Download Reactive Power Optimization in Electricity Market for Distribution Network

Reactive Power Optimization in Electricity Market for Distribution
Network
WANG Yanjun1, BAI Jie2
1. Electrical and Mechanical Engineering College Agriculture University of Hebei, Baoding, P.R.China,
071001
2.Hebei Software Institute, Baoding, P.R.China, 071000
[email protected]
，
Abstract The paper analyses the distribution network feature of our country, establishes the objective
function based on maximal economic benefit in electricity power market and node voltage is regarded as
constraint condition. According to the two significant geometric traits of reactive power current
persistent curve - "equivalent area criterion" and "equivalent length criterion", the mixed compensation
mode of fixation and switching is presented duo to the big fluctuation of the reactive power loads of
distribution network in our country. The above method is applied to the solving process of optimal
compensation mode in the lines with terminal convergence loads and distribution loads. The solving
extreme problem of integral equations for obtaining maximal economic benefit is changed into a simple
arithmetic problem.
Key words Electricity power market, Distribution network, Equivalent area criterion, Equivalent length
criterion, Reactive power optimization
1 Introduction
The distribution system in our country has the common problems of scattered loads, large undulation,
long power supply radius, low power factor, shortage of reactive power, etc. These cause serious
losses of electric energy and voltage[1-2]. The statistical datum indicates that the electric energy losses in
distribution network are about two-thirds of area line losses. Obviously, adopting effective measures to
reduce electric energy losses in distribution network as more as possible should be one of the main study
directions of electric power departments, which aim at cost lowering and benefit raising[3].
The realization of reactive power optimization is effective measures, which can guarantee the power
system security and economic operation. But in the electricity market, power plants and power network
were separated, belong to different power companies, these companies are pursuing their own economic
benefit, the traditional reactive power optimization model has been unable to meet the electricity market
in the actual situation, so to establish a reactive power optimization model considering the price of
reactive power is necessary. Moreover, with the study of reactive power optimization in depth, the
objective model is more complex and more difficult to solve. Hence it is need to establish a more
effective reactive optimization method for solving[4].
2 Mathematical Model
The objective function of reactive power compensation optimization can be various, such as the smallest
network losses or annual running costs, the smallest overall investment of compensation equipment, the
biggest economic benefits, etc.
2.1 The problem of traditional reactive power optimization model
In the traditional reactive power optimization, the objective function is usually the minimum of active
This work was supported by Hebei Science and Technology Agency of China (No.072135124,
No.072135125), Research development fund of non-life science and newly developed discipline of
Agricultural University of Hebei: security region based real and reactive power pricing of power
systems.
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power losses, also have equality constraints of power balance equations and inequality constraints such
as voltage, reactive power and transformer ratio. Before power plants and network were separated, the
system pursues the overall benefit, the reactive power optimization model that the active power losses of
the entire network is the smallest is appropriate. After power plants and network were separated the
companies are pursuing their own economic benefit, only minimal power losses in the objective
function does not guarantee their benefit, in that case, the reactive power optimization model cannot
adapt to actual power market.
2.2 Establish the objective function with the maximal economic effect
We should consider reactive power investment in power market. Take comprehensive economic benefits
as the objective function, that means economic benefit from reduced energy losses and power losses
after system compensation then deducted the cost of the compensation, in a cycle T0, it can expressed as
follow
n
n
i =1
i =1
∗
∗
+ 3 ⋅ K cs ⋅ U l ⋅ ∑ I csi
)
MaxS = K e⋅∆E L ⋅ 10 −3 + K P ⋅ ∆PL ⋅ 10 −3 − ( 3 ⋅ K cf ⋅ U l ⋅ ∑ I cfi
(1)
Where Ke is the unit price of energy (¥ /kwh), Kp is the per kilowatt energy average cost for generation
and transmission equipment (¥ /kw), Kcf is the per kvar annual average price for fixed capacitors, Kcs
is the per kvar annual average price for switching capacitor, Ul is the line voltage of distribution line
(KV), I*cfi is the optimal fixed capacity of ith group capacitor, I*cSi is the optimal switching
capacity of i-th group capacitor,
PL is the power loss reduction at peak load after capacitor
compensation, EL is the energy loss reduction at peak load after capacitor compensation.
△
△
2.3 Constraint
V i ⋅ min ≤ V i ≤ V i ⋅ max
(2)
Where Vi is the voltage at node i, Vi.min, Vi.max are the minimum and maximum voltage at node i .
