CSLB HVAC (C-20) Trade — All Questions

In residential load calculations, what does an ACCA Manual J procedure primarily determine?

• a.The size and gauge of the supply ductwork
• b.The building's heating and cooling loads✓
• c.The refrigerant charge for the condenser
• d.The electrical service size for the panel

Manual J is the industry-standard residential load calculation method. It estimates the sensible and latent heating and cooling loads by accounting for construction, insulation, windows, orientation, infiltration, and local design temperatures. Equipment is then selected to match those calculated loads.California Energy Code (Title 24)

One ton of cooling capacity is equal to how many BTU per hour?

• a.6,000 BTU/hr
• b.9,000 BTU/hr
• c.12,000 BTU/hr✓
• d.15,000 BTU/hr

A ton of refrigeration equals 12,000 BTU/hr, derived from the heat needed to melt one ton of ice in 24 hours. This unit is the standard used to size air-conditioning and heat-pump equipment. A 3-ton system therefore provides about 36,000 BTU/hr of cooling.

A common rule-of-thumb airflow value for a residential cooling system is approximately how many CFM per ton?

• a.100 CFM per ton
• b.400 CFM per ton✓
• c.800 CFM per ton
• d.1,200 CFM per ton

Standard residential cooling airflow is roughly 400 CFM per ton, though it may range from 350 to 450 depending on climate and humidity. A 3-ton system therefore typically moves about 1,200 CFM. Correct airflow is essential for proper capacity and dehumidification.

A 4-ton air-conditioning system is designed at 400 CFM per ton. What total supply airflow is required?

• a.800 CFM
• b.1,200 CFM
• c.1,600 CFM✓
• d.2,000 CFM

Multiply tons by CFM per ton: 4 tons x 400 CFM = 1,600 CFM. Sizing the blower and ductwork for this airflow ensures the coil receives adequate air for rated capacity. Undersized ducts would restrict this airflow and reduce performance.

The sensible heat formula uses which constant when air volume is in CFM and temperature difference is in degrees F?

• a.0.68
• b.1.08✓
• c.4.5
• d.0.24

Sensible heat (BTU/hr) equals 1.08 x CFM x delta-T, where 1.08 combines air density, specific heat, and a time conversion. The 4.5 factor is used with enthalpy for total heat, and 0.68 is used for latent load with grains of moisture. These formulas let technicians verify airflow and capacity.

Using the sensible heat formula, how much sensible heat is removed when 1,200 CFM of air is cooled through a 20 degree F temperature drop?

• a.12,960 BTU/hr
• b.18,000 BTU/hr
• c.25,920 BTU/hr✓
• d.32,400 BTU/hr

Sensible heat = 1.08 x 1,200 CFM x 20 F = 25,920 BTU/hr. This calculation lets a technician confirm the coil is delivering close to its rated capacity. A result far below expected suggests low airflow or an undercharged system.

What does ACCA Manual D provide guidance for?

• a.Selecting the correct condenser
• b.Sizing and designing residential duct systems✓
• c.Calculating building heat loss
• d.Selecting the proper thermostat

Manual D is the ACCA standard for residential duct design. It uses the available static pressure and required airflow to size trunks, branches, and fittings so each room receives its design CFM. Proper duct design prevents noise, high static pressure, and uneven comfort.

After a Manual J load is calculated, which ACCA procedure is used to select the specific equipment model?

• a.Manual D
• b.Manual T
• c.Manual S✓
• d.Manual RS

Manual S matches equipment to the Manual J loads using manufacturer expanded performance data at local design conditions. It ensures the unit meets both sensible and latent loads without gross oversizing. Oversized equipment short-cycles and dehumidifies poorly.

Oversizing a residential air conditioner most commonly leads to which problem?

• a.Higher heating capacity
• b.Short cycling and poor humidity control✓
• c.Lower refrigerant pressures
• d.Reduced supply air temperature

An oversized unit cools the air quickly and satisfies the thermostat before it runs long enough to remove moisture, causing short cycling. The result is a cool but clammy space and increased wear from frequent starts. Correct sizing from Manual J and S avoids this.

