Lecture Power System

9/16/2011

1

Simplified Power System Modeling

• Balanced three phase systems can be analyzed using per phase analysis

• A “per unit” normalization is simplify the analysis of systems with different voltage levels.

Energy Systems Research Laboratory, FIU

• To provide an introduction to power flow analysis we need models for the different system devices:– Transformers and Transmission lines, generators and loads

• Transformers and transmission lines are modeled as a series impedances

Load Models

• Ultimate goal is to supply loads with electricity at constant frequency and voltage

• Electrical characteristics of individual loads matter, but usually they can only be estimated– actual loads are constantly changing consisting of a large


– actual loads are constantly changing, consisting of a large number of individual devices

– only limited network observability of load characteristics

• Aggregate models are typically used for analysis

• Two common models– constant power: Si = Pi + jQi

– constant impedance: Si = |V|2 / Zi

Generator Models

• Engineering models depend upon application

• Generators are usually synchronous machines

• For generators we will use two different models:– a steady-state model, treating the generator as a constant


y , g gpower source operating at a fixed voltage; this model will be used for power flow and economic analysis

– This model works fairly well for type 3 and type 4 wind turbines

– Other models include treating as constant real power with a fixed power factor.

Per Unit Calculations

• A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformers

Thi bl i id d b li i f ll


• This problem is avoided by a normalization of all variables.

• This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity

Per Unit Conversion Procedure, 11. Pick a 1 VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3 C l l h i d b Z (V )2/S


3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unit

Note, per unit conversion on affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)

Three Phase Per Unit

1. Pick a 3 VA base for the entire system,

2. Pick a voltage base for each different voltage level,

Procedure is very similar to 1 except we use a 3VA base, and use line to line voltage bases

3BS


VB. Voltages are line to line.

3. Calculate the impedance base2 2 2, , ,3 1 1

( 3 )

3B LL B LN B LN

BB B B

V V VZ

S S S

Exactly the same impedance bases as with single phase!

9/16/2011

2

Three Phase Per Unit, cont'd

4. Calculate the current base, IB3 1 1

3 1B B

, , ,

3I I

3 3 3B B B

B LL B LN B LN

S S S

V V V


5. Convert actual values to per unit

Exactly the same current bases as with single phase!

Power Flow Analysis

• We now have the necessary models to start to develop the power system analysis tools

• The most common power system analysis tool is the power flow (also known sometimes as the load flow)


– power flow determines how the power flows in a network

– also used to determine all bus voltages and all currents

– because of constant power models, power flow is a nonlinear analysis technique

– power flow is a steady-state analysis tool

Bus Admittance Matrix or Ybus

• First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.

• The Ybus gives the relationships between all the bus


bus g pcurrent injections, I, and all the bus voltages, V,

I = Ybus V

• The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances

Ybus General Form •The diagonal terms, Yii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i.

•The off-diagonal terms, Yij, are equal to the negative


of the sum of the admittances joining the two buses.

•With large systems Ybus is a sparse matrix (that is, most entries are zero)

•Shunt terms, such as with the line model, only affect the diagonal terms.

Power Flow Analysis

• When analyzing power systems we know neither the complex bus voltages nor the complex current injections

• Rather, we know the complex power being


p p gconsumed by the load, and the power being injected by the generators plus their voltage magnitudes

• Therefore we can not directly use the Ybus equations, but rather must use the power balance equations

Power Flow Slack Bus

• We can not arbitrarily specify S at all buses because total generation must equal total load + total losses

• We also need an angle reference bus.

• To solve these problems we define one bus as the


To solve these problems we define one bus as the "slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection.

• A “slack bus” does not exist in the real power system.

9/16/2011

3

Power Balance EquationsFrom KCL we know at each bus i in an n bus system

the current injection, , must be equal to the current

that flows into the network i

n

I

I I I I


1

bus

1

Since = we also know

i Gi Di ikk

n

i Gi Di ik kk

I I I I

I I I Y V

I Y V

*iThe network power injection is then S i iV I

Real Power Balance Equations* *

i1 1

1

S ( )

(cos sin )( )

ikn n

ji i i ik k i k ik ik

k k

n

i k ik ik ik ikk

P jQ V Y V V V e G jB

V V j G jB


i1

i1

Resolving into the real and imaginary parts

P ( cos sin )

Q ( sin cos

n

i k ik ik ik ik Gi Dik

n

i k ik ik ik ik

V V G B P P

V V G B

)k Gi DiQ Q

Newton-Raphson Method (scalar)

( )

( ) ( )

