Power Systems - Ak Sinha 35302x

Indian Institute of Technology, Kharagpur

Introduction to Power Systems

Avinash K. Sinha Department of Electrical Engineering I. I. T. Kharagpur NPTEL

A.K. Sinha

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Power Systems

EE33001 Prerequsites: EE14001

3 - 0 - 3 : 5 Credits

Introduction to electric power generation, transmission and distribution. Load characteristics. Element of interconnected power system analysis, operation and control in both normal steady and abnormal transient state.

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Indian Institute of Technology, Kharagpur Modeling of Transmission Line Æ Resistance, Inductance and Capacitance calculation, Distributed parameter model, Lumped parameter model, ABCD parameters, Power Flow on line. Power Flow studies Æ Power flow equations, G-S, N-R and FDLF algorithms. Power System Operation Æ Voltage and Frequency control, Economic Dispatch, LFC and AGC for interconnected area operation, Operation in restructured scenario. NPTEL

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Indian Institute of Technology, Kharagpur Short Circuit studies Æ Symmetrical and asymmetrical faults, Symmetrical components, Sequence Networks, ZBUS algorithm for large systems Power System Protection Æ Concept of protective zones, Types of relays and their characteristics and application, protection schemes for transmission lines, generators, transformers and motors Power System Stability Æ Steady state, dynamic and transient stability, swing equation, Equal area criterion and numerical integration methods. NPTEL

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Lesson 1: Introduction Lesson Summary: ¾ Introduction to the course and subject ¾ Role of Power System in society ¾ Components of a Power System ¾Operation and Control ¾Organizational Structure NPTEL

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¾ What is a PS? ¾ functions ¾ Components ¾ How PS are operated / controlled ? ¾ How PS are organized?

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What Is a Power System? „A system that deals with the business of: „ Generation „ Transmission „ Distribution of Electrical Energy „Largest and most complex man made system. NPTEL

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Indian Institute of Technology, Kharagpur „

„

„

Power System provides a vital service to the society. Electrical power is somewhat like the air we breathe: We think about it only when it is missing. It should be operated with the goal of achieving: „ Highest reliability standards „ Lowest operation cost „ Minimum environmental impacts NPTEL

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PS Functions

SMES Storage Flywheel Compressed Air

Source of Fuel

Energy conversion

Transmission & Distribution

Energy Conversion (utilization)

Consumption devices are part of the PS & need to be modeled in PS analysis. Although they are not owned or controlled by the power utility. NPTEL

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Inverter UPS

Source of Fuel

Energy conversion

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Storage

Transmission & Distribution

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Energy Conversion (utilization)

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Power Generation

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Power Generation „ Takes place in power plants which may be geographically dispersed „ A Power plant may house more than one generating units „ Types of Generating units : NPTEL

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Energy Sources „

Hydrocarbons (oil, coal, natural gas, etc.)

„

Water

„

Nuclear

„

Solar

„

Chemical

„

Wind

„

Tidal NPTEL

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Energy Conversion Conversion processes in a thermal power plant: Chemical energy ⇒ Burners ⇒ Thermal energy Thermal energy ⇒ Boilers ⇒ Mechanical energy Mechanical energy ⇒ Turbines ⇒ Kinetic energy KE ⇒ Rotating machines ⇒ Electrical energy Overall conversion efficiency of a thermal power plant ≅ 40% NPTEL

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Indian Institute of Technology, Kharagpur Installed Generating Capacity in India Type of Installed Generation Capacity (MW) THERMAL 80201.45 (70.4%) NUCLEAR 2720.00 (2.47%) HYDRO 30135.23 (27.1%) WIND 2488.13 (2.47%) TOTAL 115544.81 Peak Demand 87906 MW Deficit 10625 MW (12.1%) NPTEL

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Power Transmission

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Indian Institute of Technology, Kharagpur „ Transmission networks : „High Voltage AC Transmission „HVDC Transmission „Connect generating plants to consumption points – Use Remote Energy Sources „interconnect power pools – Reduce Generation reserve & Cost, increase reliability NPTEL

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„High voltage transmission offers: „

Lower transmission losses / MW transfer

„

Lower line-voltage drop / Km

„

Higher transmission capacity / Km

„

„

Reduced right-of-way requirement / MW transfer Lower capital and operating costs / MW transfer NPTEL

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Power Transmission Equipment „Transformers „

Step-up transformers

„

Step-down Transformers

„

Voltage Regulators

„

Phase Shifters NPTEL

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Power Transmission Equipment „ Transmission Lines & Cables „ Circuit Breakers & Isolators „ Shunt & Series Reactors & Capacitors „ Lightning Arrestors „ Protective Relays „ FACTS Devices (SVC, Statcom, TCSC, UPFC, etc.) „ Converter / Inverter NPTEL

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Standard Transmission Voltages in India „ AC Transmission: „ 765 kV „ 400 kV „ 220 kV „ 132 kV „HVDC Transmission: „ 500 kV NPTEL

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Power Distribution

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Indian Institute of Technology, Kharagpur „ Receives electrical energy from the HV/MV levels „ Supplies energy to customers: „

At MV / LV levels

„

Single phase and/or three-phase

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Indian Institute of Technology, Kharagpur „ Main equipment in distribution system: „

Distribution transformers (DXFR)

„

Feeders (O/H or underground cables)

„

Switches, fuses, etc.

„

Protective Relays

„

Lightning Arrestors

„

CT / PT NPTEL

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Power Distribution Transmission Network Large

Industrial Consumers

Industrial Consumers

11/.4 DXFR

33/11 DXFR

Substation 33KV

Commercial & Large Consumers

11 KV

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Residential Consumers

.4/0.23 DXFR 400V

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Power System Operation/Control

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Key Operational Goals „ Power Balance: Generation must remain balanced with demand Total Generation (t) = Total Demand (t) + Losses (t)

„ System Security: Equipment power flows must not exceed equipment ratings, under normal or a single outage condition: „ |pij(t)| < pij max NPTEL

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Power Quality Considerations „ Frequency Regulation: System frequency, must remain within its operational range „ f min < f(t) < f max „ 49.5 Hz < f(t) < 50.5 Hz „ Voltage Regulation: Bus voltages must remain within their operational limits „ Vmin < V(t) < Vmax „ 0.95 pu < V(t) < 1.05 pu NPTEL

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PS Operation/Control „ Types of Operation/Control: „ Centralized (based on system-wide data) - Slow events are often handled by centralized controls „ Decentralized (based on local data) - Fast events are tackled by decentralized controls „ PS control has a Hierarchical Structure: „ Uses both centralized & decentralized control strategies NPTEL

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Indian Institute of Technology, Kharagpur „ Centralized (slow) controls: „ Dispatchers/Operators SCADA (Supervisory Control And Data Acquisition) EMS (Energy Management System) „

„

„ Decentralized (fast) controls are driven by local measurements - Protection systems are mostly decentralized NPTEL

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Indian Institute of Technology, Kharagpur „ Typical control hierarchy in PS: „ „ „ „ „ „

Device Remote Terminal Unit (RTU) Master station (MS) Utility control center Regional control centers National control center

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Centralized Controls „ Examples of centralized controls: „ Frequency control (regulation) „ Interchange control „ Generation dispatch (control of generating units) „ System security assessment & enhancement (both static and dynamic) „ Unit Commitment (units’ on/off status) NPTEL

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Decentralized Controls „ Examples of decentralized controls: Equipment protection against over-voltage and over-current „ Generator speed control (governor action) „ Generator terminal voltage control (excitation system) „

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Power System Organization

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Power Utility Structure „ Traditional „ Restructured (Deregulated)

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Indian Institute of Technology, Kharagpur „ Traditional utilities: ▪ Operate as Monopolies (have “captive customers”)

▪ Government “regulated” ▪ Have to get regulators approval for rate increases

▪ No incentive to increase generation efficiency ▪ “Vertically Integrated” (VI) business structures for improved profitability NPTEL

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Regulated Power Utility

Production Inputs

Production Vertically Integrated (VI) Business Structure

Transportation

Distribution NPTEL

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Deregulated PS Structure „

Has been legislated by governments, hoping that it results in: „ Increased competition (reduced rates) „ New technology utilization in generation, Transmission & Distribution „ Increased outside investment in PS infrastructure

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Power Industry Deregulation „

Has necessitated break-up of traditional utility VI structures into independent business units: „ Gencos (Generating Companies) „ Power Marketers „ Transco (Transmission Company – Owner) „ Independent System Operator (operating the transmission network) „ Retail Service Providers „ Discos (Distribution Companies) NPTEL

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Transmission Line Parameters

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Conductors Copper ACSR (Aluminum Conductor Steel Reinforced) AAC (Aluminum-Alloy Conductor) ACAR (Aluminum-Clad Steel Conductor) Expanded ACSR

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Indian Institute of Technology, Kharagpur Stranded Conductor: Total # of strands Æ 1, 7, 19, 37, 61, 91 Steel strands

Aluminium strands

Indian Institute of Technology, Kharagpur Stranded Conductor: Total # of strands Æ 1, 7, 19, 37, 61, 91 Steel strands

Aluminum strands


Insulator

Pin Insulator Shackle Insulator String Insulator

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Structure Transmission Towers Poles

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Shield Wires (Ground Wires)) Protects Phase Conductors from Direct Lightning Strikes Mechanical Considerations: Sag Tension Vibration NPTEL

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Electrical Parameters Resistance Inductance Capacitance

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Resistance The dc resistance of a conductor at a specified temperature T is: T

R d c ,T

ρ l = Ω A

Where ρ = conductor resistivity at temperature T T

l = conductor length (m) A = conductor cross-sectional area ( m2) NPTEL

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Conductor resistance depends on the following factors: ƒ temperature ƒ Spiraling ƒ

Frequency ( “skin effect” )

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Indian Institute of Technology, Kharagpur Resistivity of conductor metals varies linearly over normal operating temperatures according to

ρ

T

2

=ρ

T1

⎛ T2 + M ⎞ ⎜ ⎟ ⎝ T1 + M ⎠

The ac resistance or effective resistance of a conductor is

R ac

Ploss = 2 Ω I

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ρ20°C

M

Resistivity at 20°C Temperature constant Material

% Conductivity

Ωm x 10-8

°C

Copper: Annealed

100%

1.72

234.5

Hard-drawn

97.3%

1.77

241.5

61%

2.83

228.1

Iron

17.2%

10

180

Silver

108%

1.59

243

Aluminum: Hard-drawn

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λ


Inductance 1. Magnetic field intensity H

Ampere’s law

2. Magnetic flux density B ( B = μH ) 3. Flux linkages

λ

4. Inductance from flux linkages per ampere

(L = λ/I) NPTEL

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Solid Cylindrical Conductor I

Hx

x r 1m

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Internal Flux Linkage

Ñ ∫H

tan

dl = Ienclosed

Hx ( 2πx ) = Ix

for x < r

Ix Hx = A/m 2πx 2

⎛x⎞ Ix = ⎜ ⎟ I ⎝r⎠

for x < r NPTEL

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Hx =

xI 2π r

2

A/ m

μ 0 xI μ 0 xI 2 d dx Bx = μ 0 H x = Wb m φ / ; = 2 2 2π r 2π r ⎛ π x2 ⎞ μ0 I 3 dλ = ⎜ 2 ⎟ dΦ = x dx Wb − t / m 4 ⎜ πr ⎟ 2π r ⎝ ⎠ r r μ0I 3 λ in t = ∫ d λ = x dx ∫ 4 0 0 2π r μ0I 1 W b -t/m = = × 10−7 I 2 8π

Indian Institute of Technology, Kharagpur External Flux linkage D1

D2

Hx ( 2πx ) = I r

x

I Hx = A/m 2πx

x >r

HX NPTEL

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Hx =

I 2π x

(

x>r

A/ m

B x = μ 0 H x = 4π × 10

−7

d Φ = B x dx = 2 × 10

) 2π x = 2 × 10

−7

I

I dx x

−7

Wb / m

I x

Wb / m

2


d λ = d Φ = 2 × 10

−7

I dx x

Wb − t / m

Integrating between two points at distances

D1 and D2 from the conductor center will give the external flux linkage λ12 between D1 and

D2:

λ12 =

D2 ∫D 1

−7

d λ = 2 × 10 I

⎛ D2 ⎞ = 2 × 10 I In ⎜ ⎟ ⎝ D1 ⎠ −7

D2 ∫D 1

dx x

Wb-t/m


λP = λint + λ12 1 D -7 -7 λp = ×10 I + 2 ×10 I In 2 r


D⎞ ⎛ 1/4 λ p = 2 × 10 I ⎜ In e + In ⎟ r ⎠ ⎝ D -7 = 2 ×10 I In -1/4 e r -7

D = 2 ×10 I In Wb - t/m r' r' = e-1/4r = 0.7788r -7

where

λp -7 ⎛ D ⎞ LP = = 2 ×10 In ⎜ ⎟ H/m I ⎝ r' ⎠

Indian Institute of Technology, Kharagpur Composite Conductor p

DPM

M

Dpj Dpi

j

Dp2

Dij i

Dp1

2 1

Indian Institute of Technology, Kharagpur M

I1 + I2 + .... + IM =

∑ Im

=0

m =1

λ iPi

D Pi = 2 × 10 Ii In rk ' -7

-7

λ iPj = 2 × 10 I j In

DPj Dij


λiP = λiP1 + λiP2 +.....+ λiPM = 2 ×10

-7

M

∑Ij j=1

λ iP

M

In

DPj Dij M

1 = 2 × 10 ∑I j In + 2 × 10-7 ∑I j In DPj Dij j=1 j=1 -7

Indian Institute of Technology, Kharagpur M-1 ⎡M ⎤ 1 λiP = 2×10-7 ⎢∑Ij In + ∑Ij In DPj +IM In DPM ⎥ Dij j=1 ⎢⎣ j=1 ⎥⎦

(

)

M-1 M-1 ⎡M ⎤ 1 -7 λiP = 2×10 ⎢∑Ij In + ∑Ij In DPj - ∑Ij In DPM ⎥ ⎢⎣ j=1 Dij j=1 ⎥⎦ j=1


M M-1 ⎡ DPm ⎤ 1 -7 = 2 ×10 ⎢ ∑ Im In + ∑ Im In ⎥ Dkm m=1 DPM ⎥⎦ ⎢⎣m=1

M

1 λ i = 2 × 10 ∑ I j In Wb - t/m D ij j=1 -7

Indian Institute of Technology, Kharagpur Single-phase two-wire line rx

ry

⎛ 1 1 ⎞ λx =2×10 ⎜Ix In +Iy In ⎟ ⎜ ⎟ D D xx xy ⎝ ⎠ -7

= 2 × 10

-7

D

= 2 × 10

-7

⎛ 1 1⎞ - I In ⎟ ⎜ I In r' x D⎠ ⎝

D I In r' x

W b - t/m


r'x = e

-1 /4

r x = 0 .7 7 8 8 r x

λx λx D -7 Lx = = = 2 × 1 0 In Ix I r' x

H /m


⎛ 1 1 λ y = 2 × 1 0 ⎜ Ix In + Iy In ⎜ D yx D yy ⎝ ⎛ 1 1 ⎞ -7 = 2 × 1 0 ⎜ I In - I In ⎟⎟ ⎜ D r'y ⎠ ⎝ D -7 = -2 × 1 0 I In r'y -7

⎞ ⎟⎟ ⎠


λy

λy

D Ly = = = 2 ×10 In H/m per conductor Iy -I r'y -7

The total inductance of the single-phase circuit, also called loop inductance, is L = Lx +Ly

⎛ D D ⎞ = 2 × 1 0 ⎜ In + In ⎟⎟ ⎜ r'x r'y ⎠ ⎝ 2 D = 2 × 1 0 -7 I n r'xr'y -7

= 4 × 10 NPTEL

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In

D r 'xr 'y A.K. Sinha

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r ' x = r ' y = r ',

Also, if

D L = 4 × 1 0 In r' -7

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Indian Institute of Technology, Kharagpur Composite Conductor line N

M

rx

3’ 2’ g ⎛ ⎜ ⎝

k ⎛ ⎜ ⎝

⎞ ⎟ g ⎠ 2

ry ⎞ ⎟ ⎠

1’

1

Conductor y

Conductor x NPTEL

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Indian Institute of Technology, Kharagpur N M ⎡ I 1 I 1 ⎤ -7 Φk = 2 ×10 ⎢ In In ⎥ ⎣ N m=1 Dkm M m=1' Dkm ⎦ N M ⎡ Φk 1 1 1 1 ⎤ -7 λk = 2×10 I ⎢ 2 In In ∑ ⎥ N ⎣ N m=1 Dkm NM m=1' Dkm ⎦

N M ⎡ Φk 1 1 1 1 ⎤ -7 λk = 2×10 I ⎢ 2 In In ∑ ⎥ N ⎣ N m=1 Dkm NM m=1' Dkm ⎦

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Indian Institute of Technology, Kharagpur ⎛ ⎞ ⎜ mΠ= 1 ' D k m ⎟ ⎝ ⎠ M

λ x = 2 × 10

-7

N

I InΠ

k =1

Lx = 2 × 10 D

xy

=

N

MN

-7

In

D

xy

D

xx

km

k = 1m = 1 '

D

xx

=

N2

1 /N 2

M

ΠΠD N

⎛ N ⎞ ⎜ mΠ= 1 D k m ⎟ ⎝ ⎠

1 /N M

(GMD)

N

ΠΠD

km

k = 1m = 1

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Indian Institute of Technology, Kharagpur H/m per conductor

(GMR)

H/m per circuit NPTEL

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Indian Institute Institute of of Technology, Technology, Kharagpur Kharagpur Indian

Inductance Calculations Three Phase Line 1. Inductance Calculation for three phase transmission line with: a) Equilateral spacing b) Transposition c) Bundled conductor d) Ground return


Three Phase Line with Equilateral Spacing a c

D

b

1 1 1⎞ ⎛ λa = 2×10 ⎜ Ia In +Ib In +Ic In ⎟ r' D D⎠ ⎝ 1 1⎤ -7 ⎡ = 2×10 ⎢Ia In + ( Ib +Ic ) In ⎥ r' D⎦ ⎣ -7

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1 1⎞ ⎛ λ a = 2 ×10 ⎜ Ia In - Ia In ⎟ r' D⎠ ⎝ D -7 = 2 ×10 Ia In Wb - t/m r' -7

λa D -7 La = = 2 ×10 In Ia r' NPTEL

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Three phase transposed Line 1

3

2

Position 1 Position 2 Position 3

Ia Ib Ic

l 3

l 3

l

l 3


-7 ⎡

1 1 1⎤ λa1 =2×10 ⎢Ia In +Ib In +Ic In ⎥ DS D12 D31 ⎦ ⎣

Wb-t/m

-7 ⎡

1 1 1 ⎤ λa2 =2×10 ⎢Ia In +Ib In +Ic In ⎥ Wb-t/m DS D23 D12 ⎦ ⎣ -7 ⎡

1 1 1 ⎤ λa3 =2×10 ⎢Ia In +Ib In +Ic In ⎥ DS D31 D23 ⎦ ⎣

Wb-t/m


⎛l⎞ ⎛l⎞ ⎛l⎞ λa1 ⎜ ⎟ + λa2 ⎜ ⎟ + λa3 ⎜ ⎟ 3⎠ 3⎠ 3 ⎠ λa1 + λa2 + λa3 ⎝ ⎝ ⎝ λa = = l 3 ⎤ 2×10-7 ⎡ 1 1 1 = +Ib In +Ic In ⎢3Ia In ⎥ 3 ⎣ DS D12D23D31 D12D23D31 ⎦

2 × 10 -7 λa = 3

⎡ ⎤ 1 1 - Ia In ⎢ 3Ia In ⎥ D D D D S 12 23 31 ⎦ ⎣

= 2 × 10-7 Ia In

3D D D 12 23 31

DS

Wb-t/m

Indian Institute of Technology, Kharagpur 3D D D λa 12 23 31 -7 La = = 2 ×10 In Ia DS Deq = 3 D12D23D31 -7

La = 2 ×10 In

Deq DS

H/m

H/m per phase


Bundled Conductor Line d d

d

d d

d

d d

Bundling Æ Reduces Electric Field Strength on conductor surface Æ Reduces Corona Bundling Æ Increases Effective Radius (GMR) Æ Reduces Inductance


DS =

4

(r' × d)

2

d

2

= r'd 3

d

d

DS = 9 (r' × d× d) = r'd 3

2

d

DS

= 16

(

r' × d× d× d 2 -7

La = 2 ×10 In

Deq DS

)

4

H/m

4

= 1.091 r'd3 d d

d d

Indian Institute of Technology, Kharagpur 3-Φ line with earth return

Earth Conductors are connected in parallel and grounded at regular intervals along the line.


If Phase currents are not balanced return currents will flow in earth conductors and ground J. R. Carson (1923) Æ ground currents Æ earth return conductors having same GMR and located directly under the overhead conductors at a distance De (as shown in figure).

Indian Institute of Technology, Kharagpur n1(4)

n2(5) b(2)

a(1)

c(3) Earth surface

De De

De

n1’

N2’ b’

a’

De De

c’

De ∝ ρ / f; Where ρ = Earth’s resistivity f = Frequency


Dkk ' = De

Dk'k' = Dkk (m);

D e = 658.5 ρ/f m

(ρ = 100Ω − m)

-7

Rk' = 9.869 ×10 f (Ω/m) 2 ( 3+2 )

∑

Ik = 0

k=1

λk = 2 ×10

-7

( 3+2 )

Dkm' ∑ Im In D km m=1

Wb-t/m


λ = LI Where,

λ is a ( 3 + 2 ) vector I is a ( 3 + 2 ) v e c to r L is a ( 3 + 2 ) × ( 3 + 2 ) matrix whose elements are :

Dkm' Lkm = 2 ×10 In Dkm -7


⎡E Aa ⎢ E ⎢ Bb ⎢E C c ⎢ ⎢0 ⎢0 ⎣

⎤ ⎡ Ia ⎤ ⎥ ⎢ ⎥ I ⎥ ⎢b ⎥ ⎥ = ( R + jω L ) ⎢ I c ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ In 1 ⎥ ⎥ ⎢I ⎥ ⎦ ⎣ n2 ⎦


⎡(R a +Rk' ) Rk' ... Rk' ⎤ ⎢ ⎥ ⎢Rk' (Rb +Rk' ) Rk' ... ⎥ ⎢ ⎥ R=⎢ Ω (Rc +Rk' ) Rk' ... ⎥ ⎢ ⎥ Rn1 +Rk' ) Rk' ... ( ⎢ ⎥ ⎢Rk' . . . ⎥ R +R ( ) n2 k' ⎦ ⎣

Indian Institute of Technology, Kharagpur ZA

ZB

Z12 Z13 Z14 ... Z1( 3+N) ⎤ ⎡Z11 ⎡EAa ⎤ ⎢ ⎥ ⎡Ia ⎤ ⎢ ⎥ ⎢Z21 Z22 Z23 Z24 ... Z2( 3+N) ⎥ ⎢I ⎥ ⎢EBb ⎥ ⎢ ⎥⎢ b ⎥ Z32 Z33 Z34 ... Z3( 3+N) ⎥ ⎢Ic ⎥ ⎢ECc ⎥ ⎢Z31 ⎢ ⎥⎢ ⎥⎢ ⎥ Z42 Z43 Z44 ... Z4( 3+N) ⎥ ⎢In1 ⎥ ⎢0 ⎥ ⎢Z41 ⎢... ⎥ ⎢ ⎥ ⎢I ⎥ ⎢ ⎥⎢ ⎥ ⎢ n2 ⎥ ⎢⎣0 ⎥⎦ ⎢Z ⎥ ⎢⎣. ⎥⎦ Z Z Z ... Z ( 3+N)( 3+N) ⎦ ⎣ ( 3+N)1 ( 3+N)2 ( 3+N)3 ( 3+N)4

ZC

ZD


Diagonal elements (k = m) Zkk

Dkk' = Rk +Rk' + jω2 × 10 In Ω /m Dkk

Off - Diagonal elements (k ≠ m) Dkm' Ω /m Zkm = Rk +Rk' + jω2 × 10 In Dkm


EP

0

=

E Aa

ZA

Z C ZD

IP = EP = E Bb

ECc

ZB

IP In

Ia Ib Ic

In =

In1 In 2


E P = Z A IP + Z B In

0 = Z C IP + Z D I n In = Z

-1 D Z C IP

-1 ⎡ EP = Z A - Z B Z D Z C ⎤ IP ⎣ ⎦ EP = ZPIP

ZP = Z A -

-1 ZBZD ZC


⎡ Z aaeq Z abeq Z aceq ⎤ ⎢ ⎥ Z P = ⎢ Z abeq Z bbeq Z bceq ⎥ ⎢ ⎥ ⎢⎣ Z aceq Z bceq Z cceq ⎥⎦

Ωm


∧ ∧ ⎡∧ ⎤ ⎢ Z aaeq Z abeq Z abeq ⎥ ⎢∧ ⎥ ∧ ∧ ∧ ZP = ⎢ Z abeq Z aaeq Z abeq ⎥ Ωm ⎢∧ ⎥ ∧ ∧ ⎢ ⎥ Z Z Z abeq abeq aaeq ⎥ ⎢ ⎣ ⎦


∧

1 Z aaeq = Z aaeq + Zbbeq + Zcceq 3 ∧

(

)

1 Z abeq = Z abeq + Z aceq + Zbceq 3

(

)

IIT Kharagpur Indian Institute of Technology, Kharagpur Example: The conductor configuration of a completely transposed three-phase overhead transmission line with bundled conductors is shown in the next slide. All the conductors have a radius of 0.74cm with a 30cm bundle spacing. (a) Determine the inductance per phase in mH/km and in mH/m. (b) Find the inductive line reactance per phase in ohms/m at 60Hz.

IIT Kharagpur Indian Institute of Technology, Kharagpur

IIT Kharagpur Indian Institute of Technology, Kharagpur Solution: For the given configuration as shown in the figure on the last slide, we have the mutual GMD between the different phases given by Dab = (r13.r14.r23.r24)1/4 = (6 X 6.3 X 5.7 X 6)1/4 = 5.9962 m Similarly,

Dbc = 5.9962 m

IIT Kharagpur Indian Institute of Technology, Kharagpur Dca = (r15.r16.r25.r26)1/4 = (12 X 12.3 X 11.7 X 12)1/4 = 11.9981 m The equivalent equilateral spacing between the phases is given by Deq defined as Deq = (Dab X Dbc X Dca)1/3 = (5.9962 X 5.9962 X 11.9981)1/3 = 7.5559 m

IIT Kharagpur Indian Institute of Technology, Kharagpur Self GMD of all the three phases, owing to symmetry, are equal and hence the equivalent self GMD for the system can be given by Ds = (.7788 X r X 30)1/2 = 4.1580 cm. Inductance per phase for the given system is L = 2 X 10-7 X ln( Deq/ Ds) H/m/phase = 1.04049 X 10-6 H/m/phase. = 1.04049 X 10-3 mH/m/phase. = 1.04049 mH/km/phase.

IIT Kharagpur Indian Institute of Technology, Kharagpur The inductive line reactance per phase at 60Hz is given by X = 2π X 60 X 1.04049 X 10-6 ohms/m/phase. = 3.922 X 10-4 ohms/m/phase


TRANSMISSION LINE CAPACITANCE CALCULATIONS Gauss’s Law Æ Electric Field Strength (E) Voltage between conductors Capacitance Æ (C = q / V)

NPTEL

A.K. Sinha

7/32


Electric Field and Voltage Calculation Gauss’s Law

Ò ∫∫ D⊥ds = Ò ∫∫ εE⊥ds = Qenclosed NPTEL

A.K. Sinha

8/32

Indian Institute of Technology, Kharagpur P1

+ V12

D1

_ P 2 D2

X

+ ++ + r + + + + +

Ex

εEx ( 2πx )(1) = q (1) 1m NPTEL

A.K. Sinha

9/32


q Ex = 2πεx

V12 = ∫

D2

D1

V12 = ∫

D2

D1

ε=ε r ε 0

V/m

Ex dx

D2 q q volts dx = In 2πεx 2πε D1

ε 0 = 8 . 854 x 10 NPTEL

−12

F/m A.K. Sinha

10/32

Indian Institute of Technology, Kharagpur n

Multi-conductor system rk k

Dkm

j

_

Dim

1

conductor k has radius rk and charge qk

2 i

D jk qk Vki Vijk = In volts 2πε Dik + n Djk qk Vij = ∑ In volts Dik k=1 2πε NPTEL

A.K. Sinha

11/32

Indian Institute of Technology, Kharagpur rx q C/m

Single phase Line

ry

-q C/m Vxy D

Dyx Dyy 1 ⎡ = q In -q In Dxx Dxy 2πε ⎢ ⎣

⎤ ⎥⎦

DyxDxy q = In 2πε DxxDyy NPTEL

A.K. Sinha

12/32


Vxy Cxy

q D = In πε rxry

q = = Vxy

πε D In r r x y NPTEL

volts

F/m line-to-line

A.K. Sinha

13/32


πε Cxy = F/m line-to-line In (D/r ) Vxn = Vyn = Vxy / 2 q Cn = Cxn = Cyn = = 2Cxy Vxn 2πε = F/m line-to-neutral In (D/r ) n x

Cxy

y

x

Cxn = 2Cxy NPTEL

Cyn = 2Cxy

A.K. Sinha

14/32

y


Three Phase Line with Equilateral Spacing a c

D

b

qa + qb + qc = 0

NPTEL

A.K. Sinha

15/32


Dba Dbb Dbc ⎤ 1 ⎡ Vab = qa In + qb In + qc In D D D 2πε ⎢ aa ab ac ⎥ ⎣ ⎦

Vab

1 D r D = ⎡ qa In + qb In + qc In ⎤ r D D⎦ 2πε ⎣

1 D r = ⎡ qa In + qb In ⎤ r D⎦ 2πε ⎣ NPTEL

A.K. Sinha

volts

16/32


Dca Dcb Dcc ⎤ 1 ⎡ Vac = qa In + qb In + qc In D D D 2πε ⎢ aa ac ⎥ ab ⎣ ⎦

1 D D r = ⎡ qa In + qb In + qc In ⎤ r D D⎦ 2πε ⎣

1 D r = ⎡ qa In + qc In ⎤ r D⎦ 2πε ⎣ NPTEL

A.K. Sinha

17/32


Vab = 3Van +30 = 3Van ⎡ 3 + j 1 ⎤ ⎣ 2 2⎦ o

Vac = -Vca = 3Van -30 = 3Van ⎡ 3 - j 1 ⎤ ⎣ 2 2⎦ o

a b

3an n

c

NPTEL

A.K. Sinha

18/32


Vab + Vac = 3Van volts

( )

(

)

1 1 D r Van = ⎡ 2q a In + qb +qc In ⎤ r D⎦ 3 2πε ⎣ 1 D Van = qa In 2πε r qa 2πε F/m line-to-neutral Can = = Van In (D/r ) NPTEL

A.K. Sinha

19/32


Can =

2πε

(

In Deq /r

)

F/m

Deq = 3 DabDbcDac

NPTEL

A.K. Sinha

20/32

Indian Institute of Technology, Kharagpur 3-phase bundled conductor line with unequal spacing d

a

d

a’

b

d c

b’

Dab

c’

Dbc

Dac

NPTEL

A.K. Sinha

21/32


⎡ qa 2 ⎢ 1 Vab = 2πε ⎢ q b ⎢2 ⎣

In In

Dba Daa Dbb' Dab'

+ +

qa 2 qc

NPTEL

2

In In

Dba' Daa' Dbc Dac

⎤ + In + 2 Dab ⎥ qc Dbc' ⎥ + In 2 Dac' ⎥ ⎦ qb

A.K. Sinha

Dbb

22/32

Indian Institute of Technology, Kharagpur d r In + In Dab Dab

(

) (

)

Dab Dbc 1 ⎡ rd = qa In + qb In + qc In D D 2πε ⎢ rd ab ac ⎣

⎤ ⎥⎦

qb Dab ⎡ qa Dab + In + In 2 r d 1 ⎢ 2 = 2πε ⎢ q D D c bc bc + In + In ⎢ 2 D Dac ac ⎣

)

(

NPTEL

A.K. Sinha

⎤ ⎥ ⎥ ⎥ ⎦

23/32


Can =

(

In Deq /DS

DS = rd 3

2πε

)

F/m

for a two-conductor bundle

2

DS = rd

for a three-conductor bundle

4

3

DS = 1.091 rd

NPTEL

for a four-conductor bundle A.K. Sinha

24/32

Indian Institute of Technology, Kharagpur Line Charging Current and Reactive var Generation

Ichg = Yxy Vxy = jωC xy Vxy QC =

2 Vxy

Xc

2 = Yxy Vxy

2 var = ωCxy Vxy

Ichg = YVan = jωCan VLN NPTEL

A

A.K. Sinha

A

25/32


2 Q C1φ = YVan

2 = ωCan VLN var

2 2 QC3φ = 3QC1φ = 3ωCanVLN = ωCanVLL var

NPTEL

A.K. Sinha

26/32


Example: A three phase, 400kV, 50Hz, 350km overhead transmission line has flat horizontal spacing with three identical conductors. The conductors have an outside diameter of 3.28 cm with 12 m between adjacent conductors. Determine the capacitive reactance-toneutral in ohms/m/phase and the capacitive reactance for the line in ohms/phase. NPTEL

A.K. Sinha

27/32


Solution: For a fully transposed three phase line, we have the line voltage given by Vab = 1/(2πk){ qa X ln(Deq/r) + qb X ln(r/Deq)} Where Deq = (D12.D23.D31)1/3 = (12 X 12 X 24)1/3 = 15.119 m. Similarly, Vac = 1/(2πk){ qa X ln(Deq/r) + qc X ln(r/Deq)} NPTEL

A.K. Sinha

28/32

Indian Institute of Technology, Kharagpur 12 m

1(a)

12 m

3(c)

2(b)

Three phase wires in horizontal configuration

NPTEL

A.K. Sinha

29/32

Indian Institute of Technology, Kharagpur Also, we have Vab + Vac = 3Van And qb + qc = -qa Hence, adding the values of Vab and Vac we get Van = qa/(2πε) ln(Deq/r)

NPTEL

A.K. Sinha

30/32

Indian Institute of Technology, Kharagpur Hence, Cn = qa/Van = 2πε / ln(Deq/r) = 8.163 X 10-6 µF/m Yn = (2π X 50 X Cn) = 2.565 X 10-9 S/m per phase Given length of the line = 350 km. Hence, Yn = 8.978 X10-4 S per phase Xn = 1/Yn = 1.1138 X 103 Ω per phase NPTEL

A.K. Sinha

31/32


Effect of Earth on Transmission Line Capacitance Earth surface is considered as an equi potential Surface Method of Images Image conductors Æbelow the ground Ædepth equal to the height of the O/H conductor above the ground Æ opposite charge. NPTEL

A.K. Sinha

4/32


+ + + ++ + H Earth plane (a) Single conductor and earth plane

NPTEL

A.K. Sinha

5/32

Indian Institute of Technology, Kharagpur +++ + ++ (b) Earth plane replaced by image conductor

H11 =2H

NPTEL

A.K. Sinha

6/32


Capacitance of Single Phase Line Considering Effect of Earth

NPTEL

A.K. Sinha

7/32


y

x +q

-q

D

H Earth plane

Hxx

Hxy

-q x’

y’ NPTEL

+q A.K. Sinha

8/32


Vxy

Dyy Hyx Hyy q ⎡ Dyx = In - In - In + In Dxx Dxy Hxx Hxy 2πε ⎢ ⎣ HyxHxy q ⎡ DyxDxy = In - In DxxDyy HxxHyy 2πε ⎢ ⎣ Hxy q ⎡ D = In - In Hxx πε ⎢ ⎣ r

⎤ ⎥⎦

⎤ ⎥⎦

NPTEL

A.K. Sinha

9/32

⎤ ⎥⎦

Indian Institute of Technology, Kharagpur The line-to-line capacitance is

Cxy

q = = Vxy

Cn = 2Cxy =

πε

Hxy D In - In r Hxx 2πε Hxy D In - In r Hxx NPTEL

A.K. Sinha

F/m

F/m

10/32


Capacitance of Three Phase Line Considering Effect of Earth qa h11

-qa

D12 qb

D23

D31

h22

h31

h21 h12 -qb

qc h33 Earth

h32

h13 -qc

NPTEL

A.K. Sinha

11/32


⎡ ⎛ D12 h12 ⎞ ⎤ - ln ⎢ qa ⎜ ln ⎟ +⎥ r h 11 ⎠ ⎥ ⎝ ⎢ ⎢ ⎛ ⎥ 1 ⎢ r h22 ⎞ ⎥ Vab = qb ⎜ ln - ln ⎟+ 2πε ⎢ ⎝ D12 h12 ⎠ ⎥ ⎢ ⎥ h23 ⎞ ⎥ ⎢ ⎛ D23 q ln ln ⎜ ⎟ ⎥ c ⎢ D h 31 31 ⎠ ⎦ ⎝ ⎣ NPTEL

A.K. Sinha

12/32


2πε

Cn = In

Deq r

- In

3h h h 12 23 31

F/m

3h h h 11 22 33

Three Phase Line with Earth Return

NPTEL

A.K. Sinha

13/32

Indian Institute of Technology, Kharagpur n1

Dan1 a

nN

n2 b

Dab

c

Dac

Earth plane

Haa

Hab Han1 n1’

c’ b’

n2’

NPTEL

nN’ A.K. Sinha

14/332


Vkk'

1 ⎡ = 2πε ⎢

⎤ ∑ qm In D - ∑ qm In H ⎥ km m=a km ⎦ ⎣ m=a nN

Hkm

nN

Dkm

Hkm 2 nN = qm In ∑ 2πε m=a Dkm

NPTEL

A.K. Sinha

15/32


nN

Hkm 1 1 Vkn = Vkk' = qm In ∑ 2 2πε m=a Dkm

Vkn = 0

for k = n1,n2,....,nN

NPTEL

A.K. Sinha

16/32

Indian Institute of Technology, Kharagpur PA

⎡ Van ⎤ ⎡ Paa ⎢ ⎥ ⎢P Vbn ⎢ ⎥ ⎢ ba ⎢ Vcn ⎥ ⎢ Pca ⎢ ⎥=⎢ ⎢ 0 ⎥ ⎢ Pn1a ⎢: ⎥ ⎢ : ⎢ ⎥ ⎢ ⎢⎣ 0 ⎥⎦ ⎢⎣ PnNn

PB

Pab

Pac

Pan1

...

Pbb

Pbc

Pbn1

...

Pcb

Pcc

Pcn1

...

Pn1n1

...

Pn1b Pn1c

:

:

PnNb PnNc

PC NPTEL

:

:

PnNn1

...

