Lecture6 The Fast Decoupled Method

Example l

Using the Newton-Raphson PF, find the power flow solution 1 0.01 + j0.03

y12 = 10 − j 20 pu y13 = 10 − j 30 pu y23 = 16 − j 32 pu

Power Systems I

0.02 + j0.04 0.0125 + j0.025 2

3 |V3| = 1.04 200 MW

400 + j 250 S =− = − 4.0 − j 2.5 pu 100 200 P3sch = = 2.0 pu 100 sch 2

Slack Bus V1 = 1.05∠0°

400 MW 250 MVAR

Example Ybus

 20 − j 50 − 10 + j 20 − 10 + j 30 = − 10 + j 20 26 − j52 − 16 + j 32    − 10 + j 30 − 16 + j 32 26 − j 62   31.6∠1.89  53.9∠ − 1.90 22.4∠ 2.03 =  22.4∠ 2.03 58.1∠ − 1.11 35.8∠ 2.03  angles are in radians    35.8∠ 2.03 67.2∠ − 1.17  31.6∠1.89 

P2 = V2 V1 Y21 cos(θ21 − δ2 + δ1 )+ V2 Y22 cos(θ22 )+ V2 V3 Y23 cos(θ23 − δ2 + δ3 ) 2

P3 = V3 V1 Y31 cos(θ31 − δ3 + δ1 )+ V3 V2 Y32 cos(θ32 − δ3 + δ2 )+ V3 Y33 cos(θ33 ) 2

Q2 = − V2 V1 Y21 sin (θ21 − δ2 + δ1 )− V2 Y22 sin (θ22 )− V2 V3 Y23 sin (θ23 − δ2 + δ3 ) 2

Power Systems I

Example δ2    x = δ3    V2 

P2 (δ2 , δ3 ,V2 )    f (x )= P3 (δ2 , δ3 ,V2 )    Q2 (δ2 , δ3 ,V2 )

 V 1.05 22.3 cos(2.03 − δ )+ V 2 58.1 cos(− 1.11)+ V 1.04 35.8 cos(2.03 − δ + δ)  21 2 2 2 3  2  2 = V3 1.05 31.6 cos(1.89 − δ3 )+ 1.04 V2 35.8 cos(2.03 − δ3 + δ2 )+ 1.04 67.2 cos(− 1.17 )   2 ( ) ( ) ( ) − − δ − − − − δ + δ V 1 . 05 22 . 3 sin 2 . 03 V 58 . 1 sin 1 . 11 V 1 . 04 35 . 8 sin 2 . 03 2 2 2 2 2 3     P2sch  ∆P2    ∆c = ∆P3  = c − f (x )= P3sch −   Q2sch   ∆Q2    

Power Systems I

P2 (δ2 , δ3 ,V2 )  − 4.0    − = δ δ P ( , , V ) 2 . 0 3 2 3 2       − 2.5 Q2 (δ2 , δ3 ,V2 )  

P2 (δ2 , δ3 ,V2 )    δ δ P V ( , , )  3 2 3 2   Q2 (δ2 , δ3 ,V2 ) 

Example ∂P2 = ∂δ2

3

∑V

j =1, j ≠ 2

2

V j Y2 j sin (θ2 j − δ2 + δj )

= V2 V1 Y21 sin(θ21 − δ2 )+ V2 V3 Y23 sin (θ23 − δ2 + δ3 )

= V2 1.05 22.4 sin(2.03 − δ2 )+ V2 1.04 35.8 sin (2.03 − δ2 + δ3 ) ∂P2 = − V2 V3 Y23 sin(θ23 − δ2 + δ3 )= − V2 1.04 35.8 sin(2.03 − δ2 + δ3 ) ∂δ3 ∂P2 = 2 V2 Y22 cos(θ22 )+ ∂V2

3

∑V

j =1, j ≠ 2

j

Y2 j cos(θ2 j − δ2 + δj )