1. For each guess of , , define ˆ

-ˆ

v

v v

x x

x x x

G eneral fo rm of problem : F ind an x such that

( ) 0ˆf x


( )( ) ( )

2 ( ) 2( )2

-

2. Represent ( ) by a Taylor series about ( )ˆ

( )( ) ( )ˆ

1 ( )higher order terms

2

vv v

vv

x x x

f x f x

df xf x f x x

dx

d f xx

dx

Newton-Raphson Method, cont’d

( )( ) ( )

3. Approximate ( ) by neglecting all termsˆ

except the first two

( )( ) 0 ( )ˆ

vv v

f x

df xf x f x x

dx


( )

1( )( ) ( )

4. Use this linear approximation to solve for

( )( )

5. Solve for a new estim

v

vv v

x

df xx f x

dx

( 1) ( ) ( )

ate of x̂v v vx x x

Newton-Raphson Example2

1( )( ) ( )

Use Newton-Raphson to solve ( ) - 2 0

The equation we must iteratively solve is

( )( )

vv v

f x x

df xx f x

d


( ) ( ) 2( )

( 1) ( ) ( )

( 1) ( ) ( ) 2( )

1(( ) - 2)

2

1(( ) - 2)

2

v vv

v v v

v v vv

dx

x xx

x x x

x x xx

Newton-Raphson Example, cont’d

( 1) ( ) ( ) 2( )

(0)

( ) ( ) ( )

1(( ) - 2)

2

Guess x 1. Iteratively solving we get

v ( )

v v vv

v v v

x x xx

x f x x


3 3

6

v ( )

0 1 1 0.5

1 1.5 0.25 0.08333

2 1.41667 6.953 10 2.454 10

3 1.41422 6.024 10

x f x x

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4

Newton-Raphson Comments

• When close to the solution the error decreases quite quickly -- method has quadratic convergence

• f(x(v)) is known as the mismatch, which we would like to drive to zero


• Stopping criteria is when f(x(v)) < • Results are dependent upon the initial guess. What if

we had guessed x(0) = 0, or x (0) = -1?

• A solution’s region of attraction (ROA) is the set of initial guesses that converge to the particular solution. The ROA is often hard to determine

Multi-Variable Newton-Raphson

1 1

2 2

Next we generalize to the case where is an n-

dimension vector, and ( ) is an n-dimension function

( )

( )

x f

x f

x

f x

x

x


2 2( )( )

( )

Again define the solution so ( ) 0 andˆ ˆn n

x f

x f

xx f x

x

x f x

x

ˆ x x

Multi-Variable Case, cont’di

1 11 1 1 2

1 2

1

The Taylor series expansion is written for each f ( )

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order termsn

x xx x

x

x

x xx x

x


n nn n 1 2

1 2

n

g

f ( ) f ( )f ( ) f ( )ˆ

f ( )higher order terms

nn

nn

x

x xx x

xx

x xx x

x

Multi-Variable Case, cont’d

1 1 1

1 21 1

2 2 2

This can be written more compactly in matrix form

( ) ( ) ( )

( )( ) ( ) ( )

n

f f f

x x xf x

f f f

x x x

xx x x


2 2 22 2

1 2

1 2

( ) ( ) ( )( )

( )ˆ

( )( ) ( ) ( )

n

nn n n

n

f f ff x

x x x

ff f f

x x x

x x xx

f x

xx x x

higher order terms

nx

Jacobian Matrix

1 1 1

1 2

The n by n matrix of partial derivatives is known

as the Jacobian matrix, ( )

( ) ( ) ( )

n

f f fx x x

J x

x x x


1 2

2 2 2

1 2

1 2

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

x x x

f f fx x x

f f f

x x x

x x x

J x

x x x

Multi-Variable N-R Procedure

1

Derivation of N-R method is similar to the scalar case

( ) ( ) ( ) higher order termsˆ

( ) 0 ( ) ( )ˆ

( ) ( )

f x f x J x x

f x f x J x x

x J x f x


( 1) ( ) ( )

( 1) ( ) ( ) 1 ( )

( )

( ) ( )

( ) ( )

Iterate until ( )

v v v

v v v v

v

x J x f x

x x x

x x J x f x

f x

9/16/2011

5

Multi-Variable Example

1

2

2 21 1 2

2 2

xSolve for = such that ( ) 0 where

x

f ( ) 2 8 0x x

x f x

x


2 22 1 2 1 2

1 1

1 2

2 2

1 2

f ( ) 4 0

First symbolically determine the Jacobian

f ( ) f ( )

( ) =f ( ) f ( )

x x x x

x x

x x

x

x x

J xx x

Multi-variable Example, cont’d1 2

1 2 1 2

11 1 2 1

4 2( ) =

2 2

Then

4 2 ( )

x x

x x x x

x x x f

J x

x


1 1 2 1

2 1 2 1 2 2

(0)