⎤ ⎡ qa ⎤ ⎥⎢ ⎥ PbnN ⎥ ⎢ qb ⎥ PcnN ⎥ ⎢ q c ⎥ ⎥⎢ ⎥ Pn1nN ⎥ ⎢ qn1 ⎥ : ⎥⎥ ⎢⎢ : ⎥⎥ PnNnN ⎥ ⎦ ⎢⎣ qnN ⎥⎦ PanN

PD A.K. Sinha

17/32


Pk m V ⎡ P

⎣

0

Hkm 1 = In 2πε Dkm

m /F

q P P ⎡ A B ⎤⎡ p

⎤= ⎦ ⎢⎣ PC

⎤ PD ⎥ q ⎢ ⎦ ⎣ n ⎥⎦

Vp = PA qp +PBqn NPTEL

A.K. Sinha

18/32


-1 0 = PCqp +PDqn → qn = −PD PCqp

Vp =

(

-1 PA - PBPD PC

)q

qp = Vp =

(

-1 PA - PBPD PC

NPTEL

p

)

-1

F/m

A.K. Sinha

19/32


⎡ Caa ⎢ = C ⎢ ab ⎢⎣ Cac

C ab C ac ⎤ Cbb Cbc

NPTEL

⎥ Cbc ⎥ C cc ⎥ ⎦ A.K. Sinha

F/m

20/32


∧ ∧ ∧ ⎡ C aa C ab C ab

⎢ ∧ ∧ ⎢ = ⎢ C ab ⎢∧ ⎢⎣ C ab

∧ C aa ∧ C ab

NPTEL

⎤ ⎥ ∧ ⎥ C ab ⎥ ∧ ⎥ C aa ⎥ ⎦

F/m

A.K. Sinha

21/32

Indian Institute of Technology, Kharagpur ∧

1 C aa = ( C aa + C bb + C cc ) F/m 3 ∧ 1 Cab = ( Cab + Cbc + Cac ) F/m 3 Yp = jω = j ( 2πf ) ∧

∧

S/m

∧

Yp = jω = j ( 2πf ) S/m NPTEL

A.K. Sinha

22/32


Double Circuit Three Phase Line

(Ip1 + Ip2 )

+

Ep

Ip1

Ip2

NPTEL

A.K. Sinha

23/32


Series Impedance

⎡Ep ⎤ ⎡ Ip 1 ⎤ ⎢ ⎥ = Zp ⎢ ⎥ ⎢⎣ E p ⎥⎦ ⎢⎣ Ip 2 ⎥⎦

⎡Ip1 ⎤ -1 ⎡Ep ⎤ ⎡ YA ⎢ ⎥ = Zp ⎢ ⎥ = ⎢ Y ⎢⎣Ip2 ⎥⎦ ⎢⎣Ep ⎥⎦ ⎣ C

⎡( YA +YB ) ⎤ =⎢ Ep ⎥ ⎢ ⎥ YD ⎥ ⎦ ⎢⎣Ep ⎥⎦ ⎢( YC +YD ) ⎥ ⎣ ⎦ YB ⎤ ⎡Ep ⎤

NPTEL

A.K. Sinha

24/32


(Ip1 + Ip2 ) = ( YA + YB + YC + YD ) Ep

(

EP = Zp' Ip1 + Ip2

Where, ' Zp

)

= ( YA + YB + YC + YD ) NPTEL

-1

A.K. Sinha

25/32


Shunt ittance

⎡ Vp ⎤ ⎡ CA ⎡ qP1 ⎤ ⎢ ⎥ = ⎢ ⎥ = ⎢⎣ CC ⎢⎣ Vp ⎥⎦ ⎣ qP2 ⎦

⎡( C A +CB )⎤ =⎢ Vp ⎥ ⎢ ⎥ CD ⎥ ⎦ ⎢⎣ Vp ⎥⎦ ⎢⎣( CC +CD ) ⎥⎦ CB ⎤ ⎡ Vp ⎤

( qP1 + qP2 ) = eq VP eq = ( CA + CB + CC + CD ) ' Yp

= jωeq NPTEL

A.K. Sinha

26/32


Example: A three phase, 400kV, 50Hz, 350km O/H transmission line has flat horizontal spacing with three identical conductors. The conductors have an outside diameter of 3.28 cm with 12 m between adjacent conductors. The conductors are placed 20 m above ground. Consider the earth plane to be a perfect conductor. Determine the capacitive ittance-to-neutral in S/m/phase and the capacitive var generated by the line. NPTEL

A.K. Sinha

27/32

Indian Institute of Technology, Kharagpur a 20m h11

12m

b

12m

h13 h12 h22

c

h23

h33

earth

a’

b’

C’

Effect of earth on capacitance of a three phase system NPTEL

A.K. Sinha

28/32

Indian Institute of Technology, Kharagpur Solution: Let h11, h22, h33 be the distances between the conductors and their mirror images. And also, h12, h23, h31 be the distances as shown in the figure on the previous slide. We have for such systems, following the same procedure, Cn = 2πε/{ ln(Deq/r) –ln((h12h23h31)/(h11h22h33))1/3} NPTEL

A.K. Sinha

29/32

Indian Institute of Technology, Kharagpur We have, Deq = 15.119m r = 1.64 cm From the diagram, h11 = h22 = h33 = 40m h12 = h23 = (402+122)1/2 = 41.761 m h31 = (402+242)1/2 = 46.648 m Putting these values in the equation (1) we get Cn = 8.2595 X 10-6 µF/m. NPTEL

A.K. Sinha

30/32

Indian Institute of Technology, Kharagpur Hence, Yn = 2π x50 x 8.2595 X 10-12 S/m per phase = 2.595 x 10-9 S per phase Length of the line = 350 km. Therefore, Total Yn = 908 μS per phase. Now var generated by line QC = Yn x VLL2 Or QC = 908 x 10-6 x 400 x 103 x 400 x 103 = 145.358 Mvar NPTEL

A.K. Sinha

31/32


Lesson 7 Transmission Line Modeling Lesson Summary: 1. Distributed Vs. Lumped Parameter Models 2. Short Line model 3. Medium Line model 4. Long Line Model 5. Voltage regulation NPTEL

A.K. Sinha

2/30


Transmission Line Model I(x +Δx) + V(x + Δx)

zΔx

I(x) + V(x)

yΔx

x

(x + Δx)

NPTEL

A.K. Sinha

3/30


Lumped Parameter Models For Sinusoidal waves on O/H lines:

C 8 λ = ;C = 3x10 m/s f

for f = 50 Hz; λ = 6000 km If the line length is <250 km long, then one can neglect the distributed effect of the line parameters and it is sufficient to consider lumped parameter model for transmission lines NPTEL

A.K. Sinha

4/30


VS

+

IS

IR

+

Two-port network

VR

VS = AVR +BIR

volts

IS = CVR + DIR

A

NPTEL

A.K. Sinha

5/30


⎡ VS ⎤ A =⎡ ⎢ ⎥ ⎣C I ⎣S ⎦

VR ⎤ ⎡ B⎤ D⎦⎢ ⎥ I ⎣R ⎦

AD - BC = 1

NPTEL

A.K. Sinha

6/30

Indian Institute of Technology, Kharagpur Short Line Model

• Line Length < 80 kM • Generally MV / LV Lines • Capacitance can be neglected

IS +

Z = zl = (R + jωL ) l

IR +

VS

VR

NPTEL

A.K. Sinha

7/30


V S = V R + Z IR IS = IR VR ⎤ ⎡ VS ⎤ ⎡ = ⎡1 Z ⎤ ⎢ ⎥ ⎣ 0 1⎦ ⎢ ⎥ ⎣ IR ⎦ ⎣ IS ⎦

A = D = 1 per unit B=Z Ω

C=0 S NPTEL

A.K. Sinha

8/30


Phasor Diagram for Short Line VS = VRnl

VS = VRnl

VRfl

IRfl

RIRfl

IRfl jXIRfl

(a) Lagging p.f. load NPTEL

jXIRfl

VRfl

RIRfl

(b) Leading p.f. load A.K. Sinha

9/30


Medium Length Line 80 kM < Length < 250 kM

Z = zl

IS

IR +

+

VS

Y 2

Y Y' = 2 2

VR

Nominal π Circuit NPTEL

A.K. Sinha

10/30


VS = VR + Z

(

(I

VR Y

R+ 2

)

)

YZ = 1+ VR + ZIR 2

NPTEL

A.K. Sinha

11/30


VR Y VS Y IS = IR + + 2 2

(

)

VR Y Y YZ IS = IR + + ⎡ 1+ VR +ZIR ⎤ ⎣ 2 ⎦2 2

(

) (

)

= Y 1+ YZ VR + 1+ YZ IR 4 2

NPTEL

A.K. Sinha

12/30


( (

YZ ⎡ VS ⎤ ⎡ 1+ 2 =⎢ ⎢ ⎥ Y 1+ YZ I ⎣ S ⎦ ⎣⎢ 4

) Z ⎤ ⎡V ⎤ ⎥ ⎢ ⎥ YZ ) (1+ 2 ) ⎥⎦ ⎣I ⎦ R

R

YZ A = D = 1+ per unit 2 B=Z Ω YZ C = Y 1+ S 4

(

)

NPTEL

A.K. Sinha

13/30


Voltage Regulation percent VR =

VRnl - VRfl VRfl

× 100

VRnl = AVR

NPTEL

A.K. Sinha

14/30

Indian Institute of Technology, Kharagpur ABCD Matrix +

IS VS

+ VS

IR

Z

+

VR Series impedance

1 ⎡⎣ 0

Z ⎤ 1⎦

IS

IR

+

VR

Shunt ittance

1 ⎡⎣ Y NPTEL

0⎤ 1⎦ A.K. Sinha

15/30


ABCD Matrix IS +

IR

Y

VS

+ VR

T circuit

⎡ (1+YZ1) ( Z1+Z2 +YZ1Z2 ) ⎤ ⎢ ⎥ Y 1+YZ2 ) ( ⎣ ⎦ NPTEL

A.K. Sinha

16/30


ABCD Matrix

IR

IS +

VS

Y1

Y2

+

VR

Π circuit

Z ⎤ ⎡ (1+Y2Z ) ⎢ ( Y1+Y2 +Y1Y2 Z ) (1+Y1Z ) ⎥ ⎣ ⎦ NPTEL

A.K. Sinha

17/30


ABCD Matrix IS IR VS+

+

+

A 1B 1C 1D 1

A 2B2C2D2

+

VR

Cascaded networks

A1A2 +B1C2 ) ( A1B2 +B1D2 ) ⎤ ( ⎡ = ⎢⎣ C1 D1 ⎥⎢ ⎦⎣ C2 D2 ⎥⎦ ⎢⎣ ( C1A2 +D1C2 ) ( C1B2 +D1D2 ) ⎥⎦ A B A B 1 1 2 2⎤ ⎡ ⎤⎡

NPTEL

A.K. Sinha

18/30


Long Line Model I(x +Δx) + V(x + Δx)

zΔx

I(x) + V(x)

yΔx

x

(x + Δx)

NPTEL

A.K. Sinha

19/30


z = R + jωL

Ω/m

y = G + jωC S/m V ( x + Δx ) = V ( x ) + ( zΔx ) I ( x ) volts V ( x + Δx ) - V ( x ) Δx

dV ( x ) dx

= zI ( x )

= zI ( x ) NPTEL

A.K. Sinha

20/30


I ( x + Δx ) = I ( x ) + ( yΔx ) V ( x + Δx ) I ( x+Δx ) -I ( x ) Δx

dI ( x ) dx

=yV ( x+Δx ) ;lim Δx → 0

= yV ( x )

d V (x) 2

dx 2

A

=z

dI ( x ) dx

= zyV ( x )

NPTEL

A.K. Sinha

21/30


d 2 V (x) dx 2

- zyV (x) = 0

V(x) = A1e

γ = zy

γx

+ A2 e

- γx

volts

m-1

dV(x) yx -yx = γA1e - γA2 e = zI(x) dx NPTEL

A.K. Sinha

22/30


A1e

yx

-yx

- A2e I(x) = z/γ yx -yx A1e - A2 e I(x) = Zc z

z Zc =Z/γ = = Ω y zy NPTEL

A.K. Sinha

23/30


VR = V(0)

IR = I ( 0 )

VR = A1 + A2 IR

A1 - A2 = Zc

NPTEL

A.K. Sinha

24/30


VR + Z c IR

A1 =

2

A2 = V(x) =

VR - Z c IR

(

2 VR +ZcIR 2

) ( e

γx

NPTEL

+

VR -ZcIR 2 A.K. Sinha

)

e

- γx

25/30


I(x) =

(

VR +ZcIR 2Zc

(

) ( e

)

γx

-

VR -ZcIR

( ) (

2Zc

)

e

- γx

) )

γx - γx γx - γx e +e e -e V(x) = VR + Zc IR 2 2

1 I(x) = Zc

(

e γx - e- γx VR + e γx +e- γx IR 2 2 NPTEL

A.K. Sinha

26/30


V(x) = cosh( γx)VR + Zc sinh( γx)IR 1 I(x) = sinh( γx)VR + cosh( γx)IR Zc

⎡ V(x) ⎤ ⎡ A(x) = ⎢ I(x) ⎥ ⎣ C(x) ⎣ ⎦

B(x) ⎤ ⎡ VR ⎤ D(x) ⎦ ⎢ ⎥ ⎣ IR ⎦

NPTEL

A.K. Sinha

27/30


A(x) = D(x) = cosh( γx) per unit

B(x) = Zc sinh( γx) Ω

1 C(x) = sinh( γx) S Zc NPTEL

A.K. Sinha

28/30

Indian Institute of Technology, Kharagpur At the sending end X = l

⎡ VS ⎤ A =⎡ ⎢ ⎥ ⎣C ⎣ IS ⎦

VR ⎤ ⎡ B⎤ D⎦⎢ ⎥ ⎣ IR ⎦

A = D = cosh( γl) per unit

B = Zc sinh( γl) Ω 1 C= sinh( γl) S Zc NPTEL

A.K. Sinha

4/41


γ αl+jβl ) ( αl jβl αl e =e = e e = e βl γl

γl

- γl

1 αl -αl = e βl + e -βl 2

γl

- γl

1 αl -αl = e βl - e -βl 2

e +e cosh ( γl ) = 2

e -e sinh ( γl ) = 2

NPTEL

(

)

(

)

A.K. Sinha

5/41

Indian Institute of Technology, Kharagpur cosh ( αl + jβl ) = cosh ( αl ) cos(βl) + jsinh ( αl ) sin ( βl )

sinh ( αl + jβl ) = sinh ( αl ) cos(βl) + jcosh ( αl ) sin ( βl )

Y'Z' A = D = 1+ 2

B = Z' Ω

(

Y'Z' C = Y' 1+ 4

per unit

)

S

NPTEL

A.K. Sinha

6/41


IS +

VS

Z'

IR

+

Y' 2

Y' 2

VR

Z' = ZC sinh ( γl ) = ZF1 = Z

sinh ( γl ) γl

Y' tan ( γl /2 ) Y Y tanh ( γl /2 ) = = F2 = 2 Zc 2 2 ( γl /2 ) NPTEL

A.K. Sinha

7/41


Z' = Z c sinh ( γ l ) =

⎡ Z' = zl ⎢ ⎣

z sinh ( γ l ) y

sinh( γl ) ⎤ ⎤ ⎡ z sinh( γl ) zl = y zl ⎢ zyl ⎥ ⎥

= ZF1

⎣

⎦⎦

Ω NPTEL

A.K. Sinha

8/41


F1 =

sinh ( γl ) γl

per unit

Y'Z' 1+ = cosh ( γl ) 2 Y'Z' 1+ = cosh ( γl ) 2 NPTEL

A.K. Sinha

9/41


Y' cosh ( γl ) - 1 tanh ( γl/2 ) tanh ( γl/2 ) = = = 2 Zc sinh ( γl ) Zc z y

⎡ tanh( γl/2 ) ⎤ Y' yl yl ⎡ tanh( γ l/2 ) ⎤ ⎥ = ⎢ = 2 2 2 ⎢ z yl zyl/2 ⎥ ⎣ ⎦ ⎢⎣ y 2 ⎥⎦ Y = F2 2

S NPTEL

A.K. Sinha

10/41


F2 =

tanh ( γ l/2 ) γl/2

per unit

z = jωL Ω/m y = jωC S/m

z Zc = = y

jωL L = Ω jωC C NPTEL

A.K. Sinha

11/41


y = zy =

-1 jωL jωC = jω LC = jβ m ( )( )

-1

β = ω LC m

NPTEL

A.K. Sinha

12/41


LOSS LESS LINE A(x) = D(x) = cosh(γx) = cosh(jβx)

=

e

jβx

+e 2

-jβx

= cos(βx) per unit

e jβx - e-jβx sinh ( γx ) = sinh ( jβx ) = = jsin ( βx ) per unit 2

NPTEL

A.K. Sinha

13/41


L B(x) = Zc sinh ( γx ) = jZc sin ( βx ) = j sin ( βx ) Ω C

C(x) =

sinh ( γx ) Zc

=

jsin ( βx ) L C

Z' = jZc sin ( βl ) = jX' NPTEL

S

Ω

A.K. Sinha

14/41


sinh ( jβl/2 ) Y' Y tanh ( jβl/2 ) Y = = 2 2 jβl/2 2 ( jβl/2 ) cosh ( jβl/2 ) =

( )(

=

( )

jωCl 2

jωC'l 2

jsin ( βl/2 )

jβl/2 ) cos ( βl/2 )

=

( )

jωCl tan ( βl/2 ) 2 βl/2

S NPTEL

A.K. Sinha

15/41


π-Equivalent Circuit (Loss less Line) IS

VS = VS δ

Y' 2

jωCl 2

sinβl βl

NPTEL

VR = VR 0

o

= jX' Ω

tan βl /2 βl /2

IR

Y' 2

( ) ( ) ( )( )

Z' = ( jωLl ) Y' = 2

Z'

jωC't = S 2 A.K. Sinha

16/41


Wave Length (Loss less line)

2π 2π 1 λ= = = β ω LC f LC

m

= 6000 kM for 50 Hz

fλ =

1 LC NPTEL

A.K. Sinha

17/41


Surge Impedance Loading

V(x) = A(x)VR +B(x)IR

= cos ( βx ) VR + jZ c sin ( βx ) IR

I(x) = C(x)VR +D(x)IR =

jsin ( βx ) Zc

VR + cos ( βx ) IR

NPTEL

A.K. Sinha

18/41


+

IS

IR +

VS

VR

L Zc = Ω C

x =0

x=l NPTEL

A.K. Sinha

19/41


V(x) = cos ( βx ) VR + jZc sin ( βx ) IR

⎛ ⎞ = cos ( βx ) VR + jZc sin ( βx ) ⎜ Zc ⎟ ⎝ ⎠ VR

= ( cosβx + jsinβx ) VR

=e

jβx

VR

V(x) = VR

volts volts NPTEL

A.K. Sinha

20/41


VR jsin(βx) I(x) = VR + ( cosβx ) Zc Zc

VR = ( cosβx + jsinβx ) Zc

(

= e

jβx

)

VR Zc

NPTEL

A

A.K. Sinha

21/41

Indian Institute of Technology, Kharagpur *

S(x) = P(x) + jQ(x) = V(x)I (x)

e VR ⎞ ⎛ jβx = ( e VR ) ⎜ Zc ⎟ ⎝ ⎠ 2 jβx

=

*

VR

Zc NPTEL

A.K. Sinha

22/41

Indian Institute of Technology, Kharagpur Vrated (Kv) 230 345 500 765

2 SIL = Vrated /Zc

Z c = L/C

(MW)

Ω

380 285 250 257

140 420 1000 2280

VNL (x) = ( cosβx ) VRNL

VSC (x) = ( Zc sinβx ) IRSC NPTEL

A.K. Sinha

23/41


V(x) o N

d a lo

SIL

Vs

VRNL = Vs /cos(βl)

VRSIL = Vs Full l Sho oad rt ci rcui V t RFL

x =0

x=l

VRSC = 0

Receiving end

Sending end NPTEL

A.K. Sinha

24/41


Steady State Stability Limit

IR

V S - VR

Y' = VR Z' 2 jδ VS e - VR jωC'l = VR jX' 2

⎛ * SR = VRIR = VR ⎜ ⎝

VS e jδ -VR jX' NPTEL

⎞ ⎟ ⎠

*

jωC'l 2 + VR 2

A.K. Sinha

25/41


⎛ = VR ⎜ ⎝ =

VS e -jδ - VR -jX'

⎞ jω Cl 2 ⎟ + 2 VR ⎠

jVR VScosδ + VR VS sinδ - jVR2 X'

P = PS = PR = Re(SR ) =

Pmax =

VS VR X'

VR VS X'

jωCl 2 + VR 2

sinδ W

W NPTEL

A.K. Sinha

26/41

Indian Institute of Technology, Kharagpur Real power P

Pmax =

VS VR X'

δ 90° NPTEL

A.K. Sinha

27/41


⎛ =⎜ ⎝

( )

sinδ 2πl ⎞ P= ⎟ Z λ Zc sinβl sin c ⎠ 2 V V V ⎛ ⎞ sinδ S ⎞⎛ rated R ⎞ ⎛ P =⎜ Vrated ⎟ ⎜ Vrated ⎟ ⎜ Zc ⎟ 2πl ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ sin λ VS VR sinδ

VS VR

= VS.p.u. VR.p.u. (SIL) NPTEL

sinδ

( )

2πl sin λ

A.K. Sinha

( )

W

28/41


Pmax =

VS.p.u. VR.p.u. (SIL)

( )

sin 2πl λ

NPTEL

A.K. Sinha

W

29/41

Loadability in per-unit of SIL

Indian Institute of Technology, Kharagpur 4.0 SIL

3.0 SIL

Thermal limit

Th

eo

re tic

al ste a

dy

sta te s

tab ilit Pra yl ctic im al li it ne l oad abil ity

2.0 SIL

1.0 SIL

100

300

500 700 Line length (km)

NPTEL

900

A.K. Sinha

1100 30/41


Voltage (kV) SIL (MW)

Typical Thermal Rating (MW)

230

150

400

345

400

1200

500

900

2600

765

2200

5400

1100

5200

24000

NPTEL

A.K. Sinha

31/41


Maximum Power Flow (Lossy Line) IR =

VS - AVR B

=

VSe jδ - AVR e jθA Z'e jθZ

⎡ V e j( δ-θ Z ) -AV e j( θ A -θ Z ) ⎤ S R * ⎢ ⎥ SR = PR + jQR = VRIR = VR Z' ⎢ ⎥ ⎣ ⎦ NPTEL

A.K. Sinha

*

32/41


=

VR VS Z'

PR = Re(SR ) =

e

2 j( θ Z -δ )- AVR

Z'

e

j( θ Z -θ A )

AVR2 cos ( θ Z - δ ) cos ( θ Z - θ A ) Z' Z'

VR VS

NPTEL

A.K. Sinha

33/41


Example 1: A 60Hz, 138 kV, 3-phase transmission line is 200 km long. The distributed line parameters are R = 0.1 Ω/km L = 1.3 mH/km C = 0.01 µF/km G=0 The transmission line delivers 40 MW at 132kV with 95% power factor lagging. Find the sending end voltage and current, and also the transmission line efficiency. NPTEL

A.K. Sinha

34/41


Solution: For the given values of R, L and C, we have for ω = 2π. 60, z = 0.1 + j0.49 = 0.500∟78.46° Ω/km y = j3.77 X 10-6 = 3.77 X 10-6 ∟90° mho/km From the above values, Zc = √(z/y) = 364.18 ∟-5.77°Ω ‫ץ‬l = 200√(zy) = 0.2746∟84.23° = 0.0276 + j0.273

NPTEL

A.K. Sinha

35/41

Indian Institute of Technology, Kharagpur Continuing the calculations, we now find out the values of sinh ‫ץ‬l and cosh ‫ץ‬l. sinh ‫ץ‬l = (e‫ץ‬l – e-‫ץ‬l)/2 = 0.543∟84.4° cosh ‫ץ‬l = (e‫ץ‬l + e-‫ץ‬l)/2 = 0.964∟0.44°

NPTEL

A.K. Sinha

36/41

Indian Institute of Technology, Kharagpur The values of power and voltage specified in the problem refers to 3-phase line-to-line quantities. | V2 | = 132 /√3 = 76.2 kV Also, picking up the value of ∟V2 = 0°, we get V2 = 76.2 ∟0° kV

NPTEL

A.K. Sinha

37/41

Indian Institute of Technology, Kharagpur We need the value of per phase power supplied to the load. Pload = 40/3 = 13.33 MW. Given the value of power factor = 0.95, we can find I2. Pload = 0.95 |V2|.|I2| Thus, |I2| = 184.1 Also, since I2 lags V2 by cos-10.95 = 18.195°, I2 = 184.1 ∟-18.195° NPTEL

A.K. Sinha

38/41

Indian Institute of Technology, Kharagpur Finally, we have V1 = V2 cosh ‫ץ‬l + ZcI2 sinh ‫ץ‬l Putting the values, we get, V1 = 94.69 ∟18.23° kV Similarly, I1 = I2 cosh ‫ץ‬l + V2/Zc sinh ‫ץ‬l = 178.86 ∟19.42°

NPTEL

A.K. Sinha

39/41


We now calculate the efficiency of transmission. Per phase input power, Pin = Re (V1I1*) = 16.93 MW Hence, η = 13.33/16.93 = 0.79 That is, the efficiency of transmission is 79%. NPTEL

A.K. Sinha

40/41

Example 1: A 3 phase 132 kV overhead line delivers 50 MVA at 132 kV and power factor 0.8 lagging at its receiving end. The constants of the line are A = 0.98∟3° and B = 110∟75° ohms per phase. Find (a) sending in voltage and power angle (b) sending end active and reactive power (c) line losses and vars absorbed by the line (d) capacity of static compensation equipment at the receiving end to reduce the sending end voltage to 140 kV for the same load conditions

(e) The unity power load which can be supplied at the receiving end with 132 kV as the line voltage at both the ends. Solution: (a) Vr = 132000 / 30.5 = 76210 ∟0° Ir = (50 x 106) / (30.5 x 132000) = 218.7 (-cos-10.8) = 218.7 (∟-36.87°) A Vs = A . Vr + B . Ir = (0.98∟3°)(76210) + (110∟75°)(218.7∟36.87°) = 74685.8 ∟3° + 24057 ∟38.13° = 74583.4 + j 3908.86 + 18923.5 + j14853.9

= 93506.9 + j 18762.76 = 95370.62 ∟11.35° V Sending in line voltage = (30.5 x 95370.62)/1000 = 165.2 kV Power angle = 11.35°

(b) We have 3 phase power in MW given as Ss = |A||Vs|2|B|-1 ∟(β – α) - |Vr||Vs||B|-1 ∟(β + δ) = 0.98 x (165.2)2 / (110) ∟(75° - 3°) – (132 x 165.2) / (110) ∟(75° + 11.35°) = 243.14 ∟72° - 198.24∟86.35° Sending end active power Ps = 243.14 cos72° - 198.24 cos86.35° = 75.13 – 12.62 = 62.51 MW

Sending end reactive power Qs = 243.14 sin72° - 198.24 sin 86.35° = 231.24 – 197.84 = 33.40 MVar lagging (c) Line losses = Ps – Pr = 62.51 – 50 x 0.8 = 22.51 MW MVar absorbed by line = Qs – Qr = 33.40 – 50 x 0.6 = 3.4 MVar

(d)

Pr = 50 x 0.8 = 40 |Vs| = 140 |Vr| = 132

Pr = (|Vs||Vr|/|B|)cos(β – δ) – (|A||Vr|2/|B|)cos(β–α) or 40 = (140 x 132) / (110) x cos(β – δ) – 0.98 x (132)2 / (110) cos(75° - 3°) or 40 = 168 cos(β – δ) – 155.23 cos72° or cos(β – δ) = 0.524 or ∟(β – δ) = 58.4°

Qr = (|Vs||Vr|/|B|) sin(β – δ) – (|A||Vr|2/|B|)sin(β–α) = (140 x 132) / (110) sin58.4° 0.98 x (132)2 / (110) sin 72° = 143.1 – 147.6 = - 4.5 MVar Thus for Vs = 140 kV, Vr = 132 kV and Pr = 40 MW, a leading MVar of 4.5 must be drawn from the line along with the real power of 40 MW. Also, since the load requires 30 MVar lagging, the static compensation equipment must

Deliver 30 + 4.5, i.e., 34.5 MVar lagging (or must absorb 34.5 MVar leading). The capacity of static capacitors is, therefore, 34.5 MVar. (e) |Vs| = |Vr| = 132 kV, Qr = 0 Qr = (|Vs||Vr|/|B|) sin(β – δ) – (|A||Vr|2/|B|) sin(β –α) or 0 = (132 x 132) / (110) sin(β – δ) – 0.98 x (132)2 / (110) sin(75° - 3°) or ∟(β – δ) = 68.75°

Hence, Pr = (|Vs||Vr|/|B|)cos(β – δ) – (|A||Vr|2/|B|)cos(β –α) = 1322/110 cos68.75° - (0.98x1322)/110 cos72° = 57.41 – 47.97 = 9.44 MW

Indian Institute of Technology, Kharagpur POWER FLOW ON TRANSMISSION LINES

⎡ VS ⎤ A =⎡ ⎢ ⎥ ⎣C I ⎣S ⎦

VR ⎤ ⎡ B⎤ D⎦⎢ ⎥ I ⎣R ⎦

1 A Ir = Vs Vr B B D 1 A 1 Is = Vs Vr = Vs Vr B B B B NPTEL

A.K. Sinha

2/39

Indian Institute of Technology, Kharagpur SS = PS + jQS G

IS

VS

Transmission line

Generating station

Vr

Load Sr = Pr + jQr

A two bus power system

NPTEL

A.K. Sinha

3/39

Indian Institute of Technology, Kharagpur Let

Vr =

Vr

D = A = Then

Ir =

Vs B

0 , Vs = A

δ,

α, B = B

(δ - β ) -

NPTEL

Vs

A

Vr B

A.K. Sinha

β

(α - β )

4/39


Is =

A

Vs B

(α + δ - β) -

Vr B

-β

The conjugates of Ir and Is are ∗ r

I = ∗ s

I =

Vs

(β - δ) -

B

A Vs B

A

Vr B

(β - α - δ) NPTEL

(β - α)

Vr B

β

A.K. Sinha

5/39


Sr = Pr + j Q r = Vr I

0⎡ ⎣

= Vr =

Vs

Vr B

Vs B

∗ r

(β - δ) -

(β - δ) -

NPTEL

A Vr B

(β - α) ⎤ ⎦

A Vr B 2

(β - α)

A.K. Sinha

6/39


Ss = Ps + j Qs = Vs I s∗ A Vs ⎡ δ B ⎣

= Vs =

A

Vs B

(β - α - δ) -

2

(β - α) -

NPTEL

Vr

Vs B

A.K. Sinha

Vr B

β⎤ ⎦

(β + δ)

7/39


Pr =

Qr = Ps =

VS Vr

cos (β - δ) -

B VS Vr

sin (β - δ) -

B A Vs B

2

cos (β - α) NPTEL

A Vr

2

cos (β - α)

B A Vr

2

sin (β - α)

B Vr

Vs B

cos (β + δ)

A.K. Sinha

8/39


Qs =

A Vs

Pr′max = Qr = -

2

sin (β - α) -

B Vs

Vr B

A

Vr B

Vr

-

A

Vr

Vs B

sin (β + δ)

2

cos (β - α)

B

2

sin (β - α) NPTEL

A.K. Sinha

9/39

Indian Institute of Technology, Kharagpur o o ∠0 ∠θ For Short Line Æ A=D=1 ; B=Z

Pr = Qr =

Ps =

Vs

Vr Z

Vs

Vr Z

Vs Z

cos(θ - δ) sin (θ - δ) -

2

cos θ -

Vs

NPTEL

Vr Z

Vr

2

cosθ

Z Vr Z

2

sinθ

cos (θ + δ)

A.K. Sinha

10/39


Qs =

Vs Z

2

sin θ -

Vs

Vr Z

sin (θ + δ)

As R<X; |Z| ≈ X and θ ; 90°. Substituting these values in the above equations

Pr =

Vs

Vr X

sin δ

NPTEL

A.K. Sinha

11/39


Qr =

Vs

Vr X

cos δ -

Vr

2

X

As δ is normally small; cos δ ;

Qr = =

Vs

Vr X

Vr X

( Vs NPTEL

-

1

2

Vr X

- Vr

)

A.K. Sinha

12/39


1. For fixed values of Vs , Vr and X the real power depends on angle δ the phase angle by which Vs leads Vr . This angle δ is called power angle. When δ = 90° P is maximum. For system stability considerations δ has to be kept well below 90°.

NPTEL

A.K. Sinha

13/39


2. Power can be transferred over line even when | Vs |

≤ | Vr |. The phase difference

δ between Vr and Vs causes the flow of power in the line. Power systems are

operated with almost the same voltage magnitudes (i.e., 1 p u) at important busses by using methods of Voltage control.

NPTEL

A.K. Sinha

14/39


3. The maximum real power transferred over a line increases with increase in Vs and Vr. An Increase of 100% in Vr and Vs increases the power transfer by 400%. This is the reason for adopting high and extra high transmission voltages.

NPTEL

A.K. Sinha

15/39


4. The maximum real power depends on the reactance X which is directly proportional to line inductance. A decrease in inductance increases the line capacity. The line inductance can be decreased by using bundled conductors.

NPTEL

A.K. Sinha

16/39


Another method of reducing line inductance is by inserting capacitance in series with the line. This method is known as series compensation. The series capacitors are usually installed at the middle of the line.

NPTEL

A.K. Sinha

17/39

Indian Institute of Technology, Kharagpur 5. The reactive power transferred over a line is directly proportional to ( | Vs | - | Vr | ) i.e., voltage drop along the line and is independent of power angle. This means the voltage drop on the line is due to the transfer of reactive power over the line. To maintain a good voltage profile, reactive power control is necessary. NPTEL

A.K. Sinha

18/39

Indian Institute of Technology, Kharagpur VOLTAGE CONTROL Reactive Power compensation equipment has the following effects: 1. Reduction in current. 2. Maintenance of voltage profile within limits. 3. Reduction of losses in the system.

NPTEL

A.K. Sinha

19/39


4. Reduction in investment in the system per kW of load supplied. 5. Decrease in kVA loading of generators and lines. This decrease in kVA loading relieves overload condition or releases capacity for additional load growth.

NPTEL

A.K. Sinha

20/39


6. Improvement in power factor of generators. 7. Reduction in kVA demand charges for large consumers. Static Compensation

NPTEL

A.K. Sinha

21/39


Ic =

Qc =

QL =

Vs 3 Xc

3 V

kA

V

3 Xc - V XL

3

=

V

2

Xc

Mvar

2

Mvar

NPTEL

A.K. Sinha

22/39

Indian Institute of Technology, Kharagpur 400 kV

220 kV 11 kV

Static Compensation NPTEL

A.K. Sinha

23/39

Indian Institute of Technology, Kharagpur Rotating Compensators 1. A synchronous compensator can supply lagging vars up to its rating and can absorb lagging vars up to 50% of its rating. Thus a synchronous compensator of a certain rating is equvalent to a static capacitor of that rating and a shunt reactor of half its rating.

NPTEL

A.K. Sinha

24/39


2. The control of synchronous compensator is fast and continuous. 3. A synchronous compensator can be over loaded for short periods but a static capacitors do not have overload capacity.

NPTEL

A.K. Sinha

25/39


4. A drop in voltage causes the synchronous compensator to supply greater vars (because difference between generated emf and terminal voltage increase) which helps in arresting the voltage dip. In the case of static capacitor a drop in voltage causes a reduction in vars delivered to the system. NPTEL

A.K. Sinha

26/39

Indian Institute of Technology, Kharagpur 5. The inertia of synchronous compensator improves system stability and reduces the effect of sudden changes in load. 6. Static capacitor installations can be distributed in the system. However small size synchronous compensators are very uneconomical. Thus static capacitors can be located near the loads and are more effective. NPTEL

A.K. Sinha

27/39

Indian Institute of Technology, Kharagpur 7. Synchronous compensators add to short circuit currents in the system and increase the circuit ratings. 8. A failure of one unit of a static capacitor bank affects that unit only and the remaining units continue to operate. However the failure of a synchronous compensator means loss of complete unit. NPTEL

A.K. Sinha

28/39

Indian Institute of Technology, Kharagpur Voltage control of a line using tap changing transformer jx Line V1

I V2

NPTEL

A.K. Sinha

Load

R

29/39


t s V1 = tr V2 + IZ ; tr V2 + IR cos θ + I X sin θ P Q since I cos θ = and I sin θ = , we get tr V2 tr V2 RP XQ t s V1 = tr V2 + + tr V2 tr V2 Normally ts tr = 1.0 NPTEL

A.K. Sinha

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ts = or

or

1 V1

⎡ ⎣

V2 ts

+

(RP + XQ) 2

⎤ ⎦

2 t V2 2 s (RP + XQ) ts = + V1 V2

t s ⎡1 2

⎣

RP + XQ 2

NPTEL

⎤= ⎦

V2 V1

A.K. Sinha

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For complete compensation V1 = V2 and ts can be found and as tr = 1 / ts ; tr can also be found

NPTEL

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Indian Institute of Technology, Kharagpur Combined use of Tap changing Transformer and Reactive Power Injection. V1

P S

V2

T

Static var system NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur V1

I

X1

X2

V2 T12

P + jQ X3 T23 Static var system NPTEL

V3 Qs A.K. Sinha

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Indian Institute of Technology, Kharagpur REGULATING TRANSFORMER

Voltage magnitude control Voltage phase angle control

NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur a

∆ van • •

van

van + ∆van

•

b

• • •

c n

Regulating transformer

NPTEL

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Indian Institute of Technology, Kharagpur van + ∆van •

van •

∆van

• vbn

∆vbn

•

•

vcn + ∆vcn

•

•

• •

vcn

∆vcn vbn + ∆vbn

NPTEL

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Indian Institute of Technology, Kharagpur ∆van

a

van + ∆van van n c ∆vcn

NPTEL

∆vbn b

A.K. Sinha

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Indian Institute of Technology, Kharagpur POWER FLOW ON TRANSMISSION LINES

⎡ VS ⎤ A =⎡ ⎢ ⎥ ⎣C I ⎣S ⎦

VR ⎤ ⎡ B⎤ D⎦⎢ ⎥ I ⎣R ⎦

1 A Ir = Vs Vr B B D 1 A 1 Is = Vs Vr = Vs Vr B B B B NPTEL

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Indian Institute of Technology, Kharagpur SS = PS + jQS G

IS

VS

Transmission line

Generating station

Vr

Load Sr = Pr + jQr

A two bus power system

NPTEL

A.K. Sinha

3/39

Indian Institute of Technology, Kharagpur Let

Vr =

Vr

D = A = Then

Ir =

Vs B

0 , Vs = A

δ,

α, B = B

(δ - β ) -

NPTEL

Vs

A

Vr B

A.K. Sinha

β

(α - β )

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Is =

A

Vs B

(α + δ - β) -

Vr B

-β

The conjugates of Ir and Is are ∗ r

I = ∗ s

I =

Vs

(β - δ) -

B

A Vs B

A

Vr B

(β - α - δ) NPTEL

(β - α)

Vr B

β

A.K. Sinha

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Sr = Pr + j Q r = Vr I

0⎡ ⎣

= Vr =

Vs

Vr B

Vs B

∗ r

(β - δ) -

(β - δ) -

NPTEL

A Vr B

(β - α) ⎤ ⎦

A Vr B 2

(β - α)

A.K. Sinha

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Ss = Ps + j Qs = Vs I s∗ A Vs ⎡ δ B ⎣

= Vs =

A

Vs B

(β - α - δ) -

2

(β - α) -

NPTEL

Vr

Vs B

A.K. Sinha

Vr B

β⎤ ⎦

(β + δ)

7/39


Pr =

Qr = Ps =

VS Vr

cos (β - δ) -

B VS Vr

sin (β - δ) -

B A Vs B

2

cos (β - α) NPTEL

A Vr

2

cos (β - α)

B A Vr

2

sin (β - α)

B Vr

Vs B

cos (β + δ)

A.K. Sinha

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Qs =

A Vs

Pr′max = Qr = -

2

sin (β - α) -

B Vs

Vr B

A

Vr B

Vr

-

A

Vr

Vs B

sin (β + δ)

2

cos (β - α)

B

2

sin (β - α) NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur o o ∠0 ∠θ For Short Line Æ A=D=1 ; B=Z

Pr = Qr =

Ps =

Vs

Vr Z

Vs

Vr Z

Vs Z

cos(θ - δ) sin (θ - δ) -

2

cos θ -

Vs

NPTEL

Vr Z

Vr

2

cosθ

Z Vr Z

2

sinθ

cos (θ + δ)

A.K. Sinha

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Qs =

Vs Z

2

sin θ -

Vs

Vr Z

sin (θ + δ)

As R<X; |Z| ≈ X and θ ; 90°. Substituting these values in the above equations

Pr =

Vs

Vr X

sin δ

NPTEL

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Qr =

Vs

Vr X

cos δ -

Vr

2

X

As δ is normally small; cos δ ;

Qr = =

Vs

Vr X

Vr X

( Vs NPTEL

-

1

2

Vr X

- Vr

)

A.K. Sinha

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1. For fixed values of Vs , Vr and X the real power depends on angle δ the phase angle by which Vs leads Vr . This angle δ is called power angle. When δ = 90° P is maximum. For system stability considerations δ has to be kept well below 90°.

NPTEL

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2. Power can be transferred over line even when | Vs |

≤ | Vr |. The phase difference

δ between Vr and Vs causes the flow of power in the line. Power systems are

operated with almost the same voltage magnitudes (i.e., 1 p u) at important busses by using methods of Voltage control.

NPTEL

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3. The maximum real power transferred over a line increases with increase in Vs and Vr. An Increase of 100% in Vr and Vs increases the power transfer by 400%. This is the reason for adopting high and extra high transmission voltages.

NPTEL

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4. The maximum real power depends on the reactance X which is directly proportional to line inductance. A decrease in inductance increases the line capacity. The line inductance can be decreased by using bundled conductors.