= 2 V2 Y22 cos(θ22 )+ V2 Y21 cos(θ21 − δ2 + δ1 )+ V2 Y23 cos(θ23 − δ2 + δ3 ) = 2 V2 58.1 cos(2.03)+ 1.05 22.4 cos(2.03 − δ2 ) + 1.04 35.8 cos(2.03 − δ2 + δ3 )

Power Systems I

Example ∂P3 = − V3 V2 Y32 sin (θ32 − δ3 + δ2 )= − 1.04 V2 35.8 sin (2.03 − δ2 + δ3 ) ∂δ2 ∂P3 = ∂δ3

3

∑V

j =1, j ≠ 3

3

V j Y3 j sin (θ3 j − δ3 + δj )

= V3 V1 Y31 sin (θ31 − δ3 + δ1 )+ V3 V2 Y32 sin (θ32 − δ3 + δ2 )

= 1.04 1.05 31.6 sin (1.89 − δ3 )+ 1.04 V2 35.8 sin (2.03 − δ3 + δ2 ) ∂P3 = V3 Y32 cos(θ32 − δ3 + δ2 )= 1.04 35.8 cos(2.03 − δ2 + δ3 ) ∂V2

Power Systems I

Example ∂Q2 = ∂δ2

3

∑V

j =1, j ≠ 2

2

V j Y2 j cos(θ2 j − δ2 + δj )

= V2 V1 Y21 cos(θ21 − δ2 + δ1 )+ V2 V3 Y23 cos(θ23 − δ2 + δ3 )

= V2 1.05 22.4 cos(2.03 − δ2 )+ V2 1.04 35.8 cos(2.03 − δ2 + δ3 )

∂Q2 = − V2 V3 Y23 cos(θ23 − δ2 + δ3 )= − V2 1.04 35.8 cos(2.03 − δ2 + δ3 ) ∂δ3 ∂Q2 = − 2 V2 Y22 sin (θ22 )− ∂V2

3

∑V

j =1, j ≠ 2

j

Y2 j sin (θ2 j − δ2 + δj )

= − 2 V2 Y22 sin (θ22 )− V1 Y21 sin (θ21 − δ2 + δ1 )− V3 Y23 sin (θ23 − δ2 + δ3 ) = − 2 V2 58.1 sin (− 1.11)− 1.05 22.4 sin (2.03 − δ2 ) − 1.04 35.8 sin (2.03 − δ2 + δ3 )

Power Systems I

Example x [ k + 1] = x [ k ] + J − 1 ⋅∆c[ k ] [ k + 1]

δ2    = δ3  V2   

[k ]

−1

δ2  ∂P2 ∂δ2 ∂P2 ∂δ3 ∂P2 ∂V2  ∆P2    = δ3  + ∂P3 ∂δ2 ∂P3 ∂δ3 ∂P3 ∂V2  ⋅∆P3      V2    ∆Q2    ∂Q2 ∂δ2 ∂Q2 ∂δ3 ∂Q2 ∂V2   

Power Systems I

[k ]

Newton-Raphson PF Example P2sch  0.0   x [ 0 ] = 0.0 ∆c[ 0 ] = P3sch −   sch   1 . 0 Q      2  ∆x[ 0 ] = J − 1∆c[ 0 ] ∆x[ 0 ]

x [1]

P2[ 0 ]  − 4.0  [0]   − 2 . 0 = P 3    [0] Q2      − 2.5

− 1.14  − 2.86 0.562  = 1.438       − 2.28   − 0.22 

∆δ2[ 0 ]   54.28 − 33.28 24.86 − 1 − 2.86 − 0.04526  [ 0]   = ∆δ3  = − 33.28 66.04 − 16.64 1.438  = − 0.00772       [ 0 ] ∆ V2   49.72     − 0.22   − 0.02655   − 27.14 16.64

δ2[1]  0.0 + (− 0.04526) − 0.04526   = δ3[1]  = 0.0 + (− 0.00772 ) = − 0.00772     [ 1 ] V2       0.9734    1.0 + (− 0.02655)

Power Systems I

Newton-Raphson PF Example

x [1]

