Solenoid AC Inrush, Voltage, and Circuit Design 2

[RontotheB]

Hello,

I feel a little ignorant because my schooling didn't properly prepare me for simple circuits, but I have a few questions about a solenoid valve that I was hoping you guys could answer.

Here is the background in case you're interested:
I bought two solenoid valves for a pressure cycler. One is an inlet valve to a pressure supply, and one is an outlet to ambient air (0 psig); the inlet and outlet valves are joined together by a pressure vessel / piping. The valves are NC (normally closed) and the solenoid coils open the valves when they are energized (when their circuit closes). These valves were set to undergo one full cycle (open and close) every 4 seconds. The valves are rated at 5,000 psig for continuous duty, and on the pressure supply side, I'm supplying 3750 psig. The pressure vessel between the two valves cycles between 3750 psig and 2000 psig.

After 27 minutes, one of the coils shut down. Because it was extremely hot, it may have shut down because of overheating. I'm wondering if it's been wired improperly.

Now, for my questions:
We're currently supplying the solenoid valves with 115V AC, and they're designed for an AC inrush of 2.5 amps, and an AC holding of 0.2 amps, both at 120V AC. The solenoid coils have a power rating of 22 W. Do the solenoids pull only as much current as they need from our power supply, or will they pull out as much as is available? In other words, will they exceed their specified ac inrush and holding currents? What would explain the solenoids' overheating?

I'm also wondering how relevant the specified solenoid voltage is. How does it operate differently if it's rated for 120V AC instead of 48V AC, or 240V AC?

If I wanted to use DC instead of AC, what voltage would you recommend? I can get solenoids that can use 12V DC, 24V DC, 125V DC, and 250V DC.

 

[MacGyverS2000]

The coil should have a resistance around 50 ohms... 22W @ 120V suggests a roughly 2.5A draw, so things match up in that regard. Is there a temp monitor on the valve? If not, how do you know it shut down due to heat? If it's not self-regulating, it will heat up until the coil melts and opens, and it certainly won't stop on its own.

If they're rated for continuous duty, they shouldn't have a problem. Are the valves designed in a way such that pressure helps open/close them? Not common, but if so it could explain the overheating.

Dan - Owner

http://www.Hi-TecDesigns.com

 

[RontotheB]

I got around 0.18 Amps, with I = P / V. I also got R = V / I, suggesting around 650 Ohms. Am I missing something? Are solenoids modeled as a combination of inductors AND resistors, or just one or the other?

 

[RontotheB]

Oh, sorry for the double post. To answer your questions, there is no temp monitor, but it was very hot to the touch (around 150 degrees F).

They are rated for continuous duty, but from my understanding, that just means that they can stay energized indefinitely. It doesn't indicate anything as far as cycling ability.

Upon further inspection, it was not the coil that stopped functioning, but the valve itself. Still, it would be nice to know how the coil functions in the context of electronic circuits.

 

[logbook]

An inductor is always modelled as an inductor in series with a resistor. You calculations would only be correct for a DC solenoid.

>I also got R = V / I, suggesting around 650 Ohms.
In an AC circuit you would calculate the impedance Z as Z= V/I.

There is not enough information to do the calculations you are trying to do. How about a valve part number and link to a data sheet?

The holding current is the minimum current necessary to hold the valve on, which must necessarily occur at less than the minimum specified operating voltage. Once the valve is turned on it is easier to hold it on (needs less current).

An ideal solenoid could draw current and not dissipate any power, the current and voltage being out of phase. This will not occur in practice, due to losses in the core. Only the data sheet can answer these questions.

macgyvers2000; lose one house point for dodgy maths.

 

[RontotheB]

The valve P/N is EH30-043-A120. It's a Magnatrol valve (Clark Cooper Division). I'll look for a data sheet and link it if I find one. Thanks for the info, logbook!

 

[itsmoked]

Yeah, macgyvers uses [ib]special[/I][/B] math.

To answer some of your questions:
A solenoid valve needs a certain amount of power to function. That's why it consumes power. duh!

