5. energy conversion
Week 15
5.12 ALIGNMENT FORCE AND TORQUE: SINGLE EXCITATION
Fig: 2.8 Reluctance motor
The reluctance motor shown in figure above depend on the tendency of the rotor to
Align itself magnetically with the stator. A flux plot for the machine shows that,
provided there is ad equate ove4rlop, all the active flux and all the field energy can be
assumed to occupy the overlap regions. The active gap volume changes with angle Q
between the two magnetic exes, and the torgue is d e f/dQ eqtn 2.31 using equation
2.22 a basis. An angular increase Dq others the active volume of each gap by - rlg
dQ and the gap energy by -1/2 B2 Lr Dq/Uo
the torgue for the two poles is consequently
me
=
dwe
Dq
=
B2Lrlg
uo
2.24
=
4π x10-7 (const.)
The minus sign indicating that the force acts to reduce Q.
Example 1
With the rotor dimentions shown and a coil of 400 turns carrying 1.6A, calculated the
torgue acting on the rotor.
5. energy conversion
Week 15
Solution
The mmf (F) =
NI 400 X 1.6
Area fine B =UOH
Then F = 640
:.
=
640 A.t HX 21g
=>
BX2lg
Uo
B =640 x UO
2lg
640 X 4π X 190-7
2 x 10-3
= 0.40 tesla
0.43
i.e The flux density in each 1mm gap B = 0.40T.
:.
me
=
-0.42 X 0.025 X0.03 X0.001 X107 = 0.0955N-M
-B2l/g4 π
Example 2
The 4-l characteristics of a magnetic circuit are frequently described with straight
segments as shown below. The act is considered linear up to pt. a and in saturation
from a to 5 find the field energy
Example 3
A dynamic phonograph pickup consists of a 20-turn coil length of each coil =1cm)
moving normal to a field of B=0.2t. if the maximum allowable amplitude is 0.02mm,
calculated the output voltage at 100-HZ and at 100Hz
Solution
5. energy conversion
Week 15
A dynamic phonograph pickup for vertical recording the effective
conductor in moving coil is
L
=
20turns =20cm =0.2m
For a sinusoidal displacement of amplitude 2 x 10-5 m =0.02mm
X (t)
=Asinwt = 2x 10-5 sinwt
The velocity at 100HZ is d x /dt or
U
=dx =2 x10-5 x 2π x 103conwt m/s.
Neglecting internal impedance, the output voltage is
V
=
e =Blu =0.2 x 0.2 x 4π x 10-2 coswt colt
The r.m.s value of output voltage is V = vmax = 0.2 x0.2 x 4 π x10-2
2
2
= 0.0036v =3.6mv
(ii)
The velocity at 100HZ is
U
=
dx =2 x 100-5 x 2 π x 102 coswt m/s
And output voltage (neglecting internal impedance) is
V
=e bLu = 0.2 x 0.2 4 π x 10-3 coswt volt
And r.m.s value of output voltage
=
0.2 x 0.2 x 4 π x 10-3
2
=
0.00036v =0.36mv
length of
5. energy conversion
Week 15
Example4.
In a d.c machine, shown above, the armature is wound on a laminated iron cylinder
15cm long and 15cm in diameter. The N and S role faces are 15cm long (into the
paper) and 10cm along the circumference the average flux density in the air gap
under the pole faces is 1T. If there are 80 conductors in series between the brushes
and the machine turning at N =1500rpm, calculated the no-load terminal voltage
Solution
For Ns conductor (no coils) in series, the average emf is
E =NS ∆Ø wb =Ns Ø wb P poles
∆t s
pole
rev
=
Ns Øpn
n
60
rev
sec
No 2 conductor in sec)
volts
Where the flux per pole Ø = BA = 1 x0. 15 x 0.1 = 0.015 wb.
:.
Example 5.
E
=
80 x 0.015 x 2 x 1500
60
=
60 volts
5. energy conversion
Week 15
A magnetic circuit is completed through a soft –iron rotor as shown in the figure
above. Assuming (1) all the reluctance of the magnetic circuit is in the air gaps of
length L
(ii)
There is no fringing so the effective area of each gap is the area
Derive on expression for the torgue as a function of angular position.
Solution
The total reluctance for two air gaps in series is
R
=
2l
UOA
(since Ur =1 for air) R = F =1L
Q A
=
2L
Urwq
For an N-turn coil, the inductance is
L
=
NØ = NF
N2I
I
IR
IR
=
N2 trwØ
2L
Since torgue =1/2 I2 dL = Uo N2rw
dØ
2L
The torgue is independent of Ø under the assume conditions.
Example 6
(a) Wiring diagram
(b)Steady state model
A commutator machine, with the wiring diagram and steady-state model showed
above, is rated 5KW , 250V, 2000rmp. The armature resistance RA is 1. Drive from
the electrical and at 2000rmp, the no-load powder input to the armature is IA = 1.2A
at 250V with the field winding (RE =250) excited by IF =1A. Calculate the efficiency
of this machine.
5. energy conversion
Week 15
Solution
In fig (a), input power IF 2RF = 12 X 250 =250W is required to provide the necessary
magnetic flux. This is power lost and it appears as heat.
In no-load steady operation, there is no output and no change is all loss
[The armature cooper loss at no load is negligible (1.22 x 1=1.44w) and most of the
input power at no- load goes to supply air, bearing brush friction, eddy cumenty and
hystersis losses] the losses are associated with flux changes in the rotting armature
core, are dependent on speed but are nearly independent of load.
Hence, field loss
= IF2RF
=
250W
IAV
=
At full-load of 5KW, IA =5000W
=
20A
And the armature copper loss
=
IA2 Ra =202 x1 =400w
And rotational loss
=
300w p =iv :.i = p1
v
The energy balance equation gives
Mech. Energy
:input
+
field elect
input output
=
increase
energy
Field
energy
stored converted
to heat
i.e
Mech. input
+
250-5000
= 0
field arm
less less
rotat
less
Mech. input + 250 -5000 = 0 + 250 +300 + 400
(field input is all loss and appears o n both sides)
Mech. Input = 500 + 300 + 400 =5700W.
Hence efficiency
=
output
Input
=
elect output
mech. output + elect input
=
5000
5700 + 250
=
0.84. or 84%
5. energy conversion
Week 15
Example 7
In the relay shown above, the contacts are held open by the spring excerting a force
of 0.1N . The gap length is 4mm when the contacts are open
and 1mm when
closed. The coil of 5000 turns would on core 1cm2 in cross- section. Assuming
1)
All reluctance is in a uniform air gap