Transcript Wound-Rotor Motor

Wound-Rotor Motor
Torque-slip (speed) curves for a
wound-rotor motor
R R

R  (X  X )
aR
'
s
2
T D ,max
2
1
R
'
rheo
rheo
1
2
2
2
rheo
Curve 3 – TD,max @ s=1
Curves 4, 5 – TD,max @ s>1
s>1 – “plugging” – reversal
of a motor before it comes
to rest by interchanging 2 of
the 3 line leads going to the
stator. nr is negative!
Example 5.10 page 198
• A family of torque-speed curves for a yconnected, 400-hp, 2300-v, 14-pole, 60Hz, wound-rotor induction motor is shown
in Figure 5.10. Curves A and D show the
extremes of rheostat adjustment.
Determine
– a) the range of rotor speeds available by
rheostat adjustment, assuming 100% rated
torque load on the shaft:
Example 5.10 page 198 (continued)
– b) the rheostat resistance required to obtain
260% rated torque when starting.
• The ratio of stator turns per phase to rotor
turns per phase is 3.8, and the motor
parameters, in Ω/phase are R1=0.403,
R2=0.317, X1=1.32, X2=1.32, XM=35.46.
Example 5.10 page 198 (continued)
• Solution for part (a)
• nr = ns(1-s)
• Determine ns
– ns = 120f/P = (120)(60)/14 = 514.29 r/min
• Look at the intersections of the curves with
the 100% torque line
– for the low-speed curve, D, s = 0.395
– for the high-speed curve, A, s = 0.02
Torque-speed curves for Ex 5.10
Example 5.10 page 198 (continued)
• for the low-speed case,
– nr = ns(1-s) = 514.29(1-0.395) = 311 r/min
• for the high-speed case,
– nr = ns(1-s) = 514.29(1-0.02) = 504 r/min
Torque-speed curves for Ex 5.10
260%
Example 5.10 page 198 (continued)
• for part (b)
• Curve B has locked-rotor torque of 260%
• TDmax occurs at s = 0.74
sTD max 
'
R2  Rrheo
R12  ( X 1  X 2 ) 2
'
 Rrheo
 sTD max  R12  ( X 1  X 2 ) 2  R2
'
Rrheo
 0.74  (0.403) 2  (1.32  1.32) 2  0.317  1.66
Rrheo
'
Rrheo
1.66
 2 
 0.115 / phase
2
a
(3.8)