a Use the short line model to find the voltage and power at the sending end and the voltage regulation and efficiency when the line is supplying a three-phase load of 381 MVA at 0.8 pow
Trang 1• Short Line Model
• Medium-length Line Model
• Long Line Model
• Lossless Lines
• Voltage Profiles
• Complex Power Flow Through Transmission Lines
• Power Transmission Capability
S
R R
S
DI CV
I
BI AV
z = series impedance per unit length per phase
y = shunt admittance per unit length per phase
to neutral
Notation :
Line Model and Performance
l = length of line
Z = zl = total series impedance per phase
Y = yl = total shunt admittance per phase
to neutral
Trang 2Short Line Model :
• The line length is less than about 80 km
(50 miles) long
• The shunt admittance is neglected.
• The series resistance and reactance are
treated as lumped parameters.
The ABCD parameter : A =1 , B = Z , C =0 , D =A =1
L j R
Z= + ω
R R
S
I
V Z I
V
1 0
1
V
V V
Short Line Model
Trang 3Example :A 220-kV , 60 Hz , three phase transmission
line is 40 km long The resistance per phase is 0.15 Ω per
km and the inductance per phase is 1.3263 mH per km
(a) Use the short line model to find the voltage and
power at the sending end and the voltage regulation and
efficiency when the line is supplying a three-phase load of
381 MVA at 0.8 power factor lagging at 220 kV
(b) Determine the voltage regulation when the line is
supplying a three-phase load of 381 MVA at 0.8 power
factor leading at 220 kV
The series impedance per phase is
l L j r
Ω+
=6 j20
kV 012703
220∠ o= ∠ o
Short Line Model
(a)The receiving end power is
The current per phase is given by
kA 87.360.1)8.0cos(127
)3/381
S V ZI
kV 250 ( ) =
⇒ VS L −L
kV 93 4 33 144
) 87 36 1 )(
20 6 ( 0 127
o
o o
∠
=
−
∠ +
+
∠
MVA6.2288.30487.36
)8.0(cos
) 3 (
−
∠
=φ
220250
8
V
V V VR
Trang 4(b)The current per phase is given by
kA 87.360.1)8.0(cos127
)3/381
S V ZI
kV 3 210 ( ) =
⇒ VS L −L
kV 29 9 39
2203.210
) 87 36 1 )(
20 6 ( 0
Medium-length Line Model
Medium-length Line Model :
• The line length is between 80 km (50 miles) and
240 km (150 miles)
• The circuit is called a nominal π circuit
• Half of the shunt capacitance is considered to be
lumped at each end of the line
• The series resistance and reactance are treated as
lumped parameters
R
R YZ I V
2
,
, 2
=
A
L j R
Trang 5Example :A three-phase completely transposed 345-kV,
200-km line has two 795,000-cmil 26/2 ACSR conductors
per bundle and the following line constants :
S/km 10
4.2
, /km 35 0 032
.
z
Full load at the receiving end of the line is 700 MW at
0.99 p.f leading and at 95% of rated voltage , determine
the following :
(a) ABCD parameters of the nominal πcircuit
(b) Sending-end voltage , current , and real power
(c) Percent voltage regulation
(d) Transmission-line efficiency at full load
Solution :
The ABCD parameters are
S/km 10
4.2 ,
/km 35 0 032
z
S 10 4 8 200 ) 10
=
× +
= ( 0 032 j 0 35 ) 200 6 4 j 70 70 29 84 78o
Z
o159.09706.02
)704.6)(
104.8(1
4
∠
=+
×+
=
=
−
j j
704.6(1)104
8
(
4 4
Medium-length Line Model
The receiving end current is
kV 0 189.2 0
3
345 95
3 /
The receiving end voltage per phase is
The sending end voltage and current are
)11.8246.1)(
78.8429.70()02.189)(
159.09706.0
=
kV 14 26 6
=
) 11 8 246 1 )(
159 0 9706 0 ( ) 0 2 189 )(
08 90 10 277 8
⇒ VS L −L
The sending end real power is
MW 730
) 5 15 14 26 cos(
) 241 1 )(
6 199 ( 3
) 3 (
Trang 6The no-load receiving end voltage is
