ADVANCED POWER SYSTEM (Technical Elective 3)

Course Description:

This course introduce the basic theories and applications of power systems, basic structure of power systems, recent trends and innovations in power system, complex power,per unit quantities, network modeling and calculations, load flow studies, short circuit calculations and use of computer for simulation.

Course Objective:

At the end of the course, the students shall be able to:

1. Understand a broad range of topics in the power systems.
2. Learn the basic theories and applications in the power system.
3. Learn some numerical methods used in power system.
4. Solve problems about power system.
5. Emphasize the important role of computer in power system computations and advancement.

PER-UNIT SYSTEM

In power systems many transformers at various (different) voltage levels are involved. Per Unit System is a normalization procedure which provides a mathematical basis for analyzing power networks with relative ease and convenience. In addition, when various quantities are expressed in per unit ( pu ) or per cent values, they usually convey a message. For example, if a bus voltage is 0.98 pu, it means that this value is 98 % of the nominal or base value which could be at any level in the network. It also immediately conveys a message that the value is an acceptable one. On the contrary, if the voltage value is 1. 08, then it immediately conveys that the value is higher than the acceptable level of 1.05 pu. Similar conclusions can be drawn for other quantities such as current, power and impedance. The idea here is to express various variables as a fraction of their corresponding base (fixed) variables.

Quantity in per unit (p.u.) Actual Quantity

Base Value of Quantity (1)

There are several advantages offered by using per unit systems. These are listed below.

SINGLE-PHASE SYSTEMS

The basic idea is to select two electrical variables such as power and voltage as

independent base values. Then the base values for other two variables, namely, current and impedance follow by Ohm’s Law. We will illustrate this procedure for single-phase systems:

Let base value for power = S_1B (single-phase)

base value for voltage = V_B (line-neutral)

= V_B (ln)

Then I_B¹ (line) = I_B (l) = (2)

and Z_B(y) = (3)

The Z_B is on a per phase basis.

We will consider an example to illustrate the use of per unit system.

V Z = R + jX

One should realize that V, I, and Z are actual complex values. The base values used for normalization are, however, real values. Therefore by equation (1) the respective per unit values are also complex.

Now,

V_p.u. = (4)

I_p.u. = (5)

Z_p.u. = (6)

= R_p.u. + jX_pu (6a)

and S_pu = (7)

or = (8)

= (9)

= P_pu + j Q_pu (10)

Numerical Example

Let_. V = 118 0⁰ volts

Z = 5 30⁰ ohms

Then I = 23.6 -30⁰ amperes

& S = V I* = (118 0⁰)(23.6 +30⁰) va

= 2,784.8 30⁰ va

For this example, it is appropriate to choose:

S_lB = 3,000 va

V_lB = 120-volts

Then I_lB = = 25 amperes

& Z_lB = = 4.8 ohms

Three-Phase Systems

Three-phase systems may be normalized by picking appropriate three-phase bases. We will illustrate the various base choices for both Y and D systems on a comparative basis:

Wye (Y)	Delta (D )
Choose (1) S3B = 3SlB	(1) S3B = 3SlB
(2) VB (ll) = VB(ln)	(2) VB(ll)
IB (l) =	IB (l) =
Since S3B = VB(ll)IB(l)	IB(ph) =
ZB(Y) =	ZB(D ) =
ZB(Y) =	ZB(D ) =

Base Equations for D Systems

Choose S_3B as the three-phase apparent power base (rating if available) and V_B(ll) as the line-to line base. Here again if the system nominal line-to-line is available or known, choose this value as the base.

I_B (l) = or I₁(pu) =

I_B(ph) = Þ I_ph(pu) =

Z_B(D ) = Þ ZD (pu) =

Base Equations for Y Systems

Choose S_3B as the three phase power base and V_B(ll) as the line-to-line voltage base.

Then I_B(l) = =

where V_B(ln) =

and Z_B(Y) = =

This can also be shown =

Change of Base

It is often necessary to convert the base values of several pieces of equipment connected together to form a power system (or interconnected power system). Usually the name plate ratings of these individual devices are different and hence their respective individual base values will be also different. In order to refer all per unit values to a common system base, it is necessary to change all device p.u. values to the common p.u. values. An example later will illustrate this procedure. However, the key equation development is as follows:

=

or = =

ß ß

l-l values three-phase values

Advantages of Per-Unit System (P.U.)

1. Per-unit representation results in a more meaningful and correlated data. It gives relative magnitude information.

2. There will be less chance of missing up between single - and three-phase powers or between line and phase voltages.

3. The p.u. system is very useful in simulating machine systems on analog, digital, and hybrid computers for steady-state and dynamic analysis.

4. Manufacturers usually specify the impedance of a piece of apparatus in p.u. (or per cent) on the base of the name plate rating of power () and voltage (). Hence, it can be used directly if the bases chosen are the same as the name plate rating.

5. The p.u. value of the various apparatus lie in a narrow range, though the actual values vary widely.

6. The p.u. equivalent impedance (Z_sc) of any transformer is the same referred to either primary or secondary side. For complicated systems involving many transformers or different turns ratio, this advantage is a significant one in that a possible cause of serious mistakes is removed.

