Electric Power Transmission and Distribution
Solutions of exercise 2
Prof. Göran Andersson
ETH Zurich
Autumn term 2008
Exercise 1
a) Equivalent circuit diagrams
Generator
UG
a
Overhead line
ZOL
b
Load
ZL
G
UG,ph
a
ROL
jXOL
b
RL
jXL
R
UG
S
c
d
T
UR
IR
ROL
jXOL
RL
jXL
R
c
d
Figure 1: Generator-Line-Load. Top: Single phase diagram; Middle: Threephase equivalent circuit; Bottom: Single phase equivalent circuit.
Figure 1 shows the single phase diagram as well as the three-phase and
the single phase equivalent circuits. The points c and d have the same poten1
tial under balanced conditions and can therefore be linked in the one-phase
equivalent circuit.
b) Phase currents; active, reactive and apparent power; load factor
Sense of rotation
+w
0°
UR
-120°
UT
US
Figure 2: Phasor diagram of the phase voltages.
We know the absolute value of the generator phase-to-phase voltage.
√
The√generator phase voltage (voltage to neutral) is thus UG,ph = UG / 3 =
20/ 3 kV = 11.55 kV . In a balanced three-phase system, all phase voltages
have the same amplitude and each phase is shifted by ±120◦ with respect
to the other two. The reference angle of the voltage phasors and the sense
of rotation can be chosen at will. We put the voltage phasor of phase R in
the real axis (θR = 0◦ ) and define the positive sense of rotation ω as shown
in figure 2. The voltages of the other two phases S and T are consequently
shifted by −120◦ and −240◦ with respect to phase R:
U R = UG,ph ∠0◦ = 11.55∠0◦ kV
(1)
◦
◦
(2)
◦
◦
(3)
U S = UG,ph ∠ − 120 = 11.55∠ − 120 kV
UT
= UG,ph ∠ − 240 = 11.55∠ − 240 kV
Generally the phase current is
I ph =
U ph
Z ph
(4)
That means we also have to calculate the impedance. The total impedance
of one phase is
Z ph = Z OL + Z L = 2.5 + 110 + j (3.0 + 50) = 112.5 + j53 Ω
(5)
Since we assume balanced conditions, the impedance has the same value in
2
all phases. According to equation (4), we obtain for the phase currents
UR
11.55 kV∠0◦
= 92.85∠ − 25.23◦ A
=
Z ph
112.5 + j53 Ω
= 92.85∠ − 145.23◦ A
IR =
IS
IT
◦
= 92.85∠94.77 A
(6)
(7)
(8)
In a balanced system, the phase currents also have the same amplitude and
are also shifted by ±120◦ with respect to the other two currents.
The total complex power of the load corresponds to the triple value of
the complex power in one phase (R):
S L = 3 · U L,R I ∗R = 3 · I R Z L I ∗R
= 3 · (92.85∠ − 25.23◦ ) (110 + j50) (92.85∠ + 25.23◦ )
MVar}
= |2.85{zMW} +j 1.29
| {z
PL
(9)
QL
The active and reactive power of the load (PL and QL ) result from the
representation of the complex power as real and imaginary part. The values
apply to the total three-phase load. The apparent power is the absolute value
of the complex power and becomes SL = |S L | = 3.13 MVA.
The power factor of the load can be derived from the phasor diagram of
the load impedance (see figure 3):
cos ϕL =
RL
110
= 0.9104
=
|Z L |
|110 + j50|
(10)
ZL
jXL
jL
RL
Figure 3: Phasor diagram of the load impedance.
c) Compensation with capacities
To increase the power factor of the load, one can connect capacities in parallel to the load impedances. The reactive power generated by the capacity
compensates a part of the inductive reactive power consumed by the load.
As shown in figure 4, the capacities can be connected in two ways in
parallel to the load, either in Y - or ∆-connection. At first, we consider the
Y -connection; that way of connection is easier to deal with because the Y connected capacities are in parallel to the load impedances in the one-phase
equivalent circuit (the points d and e have the same potential).
3
RL
b
jXL
d
CY
e
CD
RL
jXL
CY
d
e
Figure 4: Top: Capacities for compensation of inductive reactive power in
Y - or ∆-connection. Bottom: One-phase equivalent circuit with capacities
in Y -connection.
