The square root law
When fluid accelerates through a restriction — an orifice plate, venturi or flow nozzle — the differential pressure it creates grows with the square of the flow rate. Double the flow and the DP quadruples. So the DP transmitter's signal is linear with DP but not with flow, and recovering flow requires the inverse operation:
and back:
In current terms, for a linear DP transmitter:
Worked example
A DP flow transmitter (no internal extraction) reads 8 mA. What is the flow?
- DP percentage: (8 − 4) ÷ 16 = 25% DP
- Apply the square root: √0.25 = 0.5 → 50% flow
- If a √-extracting transmitter measured the same process, it would output 4 + 0.5 × 16 = 12 mA
This is the trap: 8 mA means 25% on a level loop but 50% on this flow loop. Knowing whether extraction is configured — and where — is not optional.
Quick reference table
| DP % | DP mA (linear) | Flow % | Flow mA (√) |
|---|---|---|---|
| 0 | 4.00 | 0 | 4.00 |
| 1 | 4.16 | 10.0 | 5.60 |
| 6.25 | 5.00 | 25.0 | 8.00 |
| 25 | 8.00 | 50.0 | 12.00 |
| 50 | 12.00 | 70.7 | 15.31 |
| 75 | 16.00 | 86.6 | 17.86 |
| 100 | 20.00 | 100 | 20.00 |
Look at the first rows: between 0 and 1% DP the flow output swings a full 10%. That extreme sensitivity near zero is why every real installation uses a low-flow cutoff — below a small DP threshold the flow output is forced to zero rather than chasing noise. It is also why DP flow measurements have limited turndown, typically around 3:1 to 5:1 for a single-range transmitter.
Field notes
- Never extract twice. If the transmitter takes the square root, the DCS block must be linear — double extraction reads badly low across the range (except at 0 and 100%).
- Never extract zero times either. A forgotten extraction reads high in the low range: at 50% actual flow the indicated value is only 25%.
- Verify with the 8/12 test: inject 50% DP; a linear output shows 12 mA, an extracting one 15.31 mA. Instantly tells you what the transmitter is doing.
Frequently asked questions
Why does DP flow measurement need square root extraction?
Across an orifice or other DP element, differential pressure is proportional to the square of the flow. To recover flow from the DP signal you take the square root: 25% DP is 50% flow, 50% DP is about 70.7% flow.
What is 12 mA on a DP flow transmitter?
If the transmitter output is linear with DP, 12 mA is 50% DP which equals 70.7% flow. If the transmitter itself performs square root extraction, 12 mA means 50% flow directly.
Where should square root extraction be done — transmitter or DCS?
In exactly one place, never both. Modern practice usually configures it in the transmitter so the loop signal is linear with flow, but either location works if the other stays linear. Double extraction is a classic commissioning error.
What is low-flow cutoff?
Near zero flow the square root function amplifies noise enormously (the slope approaches infinity), so a cutoff forces the flow output to zero below a threshold, typically around 5-10% of DP.
Provided for reference and education. Verify independently before use in safety-critical work. See our disclaimer.