이상중
(SangJoong Lee)
1iD
김주철
(Ju-Chul Kim)
†iD
-
(Dept. of Electrical Engineering, Seoul National University of Science and Technology,
Korea.)
Copyright © The Korean Institute of Electrical Engineers(KIEE)
Key words
Voltage-power equation, Two-bus system, Bus angle
1. Introduction
Shown in Fig. 1 is an AC two-bus system.
Fig. 1. AC two-bus system
Let E and $θ_{E}$ be the magnitude of voltage and its bus angle at sending-end bus
1. And let V and $θ_{V}$ be the magnitude of voltage and its bus angle at load bus
2 which consumes a complex load $\dot S =P+j Q$(1). Current $\dot I_{12}$ is flowing from bus 1 to bus 2 through a transmission line
with admittance $\dot Y =Y\angle\theta_{Y}=G-j B$.
This paper proposes a new formula representing a relation between bus voltages and
the complex load of a two-bus system.
There are several existing methods concerning calculation of the voltages between
two buses. Ohmic method has been the most fundamental and commonly used method, which
is based on the following relation:
That is, sending-end voltage $E$ is equal to receiving-end voltage $V$ plus voltage
drop in transmission line with impedance $\dot z =1/\dot{Y}={R}+{j X}$. Ohmic method
yields an exact solution through a calculation with complex variables $\dot E$, $\dot
V$, $\dot z$, $\dot I_{12}$(2).
Another method is the so-called ‘Equivalent resistance method’ which uses the following
formula:
where $\theta$ is the power factor angle of the load. ‘$R\cos\theta$$+$$X\sin\theta$’
is called the ‘Equivalent resistance’ (3-7,Appendix).
$\dot I_{12}(R\cos\theta + X\sin\theta)$ in (2) equals to the voltage difference $E-V$ between bus 1 and 2, which for a long time
has been widely used for voltage drop calculation in distribution power system(8-10). In (2), the load is represented by $\dot I_{12}$ - the line current between bus 1 and 2.
If the load is denoted by $P-Q$, not by $\dot I_{12}$, however, this method requires
additional complex calculation for $I_{12}$ and its $\cos\theta$ and $\sin\theta$
terms. ‘Equivalent resistance method’ is very simple and yields an approximate but
acceptable solution.
Ajjarapu presented a formula for a lossless two-bus system that has no complex terms
as shown below(11):
$P-V$ and $Q-V$ curves can be obtained by this formula, which have been widely used
for voltage stability analysis of a power system(1,11). Let us call equation (3) the ‘Lossless formula’ from now on.
This paper presents a new formula that has no angle terms and no complex number calculation.
Proposed formula yields an exact solution and can be used for calculation of bus voltages
and the voltage drop between two buses. Sample calculations for bus voltages are demonstrated
using proposed formula and compared to existing methods.
2. An exact voltage-power equation of AC two-bus system with no angle terms
Complex power $\dot S$ of the load in Fig. 1 can be represented as follows:
Active and reactive power P and Q of complex load $\dot S$ are:
Rearranging,
Squaring both sides and adding each side of (7) and (8), we obtain a new voltage-power equation with no angle term ($\theta_{V}-$$\theta_{Y}-\theta_{E}$):
Note that equation (9) does not contain any complex terms but is represented by the magnitudes of bus voltages
$E-V$, load power $P-Q$ and line admittance $G-B$.
We also see that the load is given by active and reactive power $P-Q$ in (9), while the load is given by current $\dot I_{12}$ in the conventional formula (2). Formula (9) can be used for calculation of bus voltages as described in the next Section.
3. Example calculation of sending-end voltage
$\text{Example 1:}$ In Fig. 2, the line-to-neutral voltage $V$ at bus 2(load bus) is maintained at 13.0 kV, the
phase impedance of the distribution line is $3.64+$$j7.82$ Ω, and the load per phase
is $1,\:056 +$$j 440$ kVA. Calculate the voltage $E$ at bus 1(Substation bus)(2).
Fig. 2. Two-bus system with load bus voltage 13.0 kV
A. Solution by conventional ohmic method
Let $V=13.0\angle 0^{\circ}$ kV. Then, the current $\dot I_{12}$ in Fig. 2 is:
Thus, the vector of sending-end voltage is:
We obtain the magnitude of sending-end voltage:
B. Solution by proposed formula (9):
We have:
from which we obtain:
Substituting $V=13.0$ kV, $P=1,\:056$ kW, $Q=440$ kvar and $Y$, $G$, $B$ into (9) yields:
We see in (12) and (15) that calculation result by formula (9) is exactly the same as the result by conventional ohmic calculation.
We also see in (15) that formula (9) is composed of only real numbers and calculation by proposed formula (9) is simpler than ohmic calculation.
