Just wondering does anyone know where I could find a transfer function for a solenoid valve? This solenoid valve is being used in the control of temperature of fluid in a tank using a heat exchanger. (The solenoid valve regulates the quantity of steam entering the heat exchange coil).

Thanks a Lot!

Thanks a Lot!

Well, usually the transfer function for a solenoid valve is:

On: 0%

Off: 100% ;)

A solenoid is typically a binary on/off device. (There are exceptions)

If your valve "regulates the flow" it's probably actually a pneumatic, motorized, or otherwise proportional valve (also called a control valve).

And....that leaves us with too many unknowns to tell you anything. A lot depends on the mechanical valve the modulator is operating. By nature, the first little bit of opening of the valve is far more influential than the same amount of stroke near mid-stroke or 100%, for example, but the manufacturers know this and often profile the inside of the valves to try to linearize them.

So the bottom line is that it really just depends on what you have, that it may be pretty simple, and that it may not matter much anyway, depending on the dynamics of your process, because a control loop will work its way up and down the curve and find the spot it needs at any given load, so long as the curve is reasonably smooth with no major discontinuities.

On: 0%

Off: 100% ;)

A solenoid is typically a binary on/off device. (There are exceptions)

If your valve "regulates the flow" it's probably actually a pneumatic, motorized, or otherwise proportional valve (also called a control valve).

And....that leaves us with too many unknowns to tell you anything. A lot depends on the mechanical valve the modulator is operating. By nature, the first little bit of opening of the valve is far more influential than the same amount of stroke near mid-stroke or 100%, for example, but the manufacturers know this and often profile the inside of the valves to try to linearize them.

So the bottom line is that it really just depends on what you have, that it may be pretty simple, and that it may not matter much anyway, depending on the dynamics of your process, because a control loop will work its way up and down the curve and find the spot it needs at any given load, so long as the curve is reasonably smooth with no major discontinuities.

Thanks for the reply, I will obtain more information on what type of valve I am looking at and will update the post. Thanks for your help. I do appreciate it!

The Valve I am looking at is a proportional valve (control valve) where the output flow is proportional to the input current. To get the transfer function would I be correct to get the Laplace of the output flow (formula) divided by the input current in mA?

Declan,

The easiest way is to assume that the model just involves a simple time constant. Meaning, when I start with the valve fully closed and apply the signal to fully open the valve, how long does it take the valve to reach 100% open.

The transfer function in the Laplace domain would be a simple:

1/(1+Ts)

Where T is the time in seconds that it takes the valve to go from zero to 100% open.

But, this is not exact because T represents the time constant, and the time constant is defined as the amount of time a system takes to reach 63.2% (which is 1-1/e) of the steady state value. So, if the valve takes one second to reach 100% open, then T in your transfer function would be 1*0.632.

If you do everything based on percentage values, i.e. 50% of the current equals a 50% open valve everything should come out ok. But if you use actual current values, i.e. 12mA equals a 50% open valve, then you will need to do some scaling in the transfer function (in the numerator).

Hope that helps,

Nic

The easiest way is to assume that the model just involves a simple time constant. Meaning, when I start with the valve fully closed and apply the signal to fully open the valve, how long does it take the valve to reach 100% open.

The transfer function in the Laplace domain would be a simple:

1/(1+Ts)

Where T is the time in seconds that it takes the valve to go from zero to 100% open.

But, this is not exact because T represents the time constant, and the time constant is defined as the amount of time a system takes to reach 63.2% (which is 1-1/e) of the steady state value. So, if the valve takes one second to reach 100% open, then T in your transfer function would be 1*0.632.

If you do everything based on percentage values, i.e. 50% of the current equals a 50% open valve everything should come out ok. But if you use actual current values, i.e. 12mA equals a 50% open valve, then you will need to do some scaling in the transfer function (in the numerator).

Hope that helps,

Nic

Declan,

As NIC said the easiest way is consider the transfer function as a first order delay (1/1+T s). The correction is the assess of T.

T is the time constant and represent the time that need the output to reach 63% of the final value (supposing a step change in the imput). To reach 99,33% of final value are necessary 5xT. If you define that this time is 1 sec, then T is 0.2sec.

Regards

As NIC said the easiest way is consider the transfer function as a first order delay (1/1+T s). The correction is the assess of T.

T is the time constant and represent the time that need the output to reach 63% of the final value (supposing a step change in the imput). To reach 99,33% of final value are necessary 5xT. If you define that this time is 1 sec, then T is 0.2sec.

Regards

Sebastian is right... i should have double checked my logic and math before i submitted that one... sorry for the confusion. Thanks, Sebastian!

nic

nic

Your use of this site is subject to the terms and conditions set forth under Legal Notices and the Privacy Policy. Please read those terms and conditions carefully. Subject to the rights expressly reserved to others under Legal Notices, the content of this site and the compilation thereof is © 1999-2014 Nerds in Control, LLC. All rights reserved.

Users of this site are benefiting from open source technologies, including PHP, MySQL and Apache. Be happy.

**Fortune**

Parsley

is gharsley.

-- Ogden Nash