Transcript Electromagnetic

Per-Unit System
EE341 – Energy Conversion
Ali Keyhani
Transformer
Lecture #3
1
Per-Unit System
In the per-unit system, the voltages, currents,
powers, impedances, and other electrical
quantities are expressed on a per-unit basis by
the equation:
Quantity per unit =
Actual value
Base value of quantity
It is customary to select two base quantities to
define a given per-unit system. The ones usually
selected are voltage and power.
2
Per-Unit System
Assume:
Vb  Vrated
Sb  S rated
Then compute base values for currents and
impedances:
Sb
Ib 
Vb
2
b
Vb V
Zb 

Ib
Sb
3
Per-Unit System
And the per-unit system is:
V p.u .
Vactual

Vb
S p.u .
S actual

Sb
Z %  Z p.u . 100%
I p.u .
I actual

Ib
Z p.u .
Z actual

Zb
Percent of base Z
4
Example 1
An electrical lamp is rated 120 volts, 500 watts.
Compute the per-unit and percent impedance of
the lamp. Give the p.u. equivalent circuit.
Solution:
(1) Compute lamp resistance
V2
V2
(120) 2
P
R

 28.8
R
P
500
power factor = 1.0
Z  28.80
5
Example 1
(2) Select base quantities
Sb  500VA
Vb  120V
(3) Compute base impedance
Vb2 (120) 2
Zb 

 28.8
Sb
500
(4) The per-unit impedance is:
Z p.u .
Z 28.80


 10 p.u.
Zb
28.8
6
Example 1
(5) Percent impedance:
Z %  100%
(6) Per-unit equivalent circuit:
VS  10 p.u.
Z  10 p.u.
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Example 2
An electrical lamp is rated 120 volts, 500 watts. If
the voltage applied across the lamp is twice the
rated value, compute the current that flows
through the lamp. Use the per-unit method.
Solution:
Vb  120V
V p.u .
V 240


 20 p.u.
Vb 120
Z p.u .  10 p.u.
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Example 2
The per-unit equivalent circuit is as follows:
VS  20 p.u.
I p.u . 
V p.u .
Z p.u .
Z  10 p.u.
20

 20 p.u.
10
Sb 500
Ib 

 4.167 A
Vb 120
I actual  I p.u. I b  20  4.167  8.3340 A
9
Per-unit System for 1-  Circuits
One-phase circuits
Sb  S1  V I
where
V  Vlinetoneutral
I  I linecurrent
VbLV  VLV
VbHV  VHV
Sb

VbLV
Sb

VbHV
I bLV
I bHV
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Per-unit System for 1-  Circuits
Z bLV
VbLV (VbLV ) 2


I bLV
Sb
Z bHV
VbHV (VbHV ) 2


I bHV
Sb
S
*
S pu 
 V pu I pu
Sb
P
Ppu 
 V pu I pu cos
Sb
Q
Q pu 
 V pu I pu sin 
Sb
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Transformation Between Bases
Selection 1
Then
Sb1  S A
Vb1  VA
Vb21
Z b1 
S b1
ZL
Z pu1 
Z b1
Selection 2
Sb 2  S B
Then
Zb2
2
b2
V

Sb 2
Vb 2  VB
Z pu 2
ZL

Zb2
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Transformation Between Bases
Z pu 2
Z pu1
Z L Z b1 Z b1 Vb21 Sb 2




 2
Z b 2 Z L Z b 2 Sb1 Vb 2
2
Z pu 2
 Vb1   Sb 2 
  

 Z pu1 
 Vb 2   S b1 
“1” – old
“2” - new
2
Z pu,new
 Vb,old   Sb,new 
 

 Z pu,old 
 S

V
 b,new   b,old 
13
Transformation Between Bases
Generally per-unit values given to another base
can be converted to new base by by the
equations:
Sbase1
( P, Q, S ) pu _ on _ base_ 2  ( P, Q, S ) pu _ on _ base_1
Sbase2
V
V pu _ on _ base_ 2  V pu _ on _ base_1 base1
Vbase2
( R, X , Z ) pu _ on _ base_ 2
(Vbase1 ) 2 Sbase2
 ( R, X , Z ) pu _ on _ base_1
(Vbase2 ) 2 Sbase1
When performing calculations in a power system, every
per-unit value must be converted to the same base.
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Per-unit System for 1- Transformer
Consider the equivalent circuit of transformer
referred to LV side and HV side shown below:
RS
XS
j 2
2
a
a
RS  jX S
VLV
VHV
VLV
N1
N2
Define
S
(1) Referred to LV side
VHV
VLV
N1
a

1
VHV N 2
(2) Referred to HV side
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Per-unit System for 1- Transformer
Choose:
Vb1  VLV ,rated
Sb  S rated
Compute:
Normal choose rated
values as base values
VHV
1
Vb 2 
Vb1  Vb1
VLV
a
Vb21
Z b1 
Sb
Zb2
Vb22

Sb
Z b1 Vb21
Vb21
 2 
 a2
Z b 2 Vb 2 ( 1 V ) 2
b1
a
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Per-unit System for 1- Transformer
Per-unit impedances are:
RS  jX S
Z p.u .1 
Z b1
Z p.u .2
So:
RS jX S
RS jX S
 2
 2
2
2
RS  jX S
a
a
a
a



