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The Transactions of the Korean Institute of Electrical Engineers

ISO Journal TitleTrans. Korean. Inst. Elect. Eng.

1. (Dept. of Electrical Engineering, Seoul National University of Science and Technology, Korea.)

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:

(1)
$$\dot E =\dot V +\dot z·\dot I_{12}$$

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:

(2)
$$E=V+|\dot I_{12}|·(R\cos\theta +X\sin\theta),$$

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):

(3)
$$\left(\dfrac{PX}{EV}\right)^{2}+\left(\dfrac{QX+ V^{2}}{EV}\right)^{2}=1$$

$P-V$ 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:

(4)
\begin{aligned} \dot{S}=& \dot{V} ·\left(\dot{I}_{12}\right)^{*} \\ =& \dot{V}[\dot{Y}(\dot{E}-\dot{V})]^{*} \\ =& \dot{V} \dot{Y}^{*} \dot{E}^{*} \quad-\dot{V} \dot{Y}^{*} \dot{V}^{*} \\ =& V Y E \angle \theta_{V}-\theta_{Y}-\theta_{E} \quad-V^{2} \dot{Y}^{*} \\ =& V Y E\left[\cos \left(\theta_{V}-\theta_{Y}-\theta_{E}\right)+j \sin \left(\theta_{V}-\theta_{Y}-\theta_{E}\right)\right] \\ &-V^{2}(G+j B) \\ =&\left[-G V^{2}+V Y E \cos \left(\theta_{V}-\theta_{Y}-\theta_{E}\right)\right] \\ &+j\left[-B V^{2}+V Y E \sin \left(\theta_{V}-\theta_{Y}-\theta_{E}\right)\right] \end{aligned}

