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PE Mechanical practice questions

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Question 1Basic Engineering Practice

A pressure gauge on a pipeline reads 350 kPa. The local atmospheric pressure is 101 kPa. What is the absolute pressure in the pipeline?

  1. 249 kPa
  2. 350 kPa
  3. 101 kPa
  4. 451 kPa Correct answer.

Explanation

Absolute pressure equals gauge pressure plus the local atmospheric pressure, so 350 kPa + 101 kPa = 451 kPa. A gauge measures pressure relative to its surroundings, not relative to a perfect vacuum.

Question 2Basic Engineering Practice

How many significant figures are in the measured value 0.004560 m?

  1. 5
  2. 3
  3. 7
  4. 4 Correct answer.

Explanation

Leading zeros used only to place the decimal point are not significant, while a trailing zero to the right of the decimal point is significant. The significant digits are 4, 5, 6, and 0, giving four.

Question 3Basic Engineering Practice

A loan of $20,000 is repaid with simple interest at 6% per year for 3 years. What total interest is paid?

  1. $4,000
  2. $3,600 Correct answer.
  3. $1,200
  4. $3,820

Explanation

Simple interest accrues only on the original principal: I = P x i x n = $20,000 x 0.06 x 3 = $3,600. Compounding is not applied in a simple-interest agreement.

Question 4Basic Engineering Practice

The primary purpose of a lockout/tagout program is to:

  1. verify that electrical circuits are properly grounded before startup
  2. document the maintenance history of rotating equipment
  3. schedule routine inspections of pressure relief valves
  4. prevent unexpected release of hazardous energy during servicing Correct answer.

Explanation

Lockout/tagout isolates energy sources and applies locks and tags so equipment cannot be energized or started while someone is servicing it. The objective is protection of personnel from unexpected energization or release of stored energy.

Question 5Basic Engineering Practice

A material has a specific heat of 1.00 Btu/(lbm-deg F). What is this value in kJ/(kg-K)?

  1. 1.000
  2. 2.326
  3. 0.239
  4. 4.187 Correct answer.

Explanation

One Btu/(lbm-deg F) equals 4.187 kJ/(kg-K), because 1 Btu = 1.055 kJ, 1 lbm = 0.4536 kg, and a Fahrenheit degree interval is 1.8 times a Kelvin interval. Numerically the value becomes 1.055/(0.4536 x 1.8) = 4.187.

Question 6Basic Engineering Practice

In a feature control frame, the datum reference letters identify:

  1. the inspection frequency required for the feature
  2. the allowable variation in the size of the feature
  3. the manufacturing process that must be used
  4. the features from which the tolerance zone is located Correct answer.

Explanation

Datum reference letters tie the tolerance zone to specific datum features established on the part, fixing its location and orientation. They do not set size limits, process, or inspection frequency.

Question 7Basic Engineering Practice

A licensed engineer is asked to seal drawings for a foundation system, although her experience is limited to HVAC design. The appropriate action is to:

  1. decline unless the work is reviewed and sealed by a qualified engineer Correct answer.
  2. seal the drawings after adding a note limiting her liability
  3. seal only the sheets that contain no structural details
  4. seal the drawings because a professional engineer's license covers all disciplines

Explanation

The rules of professional conduct require engineers to practice only within their areas of competence. Work outside that competence must be performed or reviewed by an engineer qualified in that field before it is sealed.

Question 8Basic Engineering Practice

An activity in a CPM schedule has an early finish of day 16. Its only successor has a late start of day 18. What is the activity's total float?

  1. 0 days
  2. 2 days Correct answer.
  3. 18 days
  4. 16 days

Explanation

Total float is the late start of the successor minus the early finish of the activity: 18 - 16 = 2 days. It is the amount the activity can slip without delaying project completion.

Question 9Basic Engineering Practice

What is the present worth of $5,000 received at the end of each year for 5 years at 8% annual interest?

  1. $21,600
  2. $25,000
  3. $19,960 Correct answer.
  4. $17,000

Explanation

The uniform-series present worth factor (P/A, 8%, 5) equals 3.9927, so the present worth is $5,000 x 3.9927 = $19,964, or about $19,960. Discounting reduces the sum below the undiscounted total.

Question 10Basic Engineering Practice

Equipment costs $50,000, has a salvage value of $5,000, and a depreciable life of 9 years. Using straight-line depreciation, what is the annual depreciation charge?