Voltage regulation is
Transmission line efficiency is
%8.951005.730
6 199
=
=
% 7 8 100 2
189
2 189 6 205
Long Line Model
Long Line Model :
• The line length is greater than 240 km (150 miles)
• Accuracy obtained by using distributed parameters
x zI x
x V x x
∆
−
∆+
)()(
x zI dx
) ( ) (
x x yV x
x I x x I
∆ +
=
∆
−
∆ +
) ( ) (
x yV dx
Trang 7) ( ) (
x yV dx
x
) ( )
(
x zI dx
x
dx
x dI z dx
x I
2
2
= )
x x
e Z
A e Z
A x I e
A e
Long Line Model
α is the attenuation constant
β is the phase constant (rad/m)
β α
c
x x
e Z
A e Z
A x I e
A e
and 2
A Z
I V
=+
c
x x
e Z
A e Z
A x I e
A e A x
2
1 and ( ) )
(
c c R R
Z
A Z
A I A
x c R R x c R R
e I Z V e I Z V x I
e Z I V e Z I V x V
γ γ
γ γ
/)(
22
)(
Long Line Model
R x x c R x x
x c R R x c R R
I e e Z V e e
e Z I V e Z I V x V
− +
2 2
) (
γ γ γ
γ
γ γ
R x x R x x
c
x R c R x R c R
I e e V e e Z
e I Z V e I Z V x I
1
2
/ 2
/ ) (
γ γ γ γ
γ γ
Rearranging
R R
R c
R
I x V
x Z
x I
I x Z
V x x
V
)cosh(
)sinh(
1)(
)sinh(
)cosh(
)(
γγ
γγ
θ θ θ
θ
θ
Trang 8At sending end , x = l ;
The ABCD parameters :
)sinh(
1
;)sinh(
;)
Z C l Z
B l D
c S
R c
R S
I l V
l Z
I
I l Z
V l V
)cosh(
)sinh(
1
)sinh(
)cosh(
γγ
γγ
R R
c
R c
R
I x V
x Z
x I
I x Z
V x x
V
)cosh(
)sinh(
1)(
)sinh(
)cosh(
)(
γγ
γγ
+
=
+
=
Long Line Model
The computation of hyperbolic function of complex
arguments :
) sin(
) cosh(
) cos(
) sinh(
) sinh(
)
sinh(
) sin(
) sinh(
) cos(
) cosh(
) cosh(
)
cosh(
l l
j l l
l j l l
l l
j l l
l j l l
β α
β α
β α γ
β α
β α
β α γ
Example :A 50 Hz transmission line 300 km long has
a total series impedance of 40+j125 Ω and a total shunt admittance of j10-3S The receiving-end load is 50 MW
at 220 kV with 0.8 lagging power factor
Find the sending-end voltage, current, power and power factor using exact transmission line equations
Long Line Model
The total series impedance and shunt admittance are
S901010
3.722.13112540
=
−
−
j Y
j Z
Solution :
o o5.81354.035.0052.0)sinh(
2.1938.002.0938.0)cosh(
∠
=+
=
∠
=+
=
j l
j l
γγ
ZY
γ =0.0554+ j0.3577=0.362∠81.2o
Y Z
Trang 9kV 0 127 0 3
4.901077.9
)sinh(
C c
γ
Long Line Model
MW 16 52 987 0 1286 0 97 136
987 0 ) 3 15 2 6 cos(
S
R R
kA 3.15
1286
0
)9.36164.0)(
2.1938.0()0127)(
4.9010
o o
kV 2
65 72 2 128 ( ) 0 127 )(
V
o
o o
o o
The ABCD parameter :
=
=
41
,
, 2
D A
)sinh(
1
)sinh(
)cosh(
l Z
C
l Z
B
l D
A
c
c
γγγ
=
=
=
=
Long Line Model
Equating the coefficients
2 1 and )
Z
) sinh(
y
z
γ z sinh( l γ ) γ
=
l
l Z
γ
γ ) sinh(
=
) sinh(
1 ) cosh(
= ) 2 / tanh(
l
l l
′
−
= cosh( ) 1 2
) sinh(
1 ) cosh(
l Z
) 2 / tanh(
2
=
l
l Y
γ
γ )
2 /
Trang 10Equivalent ππππ Circuit
) 2 / ( ) 2 / tanh(
l Y
γ γ
l
l Z
γ
γ) sinh(
For a lossless line , R = G = 0 , and
S/m
, /m
L j
L j
ω ω
c S
R c
R S
I l V
l Z
I
I l Z
V l V
) cosh(
) sinh(
1
) sinh(
) cosh(
γ γ
γ γ
c S
R c
R S
I l V
Z
l j I
I l jZ
V l V
) cos(
) sin(
) sin(
) cos(
β β
β β
α = 0 → = j
The velocity of propagationof voltage or current wave
The wavelengthis the distance required to change the phase of the voltage or current by 2π radians or 360o
LC
v LC
=
λ
Lossless Lines
Trang 11Surge impedance loading (SIL)is the power delivered
by a lossless line to a purely resistive load equal to its
surge impedance
For a lossless line ,
R c
R jZ x I V
x x
V( )=cos(β ) + sin(β )
c
R c
R
Z
V x jZ
c
I x V
V x V
Z
x j
)cos(
= cos(β ) sin(β )
c
R x j
)
x I x V
Z
V e V
The real power flow along a lossless
line at SIL remains constant from
the sending end to the receiving end.
