Retrotec Residential Pressure & Air Leakage - Manual (Page 63)

Page 63 of 75
Appendix B: Calculate flow if required test pressure cannot be
reached
“n” setting for estimating flow @ pressure during house and duct leakage test
Houses and ducts have leaks through holes that will have both turbulent and laminar flow going through them.
The relationship between pressure and flow is as follows:
𝑓𝑙𝑜𝑤 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑛 × 𝐶
The actual flow exponent n for an enclosure can be calculated by measuring enclosure leakage at multiple
pressure differences, from 10 to 50 Pa, and determining the slope of the line when graphing log of flow versus
log of pressure. The graph of pressure versus flow will be linear if graphed on a log-log scale, and the slope will
be n. The constant C is a value depending on the flow characteristics of the opening through which the air is
moving, and can be thought of as the flow at 1 Pa. Once n is known, flow at 1 Pa can be found using the graph.
A wide open hole has an n of 0.5, meaning that when the pressure is quadrupled, the flow doubles. That is due
to completely turbulent flow going through that hole (flow = square root of pressure, a constant for that
particular hole).
𝑓𝑙𝑜𝑤 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑛
𝑓𝑙𝑜𝑤 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 0.5
An n value of 1.0 represents tiny little holes, so small that the air would not be turbulent but rather would go
through the holes as laminar flow. This means that when pressure is quadrupled, the flow will also be
quadrupled.
𝑓𝑙𝑜𝑤 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 1
𝑓𝑙𝑜𝑤 = 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
4 ∗ 𝑓𝑙𝑜𝑤 = 4 ∗ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒
Duct holes tend to be slightly larger so tend towards more laminar with less turbulent flow and larger n values,
whereas houses have more prevalent long tiny cracks, and therefore tend to have lower n values.
The n value is saved in the gauge. Set it to 0.65 for houses, and 0.60 for ductwork. Set to 0.5 for tests on the
Retrotec house simulator, while measuring air handler flow and for any large hole that is not composed of long
thin cracks.
The gauge uses the n value to extrapolate for flows at other pressures using the following formula:
𝐹𝑙𝑜𝑤 𝑎𝑡 𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑡𝑒𝑠𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = (𝐹𝑙𝑜𝑤 𝑎𝑡 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒) ∗ (
𝑑𝑒𝑠𝑖𝑟𝑒𝑑 𝑡𝑒𝑠𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
)
𝑛
Note that because the formula is using a ratio of the two pressures, and both pressures will have the same C, the
C value is not required for the extrapolation.
For example: If we guess at the n value of a duct as being 0.6 and measure 100 CFM at 20 Pa (by accident or by
design), then the gauge will complete the following calculation to estimate the flow at 25 Pa:
𝑓𝑙𝑜𝑤@25 𝑃𝑎 = 𝑓𝑙𝑜𝑤@20 𝑃𝑎 ×
25 0.6
20 0.6
If the test pressure (20 in this case) is close to the desired reference pressure (25 Pa in this case), then the
correction is small and the value of n does not play as large a role. However, if the test pressure is much higher
or lower than the reference pressure, the error can be greater.