The permeance
Курсовой проект - Физика
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2.CALCULATION OF WINDING
2.1.Determination of final MMF of a winding, taking into account safety factor
task is to calculate the winding knowing next parameters: l=250 mm; h=80 mm; (Iw)=1521.64 A; U=36 V. Winding is made of copper. ?0=4.3•10-3 K-1 is the temperature coefficient of copper; ?0=1.62•10-8 ?•m - is specific resistance of copper; q=105C - is permissible heat temperature of the winding. I determine finally MMF of a winding taking into consideration safety factor kЗ taken from the interval 1.3…1.5.
,(Iw) - is the magnetic motive force of the electromagnet A;
2.2.Calculation of the winding parameters
The resistance of copper ?q can be defined:
,?0=1.62•10-8 ?•m - is specific resistance of copper;
?0=4.3•10-3 K-1 - is the temperature coefficient of copper;=105C - is permissible heat temperature of the winding.length of one turn
,
where h - is the distance between the core and the yoke, m;- is the diameter of the core, m.
but keeping some place in reserve, let
,(Iw)final - is the magnetic motive force of the electromagnet A;
?q - is the resistance of copper, ?•m;в.ср. - is average length of one turn, m;- is voltage, V.one must choose the most suitable diameter of the wire;
Then the cross-section of the wire is to be found:
,
where dw - is the diameter of the wire, m.I find the number of turns in the coil
,
where U - is voltage, V;доп - is current density, A/m2;
?q - is the resistance of copper, ?•m;в.ср. - is average length of one turn, m.the square of the winding is to be found;
,w - is the number of turns in the coil;
q0 - is the cross-section of the wire;з - is the space factor coefficient equaled to 0.656.the width of the winding can be found:
,
where Q0 - is the square of the winding, mm2;- is the length of the yoke, mm.
equivalent circuit magnetic steel
2.3.Determination of pure resistance and MMF of the wire
first, I find the resistance of the coil:
,
where ?q - is the resistance of copper, ?•m;
w - is the number of turns in the coil;
lв.ср. - is average length of one turn, m.
q0 - is the cross-section of the wire, m2.it is possible to calculate the current flowing in the winding:
,U - is the voltage in the winding, V;- is the resistance of the coil.it is necessary to check whether the current density was chosen correctly:
where I - is the current flowing in the winding, A;- is the cross-section of the wire, m2., the density is correct., it is possible to calculate the pure MMF of the coil:
The error is:
3.Determination of the magnetic flux
3.1.Calculation of air-gaps
task is to find the magnetic flux and electromagnetic force through the operating air gap of the electromagnet, figure 1.1., knowing next parameters: a=10 mm; d=25 mm; d2=20 mm; l=250 mm; h=80 mm; ?=?1=0.2 mm; ?2=0.25 mm; winding power voltage U=36 V; magnetic circuit material is stee of quality 50HXC; (Iw)w=764.32 A.bulging fluxes will be neglected as the dimensions of the pole. According to equivalent circuit, figure 2.1., operating and parasitic air-gaps permeances will be determined.air-gap:
,0 - is magnetic constant, or vacuum permeability, ;
d - is the diameter of the core, m;
?п - is the value of the operating air-gap, m.air-gaps:
,0 - is magnetic constant, or vacuum permeability, ;
h - is the distance between the core and the yoke, m;- is the thickness of the core, m;
?п - is the value of the operating air-gap, m;- is the diameter of the core, m.
- it was found in the 1st part.permeance is equal to:
3.2.Determination of initial estimate magnetic flux through the air-gap
to the determination of MMF:
where Ф - is the magnetic flux, Wb;
?? - is the magnetic permeance, H.
In our case we may use for first approximation the next formula:
from which:
Then the induction of the core is found:
where Ф - is the magnetic flux, Wb;
?? - is the magnetic permeance, H;
d - is the diameter of the core, m.the found induction is greater than the saturation induction the saturation one is taken from the graph of magnetic curves:
,then
this graph we also must find magnetic intensity of the steel:
Then the induction of the armature and the base of the yoke is found
,
where Ф - is the magnetic flux, Wb;- is the diameter of the yoke, m;- is the thickness of the core, m.the graph of magnetic curves the magnetic intensity of steel is found:
Then one can find the induction of the yoke:
,
where Ф - is the magnetic flux, Wb;- is the diameter of the yoke, m;- is the distance between the core and the yoke, m.the graph of magnetic curves the magnetic intensity is found:
Now lets find the potential magnetic drop of the iron:
, the magnetic flux can be found:
that everything from the formula 3.1 should be repeated till the relative difference between previous and new approximation of magnetic flux is equal to 5%.
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3.3.Determination of electromagnet force
As ?=0.1 mm, then Maxwells formula will be used for finding electromagnetic pulling force:
,
where Ф - is the magnetic flux, Wb;
0 - is magnetic constant, or vacuum permeability, ;- is the square of the core, m2.
4.Defining of actuation time
calculate actuation time, knowing next parameters:
Then the average magnitude of counteracting force is to be found:
Then the inductance of the winding is to be found:
,
where ??? - is the total permeance of the electromagnet, H;
w - is the number of turns in the coil.the pick-up time of the electromagnet is to be found:
,L - is the inductance of winding, H;- is the resistance of the coil, ?;з - is the assurance factor.lets find the mass of movable parts:
,V - is the volume of the armature, m3;
?st - is the density of steel, kg/m3.
Then the moving time is determined:
,m - is the mass of the movable parts, kg;
?п - is the value of the operating air-gap.time is equal to:
.
Conclusions
electromagnet is simply a coil of wire. It is usually wound around an iron core. However, it could be wound around an air core, in which case it is called a solenoid. When connected to a DC voltage or current source, the electromagnet becomes energized, creating a magnetic field just like a permanent magnet. The magnetic flux density is proportional to the magnitude of the current flowing in the wire of the electromagnet. The polarity of the electromagnet is determined by the direction the current. The north pole of the electromagnet is determined by using your right hand. Wrap your fingers around the coil in the same direction as the current is flowing (conventional current flows from + to -). The direction your thumb is pointing is the direction of the magnetic field, so north would come out of the electromagnet in the direction of your thumb. DC electromagnets are principally used to pick up or hold objects.connected to an AC voltage or current source, the electromagnet will be changing its flux density as the current fluctuates. The polarity of the magnet will also change as the current reverses direction every half cycle. AC electromagnets can be used to demagnet