rated
Z
V SIL
2
=
Lossless Lines
23002200
266254765
1075850294
233500
425325366
280345
145134395
365230
5247405
366138
1312400
36669
)MW()
((kV)
Assume a lossless line
Determine the line phase constant β, the surge
impedance Z c , the velocity of propagation v ,
and the line wavelength λ
Lossless Lines
Trang 12The surge impedance is
Solution : For a lossless line ,
0115.0
1097.0
6 3
o64.21rad3777.0300001259
0
100115.01097.060
The line wavelength is
The velocity of propagation is
km/s 10994.2100115.01097
10994
• At SIL , the voltage profile is flat
• For a short circuit at the load ,
RSC c
R c
R jZ x I V
x x
V( )=cos(β ) + sin(β )
Sh ort circuit
SV
Lossless Lines
Trang 13c c
c c
Z
l j l
jZ l
l Z
l Z
l
YZ Y Z
YZ
Z
)sin(
)sin(
)cos(
)sinh(
1)sinh(
)cosh(
4
12
1
01
ββ
β
γγ
Transmission-line ABCD parameters
Complex Power Flow
Through Transmission Lines
R R
S AV BI
)(
S R
B
V A B
V
I = ∠δ−θ − ∠θ −θ
o 0
A A
B R S
B
V A B
V V
θθδ
V V I V
)cos(
R
B
V A B
V V
)sin(
R
B
V A B
V V
The received power will be maximum at δ =θB
)cos(
R S R
B
V A B
V V
) 1 ( φ
R
P
Through Transmission Lines
)(
R
B
V A B
V V
k
n
Complex Power Flow Through Transmission Lines
Trang 14( ) (
L R B
L L R L
V V
I
V
S
θθδ
( )
( )
L R B
L L R L
L
S
R
A B L
L R B
L L R L
V V
Q
B
V A B
V V
P
θθδ
θ
θθδ
Through Transmission Lines
Example :A 275kV transmission line has the following
line constant :
o
o , 200 755
85
A
Determine the power at unity power factor that can be
received if the voltage profile at each end is to be
L R B
L L R L L S R
B
V A B
V V
)575sin(
200
27585.0)75sin(
200
2752750
2
o o
0 =
Through Transmission Lines
)cos(
L R B
L L R L L S R
B
V A B
V V
)575cos(
200
27585.0)2275cos(
=
Complex Power Flow Through Transmission Lines
Trang 15• The thermal limit
• The voltage-drop limit
• The steady-state stability limit
For short line :the loadability is determined by the
conductor thermal limit or by ratings of line terminal
equipment such as CB
For medium-length line :the loadability is determined by
the voltage-drop limit
For long line :the loadability is determined by the
steady-state stability limit
Major line-loading limits
Through Transmission Lines
IL 1:Theoretical steady-statestability limit
2:Practical line loadability
1
2
Thermal limit
Complex Power Flow
Through Transmission Lines
Line Compensation
• Shunt Reactors (Inductors) Compensation
• Shunt Capacitor Compensation
• Series Capacitor Compensation
• Flexible AC transmission system (FACTS)Line Compensation
Line Compensation
- installed from each phase to neutral
- absorb reactive power
