313D exam, skill set A: Plan and Prepare practice questions.

Eight free practice questions on job planning, heat loss and gain, psychrometrics and equipment selection, for skill set A (Plan and Prepare) of Ontario's 313D Residential Air Conditioning Systems Mechanic Certificate of Qualification exam, which Skilled Trades Ontario (STO) sets and delivers. 313D is an Ontario trade with its own exam, separate from the 313A Refrigeration and Air Conditioning Systems Mechanic. These are original questions written by TicketPrep from public sources, Ontario regulations and the trade's working practice, not questions from the exam. Each has four options and one correct answer, as every question in our bank does. Answer each one before opening the explanation. No account, no email.

01The questions

1. Critical · Our task A-1

   A builder's row of six townhouses is to be cooled by one condensing system on the roof, piped to an indoor coil in each home. It runs on 240 V single phase, and each home's share of its capacity is about 2 tons. May a 313D install this system?

   1. No: one system for six homes is not an independent unit in a dwelling.
   2. Yes: the supply is 240 V single phase, which is inside the voltage limit.
   3. Yes: each home's share of the capacity is well under 60 000 Btu/h.
   4. No: townhouses are attached, so none of them counts as a family dwelling.
   Show the answer

   A. Correct. O. Reg. 875/21 s. 119(1) limits the 313D to a system "installed as an independent unit in a residential family dwelling". A single plant piped to six homes is a central system and belongs to the 313A mechanic. Its combined size, about 12 tons, is also far over the 60 000 Btu/h limit.

   Why not the others

   B. Meeting one limit is not enough: s. 119 requires every condition. The system must also be an independent unit in a residential family dwelling, which a shared plant for six homes is not.

   C. The capacity limit is the system's, not each home's share, and six 2-ton shares make about 12 tons. The test that fails first is that s. 119 covers a system installed as an independent unit in a residential family dwelling, and one plant serving six homes is not.

   D. Attached homes are still family dwellings: a townhouse with its own system inside the s. 119 limits is 313D work. What fails here is that one system serves six homes instead of each home having its own independent unit.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-1.02); O. Reg. 875/21

2. Calculation · Our task A-2

   A room's exterior wall has a gross area of 380 ft² (35.3 m²), including 60 ft² (5.6 m²) of windows that are calculated separately. The wall's U-value is 0.055 Btu/h·ft²·°F. Indoor design is 22 °C (72 °F) and outdoor design is −25 °C (−13 °F). What is the wall's design heat loss?

   1. 827 Btu/h
   2. 1 038 Btu/h
   3. 1 496 Btu/h
   4. 1 777 Btu/h
   Show the answer

   C. Correct. Net wall area = 380 − 60 = 320 ft² (the windows have their own U-value). ΔT = 72 − (−13) = 85 °F. Heat loss = U × A × ΔT = 0.055 × 320 × 85 = 1 496 Btu/h.

   Why not the others

   A. This uses the Celsius difference, 22 − (−25) = 47, with a U-value per °F: 0.055 × 320 × 47 = 827 Btu/h. A U-value in Btu/h·ft²·°F needs the difference in °F: 72 − (−13) = 85 °F.

   B. This drops the minus sign on the outdoor temperature: 72 − 13 = 59 °F, giving 0.055 × 320 × 59 = 1 038 Btu/h. The difference is 72 − (−13) = 85 °F.

   D. This uses the gross area: 0.055 × 380 × 85 = 1 776.5, about 1 777 Btu/h. The 60 ft² of windows are calculated with their own U-value, so the wall's net area is 320 ft², giving 1 496 Btu/h.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.01)

3. Procedural · Our task A-2

   On a dry spring day a central air conditioner runs with return air at 24 °C (75 °F) dry bulb and a chart dew point of 10 °C (50 °F). The coil's surface temperature is 12 °C (54 °F) throughout. The homeowner asks why the condensate drain is dry. What explains it?