Which factor increases a building's cooling load the most on a hot afternoon?

• a.Large west-facing glass in direct afternoon sun✓
• b.North-facing walls with continuous insulation
• c.A well-sealed and insulated attic hatch
• d.Low-wattage LED lighting in the ceilings

West-facing windows receive intense direct solar radiation in the afternoon, adding substantial sensible heat gain. This solar load is often a dominant component of the peak cooling load. Shading and low-SHGC glazing reduce this gain.

In California, plans for new HVAC systems typically must demonstrate compliance with which energy standard?

• a.The National Electrical Code
• b.Title 24, Part 6 energy efficiency standards✓
• c.The Uniform Plumbing Code
• d.The Americans with Disabilities Act

Title 24, Part 6 sets California's building energy efficiency standards, including HVAC equipment efficiency, duct sealing, and refrigerant charge verification. New and altered systems generally must show compliance through prescriptive or performance methods. Documentation is submitted with the permit application.California Energy Code (Title 24)

Under the California Mechanical Code, when is a permit generally required for HVAC work?

• a.Only for work in commercial or industrial buildings
• b.Only when a thermostat is being replaced
• c.When equipment is installed, altered, or replaced✓
• d.Only for gas-fired appliances such as furnaces

The CMC and local jurisdictions require a mechanical permit for installing, altering, or replacing regulated equipment like furnaces, air conditioners, and ductwork. Permits trigger inspection to verify code-compliant, safe installation. Minor maintenance such as filter changes does not require a permit.2022 California Mechanical Code (CMC)

What is the primary purpose of a Manual J block load versus a room-by-room load?

• a.To size the refrigerant liquid and suction lines
• b.To set the total equipment capacity for the house✓
• c.To determine the required duct insulation value
• d.To calculate the gas meter and pipe sizes

A block (whole-house) load establishes the total heating and cooling capacity the equipment must provide. A room-by-room load then distributes that capacity to size individual ducts and registers. Both are needed for a complete design.

If a home has a calculated cooling load of 30,000 BTU/hr, what nominal equipment size is the closest match?

• a.2 tons
• b.2.5 tons✓
• c.3.5 tons
• d.4 tons

30,000 BTU/hr divided by 12,000 BTU/hr per ton equals 2.5 tons. Manual S guides selecting a unit whose capacity at design conditions closely matches this load. Choosing 2.5 tons avoids the humidity and cycling problems of oversizing.

Friction rate in duct design is typically expressed in which units?

• a.CFM per square foot of floor area served
• b.BTU per hour per square foot of duct
• c.Inches of water column per 100 feet of duct✓
• d.Feet per minute per 100 feet of duct

Friction rate expresses the pressure loss per unit length of duct, in inches of water column per 100 feet. In Manual D, the available static pressure and total effective length set the design friction rate used to size ducts. A higher friction rate yields smaller, higher-velocity ducts.

Air velocity in a duct is calculated by dividing airflow (CFM) by what?

• a.The total length of the duct run in feet
• b.The cross-sectional area of the duct in square feet✓
• c.The static pressure in the duct in inches w.c.
• d.The temperature difference across the coil in F

Velocity in feet per minute equals CFM divided by the duct cross-sectional area in square feet. Excessive velocity causes noise and high friction losses, while too little velocity can reduce throw and mixing. Designers balance velocity against static pressure limits.

A branch duct must deliver 200 CFM at a velocity of about 600 feet per minute. Approximately what free area is required?

• a.0.10 square feet
• b.0.20 square feet
• c.0.33 square feet✓
• d.0.66 square feet

Area equals CFM divided by velocity: 200 / 600 = 0.33 square feet. This corresponds to roughly a 8-inch round duct. Selecting the right size keeps velocity and noise within acceptable limits.

Under California energy standards, duct leakage in many new and altered systems must be verified by which method?

