{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Power Notes 6

# Power Notes 6 - Unit 05 Magnetic Fields 7& Magnetic...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Unit 05 Magnetic Fields 7& Magnetic Circuits Electro-mechanical energy conversion and energy distribu-tion in power systems is enabled by magnetic ﬁelds created by magnetic circuits. In such a sense, these ﬁelds and circuits are the main components of electrical power systems. They are essential components of generators, transformer and motors. The creation of magnetic ﬁeld by electric current was discovered by Oersted in 1820 and the creation of the electric ﬁeld by a change of magnetic ﬂux was discovered by Faraday in 1931. These two discoveries have created the very physical fundamentals of electrical power systems. The creation of the magnetic ﬁeld, of intensity I? , is governed by two laws, Biot-Savart Law and Ampere Law. The Biot—Savart Law. When a point of space has a distance 3 ﬁ'om a conductor element oriented in direction :17 with a current i then a magnetic ﬁeld intensity dﬁ = Lian—1' =Lidl sin(dl,d)’ 472' d3 472' d2 is created at such a point. Current loop Fig. 1 The vector of the magnetic ﬁeld intensity is perpendicular to the plane of vectors :1? , and the distance ti". Orientation of this vector can be detemrined by the right-lulu ded screw rule. [if]. I' - / d id? _. displacement d . 51' h Rotation I Fig.2 Ifthe ﬁrst elementefaaeveeterpreaueuﬁhsremeed on tothe second element (a?) of the product then the result of the vector product is oriented towards the direction of displacement of the right-handed screw (Fig. 2a). This rule can be applied also in a different way. To displace a right-handed screw in the direction of the current ﬂow, it has to be rotated in the direction of the magnetic ﬁeld intensity, H, (Fig. 2b). Magnetic ﬁeld intensity, H, has dimension [A/m], but its unit does not have any name. To calculate the magnetic ﬁeld intensity H created by a current loop at some point of space, integration along the entire current loop is needed. On the condition that the current loop is on a plane and the magnetic ﬁeld intensity is calculated at same point of the same plane and is is the plane unit vector (a unit vector perpendicular to the current loop plane) a = g m = It; a} Loop Loop d Since the distance d and the angle between the loop element d1i,=Hi,. d1 and the distance vector d changes along the loop, this integral could be difﬁcult for calculation. It is easy to calculate it only if geometry of the loop is simple. Illustration 1. Calculate the magnetic ﬁeld intensity in the center of a circular current loop of diameter of D = 2R = 0.5m with the current i = 500 A. ' m Fig. 3 When the magnetic ﬁeld intensity is calculated in the center of a circuit as shown in Fig. 3, then the distance d = D/2 and the angle between the loop element 37 and the distance vector d is constant and equal to 90°. Thus, . ' 0 H: qu=4LSmg° C'Fdlé Loop 7t R Loop =_i_; =_i_=_500:_= 47: Ram 2R 2x025 1000M“ MAMA/\AAAAAAAMAMAAMA The Ampere Law. It says that the integral along a closed path of the magnetic ﬁeld intensity H is equal to the current ﬂux enclosed by this path. In this integral the symbolfdenotes the vector of current density in [Almz] and ds'is the vector of the surface element. 41%;: may: Path Area Fig. 4 The direction of integration along the path and the direction of the current ﬂux across the area satisfy the right-handed screw rule. When the integration path in the Ampere Law encloses N1 conductors with current i1, ﬂawing in direction that satisﬁes the right-handed screw rule and N2 conductors with current i; ﬂowing in the opposite direction, then the Ampere Law for any path around these conductors can be written in the form qﬁdi=NllI1—N2i2 . Path Illustration 2. Calculate the magnetic ﬁeld intensity H in the distance d = 0.25m from a conductor with the current i = 500 A. r r? Fig. 5 Solution. The vector of magnetic ﬁeld intensityH at any point of the circle with the current in its center has the same magnitude, H, and it is tangent to the circle. Hence, cjﬁdi=H cfd1=2mH = i. Path Path Thus .; ___&. = H‘ 2727 _27rx0.25 3184M” AAMMAAAAAAAAAAAAAAAAAAAAA The mechanical F exerted on electric charge q in electric and magnetic ﬁelds are governed by the Lorenz Law. The Lorenz Law. If an electric charge dq moves in the electromagnetic ﬁeld with velocity ii then a force is exerted on such a charge. If electric ﬁeld is speciﬁed by the vector of the electric ﬁeld intensity E and the magnetieﬁeld is speciﬁed by the vector of magnetic ﬂux density, B , then the force is equal to d13=dqa§+r XE). The magnetic ﬂux density is measured in Teslas [T]. It is a vector, tangent to the magnetic ﬂux, ¢, line, with the magnitude M