Solenoid force calculations - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: October 22, 2017, 10:15:19 am
content: I'm trying to build a solenoid as an actuator.The calculations that I find no the web areForce of a solenoid on a magnetic plunger as underF = (N * I)^2 * ((magnetic const * A ) / 2g^2)F= force in newtonsN=turnsI= current in coilA=g - gapI fail to understand the following1. Is this the equation for an air core solenoid or a solenoid with a ferromagnetic outer jacket. What changes in the equation if it were surrounded by a ferromagnetic jacket.2. Is A the cross sectional area of the coil or the cross sectional area of the plunger. I ask this because in certain cases I have found that, they first calculate the magnetic energy density in the air gap of the solenoid (that is the gap between the end of the solenoid plunger and the farthest end of the solenoid coil) and then multiply it by the cross sectional area of the plunger.3. What is g ? is the the gap between the front face of the solenoid and the end of the plunger or is it the gap between the back face of the solenoid and the end of the plunger (as stated in point 2)TIA

author: IanB
author_rank: Super Contributor
timestamp: October 22, 2017, 11:37:21 am
content: Calculations are useful, but nothing beats experiment. You should build experimental solenoids and measure them. You will get much better answers than that formula will give you. The formula makes various simplifying assumptions, assumes an ideal case, and as you note there are some variables that are difficult to estimate.Why would you want to put a ferromagnetic jacket around the coil? I have seen many solenoid actuators and I have never seen that done.

author: ebclr
author_rank: Super Contributor
timestamp: October 22, 2017, 12:33:18 pm
content: "Calculations are useful, but nothing beats experiment"Are you an engineer or a french chef?

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: October 22, 2017, 12:39:28 pm
content: Quote from: IanB on October 22, 2017, 11:37:21 amI was referring to something like this, as per the information I have read it tends to strengthen the magnetic field, and makes the field intensity stronger inside the solenoid

author: orolo
author_rank: Frequent Contributor
timestamp: October 22, 2017, 08:20:34 pm
content: If I understand the idea, the ferromagnetic frame is important because it closes the magnetic circuit just around the coil, leaving out of the field everything outside. So the part of plunger not yet introduced plays not part, since the magnetic field is diverted from it. I have no experience at all with these devices, but reasoning from first principles:If the lenght of the coil is L, and you have a lenght x of plunger introduced inside, so a length L-x of air remains. Since the magnetic energy density is B^2/2mu, the magnetic energy stored in the inside of the coil is:\$E \ = \ E_{air} + E_{plunger} \ = \ \frac{B^2}{2\mu_0}\cdot A\cdot (L-x) + \frac{B^2}{2\mu_p}\cdot A\cdot x \$Where A is the sectional area of the plunger (if the plunger is smaller than the coil radius, you can disregard that extra area since it doesn't take a part in the energy transfer that will follow-- A is for the plunger area, not the coil area).The variation of magnetic energy with respect to length x of plunger introduced is a force that, from Newton's third law, works against the plunger. So the plunger experiences a force:\$F \ = \ \frac{B^2}{2}\left(\frac{1}{\mu_0}-\frac{1}{\mu_p}\right)\cdot A \$If the permeability of the plunger is much higher than the permeability of vacuum, you can simplify that to:\$F \ = \ \frac{B^2}{2\mu_0}\cdot A\$That is, as the plunger gets inside the coil, it depletes the magnetic energy, which is mostly stored inside the air part of the coil. That energy is transferred as kinetic energy to the plunger. Since the energy density inside the coil is constant, the force is also constant.Now, remembering that a (very!) simplified expression for the field inside an air coil is:\$B \ = \ \frac{\mu_0 N I}{L}\$Substituting, we arrive at:\$F \ = \ \frac{\mu_0 N^2 I^2 A}{2L^2}\$So probably your constant g refers to the total length of the coil, at least in this first order approximation.Then again, I'm not experienced with this kind of devices, and there are some simplifications involved. But maybe you can work from there.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: November 15, 2017, 04:44:44 pm
content: Orolo this is amazing stuff! Ah! what deduction. A la sherlock!!Having said that I how does the inductance of the solenoid affect the force is there a definite relationship. I see that the force equation has N and I in it so that would imply force is proportional to inductance right? Also would the inductance play a part in the speed of the solenoid.?Quote from: orolo on October 22, 2017, 08:20:34 pm

