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newSchoolWorks.com › VoTechWorks › Building & Construction › HVAC/R › Load Calculation for HVAC/R Systems › Lesson 5
BLD-HVAC07-06 · Lesson 5 · 12 clock hours · theory

Infiltration, Ventilation, and Internal Gains

Objective. Students calculate infiltration heat loss and gain using the Manual J blower-door confirmed or estimated ACH method, and add correct internal gain credits for occupants, lighting, and appliances.

Why Infiltration and Internal Gains Matter

A load calculation that ignores how air leaks into a building—or forgets the heat that people and appliances release—will produce an oversized or undersized system. Both outcomes hurt comfort, efficiency, and equipment life. This lesson covers the two envelope-related loads most commonly underestimated: infiltration and internal gains, plus the minimum mechanical ventilation requirement that every residential designer must acknowledge.

Infiltration: What It Is and Where It Comes From

Infiltration is the uncontrolled movement of outdoor air into a conditioned space through gaps, cracks, and penetrations in the building envelope—around windows and doors, at sill plates, through electrical outlets on exterior walls, and wherever pipes or wires pass through the shell. It is distinct from ventilation, which is controlled, intentional fresh-air delivery.

In summer, infiltration brings in hot, humid outdoor air, adding to the cooling load. In winter, it brings in cold, dry air, adding to the heating load. Either way, the HVAC system must condition that air—it is not free.

Manual J Infiltration Method

ACCA Manual J 8th Edition is the industry-standard residential load calculation procedure. For infiltration, Manual J uses a simplified ACH (Air Changes per Hour) method tied to construction quality categories: loose, average, and tight. Each category assigns an infiltration ACH value appropriate for that level of envelope tightness.

Consult the Manual J infiltration tables for the exact ACH values assigned to each category; the tables also account for the number of stories and the local wind exposure.

Basic Infiltration Calculation Steps

1. Determine the conditioned volume of the space (floor area × ceiling height, in cubic feet).

2. Select the ACH value from the Manual J table based on construction quality.

3. Calculate CFM of infiltration: (Volume × ACH) ÷ 60

4. Calculate sensible heat load (BTU/h): CFM × 1.1 × ΔT (heating season: design indoor temp minus design outdoor temp; cooling season: reverse)

5. For latent load in cooling, a separate calculation uses the humidity ratio difference, but for this lesson we focus on sensible only.

Blower-Door Test Confirmation

A blower-door test measures actual envelope leakage. The result is reported as CFM50—the airflow (in CFM) required to depressurize the house to 50 Pascals. To convert CFM50 to a natural infiltration rate in CFM, divide by a factor (commonly 20 for single-story homes in average terrain; Manual J and ASHRAE provide specific divisors). This converted CFM can then replace the ACH-derived CFM in step 3 above. A blower-door result is more accurate than a table estimate and is required by some energy codes.

> Safety note: Blower-door tests are performed by others (energy auditors, building scientists), but as a technician you may receive the report. Read it carefully; never confuse CFM50 with natural infiltration CFM. Using the raw CFM50 figure in a load calculation inflates the infiltration load by roughly 20×.

Mechanical Ventilation: ASHRAE 62.2

Because tight homes limit natural infiltration, they need intentional fresh air. ASHRAE Standard 62.2 sets minimum whole-building ventilation rates for low-rise residential buildings. At a concept level: the required airflow depends on conditioned floor area and the number of bedrooms. The designer adds this mechanical ventilation CFM as a separate load—it is 100% outdoor air that must be heated or cooled just like infiltration air. Do not double-count: infiltration and mechanical ventilation are separate loads added together.

Internal Gains

Internal gains are heat released inside the conditioned space by occupants, lighting, and appliances. In cooling load calculations, internal gains add to the load. In heating load calculations, Manual J allows a modest credit (reduction) for internal gains because they help warm the space—but engineers apply this credit conservatively.

Occupant Gains

The human body releases both sensible heat (warming the air) and latent heat (moisture/humidity). Manual J uses standard values per occupant. A typical occupied residence is assumed to have occupants during peak cooling hours.

