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Topic 4
Lecturer : Dr. Nur Hakimah Binti Abdul Aziz...

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BEKP 3653 CHAPTER 4: PER-UNIT SYSTEM

4.1 Single Line Diagram In power engineering, a one-line diagram or single-line diagram (SLD) is a simplified notation for representing a three-phase power system. The one-line diagram has its largest application in power flow studies. Electrical elements such as circuit breakers, transformers, capacitors, bus bars, and conductors are shown by standardized schematic symbols as shown in Figure 4.1. Instead of representing each of three phases with a separate line or terminal, only one line is represented. SLD is a form of block diagram graphically depicting the paths for power flow between entities of the system. It is a common convention to organize the diagram with the same left-to-right, top-to-bottom sequence. Figure 4.2 shows the single line diagram of a simple power system.

Figure 4.1: Symbols for single line diagram

T1 G

T2

TL

M

Figure 4.2: Single line diagram of a typical power system

1. Why do we need to step up voltage for transmission and then stepping down for distribution? 2. Why -Y connection of a transformer is used in power network?

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4.2 Impedance Diagram Generator, transformer and transmission line in Figure 4.2 are represented by their own equivalent circuit in the impedance diagram in Figure 4.3. The synchronous generator at the generating station is represented by a voltage source in series with the resistance and reactance, the transformer by a nominal -equivalent circuit. The load is assumed to be passive and are represented by a resistive and inductive reactance in the series. Neutral earthing impedance does not appear in the diagram as the balanced condition is assumed.

Figure 4.3: Impedance diagram of the system in Figure 4.1

4.3 Reactance Diagram The reactance diagram gives an accurate result for many power system studies, such as shortcircuit studies, etc. The winding resistance, including the line resistance, are quite small in comparison with leakage reactance and shunt path which includes line charging and transformer magnetising circuit provide a very high parallel impedance with fault. It is considered that if the resistance is less than one-third of the reactance, and resistance is ignored, then the error introduced will be not more than 5 %. The errors mean their calculation gives a higher value than the actual value. Figure 4.4 shows the reactance diagram of the system in Figure 4.1.

Figure 4.4: Reactance diagram of the system in Figure 4.1

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4.4 Per-Unit In power system analysis, it is common practice to use per-unit quantities of voltage, current, power, and impedance. These per-unit quantities are normalized or scaled on a selected base, allowing engineers to simplify power system calculations with multiple voltage transformations. In per-unit notation, the physical quantity is expresses as a fraction of the reference value,

𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 =

𝑎𝑐𝑡𝑢𝑎𝑙 𝑏𝑎𝑠𝑒

e.g. 𝑉𝑝𝑢 (in per unit) =

𝑉𝑎𝑐𝑡 (in kV) 𝑉𝑏𝑎𝑠𝑒 (in kV)

In the most common usage for per unit calculation, the nominal or nameplate voltage rating of an individual equipment is chosen as base voltage. For example, a motor with a 100 Horsepower (HP) nameplate rating which delivers 90 HP to a dynamic load is said to be 90% loaded or 0.9 per unit. In this case, the per unit base is the nameplate horsepower rating of 100. When the motor delivers 90 HP, 90/100 equals 0.9 per unit.

4.5 Why Per-Unit? •

The per-unit system gives us a clear idea of relative magnitudes of various electrical quantities, e.g. voltage, current, power and impedance.

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The per-unit impedance of equipment of the same general type based on their own ratings fall in a narrow range regardless of the rating of the equipment.

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The per-unit values of impedance, voltage and current of a transformer are the same regardless of whether they are referred to the primary or the secondary side. The different voltage levels disappear and the entire system reduces to a system of simple impedance. This also applies to the generators and motors which have different impedance when referred to either stator or rotor winding.

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The per-unit system are ideal for the computerized analysis and simulation of complex power system problems.

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The circuit laws are valid in per-unit system. The power, current and voltage equations are simplified since the factors of √3 and 3 are eliminated in the per-unit system.

