Transcript Motor Designs A, B, C, D

Motor Designs A, B, C, D
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Cross-Sections of NEMA Motors
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Induction Motor Applications
• Design B
– Broadest field of applications
– Centrifugal pumps, fans, blowers, machine
tools
• Design A
– Same characteristics as Design B, but with
higher breakdown torque
– Higher starting current limits applications
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Induction Motor Applications continued
• Design C
– Higher locked-rotor torque but lower
breakdown torque than Design B
– Use to drive plunger pumps, vibrating
screens, and compressors
• Design D
– Very high locked-rotor torque and high slip
– Used for high-inertia loads such as elevators
and hoists
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Induction Motor Applications continued
• Design E
– High-efficiency
– Drive loads similar to Design B, but with lower
locked-rotor, breakdown, and pull-up torque
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NEMA Tables
• Tables 5.1 – 5.7 give values of lockedrotor torque, breakdown torque, and pullup torque for specific horsepower,
frequency, and synchronous speed
ratings.
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Example 5.1
• Determine the values of locked-rotor
torque, breakdown torque, and pull-up
torque that can be expected from a 3phase, 10-hp, 460-V, six-pole, NEMA
design C motor whose rated speed is 1150
r/m.
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Solution
120 f (120)(60)
ns 

 1200r / min
P
6
Tn
hp 
5252
T (1150)
10 
5252
Trated  45.67lb  ft
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Locked-Rotor Torque
• From Table 5.1, page 171 of the text, the
minimum locked-rotor torque of a 10-hp design
C motor with a synchronous speed of 1200 r/min
should be 225% of full-load torque.
Tlocked rotor  (2.25)(45.67)  102.8lb  ft
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Breakdown Torque
• From Table 5-3
Tbreakdown  (1.90)( 45.67)  86.8lb  ft
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Pull-up Torque
• From Table 5.6,
Tpullup  (1.65)( 45.67)  75.4lb  ft
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Motor Performance as a function of Machine
Parameters, Slip, and Stator Voltage
• Use the “complete” equivalent circuit model,
including both the rotor and stator.
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Equivalent Circuit for an Induction Motor
with the rotor and stator as separate circuits
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Equivalent Circuit for an Induction Motor
with all parameters referenced to the stator
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R2  a 2 Rr
X 2  a 2 X BR
Ir
a
E2  Es  aEBR
I2 
a
Ns
Nr
R fe  coreloss / phase
X M  magnetizing / phase
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I 0  exciting  current  per  phase
Rr  actual  rotor  resis tan ce  per  phase
I fe  core  loss  component  of  exciting  current
I M  magetizing  component  of  exciting  current
X BR  actual  blocked  rotor  reac tan ce  per  phase
I r  actual  rotor  current  per  phase
V  actual  voltage  per  phase  applied  to  the  stator
I1  actual  stator  current  per  phase
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Power, Torque, Speed, Losses, and
Efficiency
Z2 
Z0 
I1 
R fe  jX M
Z2 Z0
ZP 
Z2  Z0
V
Z in
Z in  Z1  Z P
E  I1 Z P
E
I2 
Z2
R2
 jX 2
s
R fe jX M
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Pscl  3 I12 R1
Prcl  3 I 22 R2
1 s
2
Prcl

3
I
s
2 R2
Pmech  Prcl
1
1 s
Pgap  Prcl
Pmech  Prcl
s
s
Pmech  Pf , w  Pstray
Pshaft 
hp
746
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2
r
21.12 I Rr
TD 
sns
Pcore
3E22

R fe
Pcore
21.12 I 22 R2
TD 
sns
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3E22

R fe
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Approximate Equivalent Circuit for an
Induction Motor
Move the
magnetizing branch
to the left of resistor
R1.
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Shaping the Torque-speed curve
21.12I R
T 
lb  ft
sn
2
r
r
D
s
I 
2
V
R
R  jX   jX
s
V
2
1
I 
1
(R  R )  (X  X ) 


s
2
2
2
2
1
T 
D
1
21.12V
1
T V
D
R
s
2
2
2
(R  R )  (X  X )  n


s
2
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2
2
2
1
2
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s
Slip at Which Maximum Torque Occurs
Take the derivative of the
expression for the
developed torque
Solve for the value of slip
that makes the derivative
equal to zero.
s
R

R  (X  X )
2
T D max
2
1
1
2
2
Slip is directly proportional to the rotor resistance.
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Slip at Which Maximum Torque Occurs
• Applications which require a very high
starting torque are designed with enough
resistance so that the maximum torque
occurs at blocked rotor (s = 1).
s
R

1
R  (X  X )
2
T D max
2
1
2
1
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Slip at Which Maximum Torque Occurs
s
R

1
R  (X  X )
2
T D max
2
1
R
2
1
2
 R  (X  X )
2
2,s 1
1
1
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2
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Maximum Torque
R
21.12V
s
T 
(R  R )  (X  X )  n


s
2
2
D
2
2
1
1
R
 R  (X  X )
s
2
2
1
1
2
2
s
at maximum torque
2
2
21.12V

2n  R  (X  X )  R 
2
T
D ,max
2
s
1
2
1
2
1
Independent ofECERotor
Resistance!
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