1(1)

( )

2 2 ( )

1Arbitrarily guess

1

1 4 2 5 2.1

1 3 1 3 1.3

f

x x x x x f

x

x

x

Multi-variable Example, cont’d

1(2) 2.1 8.40 2.60 2.51 1.8284

1.3 5.50 0.50 1.45 1.2122

Each iteration we check ( ) to see if it is below our

x

f x


(2)

specified tolerance

0.1556( )

0.0900

If = 0.2 then we wou

f x

ld be done. Otherwise we'd

continue iterating.

Real Power Balance Equations* *

i1 1

1

S ( )

(cos sin )( )

ikn n

ji i i ik k i k ik ik

k k

n

i k ik ik ik ikk

P jQ V Y V V V e G jB

V V j G jB


i1

i1

Resolving into the real and imaginary parts

P ( cos sin )

Q ( sin cos

n


n

i k ik ik ik ik

V V G B P P

V V G B

)k Gi DiQ Q

Newton-Raphson Power Flow

In the Newton-Raphson power flow we use Newton's

method to determine the voltage magnitude and angle

at each bus in the power system.

We need to solve the power balance equations


i1

We need to solve the power balance equations

P ( cosn

i k ik ikk

V V G

i1

sin )

Q ( sin cos )

ik ik Gi Di

n


B P P

V V G B Q Q

Power Flow Variables

2 2 2

Assume the slack bus is the first bus (with a fixed

voltage angle/magnitude). We then need to determine

the voltage angle/magnitude at the other buses.

( ) GP P x 2DP


2 2 2

n

2

( )

( )

G

n

P P

V

V

x

x f x

2

2 2 2

( )

( )

( )

D

n Gn Dn

G D

n Gn Dn

P

P P P

Q Q Q

Q Q Q

x

x

x

9/16/2011

6

N-R Power Flow Solution

( )

( )

The power flow is solved using the same procedure

discussed last time:

Set 0; make an initial guess of , vv x x


( )

( 1) ( ) ( ) 1 ( )

While ( ) Do

( ) ( )

1

End While

v

v v v v

v v

f x

x x J x f x

Power Flow Jacobian Matrix

1 1 1

1 2

The most difficult part of the algorithm is determining

and inverting the n by n Jacobian matrix, ( )

( ) ( ) ( )

n

f f fx x x

J x

x x x


1 2

2 2 2

1 2

1 2

( ) ( ) ( )

( )

( ) ( ) ( )

n

n

n n n

n

x x x

f f fx x x

f f fx x x

x x x

J x

x x x

Power Flow Jacobian Matrix, cont’d

i

i

Jacobian elements are calculated by differentiating

each function, f ( ), with respect to each variable.

For example, if f ( ) is the bus i real power equationn

x

x


i1

f ( ) ( cos sin )i k ik ik ik ik Gik

x V V G B P P

i

1

i

f ( )( sin cos )

f ( )( sin cos ) ( )

Di

n

i k ik ik ik iki k

k i

i j ik ik ik ikj

xV V G B

xV V G B j i

Two Bus Newton-Raphson Example

Line Z = 0 1j

For the two bus power system shown below, use the Newton-Raphson power flow to determine the voltage magnitude and angle at bus two. Assumethat bus one is the slack and SBase = 100 MVA.


Line Z = 0.1j

One Two 1.000 pu 1.000 pu

200 MW 100 MVR

0 MW 0 MVR

2

2

10 10

10 10bus

j j

V j j

x Y

Two Bus Example, cont’d

i1

General power balance equations

P ( cos sin )

Q ( i )

n


n

V V G B P P

V V G B Q Q


i1

2 1 2

22 1 2 2

Q ( sin cos )

Bus two power balance equations

(10sin ) 2.0 0

( 10cos ) (10) 1.0 0


V V G B Q Q

V V

V V V

Two Bus Example, cont’d2 2 2

22 2 2 2

2 2

P ( ) (10sin ) 2.0 0

( ) ( 10cos ) (10) 1.0 0

Now calculate the power flow Jacobian

P ( ) P ( )

V

Q V V

x

x

x x


2 2

2 2

2 2

2 2

2 2 2

2 2 2 2

P ( ) P ( )

( )Q ( ) Q ( )

10 cos 10sin

10 sin 10cos 20

VJ

V

V

V V

x x

xx x

9/16/2011

7

Two Bus Example, First Iteration(0)

2 2(0)

