Overvoltage Protection of Series Compensation Capacitors
Abstract Generated abstract
This paper examines protection of static series compensation capacitors in high-voltage transmission lines against overvoltages arising during short circuits. It proposes shunting the capacitors with a saturable reactor and analyzes the transient behavior of the resulting circuit using simplifying assumptions, a piecewise-linear approximation of the reactor magnetization curve, operational calculus, and residue evaluation. The analysis derives expressions for the currents in the capacitor, reactor, and external circuit, and indicates how time-current curves can be obtained by applying the formulas successively to each segment of the reactor characteristic.
Full Text
Protection of Series-Compensation Capacitors Against Overvoltages
A. I. Zaitsev, Ya. V. Petrov
(Presented by the scientific seminar of the Electromechanical Faculty)
The transmission capacity of high-voltage electric power transmission is determined to a considerable extent by the conditions for the stability of the parallel operation of generating stations. These conditions are determined mainly by the magnitude of the inductive reactance. On the basis of an analysis of existing and proposed methods for increasing the limit of transmitted power, it may be asserted that the most expedient method of increasing the limit of transmitted power is series compensation of the parameters of a power-transmission line. This compensation may be carried out by synchronous and asynchronous machines, as well as by static capacitors. From [1, 2, 3] the use of series compensation for purposes of voltage regulation is also known.
Static capacitors are the most natural and expedient means of series compensation of the inductance of a line. Their use eliminates the need for complex regulating devices, but along this path there are considerable difficulties. During short circuits on the line, the voltage on the capacitor plates inevitably rises.
To protect them against breakdowns and to reduce short-circuit currents, all capacitor installations are equipped with devices which, in one way or another, shunt the capacitors at the moment impermissible voltages appear. In [4] a description is given of the operating principle of a series-compensation circuit with simultaneous limitation of short-circuit currents and of the influence of this device on dynamic stability. To protect capacitors against overvoltages it is proposed to shunt them with saturable reactors. By appropriately selecting the volt-ampere characteristics of the capacitors and the saturable reactor (Fig. 1), the following operating conditions can be obtained.
Fig. 1.
- Under normal operating conditions, the load current will flow through the capacitors and series compensation will be effected.
- In the event of a short circuit, the voltage on the capacitor begins to rise sharply. As the voltage increases, the magnetizing current of the reactor increases. With a further increase in the current through the capacitor, the reactor will shunt it still more, and with the onset of saturation a current resonance occurs, or the main part of the load current will flow through the reactor if the resonance voltage is less than the value of the phase voltage.
