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# sqj.1935.0073

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```THE
INDUCTION-TYPE
FREQUENCY
CHANGER
1
Output
ONE of the outstanding developments of recent years has been the production of high-speed motors running at 5,000-18,000 r.p.m. This type
of motor is built as one unit, and has been applied successfully as an
integral part of wood-working machines, planers, sensitive drills, internal
grinders, and such like. The motors are, generally speaking, 2-pole, and
require supply frequencies varying from 80 to 300 cycles. Some form of frequency changer is thus required; and where small quantities of power are
required and a regulation of 5-10 per cent allowable, an induction type
frequency changer shows considerable capital saving over a straight motorgenerator set. For example, if 20 k\V were required, the motor-generator
set would require two 20 k\V machines, whereas the induction type would
require only two 10 kW machines.
Theory of Operation
The induction type frequency changer is really a transformer in which
the primary and secondary windings move relatively to one another.
Since the rate of cutting the mutual magnetic flux is different for the stator
and rotor conductors, it follows that the frequency generated in the
secondary is different from that of the primary. At standstill the rotor
frequency equals supply frequency and the rotor voltage varies directly as
the turn ratio. When the rotor is driven at synchronous speed in the same
direction as the revolving magnetic field, the voltage and frequency are
zero as there is no relative motion between rotor conductors and the
magnetic field, i.e. no slip. If the rotor is rotated at a speed above or
below synchronous speed, either against or with the magnetic field, the
secondary voltage and frequency will vary directly as the slip.
Let /j = supply frequency,
/ 2 = required rotor frequency,
n
i = synchronous speed of frequency changer (i.e. speed of rotating
magnetic field),
n2 = speed at which motor is driven,
_
..
_
— n2 _ synchronpus r.p.m. — running r.p.m.
"x
synchronous r.p.m.
The following simple formulae give trie relationship between speed and
frequency.
With the rotor rotating against the magnetic field
(1)
(2)
• Scottish Section.
— 71 —
Price—The Induction-Type Frequency Changer
If the rotor rotates with the magnetic field, the plus and minus signs
within the brackets are interchanged.
Typical Example
Suppose the existing supply is 50 cycles and 100 cycles are required,
we could use a 4-pole frequency changer. Substituting in the above
equations, we find:
Rotating against the magnetic field
n2 = | i 5 ? _ i~| 1,500 = 1,500 r.p.m.
1_ 50
_J
Rotating with the magnetic field, we have
n2 =
+ 1 1,500 = 4,500 r.p.m., which is too high.
50
[[_ 50
_|
Alternately we can use an 8-pole machine, rotating with the field at a
speed of 2,250 r.p.m. Generally the higher speed involves smaller frame
size and consequently a cheaper machine, but in this case the difference
would not justify the cost of the extra poles, and the 4-pole machine would
be best. In this case the frequency convertor could be driven by a 4-pole
induction motor, but the speed would not be exactly 1,500 r.p.m., so that
the frequency changer on load would not generate exactly 100 cycles, but
some slightly lower frequency depending upon the slip value.
Vector Diagram and Equivalent Circuit
Fig. 1 is the vector diagram and shows clearly the relationship between
the various quantities. However, since the induction type frequency
changer is a rotating transformer, the equivalent circuit of the transformer
can be employed, as in Fig. 2.
Let Zy = B1 + jX1 = primary impedance of stator winding,
Y = G — jB = primary magnetizing admittance per phase,
Z2 = i?2 "T" J-^2 = s e c o n dary impedance of rotor per phase,
Z = JR + jX — external load per phase,
cw
K = 2 2 = ratio of secondary to primary turns, <Sj and S2 being
"l l
the number of stator and rotor slots, and T1 and T2
the turns per slot.
5 = ^ = slip,
E2 = secondary generated e.m.f. at rotor frequency,
Ey = primary generated e.m.f. at supply frequency.
Now
E2
1
f ^
Ji
^l11
1
Secondary terminal e.m.f. = E2 — /2(-R2 + jXv) = Ei
Ji »!•* i
where a is the percentage regulation drop in the frequency changer.
— 72. —
Students' Quarterly Journal
December 1935
Total secondary impedance = Z + Z2 = (Rz + R) + j(X2 + X)
•pi
• R +j[X2 + X)
secondary terminal e.m.f.
R+jX
also To
Primary generated e.m.f.
Secondary current referred to primary = I[ = InK
where V = applied volts,
Jj = stator amps,
1'x = rotor amps referred to primary.
FIG. 1 (left) Vector diagram, and FIG. 2
(below) Equivalent circuit of frequency
changer
Equivalent secondary impedance referred to primary = -y —
1\
Conversion ratio (rotor to stator)
f
Oxi It
= ——5
Io = primary magnetizing amps = Ex(0 — jB)
C
and Regulation =
S-K2
.'
-
— JS2 rotor internal volts — terminal volts
=
:
E2
terminal
volts
Description of Set
The complete set consists of an ordinary slip-ring induction motor
driven at the required speed by a suitable motor. When a definite frequency is required the standard squirrel-cage or direct-current motor with
3 to 5 per cent inherent speed regulation is satisfactory. A wide frequency
— 73 —
Price—The Induction-Type Frequency Changer
range necessitates a variable-speed motor which may be a rheostatically
controlled slip-ring induction motor, an a.c. commutator motor, or d.c.
shunt machine with speed control by shunt field weakening.
The driving motor size depends upon the ratio of the synchronous speed
of the rotating magnetic field to the driving motor speed and whether
driven against or with the magnetic field. The total power generated in
the rotor is 3SE2I3cos<f>z. This power is divided into two parts: first
that which is transferred from stator to rotor by transformer action,
3£g72 cos <f>2; and second, that which is due to rotor rotation in the magnetic
field, i.e. 3(S — l)2?2/2 cos <f>2. The following simple formula gives the
driving motor output:
o
__
"I
Driving motor output =
(watts output + sec. copper loss)
-f rotating losses of frequency changer
The most economical set is obtained, providing the frequency ratio
permits, when rotating against the magnetic field. Consider rotation with
the revolving magnetic field between zero and synchronous speed. The
frequency changer, running as an induction motor, drives the second unit,
which becomes an induction generator. The input to the frequencychanger stator is rotor output + induction generator output + losses.
At synchronous speed it acts as a transformer and, therefore, stator input
equals rotor output plus losses. Above synchronous speed, the motor B
drives A and there is mechanical power flow along the shaft to the rotor.
Now the power from the line is flowing towards the set in both cases
and the total capacity of the set need only be equal to the rotor output plus
losses.
Points in Design
With the induction type frequency changer there is no means of voltage
regulation as the excitation voltage is constant. To keep the regulation
and rotor core losses within bounds, the frequency changer must be designed with low flux densities, which means a derated machine compared
with standard designs. With large outputs this factor makes itself
apparent and a straight motor generator set becomes competitive between
50-100 kW, especially with a frequency ratio of 2 :1 or less. Moreover, the
induction type frequency changer output voltage is limited to approximately
1,000-1,500 volts between rings for satisfactory collection.
Up to approximately 50 kW, the induction frequency changer scores
over the straight set on the score of lower capital cost, compactness, general
simplicity, and ease of operation. It has therefore found a definite market
in connection with individual motor drives.
— 74 —
```
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