Infiltration and Building Pressurization Calculator (Crack Method)

Units

The ordinary winter wind, typically 10 to 20 mph (4.5 to 8.9 m/s). For your location, the ASHRAE Climatic Design Conditions give the average wind speed for your location’s coldest month.

Find wind speed on ASHRAE's site
  1. Open ASHRAE Climatic Design Conditions.
  2. Select the edition year you want, usually the newest, or the one your energy code calls for.
  3. Click IP for mph or SI for m/s, to match the units chosen above.
  4. Right click your location on the map, or use the magnifying glass to search for it.
  5. Select the nearest weather station from the list.
  6. Scroll down to the Annual Heating, Humidification, and Ventilation Design Conditions table and note the Coldest Month, a number from 1 (January) to 12 (December).
  7. Scroll down to Monthly Climatic Design Conditions and read WSAvg, in the Wind row, under that month. That is the average wind speed for the coldest month. Enter it here.

Windows

Enter the windows as totals, or one row per size. Under Totals, crack length is the sum of every window's perimeter, 2 × (width + height) each; it sets the leakage, so total area alone is not enough. The area only takes the windows off the wall area in the pressurization check. Under Size of each, enter how many windows there are of a size, and the width and height of one. Enter the whole window, not the glass panes: double-pane glass adds no crack. For double-hung and sliding windows, add the meeting rail to the crack length (use Totals).

Doors

Exterior doors, one row per size: their cracks count with the windows', and the opening force check below checks every size. For a pair of doors, count each leaf and enter one leaf's width; a person pushes one leaf.

Building (optional)

Optional. The building's dimensions give the air changes per hour, and the wall and roof areas for the pressurization check.

Pressurization and door opening force (optional)

Is the outdoor air over-pressurizing the building? Enter the outdoor air supplied minus the air exhausted (negative if more is exhausted). The calculator finds the pressure at which that surplus leaks out through the cracks above, and checks it against the force needed to open the doors above.

Air also leaks through the walls and roof. Their areas come from the building dimensions above (the walls less the windows and doors). Rates are per unit of surface at 0.30 in. w.c. (75 Pa).

Optional. Penetrations that stay open, such as relief dampers, louvers and unsealed gaps, as free area.

Optional. The envelope leakage from a blower-door test of this building. It replaces the wall and roof estimate.

Knob to edge is the distance from the knob to the latch edge of the door. Closer force is the door closer manufacturer's published value, and is part of the total opening force; 0 for a door on plain hinges. Force limit is the total opening force allowed. The doors above are exterior doors, so the limit starts at 30 lbf (133 N), the force to set the door in motion (IBC 1010.1.3, NFPA 101 7.2.1.4.5). The same sections also allow 15 lbf (67 N) to release the latch and 15 lbf to swing the door fully open; the calculator checks the 30 lbf. Use 5 lbf (22 N) for an interior hinged door, such as one between a pressurized zone and a corridor (IBC 1010.1.3 for egress doors other than fire doors, ADA 404.2.9, ICC A117.1 404.2.8). The 5 lbf limit does not include the force to retract the latch.

How to use it
  1. Choose IP (mph, ft, cfm) or SI (m/s, m, l/s) units. Switching converts the values already entered.
  2. Enter the winter wind speed. Use the design wind speed for your location if you don’t have a site value; Find wind speed on ASHRAE’s site under the field shows where to read it.
  3. Choose how well the windows fit: tight, average or loose. The i beside the choice says which window types fall in each class.
  4. Under Totals, enter the total crack length (the sum of every window’s perimeter, plus the meeting rails of double-hung and sliding windows) and the total window area. Or choose Size of each and enter how many windows there are of each size and the overall size of one (the whole window, not its glass panes); use + Add another size window for more sizes (up to 12). Enter the exterior doors under Doors the same way, one row per size (+ Add another size door); for a pair of doors, count each leaf and enter one leaf’s width.
  5. Optionally, enter the building’s length, width and height to see the result as air changes per hour.
  6. To check whether the outdoor air over-pressurizes the building, enter the net outdoor air (outdoor air supplied minus air exhausted), how tight the walls and roof are, any openings that stay open, and the door to check. The first line of the results says whether the building is over-pressurized (or, with more exhaust than outdoor air, over-depressurized); below it are the building pressure, the force needed to open each door, and the most net outdoor air the doors allow.

The results update as you type. To see it filled in, choose Show an example: the building in the worked example below, with 500 cfm of net outdoor air. Use Copy link to these results to save or share the calculation.

Infiltration

Enter the winter wind speed.

Methodology, equation and example

The crack method estimates infiltration from the length of the gaps around operable windows and doors. The wind pressure on those gaps drives outdoor air in. The air leaking in per foot of crack depends on how well the openings fit and on the wind speed.

VHF = 0.000467 × V²

Infiltration rate (cfm per ft of crack):

Window fit k Rate
Tight 1 0.95 × VHF^0.55
Average 2 2.1 × VHF^0.64
Loose 6 6.05 × VHF^0.62

Crack length = Σ quantity × 2 × (width + height)

Q = infiltration rate × crack length

ACH = Q × 60 / (L × W × H)

Where:

  • V: winter wind speed, mph
  • VHF: velocity head factor, in. of water
  • Crack length: the perimeter of each opening (the whole window, not its glass panes), plus the meeting rail of a double-hung or sliding window, ft
  • Q: infiltration, cfm
  • L, W, H: building length, width and height, ft

Example: A 30 × 60 ft building, 16 ft high, has four 3 × 5 ft windows of average fit and one 3 × 7 ft door. The winter wind speed is 20 mph.

