Faculty of electrical and computer engineering
POWER SYSTEM I
EEng-3132
Representation of Power System Components
Compiled by: Biniyam Z.
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Chapter Two
Representation of Power System
Components
Outline
Single-phase solution of balanced three-phase networks
One-line diagram
Impedance or reactance diagram
Per unit (PU) system
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Introduction
The basic equipments of a power system are generators,
transformers, transmission lines, and loads
The interconnections among these components in the power
system may be shown in a so called one line diagram
For analysis, the equivalent circuits of the components are shown
in a reactance diagram or an impedance diagram
It is very much practical to represent a power system using simple
symbols for each component since it is very complicated to show a
complete circuit diagram of a power system for all the three
phases.
The physical quantities such as current, voltage, impedance and
power are expressed using per unit as a decimal fraction or
multiple of base quantities. 4/16/2019
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Single-phase solution of balanced three-phase networks
The solution of a three phase network under balanced conditions is easily
carried out by solving the single phase network corresponding to the
reference phase
Figure 2.1 shows a simple, balanced three phase network
The generator and load neutrals are therefore at the same potential, so that
In 0
Figure 2.1
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The neutral impedance zn does not affect network behavior
For the reference phase a
Ea ( zG zL ) I a .............(2.1)
The current and the voltage in the other phases have the same magnitude
but progressively shifted by 1200
Equation 2.1 corresponds to the single phase network of figure 2.2 below
The solution of fig. 2.2 completely determines the solution of the three
phase network
Figure 2.2
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Consider the case where a three phase transformer forms part of a three
phase system
If the transformer is Y/Y connected as shown in fig. 2.3a, in the single
phase equivalent of the three phase circuit it can be obviously represented
by a single phase transformer (as in fig. 2.3b) with primary and secondary
pertaining to phase a of the three phase transformer
Figure 2.3
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If the transformer is Y/∆ connected as in figure 2.4a below, the delta side
has to be replaced by an equivalent star connection as shown doted so as to
obtain the single phase equivalent of fig. 2.4b
When the load is balanced, the impedance of each leg of the Y is one third
the impedance of each leg of the ∆
The circuit is modeled by the single phase equivalent circuit
Since both phase voltage and line current shift through the same phase
angle from star to delta side, the transformer per phase impedance and
power flow are preserved in the single phase equivalent
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Figure 2.4
It may be noted here that irrespective of the type of connection, the
transformation ratio of the single phase equivalent of a three phase
transformer is the same as the line-to-line transformation ratio
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Delta-Wye Transformation
To simplify analysis of balanced 3 systems:
∆- connected loads can be replaced by
1
Y – connected loads with ZY Z
3
∆ - connected sources can be replaced by
VLine
Y – connected sources with phase
V
3300
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One-line diagram
Power system networks are represented by one-line diagrams using
suitable symbols for generators, motors, transformers and loads
A one-line diagram of a power system shows the main connections and
arrangements of components
Figure 2.5 shows the symbols used to represent the typical components of
a power system
Figure 2.6 is a one line diagram for a power system consisting of two
generating stations connected by a transmission line; note the use of the
symbols of fig. 2-5
The advantage of such a one-line representation is its simplicity: one phase
representation all three phases of the balanced system; the equivalent
circuit of the components are replaced by their standard symbols; and the
completion of the circuit through the neutral is omitted
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Fig. 2.6
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IMPEDANCE AND REACTANCE DIAGRAMS
In power system analysis it is necessary to draw an equivalent circuit for
the system.This is an impedance diagrams.
However, in several studies, including short-circuit analysis it is
sufficient to consider only reactances neglecting resistances. Hence, we
draw reactance diagrams.
For 3-phase balanced systems, it is simpler to represent the system by a
single line diagram without losing the identity of the 3-phase system.
Thus, single line reactance diagrams can be drawn for calculation.
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The one-line diagram may serve as the basis for a circuit representation
that includes the equivalent circuits of the components of the power
system.
Such a representation is called an impedance diagram or a reactance diagram
if resistances are neglected.
The impedance and reactance diagrams corresponding to Fig. 2-6 are shown in Fig.
2-8(a) and (b), respectively. Note that only a single phase is shown.
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The following assumptions have been incorporated into Fig. 2-8(a):
1. A generator can be represented by a voltage source in series with an
inductive reactance. The internal resistance of the generator is negligible
compared to the reactance.
2. The loads are inductive.
3. The transformer core is ideal, and the transformer may be represented
by a reactance.
4. The transmission line is a medium-length line and can be denoted by a T
circuit. An alternative representation, such as a circuit, is equally
applicable.
5. The delta-wye-connected transformer T1 may be replaced by an
equivalent wye-wye-connected transformer (via a delta-to-wye
transformation) so that the impedance diagram may be drawn on a per-
phase basis.
The reactance diagram, Fig. 2-8(b), is drawn by neglecting all resistances, the static
loads, and the capacitance of the transmission line.