   1. The coil surface is above the air's dew point, so nothing condenses.
   2. The trap is blocked, so the condensate is held back in the drain pan.
   3. The coil is cold enough that the moisture freezes onto its fins.
   4. The return air is too warm for the coil to take any moisture out of it.
   Show the answer

   A. Correct. A coil takes moisture out only where its surface is below the entering air's dew point. At 12 °C the surface is above the 10 °C dew point, so the coil cools the air sensibly and nothing condenses. On a humid day, with the dew point above the coil temperature, the drain will run again.

   Why not the others

   B. A blocked trap would leave water standing in the pan or overflowing it. Here no water forms at all, because the 12 °C coil surface is above the 10 °C dew point.

   C. Frost needs a surface below 0 °C; a 12 °C coil is well above freezing, and above the 10 °C dew point, so moisture neither freezes nor condenses on it.

   D. Warm air can carry plenty of moisture; what decides condensation is the coil surface against the air's dew point. With the surface at 12 °C and the dew point at 10 °C, the air is never cooled to its dew point.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.02)

4. Calculation · Our task A-2

   A house's design sensible heat gain is 24 000 Btu/h, and the supply air is to be 20 °F (11 K) below the room. The equipment chosen is a 2.5-ton unit. Using the sensible heat relation for standard air, what airflow per ton does the design call for, to the nearest whole CFM per ton?

   1. 107 CFM per ton
   2. 444 CFM per ton
   3. 556 CFM per ton
   4. 808 CFM per ton
   Show the answer

   B. Correct. CFM = sensible Btu/h ÷ (1.08 × ΔT) = 24 000 ÷ (1.08 × 20) = 24 000 ÷ 21.6 = 1 111.1 CFM. Per ton of equipment: 1 111.1 ÷ 2.5 = 444.4, about 444 CFM per ton. That is above the 400 CFM per ton rule of thumb; airflow above it raises sensible capacity at the cost of moisture removal, so check it against the maker's airflow range before setting the blower.

   Why not the others

   A. This uses 4.5, the total-heat factor, in place of 1.08: 24 000 ÷ (4.5 × 20) = 266.7 CFM, ÷ 2.5 = 107 CFM per ton. A temperature difference goes with 1.08, giving 444 CFM per ton.

   C. This divides by tons worked out from the sensible load alone, 24 000 ÷ 12 000 = 2.0 tons: 1 111.1 ÷ 2.0 = 555.6, about 556 CFM per ton. The airflow is per ton of the 2.5-ton equipment: 444 CFM per ton.

   D. This uses the difference in kelvins with the °F relation: 24 000 ÷ (1.08 × 11) = 2 020.2 CFM, ÷ 2.5 = 808 CFM per ton. The 1.08 factor needs the 20 °F difference, giving 444 CFM per ton.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.03)

5. Calculation · Our task A-2

   A house's heat-loss calculation takes its conditioned volume as 14 400 ft³ (408 m³) and its design infiltration as 0.30 air changes per hour. Indoor design is 21 °C (70 °F) and outdoor −30 °C (−22 °F). Using the sensible heat relation for standard air, what is the infiltration heat loss, to the nearest 10 Btu/h?

   1. 3 970 Btu/h
   2. 7 150 Btu/h
   3. 23 850 Btu/h
   4. 29 810 Btu/h
   Show the answer

   B. Correct. Air leaking in per hour = 14 400 × 0.30 = 4 320 ft³/h; per minute 4 320 ÷ 60 = 72 CFM. ΔT = 70 − (−22) = 92 °F. Sensible heat = 1.08 × CFM × ΔT = 1.08 × 72 × 92 = 7 153.9, about 7 150 Btu/h.

   Why not the others

   A. This uses the Celsius difference, 21 − (−30) = 51, with the °F relation: 1.08 × 72 × 51 = 3 965.8, about 3 970 Btu/h. The 1.08 factor works with °F, so the difference is 70 − (−22) = 92 °F.

   C. This leaves out the 0.30 air-change rate, as if the whole volume changed each hour: 14 400 ÷ 60 = 240 CFM, and 1.08 × 240 × 92 = 23 846.4, about 23 850 Btu/h. Only 0.30 of the volume changes per hour, which is 72 CFM.

