Solenoid valve inductor commutator circuit.

[Luketrician]

Would it be possible for a defective inductor commutator (resistor diode) paralleled with a solenoid coil could be not allowing the magnetic field within the coil to collapse quick enough, cause a FSV to have a slow to close stroke time?

The FSV operates as designed when energized, (Energize to OPEN)..but stroke to close time is approx 8 seconds..and spec's for this valve call for less than 4 seconds.


Also, this is the second, (brand new FSV that was installed)...it's sister valve is functioning as designed.


What wasn't changed was anything associated with the control and power circuitry of the FSV.



Would like to hear if there is any history or OE from the industry on this possible scenario thanks

 

[Luketrician]

Also should add that the coil voltage is 125VDC.

Target Rock is the manufacturer of FSV.

 

[GoldDigger]

Yes it could. A shorted or low value resistor could have this effect. As could a low breakdown voltage in the switching element itself.
What you are up against is probably an accidental application of what would be used deliberately to create a delayed opening relay.
But also check for a delay in the opening of the element controlling the coil voltage /current.

Sent from my DROIDX using Tapatalk 2

 

[petersonra]

What is a resistor diode?

I have never heard of a resistor diode combination being placed around a coil for suppression purposes. I have heard of RC networks or diodes being used for such a purpose.

I suspect there is some mechanical issue involved. is the FSV hydraulic? It may be that there is a restriction in the hydraulic line.

 

[GoldDigger]

I have heard of series resistor and diode combo used, where the resistor value is chosen for critical damping of the inductor.
I cannot give you a specific example though....

 

[Smart $]

Would it be possible for a defective inductor commutator (resistor diode) paralleled with a solenoid coil could be not allowing the magnetic field within the coil to collapse quick enough, cause a FSV to have a slow to close stroke time?

Its purpose is to mitigate damaging transient voltage spikes upon abrupt de-energizing of the solenoid. A "by-product" of a properly functioning resistor-diode suppressor is that it actually retards field collapse and extends response time. JMO, but an extra 4 seconds seems quite excessive to be a result of a snubber.

 

[petersonra]

Its purpose is to mitigate damaging transient voltage spikes upon abrupt de-energizing of the solenoid. A "by-product" of a properly functioning resistor-diode suppressor is that it actually retards field collapse and extends response time. JMO, but an extra 4 seconds seems quite excessive to be a result of a snubber.

I could see a few milliseconds, but not 8 seconds.

 

[ronaldrc]

Sounds like the circuit driving this valve is faulty.

Sounds like a solid state output so there might be a voltage while a not energized.

You said it had a sister valve which is working correctly.

I would check both valve coils voltage and current while not energized if the faulty valve measures
the same as its sister then it is probably a mechanical problem with the valve.

 

[GoldDigger]

If all else fails, you could look at the coil current waveform with shunt or DC current probe and a scope.
And looking at the applied voltage the same way will tell you whether there is a delay in the control circuit. (Trigger on the control input and view the applied voltage on the solenoid.

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[Luketrician]

This FSV installed is the second one. Both are brand new valves from Target Rock.

I appreicate the replies too.

I have recommended to lift conductors from the coil of the operating sister valve and then temporarily install test leads to tie the suspected coil FSV and attempt to stroke again.

The sister valve is in close proximity to the suspect FSV so I think if the suspected valve operates as designed using the sister valves controll circuitry, then I will be able to convince my peers that this issue is not due to any sort of mechanical binding of the valve.

They also checked the bias of the diode & resistor, but I do not feel that it was a valid test. The just used a V.O.M Fluke for this and I do not think a V.O.M is the proper way to test the resistor diode aka snubber.

Very interesting replies and I think there is something there that is not allowing that field to collapse quick enough.

Like what yall are saying, with contacts not opening properly, delaying the current interuption to the FSV, and defective snubber, causing contacts up-stream of the coil to arc while opening, and delaying the magentic field collapse of the FSV coil.

 

[ronaldrc]

Sounds like a great Idea, if that doesn't convince your your peers , you probably need to get some new peers!:rotflmao:

 

[gar]

131113-1245 EST

Luketrician:

As others have said employ some instrumentation to see what the circuit is doing.

But first, do some simple circuit analysis.

Consider a DC source (battery), mechanical switch, and the solenoid (a series resistance and ideal inductor) in a series circuit.

Close the switch and the current will slowly rise based upon the total series resistance and inductance. This rise time of current can be related to the L/R time constant. This rate of rise of the current will in part determine how fast the solenoid closes.

With no snubbing circuit, ideal components, and upon opening the switch the magnetic field will instantaneously collapse. Infinite voltage will appear across both the inductor and the switch. Only the mechanical components will determine the opening time.

In the real world there will be voltage breakdown. Hopefully only across the switch contacts. This is your shortest drop out time.

Change the circuit to a dead short across the solenoid at the instant of opening the switch from from the battery. This produces an L/R time constant that is the longest possible drop out time. For a fixed L the lower the resistance the longer the time of dropout. Here the limitation on time is the internal coil resistance. This function is approximated by the use of a reverse biased diode across the coil (longest drop out time).

