11 magnetic circuit
Post on 04-Jun-2018
228 Views
Preview:
TRANSCRIPT
-
8/13/2019 11 Magnetic Circuit
1/31
-
8/13/2019 11 Magnetic Circuit
2/31
Magnetic Flux density B
Unit for magnetic flux density is Tesla
The definition of 1 tesla is the flux density thatcan produce a force of 1 Newton per meter acting
upon a conductor carrying 1 ampere of current.
Magnetic Flux
Unit for flux is weber
The definition of 1 weber is the amount of flux
that can produce an induced voltage of 1 V in a oneturn coil if the flux reduce to zero with uniform rate.
Magnetic field strength H
Unit for magnetic field strength is Ampere/m
A line of force that produce flux
-
8/13/2019 11 Magnetic Circuit
3/31
F = B l I newton
Where F = force ; B = magnetic flux density ; l =the length of
conductor and I = current in the conductor
ABWhere = magnetic flux ; B = magnetic flux density and A =
area of cross-section
HB Where H= magnetic field strength ; B = magnetic flux density
and = permeability of the medium
o= B/H = 4x 10-7 H/mPermeability in free space
-
8/13/2019 11 Magnetic Circuit
4/31
Relative permeability is defined as a ratio of flux densityproduced in a material to the flux density produced in a
vacuum for the same magnetic filed strength. Thus
Relative Permeability (r)
r = /o
= ro = B/H
or B = roH
-
8/13/2019 11 Magnetic Circuit
5/31
H vs r
Relative
permeability
Magnetic field strength
Cast iron
Mild steel
-
8/13/2019 11 Magnetic Circuit
6/31
Relative permeability vs magnetic flux
-
8/13/2019 11 Magnetic Circuit
7/31
B vs H
-
8/13/2019 11 Magnetic Circuit
8/31
Electromagnetic Force (mmf)
turns
H
NIH mmf
Where H= magnetic field strength ; l =the path length of ; N
number of turns and I = current in the conductor
-
8/13/2019 11 Magnetic Circuit
9/31
Example 1A coils of 200 turns is uniformly wound around a wooden ring
with a mean circumference of 600 mm and area of cross-section
of 500 mm2. If the current flowing into the coil is 4 A, Calculate
(a) the magnetic field strength , (b) flux density dan (c) total flux
N = 200 turns
l = 600 x 10
-3
mA = 500 x 10-6m2
I = 4 A
(a) H = NI/l = 200 x 4 / 600 x 10-3 = 1333 A
(b) B = oH = 4x 10-7x 1333 = 0.001675 T = 1675 T
(c) Total Flux = BA = 1675 x 10-6x 500 x 10-6
= 0.8375 Wb
turns
-
8/13/2019 11 Magnetic Circuit
10/31
Ohmslaw I = V/R [A]
Where I =current; V=voltage and R=resistance
And the resistance can be relate to physical parameters as
R = l/A ohm
Where =resistivity [ohm-meter],l=
length in meter and A=areaof cross-section [meter square]
Analogy to the ohmslaw
V=NI=H l I= and R=S
Reluctance ( S )
weberS
Hl weberampere
AS
or
/
where
-
8/13/2019 11 Magnetic Circuit
11/31
Example 2
A mild steel ring, having a cross-section area of 500 mm2and a
mean circumference of 400 mm is wound uniformly by a coilof 200 turns. Calculate(a) reluctance of the ring and (b) a
current required to produce a flux of 800 Wb in the ring.