NPTEL

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Another method of reducing line inductance is by inserting capacitance in series with the line. This method is known as series compensation. The series capacitors are usually installed at the middle of the line.

NPTEL

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Indian Institute of Technology, Kharagpur 5. The reactive power transferred over a line is directly proportional to ( | Vs | - | Vr | ) i.e., voltage drop along the line and is independent of power angle. This means the voltage drop on the line is due to the transfer of reactive power over the line. To maintain a good voltage profile, reactive power control is necessary. NPTEL

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Indian Institute of Technology, Kharagpur VOLTAGE CONTROL Reactive Power compensation equipment has the following effects: 1. Reduction in current. 2. Maintenance of voltage profile within limits. 3. Reduction of losses in the system.

NPTEL

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4. Reduction in investment in the system per kW of load supplied. 5. Decrease in kVA loading of generators and lines. This decrease in kVA loading relieves overload condition or releases capacity for additional load growth.

NPTEL

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6. Improvement in power factor of generators. 7. Reduction in kVA demand charges for large consumers. Static Compensation

NPTEL

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Ic =

Qc =

QL =

Vs 3 Xc

3 V

kA

V

3 Xc - V XL

3

=

V

2

Xc

Mvar

2

Mvar

NPTEL

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Indian Institute of Technology, Kharagpur 400 kV

220 kV 11 kV

Static Compensation NPTEL

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Indian Institute of Technology, Kharagpur Rotating Compensators 1. A synchronous compensator can supply lagging vars up to its rating and can absorb lagging vars up to 50% of its rating. Thus a synchronous compensator of a certain rating is equivalent to a static capacitor of that rating and a shunt reactor of half its rating.

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2. The control of synchronous compensator is fast and continuous. 3. A synchronous compensator can be over loaded for short periods but a static capacitors do not have overload capacity.

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4. A drop in voltage causes the synchronous compensator to supply greater vars (because difference between generated emf and terminal voltage increase) which helps in arresting the voltage dip. In the case of static capacitor a drop in voltage causes a reduction in vars delivered to the system. NPTEL

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Indian Institute of Technology, Kharagpur 5. The inertia of synchronous compensator improves system stability and reduces the effect of sudden changes in load. 6. Static capacitor installations can be distributed in the system. However small size synchronous compensators are very uneconomical. Thus static capacitors can be located near the loads and are more effective. NPTEL

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Indian Institute of Technology, Kharagpur 7. Synchronous compensators add to short circuit currents in the system and increase the circuit ratings. 8. A failure of one unit of a static capacitor bank affects that unit only and the remaining units continue to operate. However the failure of a synchronous compensator means loss of complete unit. NPTEL

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Indian Institute of Technology, Kharagpur Voltage control of a line using tap changing transformer jx Line V1

I V2

NPTEL

A.K. Sinha

Load

R

29/39


t s V1 = tr V2 + IZ ; tr V2 + IR cos θ + I X sin θ P Q since I cos θ = and I sin θ = , we get tr V2 tr V2 RP XQ t s V1 = tr V2 + + tr V2 tr V2 Normally ts tr = 1.0 NPTEL

A.K. Sinha

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ts = or

or

1 V1

⎡ ⎣

V2 ts

+

(RP + XQ) 2

⎤ ⎦

2 t V2 2 s (RP + XQ) ts = + V1 V2

t s ⎡1 2

⎣

RP + XQ 2

NPTEL

⎤= ⎦

V2 V1

A.K. Sinha

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For complete compensation V1 = V2 and ts can be found and as tr = 1 / ts ; tr can also be found

NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur Combined use of Tap changing Transformer and Reactive Power Injection. V1

P S

V2

T

Static var system NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur V1

I

X1

X2

V2 T12

P + jQ X3 T23 Static var system NPTEL

V3 Qs A.K. Sinha

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Indian Institute of Technology, Kharagpur REGULATING TRANSFORMER

Voltage magnitude control Voltage phase angle control

NPTEL

A.K. Sinha

35/39

Indian Institute of Technology, Kharagpur a

∆ van • •

van

van + ∆van

•

b

• • •

c n

Regulating transformer

NPTEL

A.K. Sinha

36/39

Indian Institute of Technology, Kharagpur van + ∆van •

van •

∆van

• vbn

∆vbn

•

•

vcn + ∆vcn

•

•

• •

vcn

∆vcn vbn + ∆vbn

NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur ∆van

a

van + ∆van van n c ∆vcn

NPTEL

∆vbn b

A.K. Sinha

38/39


Example 1: A 50Hz, 138 kV, 3-phase transmission line is 200 km long. The distributed line parameters are R = 0.1 Ω/km L = 1.2 mH/km C = 0.01 µF/km G=0 The transmission line delivers 40 MW at 132kV with 0.95 power factor lagging. Find the sending end voltage and current, and also the transmission line efficiency. NPTEL

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Solution: For the given values of R, L and C, we have for ω = 2π. 50, z = 0.1 + j0.377 = 0.39 75.14° Ω/km y = j3.14 X 10-6 = 3.14 X 10-6 90° mho/km From the above values, Zc = (z/y) = 352.42 -7.43°Ω ‫ץ‬l = 200 (zy) = 0.2213 82.57° = 0.0286 + j0.2194 NPTEL

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we now find out the values of sinh ‫ץ‬l and cosh ‫ץ‬l sinh ‫ץ‬l = (e‫ץ‬l – e-‫ץ‬l)/2 = 0.2195 82.67° cosh ‫ץ‬l = (e‫ץ‬l + e-‫ץ‬l)/2 = 0.975 0.37°

NPTEL

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Indian Institute of Technology, Kharagpur The values of power and voltage specified in the problem refers to 3-phase and line-to-line quantities. | V2 | = 132 / 3 = 76.2 kV Also, using V2 as reference; V2 = 76.2

V2 = 0°, we get

0° kV

NPTEL

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Indian Institute of Technology, Kharagpur Now, per phase power supplied to the load. Pload = 40/3 = 13.33 MW. Given the value of power factor = 0.95, we can find I2. Pload = 0.95 |V2|.|I2| Thus, |I2| = 184.1 Also, since I2 lags V2 by cos-10.95 = 18.195°, I2 = 184.1 -18.195°

NPTEL

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Indian Institute of Technology, Kharagpur Finally, we have V1 = V2 cosh ‫ץ‬l + ZcI2 sinh ‫ץ‬l Putting the values, we get, V1 = 82.96 8.6° kV Similarly, I1 = I2 cosh ‫ץ‬l + (V2/Zc) sinh ‫ץ‬l = 179.46 17.79

NPTEL

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We now calculate the efficiency of transmission. Per phase input power, Pin = Re (V1I1*) = 14.69 MW Hence, η = 13.33/14.69 = 0.907 That is, the efficiency of transmission is 90.7%.

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Example 2: A 3 phase 132 kV overhead line delivers 60 MVA at 132 kV and power factor 0.8 lagging at its receiving end. The constants of the line are A = 0.98 3° and B = 100 75° ohms per phase. Find (a) sending end voltage and power angle (b) sending end active and reactive power (c) line losses and vars absorbed by the line NPTEL

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(d) capacity of static compensation equipment at the receiving end to reduce the sending end voltage to 145 kV for the same load conditions (e) The unity power factor load which can be supplied at the receiving end with 132 kV as the line voltage at both the ends.

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Solution: (a)

Vr= 132000 / 30.5 = 76210

0°

Ir =(60 x 106) /{ (30.5 x 132000) (-cos-10.8)} = 262.43 ( -36.87°) A Vs = A . Vr + B . Ir = (0.98 3°)(76210) + (100 75°)(262.43 -36.87°) = 74685.8 3° + 26243 38.13° = {74.59+ j 3.909 + 20.64 + j16.20} x103

NPTEL

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Indian Institute of Technology, Kharagpur = {95.23 + j20.11}x103 = 97.33x103 11.92° V Sending end line voltage = (30.5 x97.33)kV = 168.58 kV Power angle(δ) = 11.92°

NPTEL

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Indian Institute of Technology, Kharagpur (b) We have 3 phase power in MW given as Ss = |A||Vs|2|B|-1 (β – α) - |Vr||Vs||B|-1 (β + δ) = 0.98 x (168.58)2 / (100) (75° - 3°) – (132 x 168.58) / (100) (75° + 11.92°) = 278.49

72° - 222.53

86.92°

Sending end active power Ps = 278.49 cos72° - 222.53 cos86.92° = 86.06 – 11.96 = 74.10 MW NPTEL

A.K. Sinha

23/38

Indian Institute of Technology, Kharagpur Sending end reactive power Qs = 278.49 sin72° - 222.53 sin 86.92° = 264.89 – 222.21 = 42.65 MVar lagging (c) Line losses = Ps – Pr = 74.10 – 60 x 0.8 = 26.10 MW MVar absorbed by line = Qs – Qr = 42.65 – 60 x 0.6 = 6.65 MVar NPTEL

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Indian Institute of Technology, Kharagpur (d)

Pr = 60 x 0.8 = 48 |Vs| = 145 |Vr| = 132

Pr = |Vs||Vr||B|-1 cos(β – δ) - |A||Vr|2 |B|-1 cos(β – α) or 48 = (145 x 132) / (100) x cos(β – δ) – 0.98 x (132)2 / (100) cos(75° - 3°) or 48 = 191.4cos(β – δ) –170.75cos72° or cos(β – δ) = 0.5275 or NPTEL

(β – δ) = 58.16° A.K. Sinha

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Indian Institute of Technology, Kharagpur Qr = |Vs||Vr||B|-1 sin(β – δ) - |A||Vr|2 |B|-1 sin(β – α) = (145 x 132) / (100) sin58.16° 0.98 x (132)2 / (100) sin 72° = 162.60 – 162.40 = 0.20MVar Thus for Vs = 145 kV, Vr = 132 kV and Pr = 48 MW, a lagging MVar of 0.2 will be supplied from the line along with the real power of 48 MW. Since the load requires 36 MVar lagging, the static compensation equipment must NPTEL

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Indian Institute of Technology, Kharagpur Deliver 36 –0.2, i.e., 35.8 MVar lagging (or must absorb 35.8 MVar leading). The capacity of static capacitors is, therefore, 35.8 MVar. (e) |Vs| = |Vr| = 132 kV, Qr = 0 Qr = |Vs||Vr||B|-1 sin(β – δ) - |A||Vr|2 |B|-1 sin(β –α) or 0 = (132 x 132) / (100) sin(β – δ) – 0.98 x (132)2 / (100) sin(75° - 3°) or

(β – δ) = 68.75° NPTEL

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Hence, Pr = |Vs||Vr||B|-1 cos(β – δ) - |A||Vr|2 |B|-1 cos(β – α) = {1322/100} cos68.75° - {(0.98x1322)/100}cos72° = 63.13 – 52.77 = 10.36 MW

NPTEL

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Example 3: A 220 kV line has tap changing transformers at both ends. The transformers at the sending end has a nominal ratio 11/220 kV and at the receiving end 220/11 kV. The line impedance is 10 + j40 ohms and the load at the receiving end is 100 MW at 0.9 pf lagging. Assuming that the product of off-nominal tap settings is 1, find the tap settings to give 11 kV at load bus. NPTEL

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Solution: We have, S = (100 x 106) / (3 x 0.9) = 37.03 x 106 VA per phase And, P

= (100 x 106) / 3 = 33.33 x 106 W per phase

Hence, we can calculate Q as Q = (S2 – P2)0.5 = 16.15 x 106 Var per phase NPTEL

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Indian Institute of Technology, Kharagpur Also, we have, V1 = V2 = (220 x 103) / 30.5 V = 127.02 x 103 V R = 10 Ω X = 40 Ω

NPTEL

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Indian Institute of Technology, Kharagpur The tap setting is given as 1/ts2 = [1 – (R.P + X.Q) / (V1.V2) ] = [ 1 – (10x33.33 + 40x16.15) / (127.02)2 ] Hence, ts = 1.0317 tr = 1/ts = 0.9692

NPTEL

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Example 4: A 3 phase overhead line has A = D = 0.98 3° and B = 100 84° ohm. It is operating with Vs = 240 kV and Vr = 220 kV (line to line). Find (a) maximum power which can be received at the receiving end (b) rating of synchronous phase modifier at the receiving end if load at the receiving end is 80 MW at 0.8 p.f. lagging (c) angle δ under above conditions NPTEL

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Solution: We have, Pr = |Vs|.|Vr|.|B|-1 cos(β - δ) – |A|.|Vr|2.|B|-1 cos(β – α) (a) For maximum Pr, dPr/dδ = 0 dPr/dδ = |Vs|.|Vr|.|B|-1 sin(β - δ) = 0 or sin(β - δ) = 0 or β = δ Hence, δ = 84° NPTEL

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Indian Institute of Technology, Kharagpur Hence, Pr = (240x220)/100 – 0.98(220)2/100 cos(84°- 3°) = 528 – 74.2 = 453.8 MW (b) Pr = 80 MW, Sr = 100 MVA 80 = (240 x 220) / (100) cos(β – δ) – 0.98 x (220)2 / (100) cos(84° - 3°) or 80 = 528 cos(β – δ) – 74.2 NPTEL

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or cos(ß - δ) = 0.292 and

(ß - δ) = 73.02°

Qr = |Vs|.|Vr|.|B|-1 sin(β - δ) – |A|.|Vr|2.|B|-1 sin(β - α) = (240 x 220) / (100) sin(73.02°) – 0.98 x (220)2 / (100) sin(84° - 3°) = 504.98 – 468.48 = 36.5 MVar

NPTEL

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Now, since the load requires (1002- 802)0.5 i.e. 60 MVar lagging, the difference i.e. 60 – 36.5 = 23.5 MVar must be supplied by the phase modifier. Thus its capacity is 23.5 MVar. (c)

(ß - δ) = 73.02° δ = 84° - 73.02° = 10.98°

NPTEL

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Power System Representation

NPTEL

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Lesson 2 Lesson Summary: ¾ Introduction to Single Line Diagram ¾ Symbols used for PS components ¾ Introduction to Per Unit Representation ¾ Selection of Base Quantities ¾ Advantages of PU system NPTEL

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Instructional Objective On completion of this lesson a student should be able to: A. Explain the concept of SLD and its advantages B. Build SLD for a Power System C. Explain the concept of Per Unit system of normalisation NPTEL

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Indian Institute of Technology, Kharagpur D. Select the base values for per unit system E. Compute per unit impedance values for power system components having different base values F. Explain the advantages of per unit system representation

NPTEL

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ƒ Single Line Diagram ƒ Per – Unit Normalization

NPTEL

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Single Line Diagram ƒ

Power systems are extremely complicated electrical networks

ƒ

Three phase networks – all devices are installed in all three phases and each power circuit consists of three conductors.

ƒ

A complete conventional diagram showing all the connections is very complicated and impractical. NPTEL

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Single Line Diagram ƒ Three phase systems are designed as balanced system and are generally operated as a balanced 3φ system. ƒ Single Line Diagrams (SLD) are concise way of communicating the basic arrangement of power system components. ƒ SLDs use a single line to represent all three phases. NPTEL

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Single Line Diagram ƒ SLDs are also called One Line Diagrams. ƒ They show the relative electrical interconnections of generators, transformers, transmission and distribution lines, loads, circuit breakers, etc., used in assembling the power system.

NPTEL

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Single Line Diagram There is no universally accepted set of symbols used for single line diagrams. Some of the symbols used are shown below: Generator or Motor Transformer 3-winding Transformer Auto-transformer NPTEL

A.K. Sinha

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Single Line Diagram (Symbols) Current Transformer (CT) Potential Transformer (PT) Disconnect (Isolator) Switch Circuit Breaker Fuse Reactor NPTEL

Lightning Arrestor A.K. Sinha

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Single Line Diagram (Bus-bars) Busbar (bus) – Node in electrical circuit (one bus for each phase) Buses -

aluminum or copper bars or pipes and can be several meters long

Buses in SLDs - short straight lines perpendicular to transmission lines and to lines connecting equipment to the buses. NPTEL

A.K. Sinha

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Single Line Diagram Δ

Δ Line

Generator

Bus bars

∼

Load Δ

Line

NPTEL

Δ

A.K. Sinha

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Indian Institute of Technology, Kharagpur Single Line Diagram (Sub-Station) Bus # 1

°°

°°

°° °°

°° °°

°° °°

°° °°

° °

° ° NPTEL

Bus # 2

A.K. Sinha

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Per Unit Normalization ƒ In Power Systems electrical quantities such as POWER, VOLTAGE, CURRENT, IMPEDANCE etc. are very often expressed as per unit of a base or reference value. ƒ Per Unit value is expressed by equation:

actual quantity Per - unit quantity = base value of quantity NPTEL

A.K. Sinha

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Choice of Base Quantity Two Independent Base Values are selected ƒ Vbase –Voltage Base Value ƒ Sbase – Complex Power Base Value Single Phase System

Pb1φ = Q b1φ = Sb1φ NPTEL

Ib = A.K. Sinha

S

b1φ

VbLN 16/35

Indian Institute of Technology, Kharagpur 2 bLN

VbLN V Zb = R b = Xb = = Ib S b1φ 1 Yb = Gb = Bb = Zb

Convention for per unit system: 1. Value of Sb is same for the entire system 2. Ratio of Vb on either side of a transformer is selected to be same as the ratio of transformer voltage ratings NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur I1pu

R’pu + j X’pu Ideal

V1pu

V1pu

I2pu

Trans

V2pu

V1 N1 V2 = = × V 1b N2 V 1b

U s in g V 1b /V 2 b = V ra te d 1 /V ra te d 2 = N 1 /N 2 NPTEL

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V1pu I1 p u =

N1 V2 V2 = = = V2p.u. N1 N2 ( N2 )V2b V2b I1 I1 b

N = N

2 1

I2 I1 b

U s in g I1b = S b/V 1b = S b/{ (N 1/N 2)V = (N 2/N 1)I2b NPTEL

A.K. Sinha

2b

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}


I1pu I1pu

N2 I2 I2 = = = I2pu N2 N1 ( N1 )I2b I2b R’pu + j X’pu

I2pu V2pu

V1pu

Transformer impedance referred to primary side

I1b Z1pu = Z1 V1b NPTEL

A.K. Sinha

20/35

Indian Institute of Technology, Kharagpur Transformer impedance referred to secondary side

Z2

⎛ N2 ⎞ = Z 1⎜ ⎟ ⎝ N1⎠

2

2

Z 2pu

⎛ N 2 ⎞ ⎛ I2b ⎞ = Z1⎜ ⎟ ⎜ V 2b ⎟ 1 N ⎝ ⎠ ⎝ ⎠

⎛ = Z1⎜ ⎝ ⎛ = Z1⎜ ⎝

2

N2 ⎞ ⎛ N1⎞ .I1b ⎜ /V ⎟ ⎟ N1⎠ ⎝ N2 ⎠ I1b ⎞ = Z 1pu ⎟ V 1b ⎠ NPTEL

1b

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⎛ N2 ⎞ ⎜ N1⎟ ⎝ ⎠

21/35


Change of Base

Zp.u.new

Z actual Zp.u.old Zbold = = Zbnew Zbnew 2

Z p.u.new

Sbnew Vbold = Z p.u.old ( ( ) ) Vbnew Sbold

NPTEL

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Three Phase System

S b 3 φ = Pb 3 φ = Q b 3 φ

NPTEL

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Ib =

S

b1φ

VbLN

=

S

b3φ

3 VbLL 2 bLN

2 bLL

Vb L N V V Zb = = = Ib S b1φ S b3φ 1 Rb = Xb = Zb = Yb NPTEL

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Indian Institute of Technology, Kharagpur Example: Prepare a per phase schematic of the system shown in the figure on the next slide and show all impedances in per unit on a 100 MVA, 132 KV base in the transmission-line circuit. G1: 100 MVA, 11 KV, X = 0.15 p.u. G2: 200 MVA, 13.8 KV, X = 0.20 p.u. T1: 120 MVA, 11.0/132 KV, X = 0.10 p.u. T2: 250 MVA, 13.8/132 KV, X = 0.10 p.u. Load: 250 MVA, 0.80 PF lagging, operating at 132 KV. NPTEL

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50+j100

G1 T1

G2

25+j100 25+j100

T2

Load

Example 2: System representation in P.U.

NPTEL

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Indian Institute of Technology, Kharagpur Determine the per unit impedance of the load for the following cases: i. Load modeled as a series combination of resistance and inductance. ii. Load modeled as a parallel combination of resistance and inductance.

NPTEL

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Indian Institute of Technology, Kharagpur Solution: All the data presented are in per unit on a base specified for each component. In our analysis, we will convert all these quantities to a common system base specified in the problem. Base KV in the transmission line = 132 KV Base KV in the generator circuit G1 = 132 X 11/132 = 11 KV Base KV in the generator circuit G2 = 132 X 13.8/161 = 11.31 KV NPTEL

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Indian Institute of Technology, Kharagpur We now convert all the parameter values to per unit on the common base specified. G1: X = 0.15 X 100/100 X (11/11)2 = 0.15 p.u. G2: X = 0.20 X 100/200 X (13.8/11.31)2 = 0.1489 p.u. T1: X = 0.1 X 100/120 X (11/11)2 = 0.0833 p.u. T2: X = 0.1 X 100/250 X (13.8/11.31)2 = 0.05955 p.u. The base impedance in the transmission-line circuit is (132)2/100 = 174.24 ohms NPTEL

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Indian Institute of Technology, Kharagpur Z trans-line

= (50 + j200) / 174.24 = 0.2870 + j 1.1478 p.u.

The per unit impedance of the transmission lines connecting the load bus to the high voltage buses is given as Z = (25 + j 100) / 174.24 p.u. = 0.1435 + j 0.5739 p.u. NPTEL

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Indian Institute of Technology, Kharagpur The base impedance in the load circuit is same as the base impedance in the transmission line circuit. Also, the load is specified as 250 x (0.8 + j 0.6) = 200 + j 150 We now look for the two different types of combination for the load resistance and reactance. NPTEL

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Indian Institute of Technology, Kharagpur i. Series connection: Z load*

= =

(132)2/ (200 + j 150) 55.7568 – j 41.8176

Z load

=

55.7568 + j 41.8176 ohms

Z load,pu = =

(55.7568 + j 41.8176)/174.24 0.32 + j 0.24 p.u. NPTEL

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Indian Institute of Technology, Kharagpur ii. Parallel connection: R load = (132)2/200 = 87.12 ohms = 87.12/174.24 = 0.5 p.u. X load = (132)2/ 150 = 116.16 ohms = 116.16/174.24 = 0.66 p.u. NPTEL

A.K. Sinha

33/35

Indian Institute of Technology, Kharagpur 0.287

j 0.0833

J1.1478

j0.05955

j.1489

j 0.15 .1435

j .5739

j .5739 .1435

G2

∼

∼

G1

Load 0.5

j 0.66

Example: Impedance diagram with per unit representation where load is represented as parallel combination of resistance and reactance. NPTEL

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Advantages of PU System • Manufacturers specify the impedance of equipment in percent or PU on the base of nameplate rating. • PU impedance of same type of equipment with widely different rating lie in a narrow range. • PU impedance expressed with proper base is same referred to either side of a transformer. NPTEL

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Power Flow Lesson Summary 1. Introduction 2. Power Flow Equations 3. Formulation of YBUS Matrix 4. Classification of Busbars

NPTEL

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Instructional Objective On completion of this lesson a student should be able to: A. Explain the significance of power flow problem B. Develop YBUS Matrix for any power network C. Develop Power Flow Equations for a Power System D. Classify Different Types Network busbars. NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur 100 80 60

System Load

40 20 0

12 10 8 6 4 2 0 2 4 6 8 10 12 2 NPTEL

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3/26


Power System Network PG1, QG1

PL3, QL3

1

L1

L2

PG3, QG3 3 L6

L3 L4

L5

2

5

4

PG2, QG2 PL2, QL2

PL4, QL4 NPTEL

PL5, QL5 A.K. Sinha

3/26

Indian Institute of Technology, Kharagpur The Power Flow To bus j To bus m To bus i

Vk

= Vk δk

B us k

PGk QGk PLk QLk Gen -

Load

Pk = PGK − PLK Qk = Q GK − QLK

Indian Institute of Technology, Kharagpur The Power Flow

Pkj ,Qkj To bus m

Pkm ,Qkm

Pki ,Qki

B us k

Pk = PGK − PLK

To bus j

Pk Qk

Pk = PKi + PKj + PKm

To bus i

Vk = Vk δk

Qk = Q GK − QLK

Qk = QKi + QKj + QKm


Power Flow Study ƒ Power System Operation as Quasi-static ƒ Static Analysis of Power Network ƒ Real power balance ( ∑ Pgi - ∑ PDi - PLoss)

ƒ Reactive power balance ( ∑ Q gi - ∑ QDi - QLoss) ƒ Transmission Flow Limit ƒ Bus Voltage Limits NPTEL

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Static Analysis of Power Network Mathematical Model of the Network Transmission line - nominal

π model

Bus power injections -

SK = VKIK* = PK + jQK Pk = PGK − PLK Qk = Q GK − QLK NPTEL

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Formation of Bus ittance Matrix P1 Q1

P3 Q3 L2

1 L1

2

P2 Q2

L3

3

P1 Q1

Y120

1

Y13

Y23

Y12 Y210

Y130

Y230

2

P2 Q2

Y310

P3 Q3 3

Y320

I1 =y120V1 +y12(V1 -V2)+y130V1 +y13(V1 -V3) I2 =y210V2 +y12(V2 -V)+y 1 230V2 +y23(V2 -V3) I3 =y310V3 +y13(V3 -V)+y 1 320V3 +y23(V3 -V2)

⎡I1 ⎤ ⎡(y120 +y12 +y130 +y13) −y12 −y13 ⎤ ⎡V1 ⎤ ⎢I ⎥ =⎢−y (y +y +y +y ) −y ⎥ ⎢V ⎥ ⎢ 2 ⎥ ⎢ 21 210 12 230 23 23 ⎥ ⎢ 2 ⎥ ⎢⎣I3 ⎥⎦ ⎢⎣−y31 −y32 (y310 +y13 +y320 +y23)⎥⎦ ⎢⎣V3 ⎥⎦

⎡I1 ⎤ ⎡Y11 ⎢I ⎥ =⎢Y ⎢ 2 ⎥ ⎢ 21 ⎢⎣I3 ⎥⎦ ⎢⎣Y31

Y12 Y22 Y32

Y13 ⎤ ⎡V1 ⎤ ⎥ ⎢ ⎥ Y23 ⎥ ⎢V2 ⎥ Y33 ⎥⎦ ⎢⎣V3 ⎥⎦

Y11 = y120 + y12 + y130 + y13 Y22 = y210 + y12 + y230 + y23 Y33 = y310 + y13 + y320 + y23

Y12 = Y21 = − y12 Y13 = Y31 = − y13 Y23 = Y32 = − y23 Yii is called Self-ittance (Driving Point ittance) Yij is called Transfer-ittance (Mutual ittance)

IB U S = Y B u s V B U S ; V B U S = Z B u s IB U S

Characteristics of YBUS Matrix ƒDimension of YBUS is (n x n) Æ n = Number of Bus ƒYBUS is Symmetric Matrix ƒYBUS is a Sparse Matrix (up to 90 to 95 % sparse) ƒDiagonal Elements Yii are Obtained as Algebraic Sum of All Elements Incident to Bus i ƒOff-diagonal Elements Yij = Yji are Obtained as negative of ittance Connecting Bus i and j


Power Flow Equations

IB U S = Y B u s V B U S Ik =

N

∑

Ykn V n

n=1

S k = P k + jQ Pk + jQ k

k

= V k I k* *

⎡ ⎤ = V k ⎢ ∑ Yk n V n ⎥ k = 1 , 2 , … , N ⎣ n =1 ⎦ N

Indian Institute of Technology, Kharagpur jδ k n

Vn = Vn e jθ Ykn = Ykne

k, n, = 1, 2, …,N

kn

Pk + jQ k = V k

Pk = Vk Qk = Vk

N

∑

Yk n V n e

j(δ k - δ n - θ k n )

n =1

N

∑Y

Vncos(δk - δn - θn )

∑Y

Vnsin(δk - δn - θkn )

kn

n=1 N

kn

n= 1

Characteristics of Power Flow Equations Power Flow Equations are Algebraic – Static System Power Flow Equations are Nonlinear – Iterative Solution Relate P,Q in of V,δ and YBUS Elements – P,Q Æ f (V, δ)


Power Flow Problem Characterization of Variables – Loads (PL,QL) ÆUncontrolled (Disturbance) Variable Generation (PG,QG) Æcontrol Variable Voltage (V,δ) ÆState Variable For a Given Operating Condition Æ Loads and Generations at all buses are known (Specified) Æ Find the Voltage Magnitude and Angle (V,δ) at each bus

Problem in Power Flow Solution Æ All generation Variables (PG,QG) can not be specified as Losses are not known a priori. Solution Æ Choose one bus as reference where Voltage Magnitude and angle are specified. The losses are assigned to this bus. This bus is called “Slack Bus”.


Classification of Busbars Each bus 'k' is Classified into one of the following three bus types : 1. Swing bus - There is only one swing bus, which for convenience is numbered bus 1 The swing bus is a reference bus for w hich V1 δ 1 , typically 1.0 0 o per unit, is Specified (input data). The power - flow program computes computes P1 and Q 1 .


2. Load bus - Pk and Q k are specified (input data). The pow er - flow program com putes Vk and δ k .

Indian Institute of Technology, Kharagpur 3. Voltage controlled bus - Pk and Vk are input data The power - flow program computes Qk and δk . Examples are buses which generators,switched shunt capacitors, or static var systems are connected. Maximum and minimum var limits QG k m a x and QG k m i n that this equipment can supply are also input data. Another example is a bus to which a tap changing transformer is connected; the power - flow program then computes the tap setting


Power Flow Lesson Summary 1. Power Flow Problem 2. Computational Aspects 3. Gauss – Seidel (G-S) Method 4. Example

NPTEL

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2/26


Instructional Objective On completion of this lesson a student should be able to: A. Explain the significance of power flow problem for large system B. Develop G – S iterative technique for Power Flow solution C. Solve Power Flow problem using G – S Method. NPTEL

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Power Flow Equations


N

∑

Ykn V n

n=1

S k = P k + jQ Pk + jQ k

k

= V k I k* *

⎡ ⎤ = V k ⎢ ∑ Yk n V n ⎥ k = 1 , 2 , … , N ⎣ n =1 ⎦ N

Indian Institute of Technology, Kharagpur jδ k n

Vn = Vn e jθ Ykn = Ykne

k, n, = 1, 2, …,N

kn

Pk + jQ k = V k

Pk = Vk Qk = Vk

N

∑

Yk n V n e

j(δ k - δ n - θ k n )

n =1

N

∑Y

Vncos(δk - δn - θn )

∑Y


kn

n=1 N

kn

n= 1

Characteristics of Power Flow Equations Power Flow Equations are Algebraic – Static System Power Flow Equations are Nonlinear – Iterative Solution Relate P,Q in of V,δ and YBUS Elements – P,Q Æ f (V, δ)


Power Flow Problem Characterization of Variables – Loads (PL,QL) ÆUncontrolled (Disturbance) Variable Generation (PG,QG) Æcontrol Variable Voltage (V,δ) ÆState Variable For a Given Operating Condition Æ Loads and Generations at all buses are known (Specified) Æ Find the Voltage Magnitude and Angle (V,δ) at each bus

Problem in Power Flow Solution Æ All generation Variables (PG,QG) can not be specified as Losses are not known a priori. Solution Æ Choose one bus as reference where Voltage Magnitude and angle are specified. The losses are assigned to this bus. This bus is called “Slack Bus”.


Classification of Busbars 1. Swing Bus 2. PV Bus (Voltage Control Bus) 3. PQ Bus (Load Bus) With each bus i, 4 variables (Pi, Qi, Vi and δi) are associated. Depending on the type of bus two variables are specified (known) and two unknown variables are obtained from power flow solution.


5 Bus Power System PG1, QG1

PL3, QL3

1

L1

L2

PG3, QG3 3 L6

L3 L4

L5

2

5

4

PG2, QG2 PL2, QL2

PL4, QL4 NPTEL

PL5, QL5 A.K. Sinha

3/26


Bus Data Bus Type

V Per unit

PG QG PL QL QGmax Q Gmin δ per per per per per per deg unit unit unit unit unit unit

1 2 3 4 5

1.03 1.05 -

0 -

Swing PQ PV PQ PQ

0 0 1.4 0.8 3.0 1.2 2.2 0.8 0.4 3.0 0 0 0.6 0.3 0 0 0.5 0.2 -

NPTEL

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-1.8 -

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Line Data Bus-to-bus

R’ per unit

X’ per unit

G’ per unit

B’ per unit

Maximum MVA per unit

2–4

0.0090

0.100

0

1.72

12.0

2–5

0.0045

0.050

0

0.88

12.0

4–5

0.00225

0.025

0

0.44

12.0

1–5

0.00150

0.02

0

0

6.0

3–4

0.00075

0.01

0

0

10.0

NPTEL

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Bus Type

No. of Specified No. of Buses quantites equations

1 Swing 2 PQ 3 PV

1 3 1

V, δ P, Q P, V

Total

5

10

0 2x3 = 6 1 7

No. of (V,δ) State Variables 0 2x3 = 6 1 7

Indian Institute of Technology, Kharagpur Power Flow Solution by Gauss – Seidel Method


N

∑

Ykn V n

n=1 * k k

S k = P k + jQ k = V I * N ⎡ ⎤ Pk + jQ k = V k ⎢ ∑ Yk n V n ⎥ k = 1 , 2 , … , N ⎣ n =1 ⎦


Pk - jQ k Ik = , Also * Vk N Ik = ∑ Ykn Vn ,or n=1

Ik = Yk1V1 + Yk2 V2 + − − + Ykk Vk − − − + YkN VN k-1 N ⎡ ⎤ 1 Pk - jQ k Vk = − ( ∑ Ykn Vn + ∑ Ykn Vn ) ⎥ ⎢ * Ykk ⎣ Vk n=1 n=k +1 ⎦

k = 1,2,......,N

Iterative Procedure 1. Make an initial guess |Vi|(0) and δi(0) -- Flat Start |Vi|(0) = 1.0 and δi(0) =0.0 2. Use this solution in PFE to obtain a better first solution 3. First solution is used to obtain a better second solution and so on

k-1 N ⎡ P jQ 1 i+1 i+1 i ⎤ k k Vk = − (∑ Ykn Vn + ∑ Ykn Vn )⎥ ⎢ * Ykk ⎣ Vk n=1 n=k+1 ⎦ k = 1,2,......,N;i is iteration count

Continue iteration till

| Vki+1 − Vki |≤ ε

Algorithm Steps: 1. With Pgi, Qgi , Pdi and Qdi known bus injections Pi, Qi

Calculate

2. Form YBUS Matrix (0) i

3. Set initial voltage V

,δ

(0) i

4. Iteratively solve equation

k-1 N ⎡ P jQ 1 i+1 i+1 i ⎤ k k Vk = − (∑ Ykn Vn + ∑ Ykn Vn )⎥ ⎢ * Ykk ⎣ Vk n=1 n=k+1 ⎦

to obtain new values of bus voltages

IIT Kharagpur Example: For the system shown, ZL = j0.5, V1 = 1∟0°, SG2 = j1.0 and SD2 = 0.5 + j1.0. Find V2 using Gauss - Seidel iteration technique.

IIT Kharagpur

IIT Kharagpur Solution: Firstly, we calculate the elements of the YBUS. For ZL = j0.5, we have Y11 = -j2 Y12 = j2 Y21 = j2 Y22 = -j2 We iterate on V2 using the equation given below: V2n+1 = 1/Y22 [S2*/(V2n)* - Y21.V1] …(1)

IIT Kharagpur Given, V1 = 1∟0° S2 = SG2 – SD2 = -0.5 Putting the values of V1, S2, Y22 and Y21 in equation (1), we get V2n+1 = -j [ 0.25 / (V2n)* ] +1.0 ….(2) We start with a guess, taking V20 = 1∟0° and iterate using equation (2).

IIT Kharagpur He have,

V20 = 1 + j0

Putting in equation (2), and iterating for V2, we get V21 = -j [ 0.25 / (1+j0)* ] +1.0 = 1.0 – j0.25 = 1.030776 ∟-14.036243o V22 = -j [ 0.25 / (1.0 – j0.25)* ] + 1.0 = 1.0 –j0.25/(1.0 + j0.25) = 1.0 / (1.0 + j0.25) = 0.970143 ∟-14.036249o

IIT Kharagpur Similarly, we can iterate it further. The results of the iteration, are tabulated below. Iteration number 0 1 2 3 4 5 6

V2 1 ∟0° 1.030776 ∟-14.036243° 0.970143 ∟-14.036249° 0.970261 ∟-14.931409° 0.966235 ∟-14.931416° 0.966236 ∟-14.995078° 0.965948 ∟-14.995072°

IIT Kharagpur Since, the difference in the values for the voltage doesn’t change much between the 5th and the 6th iteration, we can stop after the 6th. Hence, we can see that starting with the value V20 = 1∟0° , convergence is reached in six steps.


Power Flow Lesson Summary 1. Computational Aspectsof Gauss – Seidel (G-S) Method 2. Power Flow Solution by N-R Method 3. Direct Solution of Linear Algeraic Equations 4. Example NPTEL

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Instructional Objective On completion of this lesson a student should be able to: A. Explain the Characteristics of G-S Method B. Develop N-R iterative technique for Power Flow solution C. Solve a Set of Linear Algebraic Equations D. Solve Power Flow problem using N-R Method. NPTEL

A.K. Sinha

3/24


Algorithm Steps for G-S Method: 1. With Pgi, Qgi , Pdi and Qdi known bus injections Pi, Qi

Calculate

2. Form YBUS Matrix 3. Set initial voltage V (0) , δ(0) i i 4. Iteratively solve equation k-1 N ⎡ P jQ 1 i+1 i+1 i ⎤ k k Vk = − (∑ Ykn Vn + ∑ Ykn Vn )⎥ ⎢ * Ykk ⎣ Vk n=1 n=k+1 ⎦

to obtain new values of bus voltages NPTEL

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Computational Characteristics of G-S Method G-S Algorithm has slow convergence Convegence rate can be accelerated using acceleration factor.