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PER-UNIT REPRESENTATION
Computations for a power system having two or more voltage levels
become very cumbersome when it is necessary to convert currents to a
different voltage level wherever they flow through a transformer (the
change in current being inversely proportional to the transformer turns
ratio).
In an alternative and simpler system, a set of base values, or base
quantities, is assumed for each voltage class, and each parameter is
expressed as a decimal fraction of its respective base.
For instance, suppose a base voltage of 345 kv has been chosen, and under
certain operating conditions the actual system voltage is 334 kV: then the
ratio of actual to base voltage is 0.97.
The actual voltage may then be expressed as 0.97 per-unit. In an equally
common practice, per-unit quantities are multiplied by 100 to obtain
percent quantities: our example voltage would then be expressed as 97
percent.
Per-unit and percent quantities and their bases exhibit the same
relationships and obey the same laws (such as Ohm's law and Kirchhoff’s
laws) as do quantities in other systems of units.
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A minimum of four base quantities is required to completely define a per-
unit system: these are voltage, current, power, and impedance (or
admittance). If two of them are set arbitrarily, then the other two become
fixed.
Base values can be determined by three ways
Taking the largest value
Taking the total sum
Any arbitrary
The per unit value of any quantity is the ratio of the actual value in any
units to the chosen base quantity of the same dimensions expressed as a
decimal.
Actual value in any units
Per unit quantity
Base or reference value in the same unit
In power systems the basic quantities of importance are voltage, current,
impedance and power. For all per unit calculations a base KVA or MVA
and a base KV are to be chosen.
Once the base values or reference values are chosen, the other quantities
can be obtained as follows:
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Where base KVA and base MVA are the total or three phase values.
If phase values are used
Base current in amperes =
base voltage
Base impedance in ohm =
base current
(base K V) 2 / 1000
base K V A per phase
Base Impedance in ohm =
In all the above relations the power factor is assumed unity, so that
Base power K W = base K VA
(actual impedance in ohm) x MVA
Now, Per unit impedance=
(base K V)2
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Selecting the total or 3-phase KVA as base KVA, for a 3-phase system
Base current in amperes =
base (KV Iine-to-line) 2 /1000
Base Impedance In ohms =
3 [(base KVA)/ 3]
(base KV (line-to-line) 2
Base Impedance in ohms = base MVA
(actual impedance in ohm) x MVA
Now, Per unit impedance = (base K V) 2
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Some times, it may be required to use the relation
actual Impedance in ohm = (Per unit impedance in ohms) (base KV) 2
(base MVA)
Very often the values are in different base values. In order to convert the
per unit impedance from given base to another base, the following relation
can be derived easily.
Per unit impedance on new base
base kVold base kVA new
2
Z(pu) new Z ( pu ) old X x
base kVnew base kVA given
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In general :
Base current = base volt amperes/base voltage (in amperes)
Base impedance = base voltage/ base current (in ohms)
Per-unit voltage = actual voltage / base voltage (per unit, or pu)
Per-unit current = actual current / base current (per unit, or pu)
Per-unit impedance = actual impedance /base impedance (per unit. or
pu)
The impedances of transmission lines are expressed in ohms, but can be
easily converted to pu values on a given voltampere base using above
expressions.
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Advantages of Per Unit System
1. While performing calculations, referring quantities from one side of
the transformer to the other side serious errors may be committed. This
can be avoided by using per unit system.
2. The p.u systems are ideal for the computerized analysis and
simulation of complex power system problems.
3. Voltages, currents and impedances expressed in per unit do not
change when they are referred from one side of transformer to the
other side.This is a great advantage'.
4. Per unit impedances of electrical equipment of similar type usually
lie within a narrow range, when the equipment ratings are used as base
values.
5.Transformer connections do not affect the per unit values.
6. Manufacturers usually specify the impedances of machines and
transformers in per unit or percent of name plate ratings.
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Example 1: - Power system loads are usually specified in terms of the
absorbed power and reactive power. In circuit analysis it is sometimes
continent such a load as a constant impedance. Two such
representations, parallel and series are possible as shown in figure
below. Determine the per unit R and X values for both the parallel
and series connections.
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Solution
Let
Parallel connection. From the parallel connection we observe that the
power absorbed depends only the applied voltage, i.e.
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Where the value subscribed u is a pu value. Substituting Rp, we
compute
Similarly, we find the expression for pu X to be
Series connection. If R and X are connected in series, the problem is
more difficult since the current in X affects the absorbed power P.
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Multiplying by its conjunction, we have
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Example 2: - Given the two-machine system of figure below we select,
quite arbitrarily a base voltage of 161kV from the transmission line and
volt-ampere of 20MVA. Find the pu impedance of all components
referred to these bases.The apparatus has ratings as follows
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Solution
For the transmission line we must convert from ohmic values to pu
values.
From the load a parallel R – X representation may be computed
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