Pie Charts
Abstract Generated abstract
This paper systematizes the theory and construction of circular, or pie, diagrams for alternating-current circuits whose currents or voltages depend on a single complex parameter through a linear fractional expression. It derives the condition for such loci to be circular, namely constancy of the parameter argument, and develops graphical procedures for constructing, superposing, adding, and using diagrams to determine currents, voltages, powers, phase angles, efficiency, and power factor. The treatment covers both calculated circuits with known impedances and experimental construction when circuit parameters or even connections are partly unknown, including applications to single-phase, polyphase, symmetric, and asymmetric cases. Worked examples, including a three-phase transformer with single-phase loading, show that the combined diagrams can reproduce measured quantities closely and clarify the limits under which superposition is valid.
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Pie Charts
R. A. Voronov
A change in the operating regime of an electric circuit, occurring in connection with a change in load resistance, or for some other reason, entails an increase or decrease in all currents and voltages, as well as a change in the phase-displacement angles between them. When these are represented by vectors, the latter will change both in magnitude and in direction.
The loci of the ends of the vectors obtained under these changes are called diagrams of currents and voltages. These diagrams, in the general case, may take the most diverse forms, beginning with a straight line and ending with complex curves of higher orders. Providing a graphical relationship between the vectors of currents and voltages and making it possible to find powers and a number of other quantities as well, these diagrams are difficult to use and, in their general form, have almost no application.
The simplest and most frequently encountered form of such diagrams is a circle and, as a special case of it, a straight line. These diagrams, called circular diagrams, are very simple to construct and at the same time possess many valuable properties that are readily usable for practical purposes. Although circular diagrams are very widespread in electrical engineering, they are still not used everywhere where this is possible and useful; and if they are used, then very incompletely. The reason for the limited application of circular diagrams may be considered to be certain difficulties encountered in their construction, which is explained by the unclear and incomplete instructions given by various sources in which their properties and methods of construction are set forth.
The diagram for the simple T-shaped equivalent circuit of single-phase or symmetrical three-phase currents is the most widely used. In this case one usually confines oneself only to finding the current entering the given circuit from the network, and the powers of the total and consumed load, i.e., the useful power. Sometimes the current and voltage of the load are also added, and very rarely any other quantities. Circular diagrams for asymmetrical polyphase circuits, although they are constructed (for example, the displacement of the neutral point of a three-phase circuit), are very rare, and they are treated somehow separately from the other diagrams, although they possess exactly the same properties.
Nevertheless, with the proper and complete construction of diagrams, they can be applied in very many cases (almost always when there is one varying quantity—a parameter), and from them all currents, voltages, powers, phase-displacement angles, etc., can be obtained. The purpose of the present work is to generalize the available material and to put it into a form that makes it possible to apply these constructions easily in various cases. Two different cases are considered: the first, when the circuit is given and all connected resistances are known; and the second, when the circuit is realized in practice, but the connected resistances, and sometimes even the connection scheme itself, are unknown.
A number of the proofs and methods of construction presented in the work are new and not borrowed from anywhere; others are taken from various sources. Diagrams of a form other than circular (bicircular, parabolic, etc.) are not included in the present work.
I. THEORY OF CIRCULAR DIAGRAMS
1. CONDITIONS FOR OBTAINING CIRCULAR DIAGRAMS
If the mode of operation of an electric circuit is determined by the value of a variable quantity $\dot K$ (a parameter), most often the load resistance, then another quantity $\dot N$, for which it is necessary to construct a circular diagram (usually current or voltage), is related to it by an equation of the form
\[ \dot N=\frac{\dot A+\dot B \dot K+\dot E \dot K^2+\cdots} {\dot C+\dot D \dot K+\dot F \dot K^2+\cdots}, \]
where $\dot A; \dot B; \dot C; \dot D; \dot E; \dot F;\ldots$ are constant coefficients.
In general, this diagram may take the most varied forms and, only under certain relations between the coefficients, will it have the form of an arc of a circle. Since an arc of a circle is easier to construct than any other curve, and since in this case all the properties of the diagram are used more easily and better, it is desirable to strive to create the conditions necessary for obtaining this form of curve.
The simplest and most frequently encountered diagram will be one for which all coefficients except $\dot A, \dot B, \dot C$ and $\dot D$ become zero, i.e. the dependence between $\dot N$ and $\dot K$ has the form
\[ \dot N=\frac{\dot A+\dot B \dot K}{\dot C+\dot D \dot K}. \tag{1} \]
Diagrams of this type only are the subject of discussion in the present work.
The variable quantity $\dot K$ may, in the general case, take any values from $+\infty$ to $-\infty$, and for the construction of all diagrams two of them play the chief role: one, when $\dot K=\pm\infty$, which corresponds to
\[ \dot N_{\infty}=\frac{\dot B}{\dot D} \tag{2} \]
and the second, when $\dot K=0$, which corresponds to
\[ \dot N_{0}=\frac{\dot A}{\dot C}. \tag{3} \]
In the case where $\dot K$ represents the resistance or conductance of the load, these values correspond to the no-load operation of the circuit ($\dot Z_n=\infty$ or $\dot Y_n=0$) and to a short circuit ($\dot Z_n=0$ or $\dot Y_n=\infty$).
Representing the values $\dot N_{\infty}$ and $\dot N_0$ in the form of segments (vectors), let us take at the same place one more segment (vector) for $\dot N$, corresponding to some-
the value \(\dot K\) (Fig. 1). The ends of these segments must lie on the desired arc of the pie chart.
