Puled Magnet Crimping Fred Niell 4/5/00 1 Magnetic Crimping Magnetoforming i a metal fabrication technique that ha been in ue for everal decade. A large capacitor bank i ued to tore energy that i ued to bend, weld, or otherwie work a piece of metal. In the following example, we will look at a magnetoforming apparatu. The apparatu ue a capacitor and a park gap to tranfer a large amount of energy into a work coil. A thin tube of Aluminum i placed in the work coil. The work coil exert ome force on the tube, and the tube i crimped radially around a plug placed in it open end (ee diagram). Below i an electrical chematic of the apparatu, and a decription of what we are going to do to analyze thi machine. Firt, we will decribe the action of the crimper, Second, we will derive an expreion for the current a a function of time in the coil, the reonant frequency, the peak current, and the kin depth of the Aluminum tube with the emitted EM radiation, Lat, our tak remain to derive an expreion for the preure a a funtion of time, the peak preure, and a imple expreion for the peak preure baed on the phyical parameter of the apparatu. 1. Crimping effect If we conider the Al tube a a um of infiniteimal look tacked up inide a larger um of current loop (the inductor), we can ue Faraday law of induction and Lenz law to explain the crimping effect. If we create a large magnetic induction B with the olenoid over a relatively hort time period with a large I, we create a large B. Faraday how that E = B (1) So an E field i created whoe direction i uch that an emf i induced in the tube. Auming the tube i roughly the ame ize a the olenoid (all of the field goe through the tube), the induced emf in the wall of the tube will have the folowing form: 1
E = M I where the I i the changing current in the inductor, and M the mutual inductance of the ytem. Thi emf (E) give rie to the induced current. Note, however, that from E, effectively Lenz Law, the induced current flow in the oppoite direction of the original current. Thi induced current lead to a new B field, oppoing the original B field. Thee two oppoing induction exert a repulive force on each other. The force exerted by the olenoid on the tube will be much greater than the force exerted by the tube on the olenoid. Thu, the tube feel a force radially inward, giving rie to the crimping effect.. Current, Peak current, Frequency, Skin Depth, etc. The differential equation for thi circuit ha the following form: V o R b i 1 1 C t 0 () i 1+ dt L i R i = 0 (3) Where V o i the initial charging voltage of the capacitor, i 1 i the current in loop 1, and i i the current through loop. We have ued Kirchoff law and Ohm law implicitly. At the peak value, the current through R b i only 0.5mA, and the current through R i imilarly derived to be everal order of magnitude higher, o we will neglect the effect of R b in the differential equation to maintain anity. Now, we have With the caviat V o R 1 C t 0 idt L i = 0 (4) L i + R i + i C = 0 (5) The general olution i found to be i(t) = e Rt V o ω 0 L inh R t 1 4L RC (6) But, we have 4L > 1, o our general olution reduce to R C i(t) = e Rt V o ω o L in ω ot (7) Note that thi ha the form we would expect- a ine function with an exponential decay. Alo notice that ω o i the reonant frequency of the ytem. The reonant frequency ha the following form:
ω o = 1 LC R 4L (8) With the familiar relation f = ωo. Plugging in the number given, we find the frequency π of ocillation to be ω o =181811.8 rad =8.94 khz. For the i max we mut take the derivative of i(t) and et it to zero- giving u the time when the current reache a maximum. Thu, we have i(t) = 0 = R e Rt V 0 ω 0 L in ω 0t + e Rt V 0 L co ω 0t (9) Thi give u a time of 8.1 10 6 econd, which correpond to a maximum current of 1.6 10 4 Amp, or 16kA. From the damping factor of a tranmitted EM wave in a diperive medium, i.e. e βz with z the direction of propagation, we find that the ditance the wave travel before falling to 1 e of it original value i given by z = β 1, which i generally referred to a the kin depth, δ. δ = c πµωo σ (10) Above, c i the peed of light, µ i permiability of the material (µµ 0, which in our cae i imply the permiability of free pace, and σ i the conductivity of the medium, or 1 ρ. Plugging in the given number, we get δ =4.97x10 4 meter, or 0.5 mm. 3. Preure a a function of time The magnetic preure can be derived in a number of way. I will outline two, a the anwer wa upriing to me (independent of the phyical dimenion). Firt we will look at virtual work method. I will work in MKS unit: B U magnetic dτ = B (volume) = B πr l (11) µ 0 µ 0 µ 0 with l, and r being the phyical dimention of the urface we want to calculate the energy inide of. Let u call the ˆr direction the radial direction. We know that force i defined a F U. Therefore, we can define a magnetic force in the following manner: F magnetic U magnetic = U magnetic ˆr = B πrl (1) r µ 0 Notice that the force i in the ˆr direction. Thi mean that the force i directed radially inward. So, the force will be puhing the tube in an inward, cruhing manner. We can define a preure 3
which i an intereting reult. f magnetic F magnetic area = F magnetic πrl Not believing it, I tarted from the tre tenor T ij. = B µ 0 (13) T ij ɛ(e i E j 1 E ij ) + 1 µ 0 (B i B j 1 B ij ) (14) For our ytem i = j and we are only intereted in the B field, o T ij reduce to: T i=j = B µ 0 (15) Force i the time derivitive of momentum of the field, and i given a p i=j Applying the divergence theorem = j v x j T i=j d 3 x (16) Thi give u p i=j = x j T i=j ˆnda (17) F magnetic = T i=j ˆnda = B µ 0 (area) (18) Which i the exact ame reult we got earlier. The magnetic preure i independent of the actual ize of the urface. The only place the phyical dimention of the urface the preure i exerted on i preent in the B field. Getting back to our problem, the magnetic field from an ideal olenoid i given a B olenoid = µ 0 ni(t)ẑ (19) where n i the number of turn per unit length, n = N. Here, the field i only dependent l on the number of turn and the total length of the coil. Now we have the preure a a function of time: f magnetic = µ 0 n I (t max ) = µ 0n e Rt Vo L ωol (in ω ot) (0) And the peak preure i calculated uing the peak current a calculated above: f peak = 40 MPa = 5805 pi. Thi i plenty of preure. It i certainly enough to cruh Coke can, if not more. Quarter placed in thi olenoid would alo probably hrink. 4
In order to get a good etimate for the actual preure, we can make a few approximation. We know that the in term of the preure will reach maximum when the t = t max. So, we let that equal 1. Alo, we know that the exponential term will be very cloe to 1, ince the t max i o mall. We are left with: f magnetic µ 0n Vo (1) ωol Now, ince we are till making approximation we can ay that ω 0 ( LC) 1. Plugging thi in we have: f magnetic = µ 0n V o C = µ ( ) 0C N ( ) l V C V0 o = () πl R for a olenoid. Notice that the radiu of the coil R i in the denominator. Any error in that meaurment could yield very different preure. Perhap the better method of predicting the preure i with the inductance meaured directly, perhap the next-tolat relation. Alo, thi i independent of the number of turn in the coil. To maximize preure, one would have to minimize the radiu firt, then the length. 5