A ball bearing from a condensate pump has been washed in clean solvent so it can be inspected and refitted. How should it be dried?

Blow it dry with clean, dry air, rings held still

Leave it uncovered on the bench, to air-dry overnight

Spin it with an air jet so the solvent is thrown out of it

Wipe it inside and out with a clean, folded shop rag

A centrifugal pump at a water-treatment plant draws from a sump below it through a foot valve. After the suction line has been drained for a repair, what must be done before the pump is started?

Fill the casing and suction line, venting the casing until water runs out free of air

Start it with the discharge valve open, so the flow it makes draws water up the suction line

Start it with the discharge valve closed, and run it until it builds enough suction to lift water

Run it in short bursts, stopping between them, until the discharge gauge shows pressure

The main gas pressure regulator in a boiler's fuel train fails wide open, and the high-gas-pressure switch locks out the burner. What hazard did the switch prevent?

Overfiring with more gas than the air supply can burn completely

The safety shutoff valves being unable to close against the pressure

Flame failure because the gas pressure fell too low to hold a flame

Flashback of the flame into the burner ports at the higher gas flow

Boiler 2 at a university central plant is being cut in to a common header. A short pipe with a drain joins its non-return valve to the header stop valve. Before the header stop valve is opened, why is that drain opened?

To clear condensate so it is not driven into the header

To vent air from boiler 2 as its pressure comes up

To relieve header pressure so the stop valve opens easily

To lower boiler 2's pressure to match the header

Condensate returning from a food plant's process area normally tests zero hardness. Today the routine test shows hardness, while the duty make-up softener's outlet still tests zero. What should the operator do with that condensate until the source is found?

Dump it to drain and make up the boiler feed with treated water

Keep returning it and increase the boiler's continuous blowdown

Keep returning it and raise the phosphate feed to match the hardness

Switch to the standby softener and regenerate the duty unit early

After a hot-water heating coil in an air-handling unit is refilled following repairs, its upper tubes stay cold while the lower tubes are hot. Water flow through the coil's control valve reads normal. What should the operator do first?

Open the manual air vent at the coil's highest point

Open the drain at the coil's lowest point to flush it

Throttle the control valve to raise the water velocity

Raise the supply water temperature at the boiler plant

10 Free 4th Class Power Engineering Exam Questions (4B)

Ten questions for paper B (4B) of the 4th Class Power Engineer exam, in the exam's format: four options, one correct answer. Paper B is the operating paper: lubrication, pumps and compressors, boiler safety devices, plant operation and maintenance, water treatment, turbines and engines, building systems, refrigeration, HVAC and heating systems. 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 4B bank and follow the units of Part B of the SOPEEC Fourth Class syllabus in order: B1 (Lubrication), B2 (Pumps and Compressors), B3 (Boiler Safety Devices), B4 (Plant (Boiler) Operations), B5 (Power Plant/Heating Plant Maintenance), B6 (Water Treatment), B7 (Prime Movers and Engines), B9 (Refrigeration), B10 (Heating Ventilating and Air Conditioning), B11 (Heating and Cooling Systems). SOPEEC does not publish 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; auxiliary building systems (B8) and types of plants (B12) 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. SOPEEC's 4B sample questions give 3.0 hours for 100 multiple-choice questions, about 1 min 48 s per question. Try these at that pace.
Data. At the sitting you are given the handbook of formulae and physical constants, the steam tables and the refrigeration tables. The questions here put the values they need in the question, apart from basic constants such as g, 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

Procedural · Unit B1 · Lubrication

A ball bearing from a condensate pump has been washed in clean solvent so it can be inspected and refitted. How should it be dried?
1. Leave it uncovered on the bench, to air-dry overnight
2. Spin it with an air jet so the solvent is thrown out of it
3. Wipe it inside and out with a clean, folded shop rag
4. Blow it dry with clean, dry air, rings held still
Show the answer

D. Correct. Clean, dry air carries off the solvent without driving dirt in, and holding the rings stops them spinning. Oil or grease the bearing straight away: a clean, dry bearing rusts quickly from finger marks and humid air.

Why not the others

A. Left bare overnight, a solvent-washed bearing has no protective film and picks up moisture and dust. Blow it dry and lubricate or wrap it at once.