0Set 0, guess

1

Calculate

(10sin ) 2.0 2.0f( )

v

V

x


(0)2

2 2 2

2 2 2(0)

2 2 2 2

(1)

f( )1.0( 10cos ) (10) 1.0

10 cos 10sin 10 0( )

10 sin 10cos 20 0 10

0 10 0Solve

1 0 10

V V

V

V V

x

J x

x1 2.0 0.2

1.0 0.9

Two Bus Example, Next Iterations(1)

2

(1)

0.9(10sin( 0.2)) 2.0 0.212f( )

0.2790.9( 10cos( 0.2)) 0.9 10 1.0

8.82 1.986( )

1.788 8.199

x

J x


1(2) 0.2 8.82 1.986 0.212 0.233

0.9 1.788 8.199 0.279 0.8586

f(

x

(2) (3)

(3)2

0.0145 0.236)

0.0190 0.8554

0.0000906f( ) Done! V 0.8554 13.52

0.0001175

x x

x

Two Bus Solved ValuesOnce the voltage angle and magnitude at bus 2 are known we can calculate all the other system values, such as the line flows and the generator reactive power output


Line Z = 0.1j


200 MW 100 MVR

200.0 MW168.3 MVR

-13.522 Deg

200.0 MW 168.3 MVR

-200.0 MW-100.0 MVR

PV Buses

• Since the voltage magnitude at PV buses is fixed there is no need to explicitly include these voltages in x or write the reactive power balance equations– the reactive power output of the generator varies to

i t i th fi d t i l lt ( ithi li it )


maintain the fixed terminal voltage (within limits)

– optionally these variations/equations can be included by just writing the explicit voltage constraint for the generator bus

|Vi | – Vi setpoint = 0

Three Bus PV Case Example

Line Z = 0 1j

2 2 2 2

3 3 3 3

2 2 2

For this three bus case we have

( )

( ) ( ) 0

V ( )

G D

G D

D

P P P

P P P

Q Q

x

x f x x

x


Line Z = 0.1j

Line Z = 0.1j Line Z = 0.1j


200 MW 100 MVR

170.0 MW 68.2 MVR

-7.469 Deg

Three 1.000 pu

30 MW 63 MVR

Solving Large Power Systems

• The most difficult computational task is inverting the Jacobian matrix– inverting a full matrix is an order n3 operation, meaning

the amount of computation increases with the cube of the size size


size size

– this amount of computation can be decreased substantially by recognizing that since the Ybus is a sparse matrix, the Jacobian is also a sparse matrix

– using sparse matrix methods results in a computational order of about n1.5.

– this is a substantial savings when solving systems with tens of thousands of buses

9/16/2011

8

“DC” Power Flow

• The “DC” power flow makes some approximations to the power balance equations to simplify the problem– completely ignore reactive power, assume all the voltages

are always 1.0 per unit, ignore line conductance, assumes l th li ll


angles across the lines are small.

– Line flow is then approximated as (i – j)/Xij

• This makes the power flow a linear set of equations, which can be solved directly = B-1P

• While the DC power flow is approximate, it is widely used to get a feel for power system MW flows.

400 MVA15 kV

T1

T2800 MVA345/15 kV

800 MVA15 kV

520 MVALine 3 345 kV

50 mi

1 4 35

Five Bus Power Flow Example


400 MVA15/345 kV

80 MW40 Mvar

280 Mvar 800 MW

Lin

e 2

Lin

e 1345 kV

100 mi345 kV 200 mi

2

Single-line diagram

DC Power Flow Example

• For the system shown in the previous slide with bus 1 as the system slack and with the below B matrix (100 MVA base) numbered from buses 2 to 5, determine the bus angles using the DC power flow approximation the and the flow on the line between bus 1 and 5 which has a pu


the flow on the line between bus 1 and 5, which has a puX of 0.02.