Connecting the obtained points by straight lines, we find the triangle \(abc\). The sides of this triangle will be found as the differences of the segments \(\dot N\), \(\dot N_0\), and \(\dot N_\infty\).
\[ \overline{cb}=\overline{ob}-\overline{oc}=\dot N_0-\dot N= \frac{\dot A}{\dot C}-\frac{\dot A+\dot B\dot K}{\dot C+\dot D\dot K} =\frac{\dot A\dot D-\dot B\dot C}{\dot C+\dot D\dot K}\cdot\frac{\dot K}{\dot C}, \tag{4} \]
\[ \overline{ac}=\overline{oc}-\overline{oa}=\dot N-\dot N\infty= \frac{\dot A+\dot B\dot K}{\dot C+\dot D\dot K}-\frac{\dot B}{\dot D} =\frac{\dot A\dot D-\dot B\dot C}{\dot C+\dot D\dot K}\cdot\frac{1}{\dot D}. \tag{5} \]
The ratio of the obtained values will be equal to
\[ \frac{\overline{ac}}{\overline{cb}} =\frac{\dot N-\dot N\infty}{\dot N_0-\dot N} =\frac{1}{\dot K}\cdot\frac{\dot C}{\dot D}. \]
If the complexes are represented in symbolic form
\[ \dot A=A\cdot e^{j\alpha};\quad \dot B=B e^{j\beta};\quad \dot C=C e^{j\gamma};\quad \dot D=D e^{j\delta};\quad \dot K=K e^{j\Psi}, \tag{6} \]
then this ratio takes the form
\[ \frac{\overline{ac}}{\overline{cb}} =\frac{C}{K.D}\cdot e^{j(\gamma-\delta-\Psi)}. \]
The angle \(\rho\) between the directions of the segments \(\overline{ac}\) and \(\overline{cb}\) will be equal to the argument of the expression obtained, i.e. equal to the difference
\[ \rho=\gamma-\delta-\Psi. \tag{7} \]
For the point \(c\) to move along the arc of a circle when the value \(\dot K\) changes, it is necessary and sufficient that the angle \(\rho\) remain constant at all times,
\[ \rho=\gamma-\delta-\Psi=\mathrm{const}. \]
Since it was stipulated above that the coefficients \(\dot C\) and \(\dot D\) are constant, their arguments \(\gamma\) and \(\delta\) must also be constant; whence it follows that the value \(\Psi\) also must not change when the magnitude \(\dot K\) changes.
Thus the single and entirely sufficient condition for obtaining a pie chart is the constancy of the argument of the variable quantity in equation (1),
\[ \Psi=\mathrm{const}. \]
Fig. 1.
In most cases, when the variable quantity is the resistance or conductance of the load, this condition reduces to the constancy of the phase-shift angle or the constancy of the load power factor.
2. BASIC RELATIONS FOR PIE CHARTS
By the property of tangents and chords of circles, the angle \(\rho\) will also lie between the chord \(\overline{ab}\) and the tangents \(aa'\) and \(bb'\) drawn at its ends; moreover, it will be measured by half the arc \(acb\). From equation (7) it follows that this angle will depend only on the values of the arguments of the variable quantity \(\dot K\) and the coefficients \(\dot C\) and \(\dot D\) entering the denominator of equation (1), and will be entirely independent of the coefficients \(\dot A\) and \(\dot B\) of the numerator of the same equation.
It follows from this that for all quantities whose denominator is the same, diagrams are obtained with equal angles \(\rho\), and consequently with similar segments \(acba\), differing only in the dimensions of the chord. ^1)
This chord can be found for each diagram as the difference
\[ \overline{ab}=\overline{ob}-\overline{oa}=\dot N_0-\dot N_\infty=\frac{\dot A}{\dot C}-\frac{\dot B}{\dot D}=\frac{\dot A\dot D-\dot B\dot C}{\dot C\dot D}, \tag{8} \]
and its length can be computed from the relation of the sides of the triangle in Fig. 1
\[ (ab)^2=(ob)^2+(oa)^2-2(ob)\cdot(oa)\cdot\cos(\overline{ob};\overline{oa}) \]
or, substituting the values of the sides from equations (2) and (3),
\[
H^2=N_0^2+N_\infty^2-2N_0N_\infty\cos(\varphi_0-\varphi_\infty)
=\left(\frac{A}{C}\right)^2+
\]
\[
+\left(\frac{B}{D}\right)^2-2\,\frac{AB}{C\cdot D}\cos(\varphi_0-\varphi_\infty),
\tag{8a}
\]
where \(H\) is the length of the chord on the same scale as the values \(N_0\) and \(N_\infty\), and the angles \(\varphi_0\) and \(\varphi_\infty\) are the arguments of the values of these quantities.
The angle \(cab\) will be determined as the argument of the ratio of the sides \(\overline{ac}\) and \(\overline{ab}\), which, taking into account equations (5) and (8), will be equal to
\[ \frac{\overline{ac}}{\overline{ab}} =\frac{\dot N-\dot N_\infty}{\dot N_0-\dot N_\infty} =\frac{\dot C}{\dot C+\dot D\dot K}. \tag{9} \]
This ratio also will not depend on the coefficients \(\dot A\) and \(\dot B\), i.e., the angle \(cab\) for a given value of the variable quantity \(\dot K\) will be the same for all diagrams having identical denominators in equations (1). It follows from this that, for any value of \(K\), similar triangles \(abc\) will be obtained for all such diagrams.
