Transcript 1. EXPERIMENT NO. 1 Schrage Motor

EXPERIMENT NO.: 1.
To Study the Variation of Speed and Load Test
on Schrage Motor
Theory
induction motor – Cheap, Rugged,
Less Maintenance
But low starting Torque
In SRIM, Test can be increased by additional r2
Or AC commutator motors are the other option.
AC commutator motors operate near to unity pf.
with wide range of speed control
Commutator frequency Converter
Consider DC Generator operation
A
N
+
ARM
AA -
S
E●
As armature rotates,
emf E is induced
For same polarity of E
across brushes
Rotate the poles,
ZN P
Φ
E
n
Anticlockwise
Dir
60 a
Condr moves
clockwise Dirn
Commutator frequency Converter
Consider DC Generator operation
S
A
+
N
S
AA -
E●
N
As armature rotates,
emf E is induced
For same polarity of E
across brushes
Rotate the poles,
ZN P
Φ
E
n
Anticlockwise
Dir
60 a
Condr moves
clockwise Dirn
Commutator frequency Converter
Consider DC Generator operation
How much is E?
S
A
+
Zero
Again rotate poles
AA -
E●
N
As armature rotates,
emf E is induced
For same polarity of E
across brushes
Rotate the poles,
ZN P
Φ
E
n
Anticlockwise
Dir
60 a
●
Condr moves
clockwise Dirn
Commutator frequency Converter
Consider DC Generator operation
S
How much is E?
A
Zero
Again rotate poles
S
N
AA
N
E●
●
Commutator frequency Converter
Consider DC Generator operation
How much is E?
A
_
Zero
Again rotate poles
ZN P
E Φ
60 a
S
N
Again rotate poles
AA +
E●
●
●
Commutator frequency Converter
Consider DC Generator operation
How much is E?
A
N_
Zero
Again rotate poles
ZN P
E Φ
60 a
S
N
Again rotate poles
AA +
S
E●
●
●
Commutator frequency Converter
Consider DC Generator operation
How much is E?
A
N_
Zero
Again rotate poles
ZN P
E Φ
60 a
Again rotate poles
E 0
AA +
S
E●
●
●
●
Commutator frequency Converter
Consider DC Generator operation
How much is E?
A
N
Zero
Again rotate poles
ZN P
E Φ
60 a
N
ARM
Again rotate poles
E 0
Again rotate poles
AA
S
S
E●
●
●
●
Commutator frequency Converter
Consider DC Generator operation
How much is E?
A
+
Zero
Again rotate poles
ZN P
E Φ
60 a
N
S
Again rotate poles
E 0
Again rotate poles
ZN P
E Φ
60 a
AA -
●
E●
●
For N rpm speed of poles
Freq of E is
fb 
PN
Hz
120
●
●
Commutator frequency Converter
Consider DC Generator operation
Rotate poles and brushes
A
+
At same speed and
same direction
N
Relative speed is zero
Freq of E will be zero
S
AA -
E●
Commutator frequency Converter
Consider DC Generator operation
Rotate poles and brushes
A
+
at same speed and
same direction
N
Relative speed is zero
Freq of E will be zero
AA Now brush stationary and rotate
armature with speed Nr rpm and
pole speed is N in opposite direction
Freq of brush emf E
will be
S
fb 
E●
Relative speed betwn air gap field and brushes = N+Nr
P( N  N r )
120
Commutator frequency Converter
Consider DC Generator operation
Now armature speed is Nr rpm and
+
A
pole speed is N
in same direction
N
Relative speed betwn
air gap field and
brushes = N - Nr
Freq of brush emf E
will be
fb 
P( N  N r )
120
S
In general emf E
is induced across
brushes
Z (N  Nr ) P
E Φ
60
a
In terms of angle
AA -
Z (N  Nr ) P

E Φ
Sin
60
a
2
Thus E can be increased by increasing θ
and axis of E can be changed by shifting
two brushes simultaneously.
ER
Axis of phase R
Pole
EB
EY
Consider inverted induction motor
3-phase supply to rotor winding
Rotor field rotates at Ns wrt rotor
As per induction motor action, rotor rotates in opposite direction
at Nr wrt stator.
Therefore, air gap speed is Ns - Nr wrt stator or wrt stationary
brushes
So brush emf frequency is
P( N s  N r )
fb 
 sf
120
Thus whatever may be the speed of rotor, brush emf freq is sf
It is suitable to inject into secondary wdg which is having slip freq
emf
Power related to sf is SLIP POWER.