• a.Visual inspection of accessible duct joints
• b.A duct leakage test under pressurization✓
• c.A refrigerant charge verification test
• d.A combustion analysis of the furnace

Title 24 requires duct leakage testing for many new installations and changeouts, typically verified through a pressurization test. Sealing ducts reduces energy waste and improves delivered airflow. Results are often confirmed by a HERS rater.California Energy Code (Title 24)

Which condition would a technician correct by performing an accurate load calculation instead of using a rule of thumb like square footage per ton?

• a.Comfort complaints from badly sized equipment✓
• b.A dirty condenser coil that restricts airflow
• c.A tripped condensate float switch
• d.A failed compressor run capacitor

Square-foot rules of thumb ignore insulation, glazing, orientation, and infiltration, often producing oversized or undersized equipment and comfort complaints. A Manual J load accounts for these variables and yields correctly sized equipment. This is the proper professional approach.

Latent load in a cooling calculation refers to the energy required to do what?

• a.Raise the dry-bulb temperature of the air
• b.Remove moisture (water vapor) from the air✓
• c.Overcome duct friction
• d.Power the compressor motor

Latent load is the heat associated with condensing water vapor out of the air, lowering humidity. Sensible load changes air temperature, while latent load changes moisture content. Both must be met for comfort, especially in humid climates.

A 3-ton system is selected for a home. At 400 CFM per ton, what is the target blower airflow?

• a.800 CFM
• b.1,000 CFM
• c.1,200 CFM✓
• d.1,500 CFM

3 tons x 400 CFM per ton = 1,200 CFM. The blower and duct system should be configured to deliver this airflow at the system's external static pressure. Verifying actual airflow after installation confirms proper capacity.

What is the main benefit of properly sizing equipment to a load calculation rather than oversizing?

• a.Faster refrigerant charging during start-up
• b.Better humidity control, comfort, and efficiency✓
• c.Elimination of the mechanical permit
• d.Higher supply air velocity at the registers

Right-sized equipment runs longer, steadier cycles that dehumidify effectively and hold temperature evenly, improving comfort and efficiency. Oversizing wastes energy through short cycling and leaves the space humid. This aligns with Title 24 efficiency goals.California Energy Code (Title 24)

You are running a Manual J load for a house in Sacramento. Which outdoor temperatures belong in the calculation?

• a.The published 1% summer and 99% winter design temperatures for that location✓
• b.The record high and record low ever measured at the airport
• c.The average high and average low for July and January
• d.The temperatures forecast for the week the system is commissioned

Manual J sizes to the 1% cooling and 99% heating design conditions, so the system meets load in nearly all hours without being sized for rare extremes. Record highs and lows occur a few hours a decade and would grossly oversize the equipment. Monthly averages sit far below the design condition and would undersize it. A weekly forecast has nothing to do with a load that must serve the house for twenty years.

A homeowner asks why you will not size the new condenser by the '400 square feet per ton' rule the last contractor used. What is the strongest technical answer?

• a.The rule ignores insulation, glazing, orientation, and infiltration, which drive the load✓
• b.Square-footage rules always undersize equipment in hot climates
• c.Square-footage rules are only valid for single-story houses
• d.The rule was written for evaporative coolers rather than refrigerated air

A per-square-foot rule assumes one construction type and one climate, while the actual load is set by envelope insulation, window area and orientation, shading, and air leakage. It is not a story-count problem, since a well-built two-story house can also be load-calculated. It does not always undersize; on a tight modern house it usually oversizes badly. And the rule is a general HVAC shortcut, not an evaporative-cooler convention.

A coil removes 30,000 BTU/hr total and 22,500 BTU/hr sensible. What is the sensible heat ratio?

• a.1.33
• b.0.25
• c.0.75✓
• d.0.60

Sensible heat ratio is sensible capacity divided by total capacity: 22,500 / 30,000 = 0.75. The 0.60 answer comes from misreading which number is total. The 1.33 figure is the ratio inverted, which can never exceed 1.0 for a cooling coil. And 0.25 is the latent fraction, the remainder of the total, not the sensible ratio.

