Air watts and pascals measure different quantities. Use airflow at the same operating point to estimate pressure difference, follow a 120 AW and 20 L/s example, and see why a sealed inlet or different flow changes the answer.
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The calculation
At a stated operating point, SI air power P≈Δp×Q, where Δp is pressure difference in Pa and Q is volume flow in m³/s; thus Δp≈P/Q. One conventional ASTM air watt is about 0.9983 SI watt, so rounded examples using 1 AW≈1 W are approximate, not a one-to-one unit conversion.
Worked example
Using the page’s rounded SI air-power convention, 120 AW≈120 W and 20 L/s=0.020 m³/s, so Δp≈120/0.020=6,000 Pa=6 kPa. Reverse: 6,000×0.020=120 W. At 30 L/s=0.030 m³/s, equal air power implies Δp≈4,000 Pa. The ASTM air-watt convention is approximately 0.9983 SI W; these kilopascal examples are rounded.
Air power is the pressure-flow product at a matching operating point, not electrical input power. An air-watt figure alone cannot determine pressure. A sealed suction condition can show high pressure with Q≈0 and air power≈0. The ASTM lab rating may not equal power at a working nozzle; hose, tool and surface conditions alter flow.
Video chapters
- 0:00 — Air power is not pressure
- 0:15 — Pressure and flow work together
- 0:26 — Build the pressure-flow rule
- 0:39 — Match the airflow units
- 0:55 — Calculate the pressure
- 1:08 — Check the result backward
- 1:20 — Same power, different pressure
- 1:37 — The sealed-inlet trap
- 1:54 — Use the rating carefully
Read the full transcript
Can one hundred twenty air watts be converted directly to pascals? No. Air watts describe power carried by moving air; pascals describe pressure difference. We also need how much air moves each second.
Picture a vacuum fan pulling air through a tube. The pressure difference drives the flow; the flow rate counts cubic meters passing each second. Neither pressure nor flow alone tells the air power.
At one operating point, air power is approximately pressure difference times volume flow rate. Pascals times cubic meters per second becomes watts. To find pressure, divide air power by the flow rate.
Suppose air power is one hundred twenty air watts, approximately one hundred twenty SI watts. Airflow at that same point is twenty liters per second. A thousand liters make one cubic meter, so the flow becomes zero point zero two cubic meters per second.
Divide one hundred twenty by zero point zero two. The pressure difference is approximately six thousand pascals, or six kilopascals. Watch the pressure gauge rise while the flow meter stays fixed.
Check backward: six thousand pascals times zero point zero two cubic meters per second returns one hundred twenty watts of air power. The units and the arithmetic both agree.
Hold power near one hundred twenty air watts, but increase the assumed flow to thirty liters per second. That is zero point zero three cubic meters per second. Pressure is now about four thousand pascals. Power alone did not choose between the two pressures.
At a sealed opening, airflow can fall toward zero even while vacuum pressure is high. With zero flow, pressure times flow gives zero air power, and dividing by zero is impossible. Pressure and flow must describe the same operating point.
Do not confuse this air power with electrical watts drawn by the motor. Losses and the nozzle change what reaches the surface. To estimate pascals, convert the matching flow to cubic meters per second, then divide air power by that flow.
Sources
- Calculator Academy — Air watts to Pa calculator
- EPA ENERGY STAR — Vacuum scoping report (PDF)
- ASTM F558-24 — Vacuum performance test method
- MIT — Fluid power and flow notes (PDF)
- NIST — SI units and prefixes