  1. $6,111
  2. $4,500
  3. $5,000 Correct answer.
  4. $5,556

Explanation

Straight-line depreciation spreads the depreciable basis, cost minus salvage, evenly over the life: ($50,000 - $5,000)/9 = $5,000 per year. Salvage value is excluded from the amount to be depreciated.

Question 11Basic Engineering Practice

A product sells for $30 per unit with a variable cost of $18 per unit and annual fixed costs of $120,000. What is the break-even quantity per year?

  1. 6,667 units
  2. 4,000 units
  3. 20,000 units
  4. 10,000 units Correct answer.

Explanation

Break-even occurs where contribution margin covers fixed costs: Q = fixed cost / (price - variable cost) = $120,000 / $12 = 10,000 units per year.

Question 12Basic Engineering Practice

A hole is specified from 25.01 to 25.03 mm and a mating shaft from 25.00 to 25.02 mm. What is the maximum interference in the fit?

  1. 0.02 mm
  2. 0.01 mm Correct answer.
  3. 0.00 mm
  4. 0.03 mm

Explanation

Maximum interference occurs with the largest shaft and the smallest hole: 25.02 - 25.01 = 0.01 mm. Interference is positive only where the shaft is larger than the hole.

Question 13Basic Engineering Practice

Which heat treatment is applied to a welded steel assembly to reduce residual stresses without appreciably changing its microstructure or strength?

  1. Normalizing
  2. Stress-relief annealing Correct answer.
  3. Case carburizing
  4. Quenching and tempering

Explanation

Stress-relief annealing heats the assembly to a temperature below the transformation range, holds it, and cools slowly so residual stresses relax without altering the microstructure. Normalizing and quenching change the microstructure, and carburizing alters the surface chemistry.

Question 14Basic Engineering Practice

Which ASME code governs the design and construction of process piping in a chemical plant?

  1. ASME B16.5
  2. ASME B31.1
  3. ASME B31.3 Correct answer.
  4. ASME Section VIII, Division 1

Explanation

ASME B31.3, Process Piping, covers piping in chemical, petroleum, and similar plants. B31.1 applies to power piping, Section VIII covers pressure vessels, and B16.5 covers pipe flanges and flanged fittings.

Question 15Basic Engineering Practice

A process has specification limits of 9.8 and 10.2 with an estimated process standard deviation of 0.05. What is the process capability index Cp?

  1. 1.33 Correct answer.
  2. 1.00
  3. 2.67
  4. 0.67

Explanation

Cp = (USL - LSL)/(6 sigma) = (10.2 - 9.8)/(6 x 0.05) = 0.4/0.30 = 1.33. Cp compares the specification width with the natural spread of the process.

Question 16Basic Engineering Practice

A pump delivers 500 gallons per minute. What is the equivalent flow rate in cubic meters per second?

  1. 0.00013 m3/s
  2. 2.05 m3/s
  3. 0.0315 m3/s Correct answer.
  4. 0.0079 m3/s

Explanation

One gallon per minute equals 6.309 x 10^-5 m3/s, so 500 gpm x 6.309 x 10^-5 = 0.0315 m3/s. This is equivalent to about 31.5 liters per second.

Question 17Basic Engineering Practice

Which arc welding process uses a continuous consumable electrode and a blanket of granular flux, with the arc completely hidden and no visible arc?

  1. Submerged arc welding Correct answer.
  2. Gas tungsten arc welding
  3. Flux-cored arc welding
  4. Shielded metal arc welding

Explanation

Submerged arc welding buries the arc under a layer of granular flux that shields the weld and produces a slag covering. The other processes leave a visible arc, and flux-cored wires carry flux inside the electrode rather than in an external blanket.

Question 18Basic Engineering Practice

A position tolerance is specified at maximum material condition (MMC). What happens as the actual feature size departs from MMC?

  1. The datum reference letters may be omitted from the frame
  2. The stated tolerance zone is reduced by the departure
  3. Bonus tolerance equal to the departure is added Correct answer.
  4. The tolerance is unchanged but inspection must be 100 percent

Explanation

The MMC modifier grants bonus tolerance equal to how far the actual size departs from the maximum material condition, because clearance increases as the feature gets smaller or larger relative to MMC. The stated tolerance is a minimum, not a maximum.

Question 19Basic Engineering Practice

An engineer evaluating vendor bids learns that her former employer is one of the bidders. The most appropriate action is to:

  1. withdraw her seal from the project until the bidding closes
  2. disclose the relationship and ask to be removed from the evaluation Correct answer.
  3. evaluate the bid but rank it lower to avoid any appearance of favoritism
  4. say nothing because she no longer works for that company

Explanation

Professional ethics require engineers to disclose any relationship that could compromise or appear to compromise objectivity and to avoid participating in decisions where a conflict exists. Intentional bias in either direction is itself improper.

