Transcript Document 7863252

Voltage Regulation
E V
reg 
V
nl
rated
 per  unit  regulation
rated
%regulation  per  unit  100%
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Voltage Regulation (continued)
• Enl = no-load output voltage
– Measure with a voltmeter when no load is
connected to the transformer
• Vrated = voltmeter reading at the output
terminals when the transformer is
supplying the rated apparent
power
• These voltages are all either High-side or
Low-side voltages!
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Voltage Regulation (continued)
• Measurements are not always easy to make,
so use the equivalent circuit.
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When the breaker is open, no current flows in Req,LS ,
jXeq,LS , or ZLOAD,LS , therefore
Vout = VLS = E’LS = Enl
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With rated load on the secondary, E’LS = ILSZeq,LS + VLS
ILS = rated low-side current at a specified power factor
VLS = rated low-side voltage
Zeq,LS = equivalent impedance of the transformer
referred to the low-side
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E’LS = no-load low-side voltage
5
Example 2.7
• The equivalent low-side parameters of a
250kVA, 4160 – 480V, 60 Hz transformer
areReq,LS = 0.00920 Ω and Xeq,LS = 0.0433
Ω. The transformer is operated in the
step-down mode and is delivering rated
current at rated voltage to a 0.840 powerfactor lagging load.
• Determine
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– the no-load voltage
– The actual input voltage on the highside
– The high-side current
– The input impedance
– The voltage regulation
– The voltage regulation if the power
factor of the load is 0.840 leading
– Sketch the tip-to-tail phasor diagram of
the secondary circuit for the 0.840
power factor lagging load. Show all
voltage drops. ECE 441
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250, 000kVA
I LS 
 520.83 A
480V
  cos 1 (0.840)  32.86  32.86(lagging )
VLS  4800V
I LS  520.83  32.86 A
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Low-Side Output
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'
ELS
 I LS Req , LS  I LS jX eq , LS  VLS
'
ELS
 520.83  32.86(0.0092)  520.83  32.86( j 0.0433)  4800
'
ELS
 4.79  32.86  22.5557.14  4800
'
ELS
 4.024  j 2.599  12.235  j18.94  480  j 0
'
ELS
 496.531.886V
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'
HS
'
LS
E
a
E
VHS 4160


 8.667
VLS
480
'
'
VT  EHS
 aELS
 8.667(496.531.886)
VT  4303.41.886V
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The high-side current
I HS
I HS
I LS 520.83  32.86


a
8.667
 60.09  32.86 A
The input impedance
VT
4304.41.886
Z 

I HS 60.09  32.86
'
in
Z  71.6234.74
'
in
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The voltage regulation
Enl  Vrated 496.53  480
reg 

 0.0344  3.44%
Vrated
480
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Voltage Regulation if the power factor is 0.840 leading
I LS  520.83  32.86
.............
'
ELS
 472.282.61
Enl  Vrated 472.28  480
reg 

 0.0161  1.61%
Vrated
480
Regulation is negative because of a voltage rise in
the transformer due to resonant effects.
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The “tip-to-tail” phasor diagram for the 0.840 power
factor lagging load.
I LS Req , LS  520.83  32.86(0.0092)  4.79  32.86V
I LS X eq , LS  520.83  32.86( j 0.0433)  22.657.14V
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Component Phasors
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Tip-to-Tail Addition
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