Which house would you expect to have the highest latent portion of its cooling load?

• a.A tight new house with triple-glazed west-facing windows
• b.A leaky older house in a high desert climate with low occupancy and no cooking
• c.A tight new house in a dry inland valley with low occupancy
• d.A leaky older coastal house with high occupancy and frequent cooking✓

Latent load comes from moisture, and the leaky coastal house pulls in humid outdoor air while people and cooking add water vapor indoors. A tight, dry-climate house with few occupants has almost no moisture source. Large west glazing raises the sensible solar load, not the latent load. A leaky high-desert house admits plenty of air, but that air is very dry, so the latent gain stays small.

You calculate 1,800 CFM of supply air cooled from 78 F to 57 F. What sensible capacity does that represent?

• a.20,412 BTU/hr
• b.32,400 BTU/hr
• c.40,824 BTU/hr✓
• d.58,320 BTU/hr

Sensible BTU/hr = 1.08 x CFM x delta-T = 1.08 x 1,800 x 21 = 40,824. The 20,412 figure uses half the airflow. The 32,400 figure drops the 1.08 constant and multiplies CFM by delta-T alone in a different form. The 58,320 figure uses the 4.5 total-heat constant with a dry-bulb difference, which mixes an enthalpy formula with a temperature reading.

Manual S is being applied after the Manual J is finished. Which capacity must be compared to the calculated sensible load?

• a.The nominal tonnage printed on the condenser nameplate
• b.The gross capacity before blower motor heat is subtracted
• c.The sensible capacity of the matched coil and condenser at design conditions✓
• d.The total capacity listed in the AHRI directory at standard rating conditions

Manual S compares the equipment's expanded-performance sensible capacity at the actual design outdoor temperature, indoor wet bulb, and airflow against the Manual J sensible load. Nominal tonnage is a marketing rounding and can differ from real output by thousands of BTU. AHRI rating conditions of 95 F outdoors and 80/67 indoors rarely match the job's design conditions. And gross capacity overstates delivery because blower heat is a real penalty on a fan-coil system.

A duct system has 0.50 in. w.c. of external static available and a total effective length of 250 feet. What friction rate should be used for sizing?

• a.0.05 in. w.c. per 100 feet
• b.0.10 in. w.c. per 100 feet
• c.0.15 in. w.c. per 100 feet
• d.0.20 in. w.c. per 100 feet✓

Friction rate = available static x 100 / total effective length = 0.50 x 100 / 250 = 0.20 in. w.c. per 100 feet. The 0.10 answer is the common default friction rate, not the one this system computes. The 0.05 and 0.15 answers come from using the wrong effective length or dividing the static pressure by the wrong factor. Sizing at too low a friction rate would make the ducts needlessly large for the fan actually installed.

In Manual D, what is 'total effective length' of a duct run?

• a.The measured length plus the equivalent lengths of all fittings in the run✓
• b.The sum of the supply and return trunk lengths on the whole system
• c.The straight length multiplied by a fixed correction factor for flex duct
• d.The measured straight run of the longest supply trunk from the plenum to the boot

Total effective length adds equivalent length credits for elbows, takeoffs, boots, and terminals to the measured straight length, because fittings cause most of the pressure loss. Measuring only the straight trunk ignores fittings entirely and yields an optimistic friction rate. Adding every trunk on the system double-counts branches that are not in the same path. And a single blanket multiplier for flex cannot represent the specific fittings a given run contains.

A round duct must carry 600 CFM at about 800 feet per minute. What cross-sectional area does that require?

• a.0.50 square feet
• b.0.60 square feet
• c.0.75 square feet✓
• d.1.33 square feet

Area = CFM / velocity = 600 / 800 = 0.75 square feet. The 0.50 and 0.60 figures come from using 1,200 or 1,000 FPM instead of the stated velocity. The 1.33 figure is the calculation inverted, dividing velocity by CFM, which produces a number with no physical meaning here. Once area is known, the duct diameter follows from the area of a circle.