Question 20Basic Engineering Practice

A $100,000 asset has a 20-year life with no salvage value and will be replaced indefinitely at the same cost. At 6% annual interest, what is the capitalized cost?

  1. $145,300 Correct answer.
  2. $127,000
  3. $100,000
  4. $183,300

Explanation

Convert the replacement cost to an annual equivalent and then to a present amount: A = $100,000 x (A/F, 6%, 20) = $2,718 per year, and capitalized cost = first cost + A/i = $100,000 + $2,718/0.06 = $145,300.

Question 21Basic Engineering Practice

Four components are stacked in series, each with a tolerance of plus or minus 0.005 in. Using root-sum-square, what assembly tolerance results?

  1. plus or minus 0.005 in.
  2. plus or minus 0.0025 in.
  3. plus or minus 0.020 in.
  4. plus or minus 0.010 in. Correct answer.

Explanation

Root-sum-square combines independent component tolerances as t_total = t x sqrt(n) = 0.005 x sqrt(4) = 0.010 in. The RSS result is smaller than the worst-case arithmetic sum because independent variations do not all reach their extremes simultaneously.

Question 22Basic Engineering Practice

A cylindrical vessel with an inside radius of 500 mm is designed for 1.5 MPa internal pressure. The allowable stress is 138 MPa and the joint efficiency is 0.85. Using t = PR/(SE - 0.6P), what minimum thickness is required?

  1. 7.6 mm
  2. 5.4 mm
  3. 6.4 mm Correct answer.
  4. 12.9 mm

Explanation

With SE = 138 x 0.85 = 117.3 MPa, the denominator is 117.3 - (0.6 x 1.5) = 116.4 MPa. Then t = (1.5)(500)/116.4 = 6.44 mm, so 6.4 mm is required.

Question 23Basic Engineering Practice

A process has a capability index Cp of 1.00 and its mean is shifted 1.5 standard deviations above the target. What is Cpk?

  1. 0.50 Correct answer.
  2. 1.50
  3. 1.00
  4. 0.75

Explanation

With Cp = 1.00 the specification half-width is 3 sigma. Cpk = (3 sigma - 1.5 sigma)/(3 sigma) = 0.50, reflecting the loss of capability caused by the off-center mean.

Question 24Basic Engineering Practice

Pump A costs $10,000 installed with $2,000 per year in operating costs and a 5-year life. Pump B costs $18,000 installed with $1,200 per year and a 10-year life. At 8% interest, which pump has the lower equivalent annual cost, and by how much?

  1. Pump B, by about $620 per year Correct answer.
  2. Pump B, by about $300 per year
  3. Pump A, by about $620 per year
  4. Pump A, by about $1,900 per year

Explanation

Pump A: $10,000 x 0.25046 + $2,000 = $4,505 per year. Pump B: $18,000 x 0.14903 + $1,200 = $3,883 per year. Pump B is lower by about $620 per year.

Question 25Psychrometrics

Moist air is cooled at constant pressure with no moisture added or removed. The temperature at which condensation begins is called the:

  1. Adiabatic saturation temperature
  2. Dry-bulb temperature
  3. Dew-point temperature Correct answer.
  4. Wet-bulb temperature

Explanation

During sensible cooling at constant pressure the water-vapor partial pressure stays fixed until saturation is reached; the saturation temperature corresponding to that fixed vapor pressure is the dew point. Any further cooling below the dew point produces condensation.

Question 26Psychrometrics

Relative humidity is defined as the ratio of:

  1. Actual specific volume to specific volume at saturation
  2. Mass of water vapor to mass of dry air
  3. Mass of dry air to mass of moist air
  4. Water-vapor partial pressure to saturation pressure at the dry-bulb temperature Correct answer.

Explanation

Relative humidity compares the actual water-vapor partial pressure with the saturation pressure evaluated at the same dry-bulb temperature. It is expressed as a ratio or a percentage and is always 100 percent at saturation.

Question 27Psychrometrics

Moist air is sensibly heated at constant pressure with no moisture added. Which change occurs?

  1. Relative humidity decreases Correct answer.
  2. Dew point increases
  3. Relative humidity increases
  4. Humidity ratio increases

Explanation

Sensible heating holds the humidity ratio and the vapor partial pressure constant while raising the dry-bulb temperature. Because the saturation pressure rises with temperature, the ratio of vapor pressure to saturation pressure falls, so relative humidity decreases.

