Why DP flow needs square root extraction
A DP flow element creates a pressure drop proportional to the square of the flow rate. Double the flow and the DP goes up four times. So the flow is proportional to the square root of the DP:
DP % = (flow % ÷ 100)² × 100
The square root can be taken in the transmitter (its transfer function set to square root), in the DCS or PLC, or in a separate square root extractor. It must be done in exactly one place. Extracting twice, or not at all, is a classic commissioning error.
Worked example
An orifice meter is sized for 25 kPa DP at a maximum flow of 100 m³/h. The transmitter reads 6.25 kPa. What is the flow?
- DP % = 6.25 ÷ 25 × 100 = 25%
- Flow % = √0.25 × 100 = 50%
- Flow = 0.50 × 100 = 50 m³/h
Half the flow gives only a quarter of the DP. A linear DP transmitter would output 8 mA, but a square-rooted output would be 12 mA.
Quick reference
| Flow % of max | DP % of span |
|---|---|
| 100% | 100% |
| 90% | 81% |
| 75% | 56.25% |
| 50% | 25% |
| 25% | 6.25% |
| 10% | 1% |
Low-flow cutoff and turndown
Near zero the square root curve is extremely steep: 1% of DP span already reads as 10% flow. Small zero shifts or noise at no-flow therefore show up as a large, false flow reading. That is why square-rooted flow signals use a low-flow cutoff that forces the output to zero below a set point, and why a single DP flowmeter is usually limited to about a 3:1 to 4:1 flow turndown before accuracy suffers.
Converting the DP signal itself? The 4–20 mA scaling calculator handles the linear part.