ECE 421:
Introduction to Power Systems
Session 14; Page 1/13
Fall 2013
ECE 421: Per Unit Examples
Define units:
MVA  1000kW
kVA  kW
MW  MVA
kVAR  kW
pu  1
Example 1: A three-phase transformer rated 5 MVA, 115/13.2 kV has per-phase series impedance of (0.007 + j0.075) per unit.
The transformer is connected to a short distribution line which can be represented by a series impedance per phase of (0.02 +
j0.10) per unit on a base of 10 MVA, 13.2 kV. The line supplies a balanced three-phase load rated 4MVA, 13.2kV, with lagging
power factor 0.85.
Srated  5MVA
Vhi  115kV
Vlo  13.2kV
2
Zbase_Hi 
Vhi
Zbase_Hi  2645 Ω
Srated
2
Zbase_Lo 
Vlo
Zbase_Lo  34.848 Ω
Srated
Zseries  0.007pu  j  0.075pu
Distribution line:
Load:
on transformer rating base.
Zline  0.02pu  j  0.10pu
Sload  4MVA
VLL_load  13.2kV
on 10MVA, 13.2kV base
pf  0.85 lagging
(a) Draw an equivalent circuit of the system indicating all impedances in per unit. Choose 10MVA, 13.2 kV as the
base of the load.
2
VBLL
SB  10MVA
VBLL  13.2kV
ZB 
ZB  17.424 Ω
SB
ECE 421:
Introduction to Power Systems
Session 14; Page 2/13
Fall 2013
Transformer impedance (change of base calculation):
2
 Vlo   SB 
Zseries_new  Zseries 
 

 VBLL   Srated 
Zseries_new  ( 0.014  0.15i)  pu
Line impedance (no change of base needed, already on the correct base)
Zline  ( 0.02  0.1i)  pu
Load:
ϕload  acos ( pf )
ϕload  31.788 deg
  VLL_load 2  j  ϕload
Zload 
 Sload  e


Zload_pu 
positive since lagging power factor
Zload  ( 37.026  22.947i) Ω
Zload
Zload_pu  2.125  1.317i
ZB
Per unit equivalent circuit:
Iload
0.014 + j 0.15 p.u.
Load
Vs = 1.0 p.u.
Line Impedance
2.125 + j 1.317 p.u.
Transformer
0.02 + j 0.1 p.u.
ECE 421:
Introduction to Power Systems
Session 14; Page 3/13
Fall 2013
(b) With the voltage at the primary side of the transformer held constant at 115kV, the load at the
receiving end of the line is disconnected. Find the voltage regulation at the load.
Vsource  115kV
Vsource_pu 
Iload 
LL
Vsource
Vhi
Vsource_pu  1 pu
Vsource_pu
Zseries_new  Zline  Zload_pu
Iload  0.375 pu
Vload_pu  Iload Zload_pu
Vload_pu  0.937 pu
Vregulation 
Use as reference angle (so 0 deg)
Iload  ( 0.303  0.22i)  pu
arg  Iload  35.971 deg
Vload_pu  ( 0.935  0.068i)  pu
arg  Vload_pu  4.183 deg
Vsource_pu  Vload_pu
Vload_pu
Vregulation  6.708 %
ECE 421:
Introduction to Power Systems
Session 14; Page 4/13
Fall 2013
Example 2: A transformer rated 200 MVA, 345 Y/20.5 kV connects a balanced load rated 180
MVA, 22.5kV, 0.8 power-factor lag to a transmission line. Determine:
Sr  200MVA
VHI  345kV
VLO  20.5kV
Sload  180MVA
Vload  22.5kV
pf  0.8
lag
(a) The rating of each of the three single phase transformers which when properly connected will be
equivalent to the above three-phase transformer.
Vrating_xfmr_HV 
VHI
3
Vrating_xfmr_LV  VLO
Sper_xfmr 
Sr
3
Vrating_xfmr_HV  199.186 kV
Vrating_xfmr_LV  20.5 kV
Sper_xfmr  66.667 MVA
(b) The complex impedance of the load in per unit in the impedance diagram if the base in the
transmission line is 100MVA, 345kV.
ϕload  acos ( 0.8)
ϕload  36.87 deg
 Vload2  j  ϕload
Zload_ohm_LV 
 Sload   e


Zload_ohm_LV  2.812 Ω
arg  Zload_ohm_LV  36.87 deg
SBASE  100MVA
ECE 421:
Introduction to Power Systems
Session 14; Page 5/13
Fall 2013
VBHI  345kV
 VLO 