Question 28Psychrometrics

Which psychrometric property is expressed in kilograms of water vapor per kilogram of dry air?

  1. Specific enthalpy
  2. Humidity ratio Correct answer.
  3. Specific volume
  4. Degree of saturation

Explanation

The humidity ratio (also called the moisture content or mixing ratio) is the mass of water vapor divided by the mass of dry air in the mixture. Basing the property on dry air keeps the mass of dry air constant through humidification and dehumidification processes.

Question 29Psychrometrics

Air at 20°C dry bulb is saturated. What is its dew-point temperature?

  1. About 20°C Correct answer.
  2. About 10°C
  3. About 30°C
  4. Indeterminate without the barometric pressure

Explanation

At saturation the vapor partial pressure equals the saturation pressure at the dry-bulb temperature, so cooling the air by any amount immediately causes condensation. The dew point therefore coincides with the dry-bulb temperature, 20°C.

Question 30Psychrometrics

Which field instrument is used to obtain the wet-bulb temperature of an air stream?

  1. A barometer
  2. A sling psychrometer Correct answer.
  3. A U-tube manometer
  4. An anemometer

Explanation

A psychrometer consists of a dry-bulb thermometer and a second thermometer with a wetted wick that is ventilated through the air stream at roughly 3 to 5 m/s. The lowered reading of the wetted thermometer is the wet-bulb temperature.

Question 31Psychrometrics

Moist air at 25°C dry bulb has a humidity ratio of 0.010 kg/kg dry air. Taking cp,air = 1.006 kJ/(kg·K), hfg at 0°C = 2501 kJ/kg, and cp,vapor = 1.86 kJ/(kg·K), what is its specific enthalpy?

  1. 50.6 kJ/kg dry air Correct answer.
  2. 25.2 kJ/kg dry air
  3. 75.9 kJ/kg dry air
  4. 38.4 kJ/kg dry air

Explanation

The moist-air enthalpy is h = 1.006·T + W(2501 + 1.86·T) per kilogram of dry air. Substituting gives 25.15 + 0.010(2501 + 46.5) = 25.15 + 25.47 = 50.6 kJ/kg dry air.

Question 32Psychrometrics

Two airstreams mix adiabatically: 1.0 kg/s of dry air at 30°C with a humidity ratio of 0.012 kg/kg, and 2.0 kg/s of dry air at 15°C with a humidity ratio of 0.006 kg/kg. What is the humidity ratio of the mixture?

  1. 0.010 kg/kg
  2. 0.008 kg/kg Correct answer.
  3. 0.006 kg/kg
  4. 0.009 kg/kg

Explanation

For adiabatic mixing the humidity ratio is a mass-weighted average of the two streams: W = (1.0×0.012 + 2.0×0.006)/3.0 = 0.024/3.0 = 0.008 kg/kg dry air.

Question 33Psychrometrics

At a fixed dry-bulb temperature, the degree of saturation differs from the relative humidity because the degree of saturation is based on:

  1. W/Ws at the same dry-bulb temperature Correct answer.
  2. Vapor pressure over saturation pressure
  3. Water-vapor mass per unit volume
  4. Enthalpy over saturation enthalpy

Explanation

The degree of saturation is the ratio of the actual humidity ratio to the humidity ratio of saturated air at the same dry-bulb temperature. It differs slightly from relative humidity because W is not exactly proportional to vapor pressure once the vapor mass is referred to dry air.

Question 34Psychrometrics

A space has a total cooling load of 20 kW, of which 6 kW is latent. What is the sensible heat ratio?

  1. 0.70 Correct answer.
  2. 0.77
  3. 0.86
  4. 0.30

Explanation

The sensible load is 20 − 6 = 14 kW, and the sensible heat ratio is sensible load divided by total load: 14/20 = 0.70. The slope of the room process line on the psychrometric chart is set by this ratio.

Question 35Psychrometrics

A cooling coil has an apparatus dew point of 10°C and a bypass factor of 0.15. Air enters the coil at 26°C dry bulb. What is the leaving dry-bulb temperature?

  1. 10.0°C
  2. 12.4°C Correct answer.
  3. 11.5°C
  4. 16.0°C

Explanation

The bypass factor is the fraction of air that passes through untouched, so the leaving state lies on a straight line between the entering state and the apparatus dew point: Tleaving = 10 + 0.15(26 − 10) = 12.4°C.