  • VHF = 0.000467 × 20² = 0.1868
  • Rate = 2.1 × 0.1868^0.64 = 0.718 cfm/ft
  • Crack length = 4 × 2 × (3 + 5) + 1 × 2 × (3 + 7) = 84 ft
  • Q = 0.718 × 84 = 60.3 cfm
  • ACH = 60.3 × 60 / (30 × 60 × 16) = 0.126 air changes per hour

Building pressurization and door opening force

A building that takes in more outdoor air than it exhausts is pressurized until the surplus leaks out as fast as it comes in. Too much pressure makes doors hard to open, and a door that opens outward is pushed shut. This part of the calculator finds the pressure the net outdoor air holds the building at, with no wind, and checks the force needed to open a door against the code limit.

Leakage. At a pressure difference ΔP (in. w.c.), the building leaks through:

  • Window and door cracks: crack length × the fit's rate above, with ΔP in place of VHF (the crack curves give leakage per foot against pressure difference).
  • Walls and roof: area × rate × (ΔP / 0.30)^0.65. Typical commercial wall leakage is 0.10, 0.30 and 0.60 cfm per ft² at 0.30 in. w.c. for tight, average and leaky walls (Tamura and Shaw 1976a, as reported in the ventilation and infiltration chapter of the ASHRAE Handbook—Fundamentals), with a flow exponent of 0.65. There is no separate roof figure, so the roof uses the same choices. The wall area is the perimeter × height less the windows and doors; the roof area is length × width.
  • Other openings that stay open, such as relief dampers and louvers: Q = 2610 × A × ΔP^0.5, with A the free area in ft² (the orifice equation with a flow coefficient of 0.65).
  • Measured leakage: a blower-door result in cfm at 75 Pa (0.30 in. w.c.) replaces the wall and roof estimate, scaled by (ΔP / 0.30)^0.65.

The building pressure is the ΔP at which the total leakage equals the net outdoor air.

Door opening force (NFPA 92 and the ASHRAE Handbook's smoke-control chapter):

F = Fdc + 5.2 × W × A × ΔP / (2 × (W − d))

  • F: force to open the door, lbf
  • Fdc: door closer force, lbf, from the manufacturer; 0 for a door on plain hinges
  • W: door width, ft; A: door area, ft²
  • ΔP: pressure difference across the door, in. w.c.
  • d: distance from the knob to the latch edge, ft

Force limits. The doors in the calculator are exterior doors, so the limit starts at 30 lbf. Exterior and other swinging doors must release the latch at no more than 15 lbf, set in motion at 30 lbf and swing fully open at 15 lbf (IBC Section 1010.1.3 and NFPA 101 Section 7.2.1.4.5). The pressure acts from the moment the door starts to move, so the calculator checks the 30 lbf set-in-motion force. Interior hinged doors are limited to 5 lbf, not counting the force to retract the latch: IBC 1010.1.3 for interior swinging egress doors other than fire doors, ADA Standards 404.2.9 and ICC A117.1 Section 404.2.8. Choose 5 lbf for an interior door that sees the pressure, such as one between a pressurized zone and a corridor. Check the code edition your jurisdiction adopts.

Example. A 3 × 7 ft door on plain hinges (Fdc = 0), knob 3 in. from the edge:

  • At the 30 lbf exterior limit: ΔP = 30 × 2 × (3 − 0.25) / (5.2 × 3 × 21) = 165 / 327.6 = 0.504 in. w.c.
  • At the 5 lbf interior limit: ΔP = 27.5 / 327.6 = 0.084 in. w.c.

This is the example that Show an example fills in: the building above, 30 × 60 × 16 ft, with average-fit windows and average walls and roof. It has 2,799 ft² of wall (after the windows and doors) and 1,800 ft² of roof. With 500 cfm of net outdoor air it settles at 0.058 in. w.c.: 472 cfm leaks through the walls and roof and 28 cfm through the cracks. The door then takes 3.4 lbf to open, within the 30 lbf exterior limit, and the door allows up to about 2,046 cfm of net outdoor air. Against the 5 lbf interior limit it would allow about 639 cfm. Change any input and the results move away from these figures.

If more air is exhausted than brought in, enter the net outdoor air as a negative number: the building is depressurized by the same amount, and the force check applies to a door that opens inward.

The force is for a door that opens against the pressure: outward from a pressurized building, inward into a depressurized one. A door that swings the other way is pushed open instead, which matters for doors that must stay closed or latched.

The pressure is for calm conditions. Wind raises the pressure across doors on the windward side and lowers it on the leeward side, and the stack effect adds to it in tall buildings in winter. The window crack curves were fitted for winds up to 36 mph, a pressure of about 0.6 in. w.c.; above that the calculator flags the crack leakage as extrapolated.

Window fit classes

  • Tight (k = 1): locked wood double-hung windows, weather-stripped, with an average gap (1/64-in. crack). Wood casement and awning windows, weather-stripped. Metal casement windows, weather-stripped.
  • Average (k = 2): locked wood double-hung windows that are not weather-stripped with an average gap (1/64-in. crack), or weather-stripped with a large gap (3/32-in. crack). All vertical and horizontal sliding windows, weather-stripped. Metal casement windows, not weather-stripped, with a large gap (3/32-in. crack).
  • Loose (k = 6): locked wood double-hung windows, not weather-stripped, with a large gap (3/32-in. crack). Vertical and horizontal sliding windows, not weather-stripped.

Limits

  • The correlations are fitted for wind speeds from 0 to about 36 mph. Above 36 mph the calculator still gives an answer, but flags it as possibly unstable.
  • The crack method's infiltration result counts only leakage around operable windows and doors. The pressurization check adds the walls, roof and other openings; a blower-door test of the building is the most accurate input where you have one. Neither part includes the stack effect, which matters in tall buildings.
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