Transcript Direct – Current Motor Characteristics and Applications

Direct – Current Motor
Characteristics and Applications
• Straight Shunt Motor
– Essentially a constant speed motor
• Compound or Stabilized – Shunt Motors
– Has both shunt and series field windings
– Series field generates mmf in the same
direction as the shunt field mmf.
ECE 441
1
Circuit Diagram of a Compound Motor
ECE 441
2
Differential Connection of Fields
• Both the series and shunt fields must
provide fluxes that are additive.
• If the series field is reversed with respect
to the shunt field, the net flux decreases,
and the speed increases.
• The time constant of the series field is
such that the current increases faster than
the shunt field current.
ECE 441
3
Differential Connection of Fields
• If the series field is reversed,
– The motor will start in the wrong direction
– Depending upon the load and the structure of
the series field, the motor could
• slow down and stop, tripping the breaker
• slow down, stop, reverse direction, and accelerate
• slow down, stop, reverse direction, slow down,
stop, reverse direction, etc. until a breaker trips
ECE 441
4
Reversing the Direction of
Compound Motors
• Reverse either the armature current or
reverse both the series and shunt fields.
– If only one field is reversed, a “differential”
connection results!
– The field mmfs will be reduced, resulting in
excessive speed!
ECE 441
5
Reversing the Armature Current
ECE 441
6
Using NEMA standard
terminal markings
ECE 441
7
Series Motor
• Series field
– Heavy windings
– Must conduct the armature current
• Potentially dangerous problem if the shaft
load is removed!
ECE 441
8
Field winding is in
series with the
armature
ECE 441
9
More Details
• When shaft load is removed, TD>Tload
– Motor speed increases
– cemf increases
– armature current decreases
– series field flux decreases
ECE 441
10
Reversing the Direction
of a Series Motor
• Reverse the current in the armatureinterpole-compensating branch
• Reverse the current in the series field
windings
ECE 441
11
Reversing the Armature Current
ECE 441
12
Using NMEA standard
Terminal Markings
ECE 441
13
Using NEMA standard
terminal markings
Reversing the series field
ECE 441
14
Effect of Magnetic Saturation on
DC Motor Performance
• Pole flux is not directly proportional to the
applied mmf due to magnetic saturation
• Net mmf is made up of the following
components, as applicable
–
–
–
–
–
Fnet = Ff + Fs - Fd
Fnet = net mmf (A-t/pole)
Ff = shunt field mmf (NfIf)(A-t/pole)
Fs = series field mmf (NsIa)(A-t/pole)
Fd = equivalent demagnetizing mmf due to armature
reaction (A-t)/pole
ECE 441
15
Effect of Magnetic Saturation on
DC Motor Performance
• Note that Fd is not exactly proportional to
the armature current, but is assumed to
be.
• If a compensating winding is used, Fd = 0.
ECE 441
16
Developed Torque and Speed
TD  B p I a k M
VT  I a Racir
n
 p kG
 p 0
Racir  Ra  RIP  RCW  Rs
ECE 441
17
Defining Parameters
•
•
•
•
Racir = resistance of armature circuit (Ω)
Ra = resistance of armature windings (Ω )
RIP = resistance of interpole windings (Ω)
RCW = resistance of compensating
windings (Ω)
• Rs = resistance of series field winding (Ω)
• Bp = air-gap flux density (T)
• Φp = pole flux (Wb)
ECE 441
18
Solve Problems with Proportions
TD1 [ B p I a ]1

TD 2 [ B p I a ]2
VT  I a Racir 


n1   p kG 1

,  0
n2 VT  I a Racir 



k


p G
2
 p  Bp  A
Bp

n1 VT  I a Racir  

 

n2 
Bp
V

I
R
1  T a acir  2
ECE 441
19
Example 11.1
• A 240-V, 40-hp, 1150 r/min stabilizedshunt motor, operating at rated conditions,
has an efficiency at rated load of 90.2%.
The motor parameters are
• Ra = 0.0680 Ω
RIP = 0.0198 Ω
Rs = 0.00911 Ω
Rshunt = 99.5 Ω
• Turns/pole series - ½ shunt - 1231
ECE 441
20
Example 11.1 (continued)
• The circuit diagram and magnetization
curve are shown on the next slide.
Determine (a) the armature current when
operating at rated conditions; (b) the
resistance and power rating of an external
resistance required in series with the shunt
field in order to operate at 125% rated
speed. Assume the shaft load is adjusted
to a value that limits armature current to
115% of rated current.
ECE 441
21
ECE 441
22
Solution for Armature Current
P
40  746
P  VT IT  IT 

