Question 1:
[7marks]
Objective: This question related to power system transformers
A single-phase transformer rated 1.2kV/120V, 7.2kVA has the following
winding parameters: r1=0.8Ω, x1=1.2Ω, r2=0.01Ω and x2=0.01Ω.
Determine:
a) The combination winding resistance and leakage reactance referred
to the primary side.
b) The values of the combined parameters referred to secondary side.
c) The voltage regulation of the transformer when it is delivering 7.5
kVA to a load at 120V and 0.8 power factor lagging.
Solution:
a )The turn ratio
a
V
1
V2

1.2  103
 10
120
Resistance referred to the primary side
R1  r1  a 2 .r2  0.8  (102  0.01)  1.8
X 1  x1  a 2 .x2  1.2  (102  0.01)  2.2
R
1.8
b) Resistance referred to the secondry side
R2  21 
 0.018
a
100
X
2.2
Reactance referred to the secondry side
X 2  21 
 0.022
a
100
V2( NL )  V2( FL )
c) voltage regulation of the transformer is
Regulation 
(%)
V2( FL )
Reactance referred to the primary side
Z1  ( R1  jX 1 )  1.8  j 2.2
cos 1 (0.8)  36.9o
I 2( FL ) 
a
S2
7200

 60  36.9o
o
V2 12036.9
I 2( FL )
I1( FL )
 I1( FL ) 
I 2( FL )
a

A
60  36.9o
 6  36.9o
10
A
a.V2( FL )  10  120  1200V
V1( FL )  a.V2( FL )  I1( FL ) .Z1  1200  6  36.9o  (1.8  j 2.2)  1216.570.19o
percentage Regulation 
Dr.Audih alfaoury
V2( NL )  V2( FL )
V2( FL )
V1( FL )
 V2( FL )
a
121.6.57  120
(%) 
(%) 
 1.38%
120
V2( FL )
Philadelphia university
Question2:
Objective: This question related to short module of TL
a)
[10marks]
Find the ABCD constants of a π circuit having a 600Ωresistor for
shunt branch al sending end, a 1kΩ resistor for the shunt branch at
the receiving end and an 80Ω resistor for series branch.
Solution:
IL
Is
IR
80Ω
Vs
1000Ω
600Ω
VR
Vs  VR  V  VR  (80  I L )
V
I L  I R  I (1000  )  I R  ( R )
1000
V
Vs  VR  (80  I L )  Vs  VR  (80  ( I R  ( R )) 
1000
80VR
 1VR  80 I R 
 1VR  0.08VR  80 I R  1.08VR  80 I R
1000
from KCL
 Vs  1.08VR  80 I R and since
A  1.08 and
And for C and D from KCL
Vs  AVR  BI R then;
B  80
I s  I L  I  600  
VR
V
V
1.08VR  80 I R
)  ( s )  (I R  R )  (
)
1000
600
1000
600
 (0.001  0.0018) VR  (1  0.133) I R
I s  I L  I  600    ( I R 
And since I s  CVR  DI R  0.0028VR  1.133I R 
C  0.0028 and
b)
D  1.133
The ABCD constant of a three phase short length of TL are:
A  D  0.936  j 0.016  0.9360.98o
B  33.5  j138  14276.4o 
C  (5.18  j 914) x106 S
Dr.Audih alfaoury
Philadelphia university
The load at the receiving end is 50MW at 220kV with a power factor of
0.9 lagging. Find the magnitude of the sending end voltage and the
voltage regulation. Assume the magnitude of the sending end voltage
remains constant.
Solution:
P
50  106
IR 

 145.8  25.84o [A]
3
3.VLL .cos 
3  220  10  0.9
cos 1 (0.9)  25.84o and since its lagging then the angle is negative,thus;
I R  145.8  25.84o
[ A]
VLL 220  103

 1270o kV and
3
3
Vs  AVR  BI R  (0.9360.98o  (1270o )  103 )  ((14276.4o )  (145.8  25.84o )) 
VR 
 (118,855  j 2,033)  (13,153  j15,990)  1332337.77 o V or 133.237.77o kV
Vs 3  3  133.23  230.8 kV
VR ( nL ) 
Vs 3
A

230.8
 246.5 kV
0.9361
(%)regulation 
VR ( nL )  VR (FL)
VR (FL)
.% 
246.5  220
 100  12%
220
Question 3:
[6marks]
Objective: This question related to TL inductance calculation
Find the inductance in each side and total inductance for a circuit of a
single-phase TL is composed of three solid 0.25cm radius wires and
return circuit is composed of 0.5cm radius wires. The arrangement of
conductors is showing in figure.
Dr.Audih alfaoury
Philadelphia university
Solution:
L  Lx  Ly
 GMD 
 GMD 
7
Lx  2  107 ln 
and
L