author: orolo
author_rank: Frequent Contributor
timestamp: November 15, 2017, 07:38:57 pm
content: Quote from: ZeroResistance on November 15, 2017, 04:44:44 pmWell, again we can make some guessing. First, remebering that the magnetic force acting on the plunger is:\$F \ = \ \frac{\mu_0 N^2 I^2 A_{p}}{2\mathit{l}^2}\$where the length of the coil is renamed to \$\mathit{l}\$ to avoid confussion with the inductance L, and we have qualified the area of the plunger with a subindex. The inductance for a long air coil is:\$L \ = \ \frac{\mu_0 N^2 A_{c}}{\mathit{l}}\$where \$A_c\$ is the area of the coil. So the inductance can be easily absorbed into the force formula:\$F \ = \ \frac{A_p}{A_c}\frac{I^2}{2\mathit{l}} L\$To get the speed of the plunger we need another reasonable guess: the restorative force of the spring is much smaller than the magnetic force acting on the plunger. After all, the spring is there to reset the plunger when there is no magnetic force, not to impede its movement.If that is the case, the plunger experiences an uniformly accelerated movement, with acceleration a = F/m, m the mass of the plunger. And so its velocity is \$v = a\cdot t.\$ This is not much help; it would be better to have the speed in terms of the positionxof the plunger. Remembering that the position in an UAM is \$x = 1/2at^2\$, we easily arrive at \$v = \sqrt{2ax}\$, or:\$v \ = \ I \cdot \sqrt{\frac{A_p}{A_c}\cdot\frac{L}{m\mathit{l}}\cdot x}\$So, we would expect the velocity of the plunger to be proportional to the current in the coil, proportional to the square root of the inductance, inversely proportional to the length of the coil, and directly proportional to the displacement of the plunger.Another interesting constant would be the time the plunger needs to move some length, to predict how much times it takes for it to switch.\$T \ = \ \sqrt{\frac{2x}{a}} \ = \ \frac{2}{I}\sqrt{\frac{A_c}{A_p}\cdot \frac{m\mathit{l}}{L}\cdot x}\$So, for example, a L=1mH coil of l=1cm length, with a plunger of m=1g mass filling the coil (Ap = Ac), at a current of I=100mA, would take T=0.2s to travel the coil (x=1cm). Its terminal velocity would be 0.1m/s.All this again remembering that we are working on an approximation, and that we are assuming the effect of the spring is small compared to the magnetic force. A lot of high school mechanics in play.

author: jcw0752
author_rank: Contributor
timestamp: November 16, 2017, 05:05:28 am
content: It makes sense that the "g" is the distance between the plunger and the back of the solenoid. As the plunger gets closer to closing the magnetic circuit the force increases (personal experience). Once the magnetic circuit is closed the current in the solenoid can be reduced as the hold current is much lower than the start current. I have enjoyed reading your posts. I had interacted with these devices over the years with the first encounter being in 4 and 8 track tape players (1960s) and more recently in the pneumatic and water control solenoids of my profession.John