TA
B 6.110500
10800
6
6
Dari graf r/B, pada B = 1.6;
r = 380
turns
47 105104380
4.0 AS or
]/[10667.1 6 WbA
(a)
-
8/13/2019 11 Magnetic Circuit
12/31
(b)
SH SH
NIAH
][134210667.110800
66mmf
][7.6200
13421342A
NI
-
8/13/2019 11 Magnetic Circuit
13/31
Magnetic circuit with different materials
11
1Aa
S l
22
2B
aS
l
BA SSS 22
2
11
1
aa
ll
and
For A: area of cross-section = a1mean length = l1absolute permeability = 1
ForB: area of cross-section= a2mean length= l2absolute permeability= 2
-
8/13/2019 11 Magnetic Circuit
14/31
Mmf for many materials in series
total mmf = HAlA + HBlB
HA=magnetic strength in material A
lA=mean length of material A
HB=magnetic strength in material B
lB=mean length of material B
In general
(m.m.f) = Hl
-
8/13/2019 11 Magnetic Circuit
15/31
Example 3
A magnetic circuit comprises three parts in series, each of
uniform cross-section area(c.s.a). They are :
(a)A length of 80mm and c.s.a 50 mm2;
(b)A length of 60mm and c.s.a 90mm2;
(c)An airgap of length 0.5mm and c.s.a 150 mm2.
A coil of 4000 turns is wound on part (b), and the flux density
in the airgap is 0.3T. Assuming that all the flux passes throughthe given circuit, and that the relative permeability r is 1300,
estimate the coil current to produce such a flux density.
WbAB CC 44 1045.0105.13.0
-
8/13/2019 11 Magnetic Circuit
16/31
mAAI 4.45104.454000
8.181 3
AtAS aor
aa 1.4410501041300
10801045.067
34
AtSSSNIcba
8.1813.1194.181.44
and
Mmf = S = H l = N I
AtA
Sbor
bb 4.18
10901041300
10601045.067
34
AtA
Scor
cc 3.119
101501041300
105.01045.067
34
Material a
Material b
airgap
Total mmf
-
8/13/2019 11 Magnetic Circuit
17/31
Leakages and fringing of flux
Some fluxes are leakage via paths a, b and c . Path d is
shown to be expanded due to fringing. Thus the usable
flux is less than the total flux produced, hence
Magnetic circuit with air-gap Leakages and fringing of flux
fluxusable
fluxtotalfactorLeakage
fringing
leakage
-
8/13/2019 11 Magnetic Circuit
18/31
Example 4
A magnetic circuit as in Figure is made
from a laminated steel. The breadth of
the steel core is 40 mm and the depth is
50 mm, 8% of it is an insulator
between the laminatings. The length
and the area of the airgap are 2 mm and2500 mm2 respectively. A coil is
wound 800 turns. If the leakage factor
is 1.2, calculate the current required to
magnetize the steel core in order to
produce flux of 0.0025 Wb across the
airgap.
-
8/13/2019 11 Magnetic Circuit
19/31
TA
Ba
a 1102500
105.26
3
]/[796000104
17 mAT
BH
o
aa
factorleakageairgapinfluxfluxTotal T
][1594002.0796000 ATHmmf
Wb003.02.10025.0
92% of the depth is laminated steel, thus the area of crosssection is
AS = 40 x 50 x 0.92 = 1840 mm2=0.00184m
-
8/13/2019 11 Magnetic Circuit
20/31
TA
BS
TS 63.1
00184.0
103 3
From the B-H graph, at B=1.63T, H=4000AT/m
mmf in the steel core = Hl = 4000 x 0.6 = 2400 AT
Total mmf. = 1592 + 2400 = 3992 AT
I = 3992/800 = 5 A
NI = 3992
-
8/13/2019 11 Magnetic Circuit
21/31
Mmf in loop C = NI = HLlL + HMlMoutside loop NI= HLlL+ HNlNAnd in loop D 0 = HMlM+ HNlNIn general (m.m.f) = Hl
Magnetic circuit applying voltage law Analogy to electrical circuit
D
applying voltage Kirchoffs law
-
8/13/2019 11 Magnetic Circuit
22/31
L = M+ N
Or L- M- N = 0
In general: = 0
At node P we can also applying current Kirchoffs law
P
L M N
E
Q
IL
IN
IM
-
8/13/2019 11 Magnetic Circuit
23/31
Example 5
340
mm
340
mm 150
mm
1mm
A magnetic circuit made of silicon steel is arranged as in the
Figure. The center limb has a cross-section area of 800mm
2
andeach of the side limbs has a cross-sectional area of 500mm2.
