400 questions
The California Electrical Code (CEC) is the governing code for what portion of HVAC work?
- a.Gas venting and flue termination clearances
- b.Wiring, disconnects, and circuit protection✓
- c.Condensate drainage and trap design
- d.Combustion air openings and sizing
The CEC governs the electrical aspects of HVAC installations, including circuit sizing, disconnecting means, grounding, and overcurrent protection. Proper electrical work protects against shock and fire. HVAC technicians must follow the CEC for all power and control wiring.
Under the California Plumbing Code, condensate waste from HVAC equipment is generally treated as what?
- a.Potable water that is suitable for reuse
- b.Fuel gas piping under pressure
- c.Drainage discharged to an approved point✓
- d.Combustion air for the appliance
HVAC condensate is drainage governed by the California Plumbing Code and must terminate at an approved point, not create a nuisance or hazard. Improper discharge onto walkways or into unapproved locations is a violation. Proper drainage prevents damage and slip hazards.2022 California Mechanical Code (CMC)
Personal protective equipment (PPE) commonly required when brazing refrigerant lines includes what?
- a.A disposable dust mask worn alone
- b.Eye protection, gloves, and ventilation✓
- c.No protection is required for short jobs
- d.Hearing protection and a hard hat only
Brazing produces intense heat, bright light, and potentially harmful fumes, so eye protection, gloves, and adequate ventilation are needed. Nitrogen purging reduces harmful phosgene-type byproducts from certain fluxes and coatings. Cal/OSHA requires appropriate PPE for the hazards present.Cal/OSHA Title 8
Why is a carbon monoxide alarm recommended or required in dwellings with fuel-burning appliances?
- a.To detect refrigerant leaks inside the home
- b.To measure indoor humidity levels
- c.To warn occupants of dangerous CO levels✓
- d.To sense high refrigerant head pressure
CO alarms detect rising carbon monoxide and warn occupants because the gas is otherwise undetectable by human senses. California law requires CO alarms in most dwellings with fuel-burning appliances or attached garages. They are a critical last line of defense against CO poisoning.
Before working inside an energized electrical panel is unavoidable, a technician should do what to reduce shock risk?
- a.Wear cotton work gloves and a long shirt
- b.Use insulated tools and appropriate PPE✓
- c.Wet the work area to reduce dust
- d.Remove all panel covers permanently
When de-energizing is not feasible, Cal/OSHA safe work practices call for insulated tools, appropriate PPE, and careful technique to reduce shock and arc-flash risk. Whenever possible, the circuit should be de-energized and locked out first. Working live is a last resort with added precautions.Cal/OSHA Title 8
Adequate ventilation air in occupied buildings is important because it does what?
- a.It raises the refrigerant charge in the system
- b.It reduces static pressure in the ducts
- c.It dilutes indoor contaminants with fresh air✓
- d.It lowers the equipment's SEER rating
Ventilation introduces outdoor air to dilute odors, carbon dioxide, and other indoor contaminants, maintaining acceptable indoor air quality. The CMC and related standards set minimum ventilation rates for occupancies. Balancing ventilation with energy use is part of good system design.2022 California Mechanical Code (CMC)
A furnace's high-limit (temperature limit) switch protects the equipment by doing what?
- a.Metering gas continuously
- b.Shutting off the burners if the furnace overheats✓
- c.Increasing blower speed permanently
- d.Recovering refrigerant
The high-limit switch senses excessive heat exchanger temperature, typically from low airflow, and shuts off the burners to prevent overheating and fire. Repeated tripping signals an airflow problem such as a dirty filter or failing blower. It is a key safety control that must never be bypassed.
When entering a permit-required confined space, atmospheric testing should be performed to check for what?
- a.The color of the refrigerant vapor
- b.The dimensions of the ductwork inside the space
- c.Oxygen level and toxic or flammable gases✓
- d.The blower speed at the air handler
Before and during entry into a permit-required confined space, the atmosphere is tested for adequate oxygen and for toxic or flammable gases to ensure it is safe. Continuous monitoring and ventilation may be required. Cal/OSHA mandates these procedures to prevent asphyxiation and explosion.Cal/OSHA Title 8
A high-pressure safety switch on refrigeration equipment protects the system by doing what?
- a.Adding refrigerant when the pressure drops
- b.Stopping the compressor on high pressure✓
- c.Draining condensate from the coil pan
- d.Boosting the condenser fan speed
A high-pressure switch opens the control circuit and stops the compressor when head pressure rises to an unsafe level, such as from a failed condenser fan or blocked airflow. This prevents component rupture and damage. It should be diagnosed and reset only after the underlying cause is corrected.
