using optimization in smart grid applicationsusing optimization in smart grid applications dr.-ing....

Informatik 7

Rechnernetze und

Kommunikationssysteme

Using Optimization in Smart Grid Applications

Dr.-Ing. Abdalkarim Awad

04.11.2015

Why Optimization

Smart Grid allows better coordination between the different entities

Optimization can be used to find the best strategy, size of different components,…

We are not going to focus on the optimization algorithms rather than on how to use them

There are a lot of tools that can be used

We are going to use some tools to solve the problems.

Dr.-Ing. Abdalkarim Awad 1

Several tools

ADMB

CONDOR

Joptimizer

NLOPT

Dr.-Ing. Abdalkarim Awad 2

R

Nlopt

+

3

What is Optimization?

Optimization is the mathematical discipline which is concerned with finding the maxima and minima of functions, possibly subject to constraints.

Dr.-Ing. Abdalkarim Awad

What do we optimize?

A real function of n variables

with or without constraints

),,,( 21 nxxxf

Dr.-Ing. Abdalkarim Awad

Unconstrained optimization

22 2),(min yxyxf

Dr.-Ing. Abdalkarim Awad

Optimization with constraints

2

2),(min

1,52

2),(min

0

2),(min

22

22

22

or

or

yx

yxyxf

yx

yxyxf

x

yxyxf

Dr.-Ing. Abdalkarim Awad

Nloptr Package

We are going to use a package called nloptr to solve non-linear optimization problems.

Dr.-Ing. Abdalkarim Awad 8

Example1

GA GB

Max=120 Mvar

LB=300LA=300

Euro/hr 0.00149PG11.83PG616.9)(PGC

Euro/hr 0.00334PG11.69PG 399.8)(PGC

2

BBBB

2

AAAA

AB

Dr.-Ing. Abdalkarim Awad 9

Example1

Total Hourly Cost :

Bus A Bus B

300.0 MWMW

199.6 MWMW 400.4 MWMW

300.0 MWMW

8459 $/hr

Area Lambda : 13.02

AGC ON AGC ON

Dr.-Ing. Abdalkarim Awad 10

Generator cost curves

PGA+PGB=600

Euro/hr 0.00149PG11.83PG616.9)(PGC

Euro/hr 0.00334PG11.69PG 399.8)(PGC

2

BBBB

2

AAAA

Dr.-Ing. Abdalkarim Awad 11

Generator cost curves

How much each generator should produce to cover the 600MW demand

The Problem can be written as:

MINIMIZE (CA(PGA)+CB(PGB))

Subject to

PGA+PGB=600

PGA>=0

PGB>=0

Dr.-Ing. Abdalkarim Awad 12

Code

Dr.-Ing. Abdalkarim Awad 13

Code

Dr.-Ing. Abdalkarim Awad 14

Run

Dr.-Ing. Abdalkarim Awad 15

Example2

Add constraints to generator 2

0<PGB<250

What is the total costs now?

Dr.-Ing. Abdalkarim Awad 16

Example3

Max=50 MvarZ=0.1j

GA GBLB=300 MW

LA=300 MW

AB

Write R script to find P1 and P2 that minimizes the operation

costs and takes into considerations line limitations

Dr.-Ing. Abdalkarim Awad 17

Including power Flow

2

1

1010

1010

3

3

PB

PA

N

k

kiiki BP1

))((

P1=(B12)(θ1-θ2)

P1=(B12)θ1-B12(θ2)

P2=B21(θ2- θ1)

P2=-B21(θ1)+(B21)(θ2)

Z=0.1j y=-10jB=-10

Dr.-Ing. Abdalkarim Awad 18

2

1

1010

1010

3

3

PB

PA

PB-3=-(10)θ2

PB-3+(10)(θ2)=0

Solve with respect

to bus #1. i.e θ1=0

(Slack bus)

