The AC two-bus system can be understood as the aggregation of the entire power system, in which the bus voltage and the load P,Q can be interpreted as the current operating voltage and the total consumers’ demand of the system. And the sending-end and receiving-end voltage of two-bus system can be the starting-point for step-by-step voltage calculation of multiple load points in a distribution line. Calculation of the voltages in two-bus system is meaningful. The voltage-power equation composed of bus voltages and load power P,Q for an AC two-bus system has been presented in publications. In addition, an explicit formula for calculating the receiving-end voltage of an AC two-bus system with line admittance G-jB recently has been announced. This formula can yield a unique and practical solution, however, the conductor parameters used in power system computation and provided by manufacturers have been mostly expressed by impedance R+jX, not by admittance G-jB. In this paper, the formulae for sending-end and receiving-end voltage of an AC two-bus system with line impedance R+jX are derived using ohmic calculation. And an explicit formula for calculating the receiving-end voltage of AC two-bus system represented by impedance R+jX is also introduced, from which a unique and feasible solution can be obtained. The line impedance data R+jX can be substituted directly into the proposed formula without any data conversion. Example calculations show that the results by the proposed formulae are exact and the same as those by existing methods.

#### Journal of the Korean Institute of Illuminating and Electrical Installation Engineers

**ISO Journal Title**J Korean Inst. IIIum. Electr. Install. Eng.

### Journal Search

## 1. Introduction

The AC two-bus system can be understood as the aggregation of an entire power system, in which the voltage and the load P,Q can be interpreted as the current operating voltage and the total consumers’ demand of the system.

In Fig. 1, $E\angle\theta_{E}$ is the voltage of sending-end bus 1. $V\angle\theta_{V}$ is
the voltage of receiving-end bus 2 with a complex load $\dot{S}=P+j Q$. $\dot{I_{12}}$
is the current flowing from bus 1 to bus 2 thru transmission line impedance $\dot{Z}=R+j
X$ (or admittance $\dot{Y}=G-j B$)^{[1, }^{2]}.

Conventional method that has been used in industrial field for voltage calculation
of distribution power systems, is the so-called ‘Equivalent resistance method’, which
uses the following approximate formula^{[3-}^{6,} Appendix A]: Voltage Formula derived by Ohmic Calculation for AC Two-bus System with
Constant Power Load P+jQ and Transmission Line Impedance R+jX

where θ is the power factor angle of the current. The term $R\cos\theta +X\sin\theta$
has no complex number and is the so-called ‘Equivalent resistance’. Equivalent resistance
method yields a fairly acceptable solution, however, it requires the load to be expressed
by constant current form^{[3-}^{6]}. Moreover, we may have a wrong answer E=V when the value of $R\cos\theta +X\sin\theta$
approaches zero^{[7]}.

The lossless formula represented by E, V, P, Q and reactance X, has been widely used to obtain P-V and Q-V curves for voltage stability analysis of large power systems[8, Appendix B].

Line conductance is not included in (2), which may result in a small error when applied to a system that has relatively low voltages.

The Voltage-Power equation(V-P equation from now on) represented by E, V, P, Q and line admittance $\dot{Y}=G-j B$, has been announced in publications[1,2, Appendix C].

Calculation results by (3) are exact, however, we have to solve the 4th order equation and need a special process for selecting one physically valid solution out of four mathematical roots.

A formula by which the receiving-end voltage V can be obtained without solving the 4th order of V-P equation, has been presented in a recent publication[9, Appendix C].

##### (4)

$V=\dfrac{1}{Y}\sqrt{-(GP+BQ-\dfrac{Y^{2}E^{2}}{2})+\sqrt{\left(GP+BQ-\dfrac{Y^{2}E^{2}}{2}\right)^{2}-Y^{2}(P^{2}+Q^{2})}}$Equation (4) yields a unique and practical solution, however, the transmission line parameters are represented by admittance $\dot{Y}=G-j B$.

Conductor data provided by manufacturers and the line parameters used in power systems are mostly expressed by impedance $R+j X$ in general, not by admittance $G-j B$.

In this paper, the exact formulae for sending-end and receiving-end voltage of AC two-bus system represented by line impedance R+jX are derived using ohmic calculation.

Calculating the voltages of two-bus system are described in the next sections.

## 2. Derivation of sending-end voltage formula represented by R, X using Ohmic calculation

In Fig. 2, the line-to-neutral voltage $V\angle\theta_{V}$ is maintained at bus 2 to supply its load with $\dot{S}=P+j Q$ via line impedance $\dot{z}=R+j X$.