Which practice best reduces the risk of heat illness for HVAC technicians working on hot rooftops?
- a.Skipping water breaks in order to finish faster
- b.Wearing heavy dark clothing at midday
- c.Providing water, rest, shade, and pacing✓
- d.Avoiding all fluids while on the roof
Cal/OSHA heat illness prevention calls for access to water, shade, and rest breaks, plus acclimatization and pacing during high temperatures. Rooftop HVAC work exposes technicians to extreme heat, raising the risk of heat exhaustion or stroke. Recognizing symptoms early and hydrating are essential.Cal/OSHA Title 8
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.
A whole-house exhaust fan removes 100 CFM continuously from a tight new home. How should the load calculation treat that air?
- a.As a latent-only load with no sensible component
- b.As irrelevant because mechanical ventilation is outside Manual J
- c.As additional outdoor air that must be conditioned✓
- d.As a reduction in load because stale air is removed
Air exhausted from a house is replaced by outdoor air entering somewhere, and that air arrives at design temperature and humidity, so it becomes load. It cannot reduce the load, because the incoming replacement air is exactly what the equipment must condition. It carries both a temperature difference and a moisture difference, so restricting it to latent is wrong. And deliberate ventilation air is specifically accounted for in a modern load calculation.
A two-story house cools unevenly, with the upstairs several degrees warmer every afternoon. Which design remedy addresses the cause rather than the symptom?
- a.Lower the thermostat setpoint until the upstairs is comfortable
- b.Install a larger condenser to raise total system capacity
- c.Calculate loads floor by floor and redistribute airflow or zone the system✓
- d.Increase the refrigerant charge to lower the supply air temperature upstairs
Upstairs carries more roof and solar gain, so the fix is to match delivered CFM to each floor's actual load, by redistribution or by zoning. Dropping the setpoint overcools the downstairs while the upstairs still lags. Overcharging raises head pressure and can flood the compressor without moving more air upstairs. And a larger condenser adds capacity to a system that is already delivering it to the wrong floor.
What is the main design risk of using a bypass duct with a zoned residential system?
- a.It prevents the zone dampers from ever closing fully
- b.It makes the outdoor unit short cycle on high head pressure
- c.It raises static pressure so high that the blower stalls
- d.It recirculates cold air to the coil and can freeze it during small zone calls✓
A bypass sends already-cooled supply air back to the return, dropping entering air temperature until the coil ices when only one small zone calls. It actually relieves static pressure rather than raising it, which is why it is installed in the first place. It has no mechanical effect on damper travel. And the low-side starvation it creates tends to lower head pressure, not raise it into a high-pressure trip.
Return air paths are being designed for bedrooms that have no dedicated return grille. What is the design objective?
- a.Keep the bedroom at a measurable negative pressure while occupied
- b.Ensure each bedroom receives more return CFM than supply CFM
- c.Limit the pressure the closed door creates between the room and the hall✓
- d.Route the transfer air path through the attic to gain extra effective length
Transfer grilles, jump ducts, or undercuts exist to keep room pressurization small when the door closes so supply air can actually enter. Deliberately returning more than is supplied would depressurize the room and pull in attic or garage air. A measurable negative pressure is the defect being avoided, not the goal. And running the path through an unconditioned attic adds heat gain and leakage to a transfer that should stay inside the envelope.
You measure total external static pressure of 1.1 in. w.c. on a system whose blower is rated for 0.5 in. w.c. What is the first consequence for the design?
- a.The compressor will draw amperage above its rated load amps
- b.The condensate trap will siphon dry on every cycle
- c.The furnace will fail to ignite on a call for heat
- d.Airflow will fall below the design CFM, cutting capacity✓
A blower operating far beyond its rated external static moves much less air than its table value, so delivered capacity and temperature split both suffer. Compressor amperage is set by refrigerant load and voltage, and low airflow typically lowers it rather than raising it above RLA. Ignition depends on the gas train and flame proving, not on duct static. And trap siphoning is caused by blower suction at the drain, a separate and much smaller pressure issue.
Which pair of measurements is used to determine total external static pressure on a residential air handler?
