Originally Posted by St-rider
Has anyone done this successfully?
You mean without noticeably shortening the service lifetime of your HVAC equipment?
If the force pushing the air, the 'pressure', through a filter at 90° to the airflow is F, then the force on this skewed filter is F[arc cos 21°] = 0.93F, other things being equal.
So the CFM through the filter will probably be less, which may affect the ΔT across your heat exchanger.
Try these guysEngineering ToolBox
to see if it will significantly affect your system.
I'm thinkin': "No", but, there is static and
Also, your local library may have Lindeburg's book on prepping for the PE exam in Mechanical Engineering. This book has at least one section on HVAC, and it may cover filter CFM vs. pressure.
I scrounged this \/ off the Internet. The author didn't believe in separate paragraphs so I added some.
"The average blower system in a residential furnace is capable of overcoming 0.50 inches of water column of static pressure in the entire airflow system while still being able to deliver its rated CFM output.
The airflow system includes all the return and supply ductwork, air grilles, dampers, cooling coil, and last but not least, the filter. The cooling coil alone uses up between 0.20 and 0.30 inches of water column of that 0.50. So we only have 0.20 to 0.30 remaining for everything else. The average duct system including the grilles uses up about 0.22.
So there is little or nothing left for the air filter. The pressure drop ratings of air filters vary from 0.07 (for an unpleated disposable filter) to 0.40 for some of the pleated models. As the velocity of the air increases, so does the pressure drop across the filter. The increase in pressure drop per increase in FPM of the velocity is usually exponential in most filters. Some of these exponential increases in pressure drop can exceed the power of 2.
So if the velocity doubles, the pressure drop quadruples. This would mean that a filter with a seemingly low pressure drop of 0.07 would have a pressure drop of 0.28 if we were to double the air velocity at which the filter was rated at for the 0.07 pressure drop. Yes, this happens all the time, because the typical rated velocity for the specs is done at a meager 300 FPM air velocity. Velocity = CFM divided by area in square feet. And when we halve the area, the velocity doubles.
So if our filter setup is only designed to accept a filter that is half the proper size for our CFM, the velocity is going to double. And in a filter with an exponent of 2, this means that the pressure drop is going to quadruple. So you’re really screwed if your filter setup is only half the proper size for the CFM moving through it.
If you’re going to use filters that are rated at 300 FPM velocity, you must use the formula: Filter size in square inches of cross sectional area = CFM divided by 2.08. This will give you the filter size required for that CFM at a velocity of 300 FPM."