X

(i+1)

= X + ΔX

X

(i+1)

= X + αΔX

(i)

(i)

(i)

(i)

α is the accelerating factor α in the range of 1.5 - 1.7 NPTEL

A.K. Sinha

5/24


Power Flow Solution Using Newton-Raphson Method Pk = Vk

N

∑Y

kn

Vncos(δk - δn - θkn )

n=1

Qk = Vk

N

∑Y

kn


n= 1

⎡ f1 ( x ) ⎤ ⎢ f (x ) ⎥ 2 ⎢ ⎥ y = f(x ) = ⎢ M ⎥ ⎢ ⎥ ⎢⎣ f N ( x ) ⎥⎦ NPTEL

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6/24

Indian Institute of Technology, Kharagpur Where

⎡ P2 ⎤ ⎡δ2 ⎤ ⎢ M⎥ ⎢ M⎥ ⎢ ⎥ ⎢ ⎥ ⎡ P ⎤ ⎢ PN ⎥ ⎡ δ ⎤ ⎢δn ⎥ y=⎢ ⎥=⎢ ⎥; x = ⎢ ⎥ = ⎢ ⎥ ⎣Q ⎦ ⎢ Q 2 ⎥ ⎣ V ⎦ ⎢ V2 ⎥ ⎢ M⎥ ⎢ M⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ Q N ⎥⎦ ⎢⎣ Vn ⎥⎦ NPTEL

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⎡ P2 (x ) ⎤ ⎢ ⎥ M ⎥ ⎢ ⎢ PN (x ) ⎥ ⎡ P (x ) ⎤ f(x ) = ⎢ = ⎢ ⎥ ⎥ ⎣ Q (x )⎦ ⎢ Q 2 (x ) ⎥ ⎢ M ⎥ ⎢ ⎥ ⎢⎣ Q N ( x ) ⎥⎦

NPTEL

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y k = Pk = Pk (x) N

= Vk ∑ Ykn Vn cos(δ k - δ n - θ kn ) n=1

y k +N = Q

k

= Q

k

(x )

N

= Vk ∑ Ykn Vn sin(δk - δn - θkn ) n=1

k = 2, 3, …, N NPTEL

A.K. Sinha

9/24

Indian Institute of Technology, Kharagpur Making an initial guess X=X0 and using Taylor series expansion for f(X) about X0

df y = f(x 0 ) + dx

x =x 0

(x - x 0 ) + L

Neglecting higher order and solving for X

⎡ df x = x0 + ⎢ ⎣ dx

NPTEL

-1

⎤ x =x 0 ⎥ [y - f(x 0 )] ⎦ A.K. Sinha

10/24


x

(i+1)

where

J

(i)

df = dx x =x (i)

=x

(i)

+J

⎡ ∂ f1 ⎢ ∂x ⎢ 1 ⎢ ∂ f2 = ⎢⎢ ∂ x 1 ⎢ M ⎢ ⎢ ∂ fN ⎢⎣ ∂ x 1 NPTEL

-1(i)

∂ f1 ∂x 2 ∂ f2 ∂x 2 M ∂ fN ∂x 2

(i)

{y - f[x ]} ∂ f1 ⎤ L ∂x N ⎥ ⎥ ∂ f2 ⎥ L ∂x N ⎥ ⎥ M M⎥ ⎥ ∂ fN ⎥ L ∂ x N ⎥⎦ x = x (i) A.K. Sinha

11/24


x

(i+1)

=x

(i)

+J

-1(i)

(i)

{y - f[x ]} -1

contains the matrix inverse J . Instead of computing J-1 one can rewrite the above equation as follows : (i)

J ΔX

(i)

=

ΔY

where ΔX and

ΔY

(i)

(i)

(i)

=

=

X

(i+1)

-

X

(i)

Y - f[x (i) ] NPTEL

A.K. Sinha

12/24


J ΔX (i)

(i)

= ΔY

(i)

is o f th e fo rm ⎡ A 11 ⎢A ⎢ 21 ⎢ M ⎢ ⎢⎣ A N 1

A 12 A 22 M A N2

A 1N ⎤ ⎡ x 1 ⎤ ⎡ y 1 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ A 2N ⎥ ⎢ x 2 ⎥ ⎢ y 2 ⎥ = M M ⎥ ⎢ M⎥ ⎢ M⎥ ⎥⎢ ⎥ ⎢ ⎥ L A N N ⎥⎦ ⎢⎣ x N ⎥⎦ ⎢⎣ y N ⎥⎦ L L

or A x = y ; a s e t o f lin e a r a lg e b r a ic e q u a tio n s NPTEL

A.K. Sinha

13/24

Indian Institute of Technology, Kharagpur Direct Solution of Linear Algebraic Equations Gauss Elimination Method

⎡ A11 ⎢ 0 ⎢ ⎢M ⎢ ⎢ 0 ⎢ 0 ⎣

A12L A22L

L L

M 0L

M AN-1,N-1

0L 0

L NPTEL

A1N ⎤ ⎡ x1 ⎤ ⎡ y1 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ A2N ⎥ x 2 y ⎢ ⎥ ⎢ 2 ⎥ ⎥ ⎢M ⎥ = ⎢M ⎥ M ⎥⎢ ⎥ ⎢ ⎥ AN-1,N ⎥ ⎢ xN-1 ⎥ ⎢ yN-1 ⎥ ANN ⎥⎦ ⎢⎣ xN ⎥⎦ ⎢⎣ yN ⎥⎦ A.K. Sinha

14/24

Indian Institute of Technology, Kharagpur Back Substitution

yN xN = A NN

x N -1 =

yk xk =

y N - 1 - A N - 1 ,N x N

N

∑A

A N - 1 ,N - 1

x

kn n

n=k+1

k = N,N- 1,…,1

Akk NPTEL

A.K. Sinha

15/24


A12 ⎡ A11 ⎢ ⎢ 0 ⎛ A - A 21 A ⎞ ⎜ 22 12 ⎟ ⎢ A11 ⎝ ⎠ ⎢ ⎢ ⎛ ⎞ A 31 A12 ⎟ ⎢ 0 ⎜ A 32 A11 ⎝ ⎠ ⎢ ⎢ M M ⎢ ⎢ ⎛ ⎞ AN1 A12 ⎟ ⎢ 0 ⎜ AN2 A11 ⎢⎣ ⎝ ⎠

⎤ ⎥ ⎛ ⎞⎥ A 21 L ⎜ A 2N A1N ⎟ ⎡ x ⎤ 1 ⎥ A ⎝ ⎠⎥ ⎢ ⎥ 11 x 2⎥ ⎢ ⎛ ⎞⎥ A 31 L ⎜ A 3N A1N ⎟ ⎥ ⎢ x 3 ⎥ A11 ⎝ ⎠ ⎥ ⎢ M⎥ ⎢ ⎥ ⎥ M M ⎥ ⎢⎣ xN ⎥⎦ ⎛ ⎞⎥ AN1 A1N ⎟ ⎥ L ⎜ ANN A11 ⎝ ⎠ ⎥⎦ L

NPTEL

A1N

A.K. Sinha

16/24


=

⎡ ⎢ ⎢ y ⎢ ⎢ ⎢ y ⎢ ⎢ ⎢ ⎢ ⎢ y ⎢⎣

2

3

N

y 1 A 21 y A 11 A A M A A NPTEL

3 1

1

y

1

y

1

1 1

N 1 1 1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

A.K. Sinha

17/24


⎡ A 1(11) ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ M ⎢ ⎢⎣ 0

(1 )

A 12

L

(1 )

L

A A

22

(1 ) 32

L

M A N( 2) 1

L

NPTEL

(1 )

A 1N ⎤ ⎥ (1 ) A 2N ⎥ (1 ) ⎥ A 3N ⎥ M ⎥ ⎥ 1 A N( N) ⎥⎦

(1 ) ⎤ ⎡ y1 ⎡ x1 ⎤ ⎢ (1 ) ⎥ ⎢x ⎥ y ⎢ 2 2 ⎥ ⎢ ⎥ ⎢ x 3 ⎥ = ⎢ y (1 ) ⎥ ⎢ 3 ⎥ ⎢ ⎥ ⎢ ⎥ M M ⎢ ⎥ ⎢ ⎥ ⎢⎣ x N ⎥⎦ ⎢⎣ y N(1 ) ⎥⎦

A.K. Sinha

18/24


(2 ) ⎡ A 11 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ M ⎢ ⎢⎣ 0

(2 )

A 11

(2 )

A 13

L

(2 )

(2 )

A 23

L

(2 )

A 33

L

(2 )

A 43 M

L

(2 )

L

A 22 0

0 M 0

A N3

NPTEL

(2 ) ⎤ ⎡ y A 1N ⎡ x ⎤ 1 1 ⎢ (2 ) ⎥ ⎥ (2 ) ⎢ A 2N ⎥ x 2 ⎥ ⎢ y 2 ⎥ ⎢ ⎥ ⎢ ⎥ (2 ) ⎥ (2 ) ⎢ x ⎥ y3 ⎥ A 3N ⎥ 3 ⎢ = ⎢ ⎥ ⎢ (2 ) ⎥ (2 ) ⎥ x 4 A 4N ⎥ ⎢ ⎥ ⎢ y 4 ⎥ ⎢ M⎥ ⎢ ⎥ M ⎢ ⎥ M⎥ ⎢ ⎥ ⎥ x 2 ( ) ⎢ (2 ) N⎥ ⎣ ⎦ A NN y ⎢⎣ N ⎥⎦ ⎥⎦ (2 ) ⎤

A.K. Sinha

19/24

Indian Institute of Technology, Kharagpur During step k, we start with A(k-1) x = y(k-1). The first k of these equations, already triangularized, are left unchanged. Also, (k-1) (k-1) /Akk equation k is multiplied by Ank and then subtracted from equation n, for n = k+1, k+2, …., N. After (N-1) steps, we arrive at the equivalent equation A(N-1) x = y(N-1), where A(N-1) is upper triangular. NPTEL

A.K. Sinha

20/24


⎡ 2 3 -1 ⎤ ⎡ x1 ⎤ ⎡5 ⎤ ⎢ -4 6 8 ⎥ ⎢ x 2 ⎥ = ⎢7 ⎥ ⎢⎣10 12 14 ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎢⎣9 ⎥⎦ 3 -1 5 ⎡2 ⎤ ⎡ x1 ⎤ ⎡ ⎤ ⎢0 6 - (-2)(3) 8 - (-2)(-1) ⎥ ⎢ x ⎥ = ⎢7 - (-2)(5)⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣0 12 - (5)(3) 14 - (5)(-1) ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎢⎣ 9 - (5)(5) ⎥⎦

NPTEL

A.K. Sinha

21/24


⎡2 3 -1 ⎤ ⎡ x1 ⎤ ⎡ 5 ⎤ ⎢0 12 6 ⎥ ⎢ x 2 ⎥ = ⎢ 17 ⎥ ⎢0 -3 19 ⎥ ⎢ x ⎥ ⎢-16 ⎥ ⎣ ⎦⎣ 3⎦ ⎣ ⎦ -1 5 ⎡2 3 ⎤ ⎡ x1 ⎤ ⎡ ⎤ ⎢0 12 ⎥ ⎢x ⎥ = ⎢ ⎥ 6 17 ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣0 0 19 - (-.25)(6) ⎥⎦ ⎢⎣ x 3 ⎥⎦ ⎢⎣-16 - (-.25)(17)⎥⎦

NPTEL

A.K. Sinha

22/24


-1 ⎤ ⎡ x1 ⎤ ⎡ 5 ⎤ ⎡2 3 ⎢0 12 6 ⎥ ⎢ x 2 ⎥ = ⎢ 17 ⎥ ⎢0 0 20.5 ⎥ ⎢ x ⎥ ⎢-11.75 ⎥ ⎣ ⎦⎣ 3⎦ ⎣ ⎦

NPTEL

A.K. Sinha

23/24


Power Flow Lesson Summary 1. Computation of Jacobian Matrix for N-R Method 2. Computational Algorithm for N-R Power Flow 3. Example

NPTEL

A.K. Sinha

2/40


Instructional Objective On completion of this lesson a student should be able to: A. Compute elements of Jacobian Matrix B. Develop N-R Power Flow Algorithm C. Solve Power Flow problem using N-R Method

NPTEL

A.K. Sinha

3/40



N

∑Y

kn


n=1

Qk = Vk

N

∑Y

kn


n= 1


A.K. Sinha

4/40



A.K. Sinha

5/40



NPTEL

A.K. Sinha

6/40


y k = Pk = Pk (x) N


y k +N = Q

k

= Q

k

(x )

N


k = 2, 3, …, N NPTEL

A.K. Sinha

7/40


df y = f(x 0 ) + dx

x =x 0

(x - x 0 ) + L

Neglecting higher order and solving for X -1

⎡ df x = x0 + ⎢ ⎣ dx

NPTEL

⎤ x =x 0 ⎥ [y - f(x 0 )] ⎦ A.K. Sinha

8/40


x

(i+1)

where

J

(i)

df = dx x = x (i)

=x

(i)

+J

⎡ ∂ f1 ⎢ ∂x ⎢ 1 ⎢ ∂ f2 ⎢ = ⎢ ∂x 1 ⎢ M ⎢ ⎢ ∂ fN ⎢ ∂x ⎣ 1 NPTEL

-1(i)

∂ f1 ∂x 2 ∂ f2 ∂x 2 M ∂ fN ∂x 2

(i)

{y - f[x ]} ∂ f1 ⎤ L ∂x N ⎥ ⎥ ∂ f2 ⎥ L ⎥ ∂x N ⎥ M M⎥ ⎥ ∂ fN ⎥ L ∂ x N ⎥⎦ x =x (i) A.K. Sinha

9/40

Indian Institute of Technology, Kharagpur (i) ⎤ ⎤ ⎡ ⎡ P−P X ⎡ ⎤ Δ P ⎣ ⎦ ⎥ (i) ⎢ Δy = ⎢ ⎥= ⎢⎣ ΔQ (i) ⎥⎦ ⎢Q − Q ⎡ X (i) ⎤ ⎥ ⎢⎣ ⎣ ⎦ ⎥⎦ ⎡ J1(i) J2 (i) ⎤ ⎡ Δ δ (i) ⎤ ⎡ Δ P (i) ⎤ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢⎣ J3 (i) J4 (i) ⎥⎦ ⎢⎣ Δ V (i) ⎥⎦ ⎢⎣ Δ Q (i) ⎥⎦ (i)

X

(i + 1)

⎡ δ(i + 1) ⎤ ⎡ δ(i) ⎤ ⎡ Δδ(i) ⎤ =⎢ ⎥=⎢ ⎥+⎢ ⎥ ⎢⎣ V (i + 1) ⎥⎦ ⎢⎣ V (i) ⎥⎦ ⎢⎣ ΔV (i) ⎥⎦ NPTEL

A.K. Sinha

10/40


J

(i)

⎡ ∂ P2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂ PN ⎢ ∂δ 2 =⎢ ⎢ ∂Q 2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂Q N ⎢⎣ ∂δ 2

J1 ∂ P2 L ∂δ N

∂ V2

M

M

L

∂ P2

∂ PN

∂δ 2

∂ V2

L

∂Q 2

∂Q 2

∂δ N

∂ V2

M

M

∂Q N

∂Q N

∂δ N

∂ V2

L J3

NPTEL

∂ P2

J2 L

L L

L J4

∂ P2 ⎤ ∂ VN ⎥⎥ M ⎥ ⎥ ∂ PN ⎥ ∂ VN ⎥ ⎥ ∂Q 2 ⎥ ∂ VN ⎥ ⎥ M ⎥ ⎥ ∂Q N ⎥ ∂ VN ⎥⎦

A.K. Sinha

X = X(i)

11/40


n≠k J1kn =

∂Pk ∂δn

J2kn = J3kn = J4kn

= Vk Ykn Vn sin ( δk - δn - θkn )

∂Pk ∂Vn ∂Q k

= Vk Ykncos ( δk - δn - θkn ) = -Vk Ykn Vncos ( δk - δn - θkn )

∂δn ∂Qk = = Vk Ykn sin ( δk - δn - θkn ) ∂Vn NPTEL

A.K. Sinha

12/40


n=k J1kk = J2kk =

J3kk = J4kk =

∂Pk ∂Vk

∂Q k ∂Vk

∂δk

N

= -Vk ∑ Ykn Vn sin ( δk - δn - θkn ) n=1 n≠k

N

= Vk Ykk cosθkk + ∑ Ykn Vncos ( δk - δn - θkn )

∂Qk ∂δk

∂Pk

n=1

N

= Vk ∑ Ykn Vn cos ( δk - δn - θkn ) n=1 n≠k

N

= -Vk Ykk sinθkk + ∑ Ykn Vn sin ( δk - δn - θkn ) n=1

NPTEL

A.K. Sinha

13/24


ing for PV Buses For each Voltage Control Bus k Vk (Voltage magnitude) is known Qk (Reactive power injection) is not specified Therefore, we omit Vk from vector X, and Qk from vector Y ∂Qk Row from Jacobian Martix ∂Vk NPTEL

A.K. Sinha

14/40

Indian Institute of Technology, Kharagpur That is the equation for reactive power injection Q at PV bus is eliminated from the N-R equation set. For a power system with n buses out of which m buses are PV buses - Number of equation to be solved for N-R Power Flow is (n-1) P equations + (n-1-m) Q equation or {2(n-1) – m} equations and the size of Jacobian Matrix will also be {2(n-1) - m} x {2(n-1) – m} NPTEL

A.K. Sinha

15/40


Algorithm for N-R Power Flow 1. Make an initial guess for state vector X(0) 2. Compute the bus power mismatches ΔY(0) 3. Compute the Jacobian Matrix J(0)

⎡Δδ ⎤ 4. Solve for voltage error vector Δ X(0) = ⎢ ⎥ Δ V ⎣ ⎦ 5. Update state vector X(1) = X(0) + Δ X(0)

(0 )

6. Increase Iteration count and Go to step 2.

NPTEL

A.K. Sinha

16/40


Example: For the 3-bus system shown find θ2, |V3|, θ3, SG1, and QG2. The transmission line is represented as nominal π equivalent network with series impedance of zl = (0.0 + j0.10) and half line charging ittance yc = j0.01.

NPTEL

A.K. Sinha

17/40

Indian Institute of Technology, Kharagpur PG2=0.6661

SG1 V1=1.0+j0

|V2|=1.04

V3 SD3=2.5+j1.0

NPTEL

A.K. Sinha

18/40


Solution: The YBUS for the system is given by j10 j10 ⎤ ⎡- j19.98 ⎥ YBUS = ⎢ j10 j19.98 j10 ⎢ ⎥ ⎢⎣ j10 j10 -j19.98⎥⎦ For this system: Bus 1 is the slack bus, bus 2 is a PV bus and bus 3 is a PQ bus. The unknown (state) variables are θ2, θ3 and |V3|. Thus the Jacobian will be a 3 X 3 matrix. NPTEL

A.K. Sinha

19/40

Indian Institute of Technology, Kharagpur Also, we have functions defined for active and reactive power injections as n

Pi(x) = ∑ | Vi | . | Vk | [ Gik cos(θi - θk) + Bik sin(θi - θk)] k=1

n

Qi(x) = ∑ | Vi | . | Vk | [ Gik sin(θi - θk) - Bik cos(θi - θk)] k=1

NPTEL

A.K. Sinha

20/40

Indian Institute of Technology, Kharagpur Real Power equations for bus 2 and 3: P2(x) = |V2|.|V1|.B21.sin(θ2-θ1) + |V2|.|V3|.B23. sin(θ2-θ3) = 10.4 sinθ2 + 10.4|V3| sin(θ2-θ3)

(1)

P3(x) = |V3|.|V1|.B31.sin(θ3-θ1) + |V3|.|V2|.B32. sin(θ3-θ2) = 10 |V3| sinθ3 + 10.4 |V3| sin(θ3-θ2) NPTEL

A.K. Sinha

(2)

21/40

Indian Institute of Technology, Kharagpur Reactive power equation for bus 3: Q3(x) = - [ |V3|.|V1|.B31cos(θ3-θ1) + |V3|.|V2|. B32cos(θ3-θ2) + |V3|2 B33] = - [ 10 |V3|.cosθ3 + 10.4 |V3|.cos(θ3-θ2) - 19.98 |V3|2 ]

(3)

We also define the unknown vector and the Jacobian for the system as ⎡θ 2 ⎤ ⎢θ 3 ⎥ X= ⎢ ⎥ ⎢⎣| V 3 |⎥⎦ NPTEL

A.K. Sinha

22/40

Indian Institute of Technology, Kharagpur and

⎡ ∂P2 ⎢ ⎢ ∂θ2 ⎢ ∂P3 J(x) = ⎢ ⎢ ∂θ2 ⎢ ∂Q3 ⎢ ⎢⎣ ∂θ2

∂P2 ∂θ3 ∂P3 ∂θ3 ∂Q3 ∂θ3

NPTEL

∂P2 ⎤ ⎥ ∂ | V3 | ⎥ ∂P3 ⎥ ⎥ ∂ | V3 | ⎥ ∂Q3 ⎥ ⎥ ∂ | V3 | ⎥⎦

A.K. Sinha

23/40

Indian Institute of Technology, Kharagpur We can find the of the Jacobian matrix by partial derivation of equations (1), (2) and (3).

dP2 = |V2||V1| B21 cos(θ2-θ1) + |V2||V3| B23 dθ2 cos(θ2-θ3)

= 10.4 cosθ2 + 10.4 |V3| cos(θ2-θ3) Similarly we have,

dP 2 = -10.4 |V3| cos(θ2-θ3) dθ 3 dP2 = 10.4 sin(θ2-θ3) d | V3 | NPTEL

A.K. Sinha

24/40

Indian Institute of Technology, Kharagpur dP 3 = -10.4 |V3| cos(θ3-θ2) dθ 2 dP3 = 10.0 |V3| cosθ3 + 10.4 |V3| cos(θ3-θ2) dθ3 dP3

= 10 sinθ + 10.4 sin(θ -θ )

3 3 2 d | V3 | dQ3 = -10.4 |V3| sin(θ3-θ2) dθ2 dQ3 = 10 |V3| sinθ3 + 10.4 |V3| sin(θ3-θ2) dθ3

dQ3 = - [10 cosθ3 + 10.4 cos(θ3-θ2) -39.96 |V3| ] d| V3 | NPTEL

A.K. Sinha

25/40

Indian Institute of Technology, Kharagpur We now start the iteration using the equation ⎡ Δ P 2 (x ) ⎤ J(x) . X = ⎢ Δ P 3 (x ) ⎥ ⎢ ⎥ ⎢⎣ Δ Q 3 (x ) ⎥⎦ We have, P2 = PG2 = 0.6661, P3 = -PD3 = -2.5, and Q3 = -QD3 = -1.0; We start with an initial guess θ20 = θ30 = 0°, |V30| = 1.0 NPTEL

A.K. Sinha

26/40

Indian Institute of Technology, Kharagpur ⎡ Δ P 2 (x ) ⎤ ⎢ Δ P 3 (x ) ⎥ ⎢ ⎥ ⎢⎣ Δ Q 3 (x ) ⎥⎦

=

0 ⎡ P 2 P (x )⎤ 2 ⎡ ⎤ ⎢ ⎢P 3 ⎥ 0 ⎥ P 3 (x ) ⎥ ⎢ ⎥ - ⎢ ⎢ Q 3 (x 0 ) ⎥ ⎢⎣ Q 3 ⎥⎦ ⎣ ⎦

The values of the P2(x0), P3(x0) and Q2(x0) can be calculated from equations (1), (2) & (3). Putting the values, we get

⎡ Δ P 2 (x ) ⎤ ⎡ 0.6661⎤ ⎡ 0 ⎤ ⎡ 0.6661⎤ ⎥ ⎢ Δ P 3 (x ) ⎥ = ⎢-2.5 ⎥ - ⎢ 0 ⎥ = ⎢-2.5 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣-1.0 ⎥⎦ ⎢⎣-0.42⎥⎦ ⎢⎣-0.58 ⎥⎦ ⎢⎣ Δ Q 3 (x ) ⎥⎦ NPTEL

A.K. Sinha

27/40

Indian Institute of Technology, Kharagpur Next, we find the jacobian matrix J0 as 0 ⎤ ⎡ 20.8 -10.4 J(0) = ⎢ ⎥ -10.4 20.4 0 ⎢ ⎥ ⎢⎣ 0 0 19.56⎥⎦ To find the changes in our unknown quantities, we need the value of the inverse of the Jacobian. 0.0329 0 ⎡0.0645 ⎤ ⎢0.0329 ⎥ (0)-1 0.0658 0 J = ⎢ ⎥ ⎢⎣0 0 0.0511⎥⎦ NPTEL

A.K. Sinha

28/40

Indian Institute of Technology, Kharagpur We have,

⎡ ΔP2(x) ⎤ ⎡Δθ2 ⎤ ⎢ ⎥ (0) ⎢ ⎥ (0)-1 . ΔP3(x) = J ΔX = Δθ3 ⎢ ⎥ ⎢ ⎥ ⎢⎣ ΔQ3(x)⎥⎦ ⎢⎣Δ| V3 |⎥⎦ o ⎡ -0.0392 2 . 2 4 6 0 ⎡ ⎤ ⎢ ⎢ -0.1426 ⎥ o 8 . 1 7 0 4 =⎢ = ⎢ ⎥ ⎢ -0 .0 2 9 6 ⎢⎣ 0.0296 ⎥⎦ ⎣

NPTEL

A.K. Sinha

⎤ ⎥ ⎥ ⎥ ⎦ 29/40

Indian Institute of Technology, Kharagpur Next, we find the value of the unknown vector for the next iteration

X (1) = X (0) + Δ X (0)

o ⎡ ⎤ 2.2460 ⎡0 ⎤ ⎢ ⎢ ⎥ o ⎥ = ⎢ 0 ⎥ + ⎢ - 8.1704 ⎥ ⎢⎣ 1 ⎥⎦ ⎢ - 0.0296 ⎥ ⎣ ⎦

⎡ - 2.2460 o ⎤ ⎢ o⎥ = ⎢ - 8.1704 ⎥ ⎢ 0.9704 ⎥ ⎣ ⎦ NPTEL

A.K. Sinha

30/40

Indian Institute of Technology, Kharagpur We proceed to the next iteration using the new values θ21 = -2.9395o, θ31 = -9.5111o, and |V31| = 0.9638. Using the equation (1), (2) and (3), we get

⎡P2 (X) ⎤ ⎢P (X) ⎥ ⎢ 3 ⎥ ⎢⎣Q 3 (X) ⎥⎦

(1)

⎡0.6341 ⎤ = ⎢⎢- 2.4208 ⎥⎥ ⎢⎣- 0.8291 ⎥⎦

NPTEL

A.K. Sinha

31/40

Indian Institute of Technology, Kharagpur Hence, we get an updated mismatch vector as

⎡ ΔP2 ⎤ ⎢ ΔP ⎥ ⎢ 3 ⎥ ⎢⎣ ΔQ 3 ⎥⎦

(1)

⎡ 0.0320 ⎤ ⎢ ⎥ = ⎢- 0.0792 ⎥ ⎢⎣- 0.1709 ⎥⎦

Next, we calculate the Jacobian matrix for the new values. NPTEL

A.K. Sinha

32/40

Indian Institute of Technology, Kharagpur We get,

⎡ 20.4303 ⎢ (1) J = ⎢-10.0382 ⎢⎣ 1.0417

-10.0382 19.6438 - 2.4218

1.0734 ⎤ ⎥ - 2.4946⎥ 18.5342 ⎥⎦

Similarly, the updated inverse matrix is

⎡0.0654 ⎢ (1)-1 J = ⎢0.0335 ⎢⎣0.0007

0.0335 0.0689 0.0071

NPTEL

0.0007 ⎤ ⎥ 0.0073 ⎥ 0.0549 ⎥⎦

A.K. Sinha

33/40


⎡ Δθ2 ⎤ ⎢ ⎥ 1 ΔX = ⎢ Δθ3 ⎥ ⎢⎣ Δ | V 3 |⎥⎦

(1)

⎡ ΔP2(x) ⎤ = J0-1 . ⎢ ΔP 3(x) ⎥ ⎢ ⎥ ⎢⎣ ΔQ3(x) ⎥⎦

o ⎡ - 0.0390 ⎤ ⎡- 0.00068 ⎤ ⎢ ⎢- 0.00564 ⎥ o ⎥ - 0.3231 ⎥ ⎢ =⎢ = ⎥ ⎢- 0.0099 ⎥ ⎢⎣- 0.00992 ⎥⎦ ⎣ ⎦

NPTEL

A.K. Sinha

34/40


Again, calculating the value of X(2)

X

(2 )

= X

=

(1 )

+ ΔX

(1 )

⎡- 2.2460 - 0.039 ⎤ ⎡- 2.2850⎤ ⎢- 8.1704 - 0.3231⎥ = ⎢- 8.4935⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0.9704 - 0.0099 ⎥⎦ ⎢⎣ 0.9605 ⎥⎦ NPTEL

A.K. Sinha

35/40

Indian Institute of Technology, Kharagpur We proceed to the next iteration using the new values θ2(2) = -2.285o, θ3(2) = -8.4935o, and |V3(2)| = 0.9605. Using the equation (1), (2) and (3), we get

⎡P2 (X) ⎤ ⎢P (X) ⎥ ⎢ 3 ⎥ ⎢⎣Q 3 (X) ⎥⎦

(2)

⎡ 0.66565 ⎤ ⎢ ⎥ = ⎢- 2.49890 ⎥ ⎢⎣- 0.99752 ⎥⎦

NPTEL

A.K. Sinha

36/40


⎡ ΔP2 ⎤ ⎢ ΔP ⎥ ⎢ 3 ⎥ ⎢⎣ ΔQ 3 ⎥⎦

(2)

⎡ 0.0004 ⎤ ⎢ ⎥ = ⎢ - 0.0011 ⎥ ⎢⎣ - 0.0025 ⎥⎦

We see that the mismatch in the values is small enough. Hence, we stop here.

NPTEL

A.K. Sinha

37/40

Indian Institute of Technology, Kharagpur For the given problem, we have, PG1 = P1 = |V1|.|V2|.B12 sin(θ1-θ2) + |V1|.|V3|.B13 sin(θ1-θ3) = 1.8333 QG1 = Q1 = - [ |V1|.|V2|.B12 cos(θ1-θ2) + |V1|.|V3|.B13 cos(θ1-θ3) + |V1|2 B11] = 0.0886 NPTEL

A.K. Sinha

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QG2 = Q2 = - [ |V2|.|V1|.B21 cos(θ2-θ1) + |V2|.|V3|.B23 cos(θ2-θ3) + |V2|2 B22] = 1.288

NPTEL

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Power Flow Lesson Summary 1. Computation of Jacobian Matrix for N-R Method 2. Computational Algorithm for N-R Power Flow 3. Example

NPTEL

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Instructional Objective On completion of this lesson a student should be able to: A. Compute elements of Jacobian Matrix B. Develop N-R Power Flow Algorithm C. Solve Power Flow problem using N-R Method

NPTEL

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N

∑Y

kn


n=1

Qk = Vk

N

∑Y

kn


n= 1


A.K. Sinha

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A.K. Sinha

5/40



NPTEL

A.K. Sinha

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y k = Pk = Pk (x) N


y k +N = Q

k

= Q

k

(x )

N


k = 2, 3, …, N NPTEL

A.K. Sinha

7/40


df y = f(x 0 ) + dx

x =x 0

(x - x 0 ) + L

Neglecting higher order and solving for X -1

⎡ df x = x0 + ⎢ ⎣ dx

NPTEL

⎤ x =x 0 ⎥ [y - f(x 0 )] ⎦ A.K. Sinha

8/40


x

(i+1)

where

J

(i)

df = dx x = x (i)

=x

(i)

+J

⎡ ∂ f1 ⎢ ∂x ⎢ 1 ⎢ ∂ f2 ⎢ = ⎢ ∂x 1 ⎢ M ⎢ ⎢ ∂ fN ⎢ ∂x ⎣ 1 NPTEL

-1(i)

∂ f1 ∂x 2 ∂ f2 ∂x 2 M ∂ fN ∂x 2

(i)

{y - f[x ]} ∂ f1 ⎤ L ∂x N ⎥ ⎥ ∂ f2 ⎥ L ⎥ ∂x N ⎥ M M⎥ ⎥ ∂ fN ⎥ L ∂ x N ⎥⎦ x =x (i) A.K. Sinha

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Indian Institute of Technology, Kharagpur (i) ⎤ ⎤ ⎡ ⎡ P−P X ⎡ ⎤ Δ P ⎣ ⎦ ⎥ (i) ⎢ Δy = ⎢ ⎥= ⎢⎣ ΔQ (i) ⎥⎦ ⎢Q − Q ⎡ X (i) ⎤ ⎥ ⎢⎣ ⎣ ⎦ ⎥⎦ ⎡ J1(i) J2 (i) ⎤ ⎡ Δ δ (i) ⎤ ⎡ Δ P (i) ⎤ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎢⎣ J3 (i) J4 (i) ⎥⎦ ⎢⎣ Δ V (i) ⎥⎦ ⎢⎣ Δ Q (i) ⎥⎦ (i)

X

(i + 1)

⎡ δ(i + 1) ⎤ ⎡ δ(i) ⎤ ⎡ Δδ(i) ⎤ =⎢ ⎥=⎢ ⎥+⎢ ⎥ ⎢⎣ V (i + 1) ⎥⎦ ⎢⎣ V (i) ⎥⎦ ⎢⎣ ΔV (i) ⎥⎦ NPTEL

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J

(i)

⎡ ∂ P2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂ PN ⎢ ∂δ 2 =⎢ ⎢ ∂Q 2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂Q N ⎢⎣ ∂δ 2

J1 ∂ P2 L ∂δ N

∂ V2

M

M

L

∂ P2

∂ PN

∂δ 2

∂ V2

L

∂Q 2

∂Q 2

∂δ N

∂ V2

M

M

∂Q N

∂Q N

∂δ N

∂ V2

L J3

NPTEL

∂ P2

J2 L

L L

L J4


A.K. Sinha

X = X(i)

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n≠k J1kn =

∂Pk ∂δn

J2kn = J3kn = J4kn


∂Pk ∂Vn ∂Q k



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n=k J1kk = J2kk =

J3kk = J4kk =

∂Pk ∂Vk

∂Q k ∂Vk

∂δk

N

= -Vk ∑ Ykn Vn sin ( δk - δn - θkn ) n=1 n≠k

N


∂Qk ∂δk

∂Pk

n=1

N


N

= -Vk Ykk sinθkk + ∑ Ykn Vn sin ( δk - δn - θkn ) n=1

NPTEL

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ing for PV Buses For each Voltage Control Bus k Vk (Voltage magnitude) is known Qk (Reactive power injection) is not specified Therefore, we omit Vk from vector X, and Qk from vector Y ∂Qk Row from Jacobian Martix ∂Vk NPTEL

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Indian Institute of Technology, Kharagpur That is the equation for reactive power injection Q at PV bus is eliminated from the N-R equation set. For a power system with n buses out of which m buses are PV buses - Number of equation to be solved for N-R Power Flow is (n-1) P equations + (n-1-m) Q equation or {2(n-1) – m} equations and the size of Jacobian Matrix will also be {2(n-1) - m} x {2(n-1) – m} NPTEL

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Algorithm for N-R Power Flow 1. Make an initial guess for state vector X(0) 2. Compute the bus power mismatches ΔY(0) 3. Compute the Jacobian Matrix J(0)

⎡Δδ ⎤ 4. Solve for voltage error vector Δ X(0) = ⎢ ⎥ Δ V ⎣ ⎦ 5. Update state vector X(1) = X(0) + Δ X(0)

(0 )

6. Increase Iteration count and Go to step 2.

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Example: For the 3-bus system shown find θ2, |V3|, θ3, SG1, and QG2. The transmission line is represented as nominal π equivalent network with series impedance of zl = (0.0 + j0.10) and half line charging ittance yc = j0.01.

NPTEL

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Indian Institute of Technology, Kharagpur PG2=0.6661

SG1 V1=1.0+j0

|V2|=1.04

V3 SD3=2.5+j1.0

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Solution: The YBUS for the system is given by j10 j10 ⎤ ⎡- j19.98 ⎥ YBUS = ⎢ j10 j19.98 j10 ⎢ ⎥ ⎢⎣ j10 j10 -j19.98⎥⎦ For this system: Bus 1 is the slack bus, bus 2 is a PV bus and bus 3 is a PQ bus. The unknown (state) variables are θ2, θ3 and |V3|. Thus the Jacobian will be a 3 X 3 matrix. NPTEL

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Indian Institute of Technology, Kharagpur Also, we have functions defined for active and reactive power injections as n

Pi(x) = ∑ | Vi | . | Vk | [ Gik cos(θi - θk) + Bik sin(θi - θk)] k=1

n

Qi(x) = ∑ | Vi | . | Vk | [ Gik sin(θi - θk) - Bik cos(θi - θk)] k=1

NPTEL

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Indian Institute of Technology, Kharagpur Real Power equations for bus 2 and 3: P2(x) = |V2|.|V1|.B21.sin(θ2-θ1) + |V2|.|V3|.B23. sin(θ2-θ3) = 10.4 sinθ2 + 10.4|V3| sin(θ2-θ3)

(1)


A.K. Sinha

(2)

21/40

Indian Institute of Technology, Kharagpur Reactive power equation for bus 3: Q3(x) = - [ |V3|.|V1|.B31cos(θ3-θ1) + |V3|.|V2|. B32cos(θ3-θ2) + |V3|2 B33] = - [ 10 |V3|.cosθ3 + 10.4 |V3|.cos(θ3-θ2) - 19.98 |V3|2 ]

(3)

We also define the unknown vector and the Jacobian for the system as ⎡θ 2 ⎤ ⎢θ 3 ⎥ X= ⎢ ⎥ ⎢⎣| V 3 |⎥⎦ NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur and

⎡ ∂P2 ⎢ ⎢ ∂θ2 ⎢ ∂P3 J(x) = ⎢ ⎢ ∂θ2 ⎢ ∂Q3 ⎢ ⎢⎣ ∂θ2

∂P2 ∂θ3 ∂P3 ∂θ3 ∂Q3 ∂θ3

NPTEL

∂P2 ⎤ ⎥ ∂ | V3 | ⎥ ∂P3 ⎥ ⎥ ∂ | V3 | ⎥ ∂Q3 ⎥ ⎥ ∂ | V3 | ⎥⎦

A.K. Sinha

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Indian Institute of Technology, Kharagpur We can find the of the Jacobian matrix by partial derivation of equations (1), (2) and (3).

dP2 = |V2||V1| B21 cos(θ2-θ1) + |V2||V3| B23 dθ2 cos(θ2-θ3)

= 10.4 cosθ2 + 10.4 |V3| cos(θ2-θ3) Similarly we have,

dP 2 = -10.4 |V3| cos(θ2-θ3) dθ 3 dP2 = 10.4 sin(θ2-θ3) d | V3 | NPTEL

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Indian Institute of Technology, Kharagpur dP 3 = -10.4 |V3| cos(θ3-θ2) dθ 2 dP3 = 10.0 |V3| cosθ3 + 10.4 |V3| cos(θ3-θ2) dθ3 dP3

= 10 sinθ + 10.4 sin(θ -θ )

3 3 2 d | V3 | dQ3 = -10.4 |V3| sin(θ3-θ2) dθ2 dQ3 = 10 |V3| sinθ3 + 10.4 |V3| sin(θ3-θ2) dθ3

dQ3 = - [10 cosθ3 + 10.4 cos(θ3-θ2) -39.96 |V3| ] d| V3 | NPTEL

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Indian Institute of Technology, Kharagpur We now start the iteration using the equation ⎡ Δ P 2 (x ) ⎤ J(x) . X = ⎢ Δ P 3 (x ) ⎥ ⎢ ⎥ ⎢⎣ Δ Q 3 (x ) ⎥⎦ We have, P2 = PG2 = 0.6661, P3 = -PD3 = -2.5, and Q3 = -QD3 = -1.0; We start with an initial guess θ20 = θ30 = 0°, |V30| = 1.0 NPTEL

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Indian Institute of Technology, Kharagpur ⎡ Δ P 2 (x ) ⎤ ⎢ Δ P 3 (x ) ⎥ ⎢ ⎥ ⎢⎣ Δ Q 3 (x ) ⎥⎦

=

0 ⎡ P 2 P (x )⎤ 2 ⎡ ⎤ ⎢ ⎢P 3 ⎥ 0 ⎥ P 3 (x ) ⎥ ⎢ ⎥ - ⎢ ⎢ Q 3 (x 0 ) ⎥ ⎢⎣ Q 3 ⎥⎦ ⎣ ⎦

The values of the P2(x0), P3(x0) and Q2(x0) can be calculated from equations (1), (2) & (3). Putting the values, we get

⎡ Δ P 2 (x ) ⎤ ⎡ 0.6661⎤ ⎡ 0 ⎤ ⎡ 0.6661⎤ ⎥ ⎢ Δ P 3 (x ) ⎥ = ⎢-2.5 ⎥ - ⎢ 0 ⎥ = ⎢-2.5 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣-1.0 ⎥⎦ ⎢⎣-0.42⎥⎦ ⎢⎣-0.58 ⎥⎦ ⎢⎣ Δ Q 3 (x ) ⎥⎦ NPTEL

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Indian Institute of Technology, Kharagpur Next, we find the jacobian matrix J0 as 0 ⎤ ⎡ 20.8 -10.4 J(0) = ⎢ ⎥ -10.4 20.4 0 ⎢ ⎥ ⎢⎣ 0 0 19.56⎥⎦ To find the changes in our unknown quantities, we need the value of the inverse of the Jacobian. 0.0329 0 ⎡0.0645 ⎤ ⎢0.0329 ⎥ (0)-1 0.0658 0 J = ⎢ ⎥ ⎢⎣0 0 0.0511⎥⎦ NPTEL

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⎡ ΔP2(x) ⎤ ⎡Δθ2 ⎤ ⎢ ⎥ (0) ⎢ ⎥ (0)-1 . ΔP3(x) = J ΔX = Δθ3 ⎢ ⎥ ⎢ ⎥ ⎢⎣ ΔQ3(x)⎥⎦ ⎢⎣Δ| V3 |⎥⎦ o ⎡ -0.0392 2 . 2 4 6 0 ⎡ ⎤ ⎢ ⎢ -0.1426 ⎥ o 8 . 1 7 0 4 =⎢ = ⎢ ⎥ ⎢ -0 .0 2 9 6 ⎢⎣ 0.0296 ⎥⎦ ⎣

NPTEL

A.K. Sinha

⎤ ⎥ ⎥ ⎥ ⎦ 29/40

Indian Institute of Technology, Kharagpur Next, we find the value of the unknown vector for the next iteration

X (1) = X (0) + Δ X (0)

o ⎡ ⎤ 2.2460 ⎡0 ⎤ ⎢ ⎢ ⎥ o ⎥ = ⎢ 0 ⎥ + ⎢ - 8.1704 ⎥ ⎢⎣ 1 ⎥⎦ ⎢ - 0.0296 ⎥ ⎣ ⎦

⎡ - 2.2460 o ⎤ ⎢ o⎥ = ⎢ - 8.1704 ⎥ ⎢ 0.9704 ⎥ ⎣ ⎦ NPTEL

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Indian Institute of Technology, Kharagpur We proceed to the next iteration using the new values θ21 = -2.9395o, θ31 = -9.5111o, and |V31| = 0.9638. Using the equation (1), (2) and (3), we get

⎡P2 (X) ⎤ ⎢P (X) ⎥ ⎢ 3 ⎥ ⎢⎣Q 3 (X) ⎥⎦

(1)

⎡0.6341 ⎤ = ⎢⎢- 2.4208 ⎥⎥ ⎢⎣- 0.8291 ⎥⎦

NPTEL

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⎡ ΔP2 ⎤ ⎢ ΔP ⎥ ⎢ 3 ⎥ ⎢⎣ ΔQ 3 ⎥⎦

(1)

⎡ 0.0320 ⎤ ⎢ ⎥ = ⎢- 0.0792 ⎥ ⎢⎣- 0.1709 ⎥⎦

Next, we calculate the Jacobian matrix for the new values. NPTEL

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Indian Institute of Technology, Kharagpur We get,

⎡ 20.4303 ⎢ (1) J = ⎢-10.0382 ⎢⎣ 1.0417

-10.0382 19.6438 - 2.4218

1.0734 ⎤ ⎥ - 2.4946⎥ 18.5342 ⎥⎦

Similarly, the updated inverse matrix is

⎡0.0654 ⎢ (1)-1 J = ⎢0.0335 ⎢⎣0.0007

0.0335 0.0689 0.0071

NPTEL

0.0007 ⎤ ⎥ 0.0073 ⎥ 0.0549 ⎥⎦

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⎡ Δθ2 ⎤ ⎢ ⎥ 1 ΔX = ⎢ Δθ3 ⎥ ⎢⎣ Δ | V 3 |⎥⎦

(1)

⎡ ΔP2(x) ⎤ = J0-1 . ⎢ ΔP 3(x) ⎥ ⎢ ⎥ ⎢⎣ ΔQ3(x) ⎥⎦

o ⎡ - 0.0390 ⎤ ⎡- 0.00068 ⎤ ⎢ ⎢- 0.00564 ⎥ o ⎥ - 0.3231 ⎥ ⎢ =⎢ = ⎥ ⎢- 0.0099 ⎥ ⎢⎣- 0.00992 ⎥⎦ ⎣ ⎦

NPTEL

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Again, calculating the value of X(2)

X

(2 )

= X

=

(1 )

+ ΔX

(1 )

⎡- 2.2460 - 0.039 ⎤ ⎡- 2.2850⎤ ⎢- 8.1704 - 0.3231⎥ = ⎢- 8.4935⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0.9704 - 0.0099 ⎥⎦ ⎢⎣ 0.9605 ⎥⎦ NPTEL

A.K. Sinha

35/40

Indian Institute of Technology, Kharagpur We proceed to the next iteration using the new values θ2(2) = -2.285o, θ3(2) = -8.4935o, and |V3(2)| = 0.9605. Using the equation (1), (2) and (3), we get

⎡P2 (X) ⎤ ⎢P (X) ⎥ ⎢ 3 ⎥ ⎢⎣Q 3 (X) ⎥⎦

(2)

⎡ 0.66565 ⎤ ⎢ ⎥ = ⎢- 2.49890 ⎥ ⎢⎣- 0.99752 ⎥⎦

NPTEL

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⎡ ΔP2 ⎤ ⎢ ΔP ⎥ ⎢ 3 ⎥ ⎢⎣ ΔQ 3 ⎥⎦

(2)

⎡ 0.0004 ⎤ ⎢ ⎥ = ⎢ - 0.0011 ⎥ ⎢⎣ - 0.0025 ⎥⎦

We see that the mismatch in the values is small enough. Hence, we stop here.

