department of electrical & communication engineering

1

Department of Electrical & Communication Engineering

CHAPTER 1 Per Unit Calculations

1. Power System Representation

Power Component

Symbol Power Component

Symbol

= Generator = Circuit breaker

M

= Transformer = Transmission line

= Motor = Feeder + load

= Busbar (substation)

Power components and symbols

POWER SYSTEMS

2


Interconnections among these components in the power system may be shown is a so-called one-line diagram or single-line diagram. Single-line diagram represents all 3- of balanced system. For the purpose of analysis, the single-line diagram of a particular power system network is represented to its equivalent reactance or impedance diagram. A sample of a interconnected of individual power component is shown in Figure 1.1. This represent a circuit diagram of a power network which is referred to as a single-line diagram.

M

Figure 1.1 – Single-line diagram

POWER SYSTEMS

3


Impedance diagramIn power system fault calculations it is often that a single-line diagram representing a typical power network in3- be converted into its per phase impedance diagram. Some assumptions for converting from single-linediagram into its equivalent impedance diagram needed to be considered.

(i) A generator can be represented by a voltage source in series with an inductive reactance. The internal resistance of the generator is assumed to be negligible compared to the reactance.(ii) The loads are usually inductive represented by resistance and inductance.(iii) The transformer core is assumed to be ideal, and the transformer may be represented by a reactance

only.(iv) The transmission line is represented by its resistance and inductance, the line-to-ground capacitance is

assumed to be negligible.

Let us consider the following on how the single-line diagram of Figure 1.2 converted into its impedance diagram counterpart.

POWER SYSTEMS

4


GeneratorG1

G2

Station A Station B

G3

G4

LoadL1

LoadL2

TransformerT1

TransmissionLineTL

TransformerT2

Figure 1.2 – Single-line diagram of a power network

POWER SYSTEMS

5


G2G1G3 G4

j X1 j X2 j X3 j X4

j XT1 j XT2

RL1

j XL1

RL2

j XL2

RTLj XTL

Transformer T1

Transformer T2

Transmission LineTL

Station A Station B

Figure 1.3 – Impedance diagram of Figure 1.2

POWER SYSTEMS

6


Per-Unit QuantitiesPer unit quantities are quantities that have been normalized to a base quantity. In general,

actualpu

base

ZZ

Z per-unit (p.u)

Choice of the base value Zbase is normally a rated value which is often one of the normal full-load operations of power component in a power network.Let us look at two of the most common per unit formula which are widely used when per unit calculations are involved.

(i) Base impedance (Zbase)

For a given single-line (one-line) diagram of a power network, all component parameters are expressed in 3- quantity whether it is the rating (capacity) expressed as MVA or voltage as kV. Let begin with 3- base quantity of

basebasebase IVS 3 ----- (i)

where Vbase = line voltage, Ibase= line or phase current

Per phase base impedance,

base

base

base I

V

Z 3 -----(ii) This is line-to-neutral impedance

POWER SYSTEMS

7


Combining (i) and (ii) yields,

base

base

base

base

VS

V

Z

3

3

base

basebase MVA

kVZ2

where kVbase and MVAbase are 3- qualtities

(ii) Changing base impedance (Znew]Sometimes the parameters for two elements in the same circuit (network) are quoted in per-unit on a different base. The changing base impedance is given as,

2

2

base OLD base NEWNEW OLD

base OLDbase NEW

kV MVAZ pu Z

MVAkV

POWER SYSTEMS

8


Determine the per-unit values of the following single-line diagram and draw the impedance diagram.

Example 1

XT1 = 0.1 p.u

5 MVAXg = 16%

100 MVA275 kV/132 kV

50 MVA132 kV/66 kVTransmission line

j 3.48

XT2 = 0.04 p.u Load

40 MW, 0.8 p.f. lagging

Solution:Chosen base: Always choose the largest rating, therefore Sbase = 100 MVA, V = 66 kV, 132 kV and 275 kV

Per-unit calculations:

Generator G1:

32.05010016.0)( puX g p.u.

Transformer T1:

1.0)(1 puXT p.u.

POWER SYSTEMS

2

2

base OLD base NEWNEW OLD

base OLDbase NEW

kV MVAZ pu Z

MVAkV

9


Transmission line TL:

0195.0132

1004.3)( 2

puXTL p.u.

Transformer T2:

08.05010004.0)(2 puXT p.u.

Inductive load:

oactualZ 87.3612.87

8.0106631040

31066

36

3

p.u. )2.16.1(87.36266

10087.3612.87)( 2 jorpuZ oo

L

POWER SYSTEMS

base

basebase MVA

kVZ2

actualpu

base

ZZ

Z

10


Now, we have all the impedance values in per-unit with a common base and we can now combine all the impedances and determine the overall impedance.

LoadG

j 0.32 p.u.

j 0.1 p.u. j 0.0195 p.u.

Transformer T1

Transformer T2

Transmission LineTL

j 0.08 p.u.

1.6 p.u..

j 1.2 p.u.

Generator

POWER SYSTEMS

XT1 = 0.1 p.u

5 MVAXg = 16%

100 MVA275 kV/132 kV

50 MVA132 kV/66 kVTransmission line

j 3.48

XT2 = 0.04 p.u Load

40 MW, 0.8 p.f. lagging

11

Department of Electrical & Communication Engineering POWER SYSTEMS

12


Load

POWER SYSTEMS

13


Load

POWER SYSTEMS

Summarise:

14


Load

POWER SYSTEMS

department of electrical & communication engineering

Documents