| FILTER
EFFICIENCY TESTING AND AIRFLOW MEASUREMENT IN RELATION
TO NEGATIVE PRESSURE UNITS.
ARTICLE
PROVIDED BY STEVE WILLIAMS, ENVIROVAC SALES AND SERVICE
LTD.
It
is generally accepted that there is a requirement for
negative pressure units and type H vacuum cleaners to
be serviced and tested every six months. As part of
this process the efficiency of the filtration should
be measured and airflow will be determined in order
to confirm the suitability of the equipment for use
with asbestos.
It
is my intention to explain in a little detail the nature
of these measurements and how they are determined.
Filter
efficiency measurements
Airflow
measurements
To measure air flow rates through a
negative pressure unit the most accurate method is to
use a pitot tube in a duct measuring the velocity pressure
in order to determine the air velocity. To determine
this there are necessary requirements to be met. These
would include long lengths of duct, air straightening
devices to produce laminar airflow, and a traverse of
the duct according to specific dimensions. In addition
the calculation would involve temperature, barometric
pressure and humidity and given all these factors the
best to be hoped for would be an accuracy of plus or
minus 2%.
Given
the situation above, air velocity measurements are made
normally using either a rotating vane or hot wire anemometer
which are obviously far less accurate.
The
method of use is varied because some anemometers will
give a single reading whilst others will average the
velocity across the duct. Generally speaking a number
of readings are taken which are then averaged out to
give a single figure. The readings may be in metres
per second or feet per minute, but are velocity not
volume readings which is where the problem lies.
In
order to convert velocity readings to volume flow readings
a multiplication factor is used which is based on the
area of the duct or inlet within which the readings
are taken.
For
example a 16” diameter inlet would have a cross
sectional area of 201 square inches or 1.396 square
feet. For example, we have a reading of 2000 linear
feet per minute from our anemometer, the volume flow
through the machine would be 2000 x 1.396 which is 2792
cubic feet per minute.
Erroneous
calculations can produce startling airflow measurements,
if for example we take a machine with two stage filtration
and an open front housing a pre filter measuring 24”
x 24” (a size commonly used.) the factor used
to calculate the volume flow would be 24” x 24”,
which is 576 square inches or 4 square feet.
If
we measure the open area of this filter what we actually
find is that the filter itself is slightly smaller than
24”x24” and that there is a cardboard frame
around the edge and supporting struts which further
reduces the open area. If we allow for this the actual
open area is equivalent to 22 x 22” which is 484
square inches or 3.36 square feet
If
we then take our example, the reading from the anemometer
which we assume is reasonably accurate and for the sake
of argument is 1000 linear feet per minute, then by
using the factor 4 we have a machine which is claimed
to have a volume flow of 4000 cubic feet per minute.
Using
the correct factor of 3.36 the machine now has the much
more realistic 3360 cubic feet per minute, a discrepancy
of 640 cubic feet per minute.
Traditionally
enclosures have been measured to calculate volume in
cubic feet, hence most negative pressure units manufactured
have been rated in cubic feet per minute. For machines
rated in cubic metres per hour it is necessary to apply
a conversion factor to determine cubic feet per minute.
The factor is 1.7 for example, 5000 metres cubed per
hour is 2940 cubic feet per minute ( 5000 / 1.7 = 2940
)
Finally
it is important to bear in mind that manufacturers figures
are given with machines fitted with clean filters and
after use dust will accumulate inside the machine reducing
rated air flows, hence the requirement for servicing
and testing every six months.
DETERMINING THE CORRECT MACHINE FOR
THE ENCLOSURE SIZE
CALCULATE
THE VOLUME OF THE ENCLOSURE IN CUBIC METRES BY MULTIPLYING
THE LENGTH HEIGHT AND WIDTH IN METRES
FOR
SOME ENCLOSURES YOU MAY HAVE TO MAKE AN APPROXIMATION
DUE TO THE COMPLEXITY OF THE STRUCTURE. THIS CAN BE
DONE BY MEASURING SECTIONS OF THE ENCLOSURE CALCULATING
VOLUMES AND ADDING THEM TOGETHER TO GIVE A TOTAL.
A
SIMPLE EXAMPLE 2.5 METRES HIGH X 5 METRES WIDE X 5 METRES
LONG
VOLUME
OF ENCLOSURE IS 2.5 X 5 X 5 = 62.5 CUBIC METRES
DETERMINE
THE REQUIRED NUMBER OF AIR CHANGES PER HOUR, NORMALLY
A MIMIMUM OF 8
THE
VOLUME FLOW OF THE EXTRACTOR REQUIRED WOULD BE 8 X 62.5
OR 500 M3/HOUR
THIS
IS PURELY A MATHEMATICAL CALCULATION AND AN ALLOWANCE
MUST BE MADE FOR ANY AIRLOCKS FITTED.
IF
THE ENCLOSURE IS NOT CONSTRUCTED CORRECTLY A NEGATIVE
PRESSURE DIFFERENTIAL BETWEEN THE INSIDE AND THE OUTSIDE
OF THE ENCLOSURE MAY NOT BE ACHIEVED
TO
CONVERT M3/HOUR TO CUBIC FEET PER MINUTE DIVIDE BY 1.7
TO
CONVERT CUBIC FEET PER MINUTE TO M3/HOUR MULTIPLY BY
1.7
TO
CONVERT CUBIC METRES TO CUBIC FEET MULTIPLY BY 35.31
TO
CONVERT CUBIC FEET TO CUBIC METRES DIVIDE BY 35.31
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