P15 = (1 – 5)/X15 = 0.072/0.02 = 3.6 pu = 360 MW

37 Bus Power Flow Example

slack

Metropolis Light and Power Electric Design Case 2SLA CK3 45

SLA CK1 38

RA Y34 5

RA Y1 38

RA Y69

FERNA 69

A

MVA

BOB1 38

WOLEN69

PET E69HISKY69

T IM6 9

T IM1 38

TIM3 45

PA I69

GROSS69

MORO13 8

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1.02 pu

1.03 pu

1.03 pu

1.01 pu

1 .0 0 pu1.01 pu

1 .0 1 pu

1 .0 1 pu

1 .0 2 pu

1 .0 0 pu

1.02 pu

1.02 pu

A

MVA

A

MVA

A

MVA

System Losses: 10.70 MW

22 0 MW 5 2 Mvar

12 MW 3 Mvar

3 7 MW

1 3 Mvar

12 MW

2 3 MW 7 Mvar

3 3 MW 1 3 Mvar

15 .9 Mvar 1 8 MW 5 Mvar

58 MW 40 Mvar

4.9 Mvar

39 MW 13 Mvar

1 7 MW 3 Mvar


DEMA R6 9

BLT 69

BLT1 38

BOB6 9

SHIMKO6 9

ROGER6 9

UIUC6 9

HA NNA H69

A MA NDA 69

HOMER69

LA UF6 9

LA UF1 38

HA LE69

PA T TEN6 9

WEBER69

BUCKY1 38

SA VOY69

SA VOY138

JO1 38 JO34 5

A

MVA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1 .00 pu

1.02 pu

1.01 pu

1.00 pu

1 .0 1 pu

1 .0 1 pu

1.00 pu0 .9 9 pu

0 .99 pu

1.00 pu

1.0 2 pu

1 .00 pu1 .0 1 pu

1.01 pu

1 .0 0 pu 1.00 pu

1 .01 pu

1 .0 2 pu 1.02 pu

1.02 pu 1.0 3 pu

A

MVA

LYNN1 38

1.02 pu

A

MVA

1.00 pu

A

MVA

2 0 MW 1 2 Mvar

124 MW 4 5 Mvar

12 MW 5 Mvar

1 50 MW 0 Mvar

56 MW

13 Mvar

1 5 MW 5 Mvar

1 4 MW

2 Mvar

3 8 MW 3 Mvar

4 5 MW 0 Mvar

25 MW 36 Mvar

36 MW 10 Mvar

1 0 MW 5 Mvar

22 MW 15 Mvar

6 0 MW 1 2 Mvar

2 0 MW 2 8 Mvar

60 MW 19 Mvar

1 4.2 Mvar

2 5 MW 1 0 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 1 4 MW

3 Mvar

7.3 Mvar

12 .8 Mvar

28 .9 Mvar

7.4 Mvar

0.0 Mvar

5 5 MW 2 5 Mvar

1 50 MW 0 Mvar

16 MW -1 4 Mvar

1 4 MW 4 Mvar

KYLE69A

MVA

Good Power System Operation

• Good power system operation requires that there be no reliability violations for either the current condition or in the event of statistically likely contingencies• Reliability requires as a minimum that there be no transmission

line/transformer limit violations and that bus voltages be within acceptable limits (perhaps 0 95 to 1 08)


limits (perhaps 0.95 to 1.08)

• Example contingencies are the loss of any single device. This is known as n-1 reliability.

• North American Electric Reliability Corporation now has legal authority to enforce reliability standards (and there are now lots of them). See http://www.nerc.com for details (click on Standards)