   D. This uses 4.5, the total-heat factor that goes with an enthalpy difference, in place of 1.08: 4.5 × 72 × 92 = 29 808, about 29 810 Btu/h. Sensible heat from a temperature difference uses 1.08, giving about 7 150 Btu/h.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.01)

6. Calculation · Our task A-2

   A room-by-room heat-gain calculation totals 30 000 Btu/h, of which the bonus room over the garage is 4 500 Btu/h; the room's heat loss is 6 000 of the house's 48 000 Btu/h. The design cooling airflow is 1 100 CFM through 11 supply outlets, divided by each room's share of the heat gain. What cooling airflow does the bonus room get?

   1. 100 CFM
   2. 138 CFM
   3. 150 CFM
   4. 165 CFM
   Show the answer

   D. Correct. The bonus room's share of the heat gain is 4 500 ÷ 30 000 = 0.15, so it gets 0.15 × 1 100 = 165 CFM. Its heating share is smaller (6 000 ÷ 48 000 = 0.125), so airflow set from the heat loss alone would leave this room warm in summer.

   Why not the others

   A. This splits the airflow equally over the outlets: 1 100 ÷ 11 = 100 CFM. Rooms get airflow by their share of the load, not by outlet count: 4 500 ÷ 30 000 × 1 100 = 165 CFM.

   B. This uses the room's share of the heat loss: 6 000 ÷ 48 000 × 1 100 = 137.5, about 138 CFM. Cooling airflow follows the heat-gain share: 4 500 ÷ 30 000 × 1 100 = 165 CFM.

   C. This applies the 400 CFM per ton rule of thumb to the room's gain: 4 500 ÷ 12 000 × 400 = 150 CFM. The system's design airflow is 1 100 CFM, so the room gets its share of that: 165 CFM.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.01)

7. Calculation · Our task A-2

   A house's design heat gain is 21 600 Btu/h sensible and 5 400 Btu/h latent. What is the sensible heat ratio of this load?

   1. 0.20
   2. 0.25
   3. 0.80
   4. 1.25
   Show the answer

   C. Correct. Total = 21 600 + 5 400 = 27 000 Btu/h; SHR = sensible ÷ total = 21 600 ÷ 27 000 = 0.80. Choose equipment whose sensible and latent capacities at design conditions, from the maker's data, each cover their part of the load.

   Why not the others

   A. This is latent over total, 5 400 ÷ 27 000 = 0.20: the latent share of the load. The sensible heat ratio is sensible over total: 21 600 ÷ 27 000 = 0.80.

   B. This divides latent by sensible, 5 400 ÷ 21 600 = 0.25. The ratio's denominator is the total load, 27 000 Btu/h, and its numerator the sensible part, giving 0.80.

   D. This is total over sensible, 27 000 ÷ 21 600 = 1.25. A sensible heat ratio can never exceed 1, since the sensible part is never more than the total; sensible over total gives 0.80.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.02)

8. Recall · Our task A-2

   Room air in a house is at 60 % relative humidity, below saturation. How do its dry-bulb, wet-bulb and dew-point temperatures compare?

   1. Dry bulb highest, dew point between, wet bulb lowest.
   2. Dry bulb highest, wet bulb between, dew point lowest.
   3. Dry bulb and wet bulb equal, dew point lower.
   4. Wet bulb and dew point equal, dry bulb higher.
   Show the answer

   B. Correct. Evaporation from the wetted bulb cools it below the dry bulb, but it cannot cool it below the dew point, so in unsaturated air dry bulb > wet bulb > dew point. The three meet only at 100 % relative humidity.

   Why not the others

   A. This swaps the wet bulb and the dew point. Evaporation cools the wetted bulb toward the dew point but never below it, so the wet bulb sits between the dry bulb and the dew point.

   C. Dry bulb and wet bulb are equal only when the air is saturated and no water can evaporate from the wick. In air below saturation the wet bulb reads lower than the dry bulb.