Let us find the sending-end voltage E using the traditional Ohmic calculation.

Let

Then, the line current $\dot{I_{12}}$ is represented by:

Thus, the vector of sending-end voltage is:

Multiplying both sides by V yields:

##### (8)

$\begin{align*} V\bullet\dot{E}=V^{2}+(R+j X)(P-j Q)\\ \\ =V^{2}+(RP+XQ)+j(XP-RQ) \end{align*} $We have the following relation from (8):

or

##### (10)

$ \begin{align*} (V\bullet E)^{2}=(V^{2}+RP+XQ)^{2}+(XP-RQ)^{2}\\ \\ =V^{4}+ 2(RP+XQ)V^{2}+(RP+XQ)^{2}+(XP-RQ)^{2}\\ \\ =V^{4}+ 2(RP+XQ)V^{2}+(R^{2}+X^{2})(P^{2}+Q^{2}) \end{align*}$From (9) and (10), we finally obtain the following formula for sending-end voltage E.

##### (11)

$ \begin{align*} E =\dfrac{1}{V}\sqrt{V^{4}+ 2(RP+XQ)V^{2}+(R^{2}+X^{2})(P^{2}+Q^{2})}\\ \\ =\sqrt{V^{2}+ 2(RP+XQ)+\dfrac{(R^{2}+X^{2})(P^{2}+Q^{2})}{V^{2}}} \end{align*} $Here, if we put:

we can have the same formula in a simpler form:

##### (14)

$ \begin{align*} E =\dfrac{1}{V}\bullet\sqrt{V^{4}+ 2\alpha V^{2}+\beta}\\ =\sqrt{V^{2}+ 2\alpha +\dfrac{\beta}{V^{2}}} \end{align*}$We see in (11)~(14) that the sending-end voltage is represented by line impedance R+jX, not by admittance G-jB.

Example 1 : In Fig. 3, the line-to-neutral voltage $V$ at bus 2(load bus) is maintained at 13.0kV. The
phase impedance of the distribution line is $3.64+$$j7.82$$ohm$, and the load per
phase is $1,\: 056+$$j440$kVA. Calculate the voltage $E$ at bus 1(Substation bus)^{[1, }^{2]}.

A. Solution by traditional ohmic calculation^{[10]}

Let $\dot{V}=13.0\angle 0^{\circ}$kV. Then, the current $\dot{I}_{12}$ is:

Thus, the vector of sending-end voltage is:

##### (16)

$\dot{E}=\dot{V}+\dot{z}\bullet\dot{I}_{12}=13,\: 000+(3.64+j7.82)(81.23-j33.84)$

$=13,\: 560.4+j512.0$

We obtain the magnitude of sending-end voltage:

B. Solution by V-P formula (3)^{[1, }^{9]}

We have:

, from which we obtain:

Substituting $V=13.0$ kV, $P=1,\: 056$ kW, $Q=440$ kvar and $G$, $B$, $Y$ into (3) yields ^{[1, }^{9]}:

##### (20)

$E =\dfrac{\sqrt{\left(P+GV^{2}\right)^{2}+\left(Q+BV^{2}\right)^{2}}}{YV}$

$=\dfrac{\sqrt{\left(1056000+.0489\cdot 13000^{2}\right)^{2}+\left(440000+.1051\cdot 13000^{2}\right)^{2}}}{.1159\cdot 13,\: 000}$

$=13,\: 570 .02$

C. Solution by proposed formula (14)

Substituting line impedance R=3.64, X=7.82 directly into (14) yields:

##### (21)

$E =\sqrt{V^{2}+ 2\alpha +\dfrac{\beta}{V^{2}}}$

$\begin{align*} =\sqrt{13,\: 000^{2}+ 2\bullet 7.28\times 10^{6}+\dfrac{97.3\times 10^{12}}{13,\: 000^{2}}}\\ =\sqrt{13^{2}+ 2\bullet 7.28 +\dfrac{97.3}{13^{2}}}\bullet 10^{3} \end{align*}$

$= 13,\: 570.02$

, where

##### (22)

$\begin{align*} \alpha =RP+XQ = 3.64\bullet 1,\: 056,\: 000 + 7.82\bullet 440,\: 000 \\ \\ =7.28464\times 10^{6} \end{align*}$##### (23)

$\begin{align*} \beta =(R^{2}+X^{2})(P^{2}+Q^{2})= 74.402\bullet 1,\: 308,\: 736\times 10^{6}\\ \\ =97.372575872\times 10^{12} \end{align*}$The same solution 13570.02 as (17) and (20) has been obtained in (21) by substituting the line impedance $\dot{z}=3.64+j7.82$ directly into the proposed formula without any specific solving process. We see that the calculation results by conventional ohmic calculation, by existing V-P formula and by the proposed formula are all the same.