VT 0.902  240
IT  137.84 A
VT
240
If 

 2.4121A
R f 99.5
I a  IT  I f  137.84  2.41  135.43 A
ECE 441
23
Solution for External Resistance
• The series field of a compound motor is
designed to be approximately equal and
opposite to the equivalent demagnetizing
mmf of armature reaction. Therefore, the
net flux is due to the shunt field alone.
Fnet  Ff  N f I f  1231 2.412  2969.2 A  t / pole
ECE 441
24
net mmf = 0.70 T
ECE 441
25
Racir  Ra  RIP  Rs
Racir  0.0680  0.0198  0.0091  0.0969
Bp

n1 VT  I a Racir  

 

n2 
Bp
V

I
R
1  T a acir  2
n1 [VT  I a Racir ]2
B p 2  B p1  
n2 [VT  I a Racir ]1
Bp 2
Bp 2
1150
240  1.15  135.43  0.0969
 0.70 

1.25  1150
240  135.43  0.0969
 0.56T
ECE 441
26
Ff = 2.3 X 1000 = 2300 A-t/pole
ECE 441
27
Ff
2300
Ff  N f I f  I f 

N f 1231
I f  1.87 A
VT
VT
If 
 Rx 
 Rf
R f  Rx
Rf
240
Rx 
 99.5  28.8
187
PRx  I 2f Rx  (1.87) 2  28.8  100.7W
ECE 441
28
Linear Approximations
• If the magnetization curve is not available
– rough approximation obtained by assuming
magnetization effects are negligible
– Do not use approximations if the motor is
operating under heavy overload or locked
rotor conditions.
• If the net mmf is to be reduced below its
rated value, approximation using the linear
assumption is OK.
ECE 441
29
Approximate Equations for
Torque and Speed
[Fnet I a ]1
TD1 [ B p I a ]1


TD 2 [ B p I a ]2
[Fnet I a ]2
Bp

n1 VT  I a Racir  

 
 
n2 
Bp
1 VT  I a Racir  2

Fnet
n1 VT  I a Racir  

 
 ,  0
n2 
Fnet
1 VT  I a Racir  2
ECE 441
30
For the Series Motor
If the range of operation is in the unsaturated
region, and armature reaction effects are
either negligible or compensated for,
TD1 [Fnet I a ]1

 TD , series  Fnet I a  N s I a
TD 2 [Fnet I a ]2
TD , series  I
2
a
The developed torque is proportional to the
square of the armature current.
ECE 441
31
Example 11.2
• Example 11.1 is re-solved using the linear
approximation, and the solution is
compared to the results obtained in
Example 11.1.
ECE 441
32
VT
240
I f1 

 2.412 A
R f 99.5
I a1  IT  I f  137.84  2.412  135.43 A
Fnet  Ff  N f I f  1231 2.412  2969.2 A  t / pole

Fnet
n1 VT  I a Racir  

 

n2 
Fnet
V

I
R
1  T a acir  2
n1 [VT  I a Racir ]2
Fnet 2  Fnet1  
n2 [VT  I a Racir ]1
Fnet 2
Fnet 2
1150
240  1.15 135.43  0.0969
 2969.2 

