Hello Faiz,

I am unfamiliar with the notion of an 'air watt'. But wikipedia tells me that it is the mechanical power developed by the motor (not the electrical power).

Regrettably that alone is not sufficient to predict the pressure differential that such a motor might create. One can imagine a terribly wide throat, a large flux of air, and for a given 'air wattage' you would have very small pressure drop.

Equally, if one were to constrict that throat, the mass flow of air would be reduced, and the pressure drop would be large.

I think that we need to know the pumping rate (kg/s) of the motor.

The conversion of vacuum motor capacity from air watt to kilopascals requires a couple of steps.

First, we need to convert air watt to air flow rate in cubic meters per hour (m^3/hr). This can be done by using the formula:

Air flow rate (m^3/hr) = Air watt / (0.5 * air density * (speed of sound)^3)

Where air density is approximately 1.2 kg/m^3 and the speed of sound is approximately 340 m/s.

Next, we can convert the air flow rate to kilopascals by using the formula:

Pressure (kPa) = (air flow rate (m^3/hr) / volume flow rate (m^3/hr)) * atmospheric pressure (kPa)

Where atmospheric pressure is approximately 101.3 kPa.

By using these formulas, we can convert the vacuum motor capacity of 350 air watt to kilopascals.

https://www.ijrdt.org/upload/6465539-DESIGN%20OF%20MULTI%20PURPOSE%20CUTTING%20MACHINE%20USING%20VACUUM%20PROCESS.pdf