NPTEL

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Indian Institute of Technology, Kharagpur For the given problem, we have, PG1 = P1 = |V1|.|V2|.B12 sin(θ1-θ2) + |V1|.|V3|.B13 sin(θ1-θ3) = 1.8333 QG1 = Q1 = - [ |V1|.|V2|.B12 cos(θ1-θ2) + |V1|.|V3|.B13 cos(θ1-θ3) + |V1|2 B11] = 0.0886 NPTEL

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QG2 = Q2 = - [ |V2|.|V1|.B21 cos(θ2-θ1) + |V2|.|V3|.B23 cos(θ2-θ3) + |V2|2 B22] = 1.288

NPTEL

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N

∑Y

kn


n=1

Qk = Vk

N

∑Y

kn


n= 1

NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur sp cal ⎡ (i) ⎤ ⎤ ⎡ P - P ⎣X ⎦ ⎡ ΔP ⎤ ⎥ ⎢ ⎥=⎢ ⎢⎣ ΔQ (i) ⎥⎦ ⎢⎢Q sp - Q cal ⎡ X (i) ⎤ ⎥⎥ ⎣ ⎦⎦ ⎣ (i)

⎡ J1(i) ⎢ (i) ⎢⎣ J3

X

(i+1)

J2 (i) ⎤ ⎡ Δ δ (i) ⎤ ⎡ Δ P (i) ⎤ ⎥⎢ ⎥=⎢ ⎥ (i) (i) (i) J4 ⎥⎦ ⎢⎣ Δ V ⎥⎦ ⎢⎣ Δ Q ⎥⎦

⎡δ(i+1) ⎤ ⎡δ(i) ⎤ ⎡ Δδ(i) ⎤ =⎢ ⎥=⎢ ⎥+⎢ ⎥ (i+1) (i) (i) ⎢⎣ V ⎥⎦ ⎢⎣ V ⎥⎦ ⎢⎣ ΔV ⎥⎦ NPTEL

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J

(i)

⎡ ∂ P2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂ PN ⎢ ∂δ 2 =⎢ ⎢ ∂Q 2 ⎢ ∂δ ⎢ 2 ⎢ M ⎢ ⎢ ∂Q N ⎢⎣ ∂δ 2

J1 ∂ P2 L ∂δ N

∂ V2

M

M

L

∂ P2

∂ PN

∂δ 2

∂ V2

L

∂Q 2

∂Q 2

∂δ N

∂ V2

M

M

∂Q N

∂Q N

∂δ N

∂ V2

L J3

NPTEL

∂ P2

J2 L

L L

L J4


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X = X(i)

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n=k J1kk =

= -Vk ∑ YknVnsin(

∂δk

J2kk =

J3kk = J4kk =

N

∂Pk ∂Pk ∂Vk

n=1 n≠k

∂Qk ∂Vk

N


∂Qk ∂δk

)

2 δk - δn - θkn = - Qk - Vk Bkk

n=1

N


N

=-VkYkksinθkk + ∑YknVnsin( δk -δn -θkn ) = Qk - Vk2Bkk n=1

NPTEL

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n≠k J1kn =

∂Pk ∂δn

J2kn = J3kn = J4kn


∂Pk ∂Vn ∂Q k



A.K. Sinha

8/42

Indian Institute of Technology, Kharagpur Comparing J 1k n and J 4 k n, the main difference is the absence of the factor | Vn | in J 4 k n ; Similarly for J 2 k n and J 3 k n the factor | Vn | in J 3 k n this time there is also a sign difference. Because of the partial derivatives with respect to the voltage magnitude variables, all the elements of the matrices J2 and J4 are missing the factors |Vn|. Suppose we replace ∂ Pk ∂ Vn by Vn ∂ Pk ∂ Vn in J 2 Similarly with for J 4 . Since these on LHS of equation is multiplied by NPTEL

A.K. Sinha

9/42

Indian Institute of Technology, Kharagpur |Vi|’s corresponding in the update vectors should be divided by the appropriate |Vi|’s in order to maintain the specified relationship.

∂Q k Vn ∂ Vn Modified Jacobian element

Δ Vn × = Vn Modified update term NPTEL

∂Q k ∂ Vn Original Jacobian element A.K. Sinha

×

Δ Vn Original update term 10/42


m

ΔP(x ) m

ΔQ(x )

m

=

H

m

M

m

N

m

L

Δδ

m m

Δ V m V

Here H = J1, M = J3, N = J2 [ V ], L = J4 [ V ], and we define

NPTEL

A.K. Sinha

11/42


V2 M [V] @ M 0

0 L 0 M M V3 , M M M L L Vn

NPTEL

Δ V V

Δ V2 V2 @

A.K. Sinha

M Δ Vn Vn

12/42


Δδ

δ

m

m+1 i

Vi

Δ Vm and m V m i

= δ + Δδ

m+1

= Vi

m

NPTEL

m i

1+

Δ Vi Vi

m

m

A.K. Sinha

13/42


Decoupled Power Flow ƒ Change in the voltage angle δ at a bus primarily affects the flow of real power P in the transmission lines and the flow of reactive power Q remains relatively unchanged. ƒ Change in the voltage magnitude |V| at a bus primarily affects the flow of reactive power Q in the transmission lines and the flow of real power P remains relatively unchanged. NPTEL

A.K. Sinha

14/42

Indian Institute of Technology, Kharagpur m

m

J(x ) =

m

J(x ) = m

J2

m

m

J4

m

m

0

J1 J3 J1 0

J4

m

m

m

J1 Δ δ = ΔP(x ) m

m

m

J4 Δ | V | = ΔQ(x ) NPTEL

A.K. Sinha

15/42


⎡ ∂P2 ⎢ ∂δ ⎡ ΔP2 ⎤ ⎢ 2 ⎢ M⎥=⎢ M ⎢ ⎥ ⎢ ⎢⎣ ΔPN ⎥⎦ ⎢ ∂PN ⎢⎣ ∂δ2

∂P2 ⎤ L ⎥ ∂δN ⎥ ⎡ Δδ2 ⎤ ⎢ ⎥ J1 M⎥⎢ M ⎥ ⎥ ∂PN ⎥ ⎢⎣ ΔδN ⎥⎦ L ∂δN ⎥⎦

NPTEL

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⎡ ⎢ V2 ⎢ ⎢ ⎢ ⎢V ⎢ 2 ⎣

∂Q 2 ∂ V2 M ∂Q N ∂ V2

L J4 L

∂Q 2 ⎤ ⎡ ΔV2 ⎤ VN ⎥ ⎢ ⎥ ∂ VN ⎥ ⎢ V2 ⎥ ⎡ ΔQ 2 ⎤ ⎢ ⎥ ⎥ ⎢ ⎥ M M =⎢ M⎥ ⎥⎢ ⎥ ∂QN ⎥ ⎢ ΔVN ⎥ ⎢⎣ ΔQN ⎥⎦ VN ∂ VN ⎥⎦ ⎢⎣ VN ⎥⎦

NPTEL

A.K. Sinha

17/42


Fast Decoupled Power Flow In a well-designed and properly operated power transmission system: • The angular differences (δi – δj) between typical buses of the system are usually so small that

(

)

cos δi - δj = 1;

NPTEL

(

) (

sin δi - δj ≈ δi - δj

A.K. Sinha

) 18/42

Indian Institute of Technology, Kharagpur • The line susceptances Bij are many times larger than the line conductances Gij so that

(

)

(

Gij sin δi - δ j = Bijcos δi - δ j

)

• The reactive power Qi injected into any bus i of the system during normal operation is much less than the reactive power which would flow if all lines from that bus were short-circuited to reference. That is, 2 Qi = Vi Bii NPTEL

A.K. Sinha

19/42


• These approximations can be used to simplify the elements of the jacobian. In equation the off-diagonal elements of J11 and J22 are given by

∂Pi ∂δ j

= Vj

∂Q i ∂ Vj

= - Vi Vj Yij sin θij + δ j - δi

(

)

NPTEL

20/42

A.K. Sinha

Indian Institute of Technology, Kharagpur Using the identity sin(α +β) = sin α cos β + U cos α sin β in equation gives us

( (

) )

⎧Bijcos δ j - δi + ⎫ ∂Pi ∂Q i ⎪ ⎪ = Vj = - Vi Vj ⎨ ⎬ ∂δ j ∂ Vj ⎪⎩Gij sin δ j - δi ⎭⎪ Where Bij = Yij sinθij and Gij = Yij cosθij The approximations listed above then yield the off-diagonal elements NPTEL

A.K. Sinha

21/42


∂ Pi ∂δ j

= Vj

∂Q i ∂ Vj

≅ - Vi V j B ij

The diagonal elements of J1 and J4 are: ∂Qi 2 = Qi - | Vi2 |Bii J1ii = - Q i - | Vi | Bii ; J4ii =| Vi | ∂ | Vi | Applying the inequality Qi = Vi those expressions yields

∂ Pi

∂δi

≅ Vi

∂Q i

∂ Vi NPTEL

≅ − Vi

2

2

Bii to

B ii

A.K. Sinha

22/42


For a 4-bus system with bus 1 as a slack bus: ⎡ - V2 V2 B22 ⎢ ⎢ - V2 V3 B32 ⎢- V2 V4 B42 ⎣

- V2 V3 B23 - V3 V3 B33 - V3 V4 B43

- V2 V4 B24 ⎤ ⎡ Δδ2 ⎤ ⎡ ΔP2 ⎤ ⎥⎢ - V3 V4 B34 ⎥ ⎢ Δδ3 ⎥⎥ = ⎢⎢ ΔP3 ⎥⎥ - V4 V4 B44 ⎥⎦ ⎢⎣ Δδ 4 ⎥⎦ ⎢⎣ ΔP4 ⎥⎦

NPTEL

A.K. Sinha

23/42


⎡ - V2 V2 B22 ⎢ ⎢ ⎢ ⎢ ⎢ - V2 V3 B32 ⎢ ⎢ ⎢ ⎢- V V B ⎣ 2 4 42

- V2 V3 B23

- V3 V3 B33

- V3 V4 B43

- V2 V4 B24 ⎤ ⎥ ⎥ ⎥ ⎥ - V3 V4 B34 ⎥ ⎥ ⎥ ⎥ - V2 V4 B24 ⎥⎦

NPTEL

⎡ Δ V2 ⎢ ⎢ V2 ⎢Δ V 3 ⎢ ⎢ V3 ⎢ ⎢ Δ V4 ⎢ ⎣ V4

A.K. Sinha

⎤ ⎡ ΔQ ⎤ 2 ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ΔQ ⎥ 3 ⎥=⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎢ ΔQ 4 ⎥ ⎥ ⎢⎣ ⎥⎦ ⎦

24/42


-B 22 Δ V2 - B 23 Δ V3 - B 24 Δ V4 =

ΔQ 2 V2

- V2 B22 Δδ2 - V3 B23 Δδ3 - V4 B24 Δδ 4 =

NPTEL

A.K. Sinha

ΔP2 V2

25/42


⎡ -B22 ⎢ ⎢ ⎢ -B32 ⎢ ⎢ ⎢⎣ -B 42

-B23 -B33 -B 43

-B24 ⎤ ⎡ Δδ 2 ⎤ ⎥⎢ ⎥ ⎥⎢ ⎥ -B34 ⎥ ⎢ Δδ 3 ⎥ ⎥⎢ ⎥ ⎥⎢ ⎥ -B 44 ⎥⎦ ⎢⎣ Δδ 4 ⎥⎦

NPTEL

⎡ ΔP2 ⎤ ⎢ ⎥ ⎢ V2 ⎥ ⎢ ΔP ⎥ 3⎥ ⎢ = ⎢ V3 ⎥ ⎢ ⎥ ⎢ ΔP4 ⎥ ⎢ V4 ⎥ ⎣ ⎦

A.K. Sinha

26/42


⎡ -B22 ⎢ ⎢ ⎢ -B32 ⎢ ⎢ ⎢⎣-B42

-B23 -B33 -B43

⎡ ΔQ 2 ⎤ -B24 ⎤ ⎡ Δ V2 ⎤ ⎢ ⎥ ⎥ ⎢ V2 ⎥ ⎥⎢ ⎥ ⎢ ΔQ ⎥ ⎥⎢ 3⎥ ⎢ ⎥ ⎢ -B34 ⎥ Δ V3 = ⎥ ⎢ V3 ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎥⎢ ΔQ 4 ⎥ ⎢ ⎥ ⎢ ⎥ -B44 ⎦ ⎣ Δ V4 ⎦ ⎢ V4 ⎥ ⎣ ⎦ NPTEL

A.K. Sinha

27/42

Indian Institute of Technology, Kharagpur Fast Decoupled Power Flow solution strategy: 1. Calculate the initial mismatches ΔP/ V , 2. Solve equation for Δδ , 3. Update the angles δ and use them to calculate mismatches ΔQ / V , 4. Solve equation for Δ V and update the magnitudes V , and 5. Return to step 1 and repeat the iteration until all mismatches are within specified tolerances. Max | ΔP |≤ ε and Max | ΔQ |≤ ε NPTEL

A.K. Sinha

28/42


Example : Consider the circuit shown on the next slide. The transmission line is represented as nominal π equivalent network with series impedance of zl = (0.0 + j0.10) and half line charging ittance yc = j0.01. Use fast decoupled power flow to find the values of θ2, θ3, and |V3|.

NPTEL

A.K. Sinha

29/42


NPTEL

A.K. Sinha

30/42


Solution: We have PG2 |V2| Bij Bii P2 P3 Q3

= = = = = = =

0.6661 1.04 10 , i ≠ j -19.98 0.6661 -2.5 -1.0

NPTEL

A.K. Sinha

31/42

Indian Institute of Technology, Kharagpur We first find B.

⎡B22 B=⎢ ⎣B32

B23 ⎤ ⎡- 19.98 10 ⎤ =⎢ ⎥ ⎥ B33 ⎦ ⎣ 10 - 19.98⎦

For the fast decoupled power flow, we have n

⎡Δθ2 ⎤ ⎡ΔP2 / | V2 |⎤ - B. ⎢ =⎢ ⎥ ⎥ ⎣Δθ3 ⎦ ⎣ΔP3 / | V3 |⎦ NPTEL

A.K. Sinha

n

32/42

Indian Institute of Technology, Kharagpur And,

- B. [ Δ | V3 |] = [ ΔQ 3 / | V3 |] n

n

We can take the inverse of the above equations to get the explicit iteration formulas n

ΔP2 / | V2 |⎤ ⎡ Δθ2 ⎤ -1 ⎡ ⎢ Δθ ⎥ = - B . ⎢ ΔP / | V |⎥ 3 ⎦ ⎣ 3⎦ ⎣ 3 NPTEL

A.K. Sinha

n

33/42

Indian Institute of Technology, Kharagpur Or n

n

⎡ Δθ2 ⎤ ⎡0.0668 0.0334 ⎤ ⎡ ΔP2 /1.04 ⎤ ⎢ Δθ ⎥ = ⎢0.0334 0.0668⎥ . ⎢ ΔP / | V |⎥ …..(1) ⎣ ⎦ ⎣ 3 3 ⎦ ⎣ 3⎦ Similarly

[ Δ | V3 |]

n

= - B . [ ΔQ 3 / | V3 |] -1

n

= 0.0501[ ΔQ 3 / | V3 |] NPTEL

A.K. Sinha

n

…..(2) 34/42

Indian Institute of Technology, Kharagpur We write the equations for the given system. P2(x) = |V2|.|V1|.B21.sin(θ2-θ1) + |V2|.|V3|.B23. sin(θ2-θ3) = 10.4 sinθ2 + 10.4|V3| sin(θ2-θ3)

(3)


A.K. Sinha

(4) 35/42


Q3(x) = - [ |V3|.|V1|.B31cos(θ3-θ1) + |V3|.|V2|. B32cos(θ3-θ2) + |V3|2 B33] = - [ 10 |V3|.cosθ3 + 10.4 |V3|.cos(θ3-θ2) - 19.98 |V3|2 ] (5)

NPTEL

A.K. Sinha

36/42

Indian Institute of Technology, Kharagpur Assuming a flat start, we have θ20 = θ30 = 0° |V30| = 1.0 The values of the P2(x0), P3(x0) and Q2(x0) can be calculated from equations (3), (4) & (5). Substituting values we get P20 = P30 = 0 Q30 = -0.42 NPTEL

A.K. Sinha

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From these values, we can find the mismatch matrix as

⎡ 0.6661⎤ ⎡ 0 ⎤ ⎡ 0.6661⎤ ⎡ Δ P 2 (x ) ⎤ ⎢- 2.5 ⎥ - ⎢ 0 ⎥ = ⎢- 2.5 ⎥ ⎢ Δ P 3 (x ) ⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎢⎣- 1.0 ⎥⎦ ⎢⎣- 0.42⎥⎦ ⎢⎣- 0.58 ⎥⎦ ⎢⎣ Δ Q 3 (x ) ⎥⎦

NPTEL

A.K. Sinha

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Also

⎡ ΔP2 ⎢ ⎢| V2 | ⎢ Δ P3 ⎢ ⎢| V3 | ⎢ ΔQ 3 ⎢ ⎢⎣ | V 3 |

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦

=

NPTEL

⎡0.6405 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢- 2.500 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣- 0.580 ⎥⎦ A.K. Sinha

39/42


Using the value of the mismatch matrix, we can calculate new values of θ2, θ3, and |V3| and continue the iteration process till we get sufficient accuracy. The results are shown next

NPTEL

A.K. Sinha

40/42


Itn. θ2 No.

θ3

|V3|

∆P2/|V2| ∆P3/|V3| ∆Q3/|V3|

0

0

0

1

0.6405

1

-2.3319 -8.3423 0.9709 0.0308

-0.0351 -0.1857

2

-2.2815 -8.4179 0.9616 0.0107

-0.0242 -0.0201

3

-2.2869 -8.4900 0.9606 0.0015

-0.0024 -0.0043

NPTEL

-2.5

A.K. Sinha

-0.58

41/42


Power Flow Solution For Large Power Systems G-S Algorithm is the simplest and requires minimum time per iteration. Number of iterations for G-S algorithm increases with system size. G-S algorithm is not suitable for large system power flow solution. NPTEL

A.K. Sinha

4/37

Indian Institute of Technology, Kharagpur N-R algorithm requires large storage for Jacobian Matrix. Computational requirement per iteration is high (Jacobian elements have to be computed every iteration). Number of iteration for N-R algorithm is independent of system size (4-5 iterations). N-R algorithm is complex to program. Reliable and preferred method for large systems. NPTEL

A.K. Sinha

5/37

Indian Institute of Technology, Kharagpur FDLF algorithm requires less storage than the N-R method but more than the G-S method. Relatively simpler to program than N-R method FDLF requires much less computation as compared to N-R method (B’ and B’’ matrices are constant). Number of iterations for FDLF does not depend on system size, but is larger (6-10 iterations) as compared to N-R method. NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur FDLF method is the most popular power flow method for large systems. Both N-R and FDLF algorithms require the solution of matrix equation of type Ax=b of large dimension at each iteration.

NPTEL

A.K. Sinha

7/37


Sparse Matrix Techniques In a large power system on an average fewer than 3 lines are connected to each bus. On an average less than 4 non-zero elements in each row of Ybus matrix. For a 1000 bus system it means less than 4000 non-zero elements in Ybus Matrix instead of 1000000 elements (996000 zeros) for a 1000 x 1000 matrix. NPTEL

A.K. Sinha

8/37

Indian Institute of Technology, Kharagpur Matrix with very few non-zero (large number of zeros) elements are called Sparse Matrix. Instead of storing them as 2-dimensional arrays they are stored as a linked list of elements to save computer storage requirement. Similar is the case for Jacobian matrix, whereever Yij =0, Hij =Mij=Nij=Lij=0. Compact storage as a linked list for a 1000 bus system with 100 PV buses (considering average of 3 lines connected to each bus) will require NPTEL

A.K. Sinha

9/37


Storage of less than 16000 elements as compared to about 4000000 elements. For FDLF algorithm the storage requirement for B’ and B” matrix will be less than 8000 elements.

NPTEL

A.K. Sinha

10/37


Linked List Storage Let us take a 4 x 4 matrix A as

⎡ 1.0 ⎢ 8.0 A=⎢ ⎢10.0 ⎢ ⎣11.0

5.0 2.0 0

6.0 0 3.0

0

0

NPTEL

7.0 ⎤ 9.0 ⎥⎥ . 0 ⎥ ⎥ 4.0 ⎦

A.K. Sinha

11/37


CE = [1.0 5.0 6.0 7.0 8.0 2.0 9.0 10.0 3.0 11.0 4.0] NXT = [ 2 ICL = [ 1

3 2

4 3

0 4

6 1

7 2

0 3

9 1

0 3

11

0]

1

4]

IRW = [ 1 5 8 10 ]

NPTEL

A.K. Sinha

12/37


Ordering Schemes Let us take the same matrix A

⎡ 1.0 ⎢ 8.0 A=⎢ ⎢10.0 ⎢ ⎣11.0

5.0 2.0

6.0 0

0

3.0

0

0

NPTEL

7.0 ⎤ ⎥ 9.0 ⎥ . 0 ⎥ ⎥ 4.0 ⎦

A.K. Sinha

13/37

Indian Institute of Technology, Kharagpur After 1st step of Gauss elimination matrix becomes:

⎡1 ⎢0 A=⎢ ⎢0 ⎢ ⎣0

5 - 38 - 50

6 - 48 - 57

- 55

- 66

7 ⎤ - 47 ⎥⎥ - 70 ⎥ ⎥ - 73 ⎦

Sparsity of A is lost completely in this case NPTEL

A.K. Sinha

14/37

Indian Institute of Technology, Kharagpur To minimize the number of fill-ins during the reduction steps, the matrix should be reordered (numbered) according to the number of non-zero elements (row’s valence). Starting with the minimum valence node. For the given example the numbering should be as follows: 4, 3, 2, 1. With this numbering the matrix becomes:

NPTEL

A.K. Sinha

15/37


⎡4 ⎢0 A(ordered) = ⎢ ⎢9 ⎢ ⎣7 NPTEL

0 3 0 6

0 0 2 5

11⎤ ⎥ 10 ⎥ 8⎥ ⎥ 1⎦

A.K. Sinha

16/37

Indian Institute of Technology, Kharagpur After 1st step the eliminated matrix becomes:

A

(1)

⎡4 ⎢0 =⎢ ⎢0 ⎢ ⎣0

0 3 0

0 0 2

6

5

⎤ ⎥ ⎥ ⎥ ⎥ - 18.25 ⎦

11 10 - 16.75

No Fill-ins in this case. NPTEL

A.K. Sinha

17/37


The ordering scheme used is called static ordering or pre-ordering as axes are ordered as per their valence before factorization. A scheme in which axes are ordered at each step in factorization as per their valence is called dynamic ordering as is more efficient than the static ordering scheme.

NPTEL

A.K. Sinha

18/37


Network solution by LU Factorization b = Ax = (LU)x :

b =Lx :

x = Ux

Triangular Factorization Method Efficient method to calculate the L and U factors of a given matrix A . NPTEL

A.K. Sinha

19/37

Indian Institute of Technology, Kharagpur Triangular Factorization Algorithm For k = 1, 2,…..(n – 1) if Akk = 0 then quit else

A ik =

Aij = Aij -

A ik A kk

Aik Akj Akk

NPTEL

i = k + 1,.....n

i, j = k + 1,.....n A.K. Sinha

20/37


A11

A12

A13

A 21 A 31 A 41

A 22 A 32 A 42

A 23 A 33 A 43

A14 x1 b1 A 24 x 2 b2 = A 34 x 3 b3 A 44 x 4 b4

NPTEL

A.K. Sinha

21/37


A A

(1) ij

(1) i1

A11 (1) A 21

A12 (1) A 22

A13 (1) A 23

A14 (1) A 24

A (1) 31 A (1) 41

A (1) 32 A (1) 42

A (1) 33 A (1) 43

(1) A 34 A (1) 44

=

= Aij -

A i1 A 11

for i = 2, 3, 4

Ai1A1j A11 NPTEL

for i, j = 2, 3, 4 A.K. Sinha

22/37


A11 A (1) 21

A12 A (1) 22

A13 A (1) 23

A14 (1) A 24

A (1) 31 (1) A 41

A (2) 32 (2) A 42

A (2) 33 (2) A 43

(2) A 34 (2) A 44

Ai2(2) =

A

(2) ij

=A

(1) ij

Ai2(1) A

-

for i = 3, 4

(1) 22

(1) i2

A A A

(1) 22

NPTEL

(1) 2j

for i, j = 3, 4 A.K. Sinha

23/37


A11 (1) A 21

A12 (1) A 22

A13 (1) A 23

A14 (1) A 24

A (1) 31 A (1) 41

A (2) 32 A (2) 42

A (2) 33 A (3) 43

(2) A 34 A (3) 44

Ai3(3) = Aij(3) = Aij(2) -

Ai3(2)

for i = 4

A (2) 33 (2) (2) Ai3 A 3j A (2) 33 NPTEL

for i, j = 4 A.K. Sinha

24/37


1 L=

A A A

(1) 21 (1) 31 (1) 41

A11 0 U= 0 0

0

0

0

1

0

0

(2) A 32 A (2) 42

1 A (3) 43

0 1

A13 (1) A 23 A (2) 33 0

A14 (1) A 24 A (2) 34 (3) A 44

A12 A (1) 22 0 0

NPTEL

A.K. Sinha

25/37


Use of Rectangular Coordinates Polar version (V∠δ ) of N-R power flow is computationally more efficient than Rectangular version (e + j f), as in case of Polar version no V equations for PV buses is required whereas in case of Rectangular version voltage magnitude equation V2 = (e + j f)2 is required for each PV bus. This increases the number of equations to be solved and therefore the size of Jacobian. NPTEL

A.K. Sinha

26/37

Indian Institute of Technology, Kharagpur However in Polar version the calculation of trigonometric functions in calculation of Jacobian matrix can be avoided by using rectangular coordinates as shown below: m ≠ i (off-diagonal elements) Him = Lim = amfi – bmei Nim = - Jim = amei + bmfi

NPTEL

A.K. Sinha

27/37

Indian Institute of Technology, Kharagpur For m = i (Diagonal elements) Hii = - Qi – Bii|Vi|2 Nii = Pi + Gii|Vi|2 Jii = Pi - Gii|Vi|2 Lii = Qi – Bii|Vi|2 Where

Yim = Gim + jBim Vi = ei + jfi (am + jbm) = (Gim + jBim) (em + jfm) NPTEL

A.K. Sinha

28/37


Regulating Transformers in power flow Regulating transformers are used in power system to control bus voltage magnitude (offnominal tap setting) or to control power flow by controlling the phase angle (phase shifting transformers). They provide small components of voltage (typically ± 10%) to the line or phase voltage. NPTEL

A.K. Sinha

29/37

Indian Institute of Technology, Kharagpur Regulating Transformer for voltage control a

∆ van van + ∆van

van

b c Series

n

transformer NPTEL

A.K. Sinha

30/37

Indian Institute of Technology, Kharagpur Phase Shifting transformer van van + ∆van

∆van

vb n∆vbn

vbn + ∆vbn

vcn

∆vcn NPTEL

vcn + ∆vcn

A.K. Sinha

31/37


∆Van

a Original Van

Shifted Van n c ∆vcn

NPTEL

∆vbn b

A.K. Sinha

32/37


i

j

+

+ Vi -

Ideal transformer 1: t

tVi

Y

-

NPTEL

+ Vj -

A.K. Sinha

33/37


Si = ViIi*

S j = tViI*j

*

Ii = − t I j The current Ij can be expressed by

I j = (Vj − tVi )Y = tYVi + YVj

NPTEL

A.K. Sinha

34/37

Indian Institute of Technology, Kharagpur Multiplying by – t* and substituting Ii for - t*Ij yield *

*

Ii = tt YVi − t YVj Setting tt* = | t |2 and rearranging equations Into Ybus matrix form, we have i

j

i

⎡ Yii Yij ⎤ ⎡ Vi ⎤ ⎢ ⎥⎢ ⎥ = j ⎢Yji Yjj ⎥ Vj ⎣ ⎦⎣ ⎦

i

j

i ⎡ t 2 Y −t* Y⎤ ⎡ Vi ⎤ ⎡Ii ⎤ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ j Y ⎥⎦ ⎣Vj ⎦ ⎣Ij ⎦ ⎢⎣ −tY

NPTEL

A.K. Sinha

35/37


i

j

tY

+

+ Vi

t(t-1)Y

(1-t)Y

-

Vj -

NPTEL

A.K. Sinha

36/37

Indian Institute of Technology, Kharagpur Example: 1

2

4

3

NPTEL

A.K. Sinha

3/37

Indian Institute of Technology, Kharagpur The line parameters are given in the table below: Line no.

From Bus

To Bus

R(pu)

X(pu)

1

1

2

0.05

0.2

2

1

3

0.1

0.3

3

2

3

0.12

0.48

4

2

4

0.09

0.25

5

3

4

0.05

0.2

NPTEL

A.K. Sinha

3/37

Indian Institute of Technology, Kharagpur Ybus matrix for this system:

YBus

⎡ Y11 Y12 Y13 Y14 ⎤ ⎢ ⎥ Y Y Y Y 21 22 23 24 ⎥ ⎢ =⎢ Y31 Y32 Y33 Y34 ⎥ ⎢ ⎥ ⎢⎣ Y41 Y42 Y43 Y44 ⎥⎦ NPTEL

A.K. Sinha

3/37


Where,

1 Y12 = -y12 = = -1.1764 + j4.705 (0.05 + j0.2) 1 Y13 = -y13 = = -1+ j3 (0.1+ j0.3) Y14 = -y14 = 0.0 Y11 = y12 + y13 + y14 = 2.1764 - j7.705 NPTEL

A.K. Sinha

3/37


YBus = 0 ⎡2.176- j7.705 -1.176+j4.705 -1+j3 ⎢ ⎢-1.176+j4.705 2.94- j10.206 -0.49+j1.96 -1.274+j3.541 ⎢-1+j3 -0.49+j1.96 2.66- j9.665 -1.176+j4.705 ⎢ -1.274+j3.541 -1.176+j4.705 2.45- j8.246 ⎢⎣0

NPTEL

A.K. Sinha

3/37

⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦

Indian Institute of Technology, Kharagpur Let us replace Line 1-2 with a regulating transformer with V / V’ = 1.05 Æ t = 1 / 1.05 i

j

ty t(t - 1)y

(1- t)y

NPTEL

A.K. Sinha

3/37

Indian Institute of Technology, Kharagpur Y V = I relationship for the regulating transformer is given by 1 2 * 1 ⎡t2 y −t y ⎤ ⎡ V1 ⎤ ⎡I1 ⎤ 12 ⎢ 2 ⎢⎣ −ty12

12

y12

⎥⎢ ⎥ = ⎢ ⎥ ⎥⎦ ⎣V2 ⎦ ⎣I2 ⎦

The modification in Ybus elements will be:

⎡(Y11(old)-y12 +t2y12) ⎢ ⎢⎣(Y21(old)+y12 -ty12) NPTEL

(Y12(old)+y12 -ty12)⎤ ⎥ (Y22(old)) ⎥⎦ A.K. Sinha

3/37


⎡2.067 - j7.267 ⎢ ⎣-1.12 + j4.48

-1.12 + j4.48 ⎤ ⎥ 2.94 - j10.206⎦

YBus = ⎡2.067-j7.267 ⎢ ⎢-1.12+j4.48 ⎢-1+j3 ⎢ ⎢⎣0

-1.12+j4.48 -1+j3 2.94-j10.206 -0.49+j1.96 -0.49+j1.96 2.66-j9.665 -1.274+j3.541 -1.176+j4.705 NPTEL

A.K. Sinha

0 ⎤ -1.274+j3.541 ⎥⎥ -1.176+j4.705 ⎥ ⎥ 2.45-j8.246 ⎥⎦ 3/37


Let us replace Line 1-2 with a regulating transformer with

V / V =e /

j 3o

⇒ t=e

- j 3o

The modification in Ybus elements will be:

⎡(Y11(old)-y12 +t2y12) ⎢ ⎢⎣(Y21(old)+y12 -ty12) NPTEL

(Y12(old)+y12 -t*y12)⎤ ⎥ (Y22(old)) ⎥⎦ A.K. Sinha

3/37


YBus(1-2) = ⎡2.176 - j7.705 ⎢ ⎣-0.845 + j4.775

- 1.501+ j4.611 ⎤ ⎥ 2.94 - j10.206 ⎦

The modified Ybus Matrix is: YBus = -1+j3 0 ⎡2.176-j7.705 -1.501+j4.611 ⎤ ⎢ ⎥ -0.845+j4.775 2.94-j10.206 -0.49+j1.96 -1.274+j3.541 ⎢ ⎥ ⎢-1+j3 -0.49+j1.96 2.66-j9.665 -1.176+j4.705 ⎥ ⎢ ⎥ -1.274+j3.541 -1.176+j4.705 2.45-j8.246 ⎥⎦ ⎢⎣0 NPTEL

A.K. Sinha

3/37

Indian Institute of Technology, Kharagpur Comparison of N-RLF and FDLF algorithms System

FDLF Time (m.sec)

N-RLF Time (m.sec)

5 Bus

0.950

1.394

14 Bus

1.577

2.954

30 Bus

2.630

6.338

57 Bus

4.910

15.443

118 Bus

11.440

52.103

NPTEL

A.K. Sinha

3/37


APPLICATION OF POWER FLOW STUDIES Power Flow analysis provides the steady state operating condition (power flow, bus voltages) of a power system. It is a fundamental and most common study of the power transmission network. It is an integral part of most system planning and operation studies. It is a starting point for many other studies Æ Short Circuit; Transient Stability; NPTEL

A.K. Sinha

3/37


System Planning and Design: Evaluate the adequacy of present and planned system expansion. Contingency evaluation Æ Any abnormal condition such as a generator or a transmission line outage Æ this will provide opportunity to the planning engineer to upgrade the existing system.

NPTEL

A.K. Sinha

3/37


System Operation and Control: Evaluate the adequacy of present operating condition Æ line overloads; bus voltage violations. Security evaluation Æ Present operating cond. may be adequate but what if Æ any abnormal condition such as a generator or a transmission line outage Æ is the system operation adequate (Secure) for contingencies also. NPTEL

A.K. Sinha

3/37


SHORT CIRCUIT ANALYSIS Lesson Summary 1. Introduction 2. Cause and Consequences 3. Types of Faults 4. Symmetrical Short Circuit Analysis

NPTEL

A.K. Sinha

2/37


Instructional Objective On completion of this lesson a student should be able to: A. Explain the significance of Short Circuit B. Explain the causes and consequences of Short Circuit C. Classify different types of faults D. Compute currents for symmetrical faults NPTEL

A.K. Sinha

3/37


Cause of Short Circuit Æ Insulation Failure ƒ Over-voltages caused by Lightning or Switching Surges ƒ Insulation contamination Æ salt spray, pollution ƒ Mechanical Causes Æ Over-heating, abrasion

NPTEL

A.K. Sinha

4/37


Faults on Transmission Lines Most Common Æ Lines are exposed to elements of nature (60-70%). Lightning strokes Æ Over voltages cause insulators to flash over Æ line to ground short circuit or line to line short circuit. High winds Æ Topple tower, tree falls on line. Winds and ice loading Æ Mechanical failure of insulator. NPTEL

A.K. Sinha

5/37

Indian Institute of Technology, Kharagpur Fog, salt spray, dirty insulator Æ Conduction path Æ insulation failure

Short circuit in other elements Æ Cables (10-15%), circuit breakers (10-12%), generators, motors, transformers etc (10-15%). Æ much less common Æ Over loading for extended periods Æ deterioration of insulation Æ Mechanical failure.

NPTEL

A.K. Sinha

6/37


Consequences of Short Circuit Currents several magnitude larger than normal operating current. Thermal damage to equipment. Windings and busbars Æ Mechanical damage due to high magnetic forces caused by high current.