If brush emf is opposite to sE2, then resultant voltage
decreases,
Current decreases, torque decrease,
Speed decreases
Motor runs at sub-synchronous speed
If brush emf is added to sE2, then resultant voltage increases,
Current increases, torque increase,
Speed increases
Motor runs at super-synchronous speed
By shifting the axis of brush voltage, power factor can be changed
The speed and power factor of slip ring induction
motor can be controlled by injecting slip
frequency voltage in the rotor circuit.
In 1911, K. H. Schrage of Sweden combined
elegantly a SRIM (WRIM) and a frequency
converter into a single unit.
This machine is known as Schrage Motor.
Schrage motor is an AC commutator Machine
Schrage motor is basically an
inverted polyphase induction motor.
Induction motor
Primary winding on Stator
Secondary winding on
Rotor
If supply is given to stator,
rotor rotates in the same
direction of rotating
magnetic field
Schrage motor
Primary winding on Rotor
Secondary winding on
Stator
If supply is given to rotor,
rotor has to rotates in the
opposite direction of
rotating magnetic field, so
that rotating magnetic
field in the air gap
becomes Ns-Nr= slip
speed.
Induction motor
Schrage motor
If supply is given to stator,
If supply is given
to rotor,
Ns
Nr
Nr
Ns
If rotor rotates in the SAME direction of rotating
magnetic field, then rotating magnetic field in the air
gap becomes N +N = Addition of speed.
Induction motor
Schrage motor
Tertiary or adjusting
winding, which is housed
in the same rotor slots of
the primary
The tertiary winding is
connected to the
commutator
On commutator there are
three sets of movable
brush pairs,
Brush pairs collect the
required emf for injection
into the secondary circuit
for speed and pf control.
Schrage motor
Supply
Slip Rings
Schrage motor
Supply
Tertiary
Winding
Primary
Winding
Slip Rings
Schrage motor
Supply
Tertiary
Winding
Primary
Winding
Slip Rings
Commutator
Secondary
Winding
Supply
a
b
Tertiary
Winding
Primary
Winding
Slip Rings
Commutator
Secondary
Winding sf
Supply
a
b
Tertiary 50Hz
Winding
Primary
Winding
50Hz
Slip Rings
Commutator
Schrage motor
Arrangement of Three Windings is slots
Stator
Rotor
Secondary
Winding
sf
Tertiary
Winding 50Hz
Primary
Winding 50Hz
The primary and tertiary windings, beings in the same
slots, are mutually coupled.
Therefore, the emfs induced in the tertiary winding are
by transformer action and are always at line frequency f,
at all the rotor speed.
The secondary phase winding on stator are not
connected to each other but are connected to
brushes a, b, c, d, e, f.
Alternate brushes a, c, e, 120 electrical degrees apart,
are mounted on one brush rocker and brushes b, d, f on
the second brush rocker.
The angle between brushes ab, cd, and ef can be
controlled by means of rack and pinion arrangement and
one hand wheel provided outside the motor frame.
Operation
Operation
At standstill, due three phase supply
to primary.
field rotates at synchronous speed Ns with respect
to primary and secondary conductors.
Rotor rotates at speed Nr, opposite direction
since primary winding is on the rotor.
air gap field speed (Ns – Nr) i.e. slip speed.
slip frequency emfs E2 in the stator winding.
slip frequency emf Ej is induced across brushes
Equal slip frequency voltages,
injection into the secondary winding i.e. stator
winding is possible
At the time of start, the brush pairs are shorted,
The shorted secondary winding (brush angle zero) is
also condition for starting the Schrage motor.
a
b
E2
a
sE2
b
Ej=0
Φ at Ns
wrt rotor
Nr < Ns
a
Φ at (Ns - Nr)
wrt stator
θ
b
E2
a
sE2
E2
b
Ej=0
Φ at Ns
wrt rotor
Nr < Ns
a
sE2
θ
b
Ej
Φ at (Ns - Nr)
wrt stator
Nr < Ns
Φ at (Ns - Nr)
wrt stator
E2 and Ej are in opposite direction
E2 - Ej, resultant voltage decreases.
speed decreases. (sub-synchronous
speed). Slip is POSITIVE.