B = g: . The magnetic ﬂux, ¢, is measured in Webers [Wb] and cones- quently, the magnetic ﬂux density, [T] = [Wb/mz]. =l‘he magnetic component ofthe force exerted on ameving charge di ~ dq dﬁm=dq17x3=quXB=E-dlx§=idlx§. The magnetic ﬂux density li’ in vacuum, and approximately in air and in a great majority of materials, is proportional to the magnetic ﬁeld intensity, H B=ﬂoH. The symbol ,uo denotes magnetic permeability of free space and it is equal to yo = 47r><10'7 [H/m]. The magnetic ﬂux density B in iron, cobalt and nickel and in alloys of these metals with each other, depends on the magnetic ﬁeld intensity H thousands time stronger than in vacuum. Such materials are called ferromagnetics. For such materials B=y,uoH, where y is referred to as a relative permeability. Its approximate value for some ferromagnetics: - annealed iron: ,u = 5 500 - iron-silicon laminations (96% Fe, 4% Si) ,u = 7 000 - permalloy(55%Fe, 45% Ni) ii = 25 000 - suppermalloy y = 100 000 Due to electron spin and their rotation around the nucleus, individual atoms form elementary magnetic dipoles. In a great majority of materials, these dipoles are oriented randomly and consequently, their net effect is zero. Ferromagnetism is caused be an alignment of elementary magnet dipoles in material microstructine. They form domains. These dipoles in a domain have the same orientation, different from the orientation in neighboring I \\ Mas- Fig. 6 External magnetic ﬁeld enlarges and rotates individual magnetic domains in the direction of the external ﬁeld intensity, causing a magnetic ﬂux to occur. This process is strongly nonlinear, speciﬁed in terms of hysteresis loop. The reorientation of magnetic domains remains to some degree permanent. Even if the external magnetic ﬁeld disappears, the ferromagnetic remains magnetized. It is speciﬁed by the residual magnetic ﬂux density, Br Negative magnetic ﬁeld is needed for material re-magnetization. It is speciﬁed by the magnetic ﬁeld intensity coersion, Hc. When all domains are oriented towards the external magnetic ﬁeld, the ferromagnetic material becomes saturated and the relative permeability p declines to unity. B 5' Fig. 7 Mechanical work is needed for changes of magnetic domains structure. This work at one eyele of ehange of ﬁeld intensity is proportional to the area of the hysteresis loop. When the ﬁeld intensity changes periodically, then energy loss in the ferromagnetic material increases with the area of the loop and with ﬁequency. This energy loss is observed as an increase of the material temperature. Therefore, magnetic circuits used for creation of variable magnetic ﬂux (for example, in transformers) should be built of ferromagnetic materials with the hysteresis loop area as small as possible. Such ferromagnetics are referred to as soft ferromagnetics. The hysteresis loop of a soﬁ ferromagnetic material is shown in Fig. 7. Ferromagnetic material for a permanent magnet should fulﬁll different requirements. It should have the residual magnetic ﬂux density, 3,, as high as possible. The higher E, the stronger magnet could be formed. A permanent magnet in an external magnetic ﬁeld can loose its magnetization, however. This unfavorable feature declines with the increase of the magnetic ﬁeld intensity coersion, Hc of the ferromagnetic material. The higher coersion intensity the better is the material for a permanent magnet. Ferromagnetics with high residual ﬂux density and high coersion intensity are referred to as hard ferromagnetics. Table l. The values of Br and Hc of ferromagnetic materials used for permanent magnets 0-4 Neod ’ 'um-iron-boron allo 1.25 Tesla 1.25 50 ~o~.—- E UI\IUI : The hysteresis loop of a hard ferromagnetic material is shown in Fig. 8. Fig 8 Magnetic circuits. At the same value of the magnetic ﬁeld inten— sity H, the ﬂux density B in ferromagnetic materials is p—time higher than in air. Consequently, magnetic ﬂux is mainly con-ﬁned to ferromagnetic material. If this material is formed to create a loop or loops for the magnetic ﬂux, a magnetic circuit is created. Winding with a current is usually the main source of the magnetic field and the ﬂux in such a magnetic circuit. A permanent magnet or external magnetic ﬁeld could serve as such a source as well. Fig. 9 The magnetic ﬂux a) in the circuit shown in Fig. 9a has the same value along the entire magnetic loop. Consequently, the ﬂux density _ A , does not change along the loop. Since the magnetic permeability changes along the ﬂux loop, changes also the magnetic ﬁeld intensity. Its value in the ferromagnetic core is Hc=——B , ##o andintheairgap B s #0 The Ampere Law for the average path along the core deI = Hcl+HgA =Ni. path When the magnetic field intensity is expressed in terms of magnetic permeability