Control Valve Sizing Formula: Liquid Cv/Kv Input Checks
Control Valve Guide
For a liquid in turbulent, non-choked flow, the basic control valve sizing formula is Kv = Q × √(SG/ΔP) when Q is in m³/h and ΔP is in bar. In US units, Cv = Q × √(SG/ΔP) uses US gpm and psi. Before using either equation, pair the flow with its actual valve pressure drop, check the pressure reference, and confirm that the simplified equation applies.
This guide explains those checks with hypothetical inputs. It estimates the required flow coefficient; it does not select a valve size, trim or actuator.
1. Match the formula to the units and service
| Result | Flow Q | Valve pressure drop ΔP | Basic equation |
|---|---|---|---|
| Kv | m³/h | bar | Kv = Q × √(SG/ΔP) |
| Cv | US gpm | psi | Cv = Q × √(SG/ΔP) |
| Cv from metric inputs | m³/h | bar | Cv = (Q/0.865) × √(SG/ΔP) |
SG is the liquid's specific gravity at the flowing condition. For consistency, the examples below use Cv = Kv/0.865, approximately 1.156 × Kv. Do not put m³/h and bar into the US-unit equation and label the result Cv.
These simplified forms assume the piping geometry factor is 1 and no non-turbulent correction is needed. The Masoneilan handbook gives the unit constants and the fuller liquid equations. Reducers or other attached fittings, significant viscosity effects and vaporization require the appropriate additional checks. Gas, steam and two-phase service need different methods. Masoneilan sizing handbook, pages 3–6 and 11–15
2. Keep each flow and pressure pair together
Use pressures at the valve inlet and outlet for the same operating condition. Do not substitute the total system pressure loss for the valve's pressure drop. Spirax Sarco's water-system example shows how the available pressure loss is divided between the valve and the rest of the circuit. Water-system sizing, Example 6.3.3
Consider three hypothetical operating states. Assume SG = 1 and that the simplified equation has been confirmed applicable in every row. The pressures represent separate operating states, not a prediction of one fixed pump curve.
| State | Q (m³/h) | Inlet / outlet (bar gauge) | ΔP (bar) | Required Kv | Required Cv |
|---|---|---|---|---|---|
| Low flow | 6 | 4 / 3 | 1 | 6 | 6.94 |
| Normal flow | 10 | 2 / 1.75 | 0.25 | 20 | 23.12 |
| High flow | 15 | 4 / 3 | 1 | 15 | 17.34 |
At normal flow, Kv = 10 × √(1/0.25) = 20. That is larger than the high-flow requirement of 15 because less pressure drop is available across the valve.
Combining the high flow of 15 m³/h with the normal-state pressure drop of 0.25 bar would give Kv = 30. That combination is a fourth operating condition; use it only if it can actually occur. Keep the real paired conditions in the calculation, including any credible start-up or upset cases.
The largest required coefficient is still not a complete valve selection. Low-flow control, valve travel and the selected trim's characteristics must also be checked. Steriflow's sizing guidance
3. Use gauge pressure for a difference, absolute pressure for the choking check
If both pressures use the same gauge reference, subtracting them gives the same ΔP as subtracting their absolute values:
ΔP = P1,gauge − P2,gauge = P1,absolute − P2,absolute.
That cancellation does not apply when inlet pressure is compared with vapor pressure. For the liquid choking check without attached fittings, the referenced method uses:
ΔP limit = FL² × (P1,absolute − FF × Pv)
FF = 0.96 − 0.28 × √(Pv/Pc)
Pv and Pc are absolute vapor and thermodynamic critical pressures, in the same pressure units as P1. FL is the liquid pressure recovery factor for the valve and relevant opening; obtain the applicable value from its manufacturer. With attached fittings, use the appropriate combined factors instead. Emerson's standardized sizing section, printed page 638 / PDF page 13
A pressure-reference check you can reproduce
Use hypothetical inputs of P1 = 1 bar gauge, P2 = 0.1 bar gauge, FL = 0.8, Pv = 0.032 bar absolute and Pc = 221 bar absolute. Assume local atmospheric pressure is 1.01325 bar for this example; use the applicable reference at the actual site.
- ΔP = 1 − 0.1 = 0.9 bar.
- P1,absolute = 1 + 1.01325 = 2.01325 bar absolute.
- FF = 0.96 − 0.28 × √(0.032/221) ≈ 0.95663.
- ΔP limit = 0.8² × (2.01325 − 0.95663 × 0.032) ≈ 1.26889 bar.
Here, 0.9 bar is below that choking limit. Substituting the gauge value 1 for P1,absolute would instead give about 0.62041 bar, incorrectly placing the same pressure drop above the limit. The subtraction for ΔP was correct; the pressure basis in the second formula was wrong.
Being below the choking limit does not establish that cavitation is absent or that the valve is suitable. Noise, cavitation, piping effects, flow regime and the selected valve's data still need evaluation.
4. Review the inputs before accepting a result
Keep a separate calculation for each operating state. Show the units beside each input, preserve gauge-versus-absolute labels, and check that ΔP and the fluid-property inputs are physically valid. If a spreadsheet or calculator accepts an entry, that alone does not establish that its unit or reference is correct.
Use the resulting Cv or Kv as an input to the manufacturer's complete sizing procedure. A flow coefficient is a capacity requirement, not a pipe diameter or a finished valve specification.