Question 36Psychrometrics

Saturated steam at about 100°C is injected into an air stream at constant pressure. Which description matches the resulting process?

  1. Humidity ratio rises while dry-bulb falls
  2. Humidity ratio rises while dry-bulb falls to saturation
  3. Humidity ratio rises with almost no dry-bulb change Correct answer.
  4. Dry-bulb rises while humidity ratio is constant

Explanation

Steam humidification adds both moisture and enthalpy to the air. The enthalpy supplied per kilogram of vapor is close to the latent heat needed to hold that vapor at the existing temperature, so the humidity ratio increases while the dry-bulb temperature stays nearly constant (rising very slightly in practice).

Question 37Psychrometrics

Air at 30°C dry bulb has a relative humidity of 50 percent, and the saturation pressure at 30°C is 4.25 kPa. What is the water-vapor partial pressure?

  1. 8.49 kPa
  2. 1.06 kPa
  3. 2.12 kPa Correct answer.
  4. 4.25 kPa

Explanation

Relative humidity is the ratio of the vapor partial pressure to the saturation pressure at the dry-bulb temperature, so pv = 0.50 × 4.25 = 2.12 kPa. This partial pressure then fixes the dew point of the mixture.

Question 38Psychrometrics

A hygrometer indicates a water-vapor partial pressure of 1.70 kPa. Saturation pressures are 1.23 kPa at 10°C, 1.71 kPa at 15°C, 2.34 kPa at 20°C, and 3.17 kPa at 25°C. What is the dew point?

  1. 25°C
  2. 20°C
  3. 10°C
  4. 15°C Correct answer.

Explanation

The dew point is the temperature whose saturation pressure equals the actual vapor partial pressure. Since 1.71 kPa is the tabulated saturation pressure at 15°C, the dew point is about 15°C.

Question 39Psychrometrics

Air is drawn through an adiabatic spray chamber in which recirculated water is sprayed continuously. Which line on the psychrometric chart does the air follow?

  1. A constant humidity ratio line
  2. A constant relative humidity line
  3. A constant dew-point line
  4. The constant wet-bulb line Correct answer.

Explanation

In an adiabatic saturator the air gives up sensible heat to evaporate the recirculated water, and the water returns that enthalpy to the air as vapor, so the wet-bulb (adiabatic saturation) temperature stays essentially constant. The state moves down and to the left along that line toward saturation.

Question 40Psychrometrics

What is the density of dry air at 20°C and 101.3 kPa? Use R = 0.287 kJ/(kg·K).

  1. 1.40 kg/m³
  2. 0.83 kg/m³
  3. 2.40 kg/m³
  4. 1.20 kg/m³ Correct answer.

Explanation

Applying the ideal-gas relation, ρ = P/(R·T) = 101.3/(0.287 × 293.15) = 1.20 kg/m³. The same calculation gives the specific volume of 0.83 m³/kg of dry air used as the basis for moist-air volume flow.

Question 41Psychrometrics

2.0 kg/s of dry air enters a cooling coil at 30°C dry bulb with a humidity ratio of 0.020 kg/kg and leaves saturated at 12°C with a humidity ratio of 0.0088 kg/kg. What mass flow of condensate is produced?

  1. 40 kg/h
  2. 161 kg/h
  3. 81 kg/h Correct answer.
  4. 24 kg/h

Explanation

The moisture removed equals the dry-air mass flow times the drop in humidity ratio: 2.0(0.020 − 0.0088) = 0.0224 kg/s. Converting to an hourly rate gives 0.0224 × 3600 ≈ 81 kg/h.

Question 42Psychrometrics

3.0 kg/s of dry air enters a cooling coil at 30°C with a humidity ratio of 0.016 kg/kg (h = 71.1 kJ/kg dry air) and leaves at 14°C with a humidity ratio of 0.010 kg/kg (h = 39.4 kJ/kg dry air). What is the total cooling rate?

  1. 47.6 kW
  2. 63.5 kW
  3. 127 kW
  4. 95.1 kW Correct answer.

Explanation

The total coil load is the dry-air mass flow times the enthalpy decrease: 3.0(71.1 − 39.4) = 95.1 kW. This total includes both the sensible cooling of the air and the latent heat of the condensed moisture.

Question 43Psychrometrics

Air at 40°C dry bulb and 24°C wet bulb flows at 101.325 kPa. Using W = [(2501 − 2.326·Twb)·Ws* − 1.006(Tdb − Twb)] / [2501 + 1.86·Tdb − 4.186·Twb], with a saturation pressure of 2.985 kPa at 24°C, what is the humidity ratio?

