introduction to power system analysis - tcipgย ยท power flow analysis g bus transmission line g...
TRANSCRIPT
Introduction to Power System Analysis
Hui Lin, Siming Guo
Agenda
โข Power System Notation
โข Power Flow Analysis
โข Hands on Matpower and PowerWorld
Simple Power System
Every power system has three major components
โ generation: source of power, ideally with a specified power, voltage, and frequency
โ load: consumes power; ideally with constant power consumption
โ transmission system: transmits power; ideally as a perfect conductor
Complications
โข No ideal voltage sources exist
โข Loads are seldom constant, and we need to balance supply and demand in real time
โข Transmission system has resistance, inductance, capacitance and flow limitations
โข Simple system has no redundancy so power system will not work if any component fails
Notation - Power
โข Power: Instantaneous consumption of energy
โข Power Units
โข Watts = voltage x current for dc (W)
โข kW โ 1 ร 103 Watt
โข MW โ 1 ร 106 Watt
โข GW โ 1 ร 109 Watt
โข Installed U.S. generation capacity is about 900 GW ( about 3 kW per person)
โข Maximum load of Champaign/Urbana about 300 MW
Notation - Energy
โข Energy: Integration of power over time; energy is what people really want (and pay for) from a power system
โข Energy Units
โข Joule = 1 Watt-second (J)
โข kWh โ Kilowatthour (3.6 x 106 J)
โข Btu โ 1055 J; 1 MBtu=0.292 MWh
โข U.S. electric energy consumption is about 3600 billion kWh (about 13,333 kWh per person, which means on average we each use 1.5 kW of power continuously)
Review of Phasors
Goal of phasor analysis is to simplify the analysis of constant frequency ac systems
v(t) = Vmax cos(wt + qv)
i(t) = Imax cos(wt + qI)
Root Mean Square (RMS) voltage of sinusoid
2 max
0
1( )
2
TV
v t dtT
Complex Power
max
max
max max
( ) ( ) ( )
v(t) = cos( )
(t) = cos( )
1cos cos [cos( ) cos( )]
2
1( ) [cos( )
2
cos(2 )]
V
I
V I
V I
p t v t i t
V t
i I t
p t V I
t
w q
w q
q q
w q q
Power
Complex Power, contโd
max max
0
max max
1( ) [cos( ) cos(2 )]
2
1( )
1cos( )
2
cos( )
= =
V I V I
T
avg
V I
V I
V I
p t V I t
P p t dtT
V I
V I
q q w q q
q q
q q
q q
Power Factor
Average