 VHI 
VBLV  VBHI 
VBLV  20.5 kV
2
ZBLV 
VBLV
ZBLV  4.202 Ω
SBASE
Zload_pu 
Zload_ohm_LV
Zload_pu  ( 0.535  0.402i)  pu
ZBLV
arg  Zload_pu  36.87 deg
Zload_pu  0.669 pu
Example 3: Sketch a per unit impedance diagram for the system shown below. Use a 100MVA impedance base, and the
generator 1 rated voltage as your reference voltage base. Use pi models for the lines.
BUS 1
G1
BUS 2
T1
BUS 3
BUS 4
Line 2
Line 1
BUS 5
T2
G2
Load




G1: 50MVA, 13.8kV
G2: 20MVA, 14.4kV
T1: 40MVA, -Y, 13.2:161kV, X = 10%
T2: 25MVA, Y-, 161kV:13.2kV, X = 10%



Load: 45MVA, 0.8pf lagging (Y connected, parallel impedances)
Line 1: 100 mile, Z = 0.28 + j0.73 ohm/mi, Y = 5.9*10-6 mho/mi
Line 2: 75 mile, Z = 0.28 + j0.73 ohm/mi, Y = 5.9*10-6 mho/mi
ECE 421:
Introduction to Power Systems
Define Base Quantities:
Session 14; Page 6/13
Fall 2013
Section I (left of T1)
SB  100MVA
VB1  13.8kV
Line to line voltage
2
ZB1 
IB1 
VB1
SB
SB
3 VB1
Section II (between T1 and T2)
161kV 

 13.2kV 
VB2  VB1 
ZB1  1.904 Ω
IB1  4183.698 A
Section II (right of T2)
VB2  168.318 kV
13.2kV 

 161kV 
VB3  VB2 
2
ZB2 
IB2 
VB2
SB
SB
3 VB2
VB3  13.8 kV
2
ZB2  283.31 Ω
IB2  343.011 A
ZB3 
IB3 
VB3
SB
SB
3 VB3
ZB3  1.904 Ω
IB3  4183.698 A
ECE 421:
Introduction to Power Systems
Session 14; Page 7/13
Fall 2013
Transmission Line Models:
Line 1:
Length1  100mi
Zline1  ( 0.28  j  0.73)

 6 mho
Yline1  j  5.9 10
mi
ohm
  Length1
mi 
Zline1  ( 28  73i)  ohm
 Length1
Yline1  5.9i  10
4
 mho
Yline1
2
4
 2.95i  10
 mho
Line 1 is in section II, so use Zbase2
Zline1pu 
Zline1
Zline1pu  ( 0.099  0.258i)  pu
ZB2
Note that Ybase is 1/Zbase:
Yline1pu  Yline1 ZB2
Yline1pu  0.167i pu
We actually need Y/2 for the pi model:
Yline1pi 
Yline1pu
Yline1pi  0.084i pu
2
Line 2:
length2  75mi
ohm
Zline2  ( 0.28  j  0.73)
  length2
mi 

 6 mho
Yline2  j  5.9 10
mi
 length2
Zline2  ( 21  54.75i)  ohm
4
Yline2  4.425i  10
 mho
Yline2
2
4
 2.212i  10
 mho
ECE 421:
Introduction to Power Systems
Session 14; Page 8/13
Fall 2013
Line 2 is also in section II, so use Zbase2
Zline2pu 
Zline2
Zline2pu  ( 0.074  0.193i)  pu
ZB2
Yline2pu  Yline2 ZB2
Yline2pu  0.125i pu
We actually need Y/2 for the pi model:
Yline2pi 
Yline2pu
Yline2pi  0.0627i pu
2
Transformer Model Calculations
Transformer 1:
ST1  40MVA
VT1Low  13.2kV
VT1hi  161kV
XT1  0.10pu
Impedance change of base calculation
 VT1Low 
XT1new  XT1  

 VB1 
2
 SB 

 ST1 

XT1new  0.229 pu
Transformer 2:
ST2  25MVA
VT2Low  13.2kV
VT2hi  161kV
Impedance change of base calculation
2
 VT2Low   SB 
XT2new  XT2  
  
 VB3   ST2 
XT2new  0.366 pu
XT2  0.10pu
ECE 421:
Introduction to Power Systems
Session 14; Page 9/13
Fall 2013
Load Model
magSload  45MVA
pfload  0.8
Vloadrated  161kV
lagging
ϕload  acos ( pfload)
ϕload  36.87 deg
j  ϕload
Sload  magSload e
Sload  ( 36  27i)  MVA
Since the load is wye connected with parallel impedances:
Rload 