For negative values of \(K\), the point \(c\) will lie on the dotted part of the circle (Fig. 1); moreover, if the angle \(\rho\) is less than \(90^\circ\), then this part of the circle will be larger than the solid part, and if it is greater than \(90^\circ\), then conversely. For most diagrams of alternating current, the quantity \(\dot K\) cannot take negative values (resistance), i.e., only part of the diagram will be used.
In the case where the angle \(\rho\) turns out to be equal to \(0^\circ\) or \(180^\circ\), the circumference of the diagram assumes the form of a straight line (the radius is equal to infinity). The diagram is also transformed into a straight line in the case where the coefficients \(\dot C\) or \(\dot D\) are equal to zero, i.e., when \(\dot N_0\) or \(\dot N_\infty\) becomes infinite.
^1) Equal angles \(\rho\) are obtained not only with identical denominators of eq. 1, but also for all those for which the condition \(\gamma-\delta=\mathrm{const}\) holds. But such diagrams will not have similar triangles \(abc\), and therefore cannot be superposed (see eq. 9).
In these cases many properties of the diagram disappear or cannot be used.
3. SUPERPOSITION OF PIE CHARTS
If there are several pie charts constructed for quantities whose denominators in equation (1) are the same, i.e.
\[ \dot N_1=\frac{\dot A_1+\dot B_1 K_1}{\dot C+\dot D K};\quad \dot N_2=\frac{\dot A_2+\dot B_2 K}{\dot C+\dot D K}, \]
and so on, then the points \(C_1\) and \(C_2\) (Fig. 2), found for one and the same value of the variable \(K\), will lie at the vertices of similar triangles \(a_1 c_1 b_1\) and \(a_2 c_2 b_2\).
To find, for some value \(N_1\) determined by the segment \(oc_1\), the corresponding value \(N_2\), it is necessary to construct on the second diagram a triangle \(c_2 a_2 b_2\) similar to the triangle \(c_1 a_1 b_1\) of the first diagram (or, equivalently, make \(\angle c_2 a_2 b_2\) equal to \(\angle c_1 a_1 b_1\)). This construction, especially when there are many diagrams or a significant number of different \(N_1\) for which such constructions must be made, takes much time and can easily lead to a number of errors.
Fig. 2.
It is much more convenient to change the mutual arrangement of the diagrams with such a calculation that their chords coincide with one another in direction, and the points \(a_1, a_2, a_3\) coincide in one point; such superposed diagrams corresponding to the diagrams of Fig. 2 are shown in Fig. 3.
With such an arrangement of the diagrams, all tangents \(a_1a_1', a_2a_2'\) and the sides \(a_1c_1, a_2c_2\) of the similar triangles coincide in direction. To find the point \(c_2\), corresponding to the point \(c_1\) of another diagram, it will be sufficient to draw the straight line \(ac_1\) and find the place of its intersection with the other corresponding circle.
The numerical values of the segments representing the quantities \(N_1\) and \(N_2\) do not change in this case, but their mutual arrangement (the phase-shift angles) will not be equal to the actual ones present in the circuit. If
Fig. 3.
to superpose the diagrams the first was rotated through an angle \(\tau_1\), and the second through an angle \(\tau_2\); then the actual angle of displacement between \(\dot N_1\) and \(\dot N_2\) will be equal to
\[ \angle(\dot N_1;\dot N_2)=\angle(\dot N'_1;\dot N'_2)-(\tau_1 \tau_2), \]
i.e., from the angle obtained in Fig. 3 it is necessary to subtract the difference of the angles of rotation of the diagrams.
Instead of superposing the points \(a_1; a_2; a_3\), the points \(b_1; b_2; b_3\) of the diagrams may be superposed. In this case the tangents \(b_1 b'_1; b_2 b'_2\) and the sides \(c_1 b_1\) and \(c_2 b_2\) of the triangles will coincide. To find the point \(c_2\), it will be necessary to draw the straight line \(b c_1\), taking its intersection with the arc of the other circle. This arrangement of the diagrams is less convenient, since with it it is more difficult to obtain the values of the variable quantity \(K\) (see below).
In many cases it proves convenient to carry out a further superposition of the diagrams, reducing all the circles to one. For this it is necessary to change the scales of the separate quantities in such a way that the chords of all the diagrams become equal to one another. Such diagrams are shown in Fig. 4. In this case there is no need at all to seek the point \(c_2\), since it will coincide with the point \(c_1\).
Fig. 4.
Although this type of diagram is very convenient for construction and use, it may reduce clarity, since, as a result of the change of scales, currents or voltages that are smaller in magnitude may turn out to be represented by longer segments than other currents or voltages that are larger in magnitude.
Below, in Figs. 16, 17, and 18, diagrams of all three types are constructed for one and the same circuit. Comparing them, it is easy to see the simplifications obtained for the superposed diagrams, especially for the last case.
4. ADDITION OF CIRCULAR DIAGRAMS
If it is necessary to construct a circular diagram for a quantity \(\dot N_3\), which is equal to the sum of two others,
\[ \dot N_3=\dot N_1=\dot N_2, \]
for which constructed diagrams already exist, then it is more convenient to carry out such a construction graphically rather than by calculation.