Looking at the Impact of Line Outages

slack

Metropolis Light and Power Electric Design Case 2SLA CK345

SLA CK138

RA Y345

RA Y138

RA Y69

FERNA 69

A

MVA

DEMA R69BOB138

WOLEN69

PET E6 9HISKY6 9

T IM69

T IM138

T IM345

PA I69

GROSS69

HA NNA H69

MORO138

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.03 pu

1 .0 2 pu

1 .0 3 pu

1.03 pu

1.01 pu

1.00 pu1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.01 pu

1.02 pu

A

MVA

1.02 pu

A

MVA

A

MVA

A

MVA


227 MW 4 3 Mvar

12 MW 3 Mvar

37 MW

13 Mvar

12 MW 5 Mvar

23 MW 7 Mvar

33 MW 13 Mvar

16.0 Mvar 18 MW 5 Mvar

58 MW 40 Mvar

60 MW19 Mvar

4 .9 Mvar

2 8.9 Mvar

3 9 MW 1 3 Mvar

17 MW 3 Mvar


BLT 69

BLT 138

BOB69

SHIMKO69

ROGER69

UIUC69

A MA NDA 69

HOMER6 9

LA UF69

LA UF138

HA LE69

P A T T EN69

WEBER69

BUCKY138

SA VOY69

SA VOY138

JO138 JO345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.00 pu

1.02 pu

1.01 pu

1.00 pu

1.01 pu

1.01 pu

1.00 pu0.90 pu

0.90 pu

0.94 pu

1 .0 1 pu

0.99 pu1.00 pu

1.00 pu

1 .0 0 pu 1.00 pu

1.01 pu

1.01 pu 1.02 pu

1.02 pu 1.03 pu

LYNN1 38

1.02 pu

A

MVA

1.00 pu

A

MVA

20 MW 12 Mvar

124 MW 45 Mvar

150 MW 4 Mvar

5 6 MW

1 3 Mvar

15 MW 5 Mvar

14 MW

2 Mvar

3 8 MW 9 Mvar

45 MW 0 Mvar

25 MW 36 Mvar

3 6 MW 1 0 Mvar

10 MW 5 Mvar

22 MW 15 Mvar

60 MW 12 Mvar

20 MW 40 Mvar

19 Mvar

11.6 Mvar

25 MW 10 Mvar

20 MW 3 Mvar

23 MW 6 Mvar 14 MW

3 Mvar

7 .2 Mvar

1 2.8 Mvar

7 .3 Mvar

0 .0 Mvar

55 MW 32 Mvar

150 MW 4 Mvar

16 MW -14 Mvar

14 MW 4 Mvar

KYLE69A

MVA

80%A

MVA

135%A

MVA

110%A

MVA

Opening one line (Tim69-Hannah69) causes an overload. This would not be allowed (i.e., we can’t operate this way when line is in.

9/16/2011

9

Contingency Analysis

Contingencyanalysis providesan automaticway of lookingat all the

i i ll


statisticallylikely contingencies. Inthis example thecontingency setIs all the single line/transformeroutages

Generation Changes and The Slack Bus

• The power flow is a steady-state analysis tool, so the assumption is total load plus losses is always equal to total generation• Generation mismatch is made up at the slack bus

• When doing generation change power flow studies


When doing generation change power flow studies one always needs to be cognizant of where the generation is being made up• Common options include system slack, distributed across

multiple generators by participation factors or by economics

Generation Change Example 1slack

SLA CK345

SLA CK138

RA Y345

RA Y138

RA Y69

FERNA 69

A

MVA

DEMA R69BOB138

BOB69

WOLEN69

UIUC6 9

PET E69

HISKY69

T IM69

T IM138

T IM345

PA I69

GROSS6 9

HA NNA H6 9

MORO138

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVAA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

0.00 pu

-0.01 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu-0.01 pu

0.00 pu

0.00 pu

0.0 0 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

A

MVA

-0 .01 pu

A

MVA

A

MVA

LYNN138

A

MVA

0.00 pu

162 MW 35 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-157 MW -45 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-0 .1 Mvar 0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

-0 .1 Mvar

-0 .1 Mvar

-0 .1 Mvar

-0 .2 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW


BLT 69

BLT 138

SHIMKO69

ROGER6 9

A MA NDA 6 9

HOMER69

LA UF69

LA UF138

HA LE69

PA T T EN69

WEBER69

BUCKY138

SA VOY69

SA VOY138

JO138 JO345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

0.00 pu

-0.03 pu

-0.0 1 pu0.0 0 pu

-0.002 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu0.0 0 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu0.00 pu

0.00 pu

A

MVA

A

MVA

0 MW 2 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

0 MW 3 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 4 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

0 .0 Mvar

0 .0 Mvar

0 .0 Mvar

0 MW 51 Mvar

0 MW 2 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

Display shows “Difference Flows” between original 37 bus case, and case with a BLT138 generation outage; note all the power change is picked up at the slack

Generation Change Example 2

slack

SLA CK345

SLA CK138

RA Y345

RA Y138

RA Y69

FERNA 69

A

MVA

DEMA R69BOB138

BOB69

WOLEN69

UIUC6 9

PET E69

HISKY69

T IM69

T IM138

T IM345

PA I69

GROSS6 9

HA NNA H6 9

MORO138

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVAA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

A

A

MVA

A

MVA

0.00 pu

-0.01 pu

0.00 pu

0.00 pu

0.00 pu

-0.03 pu0.00 pu

0.00 pu

0.00 pu

0.0 0 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu

A

MVA

0.00 pu

A

MVA

A

MVA

LYNN138

A

MVA

0.00 pu

0 MW 37 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-157 MW -45 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

-0 .1 Mvar 0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

-0 .1 Mvar

0 .0 Mvar

-0 .1 Mvar

-0 .2 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar


BLT 69

BLT 138

SHIMKO69

ROGER6 9

A MA NDA 6 9

HOMER69

LA UF69

LA UF138

HA LE69

PA T T EN69

WEBER69

BUCKY138

SA VOY69

SA VOY138

JO138 JO345

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

MVA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

0.00 pu

-0.03 pu

-0.0 1 pu-0.01 pu

-0.003 pu

0.00 pu

0.00 pu

0.00 pu

0.00 pu0.0 0 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu

0.00 pu 0.00 pu

0.00 pu0.00 pu

0.00 pu

A

MVA

A

MVA

0 MW 0 Mvar

0 MW 0 Mvar

0 MW

0 Mvar

42 MW -14 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

99 MW -20 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

0 MW 0 Mvar 0 MW

0 Mvar

0 .0 Mvar

-0 .1 Mvar

0 .0 Mvar

19 MW 51 Mvar

0 MW 0 Mvar

0 MW 0 Mvar

Display repeats previous case except now the change in generation is picked up by other generators using a participation factor approach