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

(5)
$$P= -GV^{2}+VYE\cos(\theta_{V}-\theta_{Y}-\theta_{E})$$

(6)
$$Q= -BV^{2}+VYE\sin(\theta_{V}-\theta_{Y}-\theta_{E})$$

Rearranging,

(7)
$$P+GV^{2}=VYE\cos(\theta_{V}-\theta_{Y}-\theta_{E})$$

(8)
$$Q+BV^{2}=VYE\sin(\theta_{V}-\theta_{Y}-\theta_{E})$$

Squaring both sides and adding each side of (7) and (8), we obtain a new voltage-power equation with no angle term ($\theta_{V}-$$\theta_{Y}-\theta_{E}): (9)$$(P+GV^{2})^{2}+(Q+BV^{2})^{2}=(VYE)^{2}$$Note that equation (9) does not contain any complex terms but is represented by the magnitudes of bus voltages E-V, load power P-Q and line admittance G-B. We also see that the load is given by active and reactive power P-Q in (9), while the load is given by current \dot I_{12} in the conventional formula (2). Formula (9) can be used for calculation of bus voltages as described in the next Section. 3. Example calculation of sending-end voltage \text{Example 1:} In Fig. 2, the line-to-neutral voltage V at bus 2(load bus) is maintained at 13.0 kV, the phase impedance of the distribution line is 3.64+$$j7.82$ Ω, and the load per phase is 1,\:056 +$$j 440 kVA. Calculate the voltage E at bus 1(Substation bus)(2). Fig. 2. Two-bus system with load bus voltage 13.0 kV A. Solution by conventional ohmic method Let V=13.0\angle 0^{\circ} kV. Then, the current \dot I_{12} in Fig. 2 is: (10)$$\dot{I}_{12}=\left(\frac{1,056+j 440}{13.0}\right)^{*}=81.23-j 33.84 \text{ A}$$Thus, the vector of sending-end voltage is: (11)$$\begin{aligned} \dot{E} &=V+z · \dot{I}_{12}=13,000+(3.64+j 7.82)(81.23-j 33.84) \\ &=13,560.4+j 512.0 \text { volt} \end{aligned}$$We obtain the magnitude of sending-end voltage: (12)$$|\dot{E}|=\sqrt{13,560 · 4^{2}+512 · 0^{2}}=13,570 \text { volt}$$B. Solution by proposed formula (9): We have: (13)$$\dot Y =\dfrac{1}{\dot z}=\dfrac{1}{3.64+j7.82}=0.0489-j0.1051,$$from which we obtain: (14)$$G =0.0489, \enspace B = 0.1501 \enspace \text{and} \enspace Y = 0.1159.$$Substituting V=13.0 kV, P=1,\:056 kW, Q=440 kvar and Y, G, B into (9) yields: (15)$$\begin{aligned} E&=\frac{\sqrt{\left(P+G V^{2}\right)^{2}+\left(Q+B V^{2}\right)^{2}}}{Y V} \\ &=\frac{\sqrt{\left(1056000+.0489 · 13000^{2}\right)^{2}+\left(440000+.1051 · 13000^{2}\right)^{2}}}{.1159 · 13,000} \\ &=13,570 \text { volt } \end{aligned}$$We see in (12) and (15) that calculation result by formula (9) is exactly the same as the result by conventional ohmic calculation. We also see in (15) that formula (9) is composed of only real numbers and calculation by proposed formula (9) is simpler than ohmic calculation. C. Solution by Equivalent resistance method Magnitude of the current in \dot I_{12} Fig. 2 is: (16)$$\left|\dot{I}_{12}\right|=\frac{\sqrt{1,056^{2}+440^{2}}}{13.0}=88 \text{ A}$$\cos\theta and \sin\theta of current \dot I_{12} are: (17)$$\begin{aligned} &\cos \theta=\frac{1,056}{\sqrt{1,056^{2}+440^{2}}}=0.9230\\ &\sin \theta=\frac{440}{\sqrt{1,056^{2}+440^{2}}}=0.3846 \end{aligned}$$Substituting |\dot I_{12}|, \cos\theta and \sin\theta into formula (2) yields: (18)$$\begin{aligned} E &=V+I(R \cos \theta+X \sin \theta) \\ &=13,000+88 ·(3.64 · 0.9230+7.82 · 0.3846) \\ &=13,560 \text{ volt} \end{aligned}$$D. Solution by Lossless formula Substituting V=13.0 kV, P=1,\:056 kW, Q=440 kvar and X=7.82 into (3) yields: (19)$$\left(\dfrac{1,\:056,\:000*7.82}{E *13,\:000}\right)^{2}+\left(\dfrac{440,\:000*7.82+(13,\:000)^{2}}{E *13,\:000}\right)^{2}=1$$Solving above equation w.r.t. E, we obtain: (20)$$E=13,279 \text{ volt}$$Note that solutions by ohmic method and by formula (9) are exact, while solutions by ‘Equivalent resistance method’ (2) and by ‘Lossless formula’ (3) are approximate in general. Table 1 is the comparison of each calculation result and error rate for the system in Fig. 2. Graph of error rate is illustrated in Fig. 3. Table 1. omparison of calculation methods and error rate  Method Result Remark Error rate Ohmic 13,570 Exact 0 Proposed formula (9) 13,570 Exact 0 Equivalent resistance 13,560 Approximate 0.07 % Lossless formula 13,279 Approximate 2.10 % Fig. 3. Graph of error rate for Table 1 We see in Table 1 that the ‘Equivalent resistance method’ yields an approximate but fairly acceptable solution. ‘Lossless formula’ shows a relatively big error, which is because the system is assumed to be lossless. 