Since
\[ \dot N_3=\frac{\dot A_1+\dot B_1 \dot K}{\dot C+\dot D \dot K} +\frac{\dot A_2+\dot B_2 \dot K}{\dot C+\dot D \dot K} =\frac{(\dot A_1+\dot A_2)+(\dot B_1+\dot B_2)\dot K}{\dot C+\dot D \dot K}, \]
this diagram will have the same angle \(\rho\) as the component diagrams. It is easy to see that for \(K=\infty\) and \(K=0\) the following relations hold:
\[ \dot N_{3\infty}=\dot N_{1\infty}+\dot N_{2\infty};\quad \dot N_{30}=\dot N_{10}+\dot N_{20}. \]
It follows from this that, in order to find the summation diagram, it is sufficient, by a graphical method, to add the segments corresponding to these values, and, on their difference as on a chord (Fig. 5), to construct an arc of a new circle having the same angle \(\rho\). This construction can be carried out, of course, only for diagrams brought together that have one and the same scale.
In a similar way, when computing
\[ \dot N_4=\dot N_1-\dot N_2 \]
it is necessary to find the differences of the segments of the extreme values and from them construct an arc of a new circle. In this case it is more convenient to find the difference of the segments not by subtraction, but as a sum, replacing the quantity being subtracted by its opposite:
\[ \dot N_4=\dot N_1+(-\dot N_2). \]
Fig. 5.
Such a construction is shown in Fig. 5.
If, in addition, one of the terms \(\dot M\) is constant:
\[ \dot N_2=\dot N_1+\dot M, \]
then it is sufficient to transfer the origin of coordinates from the point \(O_1\) to the point \(O_2\) by the length of the segment corresponding to the inverse value \(-\dot M\). The correctness of such a transfer is clear from Fig. 6.
In subtraction
\[ \dot N_3=\dot N_1-\dot M \]
the transfer must be made to the point \(O_3\) by the amount
\[ -(-\dot M)=+\dot M, \]
i.e. to the end of the segment representing the quantity \(\dot M\) (Fig. 6).
In inverse subtraction
\[ \dot N_4=\dot M-\dot N_1, \]
which can be replaced by
\[ (-\dot N_4)=\dot N_1-\dot M, \]
we obtain the preceding case, but the direction of the resulting segment will not be from the point \(O_3\) to the circle, but, on the contrary; that is, the origin of this vector will be movable, while its end will always be at the point \(O_3\).
Fig. 6.
A similar arrangement of vectors is also convenient when their differences are constant (for example, phase voltages). In this case, having rotated
all diagrams by 180 degrees and taking the negative directions of the segments,
\[ -(-\dot N)=\dot N, \]
we shall obtain the previous values, but with a different arrangement of the diagrams. Fig. 7a shows the usual arrangement of three phase-voltage vectors, when their ends lie on the circumference, and Fig. 7b shows the inverted diagram, in which the ends of the vectors are fixed.
5. THE VARIABLE \(K\) ON THE DIAGRAM.
The variable quantity \(K\), on which all the remaining changing quantities \(\dot N_1\), \(\dot N_2\), etc., depend, may be represented on any diagram in the form of a segment taken in some direction on a definite scale. For this purpose it is necessary, from an arbitrary point \(d\) taken on the chord \(ab\) of any of the diagrams, to draw the line \(dd'\), parallel to the tangent \(ad'\), corresponding to the value \(K=\infty\) (Fig. 8). The segment \(de\), obtained by the intersection of this line with the side \(ac\) of the triangle \(acb\) (or with the extension of this side), will be proportional to the value of \(K\).
Fig. 7.
From the similarity of the triangles \(aed\) and \(abc\) we have
\[ \frac{ed}{ad}=\frac{cb}{ac} \]
or, substituting the values from equations (4) and (5),
\[ ed=ad\cdot \frac{D}{C}\cdot K. \tag{10} \]
Since the segment \(ad\) is constant, as are also the coefficients \(C\) and \(D\), \(ed\) will, on a definite scale, correspond to the desired value of the quantity \(K\).
Fig. 8.
If, from an arbitrary point \(f\) of the chord \(ab\), one draws the line \(ff'\), parallel to the tangent \(bb'\), corresponding to the value \(K=0\), and takes on it the segment \(fg\), obtained from its intersection with the side \(bc\) of the triangle \(abc\) (Fig. 8), then this segment will be proportional to the reciprocal quantity \(K\), i.e., it will give the values
\[ \frac{1}{K}. \]
From the similarity of the triangles it follows that
\[ fg=bf\cdot \frac{C}{D}\cdot \frac{1}{K}, \tag{10a} \]
which also confirms what was said above.