Sitting New Wind Generation Examplesl

SLACK138

RAY345

RAY138

RAY69

FERNA69

A

MVA

DEMAR69BOB138

PETE69HISKY69

TIM69

TIM138

TIM345

PAI69

GROSS69

HANNAH69

MORO138

A

MVA

A

MVA

MVA

A

MVA

A

MVA

A

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

1.02 pu

1.01 pu

1.02 pu

1.03 pu

1.00 pu

0.99 pu1.01 pu

1.01 pu

1.01 pu

1.02 pu

1.00 pu

1.02 pu

A

1.02 pu

A

MVA

A

MVA


12 MW 3 Mvar

12 MW5 Mvar


BLT69

BLT138

BOB69

SHIMKO69

ROGER69

UIUC69

AMANDA69

HOMER69

LAUF69

HALE69

PATTEN69

WEBER69

A

MVA

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A

MVA

A A

A

MVA

A

MVA

A

MVA

1.00 pu

1.01 pu

1.01 pu

0.99 pu

1.01 pu

1.00 pu

1.00 pu1.00 pu

1.00 pu

1.01 pu

1.02 pu1.01 pu

1.00 pu 1.00 pu

MVA

1.00 pu

A

MVA

20 MW 12 Mvar

0 MW 0 Mvar

5 Mvar

56 MW

13 Mvar

14 MW

2 Mvar

Wind69

50 MW

-2 Mvar

What does the Future Hold for Wind?

• Wind has experienced rapid growth over the last decade; but with a large drop in installations expected for 2010.

• Nuclear and renewal generation are running into strong “head winds”

d b l


caused by lownatural gas prices.

• Expiring tax credits is acontinual concern.

• State renewable portfolio standards willhelp with growth.

9/16/2011

10

Hurricanes Impact Energy Prices

• Many oil refineries and natural gas pipelines off the coast

• Need to be shut down

Hurricane Ike, Sept. 2008


Need to be shut down and evacuated

• Takes time to get the systems back up and running afterwards http://tonto.eia.doe.gov/oog/special/hurricanes/

gustav_091308.html

Hurricanes Impact Energy Prices

Energy Systems Research Laboratory, FIUhttp://www.eia.doe.gov/emeu/steo/pub/special/pdf/2008_sp_03.pdf

What does the Future Hold for Wind?Vestas to Cut 3000 Jobs (14%) (10/26/10)

Vestas Stock Price Over Last Five Years

Closedat 176 onWednesday and 169.4 on Thursday!


Long-term the outlook for wind is probably good. On 9/1/10 EIA reduced their 2010 forecast for US capacity additions to 4.3 GW in 2010 and 6.5 GW in 2011; the totals for 2007 were 5.2 GW, 2008 8.4 GW and 2009 10.0GW

Distributed Generation (DG)

• Small-scale, up to about 50 MW

• Includes renewable and non-renewable sources

M b i l t d f th id id


• May be isolated from the grid or grid-connected

• Usually near the end user

Integrated Generation, Transmission, Buildings, Vehicles

kWh

Renewables

Grid


PHEV

N. Gas

Heat kWh

kWh

Smartmeters

Vehicle-to-Grid

CombinedHeat andPower (CHP)

Source: Masters

Pluggable Hybrid Electric Vehicles (PHEVs) as Distributed Generation

• Can charge at night when electricity is cheap


• Can provide services back to the grid

Source: www.calcars.org

Source: http://www.popularmechanics.com/automotive/new_cars/4215489.html

9/16/2011

11

San Francisco charging stations, 2009

Energy Systems Research Laboratory, FIUhttp://en.wikipedia.org/wiki/File:Charging_stations_in_SF_City_Hall_02_2009_02.jpg

DG Technologies, Excepting Solar

• Microturbines

• Reciprocating Internal Combustion Engines

• Biomass

• Micro-Hydro


y

• Fuel Cells (we’ll be skipping)

• Concentrated solar power is talked about in Chapter 4, but we’ll skip until after we cover Chapter 7.

