Jenkins N., Strbac G., Ekanayake J. Distributed Generation

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From (3.36), it may be seen that each component voltage drop only depends on

that component current and the reactance.

Figure 3.25 shows the cross section of a cable. As the three conductors are

equally spaced, the cable may be considered a symmetrical element with similar

distribution of magnetic flux around each conductor. For such a symmetrical non-

rotating element, the positive and negative sequence inductances are equal.

Conductor

Sheath

Figure 3.25 Cross section of a three-phase cable

The flux that links with the sheath (which is earthed) determines the zero

sequence inductance, with a different value to the other two sequence components.

A section of an overhead line is shown in Figure 3.26. In this case flux linkages

between different phases are not equal and therefore the positive and negative

sequence inductances are different [5].

The internal impedance of a generator, expressed in sequence values, depends

on its construction. The magnetic field produced by the positive sequence currents

rotates in the same direction as the rotor; thus with respect to the rotor it is sta-

tionary. The field produced by the negative sequence currents rotates in the

Conductor

A-phase

Shield wire

Conductor

B-phase

Conductor

C-phase

Figure 3.26 A section of an overhead line

Distributed generators and their connection to the system 77

opposite direction to the rotor. Therefore with respect to the rotor that magnetic

field rotates at twice the rotor speed. These differences lead to different positive

and negative sequence impedances.

Consider an ideal, balanced three-phase source shown in Figure 3.27 where

neutral point, N, is earthed through an impedance Z

Figure 3.27 Three-phase source

The internal generated voltages of each phase are given by the following three

equations where E is the rms value of the internal voltage and:

¼ Eﬀ0



¼ E

¼ Eﬀ120



¼ Eﬀ240



¼ l

¼ Eﬀ240



¼ Eﬀ120



¼ lE

;

ð3:37Þ

From (3.32) and (3.37) (note that from Figure IV.8 of the Tutorial Chapter IV

ð1 þ l þ l

Þ¼0, ð1 þ 2l

Þ¼3 and ð1 þ l

þ l

Þ¼0):

11 1

1 ll

1 l

ð3:38Þ

From (3.38), it is clear that the ideal three-phase source does not produce any

negative or zero sequence voltages. Therefore, a three-phase source can be repre-

sented by three sequence networks where the positive sequence network has a voltage

equal to E

¼ E and the negative and zero sequence networks have zero voltages.

Three sequence voltages are represented behind the internal impedance of the

source. With positive sequence impedance of Z

, the positive sequence network

with respect to the neutral was obtained as:

¼ E

 I

ð3:39Þ

78 Distributed generation

With negative sequence impedance of Z

the negative sequence network with

respect to the neutral was obtained as:

¼I

ð3:40Þ

Finally with zero sequence impedance of Z

, the zero sequence network with

respect to earth was obtained as:

¼ V

 I

ð3:41Þ

Since the impedance between the neutral and earth is Z

, then:

¼I

þI

½

ð3:42Þ

From (3.35), I

þI

½¼3I

and therefore from (3.42):

¼3I

ð3:43Þ

By substituting for V

from (3.43) into (3.41):

¼I

ðZ

þ 3Z

Þð3:44Þ

Three sequence networks for a three-phase source are shown in Figure 3.28. If

the source is delta connected or star connected without a neutral then Z

!1.

For a transformer, the positive and negative sequence impedances are equal.

Depending on the connection used for the primary and secondary of the transfor-

mer, the zero sequence component current is transferred from the primary to sec-

ondary or is blocked (Table 3.1).

Once the sequence component networks of the power system are known, then

the asymmetrical currents are obtained by connecting the networks in different

arrangements. The connection of three sequence networks depends on the type of

the fault:

A line-to-earth fault through a fault impedance, Z

, is shown in Figure 3.29.

+ve sequence zero sequence

′

ve sequence

Figure 3.28 Sequence networks for a three-phase source

Distributed generators and their connection to the system 79

Table 3.1 Zero sequence component and transformer connection

Transformer connection Sequence networks (see Figure IV.6)

Primary Secondary Positive and negative Zero

Zero sequence current flows

from the primary side to the

secondary side.