6. MULTIPLICATION OF THE VECTORS OF TWO DIAGRAMS
It is often necessary to be able to find graphically the product of two quantities represented by two diagrams with identical angles \(\rho\), belonging to one and the same circuit (for example, power). In this case it is required to find on the diagrams such segments which, on some scale, would give the values
\[ P N_1\cdot N_2\cdot \cos(\dot N_1;\dot N_2) \tag{11} \]
or
\[ P=N_1'\cdot N_2' + N_1''\cdot N_2'', \tag{11a} \]
where \(N_1'\) and \(N_2'\) are the real components of the complexes
\[ \dot N_1=N_1' + jN_1'';\quad \dot N_2=N_2' + jN_2'', \]
and \(N_1''\) and \(N_2''\) are the factors at the imaginary components of the same complexes. In accordance with equations (1) and (6), for the first quantity we have
\[ \dot N_1= \frac{A_1 e^{j\alpha_1}+B_1 K e^{j(\beta_1+\Psi)}} {C e^{j\gamma}+DK e^{j(\delta+\Psi)}}. \]
or, passing to trigonometric form,
\[ \dot N_1= \frac{ A_1(\cos\alpha_1+j\sin\alpha)+B_1K[\cos(\beta_1+\Psi)+j\sin(\beta_1+\Psi)] }{ C(\cos\gamma+j\sin\gamma)+DK[\cos(\delta+\Psi)+j\sin(\delta+\Psi)] }. \]
Freeing ourselves from the imaginary unit \(j\) in the denominator and carrying out the transformations, we obtain
\[ N_1'= \frac{ [A_1\cos\alpha_1+B_1K\cos(\beta_1+\Psi)]\cdot [C\cos\gamma+DK\cos(\delta+\Psi)] }{ C^2+D^2K^2+2CDK\cos\rho } + \]
\[ +\frac{ [A_1\sin\alpha_1+B_1K\sin(\beta_1+\Psi)]\cdot [C\sin\gamma+DK\sin(\delta+\Psi)] }{ C^2+D^2K^2+2CDK\cos\rho }, \]
\[ N_1''= \frac{ [A_1\sin\alpha_1+B_1K\sin(\beta_1+\Psi)]\cdot [C\cos\gamma+DK\cos(\delta+\Psi)] }{ C^2+D^2K^2+2CDK\cos\rho } - \]
\[ -\frac{ [A_1\cos\alpha_1+B_1K\cos(\beta_1+\Psi)]\cdot [C\sin\gamma+DK\sin(\delta+\Psi)] }{ C^2+D^2K^2+2CDK\cos\rho }, \]
where the angle \(\rho\) has been introduced according to equation (7).
Similar expressions are also obtained for \(N_2'\) and \(N_2''\), replacing index 1 by index 2 (instead of \(A_1\), taking \(A_2\), etc.).
Substituting these expressions into equation (11a) and carrying out the transformations and reductions (omitted, since they take up very much space), we obtain an equation of the following form
\[ P= \frac{K\left[A_1B_2\cos(\beta_2-\alpha_1+\Psi)+A_2B_1\cos(\beta_1-\alpha_2+\Psi)\right]} {C^2+D^2K^2CDK\cos\rho} + \frac{K^2B_1B_2\cos(\beta_1-\beta_2)+A_1A_2\cos(\alpha_1-\alpha_2)} {C^2+D^2K^2+2CDK\cos\rho}. \tag{12} \]
Let us prove that this value will, on a certain scale, correspond to the segment \(cd\) (Fig. 9), taken from the point \(c\) of the diagram perpendicular to a certain straight line \(mn\).
If the arc \(acb\) is constructed for the given circuit and has an angle \(\rho\) between the chord and the tangent, then from equation (9) we have
\[ \overline{ac}=\overline{ab}\cdot\frac{\dot C}{\dot C+\dot D\dot K}; \]
introducing the notation
\[ \overline{ac}=\dot E;\quad \overline{ab}=\dot H, \tag{13} \]
we have
\[ \dot E=\dot H\frac{\dot C}{\dot C+\dot D\dot K} \]
or
\[ \dot E=\dot H \frac{C(\cos\gamma+j\sin\gamma)} {C(\cos\gamma+j\sin\gamma)+DK[\cos(\delta+\Psi)+j\sin(\delta+\Psi)]}. \]
Freeing the denominator of the imaginary unit, we obtain
\[ \dot E=\dot H \frac{C^2+CDK\cos\rho+jCDK\sin\rho} {C^2+D^2K^2+2CDK\cos\rho}; \]
if we also introduce the angle \(\eta\) of inclination of the chord \(ab\) to the straight line \(mn\) and take
\[ \dot H=H(\sin\eta-j\cos\eta), \]
then the real part \(\dot E\) will be located perpendicular to the straight line \(mn\) and will be equal to
\[ E'=cf=H\frac{C^2\sin\eta+CDK\sin(\rho+\eta)} {C^2+D^2K^2+2CDK\cos\rho}. \]
Fig. 9.
Denoting also by \(F\) the distance from the point \(a\) to the straight line \(mn\),
\[ \overline{am}=\overline{fd}=F, \]
we find the expression for the segment \(cd\), which we shall denote by \(Q\),
\[ Q=\overline{cd}=\overline{cf}+\overline{fd} = H\frac{C^2\sin\eta+CDK\sin(\rho+\eta)} {C^2+D^2K^2+2CDK\cos\rho} +F. \]
After reducing to a common denominator, we obtain
\[ Q=\frac{K[H\sin(\rho+\eta)+2F\cos\rho]\cdot CD+K^2FD^2+(H\sin\eta+F)C^2}{C^2+D^2K^2+2CDK\cos\rho}. \tag{14} \]
Comparing the expressions obtained in (12) and (14), we see that their denominators are the same, while the numerators have one and the same character. For the complete identity of both equations it is necessary to choose the segment \(F\) and the angle \(\eta\) so that the coefficients of \(K\) and \(K^2\), and the constant term, in both equations are the same.