A homeowner complains of noisy registers on a system you did not install. Which design decision most likely caused it?

• a.The branch ducts were sized for a high velocity to save material✓
• b.The supply trunk was insulated to a higher R-value than required
• c.The return grille was installed in a hallway instead of a bedroom
• d.The equipment was sized from a room-by-room load rather than a block load

Noise at the register tracks air velocity, and undersized branches force the same CFM through less area, raising velocity and turbulence. Extra insulation on the trunk changes heat loss, not the sound at a register. A room-by-room load is the more accurate approach and would not create noise. Return location affects room-to-room pressure balance and comfort, but the whistling is generated at the undersized supply.

How does duct located in a vented attic affect a Manual J cooling load?

• a.It reduces the load because attic air pre-cools the supply
• b.It affects only the heating load, not the cooling load
• c.It has no effect on the load, because the ducts are inside the building footprint
• d.It adds conduction gain and leakage loss that must be included in the load✓

Ducts in an unconditioned attic gain heat through the insulation and lose conditioned air through leaks, and Manual J includes both as duct load. Being under the roofline is not the same as being inside the thermal envelope. Attic air in summer is far hotter than supply air, so it heats rather than pre-cools. And the penalty applies in both seasons, since the attic is also cold on a winter design morning.

A contractor replaces a furnace and more than 40 feet of duct in an existing California home. What compliance step is most likely triggered?

• a.Replacement of every register and grille in the house
• b.Recalculation of the electrical service load by an engineer
• c.Duct leakage testing verified by a third-party HERS rater✓
• d.A full Manual J submitted to the utility for rebate approval

California's energy standards require diagnostic duct leakage testing with independent HERS verification when a substantial portion of a duct system is replaced or the air handler is changed. A load calculation is good practice and often required for permitting, but it is not the third-party measure the alteration triggers. Registers are not required to be swapped as a condition of duct alteration. And service load recalculation belongs to electrical work, not duct alteration.California Energy Code (Title 24, Part 6)

A furnace is rated 80,000 BTU/hr input at 80% AFUE. What output should be compared to the calculated heat loss?

• a.100,000 BTU/hr
• b.80,000 BTU/hr
• c.72,000 BTU/hr
• d.64,000 BTU/hr✓

Output equals input times efficiency: 80,000 x 0.80 = 64,000 BTU/hr, and that is what actually reaches the house. Using 80,000 treats the fuel burned as if all of it were delivered, ignoring flue losses. The 72,000 figure applies 90% efficiency, which is a condensing furnace rating, not this one. The 100,000 figure divides by the efficiency, which sizes the appliance the wrong direction.

Which change to a house would reduce the design heating load the most on a cold winter night?

• a.Air sealing the attic plane and adding insulation over the ceiling✓
• b.Replacing north-facing windows with larger units of the same U-factor
• c.Painting the roof a lighter color to raise its solar reflectance
• d.Adding an exterior shade awning over the south-facing patio door

Heating load is driven by conduction and infiltration, so sealing leaks and insulating the ceiling attacks the two largest winter losses directly. Larger windows of the same U-factor increase area and therefore increase the loss. Awnings block summer sun and can even raise the winter load by blocking useful solar gain. A cool roof lowers summer attic temperature but does almost nothing on a night with no sun.

What does the balance point of an air-source heat pump represent?

• a.The outdoor temperature at which COP falls to 1.0
• b.The outdoor air temperature at which the defrost cycle first begins to initiate
• c.The outdoor temperature at which heat pump capacity equals the building load✓
• d.The indoor setpoint at which the auxiliary heat is locked out

The balance point is where the falling heat pump output line crosses the rising building heat loss line; below it, supplemental heat is needed. Defrost initiation depends on coil temperature and humidity, not on this crossing. Auxiliary lockout is a control setting the installer chooses and is not the definition. And a COP of 1.0 is the point where the heat pump is no better than resistance heat, which is a different and much colder condition.