3 Calculation Method
The reactive power loads of distribution network in our country have big fluctuation, the mixed
compensation mode of fixation and switching is presented in this paper. The node voltage is regarded as
constraint condition, the capacitor capacity is adjusted according to the loads and the configuration and
control model of parallel capacitors is determined. According to the two significant geometric traits of
reactive power current persistent curve - " equivalent area criterion" and "equivalent length criterion"
which reflect relationship of "current-time" and "current-length" in the optimal compensation mode. The
above method is applied to the solving process of optimal compensation mode in the lines with terminal
convergence loads and distribution loads. The solving extreme problem of integral equations for
obtaining maximal economic benefit is changed into a simple arithmetic problem.
3.1 Equivalent area criterion and equivalent length criterion
The loads in distribution lines are divided into convergence loads and distribution loads[5]. Taking the
optimal grading compensation with terminal convergence loads as the example to analysis the two
criteria before switching the capacitor, the power losses caused by peak current is
(3)
PM = 3( I a2 + I s2 ) R
Where Ia is the peak active current of load (A), Is is the peak reactive current of load (A).
Suppose the capacitor banks are divided into n grades. The capacitor current of each grade is Ic1,
Ic2,
,Icn respectively. The running time of each grade capacitor is T1, T2,
,Tn (hour) respectively.
After the capacitor banks are put into service, the line power losses in the peak load are
···
···
n
PMC = 3[ I a2 + ( I s − ∑ I ci ) 2 ]R
i =1
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(4)
Due to the compensation function of capacitor banks, the decrease of the peak power losses in the line
compared with uncompensated is
n
n
∆PM = PM − PMC = 6I s R⋅
∑I
∑I
ci − 3R⋅ (
i=1
ci )
2
(5)
i=1
the decrease of electric energy losses in a load cycle (a day) T0 is
T0
Tn
∆E = ∫ ∆P (T )dT = ∫ ∆P (T )dT + ∫
0
0
n
Tn −1
Tn
n
Ti
T0
∆P (T )dT + L + ∫ ∆P (T )dT
T1
n
i −1
i =2
j =1
(6)
= 6 R ⋅ ∑ [ I ci ∫ I sφ (T )dT ] − 3 R[∑ T I + 2 ⋅ ∑ (Ti ⋅ I ci ⋅ ∑ I cj )]
2
i ci
0
i =1
i =1
Because the power losses of a capacitor itself are very small, it can be neglected. Thereupon, in a load
cycle T0, deducting the capacitor cost from the economic benefits obtained by reducing the power and
energy losses of the line, we gain the economic benefit after compensation
n
S = Ke ⋅ ∆E ⋅10−3 + KP ⋅ ∆PM ⋅10−3 − Kc ⋅ ( 3Ul ∑I ci )
(7)
i=1
Where Kc is the daily average investment of unit capacity capacitor (¥ / kvar. day)
In order to get the capacitance current and in-service time for each grade that makes S get the maximum,
take
∂S
=0
∂I ci
(i=1,2,…,n)
(8)
∂S
=0
∂Ti
After the partial differentiation calculation for above equations set, get the integration equations set
containing 2n equations. Subtracting the first equation from the equations of 2 to n separately, and
ordering the equations of n+1 to 2n, then the equations set develops as follows
Tn*
∫
0
∫
n
3K c U l ⋅10 −3 K P
( I s − ∑ I ci* )
−
6R ⋅ K e
Ke
i =1
n
I s φ (T ) dT − Tn* ⋅ ∑ I ci* =
i =1
Tn*−1
T n*
n −1
I s φ (T ) dT − (T n*−1 − T n* ) ∑ I ci* = 0
i =1
M
∫
T2*
T3*
∫
T1*
T2*
I s φ (T ) dT − (T2* − T3* ) ⋅ ( I c*1 + I c*2 ) = 0
I sφ (T )dT − (T1* − T2* ) ⋅ I c*1 = 0
n
I cn* = 2 ⋅ [∑ I ci* − I s φ (T n* )]
i =1
n −1
I cn* −1 = 2 ⋅ [∑ I ci* − I s φ (Tn*−1 )]
i =1
M
I c*2 = 2 ⋅ [( I c*1+ I c 2*) − I s φ (T 2* )]
I c*1 = 2 ⋅ [ I s φ (T1* )]
(9)
Equations set (9) is a nonlinear integration equations set and is difficult to solve. However, when we
analyze the new equations set carefully, find it has two significant characters as follows.
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(1) The geometrical meanings of the equations from 1 to n in the equations set are, in Fig. 1, area
A n − A n' =
n
3 K c U l ⋅ 10 − 3 K P
−
( I s − ∑ I ci* ) , and area An-1= A’n-1,···, A1= A’1. This character is
Ke
6R ⋅ K e
i =1
called the “equal area criterion”.
