Solenoid valve thermal part

[jmex]

Hello folks,

I have a new different design of solenoid valve and would like to evaluate heat generation in the system. I am a mechanical engineer and might have less understanding of electrical components. Just trying to understand, the solenoid valve will have heat generation due to current flowing in the coil as per I^2*R*T only, correct? Is there any other heat source in this arrangement?
Is there any simulation application that might help me simulate this scenario? I can simulate Thermal analysis in different application but have doubts about heat sources and the amount of heat that is being generated.

Thanks,

 

[Baluncore]

DC or AC solenoid ?

The heat is dissipated in the coil real resistance.
For DC controlled solenoid valves, that is simply; W = I²*R.

For AC control, you need to include with the resistance, the inductance of the winding as a current limiting parameter. Heat is not dissipated in the inductive component of the solenoid, only in the coil resistive component of the impedance. With AC, you may need a laminated core to reduce magnetic losses that dissipate heat.

If you regulate solenoid current by pulse-width-modulation of the applied voltage, you may have a magnetic core loss in the armature.

You can simulate power dissipation in the solenoid with LTspice etc, or by reasonably simple calculations. You will need to simulate the flow of that heat energy using another system.

 

[jmex]

@Baluncore thank you for your reply. Yes it is a DC controlled solenoid valve. The current will be fluctuating alot (on/off due to its need of actuation really fast). I am only afraid of temperature over here. My copper wire diameter is 1 mm and the current flowing is 80 Amps. As per simulation, at this value, I am getting desired force. I need to evaluate the heat generated in this coil.
Also do I need to take into any other parameters here? Or will the heat generation will be from coil only.

 

[Baluncore]

My copper wire diameter is 1 mm and the current flowing is 80 Amps.

Take a look at the wire table;
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes
For copper wire of 1 mm diameter, the current should not average over about 16 amp. You appear to be operating nearer the "fusing in 10 seconds" current. Even if low duty, the insulation may fail due to the heat.

What is the on-time duty cycle of that 80 amps ?

 

[DaveE]

Take a look at the wire table;
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes
For copper wire of 1 mm diameter, the current should not average over about 16 amp. You appear to be operating nearer the "fusing in 10 seconds" current. Even if low duty, the insulation may fail due to the heat.

What is the on-time duty cycle of that 80 amps ?

Most ampacity tables refer to wires alone or in small bundles. You may need more derating for a large coil if wires are heating each other because of proximity. This might be a good thing to simulate. You'll be looking for the hottest spot in the coil and keeping that temperature below the wire insulation temperature rating.

You might also consider that the resistance of copper increases with temperature, which might result in greater power dissipation.

 

[DaveE]

The old rule of thumb^* for magnetics was 500 cmils/A. cmil (circular mil) is unit for area, being the area of a circle with 0.001" diameter (perhaps the stupidest unit ever conceived). So 1 cmil = 6.45 x10^-4 mm². So 500cmils/A = 0.323mm²/A means you would need AWG#3 for 80A continuous.

*Note: rules of thumb are always wrong, but often a good first guess. This one is mostly for lower power stuff, I think.

 

[Baluncore]

You might also consider that the resistance of copper increases with temperature, which might result in greater power dissipation.

That is true for a fixed current, but not for a fixed DC voltage;
An increase of +1% in absolute temperature, will raise the resistance; +1% ;
Then ohm's law, will reduce the current; -1% ;
Power; W = I²*R ; The current I is squared, so that is; -2% ;
W will fall by; -2%, and rise by; +1% due to resistance ;
So the net effect of heating by; +1%, will be a fall in power of; -1% .

 

[DaveE]

That is true for a fixed current, but not for a fixed DC voltage;
An increase of +1% in absolute temperature, will raise the resistance; +1% ;
Then ohm's law, will reduce the current; -1% ;
Power; W = I²*R ; The current I is squared, so that is; -2% ;
W will fall by; -2%, and rise by; +1% due to resistance ;
So the net effect of heating by; +1%, will be a fall in power of; -1% .

Yes, but... He needs to design for a certain H field strength to work the valve. So it doesn't really matter if the driver is a current source, voltage source, or something in between. He needs to design for a desired current value.

 

[Tom.G]

Their are three major uses of a solenoid, but you have not specified which you are asking about:

1) A coil that pulls a movable core into it
2) A coil with a fixed core that attracts magnetic material
3) Part of a resonant tuned circuit as used in radios to select a particular frequency (station)

You description indicates either 1) or 2).

Another data point:
Fusing current for that 1mm dia. Copper wire is 82A in free air.

For a worst-case thermal approximation, you could assume the solenoid as a solid cylinder dissipating the same power, with cooling being free air convection or whatever the operating environment approximates.

For the power use W=I²R, where W is Watts, I is Current, R is electrical resistance of the coil.

Here in the US, electric kitchen stoves use wire insulated with glass cloth (fiberglass), (earlier versions used asbestos, which was discontinued due to its health hazards).