To investigate the behavior of the protection circuit, let us write the transient-process equations for the circuit shown in Fig. 1. To simplify the calculations, we shall neglect the active resistance of the external circuit (line, transformer) and of the saturating reactor, and replace the reactor characteristic by a piecewise-linear approximation. Such neglect is permissible, since the inductive resistance of these elements is many times greater than the active resistance. For the transient process, taking into account the assumptions, on any rectilinear section of the magnetization curve of the saturating reactor we write the system of equations:
\[ i_1 = i_2 + i_3; \]
\[ L_1 \frac{di_1}{dt} + \frac{1}{C}\int_0^t i_2\,dt = U_m \sin \omega t; \]
\[ L_1 \frac{di_1}{dt} - L_2 \frac{di_3}{dt} = U_m \sin \omega t. \]
Let us write the equations in operational form:
\[ i_1(p)=i_2(p)+i_3(p). \tag{1} \]
\[ L_1 p i_1(p)-L_1 p i_1(0)+\frac{1}{cp}i_2(p)-U_{co} = \frac{Up\omega}{p^2+\omega^2}. \tag{2} \]
\[ L_1 p i_1(p)-L_1 p i_1(0)-L_2 p i_3(p)-L_2 p i_3(0) = \frac{Up\omega}{p^2+\omega^2}. \tag{3} \]
Substitute the value of \(i_1(p)\) from equation (1) into equations (2) and (3):
\[ L_1 p\,[i_2(p)+i_3(p)]-L_1 p\,(i_{20}+i_{30})+\frac{1}{cp}i_2(p)-U_{co} = \frac{Up\omega}{p^2+\omega^2}. \tag{4} \]
\[ L_1 p\,[i_2(p)+i_3(p)]-L_1 p\,(i_{20}+i_{30})+L_2 p i_3(p)-L_2 p i_{30} = \frac{Up\omega}{p^2+\omega^2}. \tag{5} \]
\[ i_2(p)\left(L_1p+\frac{1}{cp}\right)+L_1 p i_3(p)-L_1p\,(i_{20}+i_{30})-U_{co} = \frac{Up\omega}{p^2+\omega^2}. \tag{4a} \]
\[ i_2(p)L_1p+i_3(p)(L_1p+L_2p)-L_1p\,(i_{20}+i_{30})-L_2p i_{30} = \frac{Up\omega}{p^2+\omega^2}. \tag{5a} \]
From equation (5a) we determine \(i_3(p)\)
\[ i_3(p)= \frac{U p\omega + \left[L_1p(i_{20}+i_{30})+L_2p i_{30}\right](p^2+\omega^2)} {(p^2+\omega^2)(L_1p+L_2p)} -\frac{L_1p}{(L_1p+L_2p)}\,i_2(p). \]
We substitute the value of \(i_3(p)\) into equation (4a)
\[
i_2(p)\left(L_1p+\frac{1}{Cp}\right)
+
\frac{L_1p\left\{U p\omega+\left[L_1p(i_{20}+i_{30})+L_2p i_{30}\right](p^2+\omega^2)\right\}}
{(p^2+\omega^2)(L_1p+L_2p)}
\]
\[
-\frac{L_1^{2}p^{2}}{(L_1p+L_2p)}\,i_2(p)
-L_1p(i_{20}+i_{30})-U_{c0}
=
\frac{U p\omega}{p^2+\omega^2}.
\]
We determine the value of \(i_2(p)\)
\[
i_2(p)\left(
L_1p+\frac{1}{Cp}
-\frac{L_1^{2}p^{2}}{L_1p+L_2p}
\right)
+
\]
\[
+
\frac{
L_1p^{2}U\omega+
\left[L_1^{2}p^{2}(i_{20}+i_{30})+L_1L_2p^{2}i_{30}\right](p^2+\omega^2)
}
{(p^2+\omega^2)(L_1p+L_2p)}
-
\]
\[
-L_1p(i_{20}+i_{30})-U_{c0}
=
\frac{U p\omega}{p^2+\omega^2}.
\]
\[
i_2(p)=
\frac{
\left\{U p\omega-\left[L_1p(i_{20}+i_{30})+U_{c0}\right](p^2+\omega^2)\right\}(L_1p+L_2p)
}
{
\left(L_1L_2p^2+\dfrac{L_1}{C}+\dfrac{L_2}{C}\right)(p^2+\omega^2)
}
\]
\[
-
\frac{
\left\{L_1p^2U\omega+
\left[L_1^{2}p^{2}(i_{20}+i_{30})+L_1L_2p^{2}i_{30}\right](p^2+\omega^2)\right\}(L_1p+L_2p)
}
{
(p^2+\omega^2)
\left(L_1L_2p^2+\dfrac{L_1}{C}+\dfrac{L_2}{C}\right)
(L_1p+L_2p)
}.
\]
We define the “phi” function
\[
\varphi(p)=
\frac{
\left\{U p\omega+\left[L_1p(i_{20}+i_{30})+U_{c0}\right](p^2+\omega^2)\right\}
\left((L_1+L_2)p\right)
}
{
\left(L_1L_2p^2+\dfrac{L_1}{C}+\dfrac{L_2}{C}\right)(p^2+\omega^2)p
}
\,e^{pt}
-
\]
\[
-
\frac{
\left\{U p\omega L_1+
\left[L_1^{2}p(i_{20}+i_{30})+L_1L_2p i_{30}\right](p^2+\omega^2)\right\}
}
{
\left(L_1L_2p^2+\dfrac{L_1}{C}+\dfrac{L_2}{C}\right)(p^2+\omega^2)p
}
\,e^{pt}.