(2) The geometrical meanings of the equations from n+1 to 2n in the equations set are, dot Is φ(Ti *) in
the continuous curve of reactive current is Isφ(T) divides the line segment of Ici * into two equal parts (i =
1,2,···, n). This character is called the “equal length criterion”.
The two important geometric characters are peculiar to the optimal compensation mode. They provide a
simple and direct method for solving the optimal compensation. We needn’t solve the integration
equations set directly. Thus, the problem being difficult to solve originally is turned into a simple one of
arithmetic operation.
Fig. 1 Continuous curve of reactive current
3.2 Solving method for the mixed optimal reactive power compensation of fixation and switching
in distribution load lines
Solving process:
(1) Input lines datum, active and reactive current at valley load, reactive current at peak load of every
load point. Establish the reactive current change curve Isф(t), characterization curve Is' f(l) and reactive
current continuous curve Isф(T).
(2) Use the two criteria to calculate the optimal location of fixed compensation li* and capacitive current
I*ci in the line when excluding capacitor price (Kc=0).
(3) Obtain the optimal fixed capacity I*cf1, the optimal switching capacity I*cs1 and the running time T*s
of capacitors banks which closest to the end of the line, I*cfi( i=1,2,…n) and I*csi( i=1,2,…n-1) can be
known according to the proportional relationship, while use objective function S to solve partial
derivative equation we can get I*csn.
(4) Get the value of PL, EL and S, then obtain the optimal switching time of each capacitor bank
from the reactive current changing curve by T*s.
(5) Obtain voltage decent between the initial terminal and every load point caused by the current of
valley loads.
(6) Input the values of (2) to computer and obtain the voltage increase relative to the initial terminal due
to the capacitors compensated separately, then obtain actual voltage increase relative to the initial
terminal at every compensation point and test if the highest voltage greater than the maximum allowable
value. If not, there is no excess voltage at the valley load; otherwise, obtain the excess value.
△ △
353
(7) Obtain the value that the capacitive current should reduce in order to avoid voltage excess, get
economic benefits when adjust different capacitors and compare, the best adjustment program can get
the most economic benefits.
(8) Test if there is the compensation point of voltage exceeding started from step (6) again after
capacitor capacity adjusted. If not, calculation is end; otherwise, adjust the capacitor capacity again until
the voltage at every point less than the maximum allowable value.
4 Calculation and Analysis for An Example
A 35Kv substation with the total capacity of 1460Kvar have 10Kv distribution line, 8 distribution loads,
the route length is 22km. In a load cycle (year), peak reactive current is 60A, average power factor of
the line is 0.7, wire specification is LGJ-70, line structure and load distribution are shown in Table.1
(load numbering from the end), given Kc = 20¥/Kvar, Ke = 0.05¥/KW.h, Kp = 120¥/Kvar, the number of
compensation point n=2, the optimal compensation calculation results are shown in table 2.
Table1 Load distribution
1
2
3
The number of load point
4
5
6
7
8
Point load Capacity (KVA)
75
100
360
100
200
450
100
75
distance between the point load (Km)
1.5
2
3
2
2.5
3
3
5
Table2 The optimal compensation results
Number
of
compensation
point
2
∆PL
(kw)
∆E
(kw.h
/year)
S
(¥
/year)
average
cosφ
after
compensation
voltage
after
compensation
45
197100
14025
0.95
4.5%
Capacitor capacity
Compensation location
(kvar)
l*1
l*2
C*1
C*2
The third
point load
The sixth
point load
364
251
The results show that not only the economic benefits are considerable (the capacitors investment can be
recovered within a year) but also the power factor and voltage quality are improved notable after the
optimal compensation.
5 Conclusions
The mixed compensation mode of fixation and switching is presented because the reactive power loads
in distribution network of our country have big fluctuation. The above method is applied to the solving
process of optimal compensation mode in the lines with terminal convergence loads and distribution
loads. The solving extreme problem of integral equations for obtaining maximal economic benefits is
changed into a simple arithmetic problem. The voltage of compensation point deviating from the
allowing range have two kinds, the one is terminal voltage below the minimum value at peak load, the
other is compensation point voltage values exceed the maximum value at low load. The line voltage rise
is stringent restricted as it affects the capacitors safe operation and life. The solutions of this paper aim
at the elimination compensation point excess voltage at valley load, but also apply to eliminate the
problem of voltage below the minimum value.
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