Reasons for Distributed Generation

• Good for remote locations

• Renewable resources

• Reduced emissions


• Can use the waste heat

• Can sell power back to the grid

Terminology

• Cogeneration and Combined Heat and Power (CHP) – capturing and using waste heat while generating electricity

• When fuel is burned one product is water; if water vapor exits stack then its energy is lost (about 1060 Btu per pound of water vapor)


p p )

• Heat of Combustion for fuels– Higher Heating Value (HHV) – gross heat, accounts for latent

heat in water vapor

– Lower Heating Value (LHV) – net heat, assumes latent heat in water vapor is not recovered

– Both are used - Conversion factors (LHV/HHV) in Table 4.2

HHV and LHV Efficiency

• Find LHV efficiency or HHV efficiency from the heat rate:

HHV( )HHV( )

3412 Btu/kWh (3.16)

Heat Rate (Btu/kWh)LHVLHV


• Convert to get the other efficiency:

HHV( )LHV

HHV LHV

LHV (4.1)

HHV

Note the LHV is less than the HHV

Microturbines

• Small natural gas turbines, 500 W to 100s kW

• Only one moving part

• Combined heat and power

230 kW fuel

120 kW hot water output


Combined heat and power

• High overall efficiency

Source: http://www.capstoneturbine.com

Capstone 65 kW Microturbine

80% CHPEfficiency

65 kW electrical output

45 kW waste heat

9/16/2011

12

Microturbines

1. Incoming air is compressed

2. Moves into cool side of recuperator & is heated

3. Mixes with fuel in


combustion chamber

4. Expansion of hot gases spins shaft

5. Exhaust leaves

Figure 4.1

Microturbines and Renewable Energy

• Microturbines are not a renewable energy source since they ultimately use natural gas as their fuel. But when used for combined heat and electricity they can be quite efficient and emit significantly less CO2 than coal.

• A possible scenario: If natural gas prices remain low,


commercial/industrial entities will increasingly go “off grid.” Hence they will not be subject to state renewable portfolio standards or utility taxes. This will mean renewable energy subsidies will increasingly be born by residential customers. Average residential price per therm is $0.57 in October in Illinois; with a heat rate of 8.5 this gives electricity at 4.8 cents/kWh.

Biomass – Current Conditions

• Biomass, including waster, currently provides about 4% of the US total energy, a value that is projected to grow to about 7% by 2030.

• In 2008 the 3.85 quad of biomass was split between biofuels such as ethanol (1.37 quad), waste such as landfill gas (0.436 quad)


and wood (2.04 quad). – About 1/10 of the wood was used to produce electricity, primarily by the

paper industry; total generation capacity in US is about 11.3 GW

– We are not considering liquid fuels for non-electricity use in ECE 333; ethanol usage was growing at 30% per year

• Little to no fuel cost for wood waste but little growth

Biomass – Future Possibilities for Electricity

• Newer crops are being considered for future biomass, including various grasses (such as Miscanthus and Switch Grass) along with algae

• Potential uses include both fuel to t l t i it ( i il th


create electricity (primarily the grasses), and conversion to liquid fuels such as ethanol.– On campus 320 acres in the South

Farms are devoted to biofuel research

Biomass – Future Possibilities for Electricity

• A full consideration of biomass is beyond our scope of since it gets into agricultural economics issues

• Miscanthus can be harvested at rates of about 15 tons per acre in Illinois Once established it does not need to be replanted This


Illinois. Once established it does not need to be replanted. This allows the energy potential of about 225 Mbtu per acre; income depends on energy price, say $2/Mbtu = $450 per acre. For comparison corn can yield up to 200 bushels per acre at say $5.50/bushel = $1100 per acre– Energy yield is about 225/15 = 15 Mbtu/ton which is similar to coal

– Not a native species so containment could be an issue

Cofiring

• Burn biomass and coal

• Modified conventional steam-cycle plants

• Allows use of biomass in plants with higher ffi i i


efficiencies

• Reduces overall emissions

9/16/2011

13

Gas Turbines and Biomass

• Cannot run directly on biomass without causing damage

• Gassify the fuel first and clean the gas before combustion


combustion

• Coal-integrated gasifier/gas turbine (CIG/GT) systems

• Biomass-integrated gasifier/gas turbine (BIG/GT) systems

Biomass and Transportation

• A key issue associated with biomass is the transportation costs – these grasses are quite bulky.

• A rough estimate of the cost per ton for transportation is about $1 + 0.1 * round trip distance in miles + h ti t f b t $22 t S if th


harvesting costs of about $22 per ton. So if the power plant is 50 miles distant, total cost would be $33 per ton, or about $33/ton/($15 Mbtu/ton) = $2.2 per Mbtu

• Total US corn planting is about 87 million acres, which if planted in grass could yield about 20 quad of energy.– Won’t meet all our needs but could play a major role

A Biomass plant


Documents

Lecture Power System