\]
An image has been obtained for the current in the capacitor. We determine the original from the given image. For this we use the Bromwich integral
\[ f(t)=\frac{1}{2\pi j} \int_{s-j\infty}^{s+j\infty} \varphi(p)\,dp = \sum_{k=1}^{n} a_{-1}^{(k)}. \]
To determine the residues at the individual poles, we find the roots of the characteristic equation
\[ \left(L_1L_2p^2+\frac{L_1+L_2}{C}\right)(p^2+\omega^2)=0, \]
\[ p_1=-j\omega,\qquad p_3=-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}, \]
\[ p_2=+j\omega,\qquad p_4=+j\sqrt{\frac{L_1+L_2}{L_1L_2C}}. \]
All the poles turn out to be simple; therefore we find the residues by the following formula:
\[ a_{-1}=\lim_{p\to a}\bigl[(p-a)z(p)\bigr]. \]
The residue for the first pole \(p_1\) is
\[ a_{-1}^{(1)} = \lim_{p\to j\omega} \left\{ \frac{(p-j\omega)A} { L_1L_2 \left( p^2+\frac{L_1+L_2}{L_1L_2C} \right) (p-j\omega)(p+j\omega) } \right\} e^{j\omega t} = \]
\[ = \frac{ \left(UL_1\omega j\omega-UL_2\omega j\omega-Uj\omega\omega L_1\right)e^{j\omega t} } { L_1L_2 \left( -\omega^2+\frac{L_1+L_2}{L_1L_2C} \right) 2j\omega } -0 = \]
\[ = \frac{ UL_2\omega } { L_1L_2 \left( -\omega^2+\frac{L_1+L_2}{L_1L_2C} \right)^2 } e^{j\omega t}. \]
The residue for the second pole \(p_2\) is
\[ a_{-1}^{(2)} = \lim_{p\to -j\omega} \left\{ \frac{(p+j\omega)A} { L_1L_2 \left( p^2+\frac{L_1+L_2}{L_1L_2C} \right) (p-j\omega)(p+j\omega) } \right\} e^{-j\omega t} = \]
\[ = \frac{ U\omega(-j\omega)(L_1+L_2)e^{-j\omega t} - U\omega(-j)L_1e^{-j\omega t} } { \left( -\omega^2+\frac{L_1+L_2}{L_1L_2C} \right) (-2j\omega)L_1L_2 } = \]
\[ = \frac{ U\omega L_2 } { 2\left( \frac{L_1+L_2}{L_1L_2C}-\omega^2 \right)L_1L_2 } e^{-j\omega t}. \]
Residue for the third pole \(p_3\)
\[ a_1^{(3)}=\lim_{p\to j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}} \]
\[ \times \left\{ \frac{ \left(p-j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\right)A }{ (p^2+\omega^2) \left(p+j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\right) \left(p-j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\right)L_1L_2 } \right\} e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} \]
\[ = \frac{ U^{(\omega)}j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,(L_1+L_2) e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } \]
\[ - \frac{ L_1j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,(i_{20}+i_{30}) \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)(L_1+L_2) e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } \]
\[ + \frac{ U_{C0}\left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)(L_1+L_2) e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } \]
\[ - \frac{ U_{\omega}L_1j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\, e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_1 } \]
\[ - \frac{ L_1^2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,(i_{20}+i_{30}) \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right) e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } \]
\[ - \frac{ j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,L_1L_2i_{10} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right) e^{\,j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,t} }{ 2j\sqrt{\dfrac{L_1+L_2}{L_1L_2C}} \left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 }. \]
Residue for the fourth pole \(p_1\)
\[ \begin{aligned} a_{-1}^{(4)} &=\lim_{p\to -j\sqrt{\frac{L_1+L_2}{L_1L_2C}}} \left\{ \frac{(p-j)\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,A} {(p^2+\omega^2) \left(p+j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\right) \left(p-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\right)} \right\} e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t} \\[4pt] &= \frac{ -j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t\,(L_1+L_2) } { 2\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } \\[6pt] &\quad -\frac{ (L_1+L_2)L_1(i_{20}+i_{30}) \left(\omega^2-\frac{L_1L_2}{L_1L_2C}\right) e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t} } { 2L_1L_2\left(\omega^2-\frac{L_1L_2}{L_1L_2C}\right) } \\[6pt] &\quad -\frac{ U\omega(L_1+L_2) \left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right) e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t} } { 2j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\, t\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right)L_1L_2 } -\frac{ U\omega L_1 e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t} } { 2L_1L_2\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right) } \\[6pt] &\quad -\frac{ L_1^{2}(i_{20}+i_{30})e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}} } {2L_1L_2} + \frac{ L_1L_2i_{30}e^{-j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t} } { 2j\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,L_1L_2 } -j\sqrt{\frac{L_1+L_2}{L_1L_2C}} . \end{aligned} \]
The current in the capacitor is equal to the sum of the residues at all four poles.
