Work needed to push air through a fabric filter I am trying to compute the work needed to push air through a fabric filter. The only variable I know is the Pressure drop across the filter.
From what I gathered (mainly through online searches), the work is calculated by multiplying P and V, but I'm not sure that multiplying the pressure drop (known variable) and the volume I'd like to push through the filter makes physical sense (though a dimensional analysis does, in fact, indicate that what I would find is Work).
I also found that I can calculate ΔP by multiplying the airflow by the "resistance" of the filter, which I'd calculate using a standard flow rate (maybe this resistance is flow dependent?), dividing the pressure drop by the standard flow rate. In this way I'd find: ΔP=Q(flow)R(resistance)
and then: W=ΔPV=QRV
The problem I encounter with this second formula is that the work is dependent of flow, and thus on time, which, again, I'm not sure makes good physical sense. The second problem I have with this second derivation is that in order to calculate the Resistance I have to use a "standard flow rate", implying that the resistance is independent of flow, which, again, doesn't have me convinced...
I hope that at least some of what I wrote makes sense, if not, please let me know.
 A: For fans moving air at low velocities, the compressibility of the air is negligible. we'll leave it out of this analysis.
the power required to move air is the source pressure (which is the delta p across the filter) times the mass flow rate (m dot, not volume) of the air through it. Be sure to use consistent units when doing the calculation. In engineering units, you'll get a number with units (feet x pounds)/(seconds). Divide this by 550 to get the horsepower of the fan motor required to perform this work, for a 100% efficient fan. Multiply horsepower by 746 to get the wattage rating of the electric motor required to turn the fan, for a 100% efficient motor.
(Note added in edit: I failed to explain that the "pounds" in pressure are in pounds force (written lbf) and the pounds in mass are pounds mass (written lbm), which differ by the constant G which relates mass to force. This is like the distinction between kilograms and newtons in metric units, but in this case they gave them different names out of kindness. This awkward situation is all the fault of the engineers in our society who insist on using what are called "mixed units", which the physicists in our society call "awkward". In fact, learning to work in cgs units is one step in the 12-step program for recovering ex-engineers like me. Sadly, that program is notoriously ineffective.)
A: take the first law of thermodynamics .
u=q+pv
h=u-pv
enthalpy.
assume it is an adiabatic process.
Δh=Δpv.
this change in enthalpy is the total work done on the gas.