Faulted section must be removed from service as soon as possible (3-5 cycles). NPTEL

A.K. Sinha

7/37


Types of Short Circuit a b c

L–G

L–L

L–L–G

3φ – G

75 - 80%

5 – 7%

10 – 12%

8 – 10%

Asymmetrical Faults Symmetrical Faults NPTEL

A.K. Sinha

8/37


Short Circuit Calculations R – L Series Circuit Transients R

e(t) = 2V Sin (ϖt + α)

L SW

e(t)

Vmax

= Vmax Sin (ϖt + α) t=0

di sin ( ωt + α ) = Ri + L ; t ≥ 0 dt NPTEL

A.K. Sinha

9/37

Indian Institute of Technology, Kharagpur Solving for i(t)

Vmax i(t) = Z

⎡ sin ω t + α - θ − e −Rt /L sin α - θ ⎤ ( ) ⎥⎦ ( ) ⎢⎣

Where, Z = R + ( ωL ) and θ = tan-1 ( ωL /R ) . 2

2

or

i(t) = iac (t) + idc (t) iac (t) = Symmetrical fault current (Constant) idc (t) = DC offset current (decays with time) i(t) = Asymmetrical fault current NPTEL

A.K. Sinha

10/37

Indian Institute of Technology, Kharagpur I(t)

idc

iac

Time

NPTEL

A.K. Sinha

11/37


3-phase Short Circuit on Synchronous Machine Unloaded Machine: Symmetrical short circuit current

Subtransient period c Transient period b Steady state period

a 0

Time Actual envelope NPTEL

A.K. Sinha

12/37

Indian Institute of Technology, Kharagpur Xf

XD XI

Xa

+ Eg’’

XI

Xf

X

+ Eg’

'' d

Direct axis subtransient reactance

Xa

X

' d

Direct axis transient reactance

XI + Eg NPTEL

Xa Xd

Synchronous reactance A.K. Sinha

13/37

Indian Institute of Technology, Kharagpur ⎡⎛ 1 ⎤ 1 ⎞ - t / Td'' ⎢ ⎜ '' - ' ⎟ e ⎥ ⎝ Xd Xd ⎠ ⎢ ⎥ i a c (t) = 2 E g ⎢ ⎥ ⎛ 1 ⎞ - t / Td' 1 1 ⎢+ ⎜ ⎥ + ⎟e ' X d ⎥⎦ ⎢⎣ ⎝ X d X d ⎠ π x S in ( ϖ t + α - ) 2

idcmax (t) =

2Eg X

'' d

e

- t / TA

''

= 2I e

- t / TA

;

TA = Armature Time Const. NPTEL

A.K. Sinha

14/37


i (t) = iac (t) + idc (t) Eg

Iac (0) = O C = I = Ob = '

Eg X

' d

X ''d

= I'' (Subtransient Current)

(T ra n s ie n t c u rre n t)

Ia c ( ∞ ) = O a =

Eg Xd

(S te a d y S ta te c u rre n t)

NPTEL

A.K. Sinha

15/37

Indian Institute of Technology,Kharagpur

Short Circuit on a loaded Synchronous Machine Zext

X

'' dg

E

'' + g

'' Eg ' Eg

+

P

+

IL

Vt

S

= Vt +

'' jX dgIL

= Vt +

' jX dgIL

Vf

( + (Z

ZL

= Vf + Z ext +

'' jX dg

= Vf

' jX dg

NPTEL

ext

+

A.K. Sinha

) IL )I L

16/37

Indian Institute of Technology, Kharagpur For Motor:

E"m = Vt − jX "dm IL E'm = Vt − jX 'dm IL Example: A synchronous generator and a synchronous motor each rated 50 MVA, 11KV having 12% subtransient reactance are connected through transformers and a line as shown in figure below. The transformers are rated 50 MVA, 11/132 KV and 132/11KV with leakage reactance of 8% each. The line has a reactance of 15% on a base of 50 MVA, 132 KV. NPTEL

A.K. Sinha

17/37

Indian Institute of Technology, Kharagpur The motor is drawing 25 MW at 0.8 power factor leading and a terminal voltage og 10.6 KV when a symmetrical three-phase fault occurs at the motor terminals. Find the subtransient current in the generator, motor and fault.

Gen

T1

Line

T2

NPTEL

A.K. Sinha

Motor 18/37

Indian Institute of Technology, Kharagpur Zext

P +

(j0.31) jX”dg

E”g

I0

(j0.12)

jX”dm

(j0.12) V0

+

+ E”m

Neutral (a) Before the fault NPTEL

A.K. Sinha

19/37

Indian Institute of Technology, Kharagpur Zext (j0.31) jX”dg

E”g

(j0.12)

P I”g

I”m

jX”dm

(j0.12) I”f

+

+ E”m

Neutral (b) After the fault NPTEL

A.K. Sinha

20/37


10.6 0 V = = 0.9636 ∠ 0 pu Prefault Voltage = 11 Load = 25 MW 0.8 pf leading 0

25 = pu 0.8 pf leading 50 = 0.5 pu 0.8 pf leading Prefault Current

I0 =

0.5 ∠36.9o = 0.6486∠36.9o 0.9636×0.8

NPTEL

A.K. Sinha

21/37

Indian Institute of Technology, Kharagpur Voltage behind sub-transient reactance (generator)

E = 0.9636∠0 + j0.43 × 0.6486∠36.9 " g

o

o

= 0.7962 + j0.223pu Voltage behind sub-transient reactance (Motor)

E = 0.9636∠0 - j0.12 × 0.6486∠36.9 " m

o

o

= 1.0103 - j0.0622pu

NPTEL

A.K. Sinha

22/37

Indian Institute of Technology, Kharagpur Under faulted condition

0.7962 + j0.223 I = = 0.5186 - j1.8516pu j0.43 1.0103 - j0.0622 " Im = = -0.5183 - j8.4191pu j0.12 f " " Current in fault I = Ig + Im = 0.0003 - j10.2707pu " g

Base current (generator/motor) = 3

50 ×10 = 2624.3A 3 ×11 NPTEL

A.K. Sinha

23/37

Indian Institute of Technology, Kharagpur Contribution from Gen. and Motor

I"g =2624.3(0.5186- j1.8516) =(1360.96- j4859.15) A " Im =2624.3(-0.5183- j8.4191) =(-1360.17- j22094.24) A

If =2624.3(0.0003- j10.2707) =(0.78- j26953.39) A

NPTEL

A.K. Sinha

24/37


Short Circuit Calculations using Thevenin’s Theorem A Short Circuit Æ Structural change in network Æ addition of an impedance (ZF = fault impedance, zero for solid short circuit) at the point of fault. The change in voltage or current resulting from this structural network change can be analyzed using Thevenin’s theorem. NPTEL

A.K. Sinha

25/37


Thevenin’s Theorem – The changes in network voltages and currents due to the addition of an impedance between two points of a network are identical with those voltages and currents that would be caused by placing an emf, having a magnitude and polarity equal to the pre-fault voltage between the nodes, in series with the impedance all other voltage sources being zeroed. NPTEL

A.K. Sinha

26/37

Indian Institute of Technology, Kharagpur Post fault voltages and currents are computed by superimposing these changes on pre-fault voltages and currents. Example:

Gen

T1

Line

T2

NPTEL

A.K. Sinha

Motor 27/37


P +

(j0.31) jX”dg

E”g

I0

(j0.12)

jX”dm

(j0.12) V0

+

+ E”m

Neutral

G

(a) Before the fault NPTEL

A.K. Sinha

28/37


P

(j0.31) jX”dg

E”g

(j0.12)

+

V0

+

I0

+

V0

E”m

+ Neutral

jX”dm

(j0.12)

G

(b) After the fault NPTEL

A.K. Sinha

29/37

Indian Institute of Technology, Kharagpur Using Superposition: Zext P (j0.31) jX”dg

E”g

(j0.12)

+

V0

I0

jX”dm

(j0.12)

+

+ E”m Neutral

G

Circuit A (Pre fault Circuit) NPTEL

A.K. Sinha

30/37


ΔI

'' g

Zext

'' ΔIm

(j0.31) jX”dg

(j0.12)

(j0.12)

jX”dm

V0 P + Neutral

G Circuit B (Thevenin’s Eq. Circuit) NPTEL

A.K. Sinha

31/37


10.6 0 V = = 0.9636 ∠ 0 pu Prefault Voltage = 11 0

o 0 V 0.9636 ∠ 0 ΔI''g = = = - j2.2409 j0.31 + j0.12 j0.43 o 0 V 0.9636 ∠ 0 '' ΔIm = = = - j8.03 j0.12 j0.12

I''g = I0 + ΔI''g = 0.6486∠36.9o - j2.2409 = 0.5186 - j1.8516 '' '' Im = - I0 + ΔIm = 0.6486∠36.9o - j8.03

= - 0.5186 - j8.4194 NPTEL

A.K. Sinha

32/37


The selection of circuit breakers From the viewpoint of current the two factors that need to be considered in selecting circuit breakers are: • The maximum instantaneous current which the breaker must carry ( withstand ) and • The total current when the breaker s open to interrupt the circuit. NPTEL

A.K. Sinha

33/37


Short Circuit MVA Short-circuit MVA= √3 Χ (nominal kV) Χ | ISC | Χ 10-3 Base MVA = √3 Χ (base kV) Χ | Ibase | Χ 10-3 Short-circuit MVA in per unit = | ISC | in per unit

Z th

1.0 1.0 = per unit = per unit ISC short circuit MVA NPTEL

A.K. Sinha

34/37

Indian Institute of Technology, Kharagpur The maximum momentary current is found by calculating the ac short circuit current using sub-transient impedances of the generators and motors and then multiplying it by 1.6 to take care of the dc off-set current. The breaker interrupting current depends on the interruption time of the circuit breakers and is obtained by multiplying the sub-transient ac short circuit current by following factors:

NPTEL

A.K. Sinha

35/37

Indian Institute of Technology, Kharagpur Circuit breaker speed

Multiplying factor

8 cycle or slower

1.0

5 cycles

1.1

3 cycles

1.2

2 cycles

1.4

For CBs having short circuit MVA greater than 500 MVA the multiplying factors are increased by 0.1 NPTEL

A.K. Sinha

36/37


SYMMETRICAL COMPONENT ANALYSIS


A.K. Sinha

1/37


SYMMETRICAL COMPONENT ANALYSIS Lesson Summary 1. Introduction 2. Symmetrical Component Transformation 3. Sequence network for PS Components 4. Sequence network for Power Systems NPTEL

A.K. Sinha

2/37


Instructional Objective On completion of this lesson a student should be able to: A. Explain the significance of symmetrical component transformation B. Develop sequence network for power system components and networks C. Compute current, voltage and power in sequence networks NPTEL

A.K. Sinha

3/37


Types of Short Circuit a b c

L–G

L–L

L–L–G

3φ – G

75 - 80%

5 – 7%

10 – 12%

8 – 10%

Asymmetrical Faults Symmetrical Faults NPTEL

A.K. Sinha

8/37


Symmerical Component Transformation ƒ Introduced by C. L. Fortescue (1918) ƒ Modeling technique for analysis and design of three-phase systems ƒ Decouples a balanced three-phase network into three simpler networks ƒFor unbalanced three phase networks the three sequence networks are connected only at the point of unbalance. NPTEL

A.K. Sinha

4/37


Symmerical Component Transformation ƒ A powerful tool for analyzing three phase systems ƒ Reveals complicated phenomena during unbalanced operation in simple ƒ Sequence network results have to be superposed to obtain three phase network results

NPTEL

A.K. Sinha

5/37


SYMMETRICAL COMPONENT TRANSFORMATION An unbalanced set of n phasors can be resolved into n sets of balanced phasors (symmetrical components). The n phasors of each set of components are equal in magnitude and angles between adjacent phasors of the set are equal. Unbalanced phasors of a three phase system can be resolved into three balanced system of phasors Æ positive, negative, and zero sequence NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Positive-sequence components, consist of three phasors with equal Vc1 magnitudes with 120º phase Va1 = V1 displacement from each other, and same phase sequence as original Vb1 phasors. V Negative-sequence components, b2

Va2 = V2

consist of three phasors with equal magnitudes with 120º phase displacement from each other, and opposite phase sequence as original phasors.

Vc2

NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Zero-sequence components, consist of three phasors with equal magnitudes and zero phase displacement from each other.

NPTEL

Va0Vb0Vc0 = V0

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Va

Vc

Vb = Vb0 + Vb1 + Vb2 Vb0 Vc2 Vb Vb1 Vc Vb2

Vb Va Va0

Va1

Va2

Va = Va0 + Va1 + Va2 Phase a

Phase b NPTEL

Vc1 Vc0

Vc = Vco + Vc1 + Vc2 Phase c A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Va = V0 + V1 + V2 Vb = V0 +

a2V1

+ aV2

Vc = V0 + aV1 + a2V2

⎡ Va ⎤ ⎡1 1 ⎢ ⎥ ⎢ 2 [ VP ] = ⎢ Vb ⎥ = ⎢1 a ⎢⎣ Vc ⎥⎦ ⎢⎣1 a

NPTEL

-1 3 a = 1 120 = + j 2 2 o

1⎤ ⎥ a⎥ 2 a ⎥⎦

⎡ V0 ⎤ ⎢V⎥ ⎢ 1⎥ ⎢⎣ V2 ⎥⎦

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Vp = A

Vs

Where,

1 1 2 A = 1 a 1

1 a

and

2

a

a

-1

Vp

Vs = A

V0 Vs = V1 V2

Where,

A NPTEL

-1

1 1 1 = 1 a 3 2 1 a A.K. Sinha

1 2 a a 6/37


V0 1 1 1 V1 = 1 a 3 2 V2 1 a

1 2 a

Va Vb

a

Vc

1 V0 = (Va + Vb + Vc ) 3 1 V1 = (Va + aVb + a 2 Vc ) 3 1 2 V2 = (Va + a Vb + aVc ) 3 NPTEL

A.K. Sinha

6/37


Ia = I0 + I1 + I2 Ip = A

Is

Ib = I0 + a2I1 + aI2 Ic = I0 + aI1 + a2I2

-1

Is = A

1 I0 = (Ia + Ib + Ic ) 3 1 I1 = (Ia + aIb + a 2 Ic ) 3 1 I2 = (Ia + a 2 Ib + aIc ) 3

Ip

NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur POWER IN SEQUENCE NETWORKS

S 3φ = VagIa* + VbgIb* + VcgIc*

S 3φ

⎡Ia* ⎤ ⎢*⎥ = ⎡⎣ Vag Vbg Vcg ⎤⎦ ⎢Ib ⎥ ⎢Ic* ⎥ ⎣ ⎦

= VpT Ip* NPTEL

A.K. Sinha

6/37


S3φ = (AVs )T (AIs )*

= V ⎡⎣ A A ⎤⎦ I T s

⎡1 1 ⎢ T * 2 A A = ⎢1 a ⎢⎣1 a NPTEL

T

1⎤ ⎥ a⎥ a 2 ⎥⎦

*

T

* s

⎡1 1 ⎢1 a 2 ⎢ ⎢⎣1 a A.K. Sinha

1⎤ ⎥ a⎥ a 2 ⎥⎦

*

6/37


⎡1 1 = ⎢⎢1 a 2 ⎢⎣1 a

1⎤ a ⎥⎥ a 2 ⎥⎦

⎡1 1 ⎢1 a ⎢ ⎢⎣1 a 2

1⎤ a 2 ⎥⎥ a ⎥⎦

⎡3 0 0 ⎤ = ⎢⎢0 3 0 ⎥⎥ = 3U ⎢⎣0 0 3 ⎥⎦

NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur T * s s

S 3φ = 3V I

⎡I0* ⎤ ⎢ *⎥ = [ V0 + V1 + V2 ] ⎢ I1 ⎥ ⎢I2* ⎥ ⎣ ⎦

S 3φ = 3V0 I0* + 3V1I1* + 3V2 I2* = Sum of symmtrical component

powers NPTEL

A.K. Sinha

6/37


Sequence Network for Power System Components Ia a Impedance Load: + c Ic + Balanced Star Grounded Load Vag V cg

g NPTEL

-

b

ZY

ZY

N Ib

In

Zn

ZY

Vbg + -

A.K. Sinha

6/37


Vag = Z YIa + ZnIn

= Z YIa + Zn (Ia + Ib + Ic ) = (Z Y + Zn )Ia + ZnIb + ZnIc

Vbg = ZnIa + (Z Y + Zn )Ib + ZnIc Vcg = ZnIa + ZnIb + (Z Y + Zn )Ic NPTEL

A.K. Sinha

6/37


Vag

(Z Y + Zn )

Vbg = Vcg

Zn Zn

Zn

(Z Y + Zn ) Zn Zn (Z Y + Zn )

Vp = ZpIp

Ia Ib Ic

AVs = Zp AIs Vs = Z sIs

-1

Vs = (A Zp A)Is Where,

Zn

-1

Z s = A Zp A NPTEL

A.K. Sinha

6/37


1 1 1 Zs = 1 a 3 1 a2 1 1 × 1 a2 1 a

1 (Z Y + Zn ) Zn Zn a2 Zn (Z Y + Zn ) Zn a Zn Zn (Z Y + Zn ) 1 a a2

(Z Y + 3Zn ) 0 = 0 Zy 0 0

NPTEL

A.K. Sinha

0 0 Zy

6/37


V0

(Z Y + 3Zn )

0

0

I0

V1 =

0

ZY

0

I1

V2

0

0

ZY

I2

V0 = (Z Y + 3Zn )I0 = Z 0I0 V1 = Z YI1 = Z1I1 V2 = Z YI2 = Z 2I2 NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur +

I0

ZY

V0

3Zn

Z0 = ZY + 3Zn

Zero-sequence network

For star ungrounded load Zn = infinity Zero sequence network is open NPTEL

A.K. Sinha

6/37


+

I1

V1

ZY

Z1 = ZY

ZY

Z2 = Z1 = ZY

- Positive-sequence network +

I2

V2 - Negative-sequence network NPTEL

A.K. Sinha

6/37


Balanced Delta Connected Load: I0

ZΔ 3

a

ZΔ

ZΔ

b

ZΔ Z1 = 3

ZΔ 3 Z0 = ∞

c

I1

I2

ZΔ Z 2 = Z1 = 3

ZΔ NPTEL

A.K. Sinha

6/37


General three-phase impedance load Ia Ic

+ + Vag

g

-

Vcg

Ib +

General three-phase impedance load

Vbg

-

NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur -1

Z s = A Zp A Z0 Z10 Z 20

Z 01 Z 02 1 1 1 Z1 Z12 = 1 a 3 Z 21 Z 2 1 a2

1 a2 a

1 1 1 a2 1 a

1 a a2

NPTEL

Z aa Z ab Z ac

Z ab Zbb Zbc

A.K. Sinha

Z ac Zbc Z cc

6/37


1 Z 0 = (Z aa + Zbb + Z cc + 2Z ab + 2Z ac + 2Zbc ) 3 1 Z1 = Z 2 = (Z aa + Zbb + Z cc - Z ab - Z ac - Zbc ) 3 1 Z 01 = Z 20 = (Z aa + a2 Zbb + aZ cc - aZ ab - a2 Z ac - Zbc ) 3 1 Z 02 = Z10 = (Z aa + aZbb + a2 Z cc - a2 Z ab - aZ ac - Zbc ) 3 NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Z 12

1 = (Z aa + a 2 Z bb + aZ cc + 2aZ ab + 2a 2 Z ac + 2Z bc ) 3

1 Z 21 = (Z aa + aZ bb + a 2 Z cc + 2a 2 Z ab + 2aZ ac + 2Z bc ) 3

Z aa = Z bb = Z cc and

Z ab = Z ac = Z bc

Conditions for a symmetrical load

Z 01 = Z10 = Z 02 = Z 20 = Z12 = Z 21 = 0 Z 0 = Z aa + 2Z ab

Z1 = Z 2 = Z aa - Z ab NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Example: Three identical Y-connected resistors from a load bank with a three phase rating of 2300 V and 500 KVA. If each bank has applied voltages |Vab| = 1800V

|Vbc| = 2700V

|Vca| = 2300V

find the line voltages and currents in per unit into the load. Assume that the neutral of the load is not connected to the neutral of the system and select a base of 2300 V,100 KVA. NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Solution: On 2300 V, 100 KVA base line voltages in per unit are: Vab = 1800 2300 = .7826

Vbc = 2700 2300 = 1.174

Vca = 2300 2300 = 1.0

Let, Vca ∠1800 be taken as reference Then, Vab = .7826∠81.390 and Vbc = 1.174∠ - 41.230 Symmetrical components of the line voltages are

1 Vab1 = (Vab + aVbc + a 2 Vca ) 3

Vab1 = 1/3[.7826∠81.39 + 1.174∠(120 - 41.23) + 1.0∠(240 +180)] NPTEL

A.K. Sinha

6/37


= 1/3[.1171+j.7738+.2286+j1.152+.5+j.866] =1/3[.8454+j2.7918] = .2819+ j.9306 = 0.97236 ∠73.1470

Vab2

1 = (Vab + a 2 Vbc + aVca ) 3 0

Vab2 = 1/3[.7826∠81.39 + 1.174∠(2400 - 41.230 ) + 1.0∠(120 +180 )] 0

0

= 1/3[0.11716+j0.7737-1.111-j0.378 + 0.5 - j0.866] =1/3[-0.49339 - j0.4703] = -0.1645 - j0.1567 =.2272∠ - 136.370 =.2272 ∠223.6270 NPTEL

A.K. Sinha

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1 Vab0 = (Vab + Vbc + Vca ) 3 0 0 0 Vab0 =1/3[.7826∠81.39 + 1.174∠ - 41.23 + 1.0∠180 ]

=0

Voltages to neutral are given by

V = 0.97236∠(73.147 - 30 ) = 0.97236∠43.147 1 an

0

Vc1

0

Vca Va1 = V1

Vbc n Vb1

Vab NPTEL

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6/37

0


V = 0.2272∠(223.627 + 30 ) = 0.2272∠253.627 2 an

0

0

0 an

V =0 resistance value in p.u.=1.100/500=.2 p.u.

I = (0.97236∠43.147 )/.2 = 4.8618∠43.147 p.u. 1 a

0

0

I = (0.2272∠253.627 )/.2 = 1.136∠253.627 p.u. 2 a

0

0

0 a

I =0

NPTEL

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0


SEQUENCE NETWORKS


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SEQUEMCE NETWORKS Lesson Summary 1. Introduction 2. Sequence networks for Transmission Lines 3. Sequence network for Genetators 4. Sequence network for Transformers 5. Sequence network for Power System NPTEL

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Instructional Objective On completion of this lesson a student should be able to: A. Develop sequence network Tr. Lines B. Develop sequence network for Generators C. Develop sequence network for Transformers D. Assemble sequence networks for small power systems NPTEL

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Sequence Networks of Three-Phase Lines a b c n

Ia Van Ib Vcn Ic

Zca

Vcn In

Zaa

a’

Zbb

Zab

Zcc

Zbc

Znn NPTEL

Zcn Zbn

b’ Va’n’ Zan

c’ Vb’n’ Vc’n’

A.K. Sinha

n’ 4/36

Indian Institute of Technology, Kharagpur For a fully transposed line: Zaa = Zbb = Zcc

Zab = Zbc = Zca

Zan = Zbn = Zcn

a Van n

Ia Zab Ib Zab Ic +

-

+

Zan In -

In Zan Ia Zan Ib +

-

+

+

Zaa

-

Zan Ic -

NPTEL

a’

+

Znn

-

A.K. Sinha

Va’n’ n’ 5/36


Van = Z aa Ia + Z ab Ib + Z ab Ic + Z anIn + Va'n' - (Z nnIn + Z anIc + Z anIb + Z anIa ) Van - Va'n' = (Z aa - Z an )Ia + (Z ab - Z an )(Ib + Ic ) + (Z an - Z nn )In

Vbn - Vb'n' = (Z aa - Z an )Ib + (Z ab - Z an )(Ia + Ic ) + (Z an - Z nn )In NPTEL

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Vcn - Vc'n' = (Z aa - Z an )Ic + (Z ab - Z an )(Ia + Ib ) + (Z an - Z nn )In

In = - (Ia + Ib + Ic ) Van - Va'n' = (Z aa + Z nn - 2Z an )Ia + (Z ab + Z nn - 2Z an )Ib + (Z ab + Z nn - 2Z an )Ic Vbn - Vb'n' = (Z ab + Z nn - 2Z an )Ia + (Z aa + Z nn - 2Z an )Ib + (Z ab + Z nn - 2Z an )Ic NPTEL

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Vcn - Vc'n' = (Z ab + Z nn - 2Z an )Ia + (Z ab + Z nn - 2Z an )Ib + (Z aa + Z nn - 2Z an )Ic

Z s @Z aa + Znn - 2Z an Zm @Z ab + Znn - 2Z an

NPTEL

A.K. Sinha

8/36


Vaa'

Van - Va'n'

Zs

Zm

Zm Ia

Vbb' = Vbn - Vb'n' = Zm Vcc' Vcn - Vc'n' Zm

Zs Zm

Zm Ib Z s Ic

Vaa' @Van - Va'n'

-1

Z sy = A Zp A

Vbb' @Vbn - Vb'n' Vcc' @Vcn - Vc'n' NPTEL

Z sy

Z0 = 0

0 Z1

0 0

0

0

Z2

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Z 0 = Z s + 2Zm

Z1 = Z 2 = Z s - Z m

V0 - V0' = Z 0I0

V1 - V1' = Z1I1

V2 - V2' = Z 2I2 I0 + V0 -

Z0 = Zaa + 2Zab Zero-sequence network

NPTEL

A.K. Sinha

+ V0’ 10/36

Indian Institute of Technology, Kharagpur I1 + V1 -

Positive-sequence network

I1 + V2 -

Z1 = Zaa - Zab + V1 -

Z2 = Z1 = Zaa - Zab

Negative-sequence network

NPTEL

A.K. Sinha

+ V2 11/36


Sequence networks of Synchronous Generator c a + Ec In Zn

Ic Ia

-

+

- - Ea Eb +

b NPTEL

A.K. Sinha

Ib 12/36

Indian Institute of Technology, Kharagpur Zg1 Eg+ -

I1 + V1

Positive-sequence network Z g1 ≈ jX d

For calculating initial Sub - transient fault current Eg = E''g and Z g1 ≈ jX ''d NPTEL

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13/36

Indian Institute of Technology, Kharagpur Zg2

I2 + V2 -

Negative-sequence network Z g2 ≈ j

X q + X 'd

2 For calculating initial Sub - transient fault current Z ''g2 ≈ jX ''d as ( X q'' ≈ X d'' ) NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur Zg0

I0 + V0

3Zn

Zero-sequence network Z g0 ≈ jX l Z g1 ≈ Z g2 > Z g0 NPTEL

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Sequence networks of Synchronous Motor and Induction Motor Synchronous motor Zm1 I

Induction motor Zm1 I 1

1

+ V1 -

+

+ Em1 -

Positive-sequence network NPTEL

V1 Positive-sequence network A.K. Sinha

16/36


Synchronous motor Zm2 I

Induction motor Zm2 I

2

2

+

+

V2

V2

-

-



NPTEL

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Indian Institute of Technology, Kharagpur Synchronous motor Zm0 I

Induction motor I0

0

+

+

V0

3Zn V 0

Zm0 3Zn

-



NPTEL

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18/36

Indian Institute of Technology, Kharagpur Per-Unit Sequence Models of Three-Phase Two-Winding Transformers +

+ VH

VL

-

jXl

NPTEL

∠ ± 300

A.K. Sinha

19/36

Indian Institute of Technology, Kharagpur + VH1

+ jXl VL1 positive-sequence network

+ VH2

+ jXl VL2 negative-sequence network NPTEL

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20/36

Indian Institute of Technology, Kharagpur C

Zero Sequence Network c X3 H3 H1

N H2 B

X1 A

ZN

X2 b

a Zn

Schematic representation NPTEL

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21/36

Indian Institute of Technology, Kharagpur C

H3 N

H2 B

H1

c A X2 b

ZN

X3 a X1

Schematic representation

NPTEL

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22/36


Zn Transformer

ZN

jXl+3ZN+3Zn + VL0 -

+ VH0 -


NPTEL

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Indian Institute of Technology, Kharagpur Transformer

+ VH0

jXl+3ZN

-

+ VL0 -


NPTEL

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24/36


+ VH0

jXl

-

+ VL0 -


NPTEL

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25/36


+ VH0

jXl+3ZN

-

+ VL0 -


NPTEL

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26/36


+ VH0

jXl

-

+ VL0 -


NPTEL

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jXl + VL0 -

+ VH0 -


NPTEL

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28/36

Indian Institute of Technology, Kharagpur The per-unit sequence network of the Y- Δ transformer, shown in Figure (b), have the following features. 1. The per-unit impedances do not depend on the winding connections. That is, the perunit impedances of a transformer that is connected Y-Y. Y-Δ, Δ-Y. or Δ- Δ are the same. However, the base voltages do depend on the winding connections. NPTEL

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Indian Institute of Technology, Kharagpur 2. A phase shift is included in the per-unit positive- and negative- sequence networks. For the American standard, the positivesequence voltages and currents on the high-voltage side of the Y-Δ transformer lead the corresponding quantities on the low-voltage side by 30°. For negative sequence, the high-voltage quantities lag by 30°.

NPTEL

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3. Zero-sequence currents can flow in the Y winding if there is a neutral connection, and corresponding zero-sequence currents flow within the Δ winding. However, no zerosequence current enters or leaves the Δ winding.

NPTEL

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Indian Institute of Technology, Kharagpur Example: For the system shown in figure assemble the sequence networks

Gen

T1

Line

NPTEL

T2

A.K. Sinha

Motor

32/36


Xg1

XT1

XL1

XT1

Eg1

Xm1

Em1

Positive-sequence network

NPTEL

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Xg2

XT2

XL2

XT2

Xm2


NPTEL

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Xg0

XT0

XL0

XT0

Xm0


NPTEL

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Thank You !

NPTEL

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37/37


FAULT ANALYSIS FOR LARGE POWER SYSTEMS


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ASSYMETRICAL FAULT ANALYSIS (Contd..)

Lesson Summary 1. Introduction

2. Fault analysis using Ybus matrix 3. Fault analysis using Zbus matrix 4. Example NPTEL

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2/36


Instructional Objective On completion of this lesson a student should be able to: A. Develop algorithm for fault analysis using Ybus method B. Explain the difficulties associated with fault analysis using Ybus method C. Develop algorithm for fault analysis using Zbus method NPTEL

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Indian Institute of Technology, Kharagpur D. Develop fault impedance matrix for different types of fault E. Solve for fault current and bus voltages for a faulted power system

NPTEL

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Short-Circuit Analysis of Large Power Systems For small systems it is possible to reduce the sequence networks into its Thevenin’s equivalent at the fault point. For large systems it is extremely difficult to obtain such an equivalent. For large systems we need to develop a systematic and general procedure (algorithm) which is suitable for solution on computer. NPTEL

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Problem Statement • The system under study has n buses. • It is initially operating in a symmetrical normal state. • All prefault bus voltages and power flows are assumed to be known. • A fault occurs at bus q. • Fault can be represented as a fault impedance matrix Z f or fault ittance matrix Y f .


n-bus power system +

+

Bus q a b c

+

f f f Currents Iaq , Ibq , Icq

f aq

f f V Vbq Vcq

-

- -

Zf

Fault impedance matrix Z f

Indian Institute of Technology, Kharagpur Symmetrical Short Circuit Let the pre-fault bus voltage of the system be given as:

0 VBUS =

V10 0 V2 M Vn0

NPTEL

A.K. Sinha

8/36

Indian Institute of Technology, Kharagpur Using the Thevenin’s theorem the post-fault Bus voltages are given by: f bus

V Where,

ΔVBUS

=V

0 bus

+ Δ Vbus

ΔV1 Changes caused by ΔV2 = = the fault M ΔVn

NPTEL

A.K. Sinha

8/36


n-bus power system Bus q

+ f

I

ΔVq -

+ Zf

0 q

V

Fault impedance

Indian Institute of Technology, Kharagpur Network Equations

IBUS = YBUS VBUS VBUS = ZBUS IBUS ΔIBUS = YBUS ΔVBUS ΔVBUS = ZBUS ΔIBUS NPTEL

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⎡ΔI1 ⎤ ⎡Y11 ⎢ ⎥ ⎢ M M ⎢ ⎥ ⎢ ⎢ΔIq ⎥ = ⎢Yq1 ⎢ ⎥ ⎢ ⎢ M⎥ ⎢ M ⎢ΔI ⎥ ⎢Y ⎣ n ⎦ ⎣ n1

L

Y1q

L

L

M Yqq

L

L

M Yn2

L

NPTEL

Y1n ⎤ ⎡ΔV1 ⎤ ⎥⎢ ⎥ M ⎥⎢ M ⎥ Yqn ⎥⎢ΔVq ⎥ ⎥⎢ ⎥ M ⎥⎢ M ⎥ ⎥⎢ ⎥ Ynn ⎦ ⎣ΔVn ⎦

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⎡ΔIq ⎤ ⎡Yqq ⎢ ⎥ ⎢Y ΔI ⎢ 1 ⎥ ⎢ 1q ⎢ M⎥ = ⎢ M ⎢ ⎥ ⎢ ⎢ M⎥ ⎢ M ⎢ΔI ⎥ ⎢Y ⎢⎣ n ⎥⎦ ⎣⎢ nq

Yq1 L L Y11 L L M M Yn1 L L NPTEL

Yqn ⎤ ⎡ΔVq ⎤ ⎥⎢ ⎥ Y1n ⎥ ⎢ΔV1 ⎥ ⎥ M ⎢ M⎥ ⎥⎢ ⎥ M ⎥⎢ M ⎥ ⎥ ⎢ ⎥ Ynn ⎥ ⎢⎣ΔVn ⎥⎦ ⎦

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⎡ΔIq ⎤ ⎢ ⎥= ⎣ΔI0 ⎦ ΔI0 = 0 ;

⎡Yqq Yq0 ⎤ ⎡ΔVq ⎤ ⎢ ⎥⎢ ⎥ ⎢⎣Y0q Y00 ⎥⎦ ⎣ΔV0 ⎦ Y0q ΔVq + Y00 ΔV0 = 0 -1 00

ΔV0 = - Y Y0q ΔVq ; Substituting -1 q0 00

ΔIq = (Yqq - Y Y Y0q) ΔVq NPTEL

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8/36

Indian Institute of Technology, Kharagpur -1 00

If = - ΔIq = - (Yqq - Yq0 Y Y0q ) ΔVq f q

0 q

Also, V = V + ΔVq ; For a fault with Zero impedance V = 0; → ΔVq = - V f q

0 q

Therefore, -1 00

0 q

If = (Yqq - Yq0 Y Y0q ) V NPTEL

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8/36

Indian Institute of Technology, Kharagpur Although this algorithm looks straight forward there are some difficulties associated with this: The algorithm involves inversion of (n-1)x(n-1) matrix Y00. For finding out the Circuit breaker ratings and relay settings we require calculation of fault currents for faults at different busses. For each of these fault calculations we will require inversion of Y00 matrix which will be different in each case. This is computationally very expensive. NPTEL

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8/36

Indian Institute of Technology, Kharagpur Fault Analysis Using ZBUS Method Using the Thevenin’s theorem the postfault Bus voltages are given by f bus

V

Ibf u s

=V

⎡ 0 ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ M⎥ = ⎢ f⎥ ⎢ −I ⎥ ⎢ M⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦

0 bus

+Z

f bus bus

I

qth component

NPTEL

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⎡ΔV1 ⎤ ⎢ ⎥ M ⎢ ⎥ ⎢ΔVq ⎥ = ⎢ ⎥ ⎢ M⎥ ⎢ΔV ⎥ ⎣ n⎦

⎡Z11 L Z1q L Z1n ⎤ ⎡ 0 ⎤ ⎢ ⎥⎢ ⎥ M M M M ⎢ ⎥⎢ ⎥ ⎢Z L Z L Z ⎥ ⎢-If ⎥ q1 qq qn ⎢ ⎥⎢ ⎥ ⎢M M M ⎥ ⎢ M⎥ ⎢ ⎥ ⎢ 0⎥ ⎢⎣Zn1 L Znq L Znn ⎥⎦ ⎣ ⎦ NPTEL

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Indian Institute of Technology, Kharagpur V0bus and Zbus both are known, therefore we can find If in the following manner.

V = V − Z1qI f 1

0 1

f

.......................... V = V − Z qqI f q

0 q

f

.......................... V = V − ZnqI f n

0 n

f


V =ZI f q

f f

Z I = V − Z qqI f f

0 q

f

I = ( Z + Z qq ) V f

−1

f

0 q

Vi = V − Ziq ( Z + Z qq ) V f

0 i

−1

f

V = Z ( Z + Z qq ) V f q

f

f

−1

0 q

0 q

for i ≠ q

Indian Institute of Technology, Kharagpur Example:Two synchronous machines are connected through three-phase transformers to the transmission line. The rating and reactance of the machines and transformers are Machine 1 and 2 : 100 MVA,20KV; X d ′′ = X 1 = X 2 X 0n

''

X d = X1 = X 2 = 12%; X 0 = 5%; Xn = 4%

Transformer T1 and T2 : 100MVA,20/400KV; X=7%. On a base of 100MVA,400 KV the line reactances are X1=X2=15% and X0=50%. NPTEL

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Indian Institute of Technology, Kharagpur The system is operating at normal voltage without pre-fault currents when a bolted(Zf =0) three phase fault occurs at bus 3. Determine the sub-transient fault current.

Gen

1 T1

2

Line NPTEL

3

T2 4

A.K. Sinha

Motor 14/36

Indian Institute of Technology, Kharagpur 1

2

3 4 j0.12 j0.07 j0.15 F j0.07 j0.12 Eg1

= 1.0∠00

Ref

Em1 = 1.0∠00

Positive-sequence network 1 1

YBUS =

2 3 4

2

3

4

0 0 ⎡-22.6190j 14.2860j ⎤ ⎢ 14.2860j -20.9530j 6.6670j ⎥ 0 ⎢ ⎥ ⎢ 0 6.6670j -20.9530j 14.2860j ⎥ ⎢ ⎥ 0 0 14.2860j -22.6190j ⎣ ⎦ NPTEL

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Indian Institute of Technology, Kharagpur On inverting YBUS we get ZBUS 1 1

Zbus =

2 3 4

⎡ j.0928 ⎢ j.0769 ⎢ ⎢ j.0430 ⎢ ⎣ j.02716

2

3

4

j.0769 j.043 j.0271⎤ j.1218 j.0681 j.0430 ⎥⎥ j.0681 j.1218 j.0769 ⎥ ⎥ j.043 j.0769 j.0928 ⎦

For fault at Bus 3 fault Current If is given by −1

I = ( Z + Z 33 ) V f

f

0 3

NPTEL

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8/36

In this case Zf

= 0; Z33 = 0.1218 and V = 1.0∠0 0 3

Therefore,

0

1.0∠0 I = = 8.210 p.u. 0.1218 Bus Voltages are calculated as: 0

f

V = V − Zi3 ( Z qq ) f 1

0 1

= 0.6469 p.u.

−1

j0.043 V = 1.0 x(1.0) j0.1218 0 3


Bus Impedance (ZBUS) Matrix ZBUS can not be formed by inspection like the YBUS matrix. ZBUS formation requires assembling the impedance matrix step by step. ZBUS matrix provides an open circuit description of the network.

NPTEL

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4/39

Indian Institute of Technology, Kharagpur Network equation can be written as:

⎡V1 ⎤ ⎢ ⎥ M ⎢ ⎥ ⎢Vp ⎥ = ⎢ ⎥ ⎢ M⎥ ⎢V ⎥ ⎣ n⎦

⎡Z11 L ⎢ ⎢M ⎢Z L q1 ⎢ ⎢M ⎢ L Z n1 ⎣

Z1p L Z1n ⎤ ⎡ I1 ⎤ ⎥ ⎢ ⎥ M M ⎥ ⎢ M⎥ Zpp L Zqn ⎥ ⎢Ip ⎥ ⎥ ⎢ ⎥ M M ⎥ ⎢ M⎥ ⎥ ⎢ ⎥ Znp L Znn ⎦ ⎣ In ⎦

NPTEL

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Indian Institute of Technology, Kharagpur All the voltages are referred to a common reference and all the currents are injections into the bus. If we make all current injections zero except at bus p: ⎡Z ⎤

⎡V1 ⎤ ⎢ ⎥ ⎢ M⎥ ⎢Vp ⎥ = ⎢ ⎥ ⎢ M⎥ ⎢V ⎥ ⎣ n⎦

1p

⎢ ⎥ ⎢M ⎥ ⎢Z ⎥ x I ⎢ pp ⎥ p ⎢M ⎥ ⎢ ⎥ ⎢⎣Znp ⎥⎦

NPTEL

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Indian Institute of Technology, Kharagpur This defines the p-th column elements as:

Zpi = (Vi /Ip )Ik =0; k ≠p i = 1,.....,n Since all the impedance elements are defined with all the nodes open circuited except one the impedance elements are called open circuit driving point and open circuit transfer impedances. Zpi = Open circuit driving point impedance, p=i = Open circuit transfer impedance, p ≠ i NPTEL

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Indian Institute of Technology, Kharagpur The diagonal elements of the ZBUS matrix provides the Thevenin’s equivalent impedance of the network at that node. ZBUS matrix formation by simulation:

Ip

Vp

Vn

Vk

V1

NPTEL

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ZBUS Building Algorithm We assume that we are given ZBUS matrix for a k-bus system and we want to modify this matrix to include some more elements. In general these modification of adding new elements in the system can be of four types: 1. Adding a branch with impedance Zb from a new bus P to the reference node. NPTEL

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2. Adding Zb from a new bus P to an existing bus k. 3. Adding Zb from existing bus k to the reference node. 4. Adding Zb between two existing bus j and k.