- reduce overvoltages during light load condition
- reduce transient overvoltages due to switching and lightning surges
- reduce line loadability if they are not removedunder full-load condition
V
I =
• Shunt Reactors (Inductors) Compensation
Trang 16Example :A three-phase, 60 Hz, 765-kV transmission
line is 400 km long The line inductance is 0.88853 mH/km
per phase and its capacitance is 0.01268 µF/km per phase
The line resistance is 0.011 Ω/km
(a) Calculate the receiving end voltage when line is
terminated in an open circuit and is energized with 765 kV
at the sending end
(b) Determine the reactance and the MVAR of a
three-phase shunt reactor to be installed at the receiving end to
keep the no-load receiving end voltage at the rated value
Line Compensation
Ω +
=
×
× +
134 4
4
) 10 88853 0 60 2 011 0 (
0.0083 j
=
Solution :
S 10 912 1 ) 10 01268 0 60 2
) 5062 0 sin(
) 0083 0 sinh(
) 5062 0 cos(
) 0083 0 cosh(
) sin(
) sinh(
) cos(
) cosh(
) cosh(
∠
= +
γ
o 137 89 48495 0 4849 0 0073 0
) 5062 0 sin(
) 0083 0 cosh(
) 5062 0 cos(
) 0083 0 sinh(
) sin(
) cosh(
) cos(
) sinh(
) sinh(
∠
= +
γ
o2 88 42 128 )
o262 0 8746 0 )
0.5062 +
0.0083 j
l = γ
Line Compensation
(a) The sending end voltage per phase is
When the line is open , I R =0
kV067.44103
=
S V
R R
67
o = ∠−
∠
=The no-load receiving end line-to-line voltage is
kV 7.8745053) (L −L = × =
R
Trang 17(b) Determine the reactance and the MVAR of a three-phase
shunt reactor to be installed at the receiving end to keep
the no-load receiving end voltage at the rated value
)cos(
L R B
L L R L
V
V
)262.02.88cos(
42.128
7658746.0)2.88cos(
×
o0 ≈
o
o , 128.42 88.2262
.08746
7652
Lsh X
)sin(
L R B
L L R L
42.128
7658746.0)02.88sin(
• Shunt Capacitor Compensation
- deliver reactive power
- increase transmission voltages during heavyload condition
- always bus, rather than line, connected
Trang 18Example :A three-phase, 60 Hz, 765-kV transmission
line is 400 km long The line inductance is 0.88853 mH/km
per phase and its capacitance is 0.01268 µF/km per phase
The line resistance is 0.011 Ω/km The line supplies a load
of 1600 MW , 0.8 power factor lagging at 765 kV
Determine the MVAR and the capacitance of the shunt
capacitors to be installed at the receiving end to keep the
receiving end voltage at 765 kV when the line is energized
with 765 kV at the sending end
Line Compensation
Solution :
)cos(
L R B
L L R L
V V
)262.02.88cos(
42.128
7658746.0)2.88cos(
×
o69 20 ≈
o
o , 128.42 88.2262
.08746
A
)sin(
L R B
L L R L L S R
B
V A B
V V
MVAR
46 227
=
)262.02.88sin(
42.128
7658746.0)69.202.88sin(
42.128
C
1200 46
.