A 3-ton system moves 1,200 CFM. The blower is slowed to 900 CFM to improve dehumidification. What is the expected effect?

• a.Both sensible and latent capacity rise together
• b.Total capacity is unchanged because the compressor is unchanged
• c.Sensible capacity rises and latent capacity drops
• d.Sensible capacity drops and latent capacity rises✓

Less air over the coil makes the coil colder, so more moisture condenses while the sensible transfer falls. Both cannot rise, because the split between sensible and latent shifts rather than growing. Raising airflow, not lowering it, is what increases sensible and reduces latent. And total capacity does change, since coil performance depends on air mass flow and coil temperature, not on the compressor alone.

The latent heat formula for airflow in CFM and moisture difference in grains per pound uses which constant?

• a.0.24
• b.1.08
• c.4.5
• d.0.68✓

Latent BTU/hr = 0.68 x CFM x grains difference, where the constant folds in air density and the latent heat of water vapor. The 1.08 constant belongs to the sensible formula with a dry-bulb difference. The 4.5 constant is used with enthalpy in BTU per pound to get total heat. And 0.24 is the specific heat of air in BTU per pound per degree F, an input to those constants rather than the constant itself.

You measure 1,000 CFM entering a coil at 30 BTU/lb enthalpy and leaving at 22 BTU/lb. What total capacity is the coil delivering?

• a.8,640 BTU/hr
• b.27,000 BTU/hr
• c.36,000 BTU/hr✓
• d.52,000 BTU/hr

Total heat = 4.5 x CFM x enthalpy difference = 4.5 x 1,000 x 8 = 36,000 BTU/hr. The 8,640 figure uses the 1.08 sensible constant with the enthalpy difference, which mixes formulas. The 27,000 figure uses a 6 BTU/lb difference rather than the 8 given. The 52,000 figure would require a much larger enthalpy drop than the readings show.

Why is a room-by-room Manual J load required before a duct design, even when a block load is available?

• a.Because each branch duct must be sized to the CFM its own room needs✓
• b.Because block loads systematically understate whole-house tonnage
• c.Because permit officials will not accept a block load for equipment sizing
• d.Because latent load can only be calculated on a room-by-room basis

Manual D distributes air in proportion to each room's load, so branch sizing depends on room-level numbers that a block load never produces. A block load gives an accurate whole-house total; it simply does not break it down. Jurisdictions commonly accept a block load for equipment selection alone. And latent load is calculated at the block level routinely, so that is not the reason.

A bedroom has a calculated cooling load of 3,000 BTU/hr in a house whose total cooling load is 24,000 BTU/hr and whose blower moves 1,200 CFM. What supply airflow should the room receive?

• a.100 CFM
• b.120 CFM
• c.135 CFM
• d.150 CFM✓

The room holds 3,000 / 24,000 = 12.5% of the load, so it gets 12.5% of 1,200 CFM, or 150 CFM. The 100 and 120 figures come from using a lower total airflow or rounding the load fraction down. The 135 figure splits the difference without following the proportion. Proportional distribution is what keeps room temperatures even across the house.

Two identical houses sit side by side, one with its main glazing facing west and one facing north. How do their cooling loads compare?

• a.The west-facing house has a higher peak cooling load✓
• b.The north-facing house has a higher peak cooling load
• c.The difference appears only in the latent portion of the load
• d.The loads are identical because the glazing area is the same

West glass takes direct low-angle sun during the hottest hours, producing the largest solar gain and setting the afternoon peak. North glass in California receives mostly diffuse light, so its gain is far smaller. Equal area does not mean equal gain, because orientation controls how much solar radiation strikes the glass. And solar gain is sensible heat; it does not change the moisture in the space.

Which glazing property most directly governs the solar portion of a window's cooling load?

