Using a simple cabinet as an example
where a velocity of 0.4m/sec is required through an aperture.
Determine required
flow rate:
Flow = Velocity (m/sec) X Area (m²)
= m³/sec
Flow = 0.4 X (0.3X0.5) = 0.06 m³/sec
For m³/hr x3600 = 216 m³/hr
Determine pressure
In this simple system there will be a
filter which gives rise to two areas of restriction.
Pressure required to pass the flow through
the aperture
Pressure required to overcome the filter
resistance
With a velocity of 0.4m/sec across such a large
area there should be very little pressure required to produce this
flow. However for future reference the following formula can be
used.
Pv = [ Flow/(Area X 47,496)]² Where Flow=m³/hr
Area= m²
Pv = [ 216/(0.15 X 47,496)]² = 0.001mBar
The filter will contribute the largest restriction
to flow and this information is normally provided by the filter
manufacturer when given a desired flow rate.
Flow(m³/hr)
Clean
rp (mBar)
Dirty rp
(m Bar)
220
2
4
Ducting losses
All
ducting, bends louvres etc. will have an effect on the air passing
through that system. This will nearly always take the form of a
restriction to the flow and consequent pressure requirement.
At the earliest stage in the design process the ducting should be
carefully evaluated and made as efficient as possible. This will
reflect beneficially on the final design in the overall cost, the
compactness of the fan and the running costs.
Points to consider:
The ducting should be as large a cross section
as practicable
The total length should be kept as short
as possible
The number of bends kept to a minimum and
should have as large a bend radius as possible
Sudden expansion or contractions should be
avoided. Smooth blended taper pieces should be used to make transitional
sections.
If air is ducted outside in a total loss system
consideration should be given as to how the air will be replenished.
Ducting manufacturers should be able to provide
standard information as to the pressure losses in both straight
sections and bends. At the back of the notes there is an old imperial
table showing losses in various diameters of ducting per 100Ft run
in InWg. The loss is proportional to the length. Hence if you were
trying to pass the 216 m³/hr (130 cfm) down a 200 Ft run of
Ø4” duct a pressure drop of around 2 InWg (5mBar) would
occur.
Dwell time (Carbon filters)
For Carbon Filters
to work effectively the air passing through the filter needs to
be evenly distributed across the filter and be exposed to the carbon
for a specified period of time. This is commonly known as the dwell
time. Parameters:
Flow
(m³/sec)
Filter Depth
(m)
Dwell Time
(sec)
Filter Area
(m²)
Filter Area (m²) = Flow X Dwell Time ÷
Filter Depth
Flow (m³/sec) = Filter Area X Filter Depth
÷ Time
Dwell Time (sec) = Filter Depth X Filter Area
÷ Flow