1.25  1150
240  135.43  0.0969
 2354.8 A  t / pole
ECE 441
33
Fnet 2354.8
If 

 1.91A
Nf
1231
VT
VT
If 
 Rx   R f
R f  Rx
If
240
Rx 
 99.5  26.15
1.91
From Example 11.1, the value of resistance was
determined to be 28.8 Ω
ECE 441
34
Calculate the Percent Error
%error 
Ractual  Rapprox
Ractual
 100%
28.8  26.15
%error 
100%
28.8
%error  9.2%
This lower value of resistance would cause
a slightly higher field current, and therefore,
a speed slightly lower than 1437.5 r/min.
ECE 441
35
Comparison of Steady – State
Operating Characteristics of DC Motors
• The steady-state operating characteristics
of typical shunt, compound, and series
motors of the same torque and speed
ratings are shown on the next slide.
ECE 441
36
ECE 441
37
Comparisons (continued)
• Shunt Motor
– relatively constant speed from no-load to
full-load
– does not have high starting torque
– essentially constant flux
– torque varies linearly with armature current
– speed regulation around 5%
ECE 441
38
Relatively
Constant Speed
Linear Torque
ECE 441
39
Comparisons (continued)
• Compound Motor
– Higher torque, lower speed than shunt motor
– speed regulation between 15 and 25%
– used with loads requiring high starting torques
or have pulsating loads
• smoothes out the energy required by the pulsating
load, lowering the demand on the electrical supply
ECE 441
40
Lower Speed at
Higher Torque
Higher Torque
above base speed
than Shunt motor
ECE 441
41
Comparisons (continued)
• Series Motor
– high starting torque
– wide speed range
– REMOVING THE LOAD CAUSES IT TO RUN
AWAY!
• CONNECT LOAD BY GEARS OR SOLID
COUPLING – NO BELT DRIVES!
ECE 441
42
Wide Speed Range
High Starting Torque
ECE 441
43
Dynamic Braking, Plugging,
and Jogging
• Dynamic Braking is the deceleration of the
motor by converting the energy stored in
the moving masses into electrical energy
and dissipating it as heat via resistors.
Also called resistive braking.
ECE 441
44
Dynamic Braking (continued)
• Disconnect the armature from the
electrical supply lines and connect across
a suitable resistor while maintaining the
field at full strength.
• The motor behaves as a generator,
feeding current to the resistor, dissipating
heat.
ECE 441
45
Dynamic Braking (continued)
• Choose the resistance for current between
150 and 300% of rated current.
• The armature current is in a direction to
oppose the armature motion, producing a
negative, or, counter-torque, slowing down
the load.
ECE 441
46
Compound Motor Example
Normal Operation
Dynamic Braking
ECE 441
47
Normal Operation
Closed
Open
ECE 441
48
Dynamic - Braking
Open
Closed
ECE 441
49
Regenerative Braking
• Convert energy of overhauling loads into
electrical energy and pumps it back into
the electrical system.
• The overhauling load drives a DC motor
faster than normal, causing the cemf to
become greater than the supply voltage
and results in generator action.
• Trains, elevators, hybrid automobiles
ECE 441
50
Plugging
• The electrical reversal of a motor before it
stops
• Reverse the voltage applied to the
armature
• Current in the series and shunt fields is not
reversed
• Insert resistance in series with the
armature to limit the current
ECE 441
51
Normal Operation
ECE 441
52
Plugging
ECE 441
53
Jogging
•
•
•
•
Very brief application of power to a motor
Fraction of a revolution
Used for positioning the load
Place resistance in series with the
armature to limit the current
ECE 441
54
Example 11.7
• A 240-V, compensated shunt motor driving a 910
lb-ft torque load is running at 1150 r/min. The
efficiency of the motor at this load is 94.0%. The
combined armature, compensating winding, and
interpole resistance is 0.00707Ω, and the
resistance of the shunt field is 52.6Ω. Determine
the resistance of a dynamic-braking resistor that
will be capable of developing 500 lb-ft of braking
torque at a speed of 1000 r/min. Assume
windage and friction at 1000r/min are essentially
the same as at 1150 r/min.
ECE 441
55
Circuit for Dynamic Braking
ECE 441
56
T  n 910 1150
Pshaft 

 199.257 hp
5252
5252
Pshaft 199.257  746
Pin 

 158134W

0.940
158134
Pin  VT IT  IT 
 658.89 A
240
VT
240
If 

 4.56 A
R f 52.6
I a  IT  I f  658.89  4.56  654.33 A
VT  Ea1  I a1 Racir  Ea1  240  654.33  0.00707
Ea1  235.37V
ECE 441
57
I T
T1 [ B p I a ]1 I a1


 I a 2  a1 2
T2 [ B p I a ]2 I a 2
T1
654.33  500
Ia2 
 359.52 A
910
Ea1 [n p kG ]1
n1
n2

  Ea 2  Ea1
Ea 2 [n p kG ]2
n2
n1
Ea 2 
1000
 235.37  204.67V
1150
Ea 2  I a 2 ( Racir  RDB )  RDB
RDB
RDB
Ea 2  I a 2 Racir

Ia2
204.67  359.52  0.00707

 0.562
359.52
 0.562
ECE 441
58