227 + QC,sh =
MVAR
54 972
jX ,
sh C
Q ,
Load
Q
7 601 54 972
• Series Capacitor Compensation
- reduce the net series impedance of the line
- reduce line-voltage drops
- increase the steady-state stability limit
- increase the loadability of long lines
- installed in series with each phase conductor atselect points along a line
- automatic protection devices must be installed tobypass high current during faults and to reinsertthe capacitor banks after fault clearing
- may be excite low-frequency oscillations
Trang 19Example :A three-phase, 60 Hz, 765-kV transmission
line is 400 km long The line inductance is 0.88853 mH/km
per phase and its capacitance is 0.01268 µF/km per phase
The line resistance is 0.011 Ω/km The line supplies a load
of 1600 MW , 0.8 power factor lagging at 765 kV
Only series capacitors are installed at the midpoint of the
line providing 40 % compensation Find the sending end
voltage and voltage regulation
o137.8948495.0)sinh(γl = ∠
34.12804.4197.884
)sinh(
1)cosh(
22tanh
2
l
l Z
l Z
For 40 percent compensation
The new equivalent π model are given by
j Y
Y new′ = old′ =
o
25 0 925 0 004 0 925 0
1 2
) 00195 0 )(
77 04 4 ( 1 2
∠
= +
=
+ +
= +
Z
A new new
o8711.77
7704.4
=
−+
=
′ 4.04 j(128.34 51.34) 4.04 j77
Z new
Line Compensation
The receiving end voltage per phase is
The receiving end current is
kA 87.36509.1)8.0cos(8.07653
=
R V
The line supplies a load of 1600 MW , 0.8 power factor lagging at 765 kV
Trang 20The sending end voltage is
The sending end line-to-line voltage magnitude is
Voltage regulation is
%33.20100765
765925.0/5.851
kV 5.8516.4913) (L −L = × =
S
V
R R
S AV BI
kV 7
87 11 77 ( ) 0 7 441 )(
25
o o
∠
∠
=
Line Compensation
Flexible AC transmission system (FACTS)
• Static VAR Compensator (SVC)
• Thyristor-controlled Series Capacitor (TCSC)
• Static Synchronous Compensator (STATCOM)
• Unified Power Flow Controller (UPFC)
Line Compensation
• Static VAR Compensator (SVC)
Trang 22At system voltage up to 230 kV Lightning surge
Transmission Line Transients
Lightning surge + Switching surge
At system voltage above 700 kV Switching surge
At system voltage above 230 kV but less than 700 kV
•External (Lightning surge)
•Internal (Switching surge)
Trang 23Transient analysis : Traveling Waves
Consider only the case of a lossless line
t
i x L x x
Transmission Line Transients
x x
i
i ∆
∂
∂ +
i L x
2
2 2
t
v C t x
2 2
t
v LC x
v C x
2
2 2
t
i L t x
t
i LC
∂
∂
= 2
2
t
v LC
∂
∂
=
2 2
2 2
t
i LC x
1 x kt f x kt f
LC
k= 1
2 2
2 2
t
v LC x
2
2 1
1
) (
, ) ( ) ( )
f x
v kt
x
f x
kt x kt x
f x
2 1
1
) (
, ) (
) ( )
f k t
v kt
x
f k t
kt x kt x
f t
2 2
2
1
t
v k x
Trang 24) ( )
3 x kt f x kt f
2 2
2
2
t
i LC
v C
v dx x
v k
L i
v
=
=
Consider the forward wave v= f1(x−kt)
For a forward traveling wave
)(
: Voltage v+ = f1 x−kt
)(1 :
Z i
c
−
=+
For a backward traveling wave
c
C
L i
v Z
C
L i
v++ = = and −− =− =−
We note that
)(
: Voltage v− = f2 x+kt
)(1
:
Z i
dx= = 1
constant )
( x − kt =
) (x kt1
f
v= − v=f(x−kt2)
) (t2 t1
Z
c s
c
+
=
Trang 25We know
c
R R
c
R R
Z
v i
Z
v i
−
−
+ + = and = −
c R
Z Z
Z Z
v
R R R
R R R
i i
v v i
v
= +
+
− +
R c R c R
R R
R R
R
Z v Z v
v v i
i
v v
=
−
+
= +
+
− +
− +
− +
− +
/ /
R c R
c R
R
R
Z Z
Z Z v
At the receiving end :
The receiving-end voltage coefficient
c R
c R R
R R
Z Z
Z Z v
R
ρ
R R R c
R
c R R
R
v
v Z v
Z v i
i =− =− + =−ρ
− +
− +
−//
Transmission Line Transients
If the line is terminated in its characteristic impedance , ρR =0
For current ,
At the sending end :
c s
c s S
Z Z
Z Z
• Diagram shows the boundaries
(x =0 and x=l) and the reflection
coefficients
• Time axis shown vertically
• Slope of the line should indicate
flight time of signal
• Total voltage at any point is the
sum of all the waves that have reached that point
S
Lattice diagram
Example :A dc source of 120 V with negligible
resistance is connected through a switch S to a lossless transmission line having Z c= 30 Ω
Plot v R versus time until t = 5T where T is the time for
a voltage wave to travel the length of the line
If the line is terminated in a resistance of
(a) 90 Ω (b) 10 Ω
Transmission Line Transients
Trang 26(a)Z R = 90 Ω , Z s = 0 Ω and Z c= 30 Ω
5 0 30 90 30 90
= +
−
= +
−
= +
Trang 288 9 +
Percent Impedance %& !!+ %
, Line Compensation * Shunt Capacitor ,
%& /S $ "$ kVA $ Percent Impedance $
10.