NPTEL

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Modification of an existing Zbus Type 1. Adding Zb from a new bus P to the reference node.

Ip P

Zb

Original network with the reference node extracted 0 NPTEL

Reference A.K. Sinha

11/39


⎡ V10 ⎤ ⎡ ⎢ 0⎥ ⎢ V ⎢ 2⎥ ⎢ ⎢ M⎥ = ⎢ Z orig ⎢ 0⎥ ⎢ ⎢ VN ⎥ ⎢ ⎢ V ⎥ P ⎢0 0 L ⎣ p⎦ ⎣

P

0 ⎤ ⎡ I1 ⎤ ⎢I ⎥ ⎥ 0 ⎥ ⎢ 2⎥ M⎥ ⎢ M⎥ ⎥ ⎢ ⎥ 0 ⎥ ⎢IN ⎥ 0 Zb ⎥⎦ ⎢⎣ Ip ⎥⎦

Zbus(new) NPTEL

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Indian Institute of Technology, Kharagpur Type 2. Adding Zb from a new bus P to an existing bus k Ik

Ik + Ip

k

Ip P

Zb

Original network with bus k and the reference node extracted 0

NPTEL

Reference

A.K. Sinha

13/39


Vk = I1Zk1 + I2 Zk2 + .. + (Ik + Ip )Zkk . + IN ZkN 0 k

Vk = V + Ip Z kk 0 k

Vp = V + Ip Zkk + Ip Zb Vp = I1Zk1 + I2 Zk2 + ... + IN ZkN + Ip ( Zkk + Zb ) 0 k

V Vi = I1Z i1 + I2 Z i2 + .. + (Ik + Ip )Z ik . + IN Z iN NPTEL

A.K. Sinha

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Indian Institute of Technology, Kharagpur P

⎡ V1 ⎤ ⎡ ⎢ ⎥ ⎢ ⎢ V2 ⎥ ⎢ ⎢ M⎥ = ⎢ ⎢ ⎥ ⎢ ⎢ VN ⎥ ⎢ ⎢ V ⎥ P ⎢Z ⎣ p ⎦ ⎣ k1 Zk2

⎤ ⎥ ⎥ M ⎥ ⎥ ZNk ⎥ Zkk + Zb ⎥⎦ Z1k Z 2k

Z orig L

ZkN

⎡ I1 ⎤ ⎢I ⎥ ⎢ 2⎥ ⎢ M⎥ ⎢ ⎥ ⎢IN ⎥ ⎢ Ip ⎥ ⎣ ⎦

Zbus(new) NPTEL

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Indian Institute of Technology, Kharagpur Type 3. Adding Zb from existing bus k to the reference node. k Ik + Ip

Zb

Original network with bus k and the reference node extracted

P

0

Reference

1. Type 2 modificationÆ Add Zb from existing bus k to new bus P 2. Connect bus P to reference. NPTEL

A.K. Sinha

16/39

Indian Institute of Technology, Kharagpur P

Step 1:

⎡ V1 ⎤ ⎡ ⎢ ⎥ ⎢ ⎢ V2 ⎥ ⎢ ⎢ M⎥ = ⎢ ⎢ ⎥ ⎢ ⎢ VN ⎥ ⎢ ⎢ V ⎥ P ⎢Z ⎣ p ⎦ ⎣ k1 Zk2

⎤ ⎥ ⎥ M ⎥ ⎥ ZNk ⎥ Zkk + Zb ⎥⎦ Z1k Z 2k

Z orig L

ZkN

⎡ I1 ⎤ ⎢I ⎥ ⎢ 2⎥ ⎢ M⎥ ⎢ ⎥ ⎢IN ⎥ ⎢ Ip ⎥ ⎣ ⎦

Zbus(new) Bus P is connected to ref. Æ Vp = 0 NPTEL

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Indian Institute of Technology, Kharagpur Step 2:

⎡ V1 ⎤ ⎡ ⎢ ⎥ ⎢ ⎢ V2 ⎥ ⎢ ⎢ M⎥ = ⎢ ⎢ ⎥ ⎢ ⎢ VN ⎥ ⎢ ⎢ 0 ⎥ P ⎢⎣ Zk1 Zk2 ⎣ ⎦

P

⎤ ⎥ ⎥ M ⎥ ⎥ ZNk ⎥ Zkk + Zb ⎥⎦ Z1k Z 2k

Z orig L

ZkN

⎡ I1 ⎤ ⎢I ⎥ ⎢ 2⎥ ⎢ M⎥ ⎢ ⎥ ⎢IN ⎥ ⎢ Ip ⎥ ⎣ ⎦

Zbus(new) Eliminating last row and column, we get NPTEL

A.K. Sinha

18/39


0 = I1Zk1 + I2 Zk2 + ... + IN ZkN + Ip ( Zkk + Zb )

1 Ip = (I1Zk1 + I2 Zk2 + ... + IN ZkN ) ( Zkk + Zb ) Vi = I1Zi1 + I2 Zi2 + .. + Ik Zik + . + INZiN + Ip Zik Vi = I1Zi1 + I2 Zi2 + .. + Ik Zik + . + INZiN 1 (I1Zk1 + I2 Zk2 + ... + INZkN )Zik ( Zkk + Zb ) NPTEL

A.K. Sinha

19/39


1 1 Vi =I1(Zi1 Zk1Z1k ) +I2 (Zi2 Zk2Z2k ) ( Zkk +Zb ) ( Zkk +Zb ) 1 +..+IN(ZiN ZkNZNk ) ( Zkk +Zb ) Zik Zkj Zij(new ) = Zij(OLD) Zkk + Zb

ZBUS(new )

1 = ZBUS(OLD) Zkk + Zb NPTEL

⎡ Z1k ⎤ ⎢ M⎥ Z L Z kN ] ⎢ ⎥ [ k1 ⎢⎣ ZNk ⎥⎦ A.K. Sinha

20/39

Indian Institute of Technology, Kharagpur Case 4. Adding Zb between two existing bus j and k Ij Ik

j

k

Ij + Ib Ib

Zb Ik - Ib

Original network with Buses j , k and the reference node extracted 0

NPTEL

Reference

A.K. Sinha

21/39


Vi = Z i1I1 + ... + Z ij (I j + Ib ) L + Z ik (Ik - Ib ) + ... + Z iN IN Vi = Z i1I1 + ... + Z ijI j L + Z ik Ik + ... + Z iN IN + (Z ij - Z ik )Ib

NPTEL

A.K. Sinha

22/39


V j = Z j1I1 + ... + Z jjI j + Z jk Ik + ... + Z jN IN + ( Z jj - Z jk ) Ib Vk = Z k1I1 + ... + Z kjI j + Z kk Ik + ... + Z kNIN + ( Z kj - Z kk ) Ib = Vj + Z bIb

0 = -(Z k1I1 + ... + Z kjI j + Z kk Ik + ... + Z kNIN + ( Z kj - Z kk ) Ib ) + Z j1I1 + ... +Z jjI j + Z jk Ik + ... + Z jNIN + ( Z jj - Z jk ) Ib + Z bIb NPTEL

A.K. Sinha

23/39


0 = (Z j1 - Z k1 )I1 + L + (Z jj - Z kj )I j + L + (Z jk - Z kk )Ik + L + (Z jN - Z kN )I1 +

(Z

b

+ Z jj + Z kk - Z jk - Z kj ) Ib

NPTEL

A.K. Sinha

24/39


⎡ V1 ⎤ ⎡ ⎢M⎥ ⎢ ⎢ ⎥ ⎢ ⎢ Vj ⎥ ⎢ ⎢ ⎥ ⎢ Z orig ⎢ Vk ⎥ = ⎢ ⎢ M⎥ ⎢ ⎢ ⎥ ⎢ ⎢ VN ⎥ ⎢ ⎢ 0 ⎥ ⎢⎢( row j - row k ) of Z orig ⎣ ⎦ ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ( col.j - col.k ) ⎥ ⎥ of Z orig ⎥ ⎥ Zb + Z jj + Zkk - 2Z jk ⎥⎥⎦

⎡ I1 ⎤ ⎢M ⎥ ⎢ ⎥ ⎢ Ij ⎥ ⎢ ⎥ ⎢ Ik ⎥ ⎢ M⎥ ⎢ ⎥ ⎢ IN ⎥ ⎢I ⎥ ⎣ b⎦

Eliminating last row and column, we get NPTEL

A.K. Sinha

25/39


ZBUS(new )

1 = ZBUS(OLD) x Z jj + Zkk + Zb - 2Z jk

⎡ Z1j - Z1k ⎤ ⎢ ⎥ M ⎥ ⎡⎣(Z j1 - Zk1 ) L (Z jN - ZkN )⎤⎦ ⎢ ⎢ ZNj - ZNk ⎥ ⎣ ⎦

NPTEL

A.K. Sinha

26/39

Indian Institute of Technology, Kharagpur Modification of existing Zbus Case Add branch Zb from

Zbus(new)

Reference node to new bus P

1

P Zoring Zb 0

NPTEL

⎡ ⎢ Z oring ⎢ ⎢ ⎢ P ⎣0 L

A.K. Sinha

P 0⎤ M⎥⎥ 0⎥ ⎥ 0 Zb ⎦

27/39


Case Add branch Zb from

Zbus(new)

Existing bus k to new bus P P

2

k

Zoring

k

P

Zb

k ⎡ Zoring col.k ⎤ P ⎢row k Z + Z ⎥ kk b⎦ ⎣

NPTEL

A.K. Sinha

0

28/39


Case Add branch Zb from Existing bus k to reference node

• Repeat Case 2 and P

3

k

Zoring

Zbus(new)

Zb

• Remove row p and column p by Kron reduction

0

NPTEL

A.K. Sinha

29/39

Indian Institute of Technology, Kharagpur Case Add branch Zb from

• From the matrix

Existing bus j to existing bus k

4

Zoring

j

q q

k

Zbus(new)

Zb

Zorig col.j- col.k⎤ ⎡ q ⎢⎣row j - row k Zth.jk + Zb ⎥⎦

Where Z th.jk = Z jj + Zkk - 2Z jk and

0 (Node q is temporary)

NPTEL

• Remove row q and column q by Kron reduction A.K. Sinha

30/39

Indian Institute of Technology, Kharagpur Example: Two synchronous machines are connected through three-phase transformers to the transmission line as shown in figure. The rating and reactance of the machines and transformers are: Machine 1 and 2: 100 MVA, 20KV; X′′d = X1 = X 2 = 12% X 0 = 5%

Xn = 4%

Transformers T1 and T2: 100 MVA,

20 Δ / 400 Y kv; NPTEL

X = 8% A.K. Sinha

31/39

Indian Institute of Technology, Kharagpur On a chosen base of 100 MVA, 345 KV in the transmission line circuit the line reactances are X1=X2=15% and X0=50%. Find the positive sequence bus impedance matrix by means of the Zbus building algorithm. Machine 1

1

T1 2

3

T2

4

Machine 2

Single line diagram of the system NPTEL

A.K. Sinha

32/39

Indian Institute of Technology, Kharagpur +ve sequence network 1

1

+

j.12

2

2

j.07

3

j.15

3

4

4

j.07 j.12

Vf

Vf

−

5

+ −

reference

NPTEL

A.K. Sinha

33/39

Indian Institute of Technology, Kharagpur Formation of Zbus matrix for positive sequence network Step 1: Add branch 1 to reference node 1 1 [j0.12]

Step 2: Add branch 2 between bus1 and bus2 1 1 ⎡ j.12

⎢

2 ⎣ j.12 NPTEL

2

j.12 ⎤ j.19 ⎥⎦ A.K. Sinha

34/39

Indian Institute of Technology, Kharagpur Step 3: Add branch 3 between buses 2 and 3 1 1 2 3

2

⎡ j.12 ⎢ j.12 ⎢ ⎢⎣ j.12

j.12 j.19 j.19

NPTEL

3

j.12 ⎤ j.19 ⎥⎥ j.34 ⎥⎦

A.K. Sinha

35/39


Step 4: Add branch 4 between buses 3 and 4

1 2 3 4

1

2

3

⎡ j.12 ⎢ j.12 ⎢ ⎢ j.12 ⎢ ⎣ j.12

j.12

j.12

j.19

j.19

j.19

j.34

j.19

j.34

NPTEL

4

j.12 ⎤ j.19 ⎥⎥ j.34 ⎥ ⎥ j.41⎦

A.K. Sinha

36/39

Indian Institute of Technology, Kharagpur Step 5: Add branch 5 between bus 4(old) and reference ⎡ Z14 ⎤ ⎢Z ⎥ 1 24 ⎥ ⎢ [Z 41 Z 42 Z 43 Z 44 ] Znew = Z old × Z 44 + Zb ⎢ Z 34 ⎥ ⎢Z ⎥ ⎣ 44 ⎦ ⎡ j.12 ⎤ ⎢ j.19 ⎥ 1 ⎢ ⎥ [ j.12 = Z old j.41+ j.12 ⎢ j.34 ⎥ ⎢ j.41⎥ ⎣ ⎦ NPTEL

j.19

A.K. Sinha

j.34

j.41]

37/39

Indian Institute of Technology, Kharagpur ⎡ j.12 ⎢ j.12 =⎢ ⎢ j.12 ⎢ ⎣ j.12

j.12

j.12

j.19

j.19

j.19

j.34

j.19

j.34

⎡ j.0928 ⎢ j.077 =⎢ ⎢ j.043 ⎢ j.0272 ⎣

j.12 ⎤ ⎡ j.0272 j.19 ⎥⎥ ⎢⎢ j.0430 j.34 ⎥ ⎢ j.077 ⎥ ⎢ j.41⎦ ⎣ j.0928

j.0928 ⎤ j.0681 j.1219 j.1469 ⎥⎥ j.1219 j.2181 j.2630 ⎥ ⎥ j.1469 j.2630 j.3172 ⎦ j.0430

j.077

j.0272 ⎤ j.1219 j.0681 j.0431⎥ ⎥ j.0681 j.1219 j.077 ⎥ j.0431 j.077 j.0928 ⎥⎦ j.077

j.043

NPTEL

A.K. Sinha

38/39


Asymmetrical Fault Analysis of Large Power Systems The system under study has n buses. It is initially operating in a symmetrical normal state. All pre-fault bus voltages and power flows are assumed to be known. A fault occurs at bus q. Fault can be represented as a fault impedance matrix Zf or fault ittance matrix Yf . NPTEL

A.K. Sinha

8/36


n-bus power system +

+

Bus q a b c

+

f f f Currents Iaq , Ibq , Icq

f aq

f f V Vbq Vcq

-

- -

Zf

Fault impedance matrix Z f


Network Modeling using Sequence Bus Impedance Matrices • Network is separately modeled for each of the three sequence systems (I.e., there will be three different bus impedance matrices). • The network matrices are assembled into a Sequence bus impedance matrix, Zs,bus, of dimension 3n × 3n.


Z + bus

⎡ Z +11 ⎢Z +21 ⎢ = ⎢ M ⎢ ⎢⎣ Z +n1

Z +1n ⎤ ⎥ Z +22 L Z +2n ⎥ M M M ⎥ ⎥ Z + n2 L Z + nn ⎥⎦ Z +12 L

Z − bus = Z + bus The zero-sequence network will be vastly different from both


Z s,bus

⎡ Z s11 ⎢Z ⎢ s21 ⎢ K =⎢ ⎢ K ⎢ K ⎢ ⎢⎣ Z sn1

Z s12 Z s22

L L

K K K

K K K

L

L

Z s1n ⎤ ⎥ Z s2n ⎥ K ⎥ ⎥ K ⎥ K ⎥ ⎥ Z snn ⎥⎦

Where each element of the Z s,bus matrix is a 3X3 matrix Z s,ij

⎡ Z 0ij ⎢ = ⎢0 ⎢0 ⎣

0 Z + ij 0

0 ⎤ ⎥ 0 ⎥ ⎥ Z −ij ⎦

and voltage and current vectors are given as :


V s ,b u s

⎡ V0 1 ⎤ ⎢V ⎥ ⎢ +1 ⎥ ⎢ V−1 ⎥ ⎢ ⎥ K ⎢ ⎥ ⎢ M ⎥ ⎢ ⎥ ⎡ Vs1 ⎤ ⎢ K ⎥ ⎢ M ⎥ ⎢ V ⎥ 0i ⎢ ⎥ ⎢ ⎥ @ ⎢ V+i ⎥ = ⎢ Vsi ⎥ ⎢ ⎥ ⎢V ⎥ M ⎢ ⎥ ⎢ −i ⎥ ⎢⎣ V s n ⎥⎦ ⎢ K ⎥ ⎢ ⎥ M ⎢ ⎥ ⎢ K ⎥ ⎢ ⎥ V ⎢ 0n ⎥ ⎢V ⎥ ⎢ +n ⎥ ⎢⎣ V − n ⎥⎦

and

I s ,b u s

⎡ I0 1 ⎤ ⎢I ⎥ ⎢ +1 ⎥ ⎢ I− 1 ⎥ ⎢ ⎥ K ⎢ ⎥ ⎢ M⎥ ⎢ ⎥ ⎡ Is 1 ⎤ ⎢K ⎥ ⎢ M⎥ ⎢I ⎥ 0i ⎢ ⎥ ⎢ ⎥ @ ⎢ I+ i ⎥ = ⎢ Is i ⎥ ⎢ ⎥ ⎢I ⎥ M ⎢ ⎥ ⎢ −i ⎥ ⎢⎣ I s n ⎥⎦ ⎢K ⎥ ⎢ ⎥ M ⎢ ⎥ ⎢K ⎥ ⎢ ⎥ I ⎢ 0n ⎥ ⎢I ⎥ ⎢ +n ⎥ ⎢⎣ I − n ⎦⎥

Vs,bus = Z s,bus Is,bus Vsi = Z si1Is1 + Z si2 Is2 + ... + Z siiIi + ... + Z sinIsn Using the Thevenin’s theorem the postfault Bus voltages are given by f s,bus

V

=V

0 s,bus

+Z

f s,bus s,bus

I


Pre-fault bus voltages

Vs0,bus

⎡ 0 ⎤ ⎢ V0 ⎥ ⎢ +1 ⎥ ⎢ 0 ⎥ ⎢ ⎥ ⎢L ⎥ ⎢ 0 ⎥ ⎢ 0 ⎥ ⎢ V+ 2 ⎥ =⎢ 0 ⎥ ⎢ ⎥ ⎢L ⎥ ⎢ M⎥ ⎢ ⎥ ⎢L ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ V+0n ⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦

Indian Institute of Technology, Kharagpur Fault Current injection:

I fs ,b u s

⎡ 0 ⎤ ⎢ 0 ⎥ ⎢ ⎥ ⎢ M⎥ = ⎢ f ⎥ ⎢ − I s ,q ⎥ ⎢ M⎥ ⎢ ⎥ ⎢⎣ 0 ⎥⎦

f s,q

I

f ⎡I0q ⎤ ⎢f ⎥ = ⎢I+ q ⎥ ⎢f ⎥ ⎢⎣I− q ⎥⎦

qth component

Indian Institute of Technology, Kharagpur V0s,bus and Zs,bus both are known. We can, therefore, find Ifsq as follows:

V = V −Z f s1

0 s1

f s1q sq

I

.......................... V = V −Z f sq

0 sq

f sqq sq

I

.......................... V = V −Z f sn

0 sn

f snq sq

I


f f s sq

ZI f sq

I

= V −Z 0 sq

f sqq sq

I

= ( Z + Z sqq ) V −1

f s

0 sq

V = V − Z siq ( Z + Z sqq ) V f si

0 si

V =Z f sq

−1

f s

f s

(Z

f s

+ Z sqq ) V −1

0 sq

0 sq

for i ≠ q

Indian Institute of Technology, Kharagpur • All relations involve three-dimensional vectors and 3 × 3 matrices. The matrices Zsij are diagonal, but the matrix Zfs is not. In any case, all the involved matrix operations can be easily programmed on a digital computer. • The prefault bus voltages V0i are either obtained from a load flow analysis or, are usually set equal to 1.0 pu; i.e.,

⎡0 ⎤ ⎢ ⎥ 0 Vsi ≈ ⎢ 1⎥ ⎢⎣0 ⎥⎦

for i = 1,...,n

Indian Institute of Technology, Kharagpur • Z+bus, Z-bus, and Z0bus can be obtained either by inversion of the corresponding bus ittance matrices or by using the Zbus building algorithms.

Indian Institute of Technology, Kharagpur Fault Impedance matrix +

b

+ + f aq

f cq

f bq

V V V

f aq

I

I

I

Za

Zb

Zc

f bq

Zg -

-

a

-

f cq

f Inq

Faulted Bus q

c

1 Ya = Za

1 Yb = Zb 1 Yc = Zc

1 Yg = Zg


The Fault Impedance Matrix Z fs

V = I Za + (I + I + I f aq

f aq

f aq

f bq

f cq

V = I Zb + ( I + I + I f bq

f bq

f aq

f bq

f cq

f cq

f aq

f bq

V = I Zc + (I + I + I

)Z )Z )Z

f cq

f cq

g

g

g

Indian Institute of Technology, Kharagpur f ⎡ Vaq ⎤ ⎡ Z a + Zg ⎢ f ⎥ ⎢ ⎢ Vbq ⎥ = ⎢ Z g f ⎥ ⎢ Vcq ⎢ Zg ⎣ ⎦ ⎣

Zg Zb + Z g Zg

f ⎤ ⎤ ⎡Iaq ⎥⎢ f ⎥ Z g ⎥ ⎢Ibq ⎥ f ⎥ ⎢ ⎥ Z c + Z g ⎦ ⎣Icq ⎦

Zg

Where fault impedance matrix

⎡Z a + Zg ⎢ f Z = ⎢ Zg ⎢ Zg ⎣

Zg Zb + Z g Zg

⎤ ⎥ ⎥ Z c + Z g ⎥⎦ Zg Zg

Indian Institute of Technology, Kharagpur Eq. can be written in compact form f Pq

⎡1 T = ⎢a2 ⎢ ⎢⎣ a

f f Pq

V =Z I f sq

f

f sq

TV = Z TI

V = T Z TI @Z I f sq

-1

f

f sq

f f s sq

1 1⎤ a 1⎥ ⎥ 2 a 1⎥⎦

⎡1 a 1⎢ -1 T = ⎢1 a2 3 ⎢⎣1 1

a2 ⎤ ⎥ a⎥ 1 ⎥⎦

Indian Institute of Technology, Kharagpur ⎡ Za + Zb + Zc Za +a2 Zb +aZc Za +aZb +a2 Zc ⎤ 1⎢ ⎥ 2 2 = ⎢Za +aZb +a Zc Za + Zb + Zc Za +a Zb +aZc ⎥ 3 ⎢Za +a2 Zb +aZc Za +aZb +a2Zc Za + Zb + Zc +9Zg ⎥ ⎣ ⎦

For some type of faults some of the impedances take on infinite values. Some elements of the Zsf matrix become undefined. In such cases we must work with its inverse, the SC transformed fault ittance matrix.


I = (Z f sq

Ysf =

)

f -1 s

f sq

f s

f sq

V @Y V

1 × Ya + Yb + Yc + Yg

1 ⎡ 1 2 Y Y a Yb + aYc ) + ( Y Y Y Y + + ( ) g a g a b c ⎢ 3 ⎢ 3 − ( Yb Yc + aYa Yb + a2 Ya Yc ) ⎢ + ( Ya Yb + Yb Yc + Yc Ya ) ⎢ 1 ⎢ 1 Y Y + aY + a2 Y Yg ( Ya + Yb + Yc ) g( a b c) ⎢3 3 ⎢ 2 + ( Ya Yb + Yb Yc + Yc Ya ) ⎢ − ( Yb Yc + a Ya Yb + aYa Yc ) ⎢ 1 ⎢ 1 Yg ( Ya + a2 Yb + aYc ) Yg ( Ya + aYb + a2 Yc ) ⎢ 3 3 ⎢ ⎣⎢

⎤ 1 2 Yg ( Ya + aYb + a Yc ) ⎥ 3 ⎥ ⎥ ⎥ 1 ⎥ Yg ( Ya + a2 Yb + aYc ) ⎥ 3 ⎥ ⎥ ⎥ 1 Yg ( Ya + Yb + Yc ) ⎥ ⎥ 3 ⎥ ⎦⎥


V = V − Z sqq Y V f sq

0 sq

f s

V = (U + Z sqq Y

f −1 s

f sq

f sq

I

0 sq

V

= Y V = Y (U + Z sqq Y f s

f sq

V = V −Z I f si

)

f sq

0 si

f siq sq

)

f −1 s

f s

0 sq

V

= V − Z siq Y (U + Z sqq Y 0 si

f s

)

f −1 s

0 Vsq i≠q

Example: Two synchronous machines are connected through three-phase transformers to the transmission line as shown in figure. The rating and reactance of the machines and transformers are: Machine 1 and 2: 100 MVA, 20KV; X′′d = X1 = X 2 = 12% X 0 = 5%

Xn = 4%

Transformers T1 and T2: 100 MVA,

20Δ/400Ykv; NPTEL

X = 8% A.K. Sinha

6/37

On a chosen base of 100 MVA, 345 KV in the transmission line circuit the line reactances are X1=X2=15% and X0=50%. Draw each of the three sequence networks and find fault current and post fault bus voltages for a bolted L-G fault on bus 2 using the Zbus algorithm. Machine 1

1

T1 2

3

T2

4

Machine 2

Single diagram of the Singleline line diagram ofsystem the system NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur +ve sequence network 1

1

+

j.12

2

2

j.07

3

j.15

3

4

4

j.07 j.12

Vf

Vf

−

5

+ −

reference

NPTEL

A.K. Sinha

33/39

Indian Institute of Technology, Kharagpur Zero sequence network 1

1

j.05 j.12

2

2

j.07

3

j.15

3

4

j.07

4 j.05 5 j.12

reference

NPTEL

A.K. Sinha

33/39

Zbus for the positive sequence network was already built in the previous lesson. Also we have 1 1

Z

(1) bus

=Z

(2) bus

=

2 3 4

⎡ j.0928 ⎢ j.0769 ⎢ ⎢ j.0430 ⎢ ⎣ j.02716

2

3

4

j.0769 j.043 j.0271⎤ j.1218 j.0681 j.0430 ⎥⎥ j.0681 j.1218 j.0769 ⎥ ⎥ j.043 j.0769 j.0928 ⎦

NPTEL

A.K. Sinha

6/37

Indian Institute of Technology, Kharagpur Zbus for the zero sequence network using the building algorithm is given by

0 Z bus

⎡ j.1406 ⎢ j.1284 =⎢ ⎢ j.0417 ⎢ j.0295 ⎣

j.1284 j.1813 j.0588 j.0417

NPTEL

j.0417 j.0588 j.1813 j.1284

j.0295 ⎤ j.0417 ⎥ ⎥ j.1284 ⎥ j.1406 ⎥⎦

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur For single line to ground fault, (1) Z b = Z c = ∞ (∴ Yb = Yc = 0) (2) Z g = 0 (3)

Za = Z

(∴ Yg = ∞) f

f Y Since Yg = ∞ s

1 (∴Ya = f @Y f ) Z

Reduces to

NPTEL

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur 2 2 ⎡ Ya + Yb + Yc + + + + Ya α Yb α Yc Ya α Yb α Yc ⎤ 1⎢ ⎥ 2 2 f Ys = ⎢Ya + α Yb + α Yc Ya + Yb + Yc Ya + α Yb + α Yc ⎥ 3 ⎢⎣Ya + α 2Yb + α Yc Ya + α Yb + α 2Yc Ya + Yb + Yc ⎥⎦

Upon substitution of Ya = Y f and Yb = Yc = 0, simplifies to ⎡1 1 1⎤ Y ⎢ f 1 1 1⎥ Ys = ⎥ 3 ⎢ ⎢⎣1 1 1⎥⎦ f

NPTEL

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur ⎡ ⎡1 0 0 ⎤ ⎡ Z + qq ⎢⎢ ⎢ f ⎥ Vsq = ⎢ 0 1 0 + ⎢ 0 ⎢ ⎥ ⎢ ⎢⎣0 0 1 ⎥⎦ ⎢ 0 ⎣ ⎣

0 Z − qq 0

0 ⎤ f ⎥Y 0 ⎥ 3 Z 0 qq ⎦⎥

−1

0 ⎤ ⎡ ⎤ 1 1 1 V ⎡ ⎤ q ⎢1 1 1⎥ ⎥ ⎢ 0 ⎥ ⎢ ⎥⎥ ⎢ ⎥ ⎢⎣1 1 1⎥⎦ ⎥⎦ ⎢⎣ 0 ⎥⎦

⎡ ⎤ Yf ⎢1 + 3( Z + Z ) ⎥ − qq 0 qq ⎥ ⎢ 0 f ⎢ ⎥ V Y q f − Z − qq Vsq = ⎢ ⎥ f Y 3 ⎢ ⎥ 1+ ( Z + qq + Z − qq + Z 0 qq ) ⎢ ⎥ 3 Yf − Z 0 qq ⎢ ⎥ 3 ⎢⎣ ⎥⎦ NPTEL A.K. Sinha 1/36

Indian Institute of Technology, Kharagpur For the short circuit at the faulted bus, ⎡1⎤ f Y /3 0 f f f ⎢ ⎥ 1 I sq = Ys Vsq = Vq f ⎢ ⎥ Y 1+ ( Z + qq + Z − qq + Z 0 qq ) ⎢⎣1⎥⎦ 3 The post fault voltages other than the faulted one are obtained as Vsif = Vsi0 − Z siq I sqf

NPTEL

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur ⎡Vi 0 ⎤ ⎡ Z + iq ⎢ ⎥ ⎢ =⎢ 0 ⎥−⎢ 0 ⎢⎣ 0 ⎥⎦ ⎢⎣ 0

0 Z − iq 0

0 ⎤ ⎥ f 0 ⎥ I sq Z oiq ⎥⎦

f I Substituting sq

⎡Vi 0 ⎤ ⎡ Z + iq ⎤ f Y /3 ⎢ ⎥ ⎢ ⎥ f 0 Vsi = ⎢ 0 ⎥ − Vq Z − iq ⎥ f ⎢ Y 1+ ( Z + qq + Z − qq + Z 0 qq ) ⎢ Z 0iq ⎥ ⎢⎣ 0 ⎥⎦ ⎣ ⎦ 3 NPTEL

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur For Zf = 0 above equations simplifies to, ⎡ Z − qq + Z 0 qq ⎤ V ⎢ ⎥ f Vsq = − Z − qq ⎥ f ⎢ Y 1+ ( Z + qq + Z − qq + Z 0 qq ) ⎢ − Z 0 qq ⎥ ⎣ ⎦ 3 0 q

1 I = Ys V = Z + qq + Z − qq + Z 0 qq f sq

f

f sq

NPTEL

⎡1⎤ ⎢1⎥ ⎢⎥ ⎢⎣1⎥⎦

A.K. Sinha

1/36


⎡ Z + iq ⎤ ⎡1 ⎤ 1 ⎢ ⎥ f ⎢ ⎥ Vsi = 0 − Z − iq ⎥ for i ≠ q ⎢ ⎢ ⎥ Z + qq + Z − qq + Z 0 qq ⎢⎣ Z 0iq ⎥⎦ ⎢⎣0 ⎥⎦ Since short circuit occur at bus 2, q = 2 0 0 ⎤ ⎡ Z +22 Z sqq = Z s 22 = ⎢ 0 Z −22 0 ⎥ ⎢ ⎥ 0 Z 022 ⎥⎦ ⎢⎣ 0 ⎡ j.1218 =⎢ 0 ⎢ ⎢⎣ 0

0 j.1218 0 NPTEL

0 ⎤ 0 ⎥ p.u. ⎥ j.1813⎥⎦

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur ⎡ j.1218 + j.1813⎤ 1 f ⎢ ⎥ − j.1218 Vs 2 = ⎥ j (.1218 + .1218 + .1813) ⎢ − j.1813 ⎢⎣ ⎥⎦ ⎡ .7133 ⎤ = ⎢ −.2866 ⎥ p.u ⎢ ⎥ ⎢⎣ −.4267 ⎥⎦ ⎡1⎤ 1 f f ⎢1⎥ I sq = I s 3 = j (.1218 + .1218 + .1813) ⎢ ⎥ ⎢⎣1⎥⎦ NPTEL

⎡ 2.35⎤ = − j ⎢ 2.35⎥ p.u ⎢ ⎥ ⎢⎣ 2.35⎥⎦

A.K. Sinha

1/36


Z s12

⎡ Z +12 =⎢ 0 ⎢ ⎢⎣ 0

Z s 32

⎡ Z +32 =⎢ 0 ⎢ ⎢⎣ 0

0 Z −12 0

0 Z −32 0

0 ⎤ 0 ⎥= ⎥ Z 012 ⎥⎦

0 ⎤ 0 ⎥= ⎥ Z 032 ⎥⎦

NPTEL

0 0 ⎤ ⎡.0769 j⎢ 0 .0769 0 ⎥ ⎢ ⎥ 0 .1284 ⎥⎦ ⎢⎣ 0

0 0 ⎤ ⎡.0681 j⎢ 0 .0681 0 ⎥ ⎢ ⎥ 0 .0588⎥⎦ ⎢⎣ 0

A.K. Sinha

1/36


Z s 42

⎡ Z +42 =⎢ 0 ⎢ ⎢⎣ 0

0 Z −42 0

0 ⎤ 0 ⎥= ⎥ Z 042 ⎥⎦

0 ⎤ ⎡.043 0 j ⎢ 0 .043 0 ⎥ ⎢ ⎥ 0 .0417 ⎥⎦ ⎢⎣ 0

⎡1 ⎤ ⎡ j.0769 ⎤ 1 f ⎢ ⎥ ⎢ j.0769 ⎥ Vs1 = 0 − ⎢ ⎥ (.1218 + .1218 + .1813) ⎢ ⎥ ⎢⎣0 ⎥⎦ ⎢⎣ j.1284 ⎥⎦ ⎡ .819 ⎤ = ⎢ −.181⎥ ⎢ ⎥ ⎢⎣ −.302 ⎥⎦ NPTEL

A.K. Sinha

1/36

Indian Institute of Technology, Kharagpur ⎡ .839 ⎤ ⎡1 ⎤ ⎡ j.0681⎤ ⎢ −.160 ⎥ 1 f ⎢ ⎥ ⎢ ⎥ = Vs 3 = 0 − j.0681 ⎢ ⎥ ⎢ ⎥ (.1218 + .1218 + .1813) ⎢ ⎥ ⎢⎣ −.138⎥⎦ ⎢⎣0 ⎥⎦ ⎢⎣ j.0588⎥⎦ ⎡1 ⎤ ⎡ j.043 ⎤ 1 f ⎢ ⎥ ⎢ j.043 ⎥ Vs 4 = 0 − ⎢ ⎥ (.1218 + .1218 + .1813) ⎢ ⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ j.0417 ⎥⎦

NPTEL

A.K. Sinha

⎡ .898 ⎤ = ⎢ −.101⎥ ⎢ ⎥ ⎢⎣ −.098⎥⎦

1/36


Thank You !

NPTEL

A.K. Sinha

36/36


POWER SYSTEM STABILITY


A.K. Sinha

1/35


POWER SYSTEM STABILITY Lesson Summary 1. Introduction 2. Classification of Power System Stability 3. Dynamic Equation of Synch. Machines

NPTEL

A.K. Sinha

2/35


Instructional Objective On completion of this lesson a student should be able to: A. Explain the concept of power system stability B. Classify stability problem into different types based on the phenomenon under consideration C. Develop dynamic equation of motion for synchronous machines NPTEL

A.K. Sinha

3/35

Indian Institute of Technology, Kharagpur Power System Operates in Synchronous Mode Power System is subjected to a wide range of disturbances (small and large) Loads and generation keep changing continually Faults and Network changes also take place sometimes

NPTEL

A.K. Sinha

4/35


Power System stability involves the study of the dynamics of the power system under disturbances. In simplest term power system instability can be seen as loss of synchronism (i.e., some synchronous machines going out of step) when the system is subjected to a particular disturbance.

NPTEL

A.K. Sinha

5/35

Indian Institute of Technology, Kharagpur CIGRE-IEEE Def. Of Power System Stability:Power system stability is the ability of an electric power system, for a given initial operating condition, to regain a state of operating equilibrium after being subjected to a physical disturbance, with most system variables bounded so that practically the entire system remains intact. Stability Æ involves study of dynamics of the system about an equilibrium- initial op. cond. NPTEL

A.K. Sinha

6/35

Indian Institute of Technology, Kharagpur Power system is highly non-linear system Æ Stability study Æ study of non-linear dynamics of a large system. Stability Æ Initial operating condition Nature and magnitude of disturbance Def. applies to an interconnected power system as a whole. Most of the time we are interested in stability of a particular generator or a group of generators NPTEL

A.K. Sinha

7/35

Indian Institute of Technology, Kharagpur A typical power system will consist of large number of devices having dynamic characteristicsÆ Generators, AVRs, Speed Governors, FACTS Devices, OLTC Transformers etc. Power System Æ a large multivariable dynamic system with varying characteristics (linear / nonlinear) and response rates. Stability Æ equilibrium between the opposing forces generated by the devices as a response to the disturbance NPTEL

A.K. Sinha

8/35

Indian Institute of Technology, Kharagpur Classification of Stability: Power system stability is essentially a single problem. High dimensionality and complexity of the problem Æ Simplifying assumptions to analyze specific types of problem Æ Classification of stability problem.

NPTEL

A.K. Sinha

9/35

Indian Institute of Technology, Kharagpur For a given network topology and initial operating condition depending on nature and magnitude of disturbance, imbalance between different sets of opposing forces may lead to different types of instability. Classification is done on the basis of: Physical nature of instability Æ main variable Size of the disturbance Æ method of analysis Time span of analysis Æ devices to be modeled NPTEL

A.K. Sinha

10/35

Indian Institute of Technology, Kharagpur Power System Stability

Rotor angle stability

Frequency stability

Voltage stability Small signal stability

Small Transient signal stability stability Short term

Short term / long term NPTEL

Large signal stability

Short term / long term

A.K. Sinha

11/35


Rotor Angle Stability Analysis Assumptions: Balanced three phase system and balanced disturbances considered Deviation of machine frequencies from synchronous frequency are small, dc offset current and harmonics are neglected. Network and impedance loads are at steady state Æ Voltage current and powers can be computed from power flow equations. NPTEL

A.K. Sinha

12/35


Dynamics of a Synchronous machine The kinetic energy of the rotor at synchronous machine is

1 KE = Jω2sm ×10-6 MJ 2 Where

J = rotor moment of inertia in kg-m2 ωsm = synchronous speed in rad (mech)/s NPTEL

A.K. Sinha

13/35


But

P ωs = ω sm = rotor speed in rad (elect)/s 2

Where P = number of machine poles 2

∴

1 ⎛2⎞ -6 KE = J ⎜ ⎟ ω s ×10 ω s 2 ⎝P⎠

1 = M ωs 2 NPTEL

A.K. Sinha

14/35


Where

⎛P⎞ M = J ⎜ ⎟ ω sm ×10- 6 ⎝2⎠ = Angular Momentum in MJ-s/elect rad.

