B Closed Magnetic Circuit The strong magnetic field magnetic separator usually uses a closed magnetic circuit to generate a magnetic field. There are two main closed magnetic path: closed magnetic circuit **iron** core, and "frame" the magnet magnetic circuit.

The typical structure of the core magnetic circuit is shown in Figure 9. The magnetomotive force of the magnetic circuit is calculated by:

Where F _{m} a€”a€”a€”Magnetic potential, A;

da€”a€”a€”air gap width, m;

?1 and ?1 _{y} a€”a€”a€” is the length of the magnetic pole and yoke, m;

Aga€”the effective gap area of a€?a€?the air gap, m ^{2} ;

A, Ay and ?? _{r} , ?? _{ry} - magnetic pole and yoke breaking area, m2 and relative magnetic permeability.

It can be seen from the 44 type that in the properly designed magnetic circuit, the magnetic resistance of the magnetic pole and the yoke can preferably be neglected, and at least should be much smaller than the air gap magnetic resistance. To do this, it is necessary that A _{y} >>Ag, ?1, and ?1 _{y} be as short as possible and that ?? _{r} and ?? _{ry of} all the iron parts of the magnetic circuit are large.

(44) can also be written as follows:

F _{m} =NI=(NI)'(1+p) (45)

Where NIa€”the number of ampereps actually required for the magnetic circuit, A;

(NI)'=Hda€”the number of ampoules required for the air gap, A;

pa€”a€”a€”The ratio of ferromagnetic resistance to air gap reluctance. The magnetic flux leakage factor p is a measure of the magnetic path defect. Once p is reliably determined, the design is sure to proceed. [next]

High gradient magnetic separators often use a "window frame" magnet magnetic circuit (Fig. 10). It consists of a coil and an iron circuit. The magnetomotive force of this magnetic circuit can be calculated by the equation 45. The p value is estimated to be pa‰?0.2.

Design example: As shown in Figure 10, let D _{1} = 2 m, d = 0.5 m, background magnetic field B _{O} = 2 special, then (NI) ' = Hd = (B _{o} / ?? _{o} ) d a‰? 8 ?— 10 ^{5} A

NI=(NI)'(1+p)=8?—10 ^{5} (1+0.2)=9.6?—10 ^{5} A

The distribution of current and number of turns depends on which power supply system is used, ie high current low voltage or low current high voltage. From a technical and economic point of view, it is more reasonable to use a large current and low voltage power supply. The coil is wound with a large current carrying hollow water-cooled **copper** wire. Thus, it is more appropriate to take I=3000 amps and N=320. Taking the current density j=3.2?—10 ^{6} ampere/m ^{2} and the filling factor ??=0.75, the required coil is broken, and the area S=NI/(J??)=0.4 m ^{2} .

Substituting h=d (ie short coil, pole length 0) into equation 47, finds R=5.4?—10 ^{-2} ohm. It can be seen from the equation 47 that as h increases, R decreases and R decreases, and the power consumption and the amount of copper required for the coil decrease. On the other hand, if h is increased, a part of the magnetomotive force will be lost on the magnetization of the magnetic pole head, and the amount of iron required for the magnetic circuit will increase. In this way, the optimal design depends not only on the technical parameters, but also on the price of copper and production and the price of iron and available space. A reasonable compromise is to use a long coil, ie h = 2d. Substitute h = 2d into 47, R = 4.6 ?— 10 ^{-2} ohm. However, the magnetic flux area of a€?a€?the magnetic circuit is 1.5 to 2 times larger than the pole area. For this example, Wa‰?4 meters, Ha‰?2.5 meters.

Thus, the size of the entire magnetic separator is: D _{1} a‰? _{2} m; D _{2} a‰? 2.8 m; W a‰? 4 m; d a‰? 0.5 m; h a‰? 1 m; H a‰? 2.5. The required excitation voltage V = IR a‰? 140 volts; DC power P = I ^{2} R a‰? 420 kW. The heat generated by the coil is cooled by deionized water through a hot changer.

The magnetic force between the two poles of the magnet head is F _{m} = ?€ (B _{o} D _{1} ) ^{2} / 8 (?? _{o} ) a‰? ^{5} ?— 10 ^{5} calves 500 tons. Such a large force requires that the vertical deviation of the magnet frame structure is extremely small.

The self-inductance of the magnet L=D¤N/I=?€??o(D _{1} N) ^{2} /(4d)a‰?0.81 hen, time constant ??=L/Ra‰?17.5 seconds. The time constant indicates that it takes a considerable amount of time to reach the final stable value when the magnet is excited and demagnetized.

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