Lighting Gains

Incandescent and halogen fixtures convert most of their wattage to heat. LED fixtures are much more efficient but still release some heat. Convert watts to BTU/h using: 1 watt = 3.412 BTU/h.

Appliance Gains

Kitchen appliances, computers, and entertainment equipment all contribute heat. Manual J provides default diversity values for residential kitchens. In commercial applications (not covered here), appliance schedules become much more detailed.

> Safety note: When inspecting existing systems, confirm that combustion appliances (furnaces, water heaters) are properly vented to the outdoors. Unvented combustion is a carbon-monoxide hazard—always verify CO detector placement and function before working in a mechanical room.

Putting It Together

The total sensible cooling or heating load for a space is the sum of:

Accurate infiltration and internal gain values are the difference between a system that runs efficiently at design conditions and one that short-cycles or can't meet setpoint on the hottest or coldest day of the year.

Key terms9

InfiltrationUncontrolled leakage of outdoor air into a conditioned space through gaps and cracks in the building envelope, adding to both heating and cooling loads.
ACH (Air Changes per Hour)The number of times per hour the entire volume of air in a space is replaced by infiltrating outdoor air; used in Manual J to quantify infiltration rate.
CFM50The airflow in cubic feet per minute measured by a blower-door test at a building depressurization of 50 Pascals; must be converted by a divisor to estimate natural infiltration CFM.
Construction Quality CategoriesManual J classifications—loose, average, and tight—that describe envelope airtightness and are used to select an appropriate infiltration ACH value from the Manual J tables.
ASHRAE 62.2The ASHRAE standard that sets minimum mechanical ventilation rates for low-rise residential buildings to maintain acceptable indoor air quality.
Internal GainsHeat released inside a conditioned space by occupants, lighting, and appliances; added to the cooling load and sometimes credited against the heating load.
Sensible HeatHeat that changes the temperature of air without changing its moisture content; calculated in infiltration using CFM × 1.1 × ΔT.
Latent HeatHeat associated with a change in moisture content of air (evaporation or condensation) rather than a change in temperature; part of the total cooling load but not covered in the simplified sensible infiltration formula.
Design Temperature Difference (ΔT)The difference between the indoor design temperature and the outdoor design temperature used as the driving force in heating or cooling load calculations.

Worked examples4

1. Calculate the sensible heating infiltration load for a 1,500 sq ft single-story house with 8 ft ceilings, classified as average construction. The Manual J table assigns an ACH of 0.40 for this category. Indoor design temperature is 70°F; outdoor design temperature is 15°F.
  1. Step 1 – Find conditioned volume: 1,500 sq ft × 8 ft = 12,000 cubic feet.
  2. Step 2 – Calculate infiltration CFM: (12,000 × 0.40) ÷ 60 = 4,800 ÷ 60 = 80 CFM.
  3. Step 3 – Find ΔT: 70°F − 15°F = 55°F.
  4. Step 4 – Calculate sensible heating load: 80 CFM × 1.1 × 55°F = 4,840 BTU/h.
Answer: Infiltration sensible heating load = 4,840 BTU/h.
2. The same 1,500 sq ft house is now classified as loose construction, with a Manual J ACH of 0.80. All other conditions are identical (8 ft ceilings, 70°F indoor, 15°F outdoor). Calculate the sensible heating infiltration load and compare it to the average-construction result.
  1. Step 1 – Conditioned volume: 1,500 × 8 = 12,000 cubic feet (same as before).
  2. Step 2 – Infiltration CFM: (12,000 × 0.80) ÷ 60 = 9,600 ÷ 60 = 160 CFM.
  3. Step 3 – ΔT: 70°F − 15°F = 55°F.
  4. Step 4 – Sensible load: 160 CFM × 1.1 × 55°F = 9,680 BTU/h.
  5. Step 5 – Compare: 9,680 BTU/h (loose) vs. 4,840 BTU/h (average) — loose construction doubles the infiltration load.
Answer: Infiltration sensible heating load = 9,680 BTU/h; doubling ACH doubles the BTU/h load.
3. A blower-door test on the same 1,500 sq ft house returns a result of 1,400 CFM50. Using a conversion divisor of 20 (single-story, average terrain), calculate the natural infiltration CFM and the resulting sensible heating load (70°F indoor, 15°F outdoor).
  1. Step 1 – Convert CFM50 to natural infiltration CFM: 1,400 ÷ 20 = 70 CFM.
  2. Step 2 – ΔT: 70°F − 15°F = 55°F.
  3. Step 3 – Sensible load: 70 CFM × 1.1 × 55°F = 4,235 BTU/h.
  4. Step 4 – Note: This measured value (4,235 BTU/h) is close to but slightly lower than the Manual J average-construction estimate (4,840 BTU/h), confirming this house performs near average tightness.
Answer: Natural infiltration = 70 CFM; sensible heating load = 4,235 BTU/h.
4. A living room contains 4 occupants and 300 watts of lighting. Calculate the total internal sensible gain added to the cooling load. Use 250 BTU/h per occupant for sensible heat (a common Manual J residential default) and the standard watt-to-BTU/h conversion.
  1. Step 1 – Occupant gain: 4 occupants × 250 BTU/h = 1,000 BTU/h.
  2. Step 2 – Lighting gain: 300 watts × 3.412 BTU/h per watt = 1,023.6 BTU/h, rounded to 1,024 BTU/h.
  3. Step 3 – Total internal sensible gain: 1,000 + 1,024 = 2,024 BTU/h.
Answer: Total internal sensible cooling gain = 2,024 BTU/h.