Zero sequence current

enters the transformer from

the primary side but will not

flow to the secondary side

as the unearthed star blocks

the zero sequence current.

Zero sequence current enters

the transformer from the

primary side but will not

flow to the secondary side

as the currents circulate in

the delta loop.

Figure 3.29 Line-to-earth fault

80 Distributed generation

In this case, the fault current in the B and C phases are zero and fault current

only flows in A phase. From symmetrical components, when I

¼ I

¼ 0:

11 1

1 ll

1 l

¼ I

ð3:45Þ

From Figure 3.29:

¼ I

ð3:46Þ

Now substituting for V

and I

with their symmetrical components (see

(IV.26)):

þV

¼ I

þI

½Z

ð3:47Þ

Substituting from (3.45) into (3.47), the following equation is obtained:

þV

¼ 3I

ð3:48Þ

It is clear that if the sequence networks are connected in series, as shown in

Figure 3.30, then both (3.45) and (3.48) are satisfied.

′

Figure 3.30 Connection of sequence networks for a line-to-earth fault

The formation of the sequence networks for other types of faults are given in

Appendix (A3.1).

Distributed generators and their connection to the system 81

EXAMPLE 3.6

For the circuit shown, on 50 MVA base the sequence impedances of the

components are:

Positive

sequence

Negative

sequence

Zero

sequence

Circuit j0.6 j0.6 j1.5

Transformer j0.1 j0.1 j0.1

A single phase to earth fault occurred at the mid point of the line. Find

the fault current.

33 kV Circuit

Grid 400

MVA

Answer:

Positive sequence network:

The fault level at the grid in-feed = 400/50 pu = 8 pu.

Source impedance = 1/8 pu = 0.125 pu.

Transformer impedance = 0.1 pu.

Line impedance = 0.6/2 = 0.3 pu.

Therefore, the total positive sequence impedance to the fault:

¼ jð0:125 þ 0:1 þ 0:3Þ¼j0:525 pu

The positive sequence network is given by:

j 0.525 pu

1 pu

Fault

point

Negative sequence network:

Transformer impedance = 0.1 pu

Line impedance = 0.6/2 = 0.3 pu

The total negative sequence impedance to the fault:

¼ jð0:1 þ 0:3Þ¼j 0: 4pu

82 Distributed generation

The negative sequence network is given by:

j0.4 pu

Fault

point

Zero sequence network:

For zero sequences, it is considered that the line is connected to the earth

from both ends (due to star earth on the transformers). Therefore, the zero

sequence network is given by:

j0.75 pu

Fault

point

j0.75 pu

Therefore, the total negative sequence impedance to the fault:

¼ j0:75=2 ¼ j0:375 pu

For a single phase to earth fault, three sequence networks are connected in

series. Assuming that the fault impedance Z

= 0, the equivalent circuit for a

line-to-earth fault is given by:

j0.4 pu

j0.375 pu

j0.525 pu

1 pu

Fault

point

From the above figure and from (3.45):

¼ I

0:525 þ 0:4 þ 0:375

¼ 0 :77 pu

Distributed generators and their connection to the system 83

Therefore, the fault current = 3  0.77 = 2.31 pu.

On 50 MVA, 33 kV (line) base, the current base is given by:

50  10

3 ð33  10

ﬃﬃﬃ

¼ 874: 78 A

Therefore, the fault current = 874.78  2.31 = 2020.7 A.

3.4 Case studies

3.4.1 Steady-state voltages under peak and minimum loading

A section of a weak 33 kV network is shown in Figure 3.31. The generator con-

nected to Bus 1 represen ts the grid in-feed, which supplies power from the 132 kV

transmission system to the 33 kV distribution network. The fault level at the grid in-

feed is 1850 MVA. The generator, transformer, line and load data are given in

Tables 3.2–3.5. The 132/33 kV transformer has an on-load tap changer which

maintains Bus 2 at a voltage at 1.005 pu. The tap changer at the distributed gen-

erator (DG) is off-load and set to 0%.

