Power FlowRobert R. Krchnavek

Rowan UniversityGlassboro, New Jersey

Power Flow Calculations• Required for operation and planning.

• If all the voltages (magnitude and phase) could be measured, it would be a simple task to determine real and reactive power everywhere (linear system).

• Utilities measure voltage magnitude, real power and reactive power a certain buses.

• SCADA - supervisory control and data acquisition.

Power System – Terminology• Load Buses - P and Q are specified. These are

called PQ buses.

• Generator Buses - Voltage magnitude and real power, P, are specified. PV buses.

• Slack Bus - Voltage magnitude is specified, at 0˚ phase, and P and Q can be whatever! An “infinite” bus.

• P and Q are not specified. Neither is the voltage. Usually for transformers. Similar to PQ buses with P = Q = 0.

Example Power System(see textbook)

Example Power System

Example Power SystemAssuming a balanced 3-phase system, capable ofdelivering 100 MVA, determine the per-phase andper-unit quantities:

Vph =VLLp

3

Iph =1

3

MVA

Vph= Ibase

Zbase =Vbase

Ibase

199 kV

167 kA

1190.25 Ω

Ybase =1

Zbase

per-unit values

840 µΩ

Example Power SystemUsing a pi-circuit model for the transmission line, what are the line impedances (actual) and per-unit values?

For the 200 km long line:Zseries = 200⇥ (0.037 + |0.376) = (7.40 + |75.2) ⌦

! (0.0062 + |0.063) pu

Yshunt = �|200⇥ 4.5(10�6) f = 900 µf ! 1.07pu

Note: Be careful when working with susceptances and reactances and how they add.

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemReconfiguring for node-voltage analysis:

Example Power SystemWrite the node-voltage equations:

A linear system of equations that can easily be solved numerically. Note, in the above equations, etc.

�I1 +V1 � V2

Z12+ V1Y1G +

V1 � V3

Z13= 0

I1 = V1Y1G +V1 � V2

Z12+

V1 � V3

Z13= V1Y1G + (V1 � V2)Y12 + (V1 � V3)Y13

I2 = V2Y2G + (V2 � V3)Y23 + (V2 � V1)Y12

I3 = V3Y3G + (V3 � V1)Y13 + (V3 � V2)Y23

Y12 = Y21

Example Power SystemThe node-voltage equations can be generalized:

Ik = VkYkG +X

m,m 6=k

(Vk � Vm)Ykm

Ik = Vk

0

@YkG +X

m,m 6=k

Ykm

1

A�X

m,m 6=k

VmYkm

Example Power SystemFor an arbitrarily large system with similar properties,we can create an Admittance Matrix.

First, some nomenclature:

Ykk = YkG +X

m,m 6=k

Ykm

and redefine in the second summation on the

previous page to be instead of .

This effectively buries the minus sign inside .

Ykm

� 1

Zkm

1

Zkm

Ykm

Example Power System2

6666664

I1I2······In

3

7777775=

2

6666664

Y11 Y12 ·· ·· ·· Y1n

Y21 Y22 ·· ·· ·· Y2n

·· ·· ·· ·· ·· ···· ·· ·· ·· ·· ···· ·· ·· ·· ·· ··Yn1 Yn2 ·· ·· ·· Ynn

3

7777775

2

6666664

V1

V2

······Vn

3

7777775

Ykk = YkG +X

m,m 6=k

1

Zkm

Ykm = � 1

Zkm

where

and

A linear system of equations that can easily be solved numerically.

Power Flow Equations

• In the matrix representation, the unknowns are the voltages (node-voltage analysis).

• The current injected at each node (I1, I2, etc.) is required. Unfortunately, these are not exactly known.

• PV buses - P is known and the magnitude of V.

• PQ buses - P and Q are known.