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4.6 Per-Unit System Per-unit is defined as 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 (𝑝𝑢) =

𝑎𝑐𝑡𝑢𝑎𝑙 (𝑎𝑐𝑡) 𝑏𝑎𝑠𝑒 (𝑏)

4.1

e.g. 𝑉𝑎𝑐𝑡 𝑉𝑏

𝑉𝑝𝑢 =

𝑆𝑎𝑐𝑡 𝑆𝑏

𝑆𝑝𝑢 =

𝐼𝑎𝑐𝑡 𝐼𝑏

𝐼𝑝𝑢 =

𝑍𝑎𝑐𝑡 𝑍𝑏

𝑍𝑝𝑢 = 𝑅𝑝𝑢 = 𝑋𝑝𝑢 =

𝑅𝑎𝑐𝑡 𝑍𝑏

𝑋𝑎𝑐𝑡 𝑍𝑏

4.2 4.3

4.4 4.5

4.6 4.7

Commonly, in a power system, 𝑆𝑏 and 𝑉𝑏 will be set first. For a single power equipment e.g. generator

and transformer, the rated values 𝑆𝑟𝑎𝑡𝑒𝑑 and 𝑉𝑟𝑎𝑡𝑒𝑑 are used as the base values, 𝑆𝑏 and 𝑉𝑏 . The 𝑍𝑏

can be determined from the 𝑆𝑏 and 𝑉𝑏 values. Note that we only take the magnitude as the base value.

Thus, we can drop the conjugate from the formula. For a Y-connected system, 𝑆𝑏(3) = √3𝑉𝑏(𝐿𝐿) 𝐼𝑏(𝐿𝐿) 𝐼𝑏(𝐿𝐿) = 𝑍𝑏(1) =

4.8

𝑆𝑏(3)

√3𝑉𝑏(𝐿𝐿)

4.9

𝑉𝑏(𝐿𝐿)

√3𝐼𝑏(𝐿𝐿)

4.10

Inserting eq. 4.9 into 4.10 yields 𝑍𝑏 =

𝑉𝑏(𝐿𝐿) √3

×

√3𝑉𝑏(𝐿𝐿) 𝑆𝑏(3)

𝑉𝑏(𝐿𝐿) 2 𝑍𝑏 = 𝑆𝑏(3)

4.11

Given the 𝑉𝑏(𝐿𝐿) in kV and 𝑆𝑏(3) in MVA, the units of eq. 4.11 become =

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(kV)2 MVA

4.12

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4.6.1

Per-Unit System of A Generator

Problem 4.1 A Y-connected, three-phase, 318.75 kVA, 2300 V alternator has an armature resistance of 0.35  per phase and synchronous reactance of 1.2  per phase. Using per-unit system, determine the no-load line-to-line generated voltage at: a) Full-load kVA, 0.8 power factor lagging, and rated voltage b) Full-load kVA, 0.8 power factor leading, and rated voltage c) Full-load kVA, power factor unity, and rated voltage d) 90% of full-load kVA, 0.8 power factor lagging, and 2400 V. Solution: Choose the rated values as the base values 𝑉𝑏 = 2.3 kV

𝑆𝑏 = 0.31875 MVA Calculate the impedance in per unit. From eq. 4.12, 𝑍𝑏 =

2.32 = 16.596  0.31875

From eq. 4.5, 𝑍𝑠(𝑝𝑢) = a) The pu values for 𝑉𝑎 and S are

0.35 + 𝑗1.2 = 0.021 + 𝑗0.0723 pu 16.596 𝑉𝑎(𝑝𝑢) = 1∠0° pu

𝑆𝑝𝑢 = 1∠36.87° pu Thus 𝐼𝑎(𝑝𝑢) =

𝑆𝑝𝑢∗ ∠ − 𝑐𝑜𝑠 −1 (0.8) = 1∠ − 36.87° pu 𝑉𝑎(𝑝𝑢) ∗

The general equation for a synchronous generator is given as 𝐸𝑎 = 𝑉𝑎 + 𝐼𝑎 (𝑅𝑎 + 𝑗𝑋𝑠 ) In pu values, 𝑉𝑎(𝑛𝑜 𝑙𝑜𝑎𝑑) = 𝐸𝑎 = 1 + 1∠ − 36.87°(0.021 + 𝑗0.0723 ) = 1.06 + 𝑗0.04 pu