y no, ﬂux (D and the core cross-section areaA, then 4’ [+3441 =Ni. A ##o A .110 The term I R = —— , c ##o A is referred to as a reluctance of the magnetic core, while A R 3—!!014’ is the reluctance of the air-gap. Consequently, the Ampere Law for the magnetic circuit shown in Fig. 10a can be written in the form d>(Rc +Rg) = Ni. Consequently, the magnetic ﬂux in such a circuit can be calcu-lated from formula Ni (D = —. Rc + Rg Illustration 3. A magnetic core made of silicon-iron laminations with magnetic permeability ,u = 7000 has the length l = 0.5m and the cross-section area A = 100 cm2. The air gap A = 1 mm. The core saturates at the ﬂux density BM = 1.2 T. Calculate the maximum current in the winding with N = 100 turns that do not saturate the core. Solution. To avoid saturation the magnetic ﬂux should not be higher than em = ABsat = 0.01 [m2] x 1.2[Wb/m2] = 12 x10'3 Wb. The reluctance of the core is c=——-=—————-—_7——=5700A/Wb. ##oA 7000x47rx10 x0.01 The reluctance of the air-gap is —3 g=__4 =—__—1X1_9 =79500A/Wb. ﬂoA 471x10 x0.01 Observe, that the reluctance of 1 mm long air-gap is almost 14 times higher than the core reluctance. From the Ampere Law -3 ism = ‘13:;ch +le) = Lag—(5.7 +79.5)x1o3 = 10.2A. AMMAAAAAAAAAMAAMAAAM The Faraday Law. Change of the magnetic ﬂux (I) in a loop induces voltage e in such a loop. When the induced voltage e and the ﬂux CD are mutually oriented according to right-hand screw rule as shown in Fig 1 l, then the induced voltage is _ dd! 8 — "' '2‘— . thus, if the ﬂux in the loop increases, then the voltage induced in the loop is oriented towards terminal b, i.e., opposite to the direction marked by the arrow. Orientation of the voitage induced by a change of the magnetic ﬂux could be determined by the Lenz Rule. It says, that the voltage inducedby a change of the magnetic ﬂux has such orientation, that the current caused by this voltage creates a magnetic ﬂux ((1),) that counteract this change. Fig. 12 It means, that orientation of the magnetic ﬂax and the induced voltage satisfy a left—hand screw rule. At such orientation e=— dt ' When a loop with magnetic ﬂux is composed of N turns, as shown in Fig. 13, Fig. 13. then the induced voltage is proportional to the rate of magnetic ﬂux change and the number of turns, d (D e — N -dt— . Observe, that if the current in the coil increases, i.e., di/dt > 0, then the arrow e speciﬁes true orientation of the voltage induced in the coil. Since the magnetic ﬂux could be expressed as Ni ¢—ﬁ-. Thus, _ R dt ' dt’ where N2 =?’ is the inductance of the coil. It increases with the square of turn number N and declines with the core reluctance. For a core without air gap A = ill/:0 N2, thus, it increases with the core magnetic permeability ,u and cross- section area, A. Hysteresis power loss. When a coil with a magnetic core as shown inFig. 14 Fig. 14. is connected in AC circuit then the instantaneous power at the coil terminals is equal to ._ d H114 _ __ p(t)rul—(NE¢)(7) dt(AB)Hl—VH dt, where V=Al is the magnetic core volume. Thus, the instantaneous power is proportional to the change in the magnetic ﬂux density, dB, at magnetic intensity H, i.e., to the area HdB as shown in Fig. 10. Due to hysteresis, magnetic ﬁeld intensityH at the same ﬂux density B is higher when the core is magnetized than when it is de- magnetized. Censequently, the area of HdB is larger at dB magnetization than at de-magnetization. The difference is released as a heat of the core increase. Fig. 15 The active power at the coil terminals for one cycle T of the current variability, i.e., when B and H change in the whole range shown in Fig. 15, is equal to T T 1 1 1 P=T Jp(t)dt =V? [HdB=?VABH = ABHVf= APh, 0 0 thus it is proportional to the area ABH of the hysteresis loop, the core volume V and the supply voltage frequency, f. This power is referred to hwteresispawer loss. Since soft ferromagnetics have much smaller hysteresis loop area, ABH, than hard ferromagnetics, coils with soft ferromagne-tics have lower power loss than coils with hard ferromagnetics. Therefore, magnetics with a hysteresis loop as narrow as possible are used for AC coils. When sinusoidal voltage u=\/E’Ucoswt is applied to the coil with a magnetic core, then the ﬂux density B in the core has to satisfy F araday’s Law dd) dB u—Nw—NA—dr hence, , _ 1 _J§U _ . _ . —m udt— NA coswtdt—NAwsmwt—Bmxsmmt, then the ﬂux density B in the core has to change as a sinusoidal function, with J3 U NAa) Thus, the RMS value U of the voltage applied to the coil cannot be higher than <Bsa. Bmax = a) J5 N A 3m = 4.44NAme < 4.44NAstat Fig. 16 \ ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

Power Notes 6 - Unit 05 Magnetic Fields 7& Magnetic...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online