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Single-acting or double-acting actuator controlled by a pilot pressure in an isothermal liquid network

**Library:**Simscape / Fluids / Valve Actuators & Forces

The Pilot Valve Actuator (IL) block models a single-acting or double-acting actuator controlled by a pilot pressure for control of a connected valve or orifice in an isothermal liquid network.

For the single-acting actuator, when the control pressure,
*P*_{X} –
*P*_{atm}, exceeds the **Spring preload
force at port X**, the piston begins to actuate in the direction assigned
by the **Mechanical orientation** parameter.

For the double-acting actuator, the control pressure
*P*_{control} is the difference between
*P*_{X} –
*P*_{atm} and
*P*_{Y} –
*P*_{atm}. The piston actuates in the direction
of the larger applied pressure differential, opposing the spring force at the opposite
port. When the piston motion reverses, this spring does not extend and does not exert a
counterbalancing force on the piston position.

The pistons at port **X** and port **Y** are
attached to a single spool. Both springs restore the spool to its neutral position when
*F*_{spool} falls below the opposing spring
preload force. For single-acting actuators, the neutral position is at port
**X**. For double-acting actuators, the neutral position is at the
center of the actuator.

The force on the piston is created by the pressure differential between port
**X** and atmospheric pressure:

$${F}_{spool}=\left({p}_{X}-{p}_{atm}\right){A}_{X},$$

where *A*_{X} is the
**Piston area at port X**. When
*F*_{spool} is greater than the
**Spring preload force at port X**, the piston begins to
move.

**Schematic of a Single-Acting Actuator**

The instantaneous piston displacement is calculated as:

$${\dot{x}}_{dyn}=\frac{{x}_{steady}-{x}_{dyn}}{\tau},$$

where the steady piston position,
*x*_{steady}, is the piston position at
the current pressure differential, proportional to the spring force at maximum
piston stroke:

$${x}_{steady}=F{x}_{stroke}\epsilon =\frac{{F}_{spool}-{F}_{pre}}{{F}_{\mathrm{max}}-{F}_{pre}}{x}_{stroke}\epsilon ,$$

where:

*F*_{pre}is the**Spring preload force at port X**.*F*_{max}is the maximum spring force acting against piston displacement, $${F}_{\mathrm{max}}=K{x}_{stroke}+{F}_{pre},$$ and where*K*is the**Spring stiffness at port X**.*x*_{stroke}is the**Piston stroke from port X**.*ε*is the**Mechanical orientation**, which indicates piston movement in a positive direction (extension) or negative direction (retraction).

If the force on the piston is less than the **Spring
preload force at X**, the piston remains at the neutral position or
moves to the neutral position. If the force on the piston meets or exceeds the
maximum spring force, the piston remains at the stroke until the applied
pressure changes.

The difference between the forces at ports **X** and
**Y** dictates the piston motion:

$${F}_{spool}=\left({p}_{X}-{p}_{atm}\right){A}_{X}-\left({p}_{Y}-{p}_{atm}\right){A}_{Y}.$$

The pressure applied at port **X** shifts the
spool away from the chamber at **X** and opposes the spring at port
**Y**. Similarly, the pressure applied at port
**Y** shifts the spool away from the chamber at
**Y** and opposes the spring at port **X**.
When the spool reverses direction, the formerly extended spring compresses, exerting
a force on the spool. The formerly compressed spring, the spring at the port in the
direction of motion, does not extend and does not influence the spool
position.

**Schematic of a Double-Acting Actuator**

The piston displacement is calculated as:

$${\dot{x}}_{dyn}=\frac{{x}_{steady}-{x}_{dyn}}{\tau},$$

where the steady piston position,
*x*_{steady}, is the piston position at
the current pressure differential, proportional to the spring force at maximum
piston stroke:

$${x}_{steady}=\left({F}_{Y}{x}_{stroke,Y}-{F}_{X}{x}_{stroke,X}\right)\epsilon =\left(\frac{{F}_{spool}-{F}_{pre,Y}}{{F}_{\mathrm{max},Y}-{F}_{pre,Y}}{x}_{stroke,Y}-\frac{\left(-{F}_{spool}-{F}_{pre,X}\right)}{{F}_{\mathrm{max},X}-{F}_{pre,X}}{x}_{stroke,X}\right)\epsilon ,$$

where:

*F*_{pre,X}and*F*_{pre,Y}are the**Spring preload force at port X**and**Spring preload force at port Y**, respectively.*F*_{max,X}and*F*_{max,Y}are the maximum spring forces acting against piston displacement at ports**X**and**Y**, respectively, where:$${F}_{\mathrm{max}}=K{x}_{stroke}+{F}_{pre},$$

*K*is the spring stiffness per port.*x*_{stroke}is the piston stroke per port.*ε*is the**Mechanical orientation**, which assigns the signal for piston movement as positive (extension) or negative (retraction).

If the force on the piston is less than the respective port spring preload force, the piston remains or returns to the neutral position. If the force on the piston meets or exceeds the maximum spring force for the respective port, the piston remains at the piston stroke until the applied pressure changes.

When the actuator is close to full extension or full retraction, you can maintain numerical
robustness in your simulation by adjusting the block **Smoothing
factor**. With a nonzero smoothing factor, a smoothing function is
applied to all calculated forces, but primarily influences the simulation at the
extremes of the piston motion.

When **Actuator configuration** is set to
`Single-acting`

, the normalized force on the piston is
calculated as:

$${\widehat{F}}_{X}=\frac{{F}_{spool}-{F}_{Pre}}{{F}_{\mathrm{max}}-{F}_{Pre}}.$$

When **Actuator configuration** is set to
`Double-acting`

, the normalized force on the piston at
**X** is calculated as:

$${\widehat{F}}_{X}=\frac{-{F}_{spool}-{F}_{Pre,X}}{{F}_{\mathrm{max},X}-{F}_{Pre,X}}.$$

and the normalized force at **Y** is calculated as:

$${\widehat{F}}_{Y}=\frac{{F}_{spool}-{F}_{Pre,Y}}{{F}_{\mathrm{max},Y}-{F}_{Pre,Y}}.$$

When the **Smoothing factor**, *s*, is nonzero, each
normalized force incorporates smoothing:

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

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

Cartridge Valve Actuator (IL) | Double-Acting Actuator (IL) | Pilot-Operated Check Valve (IL) | Single-Acting Actuator (IL)