B. Spinning a clean, dry bearing with an air jet runs the balls on bare races with no lubricant, which scores them, and can spin it fast enough to be dangerous. Hold the rings still while blowing it out.

C. A rag leaves lint and fibres in the races, which the balls then roll over. Dry it with clean, dry air and keep it covered.
Procedural · Unit B2 · Pumps and Compressors

A centrifugal pump at a water-treatment plant draws from a sump below it through a foot valve. After the suction line has been drained for a repair, what must be done before the pump is started?
1. Fill the casing and suction line, venting the casing until water runs out free of air
2. Start it with the discharge valve open, so the flow it makes draws water up the suction line
3. Start it with the discharge valve closed, and run it until it builds enough suction to lift water
4. Run it in short bursts, stopping between them, until the discharge gauge shows pressure
Show the answer

A. Correct. An impeller spinning in air makes almost no suction, because air is so much lighter than water, so the casing and suction line are filled first. The vent lets the trapped air out, and the foot valve then holds the water in the line. Filling first also keeps the shaft seal from running dry.

Why not the others

B. With air in the casing the impeller makes far too little pressure difference to lift water, whichever way the discharge valve is set; the pump would just run dry and heat its seal. It must be full of water before it starts.

C. Closing the discharge does not change what an impeller can do with air: it still makes too little suction to lift water, and running dry quickly ruins a mechanical seal or burns packing. The casing must be filled first.

D. Short runs in air still make too little suction to lift the water, and each run is a dry run for the seal. Bumping a pump checks its rotation; it does not prime it.
Procedural · Unit B3 · Boiler Safety Devices

The main gas pressure regulator in a boiler's fuel train fails wide open, and the high-gas-pressure switch locks out the burner. What hazard did the switch prevent?
1. The safety shutoff valves being unable to close against the pressure
2. Flame failure because the gas pressure fell too low to hold a flame
3. Flashback of the flame into the burner ports at the higher gas flow
4. Overfiring with more gas than the air supply can burn completely
Show the answer

D. Correct. With the regulator wide open the burner receives far more gas than the air damper is set for. The flame becomes rich and unstable, making carbon monoxide and soot and risking unburned gas in the furnace and overheating. The high-gas-pressure switch shuts off the fuel before that happens.

Why not the others

A. Safety shutoff valves are chosen to close tight against the highest pressure the fuel train can see. The danger from a regulator stuck open is too much gas for the air, which the high-pressure switch guards against.

B. Too little gas pressure is the low-gas-pressure switch's concern. A regulator stuck wide open raises the pressure, and the hazard is overfiring with too little air.

C. Flashback happens when the mixture leaves the ports slower than the flame burns back, which is a low-flow problem. A higher gas flow pushes the flame away from the ports, not into them; the danger here is overfiring with too little air.
Procedural · Unit B4 · Plant (Boiler) Operations

Boiler 2 at a university central plant is being cut in to a common header. A short pipe with a drain joins its non-return valve to the header stop valve. Before the header stop valve is opened, why is that drain opened?
1. To vent air from boiler 2 as its pressure comes up
2. To relieve header pressure so the stop valve opens easily
3. To clear condensate so it is not driven into the header
4. To lower boiler 2's pressure to match the header
Show the answer

C. Correct. Steam trapped between the two valves condenses while the boiler is off line, and water collects in the pipe. Draining it first means the steam that flows when the valves open does not pick up a slug of water and drive it into the header, where it causes water hammer and wets the steam.

Why not the others

A. Air is cleared from the boiler through its drum vent early in the warm-up, long before cut-in. The drain between the valves is there to remove condensate.

B. The drain is on the boiler side of the closed header stop valve, so it cannot relieve the header. It is opened to clear condensate from the pipe between the two valves.

D. Boiler pressure is matched to the header by the firing rate and the plant's cut-in procedure, not by a small drain. The drain clears condensate lying between the two valves.
Calculation · Unit B5 · Power Plant/Heating Plant Maintenance

A 2 600 kg forced-draft fan motor at a pulp mill is lifted on a symmetric two-leg bridle, each leg at 50° above the horizontal, with the hook directly over the motor's centre of gravity. Ignoring the rigging's own weight, what is the tension in each leg?
1. 12.8 kN
2. 16.6 kN
3. 19.8 kN
4. 33.3 kN
Show the answer

B. Correct. W = 2 600 kg × 9.81 m/s² = 25 506 N = 25.5 kN. Each leg carries T = W ÷ (2 sin θ) = 25.5 ÷ (2 × 0.766) = 16.6 kN. The flatter the legs, the larger this grows.