Equation (12) gives the values of \(P\) in the dimension of a product (for example, for power, in watts), whereas equation (14) gives for \(Q\) values on the scale in which the segments \(H\) and \(F\) are measured (for example, on the scale of currents, or directly in centimeters). In order to equate them, it is necessary to introduce a factor \(M\) such that
\[ P=MQ, \]
this factor will be equal to the ratio of the scales \(M_P\) and \(M_Q\):
\[ M=\frac{M_P}{M_Q}; \]
in the case where \(F\) and \(H\) were measured in centimeters or millimeters, the scale of the product is obtained directly as
\[ M=M_P. \]
Comparing the terms of the equations, we obtain the conditions for their coincidence
\[ MCD[H\sin(\rho+\eta)+2F\cos\rho] =A_1B_2\cos(\beta_2-\alpha_1+\Psi)+ \]
\[ + A_2B_1\cos(\beta_1-\alpha_2+\Psi) \]
\[ MFD^2=B_1B_2\cos(\beta_1-\beta_2) \]
\[ MC^2(H\sin\eta+F)=A_1A_2\cos(\alpha_1-\alpha_2); \]
introducing the notation
\[ R=\frac{A_1B_2\cos(\beta_2-\alpha_1+\Psi)+A_2B_1\cos(\beta_1-\alpha_2+\Psi)}{CD}, \]
\[ S=\frac{B_1B_2\cos(\beta_1-\beta_2)}{D^2}, \tag{15} \]
\[ T=\frac{A_1A_2\cos(\alpha_1-\alpha_2)}{C^2}, \]
we obtain, solving these equations, the value of the sought angle \(\eta\)
\[ \tg\eta=\frac{(T-S)\sin\rho}{R-(T+S)\cos\rho}; \tag{16} \]
after which it is easy to find the scale \(M\)
\[ M=\frac{T-S}{H\sin\eta}=\frac{R-2S\cos\rho}{H\sin(\rho+\eta)}, \tag{17} \]
and then also the segment \(F\)
\[ F=\frac{S}{M}. \tag{18} \]
All the equations obtained make it possible to construct the straight line \(mn\) and determine the scale in which the product segments will be obtained.
If the line \(cd\) is drawn not perpendicular to the line \(mn\), but at some angle \(\theta \ne 90^\circ\), then this will affect only the scale, and the new scale will be equal to
\[ M' = M \sin \theta . \tag{19} \]
In many cases it is convenient to direct these segments parallel either to one of the tangents \(aa'\) or \(bb'\), or parallel to the chord, since this gives the most convenient scales (see below).
If in equation (18) this new scale is substituted for the scale \(M\), then the obtained value
\[ F^1 = \frac{S}{M^1} = \frac{F}{\sin \theta} \]
will give the distance from the point \(a\) to the line \(mn\) not along the perpendicular, but in the direction chosen for the segments \(cd\).
If it is necessary to find segments giving not a single product, but at once the sum of several products (for example, the power of all phases of a polyphase circuit), then, after finding for each of them the values \(R\), \(S\), and \(T\), one must substitute their sums \(\Sigma R\), \(\Sigma S\), and \(\Sigma T\) into equations (16), (17), and (18).
In some cases it is required to find the product
\[ P = N_1 N_2 \sin (\dot N_1; \dot N_2), \tag{20} \]
which differs from equation (11) by the factor \(\sin (\dot N_1; \dot N_2)\) instead of \(\cos (\dot N_1; \dot N_2)\) (for example, reactive power). This product is equivalent to
\[ P = N_1 N_2 \cos [(\dot N_1; \dot N_2) - 90^\circ], \]
i.e., it is the same as equation (11), but one of the diagrams must be rotated through an angle of \(90^\circ\). For this case all the relations obtained above may be used, but introducing for one of the quantities the angles
\[ \alpha_2^{1} = \alpha_2 - 90^\circ;\quad \beta_2^{1} = \beta_2 - 90^\circ. \]
Equations (15) for the given case take the following form:
\[ R = \frac{ A_2 B_1 \sin (\beta_1 - \alpha_2 + \Psi) - A_1 B_2 \sin (\beta_2 - \alpha_1 + \Psi) }{ C . D }; \]
\[ S = \frac{ B_1 B_2 \sin (\beta_1 - \beta_2) }{ D^2 }; \tag{21} \]
\[ T = \frac{ A_1 A_2 \sin (\alpha_1 - \alpha_2) }{ C^2 }. \]
All the remaining calculations remain unchanged.
7. SPECIAL CASES OF PRODUCTS
Let us consider several cases of products of segments often encountered in alternating-current circuits:
- Suppose that it is required to find the segments corresponding to the product
\[ P = L (N_1)^2 = L N_1 . N_1, \]
where the coefficient \(B\) in equation (1) is equal to zero. In this case the point \(a\) coincides with the origin of coordinates, since \(N_2 = 0\). The coefficients of the factors will be equal to
\[ B_1 = B_2 = 0;\quad L A_1 = A_2;\quad \alpha_1 = \alpha_2, \]
which gives, from equations (15),
\[ R=0;\ S=0;\ T=\frac{LA_1^2}{C^2}=LH^2M_1^2, \]
where \(H\) is the length of the chord in centimeters, and \(M_1\) is the scale corresponding to the quantity \(N_1\). From equations (16) and (18) we have
\[ \operatorname{tg}\eta=-\operatorname{tg}\rho;\ \eta=-\rho;\ F=0; \]
this shows that the straight line \(mn\) will pass through the point \(a\) and coincide with the tangent \(aa^1\) (Fig. 10). The scale for the product will be equal, by equation (17), to
\[ M=-L\frac{HM_1^2}{\sin\rho}. \tag{22} \]
The minus sign is not reflected in the magnitude of the scale, but indicates that the straight line \(cd\) goes not in the direction toward the chord, but away from it.