C. Solution by Equivalent resistance method
Magnitude of the current in $\dot I_{12}$ Fig. 2 is:
$\cos\theta$ and $\sin\theta$ of current $\dot I_{12}$ are:
Substituting $|\dot I_{12}|$, $\cos\theta$ and $\sin\theta$ into formula (2) yields:
D. Solution by Lossless formula
Substituting $V=13.0$ kV, $P=1,\:056$ kW, $Q=440$ kvar and $X=7.82$ into (3) yields:
Solving above equation w.r.t. $E$, we obtain:
Note that solutions by ohmic method and by formula (9) are exact, while solutions by ‘Equivalent resistance method’ (2) and by ‘Lossless formula’ (3) are approximate in general.
Table 1 is the comparison of each calculation result and error rate for the system in Fig. 2. Graph of error rate is illustrated in Fig. 3.
Table 1. omparison of calculation methods and error rate
Method
|
Result
|
Remark
|
Error rate
|
Ohmic
|
13,570
|
Exact
|
0
|
Proposed formula (9)
|
13,570
|
Exact
|
0
|
Equivalent resistance
|
13,560
|
Approximate
|
0.07 %
|
Lossless formula
|
13,279
|
Approximate
|
2.10 %
|
Fig. 3. Graph of error rate for Table 1
We see in Table 1 that the ‘Equivalent resistance method’ yields an approximate but fairly acceptable
solution. ‘Lossless formula’ shows a relatively big error, which is because the system
is assumed to be lossless.
4. Example calculation of receiving-end(load bus) voltage
When the complex load $\dot S =P+j Q$, sending-end voltage $E$ and line admittance
$Y=G-j B$ are given in a two-bus system, we can obtain the load voltage V with ease
using formula (9).
$\text{Example 2:}$ In Fig. 4, the line-to-neutral voltage $E=24$ volt is applied at bus 1 to supply a load $12+j4\sqrt{3}$
VA through a transmission line with impedance $z=1 + j\sqrt{3}$ Ω. Calculate the load
voltage $V$.
Fig. 4. AC two-node circuit with unknown load voltage $V$
A. Solution by proposed formula (9)
Since the line impedance $z=1+j\sqrt{3}$ Ω, we have:
Substituting $E=24$, $P=12$, $Q= 4\sqrt{3}$ and $Y$, $G$, $B$ into (9) yields:
Rearranging:
Solving above equation, we obtain:
B. Solution by conventional ohmic method
Let us find the load voltage $V$ in Fig. 4. using conventional ohmic method.
Let $\dot V = V\angle 0^{\circ}$.
Then, the line current $\dot I_{12}$ is represented by:
Thus, the vector of sending-end voltage is:
Since $|\dot E | = 24$, we have from (25):
Rewriting:
or
We finally obtain the solution:
We see in (24) and (30) that calculation result by formula (9) is exactly the same as the result by conventional ohmic calculation.
We also see that equation (23) by proposed formula (9) is in fact the same as equation (29) by conventional ohmic calculation.
However, we see that solving by conventional ohmic method is far more complicated
than solving by proposed formula (9).
Solution by conventional ohmic method requires several steps of circuit analyzing
with complex-numbers, however, solution by proposed formula (9) needs only a simple substitution of real-number data.
C. Solution by Equivalent resistance method
From (25), magnitude of current $\dot I_{12}$ is:
$\cos\theta$ and $\sin\theta$ of current $\dot I_{12}$ are:
Substituting $|\dot E |$, $|\dot I_{12}|$, $\cos\theta$ and $\sin\theta$ into formula
(2) yields:
or
We finally obtain the solution:
D. Solution by Lossless formula
Substituting $|\dot E |=24$, $P=12$, $Q=4\sqrt{3}$ and $X=\sqrt{3}$ into (3) yields:
Solving above equation w.r.t. $V$, we obtain:
Note that solutions by ohmic method and by formula (9) are exact, while solutions by ‘Equivalent resistance method’ (2) and by ‘Lossless formula’ (3) are approximate in general.
Table 2 is the comparison of each calculation result and error rate for the system in Fig. 4. Graph of error rate is illustrated in Fig. 5.
Table 2. Comparison of calculation methods and error rate
Method
|
Result
|
Remark
|
Error rate
|
Ohmic
|
22.94649
|
Exact
|
0
|
Proposed formula (9)
|
22.94649
|
Exact
|
0
|
Equivalent resistance
|
$22.9 5445$
|
Approximate
|
0.034 %
|
Lossless formula
|
$23.47247$
|
Approximate
|
2.29 %
|
Fig. 5. Graph of error rate for Table 2
We see in Table 2 that the ‘Equivalent resistance method’ yields an approximate but fairly acceptable
solution, while ‘Lossless formula’ shows a relatively big error.