- a.Velocity pressure at the register and velocity pressure at the grille
- b.Supply plenum pressure and outdoor barometric pressure
- c.Filter pressure drop and coil pressure drop added together
- d.Supply plenum pressure and return pressure at the equipment✓
Total external static is the pressure the blower works against outside the cabinet, read as supply-side positive plus return-side negative at the equipment. Barometric pressure is an ambient reading and is not part of a duct measurement. Filter and coil drops are internal component losses that many manufacturers exclude from external static. And velocity pressure measures air speed, which is not the static resistance the fan table uses.
A designer wants to reduce the friction loss in a long branch run without changing the duct diameter. What is the most effective option?
- a.Wrap the duct with a higher R-value insulation blanket
- b.Replace sharp fittings with long-radius elbows and smooth boots✓
- c.Add a manual balancing damper near the takeoff
- d.Increase the blower speed to push more air through the branch duct
Fittings contribute most of the effective length in a branch, so long-radius elbows and better boots cut pressure loss substantially without resizing. Insulation changes heat transfer, not friction. A balancing damper deliberately adds resistance to reduce flow, which is the opposite of the goal. And more blower speed does not lower friction; it raises pressure loss further while burning more fan power.
Flexible duct is installed with visible sag and slack between supports. How does this affect the Manual D design?
- a.Effective friction rises well above the tables, so delivered CFM drops✓
- b.It matters only on return ducts, since supply ducts are pressurized anyway
- c.Capacity improves because the longer path increases heat exchange
- d.Only noise increases, because friction depends on diameter alone
Manual D friction values for flex assume the duct is pulled taut, and sagging inner liner adds turbulence that can cut delivered airflow sharply. Friction depends on interior surface condition and path shape, not diameter alone. Extra path length in an unconditioned space adds losses rather than useful heat exchange. And the friction penalty applies equally to return ducts, which are just as sensitive to added resistance.
Which measurement pair lets a technician determine the latent capacity a coil is actually producing?
- a.Suction pressure and suction line temperature
- b.Entering and leaving wet-bulb temperature✓
- c.Supply plenum static and return plenum static
- d.Entering and leaving dry-bulb temperature
Wet-bulb readings translate to enthalpy, and the drop in enthalpy beyond the sensible portion is the latent work the coil is doing. Dry-bulb alone captures only sensible heat and says nothing about moisture removed. Suction pressure and line temperature give superheat, a refrigerant-side charge check. And static pressures describe duct resistance, not the moisture content of the air.
A house has a design cooling load of 22,000 BTU/hr. Available equipment comes in 1.5, 2, 2.5, and 3 ton sizes. Which selection best follows Manual S?
- a.1.5 ton, because smaller equipment runs longer and dehumidifies much better
- b.2 ton, verified against the expanded performance data at design conditions✓
- c.2.5 ton, to leave headroom for hotter than design summers
- d.3 ton, because the next size up protects against future additions
A 2-ton unit sits closest to the 22,000 BTU/hr load, and Manual S then confirms its sensible and latent split at the actual design conditions. A 1.5-ton unit is roughly 4,000 BTU short and will not hold setpoint on a design day. A 2.5-ton unit exceeds the Manual S oversizing allowance and short cycles. And sizing for a hypothetical future addition penalizes comfort and efficiency for every year until that addition exists.
Two houses have the same floor area, but one has 12 air changes per hour at 50 pascals and the other has 3. What does that tell you about their loads?
- a.The loads are equal because area drives infiltration
- b.The tighter house has the higher heating load
- c.The difference shows up only in the summer cooling load
- d.The leakier house has the higher infiltration load in both seasons✓
A blower-door result of 12 ACH50 means far more air exchange than 3 ACH50, and that unconditioned air is load in both winter and summer. Tightness reduces load, so the tighter house cannot be the higher one. Floor area is only one input; leakage area is what drives infiltration. And infiltration is driven by temperature and wind differences year round, not summer only.
Why does Manual J use a cooling load temperature difference rather than the plain indoor-outdoor difference for roofs and walls?
- a.To convert the BTU per hour result into tons of refrigeration directly
- b.To correct the calculation for elevation above sea level
- c.To include the latent gain absorbed by the roof sheathing
- d.To account for solar absorption and the mass delay of the assembly✓
Sunlit opaque surfaces run hotter than the air and release that heat on a lag governed by mass, so an adjusted temperature difference represents the real driving force. Converting to tons is a simple division at the end and needs no such adjustment. Elevation corrections apply to air density in the airflow constants, not to wall conduction. And roof sheathing does not add moisture to the space, so no latent term is involved.