BUS ADMITTANCE MATRIX - PART – 02 -FORMATION OF Y-BUS AND TWO PROBLEMS

PROBLEM - 01

Calculate the elements of Y-bus matrix. The relevant per unit line admittances on 100 MVA base are indicated on the diagram.

PROBLEM - 02

For the following network shown in figure find the elements of Y-bus directly by inspection method.

BUS ADMITTANCE MATRIX - PART - 01 - FORMATION OF Y-BUS - ADVANTAGES

I-bus = Y-bus x V-bus       ----- 01

The voltages are measured with respect to ground. The equation - 01 is called nodal current equation.

It is a vector equation consisting of n scalar equations.

I-bus is the vector of the injected bus current (i.e. external current sources)

Y-bus is the bus admittance matrix

The diagonal element of each node is the sum of the admittances connected to it.

It is known a self-admittance matrix or driving point admittance.

The off-diagonal element is equal to the negative of the admittance between the nodes.

It is known as the mutual admittance or transfer admittance.

SPARSE MATRIX – Sparse mean not dense

The Y-bus is a sparse matrix. For a large power system, more than 90 % of off-diagonal elements are zero. This is due to the fact that in power system network networks each node (bus) is connected to not more than three nodes.
V-bus is the vector bus voltages measured from the reference node (i.e. node voltages.

ADVANTAGES OF Y-BUS

1. Y-bus is a sparse matrix hence (i.e. most of its elements are zero) computer memory requirements are less.

2. It is very useful in load flow and stability analysis.

3. Its formation and modification are easy.

4. Data preparation is simple.

PER UNIT REPRESENTATION - PART - 12 - ADVANTAGES OF PER UNIT SYSTEM - ONE PROBLEM

PROBLEM - 01

A three phase three winding 50 Hz, transformer rated 110/11/6.6kV as the measured impedances between windings referred to 30 MVA as Zps = j0.16 p.u, Zpt = j0.1 p.u and Zst = j0.09 p.u. A balanced load supplied from 6.6kV secondary winding draws a current of 2kA, 0.8 power factor lagging, and a star connected capacitor bank with 50 ohms reactance per phase is supplied from 11kV tertiary winding. Find the voltage required at the 110kV primary terminals to maintain 6.6kV across the secondary winding. Draw the equivalent single phase circuit marking all parameters thereon.

ADVANTAGES OF PER UNIT SYSTEM

1. Manufacturers usually specify the impedance value of equipment in per unit of the equipment rating. If any data is not available, it is easier to assume it’s per unit value than its numerical value.

2. The circuit laws are valid in p.u. systems and the power and voltage equations are simplified since the factors of root 3 and 3 are eliminated.

3. The per-unit system gives us a clear idea of the relative magnitude of various quantities, such as voltage, current, power, and impedance.

4. The per-unit value of impedance, voltage and current of a transformer are same regardless of whether they are referred to the primary or the secondary side.

5. The p.u. impedance is the same on the two sides of the transformers. i.e., Star or Delta.

6. For computerized analysis and simulation of complex power problems, per-unit system representations are ideal.

PER UNIT REPRESENTATION - PART - 11 - THREE-WINDING TRANSFORMERS- ONE PROBLEM

THREE-WINDING TRANSFORMERS

In some high rating transformer, one winding in addition to its primary and secondary winding is used.

This additional winding, apart from primary and secondary windings, is known as a Tertiary winding of the transformer. Because of this third winding, the transformer is called three winding transformer or 3 winding transformers.

TERTIARY WINDING

Tertiary-winding is generally connected in delta. Delta connected tertiary winding reduced the impedance offered to zero sequence current. A large unbalance of phase voltage may be produced, which is compensated by large tertiary winding circulating currents. Third winding is also known as stabilizing winding.

ADVANTAGES

1. The main advantage of used three-winding transformer is that when any fault and the short circuit occur on the primary and secondary side. So large earth fault current flows for proper operations of protective devices.

2. When the load is unbalanced, tertiary winding reduced third harmonic voltage and limits the unbalanced voltage. The magnetizing current is neglected.

3. The tertiary winding may be used for connecting shunt capacitors.

4. In the testing transformer, it is used as the voltage coil.

5. To supplying the voltage to the device which is operated different voltage from secondary and primary voltage.

DISADVANTAGES

1. The additional winding is used hence more copper is require, so cost is high.

2. Construction is a little complex.

PER UNIT REPRESENTATION - PART - 10 - REACTANCE DIAGRAM - ONE PROBLEM

ONE PROBLEM - REACTANCE DIAGRAM 

PROBLEM - 01

For the system shown in the figure, draw the reactance diagram.

PER UNIT REPRESENTATION - PART - 09 - CALCULATION OF BUS VOLTAGES - TWO PROBLEMS.

PROBLEM - 01

For the system shown in the figure, calculate the bus voltage V2 and V3.

PROBLEM - 02

For the system shown in the figure, calculate the bus voltage V1 and V2.

PER UNIT REPRESENTATION - PART - 08 - ONE PROBLEM ON CALCULATION OF GENERATOR BUS VOLTAGE

PROBLEM - 01
For the system shown in figure draw the impedance diagram and calculate the generator bus voltage.

PER UNIT REPRESENTATION - PART - 07 - PER UNIT REPRESENTATION OF SERIES AND PARALLEL LOAD IMPEDANCE

PER UNIT LOAD IMPEDANCE

(a) Series Representation of Load Impedance

(b) Parallel Representation of load Impedance
PROBLEM - 01
For the system shown in figure, draw the reactance diagram.

PER UNIT REPRESENTATION - PART - 06 - TWO PROBLEMS IN PER UNIT SYSTEMS

PROBLEM - 01

For the system shown in the figure, draw the reactance diagram.

PROBLEM - 02

For the system shown in the figure, draw the reactance diagram.

PRESENTATION - PART - 05 - TWO PROBLEMS IN PER UNIT SYSTEMS

PROBLEM - 01

For the system shown in the figure draw the reactance diagram.

PROBLEM - 02

For the system shown in the figure determine the generator voltage.