If the straight line \(cd_1\) is drawn not perpendicular to the straight line \(mn\), but parallel to the chord \(ab\) (in this case \(\theta=\rho\)), then we obtain a new scale from equation (19)
\[ M^1=-LHM_1^2. \tag{22a} \]
Fig. 10.
- Let us consider the second case, when it is required to find the same product
\[ P=L(N_1)^2=LN_1\cdot N_1, \]
where the coefficient \(A_1\) turns out to be equal to zero. In this case the point \(b\) coincides with the origin of coordinates, i.e. \(N_0=0\). The coefficients of the factors will be equal to
\[ A_1=A_2=0; \]
\[ LB_1=B_1;\ \beta_1=\beta_2, \]
which gives, from equations (15),
\[ R=0;\ T=0; \]
\[ S=\frac{LB_1^2}{D^2}=LH^2M_1^2. \]
From equations (16), (17) and (18) it follows that
\[ \operatorname{tg}\eta=\operatorname{tg}\rho;\ \eta=\rho; \]
\[ F=-H\sin\rho; \]
\[ M=-L\frac{HM_1^2}{\sin\rho} \tag{23} \]
Fig. 11.
\(^1\) When \(A=0\) and \(B=0\), this straight line will be the semipolar of the circle; \(M\) can be found by a graphical construction.
From Fig. 11 it is easy to see that the line \(mn\) will pass through the point \(b\) and coincide with the tangent \(bb^{1}\). As before, if the line \(cd_{1}\) is drawn parallel to the chord \((\theta=\rho)\), we obtain the scale
\[ M'=-LHM_{1}^{2}. \tag{23a} \]
In both cases considered, the line passes through the origin and coincides with the tangent.
- Let us find the product \(N_{1}\) and \(N_{2}\) in the case when the relation
\[ \dot N_{2}=K\dot N_{1} \]
holds, with \(B_{1}=O\). In this case \(A_{2}=O\) and \(B_{2}=A_{1}\). From equations (15) we have
\[ R=\frac{A_{1}B_{2}\cos\Psi}{C\cdot D}; \]
\[ S=O;\quad T=O. \]
Since for \(N_{1}\) the origin coincides with the point \(a\), and for \(N_{2}\) with the point \(b\), the chord will be equal to
\[ HM_{1}=\frac{A_{2}}{C};\quad HM_{2}=\frac{B_{2}}{D}, \]
which gives for \(R\) the value
\[ R=H^{2}M_{1}M_{2}\cos\Psi. \]
Fig. 12.
From equations (16) and (18) we find
\[ \tg\eta=O;\quad \eta=O;\quad F=O, \]
i.e. the line \(mn\) coincides with the chord \(ab\) (Fig. 12).
The scale of the product is determined from equation (17)
\[ M=\frac{R}{H\sin\rho}= \]
\[ =HM_{1}M_{2}\frac{\cos\Psi}{\sin\rho}. \tag{24} \]
Fig. 13.
If, however, the segments are taken parallel to one of the tangents \(aa^{1}\) or \(bb^{1}\) (the segments \(cd_{1}\) and \(cd_{2}\)), then the new scale will be equal to
\[ M'=HM_{1}M_{2}\cos\Psi, \tag{24a} \]
since in this case \(\theta=\rho\).
- Finally, let us consider the case when, in the product, one of the factors \((\dot N_2)\) is constant. In this case, having reduced it to the denominator of the second factor,
\[ \dot N_2=\frac{\dot N_2\dot C+\dot N_2\dot D K}{\dot C+\dot D K}, \]
one may use the relations derived above. If the diagram is arranged so that \(\dot N_2\) is directed along the axis of real values (Fig. 13), then equations (15), (16), (17), and (18), after transformations (omitted, since they take up much space), give
\[ \operatorname{tg}\eta= \frac{ \dfrac{A}{C}\cos(\alpha-\gamma)-\dfrac{B}{D}\cdot\cos(\beta-\delta) }{ \dfrac{A}{C}\sin(\alpha-\gamma)-\dfrac{B}{D}\sin(\beta-\delta) } \]
\[ M=N_2M_1;\qquad F=\frac{B}{D}\cdot\cos(\beta-\delta). \tag{25} \]
To clarify the meanings of the expressions obtained, let us consider this case in a somewhat different way. Arranging the diagram as was done in Fig. 13, we obtain
\[ \dot N_2=\dot N_2^{\,1}=\dot N_2, \]
which gives for the product
\[ P=N_1^{\,1}N_2^{\,1}=N_1^{\,1}\cdot N_2; \]
from this equation we obtain that the product is determined by the real component \(\dot N_1\) and will be equal to the segment \(cd\) up to the horizontal axis, taken on the scale \(M=\dot N_2M_1\).
Since for the limiting values
\[ oa=N_{1\infty}=\frac{B}{D};\qquad \angle N_2oa=\beta-\delta; \]
\[ ob=N_{10}=\frac{A}{C};\qquad \angle N_2ob=\alpha-\gamma, \]
equations (25) are completely justified. It follows from this that the straight line \(mn\) must be drawn from the origin perpendicular to the direction of the segment \(\dot N_2\).
If the product is found from equation (20), then it is necessary mentally to rotate the segment \(\dot N_2\) by \(90^\circ\), i.e., to draw the straight line \(mn\) along the direction of \(\dot N_2\) and to take the segments \(cd_1\) perpendicular to it (Fig. 13).