P2sch  − 0.04526   = − 0.00772 ∆c[1] = P3sch −   Q2sch    0.9734    

P2[1]  − 4.0  [1]   − = 2 . 0 P 3    [1] Q2      − 2.5 −1

∆x [1]

 51.72 − 31.77 21.30  = − 32.98 65.66 − 15.38   48.10   − 28.54 17.40 

− 3.901 − 0.099  1.978  = 0.0217       − 2.449   − 0.051 

− 0.099 − 0.001795 0.0217  = − 0.000985      − 0.051   − 0.001767 

δ2[ 2 ]  − 0.04526 + ( − 0.001795) − 0.04706  [2]   [2] x = δ3  = − 0.00772 + ( − 0.000985) = − 0.00870     [ 2 ] V2      0.9717      0.9734 + ( − 0.001767)  Power Systems I

Newton-Raphson PF Example x [2]

P2sch  − 0.04706   = − 0.00870 ∆c[ 2 ] = P3sch −   sch   0 . 9717 Q      2 

P2[1]  − 4.0  [1]   − = 2 . 0 P 3    [1] Q2      − 2.5 −1

∆x [ 2 ]

x [ 3]

 51.60 − 31.69 21.14  = − 32.93 65.60 − 15.35   47.95   − 28.55 17.40 

− 3.999 − 0.0002  1.999  = 0.00004       − 2.499   − 0.0001 

− 0.000216 − 0.000038  0.000038  = − 0.000002      − 0.000143   − 0.000004 

δ2[ 3]  − 0.04706 + ( − 0.000038)  − 0.04706    = δ3[ 3]  = − 0.00870 + ( − 0.000002) = − 0.008705     [ 3 ] V2      0.97168      0.9717 + ( − 0.000004) 

Power Systems I

Newton-Raphson PF Example P2sch   − 0.04706    x [ 3] = − 0.008705 ∆c[ 2 ] = P3sch −   Q2sch    0.97168     εmax = 2.5 ×10− 4

P2[1]  − 4.0  [1]   − P 2 . 0 = 3     [1] Q2      − 2.5

− 4.0 0.0000  2.0  = 0.0000      − 2.5   0.0000 

P1 = V1 Y11 cos(θ11 )+ V1 V2 Y12 cos(θ12 − δ1 + δ2 )+ V1 V3 Y13 cos(θ13 − δ1 + δ3 ) 2

Q1 = − V1 Y11 sin (θ11 )− V1 V2 Y12 sin (θ12 − δ1 + δ2 )− V1 V3 Y13 sin (θ13 − δ1 + δ3 ) 2

Q3 = − V3 V1 Y31 sin (θ31 − δ3 + δ1 )− V3 V2 Y32 sin (θ32 − δ3 + δ2 )− V3 Y33 sin (θ33 ) 2

P1 = 2.1842 pu Q1 = 1.4085 pu Q3 = 1.4617 pu Power Systems I

Fast Decoupled Power Flow l

Transmission lines and transformers have high X/R ratios u

Real power change, ∆P n n

u

Reactive power changes, ∆Q n n

u

l

is less sensitive to changes in the voltage magnitude, ∆|V| is more sensitive to changes in the phase angle, ∆δ is less sensitive to changes in the phase angle , ∆δ is more sensitive to changes in the voltage magnitude, ∆|V|

Jacobian submatrices JQd and JPV tend to be much smaller in magnitude compared to JPd and JQV

Jacobian submatrices JQd and JPV can be set to zero ∂P ∆ P = J ⋅ ∆ δ = J 0   ∆ δ ∂δ∆δ P ∆ P δ     = Pδ ∂Q   V ∆ ∆ Q   0 J ∆ Q = J ⋅ ∆ V =      QV   ∂V ∆ V QV 

Power Systems I

Fast Decoupled Power Flow l

JPV elements

∂Pi = Vi Yij cos(θij − δi + δj ) ∂V j θij ≈90° δi ≈δj ∂Pi ≈Vi Yij cos(90°)= 0.0 ∂V j

l

JQd elements

∂Qi = − Vi V j Yij cos(θij − δi + δj ) ∂δj θij ≈90° δi ≈δj ∂Qi ≈− Vi V j Yij cos(90°)= 0.0 ∂δj