Z b1
Zb2
Z b1
a2
Z p.u .1  Z p.u .2
Per-unit equivalent
circuits of transformer
referred to LV side and
HV side are identical !!
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Per-unit Eq. Ckt for 1- Transformer
RS  jX S
S
a
Z b1
N1
VLV
Vb1
N2
VHV
VLV
N
 1 1
VHV N 2
Vb 2
Fig 1. Eq Ckt referred to LV side
Z S , pu
Z S , pu
Vb1
Vb 2
Vb 2
Vb1
1:1
Fig 2. Per-unit Eq Ckt referred to LV side Fig 3.
Sb
18
Per-unit Eq. Ckt for 1- Transformer
RS
XS
j 2
2
a
a
S
a
N1
Zb2
N2
Vb1 VLV VHV
VLV
N
 1 1
VHV N 2
Vb 2
Fig 4. Eq Ckt referred to HV side
Z S , pu
Vb1
Z S , pu
Vb 2
Vb 2
Vb1
1:1
Fig 5. Per-unit Eq Ckt referred to HV side
Sb
Fig 6.
19
Voltage Regulation
Voltage regulation is defined as:
VR 
Vnoload  V fullload
V fullload
100%
In per-unit system:
VR 
V pu,noload  V pu, fullload
V pu, fullload
100%
Vfull-load: Desired load voltage at full load. It may be equal
to, above, or below rated voltage
Vno-load: The no load voltage when the primary voltage is
the desired voltage in order the secondary voltage
be at its desired value at full load
20
Voltage Regulation Example
A single-phase transformer rated 200-kVA, 200/400-V, and
10% short circuit reactance. Compute the VR when the
transformer is fully loaded at unity PF and rated voltage
400-V.
Solution:
Vb 2  400V
Sb  200kVA
Sload, pu  10 pu
X S , pu  j 0.1 pu
VP
j 0 .1
XS
VS
Sload
Fig 7. Per-unit equivalent circuit
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Voltage Regulation Example
Rated voltage:
VS , pu  1.00 pu
*
I load, pu
 Sload, pu   1.00 *
 

 1.00 pu

 V
  1.00 
 S , pu 
VP , pu  VS , pu  I pu X S , pu
 1.00  1.00  j 0.1  1  j 0.1
 1.0015.7 o pu
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Voltage Regulation Example
Secondary side:
V pu , fullload  VS , pu  1.00 pu
Vpu,noload  VP, pu  1.0015.7o pu
Voltage regulation:
VR 

V pu ,noload  V pu , fullload
V pu , fullload
 100%
1.001  1.0
 100%  0.1%
1.0
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Problem 1
j100
G
20 kV
22kV/220kV
80MVA
14%
220kV/20kV 50MVA
50MVA
0.8 PF
10%
lagging
Select Vbase in generator circuit and Sb=100MVA,
compute p.u. equivalent circuit.
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Per-unit System for 3- Circuits
Three-phase circuits
Sb  S3  3S1  3V I
where
V  Vlinetoneutral  VL (line) / 3
I  I linecurrent  I L
Sb  3VL I L
VbLV  VL , LV
VbHV  VL , HV
Sb  3VbLV I bLV  3VbHV I bHV
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Per-unit System for 3- Circuits
Sb

3VbLV
I bLV
Z bLV 
Z bHV
S pu
VLV
ILV
VbLV

3
I bHV
Sb

3VbHV
3VbLV (VbLV ) 2

Sb
Sb
(VbHV ) 2

Sb
3VL I L*


 V pu I *pu
Sb
3Vb I b
S3
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Per-unit System for 3- Transformer
Three 25-kVA, 34500/277-V transformers
connected in -Y. Short-circuit test on high voltage
side:
VLine, SC  2010V
I Line, SC  1.26 A
P3 , SC  912W
Determine the per-unit equivalent circuit of the
transformer.
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Per-unit System for 3- Transformer
(a) Using Y-equivalent
I SC  1.26
VSC
RS  jX S
2010

3
Sb  25000VA
34500
277
3
2010
 1160.47V
3
1160.47

 921.00
1.26
VSC 
Z SC
28
Per-unit System for 3- Transformer
912
P 
 304W
3
XS 
RS 
P
2
I SC

304
 191.48
2
1.26
Z SC  RS2  9212  191.482  900.86
2
So Z SC  191.48  j900.86
Sb  25000VA
Z b,HV
Vb, HV
34500

 19918.58V
3
19918.582

 15869.99
25000
Z SC , pu,Y
191.48  j 900.86

 0.012  j 0.0568 pu
15869.99
29
Per-unit System for 3- Transformer
(b) Using -equivalent
I SC
1.26

3
Z SC ,
Sb  25000VA
VSC  2010
34500 277
VSC  2010V
Z SC , 
I SC 
1.26
 0.727 A
3
2010
 2764.79
0.727
30
Per-unit System for 3- Transformer
912
P 
 304W
3
RS , 
P
2
I SC

304
 575.18
2
0.727
2
X S , 
Z SC ,  RS2,  2764.79 2  575.182  2704.30
So Z SC  191.48  j900.86
Sb  25000VA
Z b,HV
Z SC , pu ,
Vb , HV  34500V
34500 2

 47610
25000
575.18  j1704.30

 0.012  j 0.0568 pu
47610
31
Related Materials in Textbook
(1) Section 2.6 and 2.7, page 83~90, Chapman
book
(2) Section 2.10, page 113~116, Chapman book
32