Add some resistance in series with the shunt diode and the dropout time diminishes. But the peak turn off voltage across the coil and switch increases as this snubbing resistance increases.

It will take a very large inductance relative to coil resistance to get into the 1 second range. I have worked with a large magnetic clutch were our goal was a short drop out time. Excitation was about 1 A at 110 V. With an RC snubber and increased magnetic air gap we got dropout down to the range of 100 mS.

I suggest you connect a DC current probe to an oscilloscope and look at the drop out current to the solenoid, and trigger the scope on the command signal to open the solenoid. You might simultaneously monitor the voltage across the solenoid coil. Note: I said trigger on the command signal to the switch. This means not on the output voltage of the switch.

.

 

[Luketrician]

udpate

udpate

We were able to get some (i believe) some valid data.

To measure the current to the solenoid during the drop out time, to monitor directly at the coil.

Once the circuit was interrupted by hand switch up stream in the control circuit, current and voltage went to 0 almost instantaneously. (We were not able to get a specific time)

Intersting to note about this solenoid valve, with no measurable current or voltage at the coil, the valve was still not in the closed position, only in a mid-travel state.

I believe we have a design issue with these type of target rock valves, with no measured current/potential accross the solenoid coil, I think it is safe to say that we have a mechanical design issue with the FSV valve itself.

I think I can refute any claims that our control circuit is causing this issue now. Also, this valve, takes in upwards to one minute, before we get to full closed.

The system line is only a 4" steam line also.

 

[just the cowboy]

Can you feel the mag field with a metal object or maybe a compass moving. If you can't feel it it should not hold it open.

 

[GoldDigger]

Intersting to note about this solenoid valve, with no measurable current or voltage at the coil, the valve was still not in the closed position, only in a mid-travel state.

Very interesting indeed.

I can see a couple of potential causes:
1. The valve is designed to rely on spring pressure to close and the spring has failed.
2. The valve is designed to close with gravity assist and it is mounted in the wrong orientation.
3. The valve is designed to close and seal with assist from the material flow through it and the resulting pressure differential. In that case a valve removed from the system might not close properly when tested with no flow.
4. The valve design assumes that there is a separate spring in the actuator which will drive the valve shaft positively in both directions, and the wrong type of solenoid actuator was used.

 

[Smart $]

We were able to get some (i believe) some valid data.

To measure the current to the solenoid during the drop out time, to monitor directly at the coil.

Once the circuit was interrupted by hand switch up stream in the control circuit, current and voltage went to 0 almost instantaneously. (We were not able to get a specific time)

We could get into the specifics of how and where volatge and current should be measured, but I believe it is pointless given the valve's performance.

Intersting to note about this solenoid valve, with no measurable current or voltage at the coil, the valve was still not in the closed position, only in a mid-travel state.

I believe we have a design issue with these type of target rock valves, with no measured current/potential accross the solenoid coil, I think it is safe to say that we have a mechanical design issue with the FSV valve itself.

I think I can refute any claims that our control circuit is causing this issue now. Also, this valve, takes in upwards to one minute, before we get to full closed.

This being the second replacement, the first consideration that should be made is what is different.

Is this a long standing process control or a revised or otherwise relatively new implementation?

How long did the original function?

Same make and model?

If "sister" valve is an identical make and model, is there anything different about the process lines and/or parameters on this one, or is it a redundancy valve (i.e. A train, B train, or such where all parameters are identical also)?

Have you contacted the manufacturer for their recommendation on, or to investigate the failing performance issue?

 

[Luketrician]

Good questions..

This is a new design..electrically it is the same operating coil and control. Old solenoids were welded in, completely sealed system.

New design has bolt on "bonnets" and flanges.

Suprisingly even with Vendor support on-site, no determination has been made, just more speculations at this piont.

These FSVs operate as pressure relief valves for our reactor coolant system.

Intersting to note, is that these valves close by two springs sealed within the stem and by system pressure.

Normal operating pressure for our Reactor coolant system is 2200 psig.

The fsv failed while system pressure was around 125-200 psig.

Both valves are located off the same system line also.

Both valves were sent to lab for testing and passed testing through a wide range of pressures...100 psig up to 2200 psig.

The existing FSVs have been in service for more than 5 years. Also interesting to note..that one of the existing valves was also sent for testing and failed.

I will update more when time allows, and I appreciate the responses also.

 

[pelikan]

@ Luketrician, a resistor diode combination is used for a variety of purposes, it can either be used to prolong the charging time or reduce it depending upon the circuit design. For example, when the diode is working in forward bias, the equivalent resistance of the combination is very less as compared to the resistance of the combination when diode is reverse biased. Basically, the answer to your question lies in the actual circuit design as well as the value of the resistance used in parallel with the diode. The easiest way to understand a circuit like this is to simply simulate it in MultiSim, proteus or whatever tools are available to you. This way you'll not only be able to analyze the circuit in detail but you can also see a clearer picture.