The math in this lesson2

Learners compute infiltration CFM from house volume and ACH, then sensible load with 1.08 × CFM × ΔT.
A-SSE.1G-GMD.3
Brush up in the math halls at newSchoolWorks.com:
In class: Lecture and worked example: instructor walks through the Manual J infiltration calculation using construction quality categories and the simplified ACH method, then shows how a blower-door test result (CFM50) translates to natural infiltration rate. · Students complete a structured infiltration worksheet for a sample 1,500 sq ft single-story house using three different construction quality assumptions (loose, average, tight) and compare the BTU/h differences. · Students review ASHRAE 62.2 ventilation rate requirements for residential buildings (concept level, not full code compliance calculation) and add a fixed mechanical ventilation load to their infiltration worksheet.
Materials: ACCA Manual J 8th Ed. infiltration tables and worksheets, ASHRAE 62.2 summary reference sheet, scientific calculators, printed worksheet packet, blower-door test result sample reports (instructor-provided)
Check: Graded infiltration worksheet for the sample house; pass = infiltration BTU/h values within 10% of instructor key for all three construction quality levels.

Practice check7

Answer each question, then check it. Nothing is recorded; this is your own practice.

HVAC Excellence – Heating and Cooling Load Calculation
1. In the Manual J simplified ACH infiltration method, what does 'ACH' stand for and what does it measure?
NATE – Air Distribution/Building Science
2. A technician receives a blower-door report showing CFM50 = 2,000. Using a divisor of 20, what is the estimated natural infiltration rate?
HVAC Excellence – Residential Load Calculation
3. Which Manual J construction quality category would best describe a newer home built to current energy codes with foam air-sealing, double-pane low-e windows, and continuous exterior insulation?
NATE – Air Distribution/Indoor Air Quality
4. ASHRAE Standard 62.2 applies to which type of building and addresses which requirement?
NCCER HVAC – Load Calculations
5. When adding infiltration and mechanical ventilation loads to a Manual J cooling calculation, how should they be treated?
HVAC Excellence – Cooling Load Calculation
6. A room has 500 watts of lighting. Using the conversion factor 1 watt = 3.412 BTU/h, what is the sensible cooling gain from those lights?
NATE – Safety/Indoor Air Quality
7. A technician is performing a blower-door test and notices the home has an older gas water heater in the utility room with no CO detector present. What is the correct immediate action?
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