So. The coil will be modified for each type of power source. Hence 24VAC, 240VAC, or 120VAC coils will all be changed so when all are appropriately energized they will deliver the SAME force.
This means different coils that will have the same wattage.

Next point. AC coils are often considered better than DC coils. They are essentially current limited by their inductance NOT the coil resistance. The coils are designed so as they pull in the inductance climbs. It climbs a huge amount! So the ultimate holding power becomes the pulled in current. 5-30W would be typical.
When they aren't pulled in they can consume 500W. This also means they will smartly apply, once applied, their power drops to the holding power automatically.

Downside. If an AC solenoid is prevented from pulling in that 500W will surely snuff them out. Usually about 2 seconds is more than enough to let out the smoke.

A DC solenoid never has this problem because there is no functional inductance involved. The initial current is the same as the holding current. This wrecks the ability of the valve to actuate quickly with high power then drop the power automatically to the required holding power. Hence DC solenoids may always consume more power than an AC counter part. Often not a lot though, maybe, a few percent, 10%.

Now if your valve isn't completely closing it will run hot. VERY HOT! If it can't get its inductance up by successfully seating then it will certainly fry sooner than later.

Furthermore, since all solenoids have to deal with the heat of holding power, that ~22W will heat up the valve/body.

You need to picture a ~22W light bulb being held in your hand. If the volume is the same as a 100W bulb you will be able to hold it all day. If it is the size of a small appliance bulb it will likely be too hot to hold. If it's the size of a Christmas tree bulb you will be promptly burnt.

This size, temperature, situation is why the solenoid valves have temperature specs for the liquid/gas they are controlling. If your controlled substance is too warm then the valve will overheat when the conducted media temp and the coil power combine to give the resultant valve temperature.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[MacGyverS2000]

<sigh> and this wouldn't be the first time using my own special math <shake head>



Dan - Owner

http://www.Hi-TecDesigns.com

 

[itsmoked]

Buck up man that special math must be where we get those cool First Down lines in football games.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[logbook]

http://www.clarkcooper.com/eh30.html

 

[sreid]

As others point out, applying 120 VAC to the coils should not cause the coil to burn out. Also as others point out, a coil will likely pull more current if the plunger does not pull in all the way.

Coils burn up from over current. I would connect a current meter (Fluke VOM) in series with the coil and see what the actual current is. With any luck, it might have been one sticky valve.

 

[RontotheB]

Hey logbook, I had already found that sheet, but it didn't contain any other useful information besides what I had provided. What do I need to know in order to find the coil's impedance?

 

[logbook]

Well having seen the data sheet, and taken on-board itsmoked’s insights, let’s now work through the maths.

The initial (inrush) current is 2.5A @ 120V. When it is nicely settled down the current has magically dropped to 0.2A @ 120V. So let’s start from the holding current.

We know that we have 22W dissipation at 0.2A. We have to assume that the current is sinusoidal if we are going to get very far.

The current squared times the resistance equals the power dissipation. Hence the equivalent series resistance is 22/(0.2*0.2)= 550 ohms.

But wait a minute, then how did we get 2.5A at 120V initially? Clearly this surge current tells us the maximum impedance was not more than 48 ohms initially. This is precisely why I used the very explicit term “equivalent series resistance”. The resistance of the copper wire is less than 48 ohms and the extra resistance to make up the 550 ohms is coming from “core loss” (iron loss) in the magnetic circuit.

Let’s continue from the holding current. We apply 120V and get 0.2A. This tells us the impedance is 120/0.2= 600 ohms. The inductive part is not 50 ohms though! The inductive reactance is phase shifted relative to the resistance so the impedances add as the square root of the sum of the squares.

550*550 + ZL*ZL = 600*600

ZL, the reactance of the inductive part of the impedance, is therefore 240 ohms.

So, when the valve is holding, the equivalent circuit is a series 550 ohms resistor and 240 ohms of inductive reactance. When the valve is first energised the series resistance is less than 48 ohms and the inductive reactance is also less than 48 ohms. More than that we can’t say. Analog stuff is deceptively complicated!