/ ,+!9 +
Trang 29• Single-line Diagram
• Bus Admittance Matrix , YBUS
• Power Flow Study
• Gauss-Seidel Power Flow Solution
• Newton-Raphson Power Flow Solution
• Fast Decoupled Power Flow Solution
• Control of Power Flow
Single-line Diagram
Single-line diagram (or one-line diagram)is a simplified
notation for representing a three-phase power system
Electrical elements such as generators, transformers ,
transmission lines, loads, circuit breakers, capacitors, etc.,
are shown by standardized schematic symbols
Single-line Diagram
Single-line diagram symbols :
Trang 30Real power systems are convenient to analyze using their
per-phase(since the system is three-phase) per-unit
(since there are many transformers) equivalent circuits
In the per-unit system, all quantities are represented as a
fraction of the base value:
Usually, base apparent powerand base voltageare specified at a point in the circuit, and the other values are calculated from them
Single-line Diagram
The per-unit impedance may be transformed from one
base to another as:
The base voltage varies by the voltage ratio of each
transformer in the circuit but the base apparent power
stays the same through the circuit
Bus Admittance Matrix
Trang 31Applying KCL to the
independent buses 1 through 4.
) ( ) ( 1 2 13 1 3
−
−
− +
+
−
−
− +
34 34
34 34 23 13 23
13
23 23
12 20 12
13 12
13 12 10
2 1
0 0
) (
0 )
(
0 )
(
0
0
V V V V
y y
y y y y y
y
y y
y y y
y y
y y y I I
Bus 4 :
Bus 1 : Bus 2 : Bus 3 :
4 34 3 34
4 34 3 34 23 13 2 23 1 13
3 23 2 23 12 20 1 12 2
3 13 2 12 1 13 12 10 1
0
)(
0
)(
)(
V y V y
V y V y y y V y V y
V y V y y y V y I
V y V y V y y y I
+
−
=
−+
++
−
−
=
−+
++
−
=
−
−+
+
=
Matrix formation of the equations
Bus Admittance Matrix
−
−
− +
+
−
−
− +
34 34
34 34 23 13 23
13
23 23
12 20 12
13 12
13 12 10
2
1
0 0
) (
0 )
(
0 )
(
0
0
V V V V
y y
y y y y y
y
y y
y y y
y y
y y
nn ni
n n
in ii
i i
n i
n i
n i
V V
V V
Y Y
Y Y
Y Y
Y Y
Y Y
Y Y
Y Y
Y Y
I I
I I
M M
L L
M M
M M
L L
M M
M M
L L
L L
M
M
2 1
2 1
2 1
2 2
22 21
1 1
12 11
2 1
bus bus bus Y V
bus bus bus bus
Trang 32Example : Determine bus admittance matrix Ybus
0.025 0.24 0.08 5 4 7
0.010 0.03 0.01 4 3 6
0.015 0.12 0.04 5 2 5
0.020 0.18 0.06 4 2 4
0.020 0.18 0.06 3 2 3
0.025 0.24 0.08 3 1 2
0.030 0.06 0.02 2 1 1
B/2 (p.u.)
X
(p.u.)
R
(p.u.) To From Line No.
0.025 0.24 0.08 5 4 7
0.010 0.03 0.01 4 3 6
0.015 0.12 0.04 5 2 5
0.020 0.18 0.06 4 2 4
0.020 0.18 0.06 3 2 3
0.025 0.24 0.08 3 1 2
0.030 0.06 0.02 2 1 1
B/2 (p.u.)
X
(p.u.)
R
(p.u.) To From Line No.
15 5 06 0 02 0
1
Line admittance (p.u.)
Bus Admittance Matrix
0.025 0.24 0.08 5
4
7
0.010 0.03 0.01 4
3
6
0.015 0.12 0.04 5
2
5
0.020 0.18 0.06 4
2
4
0.020 0.18 0.06 3
2
3
0.025 0.24 0.08 3
1
2
0.030 0.06 0.02 2
1
1
B/2 (p.u.)
X
(p.u.)
R
(p.u.) To
) 055 0 ( ) 75 3 25 1 ( ) 15 5
=
p.u.
695 18 25