4. Example calculation of receiving-end(load bus) voltage When the complex load \dot S =P+j Q, sending-end voltage E and line admittance Y=G-j B are given in a two-bus system, we can obtain the load voltage V with ease using formula (9). \text{Example 2:} In Fig. 4, the line-to-neutral voltage E=24 volt is applied at bus 1 to supply a load 12+j4\sqrt{3} VA through a transmission line with impedance z=1 + j\sqrt{3} Ω. Calculate the load voltage V. Fig. 4. AC two-node circuit with unknown load voltage V A. Solution by proposed formula (9) Since the line impedance z=1+j\sqrt{3} Ω, we have: (21)$$ \begin{aligned} &\dot Y =\dfrac{1}{1+j\sqrt{3}}=\dfrac{1}{4}-j\dfrac{\sqrt{3}}{4}\\ &\text{or}\\ &G=\dfrac{1}{4}, B=\sqrt{\dfrac{3}{4}}, Y=|\dot Y |=\dfrac{1}{2} \end{aligned} $$Substituting E=24, P=12, Q= 4\sqrt{3} and Y, G, B into (9) yields: (22)$$\left(12+\dfrac{1}{4}· V^{2}\right)^{2}+\left(4\sqrt{3}+\dfrac{\sqrt{3}}{4}V^{2}\right)^{2}=\left(V·\dfrac{1}{2}· 24\right)^{2}$$Rearranging: (23)$$\dfrac{V^{4}}{4}-132 V^{2}+ 192=0$$Solving above equation, we obtain: (24)$$V = 22.94 \text{ volt}$$B. Solution by conventional ohmic method Let us find the load voltage V in Fig. 4. using conventional ohmic method. Let \dot V = V\angle 0^{\circ}. Then, the line current \dot I_{12} is represented by: (25)$$\dot I_{12}=\left(\dfrac{12+j4\sqrt{3}}{V}\right)^{*}=\dfrac{12 - j4\sqrt{3}}{V}$$Thus, the vector of sending-end voltage is: (26)$$ \begin{aligned} \dot E &= V + z·\dot I_{12}=V+(1+ j\sqrt{3})·\dfrac{12 - j4\sqrt{3}}{V}\\ &=\left(V+\dfrac{24}{V}\right)+ j\dfrac{8\sqrt{3}}{V} \end{aligned} $$Since |\dot E | = 24, we have from (25): (27)$$|\dot E | =24 =\sqrt{\left(V+\dfrac{24}{V}\right)^{2}+\left(\dfrac{8\sqrt{3}}{V}\right)^{2}}$$Rewriting: (28)$$24^{2}= V^{2}+\dfrac{768}{V^{2}}+48$$or (29)$$V^{4}-528V^{2}+768=0$$We finally obtain the solution: (30)$$V = 22.94 \text{ volt}$$We see in (24) and (30) that calculation result by formula (9) is exactly the same as the result by conventional ohmic calculation. We also see that equation (23) by proposed formula (9) is in fact the same as equation (29) by conventional ohmic calculation. However, we see that solving by conventional ohmic method is far more complicated than solving by proposed formula (9). Solution by conventional ohmic method requires several steps of circuit analyzing with complex-numbers, however, solution by proposed formula (9) needs only a simple substitution of real-number data. C. Solution by Equivalent resistance method From (25), magnitude of current \dot I_{12} is: (31)$$\left |\dot I_{12}\right | =\dfrac{\sqrt{12^{2}+(4\sqrt{3})^{2}}}{V} \text{ A}$$\cos\theta and \sin\theta of current \dot I_{12} are: (32)$$\cos θ =\dfrac{12}{\sqrt{12^{2}+(4\sqrt{3})^{2}}}=\dfrac{\sqrt{3}}{2}\\ \sin θ =\dfrac{4\sqrt{3}}{\sqrt{12^{2}+(4\sqrt{3})^{2}}}=\dfrac{1}{2}$$Substituting |\dot E |, |\dot I_{12}|, \cos\theta and \sin\theta into formula (2) yields: (33)$$24=V+\dfrac{\sqrt{12^{2}+(4\sqrt{3})^{2}}}{V}\left(1·\dfrac{\sqrt{3}}{2}+\sqrt{3}·\dfrac{1}{2}\right)$$or (34)$$V^{2}-24 V + 24=0$$We finally obtain the solution: (35)$$V = 22.95 \text{ volt}$$D. Solution by Lossless formula Substituting |\dot E |=24, P=12, Q=4\sqrt{3} and X=\sqrt{3} into (3) yields: (36)$$\left(\dfrac{12*\sqrt{3}}{24*V}\right)^{2}+\left(\dfrac{4\sqrt{3}*\sqrt{3}+ V^{2}}{24*V}\right)^{2}=1$$Solving above equation w.r.t. V, we obtain: (37)$$V = 23.47 \text{ volt}$$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. Fig. A2. Vector diagram of Fig. A1 In Fig. A2, we have the following relation: (A1)$$E^{2}=(V+IR\cos θ +IX\sin θ)^{2}+(IX\cos θ- IR\sin θ)^{2}$$and we obtain: (A2)$$E=\sqrt{(V+IR\cos θ +IX\sin θ)^{2}+(IX\cos θ- IR\sin θ)^{2}}$$In general, (A3)$$(V+IR\cos θ +IX\sin θ)≫(IX\cos θ- IR\sin θ)$$hence, (A2) can be approximated as follows: (A4)$$E\approx V+I(R\cos\theta +X\sin\theta)$$The term (A5)$$R\cos\theta +X\sin\theta\$

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

이상중 (SangJoong Lee)

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

김주철 (Ju-Chul Kim)

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