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Gate valve in an isothermal liquid system

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

The Gate Valve (IL) block models flow control by a gate valve in an isothermal liquid
network. The valve comprises a round, sharp-edged orifice and a round gate with the same
diameter. The gate opens or closes according to the displacement signal at port
**S**. A positive signal retracts the gate and opens the valve.

**Gate Valve Opening Schematic**

The area open to flow as the gate retracts is:

$${A}_{open}=\frac{\pi {d}_{0}^{2}}{4}-{A}_{shielded},$$

where *d*_{0} is the
**Valve orifice diameter**. The area shielded by the gate in a
partially open valve is:

$${A}_{shielded}=\frac{{d}_{0}^{2}}{2}{\mathrm{cos}}^{-1}\left(\frac{\Delta l}{{d}_{0}}\right)-\frac{\Delta l}{2}\sqrt{{d}_{0}^{2}-{\left(\Delta l\right)}^{2}},$$

where *Δl* is the gate displacement, which is
the sum of the signal at port **S** and the **Gate position
when fully covering orifice**.

If displacement exceeds the **Orifice diameter**, the valve area
*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 gate offset that is less than 0, the minimum valve area is the
**Leakage area**.

At the extremes of the gate displacement 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 gate displacement is:

$$\Delta \widehat{l}=\frac{\Delta l}{\Delta {l}_{\mathrm{max}}},$$

where *Δl _{max}* is the

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

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

The smoothed displacement is:

$$\Delta {l}_{smoothed}=\Delta {\widehat{l}}_{smoothed}\Delta {l}_{\mathrm{max}}.$$

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 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}, which is 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.