We define the inertia constant H: KE stored at ω s (MJ) 1 H= ; GH = KE = M ω s MJ G (MVA rating of Mc.) 2

Where G = machine rating (base) in MVA (3phase). H = inertia constant in MJ/MVA or MWs/MVA NPTEL

A.K. Sinha

15/35


2GH GH M= = MJ- s/elect rad ωs πf GH = MJ- s/elect degree 180f

M is also called the inertia constant. Taking G as base, the inertia constant in pu is H 2 M(pu) = s /elect rad πf H = s2 /elect degree 180f NPTEL

A.K. Sinha

16/35

Indian Institute of Technology, Kharagpur Type of machine

Inertia Constant H Stored Energy in MW Sec MVA

Turbine Generator Condensing Non-condensing

1,800 rpm 3,000 rpm 3,000 rpm

Water wheel Generator Slow-speed (< 200 rpm) High-speed (> 200 rpm)

NPTEL

9–6 7–4 4–3 2–3 2–4

A.K. Sinha

17/35


Type of machine

Inertia Constant H Stored Energy in MW Sec MVA

Synchronous Condenser Large Small

1.25 1.00

Synchronous Motor with load varying from 1.0 to 5.0 and higher for heavy wheels

2.00

NPTEL

A.K. Sinha

18/35


Synchronous Machine Rotor Dynamics (The Swing Equation) d2θm J 2 = Tm - Te Nm dt

Where θm = angle in rad (mech)

Tm = turbine torque in Nm; it acquires a negative value for a motoring machine. Te = electromagnetic torque developed in Nm; it acquires negative value for a motoring machine. NPTEL

A.K. Sinha

19/35

Indian Institute of Technology, Kharagpur Pe

Te Pm

Generator Tm ωs

(a)

Pe

Tm Pm

Motor

(b)

ωs Ts Fig. Flow of mechanical and electrical powers in a synchronous machine NPTEL

A.K. Sinha

20/35


d2θm 6 Jω sm ×10 = Pm - Pe MW 2 dt Where Pm = mechanical power input in MW Pe = electrical power output in MW; stator copper loss is assumed negligible.

NPTEL

A.K. Sinha

21/35


⎛2⎞ J ⎜ ⎟ ω s ×10- 6 ⎝P⎠

d2θ e = Pm - Pe MW 2 dt

Where θe = angle in rad (elect) or

d2θ e M 2 = Pm - Pe dt

NPTEL

A.K. Sinha

22/35

Indian Institute of Technology, Kharagpur It is more convenient to measure the angular position of the rotor with respect to a synchronously rotating frame of reference. Let δ = θe – ωst; rotor angular displacement from synchronously rotating reference frame (called torque angle/power angle) d θe d2θ e dδ d2δ = + ωs ; = 2 dt dt dt dt 2 d2δ M 2 = Pm - Pe dt NPTEL

A.K. Sinha

23/35

Indian Institute of Technology, Kharagpur Dividing throughout by G, the MVA rating of the machine, 2

dδ M(pu) 2 = Pm - Pe ; dt

in pu of machine rating as base

Where

H M(pu) = πf 2

or

H dδ = Pm - Pe pu (Swing Equation) 2 πf dt NPTEL

A.K. Sinha

24/35

Indian Institute of Technology, Kharagpur Example: A 50 Hz. 20-pole hydro-electric generator rated at 500 MVA, 20 kV has H=2 MJ/MVA. (i) Determine its ωs and ωsm. (ii) Write the swing equation for this generator. (iii) The generator is initially working at Pm=Pe=1.0 pu with δ0=100 when a 3-phase short circuit occurs at its terminals which results in its electrical output reducing to 0 for t > 0. (iii) Determine its power angle δ, 2 cycles after the short circuit. Assume the mechanical input power Pm remains constant at 1.0 pu during this time. NPTEL

A.K. Sinha

25/35

Indian Institute of Technology, Kharagpur Solution: (i) For a 50 Hz. Generator ωs = 2π50 = 314 rad/s ωsm = (2/P) ωs = (2/20)314 = 31.4 rad/s (ii) or

H d2δ = Pm - Pe pu 2 πf dt 2 d2δ = Pm - Pe pu 2 π50 dt

(iii) Initial power angle d0=100 = 0.1745 rad dδ Also at t = 0 , = 0 and dt Pm (0+ ) = 1.0 pu and Pe (0+ ) = 0.0 pu +

NPTEL

A.K. Sinha

26/35

Indian Institute of Technology, Kharagpur Substituting the values in swing equation 2 d2δ = 1.0 - 0 pu; Integrating 2 π50 dt dδ 2π50 = 4 t +0 dt 2 δ(t) = 2π50 t + 0.1745 4 t = 2 cycles at 50Hz. = 40 ms 2 δ(0.04) = 2π50 (0.04) + 0.1745 4

= 0.30 rad = 17.20 NPTEL

A.K. Sinha

27/35

Indian Institute of Technology, Kharagpur Machine Connected to Infinite Bus

X transfer = X′d + X e E′ V Pe = sinδ = Pmax sinδ X transfer 2

H dδ = Pm - Pe pu 2 πf dt

NPTEL

A.K. Sinha

28/35


X′d

E′ ∠δ

Xe

Infinite bus Pe

+

o ∠ 0 V

Fig. Machine connected to infinite bus

NPTEL

A.K. Sinha

29/35


The electrical power transfer is given by expression

E′1 E′2 Pe = sinδ X′d + X e

NPTEL

A.K. Sinha

30/35


X′d1

Xe

X′d2

+ Pm1

+ 1

Pe1

Pe2

E′1 ∠δ1

2

Pm2 E′2 ∠δ2

Fig. Two-machine system NPTEL

A.K. Sinha

31/35


Pm1 = - Pm2 = Pm Pe1 = - Pe2 = Pe Pm1 - Pe1 d2δ 1 = πf 2 dt H1 2

d δ2 = πf 2 dt

Pm2 - Pe2 H2 NPTEL

Pm - Pe = πf H1 = πf

Pe - Pm H2

A.K. Sinha

32/35


d2 (δ1 - δ2 ) H1 + H2 = πf (Pm - Pe ) 2 dt H1H2 Heq d2δ = Pm - Pe 2 πf dt δ = δ1 - δ 2

NPTEL

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33/35


H1H2 Heq = H1 +H2 The electrical power transfer is given by expression

Pe =

E′1 E′2 X′d1 + X e + X′d2

sinδ

NPTEL

A.K. Sinha

34/35


Synchronous Machine Rotor Dynamics (The Swing Equation) d2θm J 2 = Tm - Te Nm dt

Where θm = angle in rad (mech)

Tm = turbine torque in Nm; it acquires a negative value for a motoring machine. Te = electromagnetic torque developed in Nm; it acquires negative value for a motoring machine. NPTEL

A.K. Sinha

19/35


Te Pm

Generator Tm ωs

(a)

Pe

Tm Pm

Motor

(b)

ωs Ts Fig. Flow of mechanical and electrical powers in a synchronous machine NPTEL

A.K. Sinha

20/35


d2θm 6 Jω sm ×10 = Pm - Pe MW 2 dt Where Pm = mechanical power input in MW Pe = electrical power output in MW; stator copper loss is assumed negligible.

NPTEL

A.K. Sinha

21/35


⎛2⎞ J ⎜ ⎟ ω s ×10- 6 ⎝P⎠

d2θ e = Pm - Pe MW 2 dt

Where θe = angle in rad (elect) or

d2θ e M 2 = Pm - Pe dt

NPTEL

A.K. Sinha

22/35

Indian Institute of Technology, Kharagpur It is more convenient to measure the angular position of the rotor with respect to a synchronously rotating frame of reference. Let δ = θe – ωst; rotor angular displacement from synchronously rotating reference frame (called torque angle/power angle) d θe d2θ e dδ d2δ = + ωs ; = 2 dt dt dt dt 2 d2δ M 2 = Pm - Pe dt NPTEL

A.K. Sinha

23/35

Indian Institute of Technology, Kharagpur Dividing throughout by G, the MVA rating of the machine, 2

dδ M(pu) 2 = Pm - Pe ; dt

in pu of machine rating as base

Where

H M(pu) = πf 2

or

H dδ = Pm - Pe pu (Swing Equation) 2 πf dt NPTEL

A.K. Sinha

24/35


Including Damping term in Swing Equation Damping torque produced by the damper winding is proportional to slip speed

Td α (ωe - ωs ) dθe d dδ (ωe - ωs ) = - ωs = (ωs t + δ) - ωs = dt dt dt dδ dδ Pd = Tdωe = k (ωe - ωs )ωe = k(ωs + - ωs )(ωs + ) dt dt 2

dδ dδ ⎛ dδ ⎞ Pd = k(ωs ) + k ⎜ ⎟ ≈ D dt dt ⎝ dt ⎠ NPTEL

A.K. Sinha

24/35

Damping power due to friction is usually small and it is ed for in the mechanical input power Pm. Therefore, the swing equation with damping included is given by : 2

H dδ dδ +D = (Pm - Pe ) pu 2 πf dt dt

Indian Institute of Technology, Kharagpur Example: A 50 Hz. 20-pole hydro-electric generator rated at 500 MVA, 20 kV has H=2 MJ/MVA. (i) Determine its ωs and ωsm. (ii) Write the swing equation for this generator. (iii) The generator is initially working at Pm=Pe=1.0 pu with δ0=100 when a 3-phase short circuit occurs at its terminals which results in its electrical output reducing to 0 for t > 0. Determine its power angle δ, 3 cycles after the short circuit. Assume the mechanical input power Pm remains constant at 1.0 pu during this time. NPTEL

A.K. Sinha

25/35

Indian Institute of Technology, Kharagpur Solution: (i) For a 50 Hz. Generator ωs = 2π50 = 314 rad/s ωsm = (2/P) ωs = (2/20)314 = 31.4 rad/s (ii) or

H d2δ = Pm - Pe pu 2 πf dt 2 d2δ = Pm - Pe pu 2 π50 dt

(iii) Initial power angle δ0=100 = 0.1745 rad dδ Also at t = 0 , = 0 and dt Pm (0+ ) = 1.0 pu and Pe (0+ ) = 0.0 pu +

NPTEL

A.K. Sinha

26/35

Indian Institute of Technology, Kharagpur Substituting the values in swing equation 2 d2δ = 1.0 - 0 pu; Integrating 2 π50 dt dδ 2π50 = 4 t + 0 ; Integrating again dt 2 δ(t) = 2π50 t + 0.1745 8 t = 3 cycles at 50Hz. = 60 ms 2 δ(0.06) = 2π50 (0.06) + 0.1745 8

= 0.3158 rad = 18.10 NPTEL

A.K. Sinha

27/35

Simplified machine model E' = V + rI + jX ′d Id + jX ′q Iq E' = V + jX ′d I

E′ = E′ δ

+

I

X′d V

Machine Connected to Infinite Bus i

X′di + X e

V∠00

+

Ei' ∠δ

System network

H=∞ r

Indian Institute of Technology, Kharagpur Machine Connected to Infinite Bus

X′d

E′ ∠δ

Xe

Infinite bus Pe V∞ ∠0o

+

NPTEL

A.K. Sinha

29/35


⎛ ⎞ V∞ ⎜ 1 1 ⎟ E′ V∞ Pe = sinδ + - ~ sin2δ ~ Xt 2 ⎜⎜ X' X' ⎟⎟ d ⎠ ⎝ q 2

~

~

X t = X′d + X e = X′d ; X′q + X e = X′q E′ V∞ Pe = sinδ = Pmax sinδ Xt 2

H dδ dδ +D = Pm - Pe pu 2 πf dt dt NPTEL

A.K. Sinha

28/35

Indian Institute of Technology, Kharagpur Two-machine system

X′d1

Xe

X′d2

+ Pm1

+ 1

Pe1

Pe2

E′1 ∠δ1

NPTEL

A.K. Sinha

2

Pm2 E′2 ∠δ2

31/35


Pm1 = - Pm2 = Pm Pe1 = - Pe2 = Pe Pm1 - Pe1 d2δ 1 = πf 2 dt H1 2

d δ2 = πf 2 dt

Pm2 - Pe2 H2 NPTEL

Pm - Pe = πf H1 = πf

Pe - Pm H2

A.K. Sinha

32/35


d2 (δ1 - δ2 ) H1 + H2 = πf (Pm - Pe ) 2 dt H1H2 Heq d2δ = Pm - Pe 2 πf dt δ = δ1 - δ 2

NPTEL

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H1H2 Heq = H1 +H2 The electrical power transfer is given by expression

Pe =

E′1 E′2 X′d1 + X e + X′d2

sinδ

NPTEL

A.K. Sinha

34/35

Multimachine System Gmach

= machine rating (base)

Gsystem = system base

Gmach Gsystem Or

2

Hmach d δ Gmach = (Pm - Pe ) 2 f dt Gsystem

Hsystem d2δ = Pm - Pe pu in system base 2 πf dt

Where

Hsystem = Hmach

Gmach Gsystem

= machine inertia constant in system base

Machines Swinging Coherently Consider the swing equations of two machines on a common system base.

H1 d2δ1 = Pm1 - Pe1 pu 2 πf dt H2 d2δ2 = Pm2 - Pe2 pu 2 πf dt

Since the machine rotors swing together (coherently or in unison)

δ1 = δ 2 = δ Adding the two equations

Heq d2δ = Pm - Pe 2 πf dt Where Pm = Pm1 + Pm2 Pe = Pe1 + Pe2 Heq = H1 + H2

Small Signal Stability 2

dδ dδ M 2 +D = Pm - Pe pu dt dt H in pu system M = πf E V Pe = sinδ = Pmax sinδ Xd

For small disturbance the swing eqn. can be linearized about the initial operating point → δ = δ0 + Δδ

d2δ dδ M 2 +D = Pm - Pe (δ) dt dt d2 (δ0 + Δδ) d(δ0 + Δδ) M +D = Pm - Pmax Sin(δ0 + Δδ) 2 dt dt Pmax Sin(δ0 + Δδ) = Pmax (Sinδ0CosΔδ + Cosδ0 SinΔδ) SinΔδ ; Δδ ; CosΔδ = 1 2 d2 δ0 dδ0 d Δδ dΔδ M 2 +M +D +D 2 dt dt dt dt = Pm - (Pmax Sinδ0 + Pmax Cosδ0 Δδ)

d2 Δδ dΔδ M +D = Pmax Cosδ0 Δδ 2 dt dt dPe (δ0 ) Pmax Cosδ0 = =ℑ dt ℑ = Synchronizing Coefficient Stiffness Coefficient d Δδ dΔδ M +D +ℑ=0 2 dt dt 2

d2 Δδ dΔδ M +D + ℑΔδ = 0 2 dt dt Describes the dynamics of the single machine connected to infinite bus system for small disturbances about any arbitrary operating point. Dynamics of this system can be analyzed by finding the roots of the characteristic polynomial Ms2 +Ds + ℑ = 0

roots of the characteristic polynomial Ms 2 + Ds + ℑ = 0 are given by D 2 - 4M ℑ s 1,2 = 2M For normal operating condition M ℑ ? D 2 -D±

s 1,2 =

ω=

ℑ

W ith D small → lightly damped system δ = δ 0 + Δδ → δ 0 How ever w ith negative value of ℑ or D, one of the roots have →

Stable

Unstable


Small Signal Stability d2δ dδ M 2 +D = Pm - Pe pu dt dt H in pu system M = πf E V Pe = sinδ = Pmax sinδ Xd

For sm all disturbance the sw ing eqn. can be linearized about the initial operating point → δ = δ 0 + Δδ NPTEL

A.K. Sinha

4/30


dδ dδ M 2 +D = Pm - Pe (δ) dt dt d2 (δ0 + Δδ) d(δ0 + Δδ) M +D = Pm - Pmax Sin(δ0 + Δδ) 2 dt dt Pmax Sin(δ0 + Δδ) = Pmax (Sinδ0CosΔδ + Cosδ0 SinΔδ) SinΔδ ; Δδ ; CosΔδ = 1 d2 δ0 dδ0 d2 Δδ dΔδ M 2 +M +D +D 2 dt dt dt dt = Pm - (Pmax Sinδ0 + Pmax Cosδ0 Δδ) NPTEL

A.K. Sinha

5/30


d2 Δδ dΔδ M +D = Pmax Cosδ0 Δδ 2 dt dt dPe (δ0 ) Pmax Cosδ0 = = dt = Synchronizing Coefficient Stiffness Coefficient d Δδ dΔδ M +D + 2 dt dt 2

NPTEL

=0

A.K. Sinha

6/30


d Δδ dΔδ M +D + Δδ = 0 2 dt dt Describes the dynamics of the single machine connected to infinite bus system 2

for small disturbances about any arbitrary operating point. Dynamics of this system can be analyzed by finding the roots of the characteristic polynomial Ms2 +Ds +

=0 NPTEL

A.K. Sinha

7/30

Indian Institute of Technology, Kharagpur roots of the characteristic polynomial Ms 2 + Ds +

= 0 are given by

-D±

D 2 - 4M s 1,2 = 2M For normal operating condition M s 1,2 =

±

? D2

ω=

W ith D small → lightly damped system δ = δ 0 + Δδ → δ 0 How ever w ith negative value of of the roots have → NPTEL

or D, one

A.K. Sinha

8/30

Indian Institute of Technology, Kharagpur ⎛ ⎞ V V 1 1 ⎟ E′ ∞ ∞ ⎜ Pe = sinδ + - ~ sin2δ ~ Xt 2 ⎜⎜ X' X' ⎟⎟ d ⎠ ⎝ q E′ V∞ Pe = sinδ Pe Xt 2

δ -180

-90

0

NPTEL

90

180

A.K. Sinha

9/30


H d2 Δδ dΔδ +D + 2 πf dt dt

Δδ = 0

d2 Δδ dΔδ 2 + 2 ς ω + ω n n Δδ = 0 2 dt dt

πf0 = H

and

D = 2

πf0 H

damped freq. of oscillation ωd = ωn 1- ς

2

NPTEL

A.K. Sinha

10/30

Indian Institute of Technology, Kharagpur F o r a s m a ll p u rtu rb a tio n Δ δ 0 a b o u t th e in itia l o p e ra tin g p o in t δ 0 th e d yn a m ic re s p o n s e o f th e s ys te m is

τ NPTEL

A.K. Sinha

11/30


Stable

NPTEL

Unstable

A.K. Sinha

12/30

Indian Institute of Technology, Kharagpur Problem: A 50 Hz synchronous generator having inertia constant H =5 MJ /MVA and a transient reactance Xd’ = 0.2 p.u. is connected to an infinite bus through a purely reactive circuit as shown in figure. Reactance values are marked on the diagram on a common system base. The generator is delivering real power of a) 0.6p.u. b) 2.0 p.u, at 0.8 power factor lagging to the infinite bus at a voltage of V=1 p.u. X = 0.4 X = 0.1

X = 0.4

' d

X = 0.2 NPTEL

A.K. Sinha

V∞

13/30

Indian Institute of Technology, Kharagpur The transfer reactance between the generated voltage and the infinite bus is 0.4 X = 0.2 + 0.1+ = .5 2 Apparent power in p.u 0.6 S= cos-1(0.8) = 0.75 36.87o 0.8 The current is S* 0.75 - 36.87o I= * = = 0.75 - 36.87 o V 1.0 0 NPTEL

A.K. Sinha

14/30

Indian Institute of Technology, Kharagpur Excitation voltage is E′ = V + jXI = 1 0o + (j0.5) × 0.75 - 36.870 = 1.225 + j0.3 = 1.261 13.76o

Initial operating power angle is 13.76 degree The synchronizing power coefficient is given by

= Pmaxcosδ0 1.261×1.0 = cos(13.76o ) = 2.449 0.5 NPTEL

A.K. Sinha

15/30

Indian Institute of Technology, Kharagpur The undamped angular frequency of oscillation and damping ratio are πf0 ωn = H

π × 50 = × 2.449 = 8.771 rad/sec 5

D πf0 0.15 π × 50 ς= = = 0.2686 2 H 2 5 × 2.449

The linearized force-free equation which determines the mode of oscillation is given by NPTEL

A.K. Sinha

16/30

Indian Institute of Technology, Kharagpur d2 Δδ dΔδ 2 + 2 ς ω + ω n n Δδ = 0 2 dt dt d2 Δδ dΔδ or, + 4.71 + 76.93Δδ = 0 2 dt dt

The damped angular frequency is =

ς =

Frequency of damped oscillation ωd fd = = 1.345Hz 2π NPTEL

A.K. Sinha

17/30

Indian Institute of Technology, Kharagpur The motion of rotor relative to the synchronously revolving field is

δ = δ0 +

Δδ0 1- ς 2

e- ςωnt sin(ωdt + θ)

δ = 13.76 +10.38e-2.356t sin(8.448t + 74.419) Frequency excursion

f = 50 - .2528e

-2.356t

sin(8.448t)

NPTEL

A.K. Sinha

18/30

Indian Institute of Technology, Kharagpur Case (b): The generator is delivering real power of 2.0 p.u, at 0.8 power factor lagging to the infinite bus. Apparent power in p.u

The current is

NPTEL

A.K. Sinha

19/30

Indian Institute of Technology, Kharagpur Excitation voltage is

′ Initial operating power angle is 29.74 degree The synchronizing power coefficient is given by

= Pmaxcosδ0 2.015 ×1.0 = cos29.74o = 3.5 .5 NPTEL

A.K. Sinha

20/30

Indian Institute of Technology, Kharagpur The undamped angular frequency of oscillation and damping ratio are

ω


A.K. Sinha

21/30


The damped angular frequency is

ς Frequency of damped oscillation

NPTEL

A.K. Sinha

22/30


Dynamic response

NPTEL

A.K. Sinha

23/30

Indian Institute of Technology, Kharagpur Case(c): The generator is delivering real power of 0.6p.u, at 0.8 power factor lagging to the infinite bus and D= - 0.01pu Apparent power in p.u 0.6 S= cos-1(0.8) = 0.75 36.87o 0.8 The current is S* 0.75 - 36.87o I= * = = 0.75 - 36.87 o V 1.0 0 NPTEL

A.K. Sinha

24/30

Indian Institute of Technology, Kharagpur Excitation voltage is E′ = V + jXI = 1 0o + (j0.5) × 0.75 - 36.870 = 1.225 + j0.3 = 1.261 13.76o

Initial operating power angle is 13.76 degree The synchronizing power coefficient is given by Ps = Pmaxcosδ0 1.261×1.0 = cos(13.76o ) = 2.449 0.5 NPTEL

A.K. Sinha

25/30

Indian Institute of Technology, Kharagpur The undamped angular frequency of oscillation and damping ratio are

πf0 ωn = H

π × 50 = × 2.449 = 8.771 rad/sec 5

D πf0 -0.01 π × 50 = = - 0.179 ς= 2 H 2 5 × 2.449


A.K. Sinha

26/30

Indian Institute of Technology, Kharagpur d2 Δδ dΔδ 2 + 2 ω + ω ς n n Δδ = 0 2 dt dt d2 Δδ dΔδ or, - 0.942 + 76.93Δδ = 0 2 dt dt The damped angular frequency is ς

Frequency of damped oscillation ωd fd = = 1.396Hz 2π NPTEL

A.K. Sinha

27/30

Indian Institute of Technology, Kharagpur The motion of rotor relative to the synchronously revolving field is δ = δ0 +

Δδ0 1- ς 2

e- ςωnt sin(ωdt + θ)

δ = 13.76 +10.38e0.471t sin(8.769t + 74.419) Frequency excursion

f = 50 - .2528e0.471t sin(8.769t) NPTEL

A.K. Sinha

28/30


Dynamic response

NPTEL

A.K. Sinha

29/30


Large Signal Rotor Angle Stability (Transient Stability) Large disturbances such as Faults, Outages of heavily loaded lines, generators etc. Æ sudden large changes in power outputs of generators Æ acceleration / deceleration of generator rotor Æ large deviation of rotor angle and rotor oscillations about the operating point. Linearized dynamic model can not be used. NPTEL

A.K. Sinha

4/33


Nonlinear swing equation has to be solved. 2

dδ dδ M 2 +D = Pm - Pe dt dt

Pe = Pmax sinδ D = 0 → Pessimistic result d2δ M 2 = Pm - Pmax sinδ dt NPTEL

A.K. Sinha

5/33

Indian Institute of Technology, Kharagpur Pe P’m1 Pm1 Pm0

0

δ0 ωs

δ1 δ’1

δ2 ω > ωs ωs ω < ωs NPTEL

π

A.K. Sinha

δ

6/33

Indian Institute of Technology, Kharagpur Assumptions for Transient Stability Study 1. Transmission line as well as synchronous machine resistance are ignored. 2. Damping term contributed by synchronous machine damper windings is ignored. 3. Rotor speed is assumed to remain constant. Note: Since rotor speed and hence frequency vary insignificantly, the network parameters remain fixed during a stability study. NPTEL

A.K. Sinha

7/33


4. Mechanical input to machine is assumed to remain constant during the transient. 5. Voltage behind transient reactance is assumed to remain constant. 6. Loads are ittances.

modeled

NPTEL

as

constant

A.K. Sinha

8/33


Unstable dδ =0 dt

Stable t1 t 2 t3 NPTEL

A.K. Sinha

9/33


Equal Area Criterion 2

Pa dδ 1 = (Pm - Pe ) = ; Pa = accelerating power 2 dt M M H in pu system M = πf

NPTEL

A.K. Sinha

10/33

Indian Institute of Technology, Kharagpur Multiplying both sides of the swing equation by

dδ 2 , we get dt 2

dδ d δ 2Pa dδ 2 = 2 dt dt M dt 2

2 ⎛ dδ ⎞ ⎜ ⎟ = ⎝ dt ⎠ M NPTEL

δ

∫ P dδ a

δ0

A.K. Sinha

11/33


dδ ⎛ 2 =⎜ dt ⎜⎝ M

1

⎞2 ∫δ Padδ ⎟⎟ 0 ⎠ δ

dδ For the system to be stable =0 dt 1

⎛2 ⎜⎜ ⎝M

⎞2 ∫δ Padδ ⎟⎟ = 0 0 ⎠ δ

δ

∫ P dδ = 0 a

δ0

NPTEL

A.K. Sinha

12/33


Sudden Change in Mechanical Input E′ ∠δ

X′d

Xe

Infinite bus bar

Pm

o ∠ 0 V

NPTEL

A.K. Sinha

13/33


E′ V Pe = sinδ = Pmax sinδ X′d + X e

Under steady operating condition Pm0 = Pe0 = Pmax sinδ

NPTEL

A.K. Sinha

14/33


c b

Pm1

A2 Fig. Pe – δ diagram for sudden increase in mechanical input to generator

A1

Pm0

a

0

δ0 ωs

δ1 ω > ωs ω < ωs

δ2 ωs NPTEL

π

A.K. Sinha

δ

15/33

Indian Institute of Technology, Kharagpur δ2

∫ P dδ = 0 a

δ0

Since the rotor is decelerating

Pm1 = Pe = Pmax sin δ1 δ1

A1 = ∫ (Pm1 - Pe )dδ δ0

δ2

A 2 = ∫ (Pe - Pm1 )dδ δ1

δ2 = δmax

Pm1 = π - δ1 = π - sin Pmax -1

NPTEL

A.K. Sinha

16/33

Indian Institute of Technology, Kharagpur Pe b

Pm1

A1

Pm0

a

A2

c

0

δ δ2 π δ0 δ1 Fig. Limiting case of transient stability with mechanical input suddenly increased NPTEL

A.K. Sinha

17/33


Effect of Clearing Time on Stability

∞

Pm

P

NPTEL

A.K. Sinha

18/33


e

Pmax

d A2

Pm

a A1 b δ0

Pe = 0 (3-phase fault)

c δc

π

δ2

δ

Clearing angle NPTEL

A.K. Sinha

19/33

Indian Institute of Technology, Kharagpur Pe Fig. Critical clearing angle

Pmax A2 Pm A1

δ0

δmax

δcr

δ

Critical clearing angle NPTEL

A.K. Sinha

20/33


δmax = π - δ0 Pm = Pmax sinδ0 δcr

A1 =

∫ (P

m

- 0)dδ = Pm (δcr - δ0 )

δ0

δmax

A2 =

∫

(Pmax sin δ - Pm )dδ

δcr

= Pmax (cosδcr - cosδmax ) - Pm (δmax - δcr ) NPTEL

A.K. Sinha

21/33

Indian Institute of Technology, Kharagpur For the system to be stable, A2 = A1, which yields

Pm cos δcr = (δmax - δ0 ) + cos δmax Pmax Where

δcr = critical clearing angle

δcr = cos-1[(π - 2δ0 ) sin δ0 - cos δ0 ]

NPTEL

A.K. Sinha

22/33


d2δ πf = Pm ; Pe = 0 2 dt H Integrating twice

πf δ= Pm t 2 + δ0 2H or

πf 2 δcr = Pm t cr + δ0 2H NPTEL

A.K. Sinha

23/33


Where tcr = critical clearing time δcr= critical clearing angle

t cr =

2H (δcr - δ0 ) πfPm

NPTEL

A.K. Sinha

24/33

Indian Institute of Technology, Kharagpur Problem: For the system shown in fig, both the terminal voltage and infinite bus voltage are 1.0 per unit and the generator is delivering 1.0 p.u. power. Calculate the critical clearing angle and the critical clearing time when the system is subjected to a 3 ph. Fault at point P on the short transmission line. H=5 MJ/MVA j0.1 j0.5

X′d = j0.15

A P NPTEL

8

j0.5 B

open

A.K. Sinha

25/33

Indian Institute of Technology, Kharagpur j0.1

j0.5

X′d = j0.15

A P

8

j0.5 B

open The series reactance between the terminal voltage and the infinite bus is .5 X = j.1+ j = j.35 2 NPTEL

A.K. Sinha

26/33

Indian Institute of Technology, Kharagpur Generator is delivering 1p.u power, so Vt V 1×1 1.0 = sinδ = sinδ X .35 or, sinδ = .35 i.e., δ = sin-1.35 = 20.49o So, the terminal voltage is given by, Vt = 1.0 20.49o = .937 + j.35

The output current from the generator is now calculated as, 1.0 δ2 - 1.0 0o I= = 1+ j.18 = 1.016 10.2o j.35 NPTEL

A.K. Sinha

27/33

Indian Institute of Technology, Kharagpur And the transient internal voltage is then found to be E′1 = .937 + j.35 + (j.15)(1+ j.18)

= .91+ j.5 = 1.038 28.786o p.u Series reactance between transient internal voltage and infinite-bus j.5 X = j.15 + j.1+ 2 Hence power angle equation can be written as 1.038 ×1.0 Pe = sinδ = 2.076sinδ .5 NPTEL

A.K. Sinha

28/33

Indian Institute of Technology, Kharagpur Initial rotor angle

δ0 = 28.786o

δcr

A1 = ∫ Pmdδ = Pm (δcr - δ0 ) δ0

A2 = ∫

δmax

δcr

(Pmax sinδ - Pm )dδ

= Pmax (cosδcr - cosδmax ) - Pm (δmax - δcr )

Equating A1 and A2 cosδcr = (Pm /Pmax )(δmax - δ0 ) + cosδmax NPTEL

A.K. Sinha

29/33


δmax = π - δ0 Pm = Pmax sinδ0 Substituting for δmax and Pm δcr = cos-1[(π - 2δ0 )sinδ0 - cosδ0 ] -1

= cos [(π - 2 × .5024)sin28.786 - cos28.786] = cos-1(.1525) = 81.226o = 1.4176 elec rad NPTEL

A.K. Sinha

30/33

Indian Institute of Technology, Kharagpur If time to clear the fault is denoted by tc Then for time t less than tc, d2δ w s = Pm (as Pe = 0) 2 dt 2H dδ t w s ws ⇒ =∫ Pmdt = Pm t dt 0 2H 2H w sPm 2 ⇒δ= t + δ0 4H w sPm 2 δ(t) |t=tc = t c + δ0 4H NPTEL

A.K. Sinha

31/33


w sPm 2 δcr = t cr + δ0 4H So, critical clearing time,

4H(δcr - δ0 ) t cr = w sPm

4 × 5(1.4176 - .5024) t cr = = .241sec 314 ×1 NPTEL

A.K. Sinha

32/33


POWER SYSTEM STABILITY Lesson Summary 1. Equal Area Criterion 2. Examples

NPTEL

A.K. Sinha

2/30


Effect of Clearing Time on Stability

∞

Pm

P

NPTEL

A.K. Sinha

18/33


e

Pmax

d A2

Pm

a A1 b δ0

Pe = 0 (3-phase fault)

c δc

π

δ2

δ

Clearing angle NPTEL

A.K. Sinha

19/33

Indian Institute of Technology, Kharagpur Pe Fig. Critical clearing angle

Pmax A2 Pm A1

δ0

δmax

δcr

δ

Critical clearing angle NPTEL

A.K. Sinha

20/33


δmax = π - δ0 Pm = Pmax sinδ0 δcr

A1 =

∫ (P

m

- 0)dδ = Pm (δcr - δ0 )

δ0

δmax

A2 =

∫

(Pmax sin δ - Pm )dδ

δcr

= Pmax (cosδcr - cosδmax ) - Pm (δmax - δcr ) NPTEL

A.K. Sinha

21/33

Indian Institute of Technology, Kharagpur For the system to be stable, A2 = A1, which yields

Pm cos δcr = (δmax - δ0 ) + cos δmax Pmax Where

δcr = critical clearing angle

δcr = cos-1[(π - 2δ0 ) sin δ0 - cos δ0 ]

NPTEL

A.K. Sinha

22/33


d2δ πf = Pm ; Pe = 0 2 dt H Integrating twice

πf δ= Pm t 2 + δ0 2H or

πf 2 δcr = Pm t cr + δ0 2H NPTEL

A.K. Sinha

23/33


Where tcr = critical clearing time δcr= critical clearing angle

t cr =

2H (δcr - δ0 ) πfPm

NPTEL

A.K. Sinha

24/33

Indian Institute of Technology, Kharagpur Problem: For the system shown in fig, both the terminal voltage and infinite bus voltage are 1.0 per unit and the generator is delivering 1.0 p.u. power. Calculate the critical clearing angle and the critical clearing time when the system is subjected to a 3 ph. Fault at point P on the short transmission line. H=5 MJ/MVA j0.1 j0.5

X′d = j0.15

A P NPTEL

8

j0.5 B

open

A.K. Sinha

25/33

Indian Institute of Technology, Kharagpur j0.1

j0.5

X′d = j0.15

A P

8

j0.5 B

open The series reactance between the terminal voltage and the infinite bus is .5 X = j.1+ j = j.35 2 NPTEL

A.K. Sinha

26/33

Indian Institute of Technology, Kharagpur Generator is delivering 1p.u power, so Vt V 1×1 1.0 = sinδ = sinδ X .35 or, sinδ = .35 i.e., δ = sin-1.35 = 20.49o So, the terminal voltage is given by, Vt = 1.0 20.49o = .937 + j.35

The output current from the generator is now calculated as, 1.0 δ2 - 1.0 0o I= = 1+ j.18 = 1.016 10.2o j.35 NPTEL

A.K. Sinha

27/33

Indian Institute of Technology, Kharagpur And the transient internal voltage is then found to be E′1 = .937 + j.35 + (j.15)(1+ j.18)

= .91+ j.5 = 1.038 28.786o p.u Series reactance between transient internal voltage and infinite-bus j.5 X = j.15 + j.1+ 2 Hence power angle equation can be written as 1.038 ×1.0 Pe = sinδ = 2.076sinδ .5 NPTEL

A.K. Sinha

28/33

Indian Institute of Technology, Kharagpur Initial rotor angle

δ0 = 28.786o

δcr

A1 = ∫ Pmdδ = Pm (δcr - δ0 ) δ0

A2 = ∫

δmax

δcr

(Pmax sinδ - Pm )dδ

= Pmax (cosδcr - cosδmax ) - Pm (δmax - δcr )

Equating A1 and A2 cosδcr = (Pm /Pmax )(δmax - δ0 ) + cosδmax NPTEL

A.K. Sinha

29/33


δmax = π - δ0 Pm = Pmax sinδ0 Substituting for δmax and Pm δcr = cos-1[(π - 2δ0 )sinδ0 - cosδ0 ] -1

= cos [(π - 2 × .5024)sin28.786 - cos28.786] = cos-1(.1525) = 81.226o = 1.4176 elec rad NPTEL

A.K. Sinha

30/33

Indian Institute of Technology, Kharagpur If time to clear the fault is denoted by tc Then for time t less than tc, d2δ w s = Pm (as Pe = 0) 2 dt 2H dδ t w s ws ⇒ =∫ Pmdt = Pm t dt 0 2H 2H w sPm 2 ⇒δ= t + δ0 4H w sPm 2 δ(t) |t=tc = t c + δ0 4H NPTEL

A.K. Sinha

31/33


w sPm 2 δcr = t cr + δ0 4H So, critical clearing time,

4H(δcr - δ0 ) t cr = w sPm

4 × 5(1.4176 - .5024) t cr = = .241sec 314 ×1 NPTEL

A.K. Sinha

32/33

Sudden Loss of One of Parallel Lines E′ V PeI = sinδ = PmaxI sinδ X d + X1 X 2

Immediately on switching off line 2, power angle curve is given by E′ V PeII = sinδ = PmaxII sinδ X′d + X1

Infinite bus Pm

V ∠0 (a)

o

X1

E′ ∠δ X′d Pm

Switched off

o ∠ 0 V

X2 (b) Fig. Single machine tied to infinite bus through two parallel lines

PmI (both lines in)

Pe A2 a

Pm

c b

0

d

δ0

PmII (line 2 out)

A1

δ1

δ2

π

δ

Fig. Equal area criterion applied to the opening of one of the two lines in parallel

δ1 = δmax = π - δc Short circuit at one end of line X1

E′ ∠δ X′d

Infinite bus

Pm

o ∠ 0 V

X2 (a)

X1

E′ ∠δ X′d

o ∠ 0 V

Pm (b)

X2

Fig. Short circuit at one end of the line.

E′ V PeI = sinδ = PmaxI sinδ X′d + X1 X 2 PeII = 0

E′ V PeIII = sinδ = Pmax II sinδ X′d + X1

Pe

Pe I, prefault (2 lines) A2

Pm

Pe III, postfault (1 line)

A1 Pe II 0

δ0

δc

δ2 δmax

π

δ

Fig. Equal area criterion applied to the system of Fig. , I system normal, II fault applied, III faulted line isolated.

Case b: Short circuit away from line ends E′ V PeII = sinδ = Pmax II sinδ X II

X′d

G

X2/2

X1

Xa

Xb Xc (a)

X2/2

∞

X1

Xa

Xb

Xc

G

(b)

∞

XII

∞

G

(c)

Pe

Pe I, prefault (2 lines) Pe III, postfault (1 line)

A2

Pm

Pe II, during fault

A1

0

δ0

δc

δ2

δmax

π

δ

Fig. Fault on middle of one line of the system of Fig. , with δc < δcr.

Pe

Pe I, prefault (2 lines) Pe III, postfault (1 line) A2

Pm

Pe II, during fault

A1

δ0 δcr 0 Critical clearing angle

δ2= δmax

π

δ

Fig. Fault on middle of one line of the system of Fig. , case of critical clearing angle.

δcr

∫ (P

m

δmax

- Pmax II sin δ)dδ =

δ0

∫

(Pmax III sin δ - Pm )dδ

δcr

Pm1 Pmax III

-1

δmax = π - sin

(Pmδ +Pmax II cosδ)

δcr δ0

+ (Pmax III cosδ + Pmδ)

δmax δcr

Pm (δcr - δ0 ) +Pmax II (cosδcr - cosδ0 ) + Pm (δmax - δcr ) +Pmax III (cosδmax - cosδcr ) = 0

=0

Pm (δmax - δ0 ) - Pmax IIcosδ0 +Pmax III cosδmax cos δcr = Pmax III - Pmax II

π Pm (δmax - δ0 ) - Pmax IIcosδ0 +Pmax III cosδmax cos δcr = 180 Pmax III - Pmax II

-1

δ1 = δmax = π - sin (Pm1 Pmax I ) δcr

∫ (P

m

δ0

δrc

- Pmax II sin δ)dδ =

∫ (P

max III

sin δ - Pm )dδ

δcr

δmax

+

∫

δrc

(Pmax I sin δ - Pm )dδ

Pe Pe I Pe II

A2

Pm A1

Pe III δrc

0

δ0

δc

δ2 δmax

π

δ

(Clearing angle) (Angle of reclosure) Fig. Fault in middle of a line of the system of Fig. Where trc = tcr + τ; τ = time between clearing and reclosure.


Thank You !

NPTEL

A.K. Sinha

35/35

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