Design of Magnetic Circuits
3.3 Magnetic Voltage of a Tooth and a Salient Pole .1 Magnetic Voltage of a Tooth
=0.571, bve=κbv=0.571·10 mm=5.71 mm, l≈l−nvbve+2δ =990−24·5.71+2·3 mm=859 mm.
(c) As in (a),κ =0.40,bve=4.0 mm for the stator and rotor ducts,
l≈l−nvsbves−nvrbver+2δ =990−24·4.0−25·4.0+2·3 mm=800 mm.
3.3 Magnetic Voltage of a Tooth and a Salient Pole 3.3.1 Magnetic Voltage of a Tooth
When there areQs slots in the stator, we obtain the slot pitch of the stator by dividing the air-gap periphery by the number of slots
τu =πD Qs
. (3.39)
Figure 3.13a illustrates the flux density distribution in an air gap, the other surface of which is smooth, and Figure 3.13b illustrates a tooth and a slot pitch.
The magnetic voltage of a tooth is calculated at a peak of the air-gap fundamental flux density. When a tooth occurs at a peak value of the air-gap flux density, an apparent tooth flux passes the slot pitch
Φˆd =lτuBˆδ. (3.40)
bd hd
τu
B(θ)
B1
(a) (b)
Figure 3.13 (a) Semi-closed slots and the flux density in the air gap. (b) The dimensions of a tooth and a slot: the heighthdof the tooth and the slot, slot pitchτu, and the width of the toothbd
If the teeth of the machine are not saturated, almost the complete flux of the slot pitch flows along the teeth, and there is no flux in the slots and the slot insulations. Neglecting the slot opening and taking into account the space factorkFe of iron, we obtain for a tooth with a uniform diameter and cross-sectional areaSd
Sd =kFe(l−nvbv)bd. (3.41) Here nv andbv are the number of ventilating ducts and their width (see Figure 3.12), and lis the total length of the machine stack. Punching influences the crystal structure of iron, and therefore the permeability on the cutting edges of the tooth is low. Thus, in the calcula-tion of the flux density in a tooth, 0.1 mm has to be subtracted from the tooth width, that is bd=breal−0.1 mm in Equation (3.41) and the following equations. During the running of the motor, a relaxation phenomenon occurs, and the magnetic properties recover year by year.
The space factor of ironkFedepends on the relative thickness of the insulation of the electric sheet and on the press fit of the stack. The insulators are relatively thin, their typical thickness being about 0.002 mm, and consequently the space factor of iron can in practice be as high as 98%. A space factor varies typically between 0.9 and 0.97. Assuming that the complete flux is flowing in the tooth, we obtain its apparent flux density
Bˆd = Φˆd
Sd = lτu
kFe(l−nvbv)bdBˆδ. (3.42) In practice, a part of the flux is always flowing through the slot along an areaSu. Denoting this flux by ˆΦu, we may write for a flux in the tooth iron
Φˆd =Φˆd−Φˆu =Φˆd−SuBˆu. (3.43)
By dividing the result by the area of the tooth ironSdwe obtain the real flux density of the tooth iron
Bˆd=Bˆd −Su Sd
Bˆu, where Su Sd
= lτu
kFe(l−nvbv)bd
−1. (3.44)
The apparent flux density of the tooth iron ˆBd can be calculated when the peak value ˆBδ of the fundamental air-gap flux density is known. To calculate the flux density in the slot, the magnetic field strength in the tooth is required. Since the tangential component of the field strength is continuous at the interface of the iron and the air, that isHd =Hu, the flux density of the slot is
Bˆu =µ0Hˆd. (3.45)
The real flux density in the tooth is thus
Bˆd =Bˆd−Su
Sdµ0Hˆd. (3.46)
Now, we have to find a point that satisfies Equation (3.46) on theBHcurve of the electric sheet in question. The easiest way is to solve the problem graphically as illustrated in Figure 3.14. The magnetic voltage ˆUm,din the tooth is then approximately ˆHdhd.
When a slot and a tooth are not of equal width, the flux density is not constant, and therefore the magnetic voltage of the tooth has to be integrated or calculated in sections:
Uˆm,d=
hd
0
Hd·dl.
B
H
d d 0
' u
d ˆd ˆ
ˆ H
S S
B µ
B = –
ˆd
B
d' Bˆ
d
Hˆ
Figure 3.14 Definition of the flux density ˆBdof the tooth with theBHcurve of the electrical sheet and the dimensions of the tooth
Example3.3: The stator teeth of a synchronous machine are 70 mm high and 14 mm wide. Further, the slot pitch τu = 30 mm, the stator core lengthl =1 m, there are no ventilation ducts, the space factor of the corekFe=0.98, the core material is Surahammars Bruk electrical sheet M400-65A (Figure Example 3.3), the air gap δ = 2 mm and the fundamental flux density of the air gap ˆBd=0.85 T. Calculate the magnetic voltage over the stator tooth.
2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0
0 2 4 6 8 10 12 14 16 18
B/T
H/kA/m 13 .3kA /m
ˆd= H
Figure Example 3.3 BHcurve of the material used in Example 3.3 and its solution
Solution: The effective core lengthl=1000+2·2 mm=1004 mm. The apparent flux density of the tooth Equation (3.42) is
Bˆd = 1004·30
0.98·1000·(14−0.1)0.85 T=1.88 T.
The intersection of theBHcurve of the electrical sheet M400-65A and the line (3.46) gives Bˆδ=1.88− 1004·30
0.98·1000·(14−0.1)−1
4·π·10−7Hˆd.
The figure above gives the field strength of the teeth ˆHd=13.3 kA/m and the magnetic voltage
Uˆm,d=
hd
0
Hd·dl=13 300·0.07 A=931 A.
3.3.2 Magnetic Voltage of a Salient Pole
The calculation of the magnetic voltage of a salient pole is in principle quite similar to the procedure introduced for a tooth. However, there are certain particularities: for instance, the magnetic flux density of the pole shoe is often so small that the magnetic voltage required by it can be neglected, in which case only the magnetic voltage required by the pole body has to be defined. When determining the magnetic voltage of the pole body, special attention has to be paid to the leakage flux of the pole body. Figure 3.7 shows that a considerable amount of the flux of the pole body is leaking. The leakage flux comprises 10–30% of the main flux. Because of the leakage flux, the flux density ˆBp at the foot of a uniform pole body of cross-sectional areaSpis thus written with the main flux ˆΦmas
Bˆp=(1.1. . .1.3)Φˆm
Sp . (3.47)
The calculation of the peak value of the flux ˆΦmwill be discussed later at the end of the chapter. Because of the variation of the flux density, the magnetic voltage ˆUm,p of the pole body has to be integrated:
Uˆm,p=
hdr
0
H·dl.