\[ i_2=a_{-1}^{(1)}+a_{-1}^{(2)}+a_{-1}^{(3)}+a_{-1}^{(4)} . \]
\[ \begin{aligned} i_2 &= \frac{U\omega(L_1+L_2)\cos\omega t} {L_1L_2\left(\frac{L_1+L_2}{L_1L_2C}-\omega^2\right)} + \left\{ \frac{U\omega(L_1+L_2)} {L_1L_2\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right)} + \frac{L_1(L_1+L_2)(i_{20}+i_{30})}{L_1L_2} \right. \\[6pt] &\qquad\left. -i_{30} - \frac{U\omega L_1} {L_1L_2\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right)} - \frac{L_1^2(i_{20}+i_{30})}{L_1L_2} \right\} \cos\left[ \sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t+ \right. \end{aligned} \]
\[ -\left\{ \frac{U_{c0}(L_1+L_2)} {\sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,L_1L_2} \right\} \sin \sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t . \]
From equation (5a) we determine the value of \(i_2(p)\)
\[ i_2(p)= \frac{ Up\omega+\left[L_1p(i_{20}+i_{30})+L_2pi_{30}\right](p^2+\omega^2) } {(p^2+\omega^2)L_1p}, \qquad i_3(p)=\frac{(L_1p+L_2p)}{L_1p}. \]
We substitute the value of \(i_2(p)\) into equation (4a) and determine the current in the reactor
\[ i_3(p)= \frac{ \left\{Up\omega+\left[L_1p(i_{20}+i_{30})+L_2pi_{30}\right](p^2+\omega^2)\right\} \left(L_1p+\dfrac{1}{Cp}\right)L_1p } { L_1p(p^2+\omega^2) \left(p^2+\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_1 } \]
\[ - \frac{ \left\{Up\omega+\left[L_1p(i_{20}+i_{30})+U_{c0}\right](p^2+\omega^2)\right\}L_1p } { (p^2+\omega^2) \left(p^2+\dfrac{L_1+L_2}{L_1L_2C}\right)L_1L_2 }. \]
Next, by analogy, determining the “\(\xi\)” function, the roots of the characteristic equation, and the values of the residues for these roots, we find the value of the current \(i_3\) flowing through the reactor.
\[ i_3= -\frac{ U\dfrac{1}{\omega C}\,p } { \left(\dfrac{L_1+L_2}{L_1L_2C}-\omega^2\right)L_1L_2 } \cos\omega t - \left\{ \frac{U\omega} {(L_1+L_2)\left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right)} \right. \]
\[ \left. + \frac{L_1(i_{20}+i_{30})+L_2i_{30}}{2(L_1+L_2)} \right\} \cos\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t - \frac{ U_{c0} } { \sqrt{\dfrac{L_1+L_2}{L_1L_2C}}\,L_2 } \sin\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t . \]
The current in the external circuit is \(i_1=i_2+i_3\)
\[ i_1= \frac{ U\left(\omega L_2-\dfrac{1}{\omega C}\right) } { L_1L_2\left(\dfrac{L_1+L_2}{L_1L_2C}-\omega^2\right) } \cos\omega t + \left\{ \frac{ U\omega(L_1+L_2) } { L_1L_2\left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right) } + \right. \]
\[ \left. \frac{ L_1(L_1+L_2)(i_{20}+i_{30}) } {L_1L_2} -i_{30} - \frac{ U\omega } { L_2\left(\omega^2-\dfrac{L_1+L_2}{L_1L_2C}\right) } \right\} \tag{6} \]
\[ -\frac{L_1^{2}(i_{20}+i_{30})}{L_1L_2} \left| \cos \sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t + \frac{U_{c0}\cdot L_2}{\sqrt{\frac{L_1+L_2}{L_1L_2C}}\,L_1L_2} \cdot \sin \sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t \right. \]
\[ -\left\{ \frac{U_0}{(L_1-L_2)\left(\omega^2-\frac{L_1+L_2}{L_1L_2C}\right)} + \frac{L_1(i_{20}+i_{30})+L_2i_{30}}{2(L_1+L_2)} \right\} \cos \sqrt{\frac{L_1+L_2}{L_1L_2C}}\,t . \]
To determine the curves of variation of the currents \(i_1\), \(i_2\), and \(i_3\) with time, it is necessary, in the corresponding expressions for the currents, to substitute successively the values of \(L_2\) for each segment of the piecewise-linear approximation of the magnetization curve of the saturating reactor. In this case, the final values of the currents \(i_1\), \(i_2\), and \(i_3\) for the preceding segment will be the initial values for the subsequent segment.
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-
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