• a.The solar heat gain coefficient of the window✓
• b.The U-factor of the window assembly
• c.The visible transmittance of the glass
• d.The air leakage rating of the operable window sash

The solar heat gain coefficient is the fraction of incident solar energy that ends up as heat in the room, so it scales the solar gain term directly. U-factor governs conduction driven by the indoor-outdoor temperature difference, which is a separate term. Visible transmittance describes daylight, and a glass can pass light while blocking heat. Air leakage feeds the infiltration term, not the solar term.

A wall measures 400 square feet with a U-factor of 0.075 and a design temperature difference of 40 F. What is its heat loss?

• a.600 BTU/hr
• b.1,200 BTU/hr✓
• c.2,400 BTU/hr
• d.16,000 BTU/hr

Heat loss = U x Area x delta-T = 0.075 x 400 x 40 = 1,200 BTU/hr, and every opaque assembly in the house is totaled the same way. The 600 figure uses half the temperature difference. The 2,400 figure doubles either the area or the delta-T. The 16,000 figure treats 0.075 as an R-value and divides by it, which inverts the relationship between thermal resistance and conductance.

A wall assembly has a total R-value of 20. What is its U-factor?

• a.0.20
• b.0.10
• c.0.08
• d.0.05✓

U-factor is the reciprocal of R-value, so 1 divided by 20 gives 0.05, and that number is what the heat loss formula multiplies by area and temperature difference. The 0.20 answer simply moves the decimal point on the R-value. The 0.10 answer is the reciprocal of R-10, a much poorer wall. The 0.08 answer corresponds to roughly R-12.5, which is also lower performance than the assembly described.

Which internal gain is normally the largest single line item in a residential Manual J for an occupied kitchen and living area?

• a.Appliances and lighting operating during the peak hour✓
• b.The blower motor heat added by the furnace fan
• c.The sensible heat given off by occupants who are sleeping
• d.Heat conducted through interior partition walls

Cooking appliances and lighting release substantial sensible heat in the same afternoon hour when the envelope load peaks, so they dominate internal gains. Sleeping occupants give off the lowest metabolic heat of any activity level and are counted at a reduced rate. Interior partitions separate two conditioned rooms at nearly the same temperature, so conduction across them is close to zero. Blower heat is real but is handled as an equipment correction, not a room internal gain.

A 90,000 BTU/hr input furnace must hold a 50 F temperature rise. Roughly what airflow is required at 80% efficiency?

• a.About 900 CFM
• b.About 1,330 CFM✓
• c.About 1,670 CFM
• d.About 2,000 CFM

Output is 90,000 x 0.80 = 72,000 BTU/hr, and CFM = 72,000 / (1.08 x 50) = about 1,330. The 900 figure would give a rise near 74 F and would trip the limit switch. The 1,670 and 2,000 figures use the full input rather than the output, which overstates the air the furnace actually needs and would push the rise below the rating plate range.

Where does a technician find the acceptable temperature rise range for a specific gas furnace?

• a.In the ACCA Manual J tables for the climate zone
• b.In the California Energy Code residential compliance forms
• c.On the equipment rating plate or in the installation instructions✓
• d.In the AHRI certified ratings directory listing for that model number

Temperature rise is a manufacturer-specified range printed on the rating plate and repeated in the installation manual, and the blower must be set to land inside it. Manual J produces building loads, not appliance operating limits. Title 24 compliance forms document energy features, not the rise band. And the AHRI directory reports efficiency and capacity ratings, not the allowable rise for a given model.

Fan laws describe blower behavior. If blower speed increases 25%, what happens to the power the motor must draw?

• a.It increases about 25%
• b.It stays the same because the motor is constant speed
• c.It increases about 56%
• d.It increases about 95%✓

Power varies with the cube of speed, so 1.25 cubed is about 1.95, roughly a 95% increase. Airflow, not power, varies directly with speed, which is where the 25% answer comes from. Static pressure varies with the square of speed, which gives the 56% figure. And a blower whose speed was just changed is by definition not running at constant speed.