Power Systems I

Fast Decoupled Power Flow l

The matrix equation is separated into two decoupled equations u

u

requires considerably less time to solve compared to the full Newton-Raphson method JPd and JQV submatrices can be further simplified to eliminate the need for recomputing of the submatrices during each iteration n n n n

some terms in each element are relatively small and can be eliminated the remaining equations consist of constant terms and one variable term the one variable term can be moved and coupled with the change in power variable the result is a Jacobian matrix with constant term elements

Power Systems I

Jacobian JPd Diagonal Terms n ∂Pi = ∑ Vi V j Yij sin (θij − δi + δj ) ∂δi j =1 j ≠i

= − Vi Yii sin (θii )+ 2

n

∑VV j =1

∂Pi 2 = − Vi Yii sin (θii )− Qi ∂δi Yii sin (θii )= Bii Vi ≈Vi 2

Power Systems I

⇒

Bii > > Qi

i

j

Yij sin (θij − δi + δj )

Qi = −

n

∑VV j =1

i

∂Pi 2 = − Vi Bii ∂δi

∂Pi = − Vi Bii ∂δi

j

Yij sin (θij − δi + δj )

Jacobian JPd Off-diagonal Terms ∂Pi = − Vi V j Yij sin (θij − δi + δj ) ∂δj δj − δi ≈0 ∂Pi = − Vi V j Yij sin (θij ) ∂δi

Yij sin (θij )= Bij ∂Pi = − Vi Bij ∂δi

Power Systems I

V j ≈1

Jacobian JQV Diagonal Terms ∂Qi = − 2 Vi Yii sin (θii )− ∂Vi

n

∑V j =1 j ≠i

j

Yij sin (θij − δi + δj )

n ∂Qi −1 = − Vi Yii sin (θii )− Vi ∑ Vi V j Yij sin (θij − δi + δj ) ∂Vi j =1

∂Qi −1 = − Vi Yii sin (θii )+ Vi Qi ∂Vi Yii sin (θii )= Bii

Power Systems I

Bii > > Qi

⇒

Qi = −

n

∑VV i

j

j =1

∂Qi = − Vi Bii ∂Vi

Yij sin (θij − δi + δj )

Jacobian JQV Off-diagonal Terms ∂Qi = − Vi Yij sin (θij − δi + δj ) ∂V j δj − δi ≈0 ∂Qi = − Vi Yij sin (θij ) ∂V j

Yij sin (θij )= Bij ∂Qi = − Vi Bij ∂V j

Power Systems I

Fast Decoupled Power Flow l

Individual power change equations in JPd and JQV n

∆Pi = ∑ − Vi Bij ∆δj

⇒

n ∆Pi = ∑ − Bij ∆δj Vi j =1

⇒

n ∆Qi = ∑ − Bij ∆ V j Vi j =1

j =1 n

∆Qi = ∑ − Vi Bij ∆ V j j =1

l

Matrix equation for JPd and JQV

∆P = − B′ ∆δ Vi ∆Q ′ = − B′ ∆V Vi Power Systems I

∆P V

⇒

]− 1 B′ ∆δ= − [

⇒

∆Q ′ ] B′ ∆V = − [ V −1

Example l

Using the fast decoupled PF, find the power flow solution 1 0.01 + j0.03

y12 = 10 − j 20 pu y13 = 10 − j 30 pu y23 = 16 − j 32 pu

Power Systems I

0.02 + j0.04 0.0125 + j0.025 2

3 |V3| = 1.04 200 MW

400 + j 250 S =− = − 4.0 − j 2.5 pu 100 200 P3sch = = 2.0 pu 100 sch 2

Slack Bus V1 = 1.05∠0°

400 MW 250 MVAR

Example − 52 32  B′ =  − 32 62   − 0.028182 − 0.014545 −1 [B′ ] =  0 . 014545 0 . 023636 − −   ′ B′ =[ − 52]