As an additional constraint, keep in mind that metals lose their magnetic properties at high temperatures; and it is different for different alloys.

You may be able to reduce the needed current by:
1) Changing the dimensions of the solenoid
2) Changing its core (if present), or the material it is attracting
3) Adding a Permanent magnet to either its core or what it is attracting

Although not of much help for your particular situation, their are several on-line calculators for designing coils or solenoids:
https://www.google.com/search?hl=en&=multilayer+solenoid+inductance+calculator

Cheers,
Tom

p.s. You have got me curious about that 80A requirement. Without giving away any secrets, could you fill us in about what this is?

 

[DaveE]

Here in the US, electric kitchen stoves use wire insulated with glass cloth (fiberglass), (earlier versions used asbestos, which was discontinued due to its health hazards).

For magnetics, he'll want to use magnet wire with a "enamel" coating. Probably MW35 or MW16. There typically isn't space for other types since there is a significant tradeoff between filling the winding area with copper or insulation. Plus you are limited to high temperatures for this sort of design. Care must be taken with the temperature ratings of all of the coil materials (tapes and such).

https://ccoils.com/wp-content/uploads/WireInsulationGuide.pdf

 

[Baluncore]

Yes it is a DC controlled solenoid valve. The current will be fluctuating alot (on/off due to its need of actuation really fast). I am only afraid of temperature over here. My copper wire diameter is 1 mm and the current flowing is 80 Amps. As per simulation, at this value, I am getting desired force.

The fundamental design problem now, is insufficient space for copper windings on the solenoid coil.
1. How many turns are now on the coil?
2. What is the DC drive voltage now?
3. Approximately how often is the coil turned on?
4. How long does it typically stay on for?
5. What is the minimum required magnetic field rise time?
6. What is the maximum acceptable magnetic field decay time?

 

[jmex]

Thank you everyone for your inputs. I am currently in design phase and using DoSA2D to find the desired force to work for my system to run. I only know for now is that having current 80Amps, my system works. I have not looked for magnetic field decay time as well (which i should look into but not sure where can I get those data). There are so many unknown factors for now. I only have to look into temperature for now. So that I can find thermal impact because of such high temperature. Let's assume that the coil turns on/off for 5 milliseconds and it keeps doing that for long period of time. The time is crucial here and the amount of force required is huge for the solenoid to travel (hence such high current). I am only aware that there will be heat problem, but I will certainly consider your feedback and work on those too.

 

[Baluncore]

I only know for now is that having current 80Amps, my system works.

But you have not stated how many turns there are now on the solenoid.
It is amps*turns, that determines the magnetic field.

The solenoid will "hold in" with a lower voltage and current, than was needed to initially actuate it. By lowering the hold current, you reduce heat generation.

You can cause the magnetic field to rise faster, by discharging a higher voltage capacitor into the winding during the initial turn on phase.

You can remove the magnetic field faster by placing a resistor in series with the switch flyback protection diode.

 

[jmex]

But you have not stated how many turns there are now on the solenoid.
It is amps*turns, that determines the magnetic field.

The solenoid will "hold in" with a lower voltage and current, than was needed to initially actuate it. By lowering the hold current, you reduce heat generation.

You can cause the magnetic field to rise faster, by discharging a higher voltage capacitor into the winding during the initial turn on phase.

You can remove the magnetic field faster by placing a resistor in series with the switch flyback protection diode.

the number of turns is 1000. Also thank you once again, this is a very good suggestion to remove magnetic field faster. Will keep that in mind

 

[Baluncore]

80,000 amp⋅turns. That valve must be massive. The solenoid will be of similar size to the industrial magnets used to pick up scrap in the steel industry. The electronic control systems used for those magnets, or on your valve, is certainly not for a beginner. All that stored energy has to go somewhere.

 

[jmex]

yes, motivated from industrial magnets to pick up scrap, we need that strong. Right now, my only concern is the thermal management of that system. Not sure how much heat will be generated and how should I calculate. Rest I can provide the boundary condition accordingly to see the thermal effects into the system.

 

[Tom.G]

Not sure how much heat will be generated and how should I calculate.

(The following examples are based on 1000 feet of wire because the look-up tables here in the States use that as a standard length.)

If you know the electrical resistance of the solenoid, the power it dissipates is shown in post 9 above (https://www.physicsforums.com/posts/7128278).

Or if you do not have a way to measure the resistance, it can be calculated if you know the wire length. Your Copper wire of 4mm dia. has a resistance of 6.385 Ohm per 1000 feet.

With 80A thru it, that is 80×80×6.385 = 40864 Watts per 1000 feet of wire. Of course this power will scale proportionally with the wire length.

To get 80A thru 1000 feet (6.385 Ohms) you will need 511 Volts just to overcome the resistance. This will also scale proportionally with wire length. Though since your coil will have some Inductance, a rather higher voltage will be needed.

For comparison, a high-end electric stove may have their large burners using 3000 Watts each.

Cheers,
Tom