Calculate the m.m.f required to produce a flux of 1mWb in the
center limb, assuming the magnetic leakage to be negligible.
-
8/13/2019 11 Magnetic Circuit
24/31
aBAA SSSfmm 21..
AB TA
B 25.110800
1016
3
159151050010434000
1034067
3
1
11
AS
or
Looking at graph at B=1.25T r=34000
Apply voltage law in loops A and B 340mm
340
mm 150
mm
1mm
AB
43881080010434000
10150
67
3
2
S
99471810800104
10167
3
aS
-
8/13/2019 11 Magnetic Circuit
25/31
10079998
994718438810115915105.0.. 33 fmm
Since the circuit is symmetry A=B
In the center limb , the flux is 1mWb which is equal to 2
Therefore =0.5mWb
aSSSfmm 21 2..
A
-
8/13/2019 11 Magnetic Circuit
26/31
Weight
Iron yoke
KeeperAir gap
Example 6
A U-shaped electromagnet shown in
Fig. is designed to lift a mass. The
material for the yoke has a relativepermeability of 2900. The yoke has a
uniform cross-sectional area of 4000
mm2and a mean length of 600 mm.
Each of the air gaps is 0.1 mm long.
The number of turns of the coil (N) is
240. Assuming that the reluctance of
the keeper is negligible, calculate the
maximum mass in kg, which can be
lifted by the system if a current of 1.5 Ais passed through the coil. You may
neglect the fringing effect and flux
leakage; and assume that
-
8/13/2019 11 Magnetic Circuit
27/31
aB
a
a
o
aa
aa
B
B
lBlH
2
10
104
101.02
22
3
7
3
aa
or
iaii B
BlBlH
6.11
106
1042900
10600 3
7
3
Calculation of maximum weight lifted by and electromagnet.
Let the flux density in the air gap be
For the air gap
For the iron yoke
-
8/13/2019 11 Magnetic Circuit
28/31
NIBBlHlHHl aaiiaa
8.32310
6.11
6
2
1 3
T11.18.323
360
8.323
NIBa
N3922104
10400011.1
22
7
6222
o
a
o
a ABABF
3922mgkg400
81.9
39223922
gm
Total mmf;
Since there are two air gaps;
-
8/13/2019 11 Magnetic Circuit
29/31
Hysteresis loss
Materials before applying m.m.f (H), the polarity of
the molecules or structures are in random.
After applying m.m.f (H) , the polarity of themolecules or structures are in one direction, thus the
materials become magnetized. The more H applied
the more magnetic flux (B )will be produced
-
8/13/2019 11 Magnetic Circuit
30/31
When we plot the mmf (H) versus the magnetic flux will produce a
curve so called Hysteresis loop
1. OAC when more H applied, B
increased until saturated. At this
point no increment of B when we
increase the H.
2. CD- when we reduce the H the B
also reduce but will not go to zero.
3. DE- a negative value of H has to
applied in order to reduce B to zero.
4. EF when applying more H in the
negative direction will increase B in
the reverse direction.
5. FGC- when reduce H will reduce B
but it will not go to zero. Then byincreasing positively the also
decrease and certain point it again
change the polarity to negative until
it reach C.
-
8/13/2019 11 Magnetic Circuit
31/31
Eddy current
metal insulator
When a sinusoidal current enterthe coil, the flux also varies
sinusoidally according to I. The
induced current will flow in the
magnetic core. This current is
called eddy current. This
current introduce the eddy
current loss. The losses due to
hysteresis and eddy-core totally
called core loss. To reduce eddycurrent we use laminated core
top related