2

6666664

I1I2······In

3

7777775=

2

6666664

Y11 Y12 ·· ·· ·· Y1n

Y21 Y22 ·· ·· ·· Y2n

·· ·· ·· ·· ·· ···· ·· ·· ·· ·· ···· ·· ·· ·· ·· ··Yn1 Yn2 ·· ·· ·· Ynn

3

7777775

2

6666664

V1

V2

······Vn

3

7777775

Power Flow EquationsSk = Pk + |Qk = VkI

⇤k

I1 = Y11V1 +Y12V2 + · · ·

Ik =nX

m=1

YkmVm

Ykm = Gkm + |Bkm

Sk = Pk + |Qk = VkI⇤k = Vk

nX

m=1

Y⇤kmV⇤

m

= Vk

nX

m=1

(Gkm � |Bkm)V⇤m

Power Flow EquationsSk = Pk + |Qk =

nX

m=1

(Gkm � |Bkm)VkV⇤m

Pk + |Qk =nX

m=1

(Gkm � |Bkm)Vk ✓kVm �✓m

Pk + |Qk =nX

m=1

(Gkm � |Bkm)VkVm ✓k � ✓m

Pk + |Qk =nX

m=1

(Gkm � |Bkm)VkVme✓k�✓m

Pk + |Qk =

nX

m=1

(Gkm � |Bkm)VkVm [cos (✓k � ✓m) + | sin (✓k � ✓m)]

Power Flow Equations

nX

m=1,m 6=k

VkVm [Gkm cos (✓k � ✓m) +Bkm sin (✓k � ✓m)]

Pk = GkkV2k +

Qk = BkkV2k +

nX

m=1,m 6=k

VkVm [Gkm sin (✓k � ✓m)�Bkm cos (✓k � ✓m)]

Power Flow Equations

• Assume we have the following types of buses: one slack bus, PV-buses, and buses.

• equations where P is specified.

• equations where P and Q are specified.

• equations for Pk and/or Qk.

Consider an n-bus system:

nPV

nPQ

nPQ

1 + nPV + nPQ = n

nPV + nPQ

nPV + 2nPQ

Power Flow Equations

• Unknowns:

• unknown voltage magnitudes.

• unknown voltage phase angles.

• Total unknowns:

• Equal number of unknowns and equations.

nPQ

nPV + nPQ

nPV + 2nPQ

Power Flow EquationsP spk � Pk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0

Qspk �Qk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0

specified values of real and reactive power

All buses exceptthe slack bus.

All PQ buses.

Newton-Raphson Method

• Common method of solving a nonlinear system of equations.

• Relatively quick and likely to converge. Convergence usually occurs within 10 iterations regardless of matrix size.

• Other techniques are available, e.g., Gauss-Seidel.

Newton-Raphson Method

• c is a constant and f(x) is nonlinear.

• an initial guess of the solution.

• an adjustment in x gets closer to the solution. What should be?

• A Taylor series expansion is used to calculate a and a new x value.

c� f(x) = 0

x

(0)

c� f(x(0) +�x) ⇡ 0

f(a+ x) =NX

n=0

x

n

n!f

(n)(a) +RN

�x

�x

Newton-Raphson Methodf(a+ x) =

NX

n=0

x

n

n!f

(n)(a) +RN

f(x(0) +�x) = f(x(0)) +�x

df

dx

����x

(0)

+ · · ·

c� f(x(0) +�x) ⇡ 0 ) c�f(x(0)) +�x

df

dx

����x

(0)

�= 0

�x =c� f(x(0))

df

dx

���x

(0)

x

(1) = x

(0) +�x

Newton-Raphson Methodx

(1) = x

(0) +�x

�x =c� f(x(1))

df

dx

���x

(1)

x

(2) = x

(1) +�x

calculate a new :�x

and then a new x value:

and again, ……

Newton-Raphson Method

See Example 5.2 in the Textbook

and understand it !

N-R Method forPower Flow Equations

• Apply the Newton-Raphson method to the matrix of Pk and Qk equations.

• Pk and Qk are in estimates of voltage magnitudes and phases, or totally unknown voltages.

c� f(x) = 0

P spk � Pk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0

Qspk �Qk (V1, · · · , Vn, ✓1, · · · , ✓n) = 0

N-R Method forPower Flow Equations

c� f(x(0)) = �x

df

dx

����x

(0)

P sp � PQsp �Q

�

| {z }(2nPQ+nPV )⇥1

=

2

4@P@✓

@P@V

@Q@✓

@Q@V

3

5

| {z }[J]

�✓�V

�

| {z }(2nPQ+nPV )⇥1

J is the Jacobian and consists of submatrices.

N-R Method forPower Flow Equations

nX

m=1,m 6=k

VkVm [Gkm cos (✓k � ✓m) +Bkm sin (✓k � ✓m)]

Pk = GkkV2k +

Recall:

@Pk

@✓k=

nX

m=1,m 6=k

VkVm

[�Gkm sin (✓k � ✓m) +Bkm cos (✓k � ✓m)]