which measures neither flow. Metered control measures both and, with cross-limiting, moves them in a set order: fuel first on a falling load.

C. O₂ trim only makes a small correction to the air flow from the flue-gas oxygen reading; it is too slow and too limited to carry the air through a load change. The air flow controller brings the air down, following the fuel.

03Answer key

Question	Answer	Syllabus unit
Q1	C	A · Elementary Mechanics and Dynamics
Q2	C	B · Elementary Chemistry and Thermodynamics
Q3	A	D · Power Plant/Heating Plant Safety
Q4	B	E · Environment
Q5	D	F · Material and Welding
Q6	D	G · Piping and Valves
Q7	A	H · Electricity
Q8	B	I · Energy Plant Instrumentation and Controls
Q9	B	K · Boilers
Q10	D	L · Boiler Systems

04More free questions

For paper A itself, the 4th Class Power Engineering exam guide covers the twelve units, the exam format by province and what you may bring into the room, and the 4th Class Power Engineering practice exam page describes the full TicketPrep 4A question bank. If you also write a Red Seal exam, the same free format covers sixteen Red Seal trades, including the ones closest to plant work.

How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it. TicketPrep is an independent study site; it is not affiliated with SOPEEC or any provincial or territorial regulator, and nothing here is taken from the exam, the exam handbook, the textbooks or the codes.