′ [B′ ]− 1 = [− 0.019231]

Power Systems I

Example Initial values:

V [ 0] First iteration:

1.05∠0°  = 1 . 00 ∠ 0 °    1.00∠ 0° 

P2sch  − 4.0 δ2[ k ]  0.0    x [ k ] = δ[ k ]  x [ 0 ] = 0.0 y = P3sch  =  2 . 0 3      [k ]  Q2sch      2 . 5 V −  1.0      2  Pinj 2 (x ) Pinj i = ∑ Vi V j Yij cos(θij − δi + δj )   j =1 f (x )= Pinj 3 (x ) n Qinj 2 (x ) Qinj i = − ∑ Vi V j Yij sin (θij − δi + δj )   n

j =1

Power Systems I

Example V2 2 Y22 cos(θ22 )+ V2 V1 Y21 cos(θ21 − δ2 + δ1 )+ V2 V3 Y23 cos(θ23 − δ2 + δ3 )  2  V Y cos(θ33 )+ V3 V1 Y31 cos(θ31 − δ3 + δ1 )+ V3 V2 Y32 cos(θ32 − δ3 + δ2 ) f (x )=  3 2 33 − V Y sin (θ )− V V Y sin (θ − δ + δ)− V V Y sin (θ − δ + δ) 22 2 1 21 21 2 1 2 3 23 23 2 3  2 22       V2 2 58.1 cos(− 1.11)+ V2 1.05 22.4 cos(2.03 − δ2 )+ V2 1.04 35.8 cos(2.03 − δ2 + δ3 )    2 ( ) ( ) ( ) 1 . 04 67 . 2 cos 1 . 17 1 . 04 1 . 05 31 . 6 cos 1 . 89 δ 1 . 04 V 35 . 8 cos 2 . 03 δ δ − + − + − + 3 2 3 2  =  − V 2 58.1 sin (− 1.11)− V 1.05 22.4 sin (2.03 − δ)− V 1.04 35.8 sin (2.03 − δ + δ)  2 2 2 2 2 3      

Power Systems I

Example P2sch    ∆y [ 0 ] = P3sch − Q2sch   

P2[ 0]  − 4.0  [ 0]   − P = 2 . 0 3    [ 0] Q2      − 2.5

− 1.14  − 2.86 0.562  = 1.438        − 2.28   − 0.22

∆δ2[ 0 ]  0.028182 0.014545 − 2.86 1.0  − 0.06048 =  [0]  =      − 0 . 014545 0 . 023636 1 . 438 1 . 04 0 . 00891 δ ∆      3  

[∆ V ]= [0.019231][− 0.22 1.0]= [− 0.004231] [0] 2

δ2[1] = 0.0 + (− 0.06048)= − 0.06048 δ3[1] = 0.0 + (− 0.00891)= − 0.00891

V2[1] = 1.0 + (− 0.004231)= 0.995769 Power Systems I

Example Remaining iterations: Iter 1 2 3 4 5 6 7 8 9 10

δ2 -0.060482 -0.056496 -0.044194 -0.044802 -0.047665 -0.047614 -0.046936 -0.046928 -0.047087 -0.047094

Power Systems I

δ3 -0.008909 -0.007952 -0.008690 -0.008986 -0.008713 -0.008645 -0.008702 -0.008720 -0.008707 -0.008702

|V2| 0.995769 0.965274 0.965711 0.972985 0.973116 0.971414 0.971333 0.971732 0.971762 0.971669

∆ P2 -2.860000 0.175895 0.640309 -0.021395 -0.153368 0.000520 0.035980 0.000948 -0.008442 -0.000470

∆ P3 1.438400 -0.070951 -0.457039 0.001195 0.112899 0.002610 -0.026190 -0.001411 0.006133 0.000510

∆ Q2 -0.220000 -1.579042 0.021948 0.365249 0.006657 -0.086136 -0.004067 0.020119 0.001558 -0.004688