## 3. Derivation of receiving-end voltage formula represented by R, X using Ohmic calculation

Rephrasing equation (10):

##### (24)

$\begin{align*} (V\bullet E)^{2}\\ \\ =V^{4}+ 2(RP+XQ)V^{2}+(R^{2}+X^{2})(P^{2}+Q^{2})\\ \\ =V^{4}+ 2\alpha V^{2}+\beta \end{align*} $we obtain the following relation:

, where

Note that α and β in (26), (27) are the same as those of (12), (13) in Section 2.

Let $v=V^{2}$.

Then, from (25), we get the following quadratic equation w.r.t. $v$:

We take the high-voltage solution of (28) on P-V curve for the voltage-stable operation of the system^{[11-}^{13]}:

We finally obtain the following formula for the receiving-end voltage V :

##### (30)

$V =\sqrt{v}=\sqrt{-(\alpha -\dfrac{E^{2}}{2})+\sqrt{\left(\alpha -\dfrac{E^{2}}{2}\right)^{2}-\beta}}$or

##### (31)

$V=\sqrt{-(RP+XQ-\dfrac{E^{2}}{2})+\sqrt{\left(RP+XQ-\dfrac{E^{2}}{2}\right)^{2}-(R^{2}+X^{2})(P^{2}+Q^{2})}}$We see in (26), (27) and (30), (31) that the receiving- end voltage is represented by line impedance R+jX, not by admittance G-jB.

And we also see that formulae (30) and (31) are represented by explicit functions with a form of V=V(E,P,Q)^{[14]}. There are no V terms in RHS.

Example 2 : In Fig. 4, the line-to-neutral voltage $E=24$ V is applied at bus 1 to supply a load $12+j4\sqrt{3}$
VA through a transmission line with impedance $\dot{z}=1+j\sqrt{3}$ $ohm$. Calculate
the load voltage $V$ ^{[1, }^{9]}.

A. Solution by classical ohmic calculation

Let $\dot{V}=V\angle 0^{\circ}$ and $\dot{E}=E_{Re}+j E_{im}$, where $E_{Re}$ and $E_{im}$ are the real and imaginary part of $\dot{E}$. The line current $\dot{I}_{12}$ is:

And the vector of $\dot{E}$ is represented by:

##### (33)

$ \begin{align*} \dot{E}=E_{Re}+j E_{im}=V+\dot{z}\bullet\dot{I}_{12}\\ =V+(1+j\sqrt{3})\bullet\dfrac{12-j4\sqrt{3}}{V}\\ =V+\dfrac{12+12-j4\sqrt{3}+j12\sqrt{3}}{V}\\ =V+\dfrac{24+j8\sqrt{3}}{V} \end{align*} $Then, we have:

Squaring each side of (34) and (35) yields:

Adding each side of (36) and (37) yields:

Since the magnitude of $vert\dot{E vert}=\sqrt{E_{Re}^{2}+E_{im}^{2}}$ $=24$ is given, we get:

, from which we have:

##### (40)

$V^{4}-528V^{2}+768=0$

$ \begin{align*} V^{2}-528+\dfrac{768}{V^{2}}=0\\ \Rightarrow \\ \\ \end{align*} $

Solving (40), we have four mathematical roots as shown below:^{[9, }^{15]}

##### (41)

$ \begin{align*} V_{1}= 2\sqrt{2(33+\sqrt{1077})}\\ \\ V_{2}= -2\sqrt{2(33+\sqrt{1077})}\\ V_{3}= 4\sqrt{\dfrac{6}{33+\sqrt{1077}}}\\ V_{3}= - 4\sqrt{\dfrac{6}{33+\sqrt{1077}}} \end{align*} $Discarding unrealistic roots and selecting one practical solution, we finally obtain the following solution:

B. Solution by existing V-P formula

Since the line impedance $z=1+j\sqrt{3}$ $ohm$, we have:

##### (43)

$\dot{Y}=\dfrac{1}{1+j\sqrt{3}}=\dfrac{1}{4}-j\dfrac{\sqrt{3}}{4}$

or

$G=\dfrac{1}{4}$, $B=\dfrac{\sqrt{3}}{4}$, $Y= vert\dot{Y}vert =\dfrac{1}{2}$

Substituting $E=24$, $P=12$, $Q=4\sqrt{3}$ and $G$, $B$, $Y$ into (3) yields:

##### (44)

$\left(12+\dfrac{1}{4}\cdot V^{2}\right)^{2}+\left(4\sqrt{3}+\dfrac{\sqrt{3}}{4}V^{2}\right)^{2}=\left(V\cdot\dfrac{1}{2}\cdot 24\right)^{2}$Rearranging:

We see equation (45) above exactly matches equation (40) obtained by ohmic calculation.