Why not the others

A. This is half the load's weight, 25.5 ÷ 2 = 12.8 kN, as if the legs hung straight down. Angled legs pull sideways as well as up, so each carries more: T = W ÷ (2 sin 50°) = 16.6 kN.

C. This uses cos 50° in place of sin 50°, as if the angle had been measured from the vertical: 25.5 ÷ (2 × 0.643) = 19.8 kN. With the angle above the horizontal, T = W ÷ (2 sin 50°) = 16.6 kN.

D. This drops the 2, putting the whole load on one leg: 25.5 ÷ 0.766 = 33.3 kN. Two legs share the load: T = W ÷ (2 sin 50°) = 16.6 kN.
Procedural · Unit B6 · Water Treatment

Condensate returning from a food plant's process area normally tests zero hardness. Today the routine test shows hardness, while the duty make-up softener's outlet still tests zero. What should the operator do with that condensate until the source is found?
1. Keep returning it and increase the boiler's continuous blowdown
2. Keep returning it and raise the phosphate feed to match the hardness
3. Dump it to drain and make up the boiler feed with treated water
4. Switch to the standby softener and regenerate the duty unit early
Show the answer

C. Correct. Steam carries no hardness, so hardness in returned condensate means raw or cooling water is leaking in, usually at a process exchanger. The condensate is dumped so the hardness and whatever else is leaking cannot reach the boiler, and exchangers are isolated one at a time to find the leak.

Why not the others

A. Extra blowdown removes some of what enters but does not stop scale-forming water being fed to the boiler. The contaminated condensate is kept out of the boiler instead.

B. More phosphate would precipitate some of the hardness in the boiler, but the leak's other contaminants still enter and the sludge load climbs. Contaminated condensate is dumped until the leak is found.

D. The duty softener's outlet tests zero, so it is not the source, and switching softeners leaves the hardness-bearing condensate going to the boiler. The condensate is dumped until the leaking exchanger is found.
Calculation · Unit B7 · Prime Movers and Engines

A refinery cooling tower receives water at 37.0 °C and returns it to the condensers at 28.0 °C. The entering air is at 31.0 °C dry bulb and 22.0 °C wet bulb. What is the tower's approach?
1. 3.0 °C
2. 6.0 °C
3. 9.0 °C
4. 15.0 °C
Show the answer

B. Correct. Approach = cold-water temperature − entering-air wet-bulb temperature = 28.0 − 22.0 = 6.0 °C. It shows how close the tower brings the water to the lowest temperature evaporation could reach; a widening approach at the same load and wet bulb points to a problem in the tower.

Why not the others

A. 3.0 °C is 31.0 − 28.0, measured to the dry-bulb temperature. Evaporative cooling is limited by the wet bulb, so approach = cold water − wet bulb = 28.0 − 22.0 = 6.0 °C; the cold water can be below the dry bulb, as it is here.

C. 9.0 °C is 37.0 − 28.0, the range (hot water in minus cold water out), which measures the heat taken off the water. Approach compares the cold water with the wet bulb: 28.0 − 22.0 = 6.0 °C.

D. 15.0 °C is 37.0 − 22.0, the hot water against the wet bulb, which is range plus approach. Approach uses the cold water: 28.0 − 22.0 = 6.0 °C.
Calculation · Unit B9 · Refrigeration

A cold-storage plant runs a single-stage ammonia compressor at a suction pressure of 30 kPa(g) and a discharge pressure of 1 060 kPa(g), with atmospheric pressure at 101.3 kPa. What is its compression ratio, to one decimal place?
1. 8.1 to 1
2. 8.8 to 1
3. 35.3 to 1
4. 38.7 to 1
Show the answer

B. Correct. Compression ratio uses absolute pressures: discharge = 1 060 + 101.3 = 1 161.3 kPa absolute; suction = 30 + 101.3 = 131.3 kPa absolute; ratio = 1 161.3 ÷ 131.3 = 8.8 to 1. A ratio this high in one stage means hot discharge gas and poor volumetric efficiency, which is why low-temperature ammonia plants often compress in two stages.