Bus 1 Bus 2 Bus 3 Bus 4 Bus 5 Bus 6

Bus 7

Distributed generator

Figure 3.31 33 kV distribution system with a distributed generator

Table 3.2 Distributed generator data

Name MVA rating X

(pu) X

(pu) T

(s)

DG 8 0.29 0.16 0.29 4.5

84 Distributed generation

1. Steady-state voltages, power flows and losses

A load flow study was carried out using the IPSA package and voltages; power

flows and losses were obtained for peak and minimum loading conditions. Figure

3.32 shows the load flow under peak and minimum loading conditions when DG is

importing VArs (0.95 leading power factor). It is clear from the diagram that during

the peak loading condition power flows from upstream to downstream via the line

between Buses 2 and 3. However, during the minimum loading condition excess

power in the distribution network is fed back to the grid.

Table 3.6 shows the voltage on busbars 3, 4, 5 and 6 and losses when a dis-

tributed generator is not connected, when it is operating at unity, 0.95 lagging

(exporting VArs

) and 0.95 leading (importing VArs) power factors. As the network

is weak (with a high impedance), under peak loading condition the voltage at some

busbars are below the limit (6%) when the distributed generator is not connected.

The distributed generator reduced the power flow in the line between Bus 2 and 3,

thus reducing the losses. When the distributed generator operates at unity power

Table 3.4 Circuit data

From bus To bus Length (km) R (pu/km) X (pu/km)

Bus 2 Bus 3 17.2

Bus 3 Bus 4 12.0

Bus 3 Bus 7 0.5 0.016250 0.034435

Bus 4 Bus 5 0.8

Bus 5 Bus 6 1.4

Table 3.5 Load data (all loads have a lagging power factor)

Busbar Peak Minimum

P (MW) Q (MVAr) P (MW) Q (MVAr)

3 2.0 0.48 0.25 0.06

4 2.0 0.48 0.25 0.06

5 2.4 0.56 0.30 0.07

6 3.2 0.80 0.40 0.10

Table 3.3 Transformer data

From bus To bus MVA rating Winding Reactance (%)

Bus 1 Bus 2 31.50 Yd1 10.5

Bus 7 DG 11.00 Yd1 8.8

Note that for a load, lagging power factor means it absorbs reactive power.

Distributed generators and their connection to the system 85

For peak loading condition

Bus 7

Bus 1 Bus 2 Bus 3 Bus 4 Bus 5 Bus 6

For minimum loadin

condition

P = 1.87

Q = 6.01

P = 7.75

Q = 2.16

P = 5.61

Q = 1.38

P = 3.21

Q = 0.81

P = 8.0

Q = −2.5

P = 6.79

Q = −3.73

P = 0.95

Q = 0.23

P = 0.70

Q = 0.17

P = 0.4

Q = 0.1

P = 7.99

Q = −2.5

Figure 3.32 Power flows in peak and minimum loading conditions (P in MW and

Q in MVAr)

Table 3.6 8 MVA distributed generator is connected to Bus 3

Voltage (pu) Losses

Bus 3 Bus 4 Bus 5 Bus 6 MW MVAr*

Peak loading

No distributed generator 0.959 0.934 0.933 0.932 0.46 1.15

Distributed generator at unity pf 0.981 0.957 0.956 0.955 0.18 0.94

Distributed generator at 0.95 pf leading 0.964 0.940 0.939 0.938 0.26 1.25

Distributed generator at 0.95 pf lagging 0.996 0.973 0.972 0.970 0.15 0.88

Minimum loading

No distributed generator 1.000 0.997 0.997 0.997 0.01 0.02

Distributed generator at unity pf 1.018 1.015 1.015 1.015 0.13 0.85

Distributed generator at 0.95 pf leading 1.003 1.000 1.000 1.000 0.17 1.04

Distributed generator at 0.95 pf lagging 1.032 1.030 1.030 1.030 0.14 0.87

*Reactive power losses are the reactive power absorbed by the circuits.

86 Distributed generation