Vloadrated
Re  Sload
2
Rload  720.028 Ω
Xload 

Vloadrated
Im  Sload
2
Xload  960.037 Ω
As a check:
Zequivload
1
1 
 


 Rload j  Xload 
1
Zequivload  576.022 Ω
arg  Zequivload   36.87 deg
Zequivload  ( 460.818  345.613i) Ω
Scheck 

Vloadrated 

Zequivload
2
Scheck  ( 36  27i)  MVA
note complex conjugate in equation
Convert to per unit (load in section II):
Rloadpu 
Rload
ZB2
Rloadpu  2.541 pu
Xloadpu 
Xload
ZB2
Xloadpu  3.389 pu
ECE 421:
Introduction to Power Systems
Session 14; Page 10/13
Fall 2013
Per Unit Equivalent Circuit:
j0.229pu
0.099 + j 0.258 pu
Y/2=j 0.084 pu
2.54 pu
Ea1
Line 1
j 3.39pu
Transformer1
0.074 + j 0.193 pu
Load
Line 2
Y/2=j 0.0627 pu
j0.366pu
Transformer2
Ea2
Generator2
Generator1
(b) Suppose G1 is operating at 13.6kV and G2 is set to operate at the same magnitude. Suppose also, that the two generators are
in phase with each other. Determine the phase A line to neutral voltage in Volts and in per unit and the phase A current at the load
in Amperes and in per unit. Determine magnitude and phase is each case. Use the generator 1 voltage as your reference angle.

Create a Thevenin equivalent circuit looking back to the two sources from the load:
Circuit to left of the load:
Zleft  j  XT1new  Zline1pu
Zleft  ( 0.099  0.486i)  pu
13.6kV  j  0deg
Vgen1pu  
e
 VB1 
Vgen1pu  0.986 pu
Create a Norton equivalent:
Ileft 
Vgen1pu
Zleft
Ileft  1.986
arg  Ileft  78.514 deg
ECE 421:
Introduction to Power Systems
Session 14; Page 11/13
Fall 2013
Circuit to right of load:
Zright  j  XT2new  Zline2pu
Zright  ( 0.074  0.559i)  pu
13.6kV  j  0deg
Vgen2pu  
e
 VB3 
Vgen2pu  0.986 pu
Vload
2.54 pu
Zleft
j 3.39pu
Ea
Zleft
Ea
Zright
Zright
Zleft Zright
Vload
(
Ea
Zleft
E
+ Za
)
Zleft Zright
right
Create a Norton equivalent:
Iright 
Vgen2pu
Zright
Iright  1.747
arg  Iright  82.45 deg
2.54 pu
Zleft Zright
j 3.39pu
right
2.54 pu
E
+ Za
j 3.39pu
Ea
Zleft
Vload
ECE 421:
Introduction to Power Systems
Session 14; Page 12/13
Fall 2013
Combine the two parallel impedances from the sources:
1
1 
Zpara  


 Zleft Zright 
1
Zpara  0.264
arg  Zpara  80.356 deg
Combine the two current sources:
Ipara  Ileft  Iright
arg  Ipara  80.356 deg
Ipara  3.73
Find the Thevenin equivalent source voltage:
Vthev  Ipara Zpara
Vthev  0.986 pu
Zthev  Zpara
Zloadequiv
1
1

 


 Rloadpu j  Xloadpu 
Now we can find the load current:
Iloadpu 
Vthev
Zthev  Zloadequiv
1
Zloadequiv  2.033 pu
arg  Zloadequiv   36.87 deg
Zloadequiv  1.627  1.22i
Iloadpu  0.441 pu
arg  Iloadpu   41.542 deg
Vloadpu  0.898 pu
arg  Vloadpu   4.672 deg
And the voltage across the load:
Vloadpu  Iloadpu  Zloadequiv
ECE 421:
Introduction to Power Systems
Session 14; Page 13/13
Fall 2013
Now convert to Ampere and Volts:
IloadAmps  Iloadpu  IB2
arg  IloadAmps  41.542 deg
IloadAmps  151.433 A
VANload  Vloadpu 
VB2
Note that we are using a line to neutral voltage base, since the angles in per unit
correspond to the line to neutral voltages.
3
VANload  87.229 kV
VABload 
arg  VANload  4.672 deg
j  30deg
3 VANload e
VABload  151.084 kV
arg  VABload   25.328 deg