2

10
ln


y

 GMRy 
 GMRx 
GMD  9 ( Dad  Dae  Daf )  ( Dbd  Dbe  Dbf )  (Dcd Dce  Dcf  10.940m
GMRx  9 ( Daa  Dab  Dac)  ( Dba  Dbb  Dbc)  ( Dca  Dcb  Dcc)  0.480 m
GMRy  9 ( Ddd  Dde  Ddf )  ( Ded  Dee  Def )  ( Dfd  Dfe  Dff )  0.606 m
note Dnn  r '  r for solid conductors and Dnn  r for stranded conductors
Daa  Dbb  Dcc  0.7788  (0.25  102 )  1.947  103 m
Ddd  Dee  Dff  0.7788  (0.5  102 )  3.894  103 m
  GMD 
 GMD  
10.940 
10.940  
7 
L  Lx  Ly  2  107  ln 

ln

2

10
ln

ln



  0.480 
 0.606   
  GMRx 

G
M
R






y



 1.2  106 ( H / m)
Question4:
Objective: This question related to capacitance of TL
[10marks]
A three phase transposed line is composed of ACSR Bobolink conductor per
phase with flat horizontal spacing of 11 meters as in figure .The conductors
have a diameter of 3.625 cm and GMR of 1.439cm. The line is to be placed
by a three conductor bundle of ACSR Hawk conductor .The conductors have
diameter of 2.1793cm and GMR of 0.8839cm .The new line will also have a
flat horizontal configuration, but it is to be operated at higher voltage and
therefore the phase spacing is increased to 14 m as measured from the center
of the bundles as in figure .The spacing between the conductors in the bundle
is 45cm .Determine the percentage change in the capacitance.
Dr.Audih alfaoury
Philadelphia university
Solution:
11m
a
11m
b
c
The capacitance per unit length is
2 o
,( F / m) (We propose to define a capacitance C between
 GMD 
ln 

 GMR 
each conductor and neutral as in figure).
C
From electromagnetic cours
 .l
C Q

 l2 
l V .l E.l
l

l
.l
2 o r
l
b

a
l
.dr
2 o r

2 o
b

a
1
dr
r

2 o
r 
ln  b 
 ra 
, ( F / m)
GMD  3 Dab .Dbc .Dca  3 112  22  13.859m
GMR=r (for single bundled stranding conductor), and since we have diameter
of conductor is 3.625cm, then
r
d 3.625

 1.8125  102 m
2
2
The capacitance per unit length of old conductors is
2  8.85  1012
5.5606  10 11
C

 8.375  1012 F / m
6.63939
 13.859 
ln 
2 
 1.8125  10 
or
C  8.375  1012  1000  8.375  109 F / km  0.008375 F / k m
For new bundled stranded conductors with new spacing as in figure
Bundled spacing
14m
Dr.Audih alfaoury
14m
14m
Philadelphia university
45cm
The new GMD= 3 142  28  17.63889m
and GMR for tree bundled
stranded conductors is found by the following relation:
GMRb3  3 r.d 2 Where r is radius of conductor and d is the spacing of bundled
GMRb3  3
2.1793
 452  13.01879  102 m
2
The new capacitance per unit length is:
2 o
2  8.85  1012
Cnew 

 1.1327766  1011 F / m
 GMD 
 17.63889 
ln 
ln 
2 
3 
 13.01879  10 
 GMRb 
Or capacitance per unit length for new
Cnew  1.1327766  1011  1000  1.1327766  108 F / k m
or Cnew  0.01132766  F / k m
The percentage change of capacitance when replacing conductor by using
bundled is:
percentage 
Cnew  Cold
0.01132766  0.00837
.100% 
.100%  35.25%
Cold
0.00837
Dr.Audih alfaoury
Philadelphia university
From above question we note:
2 o
GMD
ln(
)
GMR
for solid conductor GMR  DS  0.7788  r
C
bundled stranded conductors GMR is :
 for one stranded conductors GMR  Ds  r
 for two stranded conductors GMR  Ds  4 (r  d ) 2  r  d
 for three stranded conductors GMR  Ds  9 (r  d  d )3  3 r  d 2
 for four stranded conductors GMR  Ds  16 (r  d  d  d  2) 4  1.09 4 r  d 3
d  distance between bundled conductor per phase is meter
r  conductor radius is meter
Question5:
Objective: This question related to symmetrical fault
[10marks]
A60-Hz generator is rated 500MVA,20kV with sub-transient reactance 0f
0.2pu.Supply a purely resistive load of 400MW at 20kV .The load is
connected directly to generator terminals ,if all three phases of load are
short-circuit simultaneously, find the initial symmetrical rms current in
the generator in per unit on base of 500MVA and 20kV.
Solution:
I 
''
g
Eg''
Z
''
g
where E g''  Vt  I L .Z''L
In per unit we chose 500MVA as power base and 20kV as voltage base
(note always chose the source as power base and TL as voltage base since
they are greater value) and we need to find the current IL in per unit.
Dr.Audih alfaoury
Philadelphia university
I pu 
I actual

I base
P
3VLL cos 
S
3VLL
and since the load is resistiv that means power
factor is unity cos   1 and R is neglected Z ''  jX ''
I pu 
P
3VLL P 400
.
 
 0.8 pu
S 500
3VLL S
and V pu 
Vactual 20000

 1 pu 
Vbase 20000
E g''  Vt  I L . jX ''  10o  (0.8  j 0.2)  1  j0.16 pu 
I g'' 
Eg''
jX
''

1  j0.16
 0.8  j 0.5 pu
j 0.2
or
I g''  0.82  0.52  5.06 pu
 (optional )
I g'' (actual )  I g'' ( pu )  I g'' (base)  5.06  (
Dr.Audih alfaoury
500  106
)  5.06  14433.7  73.0348kA
3  20  103
Philadelphia university