By the same way as (41) and (42), we obtain:

C. Solution by exisiting explicit formula (4)

Since the line impedance $z=1+j\sqrt{3}$ $ohm$, we have to convert the impedance into admittance:

##### (43)

$\dot{Y}=\dfrac{1}{\dot{z}}=\dfrac{1}{1+j\sqrt{3}}=\dfrac{1}{4}-j\dfrac{\sqrt{3}}{4}$

or

$G=\dfrac{1}{4}$, $B=\dfrac{\sqrt{3}}{4}$, $Y= vert\dot{Y}vert =\dfrac{1}{2}$

Substituting $E=24$, $P=12$, $Q=4\sqrt{3}$ and $G$, $B$, $Y$ into (4) yields:

##### (47)

$V =$

$\dfrac{1}{Y}\sqrt{-(GP+BQ-\dfrac{Y^{2}E^{2}}{2})+\sqrt{\left(GP+BQ-\dfrac{Y^{2}E^{2}}{2}\right)^{2}-Y^{2}(P^{2}+Q^{2})}}$

$\begin{align*} \\ = 2\sqrt{-(-66)+\sqrt{(-66)^{2}-48}}\\ \end{align*}$

$= 22.94649$

D. Solution by proposed formula (30)

Substituting $E=24$, $P=12$, $Q=4\sqrt{3}$ and line impedance $z=1+j\sqrt{3}$ directly into (30) yields:

##### (48)

$ \begin{align*} V =\sqrt{-(\alpha -\dfrac{E^{2}}{2})+\sqrt{\left(\alpha -\dfrac{E^{2}}{2}\right)^{2}-\beta}}\\ \\ =\sqrt{-(24 -\dfrac{24^{2}}{2})+\sqrt{\left(24-\dfrac{24^{2}}{2}\right)^{2}-768}}\\ \\ = 22.94649 \end{align*}$, where

We see that a unique solution 22.94649 has been directly obtained by substituting the line impedance $\dot{z}=1+j\sqrt{3}$ directly into the proposed formula without any specific solving process. We see in (42),(46),(47) and (48) that the calculation results by conventional ohmic calculation, by existing V-P formula (3), by exisiting explicit formula (4) and by the proposed formula (30) are all the same.

## 4. Advantages of proposed formulae

Ohmic calculation and existing V-P equation (3) yield exact solution, however, they require a special process for solving the 4th order equation and selecting one feasible solution out of four mathematical roots.

Recently announced formula (4) yields a unique, exact and practical solution, however, the transmission line is represented by admittance G-jB. It has to be converted into R+jX when impedance data are required.

Proposed explicit formulae yield a unique and exact solution. And the transmission line is represented by impedance. R+jX data of the transmission line can be substituted directly into the proposed formulae without any data conversion.

By introducing intermediate variables $\alpha$ and $\beta$^{[16]}, the proposed formulae for both the sending-end voltage E and receiving-end voltage
V can be expressed in a simpler form as follows:

Sending-end voltage

##### (14)

$E =\dfrac{1}{V}\bullet\sqrt{V^{4}+ 2\alpha\bullet V^{2}+\beta}$

$=\sqrt{V^{2}+ 2\alpha +\dfrac{\beta}{V^{2}}}$

Receiving-end voltage

Note that the same intermediate variables $\alpha$ and $\beta$ are applied to the formulae (14) and (30) for receiving-end voltage V as well as sending-end voltage E.

## 5. Conclusion

The line parameters in power systems are given by impedance R+jX in general, not by admittance G-jB.

In this paper, the formulae for sending-end and receiving-end voltage of AC two-bus system represented by line impedance R+jX have been derived using ohmic calculation. The line impedance data R+jX can be substituted directly into the proposed formulae without any data conversion.

The proposed formulae are expressed by explicit form, from which a unique and feasible solution can be obtained.

The proposed formulae for both the sending-end and receiving-end voltage can also be expressed in a simpler form by introducing the same intermediate variables.

Example calculations have illustrated the solving process - by traditional ohmic method, by existing V-P formula with G-jB and by the proposed formulae with R+jX. The results have turned up to be all the same, through which the effectiveness of the proposed formulae has been proven.