  1. 0.0189 kg/kg
  2. 0.0095 kg/kg
  3. 0.0154 kg/kg
  4. 0.0122 kg/kg Correct answer.

Explanation

First Ws* = 0.622(2.985)/(101.325 − 2.985) = 0.0189 kg/kg. The numerator is (2501 − 55.8)(0.0189) − 1.006(16) = 46.2 − 16.1 = 30.1, and the denominator is 2501 + 74.4 − 100.5 = 2474.9, giving W ≈ 0.0122 kg/kg dry air.

Question 44Psychrometrics

A psychrometer reads 30°C dry bulb and 22°C wet bulb at 101.325 kPa. Using pv = pws,wb − A·P·(Tdb − Twb) with A = 0.00066(1 + 0.00115·Twb), and saturation pressures of 2.645 kPa at 22°C and 4.246 kPa at 30°C, what is the relative humidity?

  1. 62 percent
  2. 43 percent
  3. 35 percent
  4. 50 percent Correct answer.

Explanation

A = 0.00066(1 + 0.00115 × 22) = 0.000677, so pv = 2.645 − 0.000677(101.325)(8) = 2.645 − 0.549 = 2.10 kPa. Relative humidity is then 2.10/4.246 ≈ 0.49, or about 50 percent.

Question 45Heating and Cooling Loads

Which of the following is classified as an internal sensible heat gain when computing a space cooling load?

  1. Heat carried in with outdoor ventilation air
  2. Heat passing through a sunlit window
  3. Heat released by the lighting fixtures Correct answer.
  4. Heat transmitted through an exterior wall

Explanation

Internal gains are heat sources located inside the conditioned space, such as lighting, people, and appliances. Envelope conduction, outdoor-air loads, and solar transmission all bring heat in from outside the space, so they are treated as external gains.

Question 46Heating and Cooling Loads

Which item contributes to the latent cooling load of a space?

  1. Heat given off by the space lighting
  2. Moisture released by occupants and by processes in the space Correct answer.
  3. Solar radiation absorbed by the floor slab
  4. Heat conducted through the roof on a hot afternoon

Explanation

Latent load is the energy needed to condense water vapor, so it is driven by moisture sources and by the difference in humidity ratio between indoor and outdoor air. Roof conduction, solar absorption, and lighting all add sensible heat only.

Question 47Heating and Cooling Loads

A wall assembly has a total thermal resistance of 13 h·ft²·°F/Btu. Its overall heat transfer coefficient is closest to:

  1. 13.0 Btu/h·ft²·°F
  2. 0.0077 Btu/h·ft²·°F
  3. 3.6 Btu/h·ft²·°F
  4. 0.077 Btu/h·ft²·°F Correct answer.

Explanation

The U-factor is the reciprocal of the total resistance of the assembly, U = 1/R. For R = 13, U = 1/13 ≈ 0.077 Btu/h·ft²·°F.

Question 48Heating and Cooling Loads

An exterior wall has U = 0.077 Btu/h·ft²·°F, an area of 200 ft², and sees a design temperature difference of 40°F. The conduction heat gain is:

  1. 616 Btu/h Correct answer.
  2. 308 Btu/h
  3. 1,232 Btu/h
  4. 154 Btu/h

Explanation

Steady-state conduction through an opaque assembly is Q = U × A × ΔT. Multiplying 0.077 × 200 × 40 gives about 616 Btu/h.

Question 49Heating and Cooling Loads

Infiltration is best described as:

  1. Air moving between zones through interior openings
  2. Conditioned air escaping from the building
  3. Outdoor air delivered by the ventilation system
  4. Air leaking in through the envelope Correct answer.

Explanation

Infiltration is uncontrolled inward leakage of outdoor air through cracks, joints, and penetrations in the envelope, driven by wind and stack effect. Mechanical outdoor air is deliberate ventilation, outward leakage is exfiltration, and interior transfer is not envelope leakage.

Question 50Heating and Cooling Loads

Conduction heat gain through a window is calculated from which property of the glazing assembly?

  1. U-factor Correct answer.
  2. Condensation resistance
  3. Solar heat gain coefficient
  4. Visible transmittance

Explanation

Conduction through glass and frame depends on the temperature difference across the assembly and is computed with Q = U × A × ΔT. The solar heat gain coefficient governs the radiant solar portion instead.