Dr.-Ing. Abdalkarim Awad 19

P12=B12*(θ1-θ2)

|P12|=|10(θ1-θ2)|<0.5

-0.5<10*θ2<0.5

-0.05<θ2<0.05

Example 4

20

P1 P2

BUS2

BUS3

500 MW

Dr.- Ing. Abdalkarim Awad

100MW

L1=50

Generator cost curves

C1(P1) = 400 + 9 x P1 + 0.0015 x (P1)2

100 MW ≤ P1 ≤ 600 MW

C3(P2) = 100 + 8 x P2 + 0.0048 x (P2)2

50 MW ≤ P2 ≤ 400 MW

Minimize (C1(P1)+C3(P2))

Dr.- Ing. Abdalkarim Awad 21

Optimization Problem

Minimize (C1(P1)+C2(P2))

Subject to:

100 MW ≤ P1 ≤ 600 MW

50 MW ≤ P2 ≤ 400 MW

Dr.- Ing. Abdalkarim Awad 22

Solution

9 + 2*0.0015 x (P1)=λ

8+ 2*0.0048 x (P2) =λ

P1+P2=500+50-100=450

P1=263.5

P2=186.5

Minimum Cost = 4634.60

Dr.- Ing. Abdalkarim Awad 23

Do they satisfy all constraints?

24

P1 P2

BUS2

BUS3

BUS1

y12=-10j

y13=-10jy23=-10j

L3=500= 5 pu

P13<200=2 pu

SB=100 MVA

100MW= 1 pu

P13=221.2MW

P12=42.3

P23=178.8

P1=263.5MW 186.5MWL1=50= 0.5 pu

Including power Flow

Dr.- Ing. Abdalkarim Awad 25

3

2

1

32213231

23232121

13121312

3

2

1

BBBB

BBBB

BBBB

P

P

P

N

k

kiiki BP1

))((

P1=(B12)(θ1-θ2)+B13(θ1-θ3

P1=(B12+B13)θ1-B12(θ2)-B13(θ3)

P2=-B21(θ1)+(B21+B23)(θ2)-B23(θ3)

P3=-B31(θ1)-B32(θ2)+(B31+B32)(θ3)

Compare to Y matrix!

Dr.- Ing. Abdalkarim Awad 26

3

2

1

201010

102010

101020

51

5.02

1

P

P

singular P1 P2

BUS2

BUS3

BUS1

y12=-10j

y13=-10jy23=-10j

L3=500= 5 pu

P13<200=2 pu

SB=100 MVA

100MW= 1 pu

L1=50= 0.5 pu

Dr.- Ing. Abdalkarim Awad 27

3

2

2010

1020

4

5.02

P

P13<200 B13(θ1-θ3)<2 (pu) θ3<0.20

(B13=-10)

Solve with respect

to bus #1. i.e θ1=0

(Slack bus)

3

2

1

20100

102010

01020

4

5.02

1

P

P

Optimization Problem

Minimize (C1(P1)+C2(P2)) (in pu)

Subject to:

1 pu ≤ P1 ≤ 6 pu

0.5 pu ≤ P2 ≤ 4 pu

P1+P2=4

Dr.- Ing. Abdalkarim Awad 28

3

2

2010

1020

4

5.02

P

Θ3<0.20

Solution (using R)

Dr.- Ing. Abdalkarim Awad 29

2.0

0

0

3

2

1

P1=200 MW

P2=250 MW

P13=-B13(θ1-θ3)=-10*(0-0.2)=2 pu=200 MW

Costs=4659.95>4634.60 Euro

30

P1 P2

BUS2

BUS3

BUS1-10j

-10j -10j

5

P13<1.7

1

0.5 pu

Optimization Problem

Minimize (C1(P1)+C2(P2)) (in pu)

Subject to:

1 pu ≤ P1 ≤ 6 pu

0.5 pu ≤ P2 ≤ 4 pu

Dr.- Ing. Abdalkarim Awad 31

P13<170 B13(θ1-θ3)<1.70 θ3<0.17

3

2

2010

1020

4

2

P

Solution (using R)

The total generation cost becomes: 4783.03> 4634.60

Here, we had to sacrifice “cost” for “implementation”.

Dr.- Ing. Abdalkarim Awad 32

P1=110

P2=340

θ2=-0.06

θ3=0.17