= (1.06 + 𝑗0.04 ) × 2300 = 2.44∠2.44° kV

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b) 𝐼𝑎(𝑝𝑢) = 1∠36.87° pu

𝑉𝑎(𝑛𝑜 𝑙𝑜𝑎𝑑) = 𝐸𝑎 = 1 + 1∠36.87°(0.021 + 𝑗0.0723) = 0.9735 + 𝑗0.0705 pu

= (0.9735 + 𝑗0.0705 ) × 2300 = 2.245∠4.142° kV

c) 𝐼𝑎(𝑝𝑢) = 1∠0° pu

𝑉𝑎(𝑛𝑜 𝑙𝑜𝑎𝑑) = 𝐸𝑎 = 1 + 1∠0°(0.021 + 𝑗0.0723) = 1.021 + 𝑗0.0723 pu

= (1.021 + 𝑗0.0723 ) × 2300 = 2.354∠4.05° kV d) The pu values 𝑉𝑎(𝑝𝑢) =

2400 = 1.0435∠0° pu 2300

𝑆𝑝𝑢 = 0.9∠36.87° pu Thus 𝐼𝑎(𝑝𝑢) =

𝑆𝑝𝑢∗ 0.9∠ − 36.87° = 0.8625∠ − 36.87° pu ∗ = 𝑉𝑎(𝑝𝑢) 1.0435

In pu values, 𝑉𝑎(𝑛𝑜 𝑙𝑜𝑎𝑑) = 𝐸𝑎 = 1.0435 + 0.8625∠ − 36.87°(0.021 + 𝑗0.0723 ) = 1.095 + 𝑗0.039 pu

= (1.095 + 𝑗0.039) × 2300 = 2.521∠2.038° kV 4.6.2

Per-Unit System of A Transformer

Problem 4.2 40 MVA, 20 kV/400 kV, single-phase transformer has the following series impedances: 𝑍1 = 0.9 + 𝑗1.8 

𝑍2 = 128 + 𝑗288  i.

Determine the pu impedance of the transformer using a) ohmic values of both primary and secondary sides b) ohmic values referred to the primary side c) ohmic values referred to the secondary side

ii. Comment on the pu values in (a), (b), and (c). iii. Using the pu values, find the primary and secondary voltages when the transformer operating at full load 0.85 PF lagging.

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Solution: i. a) For a transformer, 𝑉𝑏 at the primary is not the same as 𝑉𝑏 at the secondary. So, we can name them as 𝑉𝑏1 and 𝑉𝑏2.

𝑉𝑏1 = 20 kV

𝑉𝑏2 = 400 kV

Next, calculate 𝑍𝑏 accordingly. Noted that 𝑆𝑏 of a transformer is the same for both sides. 202 = 10  40 4002 = = 4000  40

𝑍𝑏1 = 𝑍𝑏2 The total pu impedance, 𝑍𝑒𝑞(𝑝𝑢) =

0.9 + 𝑗1.8 128 + 𝑗288 = 0.122 + 𝑗0.252 pu + 4000 10

b) The 𝑍𝑒𝑞 referred to primary/ low voltage side 𝑍𝑒𝑞1 = 0.9 + 𝑗1.8 + (128 + 𝑗288) ( 𝑍𝑏1 = 10 

𝑍𝑒𝑞1(𝑝𝑢) =

20 2 ) = 1.22 + 𝑗2.52  400

1.22 + 𝑗2.52 = 0.122 + 𝑗0.252 pu 10

c) The 𝑍𝑒𝑞 referred to secondary/ high voltage side

𝑍𝑒𝑞2 = 128 + 𝑗288 + (0.9 + 𝑗1.8) (

400 2 ) = 488 + 𝑗1008  20

𝑍𝑏2 = 4000 

488 + 𝑗1008 = 0.122 + 𝑗0.252 pu 4000 ii. The pu impedance of a transformer is the same when referred to the primary and secondary 𝑍𝑒𝑞1(𝑝𝑢) =

sides. The difference between the two sides is represented by the 𝑍𝑏1 and 𝑍𝑏2 . iii. The general equation of a transformer is given as 𝑉1 = 𝑉2 + 𝐼𝑍𝑒𝑞

𝑉1 = 1 + 1∠ − 36.87°(0.122 + 𝑗0.252) = 1.236 + 𝑗0.15 pu 𝑉1𝑝𝑟𝑖𝑚 = (1.236 + 𝑗0.15 ) × 20 k = 24.91∠6.91° kV

𝑉1𝑠𝑒𝑐 = (1.236 + 𝑗0.15 ) × 400 k = 498.2∠6.91° kV

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EXERCISE 4.1

A 60-MVA, 69.3-kV, three-phase synchronous generator has a synchronous reactance of 15 Ω/phase and negligible armature resistance. The generator is delivering rated power at 0.8 power factor lagging at the rated terminal voltage to an infinite bus bar. Determine the magnitude of the generated emf per phase and the power angle using

4.2

i.

actual value calculation

ii.

per unit system

Three identical 9-MVA, 7.2-kV/4.16-kV, single-phase transformers are connected in Y on the high-voltage side and  on the low voltage side. The equivalent series impedance of each transformer referred to the high-voltage side is 0.12 + j0.82 Ω per phase. The transformer supplies a balanced three-phase load of 18 MVA, 0.8 power factor lagging at 4.16 kV. Determine the line-to-line voltage at the low-voltage and high-voltage terminals of the transformer using

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actual value calculation

ii.

per unit system

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