7. Though the type of transformer in 3-phase system, determine the ratio of voltage bases, the p.u. impedance is the same irrespective of the type of 3-phase transformer. (Yç D , D ç Y, D ç D , or Yç Y)

8. Per-unit method allows the same basic arithmetic operation resulting in per-phase end values, without having to worry about the factor ‘100’ which occurs in per cent system.

Experience will, definitely, show the usefulness of the p.u. system.

Per Unit Representation

In a power system different power equipment with different voltage and power levels are connected together through various step up or step down transformers. However the presence of various voltage and power levels causes problem in finding out the currents (or voltages) at different points in the network. To alleviate this problem, all the system quantities are converted into a uniform normalized platform. This is called the per unit system . In a per unit system each system variable or quantity is normalized with respect to its own base value. The units of these normalized values are per unit (abbreviated as pu) and not Volt, Ampere or Ohm. The base quantities chosen are:

• VA base ( P_base ): This is the three-phase apparent power (Volt-Ampere) base that is common to the entire circuit.
• Voltage Base ( V_base ): This is the line-to-line base voltage. This quantity is not uniform for the entire circuit but gets changed by the turns ratio of the transformer.

Fig. 1.20 Three balanced sources supplying two balanced load through balanced source impedances.

Fig. 1.21 Per phase equivalent circuit of the network of Fig. 1.20.

Based on the above two quantities the current and impedance bases can be defined as

	(1.119)
	(1.120)

Assume that an impedance Z is defined as Z₁ per unit in a base impedance of Z_{base _ old} . Then we have

(1.121)

The impedance now has to be represented in a new base value denoted as Z _{base_new} . Therefore

(1.122)

From (1.120) Z₂ can be defined in terms of old and new values of VA base and voltage base as

Example 1.1:

Let us consider the circuit shown in Fig. 1.19 (a) which contains the equivalent circuit of a transformer. Let the transformer rating be

500 MVA, 220/22 kV with a leakage reactance of 10%.

The VA base of the transformer is 500 MVA and the voltage bases in the primary and secondary side are 200 kV and 22 kV respectively. Therefore the impedance bases of these two sides are

Ω and W

where the subscripts 1 and 2 refer to the primary (high tension) and secondary (low tension) sides respectively. Assume that the leakage reactance is referred to the primary side. Then for 10%, i.e., 0.1 per unit leakage reactance we have

Ω

The above reactance when referred to the secondary side is

Ω

Hence the per unit impedance in the secondary side is 0.0968/0.968 = 0.1. Therefore we see that the per unit leakage reactance is the same for both sides of the transformer and, as a consequence, the transformer can be represented by only its leakage reactance. The equivalent circuit of the transformer is then as shown in Fig. 1.22. Since this diagram only shows the reactance (or impedance) of the circuit, this is called the reactance (or impedance ) diagram .

Fig. 1.22 Per unit equivalent circuit of a transformer.

Example 1.2:

Consider the 50 Hz power system the single-line diagram of which is shown in Fig. 1.23. The system contains three generators, three transformers and three transmission lines. The system ratings are

Generator G1	200 MVA, 20 kV, Xd = 15%
Generator G2	300 MVA, 18 kV, Xd = 20%
Generator G3	300 MVA, 20 kV, Xd = 20%
Transformer T 1	300 MVA, 220Y/22 kV, Xd = 10%
Transformer T 2	Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10%
Transformer T 3	300 MVA, 220/22 kV, X = 10%

The transmission line reactances are as indicated in the figure. We have to draw the reactance diagram choosing the Generator 3 circuit as the base.

Fig. 1.23 Single-line diagram of the power system of Example 1.2.

As we have chosen the circuit of Generator3 as the base, the base MVA for the circuit is 300. The high voltage side of transformer T₂ is connected wye. Therefore its ratedline to line voltage is √3 X 130 = 225 kV. Since the low voltage side is connected in D , its line to line voltage is 25 kV. The base voltages are chosen as discussed below.

Since the base voltage of G₃ is 20 kV, the base voltage between T₃ and bus 1 will be 20 X 10 = 200 kV. Also as there is no transformer connected in bus 1, the base voltage of 200 kV must be chosen for both the lines that are connected to either side of bus 1. Then the base voltage for the circuit of G₁ will also be 20 kV. Finally since the turns ratio of T₂ is 9 (= 225 ÷ 25), the base voltage in the G 2 side is 200 ÷ 9 = 22.22 kV. The base voltages are also indicated in Fig. 1.23.

Once the base voltages for the various parts of the circuit are known, the per unit values for the various reactances of the circuit are calculated according to (1.123) for a base MVA of 300. These are listed below.

Generator G1
Generator G2
Generator G3
Transformer T 1
Transformer T 2
Transformer T 3

The base impedance of the transmission line is

Ω

Therefore the per unit values of the line impedances are

pu and pu

The impedance diagram is shown in Fig. 1.24.

Fig. 1.24 The impedance diagram of the system of Fig. 1.23.

Closure

This completes our discussion on the modeling of power system components. In the subsequent portion of this course we shall use these models to construct a power system and use the per unit notation and the impedance diagram to represent the system.

Fig. 1.24 The impedance diagram of the system of Fig. 1.23.