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Needle valve in an isothermal liquid system

**Library:**Simscape / Fluids / Isothermal Liquid / Valves & Orifices / Flow Control Valves

The Needle Valve (IL) block models flow reductions by a needle valve. The valve
comprises a conical needle and a round, sharp-edged seat. The needle opens or closes
according to the displacement signal at port **S**. A positive signal
retracts the needle and opens the valve.

**Needle Valve Schematic**

**Needle Valve Top View**

The opening area of the valve is calculated as:

$${A}_{open}=\pi h\mathrm{sin}\left(\frac{\theta}{2}\right)\left[{d}_{0}-\frac{h}{2}\mathrm{sin}\left(\theta \right)\right]+{A}_{leak},$$

where:

*h*is the vertical distance between the outer edge of the needle and the seat, as indicated in the schematic above.*θ*is the seat cone angle, which always matches the**Needle cone angle**.*d*_{0}is the**Seat orifice diameter**.*A*_{leak}is the**Leakage area**.

The opening area is bounded by the maximum displacement
*h*_{max}:

$${h}_{\mathrm{max}}=\frac{{d}_{0}\left[1-\sqrt{1-\mathrm{cos}\left(\frac{\theta}{2}\right)}\right]}{\mathrm{sin}\left(\theta \right)}.$$

For any needle displacement larger than
*h*_{max},
*A*_{open} is the sum of the maximum
orifice area and the **Leakage area**:

$${A}_{open}=\frac{\pi}{4}{d}_{0}^{2}+{A}_{leak}.$$

For any combination of the signal at port **S**
and the needle offset that is less than 0, the minimum valve area is the
**Leakage area**.

At the extremes of the orifice opening range, you can maintain numerical robustness in your
simulation by adjusting the block **Smoothing factor**. When the
smoothing factor is nonzero, a smoothing function is applied to every calculated
displacement, but primarily influences the simulation at the extremes of this
range.

The normalized orifice opening is:

$$\widehat{h}=\frac{h}{{h}_{\mathrm{max}}}.$$

The **Smoothing factor**, *s*, is applied to the
normalized opening:

$${\widehat{h}}_{smoothed}=\frac{1}{2}+\frac{1}{2}\sqrt{{\widehat{h}}_{}^{2}+{\left(\frac{s}{4}\right)}^{2}}-\frac{1}{2}\sqrt{{\left(\widehat{h}-1\right)}^{2}+{\left(\frac{s}{4}\right)}^{2}}.$$

The smoothed opening is:

$${h}_{smoothed}={\widehat{h}}_{smoothed}{h}_{\mathrm{max}}.$$

Mass is conserved through the valve:

$${\dot{m}}_{A}+{\dot{m}}_{B}=0.$$

The mass flow rate through the valve is calculated as:

$$\dot{m}=\frac{{C}_{d}{A}_{valve}\sqrt{2\overline{\rho}}}{\sqrt{P{R}_{loss}\left(1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\right)}}\frac{\Delta p}{{\left[\Delta {p}^{2}+\Delta {p}_{crit}^{2}\right]}^{1/4}},$$

where:

*C*_{d}is the**Discharge coefficient**.*A*_{valve}is the instantaneous valve open area.*A*_{port}is the**Cross-sectional area at ports A and B**.$$\overline{\rho}$$ is the average fluid density.

*Δp*is the valve pressure difference*p*_{A}–*p*_{B}.

The critical pressure difference,
*Δp*_{crit}, is the pressure differential
associated with the **Critical Reynolds number**,
*Re*_{crit}, the flow regime transition
point between laminar and turbulent flow:

$$\Delta {p}_{crit}=\frac{\pi \overline{\rho}}{8{A}_{valve}}{\left(\frac{\nu {\mathrm{Re}}_{crit}}{{C}_{d}}\right)}^{2}.$$

*Pressure loss* describes the reduction of pressure in the
valve due to a decrease in area. *PR*_{loss} is
calculated as:

$$P{R}_{loss}=\frac{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}-{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}{\sqrt{1-{\left(\frac{{A}_{valve}}{{A}_{port}}\right)}^{2}\left(1-{C}_{d}^{2}\right)}+{C}_{d}\frac{{A}_{valve}}{{A}_{port}}}.$$

*Pressure recovery* describes the positive pressure change in
the valve due to an increase in area. If you do not wish to capture this increase in
pressure, set **Pressure recovery** to
`Off`

. In this case,
*PR*_{loss} is 1.