The power factor shall be increased to 0.98. The corresponding phase
angle ϕ′L can be calculated from the compensated load impedance (load k
capacity):
¶
µ
ℑ{Z ′L }
′
= arccos(0.98)
(11)
ϕL = arctan
ℜ{Z ′L }
where
Z ′L = jXCY k Z L
(12)
Z ′L is the compensated load impedance, XCY = −1/(ωCY ) is the reactance
of the capacity. We can determine the value of the capacity by calculating
4
the real and imaginary part of Z ′L :
Z ′L
=
jXCY Z L
jXCY (RL + jXL )
=
·
jXCY + Z L
RL + j(XL + XCY )
| {z }
X
RL XCY (X − XL ) + jXCY (XXL +
2 + X2
RL
RL XCY (X − XL )
⇒ ℜ{Z ′L } =
2 + X2
RL
=
⇒ ℑ{Z ′L } =
RL − jX
R − jX
| L {z }
conj. compl. expansion
2)
RL
2)
XCY (XXL + RL
2 + X2
RL
(13)
(14)
(15)
From equation (11) we obtain
tan(ϕ′L ) = tan(arccos(0.98)) =
2)
XCY (XXL + RL
RL XCY (X − XL )
(16)
We solve this equation for XCY and obtain
2 + X2
RL
1102 + 502
L
=
= −527.86 Ω
RL · tan(ϕ′L ) − XL
110 · tan(arccos(0.98)) − 50
(17)
Eventually, the capacity becomes
XC Y =
CY = −
1
1
=
= 6.03 µF
ωXCY
2π50 · 527.86
(18)
As one can see in figure 2.10 on p. 34 in the lecture notes, the value of the
capacity in ∆-connection is
C∆ =
CY
= 2.01 µF
3
(19)
We again consider figure 4. The phase-to-phase voltage is applied to
the capacities C∆ , whereas the voltage to neutral is applied to CY . The
capacities are smaller in the case of the ∆-connection. In that way, the
1
reactance becomes bigger (XC = − ωC
). Due to the higher phase-to-phase
voltage, the smaller capacities C∆ generate the same reactive power in ∆connection as the bigger capacities CY in Y -connection.
d) Comparison of the line losses
To determine the line losses, at first we have to calculate the current in the
balanced state. For that we need the impedance of the compensated load:
Z ′L =
−j527.86(110 + j50)
jXCY Z L
=
= 127.47 + j25.89 Ω
jXCY + Z L
110 + j(50 − 527.86)
5
(20)
The absolute value of the phase current then results as
UR
11.55 kV
=
127.47 + 2.5 + j(25.89 + 3.0) Ω
+ Z OL
= 84.68 − j18.82 A
I ′R =
′
IR
=
Z ′L
|I ′R |
= 86.75 A
(21)
(22)
The active and reactive power losses on the overhead line are then
′
′ 2
POL
= 3 · (IR
) R = 3 · 86.752 · 2.5 = 56.44 kW
| {z OL}
Q′OL
= 3·
per phase
′ 2
(IR
) XOL
|
{z
per phase
}
= 3 · 86.752 · 3.0 = 67.73 kVar
(23)
(24)
In the uncompensated state, i.e. without capacities, the losses are higher:
2
= 3 · 92.852 · 2.5 = 64.66 kW
POL = 3 · IR
R
| {zOL}
QOL = 3 ·
per phase
2
IR
XOL
| {z }
= 3 · 92.852 · 3.0 = 77.59 kVar
(25)
(26)
per phase
e) Comparison of the load voltages
The load phase-to-phase voltage in the compensated state results from
√
√
U ′L =
3 · I ′R Z ′L = 3 · (84.68 − j18.82 A)(127.47 + j25.89 Ω)
| {z }
phase voltage
◦
= 19.54∠ − 1.05 kV
(27)
And without compensation:
√
√
3 · I R Z L = 3 · (92.85∠ − 25.23◦ A)(110 + j50 Ω)
UL =
| {z }
phase voltage
◦
= 19.43∠ − 0.79 kV
(28)
f) Effects of the compensation
• The phase current is reduced, by that
• line losses are reduced and
• the amplitude and phase angle of the load voltage increase
Exercise 2
We consider the transmission system from figure 5 consisting of a generator,
transformer-line-transformer and a load.
6
Generator Transformer 1
UG
Overhead line
ZOL
Transformer 2
Load
ZL
G
1
2
3
Figure 5: Power transmission with transformer-load-transformer.
a) Impedances in p.u.
