Gas Control Valve: Worked Example

1. Calculator Inputs

Select units before entering values.

2. Step-by-step Math

Pressure-drop ratio and expansion

x = (100−80)/100 = 0.2
Fγ = k/1.4 = 1
xcritical = Fγ xT = 0.7
Y = 1−x/(3xcritical) ≈ 0.905

Because x is below the critical ratio, use x = 0.2 in the flow equation.

Calculate required Cv

T = 60+459.67 = 519.67 °R
Q = 1360 Fp Cv P1 Y √[x/(SG T Z)]
Cv = 10000 / [1360×1×100×0.904761905
     × √(0.2/(1×519.67×1))]
Cv ≈ 4.14

The 1360 coefficient requires scfh at the specified reference conditions, psia and degrees Rankine.

Convert capacity coefficient

Kv = Cv/1.156099 ≈ 3.58

Cv and Kv are different unit conventions for valve capacity.

3. Compare Your Answers

Keep full precision until the final result. Display rounding may differ; compare answers rounded to three significant figures; retain unrounded intermediate values.

Second check

Select the flow solve mode and enter Cv = 4.142624894, keeping the other inputs unchanged. The result returns to 10000 scfh.

4. Assumptions and Method

Single-phase gas using supplied SG, Z, k, Fp and xT. The example assumes Fp = 1 and a hypothetical xT; actual valve selection requires appropriate manufacturer data. Use absolute pressures, not gauge pressures. Standard flow is tied to 60°F and 14.7 psia.

Equation definitions and implementation: open calculator. See also calculation methodology.