E2
b
sE2
(-s)
a
a
Ej
Φ at Ns
wrt rotor
Nr > Ns
sE2
θ
b
Ej
Φ at (Nr - Ns)
wrt stator
Change the position of a & b.
Axes of voltages are
coincidence
But E2 and Ej are in same
direction.
E2
Nr < Ns
Φ at (Ns - Nr)
wrt stator
E2 and Ej are in opposite direction
E2 - Ej, resultant voltage decreases.
speed decreases. (sub-synchronous
speed). Slip is POSITIVE.
E2 + Ej, resultant voltage increases.
speed increases. (super-synchronous
speed), Slip is NEGATIVE.
Speed
θ=-1800
θ=0
θ=1800
θ
E2
a
θ
a
sE2
Now move a, b brushs
bodily towards left
bb
There is angle ρ between
axes of voltages.
Ej
Φ at Ns
wrt rotor
Nr < Ns
I1
Φ at (Ns - Nr)
wrt stator
Φ Power factor angle
ρ
ρ
Ej
sE2
Thus power factor is controlled.
E2
I2
I 2 z2
Circuit Diagram
L1
10A
M
C
L
V
+ A-
A
V
300V
10A
R
A
N
F
G
L3
B
Y
L2
Schrage Motor
FF
AA
+
V
-
300V
1000Ω
2A
DC Generator
Experimental set-up for studying the Variation of
Speed and Load Test on Schrage Motor
L
O
A
D
Apparatus Required
1. Schrage Machine: Rating:…………………
2. DC Shunt Generator: Rating……………………
3. AC Ammeter: one, 0-10 amps
4. AC Voltmeter: one, 0-300 volts.
5. Wattmeter: one, 300V, 10A
6. DC Ammeter: one, 0-10 amps.
7. DC Voltmeter: one, 0-300 volts.
8. Rheostat: one, 1000 ohms, 2 amps.
9. Tachometer or speedometer: one
10. Rheostatic Load: 7.5kW
Speed Test:
Procedure
1. Note down ratings of the machines and
make the connection as shown in Fig.
2. Put the DC generator field rheostat at maximum
resistance point.
3. Keep the brush angle pointer at zero (condition for
starting the Schrage motor).
4. Switch on AC supply to Schrage motor by I. L. T. P.
switch. Press green button of ILTP to start motor.
5. Note down speed for zero brush angle.
6. Now increase brush angle from zero to
3600 gradually at regular intervals and note
down the speed for each interval.
Procedure
Load Test:
7. Run the Schrage motor at rated speed of DC shunt
generator.
8. Excite the DC shunt generator by decreasing the
field rheostat resistance and build up to its rated
voltage. Maintain this voltage CONSTANT through
out the experiment.
9. Increase the load gradually and note down the
speed and meters readings for each load.
Observations and
Calculations
Speed Test:
Angle θ
0
20
40
Speed N
Observations and
Calculations
DC Generator
Schrage Motor
Nm
Vm
Im
Wm
PFm, (to be
Calculated)
Ig
1A
2A
3A
Vg
CONSTANT
Load Test:
Output of
Gen= VgIg
Efficiency
of Motor
ηm
Results and Conclusions
Speed Test:
Speed
θ=00
θ=3600
θ
1. The speed of motor increases as the brush angle is
increased. Speed is directly proportional to brush
angle.
Results and Conclusions
Load Test:
Power
factor
N
Pf
Im
η
Speed
Im
Efficiency
0
Motor
Output
2. The speed of motor decreases as the load on
motor is increased.
Results and Conclusions
Load Test:
Power
factor
N
Pf
Im
η
Speed
Im
Efficiency
0
Motor
Output
3. The power factor of motor increases or
improves as the load on motor is
increased.
Results and Conclusions
Load Test:
Power
factor
N
Pf
Im
η
Speed
Im
Efficiency
0
Motor
Output
4. The current of motor increases as the load on
motor is increased.
Results and Conclusions
Load Test:
Power
factor
N
Pf
Im
η
Speed
Im
Efficiency
0
Motor
Output
5. The efficiency is zero at no load. It increases as the
load increases and is maximum when variable
losses are equal to constant losses.
Results and Conclusions
Load Test:
Power
factor
N
Pf
Im
η
Speed
Im
Efficiency
0
Motor
Output
6. The efficiency at rated output is less than maximum
value and the rated operating point is after maximum
efficiency point.
Write down all the precautions
which are listed in Laboratory
and attach after Index page