### References

### Appendix A quivalent resistance method^{[3-}^{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 $\cos\theta$.

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 the so-called ‘Equivalent resistance’.

### Appendix B Two Bus Example^{[8]}

Consider a generator connected to a load bus through a lossless-transmission line X as shown in Fig A3.

We have:

##### (A6)

$\dot{S}=\dot{V}\bullet I^{*}$

$ \begin{align*} =\dot{V}\left(\dfrac{\dot{E}-\dot{V}}{j X}\right)^{*}= V\angle\theta\bullet\dfrac{E\angle 0-V\angle -\theta}{-j X}\\ \\ =\dfrac{EV}{X}\sin\theta + j(\dfrac{EV}{X}\cos\theta -\dfrac{V^{2}}{X}) \end{align*} $

Separating real and imaginary parts:

##### (A7)

$ \begin{align*} P=\dfrac{EV}{X}\sin\theta \\ Q=\dfrac{EV}{X}\cos\theta -\dfrac{V^{2}}{X} \end{align*}$

$ \begin{align*} \Rightarrow \\ \sin\theta =\dfrac{PX}{EV}\\ \cos\theta =\dfrac{QX+V^{2}}{EV} \end{align*} $

We have:

### Appendix C An exact voltage-power equation of AC two-bus system with no angle terms^{[1]}>

Complex power $\dot{S}$ of the load in Fig. A4 can be represented as follows:

##### (A9)

$\dot{S}=\dot{V}\bullet(\dot{I}_{12})^{*}$

$=\dot{V}[\dot{Y}(\dot{E}-\dot{V})]^{*}$

$=\dot{V}\dot{Y}^{*}\dot{E}^{*}-\dot{V}\dot{Y}^{*}\dot{V}^{*}$

$=VYE\angle\theta_{V}-\theta_{Y}-\theta_{E}-V^{2}\dot{Y}^{*}$

$=VYE[\cos(\theta_{V}-\theta_{Y}-\theta_{E})+j\sin(\theta_{V}-\theta_{Y}-\theta_{E})]$

$- V^{2}(G+j B)$

$=[-GV^{2}+VYE\cos(\theta_{V}-\theta_{Y}-\theta_{E})]$

$+j[-BV^{2}+VYE\sin(\theta_{V}-\theta_{Y}-\theta_{E})]$

Active and reactive power P and Q of complex load $\dot{S}$ are:

##### (A10)

$P= -GV^{2}+VYE\cos(\theta_{V}-\theta_{Y}-\theta_{E})$

$Q= -BV^{2}+VYE\sin(\theta_{V}-\theta_{Y}-\theta_{E})$

Rearranging,

##### (A11)

$P+GV^{2}=VYE\cos(\theta_{V}-\theta_{Y}-\theta_{E})$

$Q+BV^{2}=VYE\sin(\theta_{V}-\theta_{Y}-\theta_{E})$

Squaring both sides and adding each side, we obtain a new voltage-power equation with no angle term ($\theta_{V}-$$\theta_{Y}-\theta_{E}$):

Extracting the parentheses of equation (A12), we have:

Arranging w.r.t. V, we get:

Then, we obtain a positive solution for V as follows^{[9]}:

### Appendix D Limitations and Applicability

The proposed formulae are exact under the assumption that the system is an AC two-bus system with a P,Q load.

Shunt capacitance $Y_{C}$ between bus 1 and 2 can be represented by Fig. A5 and Fig. A6, where $Q_{C}$ in Fig. A6 is:

A system with two load points A,B as shown in Fig. A7 can be reduced to a two-bus system^{[2]}. The load $P_{B}$, $Q_{B}$ at bus B and the line loss $P_{LOSS}$, $Q_{LOSS}$ between
bus A and B can be aggregated at bus A as shown in Fig. A8.

## Biography

He has worked for SD E&GC Co., Ltd, for 12 years since 2002 as Chief Executive of R&D Center. He has been a Professor of Chuncheon Campus of Korea Polytechnic 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, Improvement of SPD, Quiescent power cut-off, and etc. He received Ph.D. at Seoul National University of Science and Technology.

He received Ph.D. at Chungnam National University in 1995. He worked for Korea Electric Power Corporation(KEPCO) for 22 years since 1976, mostly at Power System Research Center. He also completed the Power System Engineering Course at GE Research Center in Schenectady, NY, during 1987∼1988. 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 proposed ‘Angle reference transposition(ART) in power system 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 the loss sensitivities derived by ART to ‘Penalty factor calculation in ELD computation’, ‘Optimal MW generation for system loss minimization’ and etc.