author: orolo
author_rank: Frequent Contributor
timestamp: November 16, 2017, 11:34:24 am
content: Of course, I thought about g being the "gap" of air remaining, but it didn't make sense to me, for two reasons:1) If g goes to zero, force goes to infinity. Moreover, if you integrate force times space, energy also goes to infinity. That doesn't match the fact that the energy inside the coil is finite.2) N is the total number of turns in the coil. If the gap changes, the number of turns should change also. In fact, you can rewrite the formula using "g" for gap, and "n" for number of turns per unit length.Anyway, I have no experience with these devices. However, I got some big coils, some iron for the plug and a dynamometer. I will try some measurements and come back with the results. I don't know what resolution I can get, but I'll try some experiment.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 16, 2017, 12:07:38 pm
content: First of all, note the Maxwell stress: sigma = B^2 / (2*mu)This has units of pressure (force per area, much as B is flux per area), logical enough. This is also in units of energy density (force/area == energy/volume), so by summing up sigma throughout a volume, you get the magnetic energy input into the system. Which is, of course, energy that is able to do work, and therefore pull on an armature.Electrically, the consequence is this: when the armature is out, the inductance is small and current builds up rapidly. When the armature pulls in, magnetic energy is converted to mechanical energy, inductance shoots up, and current builds up slowly.Normally, DC solenoids are designed so that winding resistance limits DC current, allowing safe operation, in steady state, whether the armature pulls in or not.AC solenoids are not designed this way; instead, they are impedance limited, but only when the armature is able to pull in. This gives increased current draw when open, allowing the armature to close more quickly, but also means the coil burns up when the armature gets stuck. (A common fault in industrial machinery.)Back to design: so, the armature is actuated by a pressure, the Maxwell stress. Where is this applied? Well, in the above figure, the pressure lines follow the flux lines, and the armature gets a radial pressure where it pierces the frame (flux flows sideways, from the frame to the armature shank). No lengthwise component, so the magnetic force here is zero. At the butt end, however, the flux is lengthwise, and so is the force.If you had a coil (no frame) and a slug of iron, the force is much less, for two reasons: 1. the coil is very poor at generating flux density in the first place; 2. the armature acts like a bar magnet, funneling flux from the coil down its length. There's still a difference in flux density at the ends of the armature, but it's less than if it were in a frame.How big of a solenoid is needed? Assuming you're using steel parts, you won't be doing much more than about 1.5T flux density. This is a pressure of 0.895 MPa (or about 9 atm). Say you need 10N (~1 kgf) of force: then you need about 12 mm^2 of cross sectional area.In practice, you might achieve less than half that flux density, particularly when the armature is not fully seated (because of air gap). To ensure pull-in, you need to guard-band the design, both in how much flux is available when fully open, and in having enough excess force to guarantee it will pull in.HTH,Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 16, 2017, 12:09:06 pm
content: Quote from: orolo on November 16, 2017, 11:34:24 amIt may be useful to consider the magnetic circuit, as if it were 100% air. In that case, the magnetic path length of the core, l_e, is effectively reduced by its relative permeability. Thus, for l_g = 0 (no physical air gap), the gap is still l_e/mu_r. For nonzero gap, the total (mu_r = 1 equivalent) length is l_e / mu_r + l_g.

author: orolo
author_rank: Frequent Contributor
timestamp: November 16, 2017, 04:59:13 pm
content: Okay, I have tried the dynamometer but results are not good because of lack of armature. So I decided to usefemmto simulate the situation, slightly altering a coilgun example with LUA script (this is the reason the files are still called "coilgun").Of course, jcw0752 experience is right, and the acceleration greatly increases with gap reduction. I will attach the graphics because I believe they are interesting. The femm file and lua script are attached at the bottom.The simulation has axial symmetry, so only half plane is modeled:The armature is all-important, including closing the gap at the other side of the plunger. Otherwise the plunger doesn't accelerate as per the formula. Here is where my simplistic model broke down: I used the formula for the inductance of whole air coil, while things are much more complex than that: the plunger is closing the magnetic circuit, compressing all the energy inside the gap. By the way, I also assumed that the current was turned off when the plunger started moving: that is clearly wrong. The current stays on, and replenishes the energy inside the gap, in an ever decreasing volume, so the plunger gets more and more energy as it advances.All in all, the equations in the posts above work only when the plunger starts to move, and the gap is very large. They are valid not much longer.So here is the femm computed acceleration versus gap size (force in newtons, gap in inches), showing that hard earned experience beats idle theoretics every time:The coil characteristics, turns, current, etc. are in the femm file and are more or less irrelevant: the 1/g^2 curve is clear enough.I will try and modify the reasoning above, using a better flux estimate, to arrive at the correct force relation. The speed question too, with this new situation.This is a very instructive and subtle problemEdit:Okay, the problem was in the flux computation part. One cannot use the flux of an air core coil. A complete magnetic circuit has to be considered.Assuming the armature has zero reluctance (that's the reason it is there), the only paths of reluctance are the plunger and the air gap. The reluctance of the air gap is \$\frac{g}{\mu_0 A}\$, where g is the length of the gap. And the reluctance of the plunger is \$\frac{\mathit{l}-g}{\mu A}\$. The magnetomotive force is N·I, and the reluctances are in series, so the flux is:\$\Phi \ = \ \frac{NI}{\frac{g}{\mu_0 A} + \frac{\mathit{l}-g}{\mu A}} \ = \ \frac{\mu_0 N I A}{g + \frac{\mathit{l}-r}{\mu_r}} \ \approx \ \frac{\mu_0 N I A}{g}\$From there, the approximate flux is \$B \ = \ \frac{\mu_0 N I }{g}\$ and, introducing that into the energy formula, we arrive at the force: \$F \ = \ \frac{\mu_0 N^2 I^2 A}{2g^2}\$, with the gap length instead of the coil length in the divisor.So the trouble was with taking the flux of a whole air coil, instead of a mixed air/plunger coil. The correct force is inversely proportional to the gap length, as jcw0752 said.