As demanded, we choose one single base power SB = 10 MVA for the entire
system. Then we choose a base voltage for section 2: UB2 = 138 kV. According to the transformation ratios of the transformers, we define UB1 = 13.8 kV
and UB3 = 69 kV. The base values for the impedances now result from the
base values for power and voltages:
ZB3 =
ZB2 =
2
UB3
690002
= 476 Ω ⇒ z L = 0.63 p.u.
=
(29)
SB
107
2
UB2
1380002
= 1904 Ω ⇒ z OL = 0.00525 + j0.0525 p.u.(30)
=
SB
107
Transformer 1 has a nominal voltage of 13.2 kV (in section 1) and a nominal
power of 5 MVA; we now want to calculate its impedance in p.u. related to
UB1 = 13.8 kV and SB = 10 MVA:
µ
¶ µ
¶
µ
¶ µ ¶
13.2 kV 2
SB
13.2 2 10
xT 1 = XT 1
= 0.1
= 0.183 p.u.
UB1
5 MVA
13.8
5
(31)
The inductivity of transformer 2 does not have to be recalculated because
the chosen base values for voltage and power correspond to the nominal
quantities:
xT 2 = 0.08 p.u.
(32)
Eventually, we have to relate the generator voltage to the base voltage:
13.2
= 0.96 p.u.
(33)
13.8
The angle of uG can be chosen arbitrarily; for the sake of simplicity we
choose 0◦ : uG = 0.96∠0◦ p.u.
|uG | =
b) Single phase equivalent circuit
The deduction of the single phase equivalent circuit of a balanced threephase system is discussed on p.33 of the lecture notes. Figure 6 shows the
single phase equivalent circuit of the system from figure 5. The points k, l, m
and n correspond in both figures to the transformer primary and secondary.
7
Generator
iph
uph
Transformer 1
jxT1
k
Overhead line
zOL
Transformer 2
jxT2
m
l
Load
n
zL
Figure 6: Single phase equivalent circuit with p.u. values.
c) Phase currents
With the help of figure 6, we can now calculate the phase current in p.u.:
iph =
uph
0.96
= 1.35∠ − 26.4◦ p.u.
=
jxT 1 + z F + jxT 2 + z L
0.709∠26.4◦
(34)
To obtain the actual phase currents (in A) in the sections 1, 2 and 3, we have
to multiply with the respective base value; this base value is the quotient
of the base power and the base voltage. However, we have to take into
consideration that we now calculate per phase. This means we have to take
phase values for the base power and the base voltage:
IBi =
SB /3
S
√ =√ B
UBi / 3
3UBi
(i = 1, 2, 3)
(35)
We obtain for the phase currents in sections 1, 2 and 3:
¯ ¯
107
· 1.35 = 564.4 A
Iph1 = IB1 ¯iph ¯ = √
3 · 13800
¯ ¯
107
Iph2 = IB2 ¯iph ¯ = √
· 1.35 = 56.4 A
3 · 138000
¯ ¯
107
Iph3 = IB3 ¯iph ¯ = √
· 1.35 = 112.9 A
3 · 69000
(36)
(37)
(38)
d) Complex power
The complex load power in p.u. is calculated as follows:
sL = iph i∗ph · z L = 1.352 · 0.63 = 1.147 p.u.
(39)
By multiplication with the base value, we obtain
S L = sL · SB = 1.147 · 107 = 11.47 MW
(40)
As the load impedance is a purely ohmic resistance, ℑ (S L ) = 0, i.e. only
active power is consumed.
8
Exercise 3
How can the electric field strength on the conductor surface be reduced?
Conductor voltage: The higher the voltage, the higher the electric field
strength on the conductor surface. To reduce the electric field strength,
one would have to reduce the voltage.
Geometry of the line towers, arrangement of the conductors: The longer
the distance between the conductors and between earth and conductors, the smaller the electric field strength.
Cross section and form of the conductors: The smoother and the cleaner
the conductor surface, locally high electric field strengths will occur to
a lesser extent. Sharp edges stemming from the production process and
dirt on the conductor surface (e.g. metal cuttings) can lead to locally
high electric field strengths and can thus favor corona discharges.
The distribution of the whole conductor cross section on several bundle
conductors leads to an increase of the equivalent conductor radius and
thus to a reduction of the electric field strength.
A further possibility is to treat the conductor with a special coating
that accelerates the draining of water; with that, local rises of the
electric field strength can be prevented.
See also:
http://www.eeh.ee.ethz.ch/hvl/forschung/conor.html.
9