P
Angle
ower
Phasor Representation
Eulerโs identity: ๐๐๐ = cos ๐ + ๐ sin ๐
Phasor notation is developed by
rewriting using Eulerโs identity
๐ฃ ๐ก = 2๐cos(๐๐ก + ๐๐)
๐ฃ ๐ก = ๐๐[ 2๐๐๐๐๐ก๐๐๐]
Phasor Representation, contโd
The RMS, cosine-reference phasor is:
๐ = ๐๐๐๐ = ๐โ ๐๐
๐ = ๐(cos ๐๐ + ๐sin ๐๐)
๐ผ = ๐ผ(cos ๐๐ผ + ๐ sin ๐๐ผ)
Complex Power
๐ = ๐ ๐ผ โ
= ๐๐ผ๐๐(๐๐โ๐๐ผ)
= ๐๐ผ cos ๐๐ โ ๐๐ผ + ๐ sin ๐๐ โ ๐๐ผ
= ๐ + ๐๐
๐: real power (W, kW, MW)
Q: reactive power (var, kvar, Mvar)
S: complex power (va, kva, Mva)
Power factor (pf): cos (๐๐ โ ๐๐ผ)
If current leads voltage, then pf is leading
If current lags voltage then pf is lagging
Power Flow Analysis
G
Bus Transmission Line
G Generators Loads
(๐1, ๐1) (๐2, ๐2)
(๐3, ๐3)
โข The power flow analysis is the process of solving the steady state of the power system
โ Steady state: voltage magnitude and angle for each bus
โ Generator: modeled as constant power delivery
โ Loads: modeled as constant power consumption
โ Transmission line: modeled as constant impedance
Power Flow Analysis
๐๐บ โ ๐๐ท = ๐ ๐๐ผ ๐โ
(๐ , ๐ผ : the voltage and the current that injects into bus i)
๐ผ ๐ = ๐ผ ๐๐๐
๐ผ ๐๐ =๐ ๐โ๐ ๐
๐๐๐
(k takes indices of all buses that connected to bus i; Zik
specifies the impedance of transmission line connecting
bus i and bus k)
Bus i
To other buses
SGSD
Y-Bus (admittance matrix)
๐ผ๐ = ๐๐โ๐๐
๐๐๐๐ = ๐๐ โ ๐๐ ๐ฆ๐๐๐
= โ๐ฆ๐1๐1 + โ๐ฆ๐2๐2 +,โฆ , ๐ฆ๐๐ ๐๐๐ +,โฆ+ โ๐ฆ๐๐ ๐๐
= โ๐ฆ๐1 โ๐ฆ๐2 โฆ ๐ฆ๐๐๐ โฆ โ๐ฆ๐๐
๐1
๐2
โฎ๐๐
โฎ๐๐
Y-Bus (admittance matrix), contโd
Write ๐ผ๐ for all buses together: ๐ผ = ๐๐ , where
(๐ผ = [๐ผ1, ๐ผ2, โฆ , ๐ผ๐], ๐ = [๐1, ๐2, โฆ , ๐๐])
Construction of Y:
๐๐๐ = ๐ฆ๐๐๐ ๐๐๐ = ๐๐๐ = โ๐ฆ๐๐
๐ = ๐บ + ๐๐ต
So we have: ๐ผ๐ = (๐๐๐๐๐)๐
Power Flow Equation at Bus j
๐๐บ โ ๐๐ท = ๐๐ ๐ผ๐
โ
๐๐บ โ ๐๐ท = ๐๐ ๐ผ๐
โ= ๐๐
๐๐๐๐๐๐โ
= ๐๐๐๐๐๐ ๐บ๐๐ + ๐๐ต๐๐ ๐๐๐ ๐๐๐๐
โ
= ๐๐๐๐๐๐ ๐๐ ๐บ๐๐ โ ๐๐ต๐๐ ๐โ๐๐๐
๐
= ๐๐๐๐๐ ๐บ๐๐ โ ๐๐ต๐๐ ๐๐(๐๐โ๐๐)
= ๐๐๐๐๐ ๐บ๐๐ cos ๐๐ โ ๐๐ + ๐ต๐๐sin (๐๐ โ ๐๐)
+๐ ๐๐๐๐๐ ๐บ๐๐sin (๐๐ โ ๐๐) โ ๐ต๐๐cos (๐๐ โ ๐๐)
Bus i
To other buses
SGSD
Power Flow Equation at Bus j
๐๐บ โ ๐๐ท = ๐๐๐๐๐ ๐บ๐๐ cos ๐๐ โ ๐๐ + ๐ต๐๐sin (๐๐ โ ๐๐)
Q๐บ โ ๐๐ท = ๐๐๐๐๐ ๐บ๐๐sin (๐๐ โ ๐๐) โ ๐ต๐๐cos (๐๐ โ ๐๐)
โข Slack bus:
โ ๐ and ๐ are known, used as a reference
โข PV bus, with generators connected
โ P and V are known
โข PQ bus, with only load units connected
โ P and Q are known
Solving Power Flow Equations
โข Assuming m-1 PV buses
โ Given: ๐1, ๐1, ๐๐บ,2, ๐2, โฆ, ๐๐บ,๐, ๐๐, ๐๐ท,๐+1, ๐๐ท,๐+1,โฆ,
๐๐ท,๐, ๐๐ท,๐
โ Unknown: ๐๐บ,1, ๐๐บ,1, ๐๐บ,2, ๐2, โฆ, ๐๐บ,๐, ๐๐, ๐๐+1,
๐๐+1,โฆ, ๐๐, ๐๐
Newton-Raphson Methods
Assume (m-1) PV buses among n buses
๐ฅ =
๐2
๐3
โฎ๐๐
๐๐+1
๐๐+2
โฎ๐๐
๐ ๐ฅ =
๐2(๐ฅ) โ ๐๐บ,2 + ๐๐ท,2
๐3(๐ฅ) โ ๐๐บ,3 + ๐๐ท,3
โฎ๐๐(๐ฅ) โ ๐๐บ,๐ + ๐๐ท,๐
๐๐+1(๐ฅ) โ ๐๐บ,๐+1 + ๐๐ท,๐+1
๐๐+2(๐ฅ) โ ๐๐บ,๐+2 + ๐๐ท,๐+2
โฎ๐๐(๐ฅ) โ ๐๐บ,๐ + ๐๐ท,๐
Multi-Variable Example
1
2
2 21 1 2
2 22 1 2 1 2
1 1
1 2
2 2
1 2
xSolve for = such that ( ) 0 where
x
f ( ) 2 8 0
f ( ) 4 0
First symbolically determine the Jacobian
f ( ) f ( )
( ) =f ( ) f ( )
x x
x x x x
x x
x x
x f x
x
x
x x
J xx x
๐ฅ ๐ก + 1 = ๐ฅ ๐ก โ ๐ฝโ1(๐ฅ)f(x[t])
Multi-variable Example, contโd
1(2)
(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
specified tolerance
0.1556( )
0.0900
If = 0.2 then we wou
x
f x
f x
ld be done. Otherwise we'd
continue iterating.
N-R Power Flow Solution
1 1 1
1 2
2 2 2
1 2
1 2
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, ( )
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( ) ( )
n
n
n n n
n
f f f
x x x
f f f
x x x
f f f
x x x
J x
x x x
x x x
J x
x x x
Other Power Flow Solution
Divide the Jacobian matrix into four sub-matrices:
๐ฝ ๐ฅ =
๐๐
๐๐
๐๐
๐๐๐๐
๐๐
๐๐
๐๐
Decoupled power flow:
๐ ๐ฅ =
๐๐
๐๐0
0๐๐
๐๐
Other Power Flow Solution
Fast decoupled power flow, assuming ๐๐ โ ๐๐ โ 0 ,
๐๐ โ 1, ๐บ๐๐ โช ๐ต๐๐
So we have: ๐๐
๐๐โ โ๐ต = โ
๐ต22 โฆ ๐ต2๐
โฎ โฎ โฎ๐ต๐๐ โฆ ๐ต๐๐
๐๐
๐๐โ โ๐ต =
๐ต๐+1,๐+1 โฆ ๐ต๐+1,๐
โฎ โฎ โฎ๐ต๐,๐+1 โฆ ๐ต๐๐
๐ ๐ฅ =โ๐ต 0
0 โ๐ต
DC Power Flow Analysis
Assumption: small deviation flat voltage profile, ๐๐ โ 1
and ๐๐ โ 0 ๐๐
๐๐โ โ๐ต
The ultimate steady state: ๐ = ๐0 + ๐๐, ๐ = ๐0 + ๐๐.
In the flat voltage profile, ๐0 = 0, ๐0 = 0
๐ โ โ๐ต ๐
Matpower
โข http://www.pserc.cornell.edu/matpower/
โข runpf: run power flow analysis
โข runopf: solves an optimal power flow
โข makeYbus: Builds the bus admittance matrix and branch admittance matrices.
PowerWorld
PowerWorld, contโd
Thanks