   D. Wet bulb equals dew point only at saturation, when all three are equal. Below saturation the wet bulb sits above the dew point.

   Source: STO 313D counselling sheet, skill set A: Plan and Prepare (our sub-task A-2.02)

02Answer key

03What skill set A covers

Skill set A is the work done before anything is set or piped. Our four tasks for it are reading the job documents and organizing the work, working out the loads and design conditions, choosing the system, its components and controls, and planning for site safety and the instruments the job needs. Most of our questions here are calculations a 313D mechanic is expected to run by hand: heat loss through walls and by infiltration, heat gain split room by room, airflow from the sensible load, and the sensible heat ratio that steers equipment selection. The rest test the psychrometrics behind those numbers and the limits of the trade itself, because a job outside the 313D scope should be caught at the planning stage, not at start-up.

STO's 313D exam counselling sheet lists six weighted skill sets and gives Plan and Prepare 7 per cent of the exam, with a pass mark of 70 per cent. The sheet's body says “Updated as of December 5, 2016” and follows the older training standard, not the one in force since December 2023, so its weights may not match today's exam. STO publishes no question count for the 313D exam, only that its exams have between 90 and 150 questions. Our mock is 130 questions in 4 hours, and its draw of 9 questions from skill set A is ours, worked out from the sheet's percentages. The four tasks below and the right-hand column are TicketPrep's own grouping: the column is how many questions our bank holds on each task, Standard and Advanced together, not a count from the exam.

04What the questions turn on

Scope comes first

O. Reg. 875/21 s. 119(1) limits the 313D to a system "installed as an independent unit in a residential family dwelling", at no more than 240 V single phase on a branch circuit of 60 A or less, with no more than 60 000 Btu/h of cooling and a saturated suction temperature above 2 °C (35 °F). Every one of those conditions has to hold. A single roof-top plant piped to six townhouses fails on the first one even though its supply is 240 V single phase, and its combined capacity is over the limit as well.

Heat loss on the net area, and infiltration in CFM

Conduction loss is U × A × ΔT, where A is the net wall area once the windows (which carry their own U-value) are taken out, and ΔT runs from the indoor to the outdoor design temperature. Infiltration is a sensible load on air: convert the house volume times its air changes per hour into CFM by dividing by 60, then apply 1.08 × CFM × ΔT in °F. Using the gross wall area or forgetting the 60 are the slips the wrong options are built on.

Airflow follows the heat gain

Design airflow comes from the sensible heat relation, CFM = sensible Btu/h ÷ (1.08 × ΔT), and each room's share of that airflow follows its share of the heat gain, not of the heat loss: a bonus room over a garage may need more air in summer than its winter share suggests. Dividing the design airflow by the equipment's tons gives a CFM per ton figure to check against the maker's airflow range before the blower is set.

Dew point decides whether the coil dehumidifies

In air below saturation, dry bulb is highest, wet bulb sits between and dew point is lowest; the three meet only at 100 % relative humidity. A coil removes moisture only where its surface is colder than the entering air's dew point, so on a dry day a coil at 12 °C (54 °F) in air with a 10 °C (50 °F) dew point leaves the drain dry. The sensible heat ratio, sensible load divided by total load, is what the selected equipment's sensible and latent capacities have to cover.

05The other skill sets

STO publishes no minimum score for any one skill set, only the 70 per cent pass mark on the whole exam, so every skill set's marks count toward it. Each of the six skill sets has its own page of free questions:

The 313D practice exam page describes the full question bank. Working toward the full-scope trade as well? See the 313A practice exam; the two trades share their first-level training. More free questions for every exam we cover: free practice questions. How these questions are written and checked: how our questions are made. Found a mistake? Tell us and we will fix it. TicketPrep is not affiliated with Skilled Trades Ontario or CSA Group. CSA B52, CSA F280 and the Ontario Electrical Safety Code are cited by name only and never reproduced; Ontario regulations such as O. Reg. 875/21 and O. Reg. 463/10 are quoted only briefly.