Why not the others

A. This converts only the suction to absolute: 1 060 ÷ 131.3 = 8.1. Both pressures must be absolute: 1 161.3 ÷ 131.3 = 8.8 to 1.

C. This divides the gauge readings: 1 060 ÷ 30 = 35.3. Gauge pressures exaggerate the ratio badly when the suction is near atmospheric; with absolute pressures it is 1 161.3 ÷ 131.3 = 8.8 to 1.

D. This converts only the discharge to absolute: 1 161.3 ÷ 30 = 38.7. The suction must be absolute too, 131.3 kPa, giving 8.8 to 1.
Procedural · Unit B10 · Heating Ventilating and Air Conditioning

After a hot-water heating coil in an air-handling unit is refilled following repairs, its upper tubes stay cold while the lower tubes are hot. Water flow through the coil's control valve reads normal. What should the operator do first?
1. Open the manual air vent at the coil's highest point
2. Open the drain at the coil's lowest point to flush it
3. Throttle the control valve to raise the water velocity
4. Raise the supply water temperature at the boiler plant
Show the answer

A. Correct. Refilling leaves air trapped in the top of the coil, where it blocks water from the upper tubes, and normal flow simply bypasses them through the lower tubes. Venting at the high point lets water fill the whole coil; the vent is then closed once water runs clear of air.

Why not the others

B. Draining at the bottom removes water, not the air pocket at the top, and debris would not leave only the upper tubes cold after a refill. Air is vented at the coil's highest point.

C. Throttling reduces the flow and the velocity in the tubes; it does not clear an air pocket. The upper tubes are air-bound after the refill, so the coil's high-point vent must be opened.

D. Hotter water heats only the tubes it can reach; the air pocket in the upper tubes still keeps them cold. Vent the air at the coil's highest point first.
Calculation · Unit B11 · Heating and Cooling Systems

A greenhouse's hot-water heating loop must deliver 850 kW. The water leaves the boiler plant at 87 °C and returns at 71 °C, and the greenhouse air is held at 18 °C. Taking c = 4.19 kJ/kg·K for the water, what water flow must the circulating pumps deliver?
1. 2.33 kg/s
2. 2.94 kg/s
3. 12.7 kg/s
4. 53.1 kg/s
Show the answer

C. Correct. Each kilogram of water gives up c × ΔT = 4.19 × (87 − 71) = 67.04 kJ. Flow = 850 kW ÷ 67.04 kJ/kg = 12.7 kg/s. A smaller design temperature drop would need even more flow.

Why not the others

A. This uses the supply temperature itself as the temperature difference: 850 ÷ (4.19 × 87) = 2.33 kg/s. The heat each kilogram gives up is set by its drop from supply to return, 16 K.

B. This takes the difference between the supply water and the greenhouse air: 850 ÷ (4.19 × 69) = 2.94 kg/s. The air temperature governs how fast the pipes give off heat, but each kilogram of water carries c times its own drop, 87 − 71 = 16 K.

D. This leaves out the specific heat: 850 ÷ 16 = 53.1. Flow = load ÷ (c × ΔT), and c = 4.19 kJ/kg·K is what turns a temperature drop into kilojoules per kilogram.

03Answer key

Question	Answer	Syllabus unit
Q1	D	B1 · Lubrication
Q2	A	B2 · Pumps and Compressors
Q3	D	B3 · Boiler Safety Devices
Q4	C	B4 · Plant (Boiler) Operations
Q5	B	B5 · Power Plant/Heating Plant Maintenance
Q6	C	B6 · Water Treatment
Q7	B	B7 · Prime Movers and Engines
Q8	B	B9 · Refrigeration
Q9	A	B10 · Heating Ventilating and Air Conditioning
Q10	C	B11 · Heating and Cooling Systems

04More free questions

For paper B itself, the 4th Class Power Engineering exam guide (4B) 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 (4B) page describes the full TicketPrep 4B question bank. For paper A, there are ten free 4A questions in the same format. 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.