Retrotec DucTester 341 - Manual (Page 70)

Page 70 of 83
©Retrotec Inc. 2014
Appendix C: 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.