5. Advantages of Proposed Formula
Advantages of the proposed formula (9) compared with existing methods, especially with conventional formula (2), can be summarized as follows:
1. Proposed formula (9) has no angle term $\theta$. No calculation for angle $\theta$ is required in (9).
2. Calculation result of proposed formula (9) is exact while the result of conventional formula (2) is approximate.
3. In proposed formula (9), the load is given by active and reactive power $P$, $Q$. Loads in the power system
are represented by $P$, $Q$ in general. In conventional formula (2), however, the load is represented by current $\dot I_{12}$ between bus 1 and 2. If
the load is denoted by $P$, $Q$, not by $\dot I_{12}$, formula (2) requires additional complex calculation to obtain the line current $I_{12}$ and its
$\cos\theta$ and $\sin\theta$ terms.
4. Proposed formula (9) does not contain any complex terms but is represented by the magnitudes of bus voltages
and load power $P$, $Q$. Therefore, only a simple substitution of real-number data
into formula (9) is needed to obtain the solution without any complex-number calculation.
6. Conclusion
This paper has presented a new voltage-power equation for an AC two-bus system with
complex load $\dot S = P + j Q$ connected through admittance $\dot Y = G-j B$. The
proposed formula has no angle terms and no complex terms but is represented only by
the magnitudes of voltage and load powers. The formula yields exact solutions and
can be applied to the following calculations for a two-bus system.
- Calculation of sending-end voltage $E$ when load voltage $V$, complex load $\dot
S$ and line admittance $Y$ are given.
- Calculation of load voltage $V$ when sending-end voltage $E$, complex load $\dot
S$ and line admittance $Y$ are given.
- Calculation of voltage drop between two buses
Example calculations for sending-end and receiving-end bus voltages using proposed
formula have been demonstrated and compared with existing three methods - ohmic method,
‘Equivalent resistance method’ and ‘Lossless formula’. Calculations by proposed formula
have shown the same results as the conventional ohmic method. Solution by conventional
ohmic method requires several steps of circuit analyzing with complex numbers, however,
solution by proposed formula needs only a simple substitution of real-number data
without any complex-number calculation.
The proposed formula can be used together with conventional ohmic calculation and
‘Equivalent resistance method’. They can be useful checks on each other for calculation
of two-bus system.
Appendix A
A. 1 Equivalent resistance method(4-7)
Shown in Fig. A1 is the short-length transmission system with sending-end voltage $E$, receiving-end
voltage $V$ and line impedance $R+j X$. Let the current and power factor(lagging)
of the load be $I$ and $\cos\theta$.
Fig. A1. Two-bus system with short-length transmission line
Shown in Fig. A2 is the vector diagram of Fig. A1.
In Fig. A2, we have the following relation:
and we obtain:
In general,
hence, (A2) can be approximated as follows:
The term
has no complex number, and is so-called the ‘Equivalent resistance’.
Calculation by formula (A4) yields an approximate but fairly reliable solution.
Acknowledgements
This study was supported by the Research Program funded by the SeoulTech.(Seoul National
University of Science and Technology)
Acknowledgements
This study was supported by the Research Program funded by the SeoulTech.(Seoul National
University of Science and Technology)
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저자소개
He proposed ‘Angle reference transposition in power flow computation’ on IEEE Power
Engineering Review in 2002, which describes that the loss sensitivities for all generators
including the slack bus can be derived by specific assignment of the angle reference
on a bus where no generation exists, while the angle reference has been specified
conventionally on the slack bus.
He applied these loss sensitivities derived by ‘Angle reference transposition’ to
‘Penalty factor calculation in ELD computation’ [IEEE Power Engineering Review 2002],
‘Optimal MW generation for system loss minimization’ [IEEE Trans 2003, 2006] and etc.
He worked for Korea Electric Power Corporation(KEPCO) for 22 years since 1976, mostly
at Power System Research Center.
He has been a professor of Seoul National University of Science and Technology since
1998.
His research interest includes power generation, large power system and engineering
mathematics.
He received Ph.D. at Chungnam National University in 1995.
E-mail : 85sjlee@seoultech.ac.kr
He has worked for S-D E&GC Co., Ltd, for 12 years since 2002 and used to be the Chief
Executive of R&D Center.
He has been a professor of Chuncheon Campus of Korea Poly- technic University since
2014.
His research interest includes Power system optimization, Quiescent power cut-off
and Human electric shock.
He published many papers on ELCB (Earth Leakage Circuit-Breakers), Human body protection
against electric shock, Im- provement of SPD, Quiescent power cut-off, and etc.
E-mail : cjfwnxkq@hanmail.net