8. RATIO OF TWO PRODUCTS
When using pie charts it may be necessary to know the values of the ratios of two products, as, for example, when finding the efficiency coefficient
\[ \eta=\frac{V_2 I_2\cos\varphi_2}{V_1 I_1\cos\varphi_1}. \]
or the phase-shift angle
\[ \tan \varphi=\frac{VI\sin \varphi}{VI\cos \varphi};\quad \tan (I_1;I_2)=\frac{I_1 I_2 \sin (I_1;I_2)}{I_1 I_2 \cos (I_1;I_2)}. \]
Let us prove that the ratio of two products can be obtained in the form of a segment taken in a definite direction. Suppose that for a certain reduced diagram (Fig. 14) the first product \(P_1\) is represented by the segment \(cd_1\), perpendicular to the line \(m_1 n_1\), and the second product \(P_2\) by the segment \(cd_2\), perpendicular to \(m_2 n_2\), and that it is necessary to find the ratio \(\dfrac{P_2}{P_1}\).
Find the point \(k\) of intersection of the straight lines \(m_1 n_1\) and \(m_2 n_2\), and from an arbitrary point \(e\), lying on \(m_2 n_2\), draw the straight line \(ef\), parallel to the straight line \(m_1 n_1\). Joining the points \(k\) and \(c\) and extending this straight line to its intersection with the straight line \(ef\), we obtain the segment \(eg\), which will be proportional to the required ratio.
Fig. 14.
For the proof, draw the straight line \(eh\), perpendicular to the straight line \(kc\). From the similarity of triangles
\[ \Delta egh \sim \Delta ckd;\quad \Delta ehk \sim \Delta cd_2 k \]
we obtain
\[ \frac{eh}{cd_1}=\frac{eg}{kc};\quad \frac{cd_2}{eh}=\frac{kc}{ke}; \]
multiplying both proportions term by term, we find
\[ \frac{cd_2}{cd_1}=\frac{eg}{ke}. \]
If the scales of the products are equal to \(M_1\) and \(M_2\), i.e.
\[ P_1=M_1\cdot cd_1;\quad P_2=M_2\cdot cd_2, \]
then the segment \(eg\) will give the required ratio
\[ \frac{P_2}{P_1}=\frac{M_2\cdot cd_2}{M_1\cdot cd_1} =\frac{M_2}{M_1}\cdot \frac{1}{ke}\cdot eg \]
on the scale
\[ M_3=\frac{M_2}{M_1}\cdot \frac{1}{ke}. \tag{26} \]
If the segments of the products are taken at the angles \(\Theta_1\) and \(\Theta_2\) to the corresponding lines, then the scale of the ratios will be equal to
\[ M_3=\frac{1}{ke}\cdot \frac{M_2}{M_1}\cdot \frac{\sin \Theta_1}{\sin \Theta_2}. \tag{26a} \]
Taking the point \(e\) on the straight line \(m_1 n_2\) and drawing the straight line \(ef\) parallel to the direction \(m_2 n_2\), we obtain in the same way the ratio \(\dfrac{P_1}{P_2}\), the reciprocal of that obtained earlier.
These constructions cannot be carried out if the straight lines \(m_1n_1\) and \(m_2n_2\) are parallel to each other, or if the point of their intersection lies far outside the limits of the drawing.
II. EXAMPLES OF CONSTRUCTING PIE CHARTS
1. T-SHAPED EQUIVALENT CIRCUIT
As a first example of the application of the equations obtained above, let us consider the simple T-shaped equivalent circuit shown in Fig. 15. Let us take the resistances included in this circuit as equal to
\[ \dot Z_1=\dot Z_2=1+j2;\qquad \dot Z_0=-j20. \]
Let the network voltage be \(200\ \mathrm{V}\), and let the load be regarded as ohmic, i.e. having a phase-shift angle \(\varphi_{\mathrm{н}}=0\); \((\cos\varphi_{\mathrm{н}}=1)\).
Fig. 15.
Solving this circuit, we find in the usual way the equations relating all currents and voltages to the load resistance.
The total current is
\[ \dot I_1=\dot V_1\frac{(\dot Z_0+\dot Z_2)+\dot Z_{\mathrm{н}}}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
the load current
\[ \dot I_2=\dot V_1\frac{\dot Z_0}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
the current in the branch
\[ \dot I_0=\dot V_1\frac{\dot Z_2+\dot Z_{\mathrm{н}}}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
the voltage on the load
\[ \dot V_2=\dot V_1\dot Z_2\frac{\dot Z_0+\dot Z_{\mathrm{н}}}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
the voltage at the branching point
\[ \dot V_0=\dot V_1\dot Z_0\frac{\dot Z_2+\dot Z_{\mathrm{н}}}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
and the voltage drops in the resistances \(\dot Z_1\) and \(\dot Z_2\), before and after the branching,
\[ \dot V_{\mathrm{н}1}=\dot V_1\dot Z_1 \frac{(\dot Z_0+\dot Z_2)+\dot Z_{\mathrm{н}}}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \qquad \dot V_{\mathrm{н}2}=\dot V_1\dot Z_2 \frac{\dot Z_0}{\dot C+\dot D\dot Z_{\mathrm{н}}}; \]
where in all these equations the coefficients \(\dot C\) and \(\dot D\) of the denominators have the expressions
\[ \dot C=\dot Z_1\dot Z_2+\dot Z_1\dot Z_0+\dot Z_2\dot Z_0; \qquad \dot D=\dot Z_1+\dot Z_0. \]
From the equations obtained we see that, for all currents and voltages, the denominators have proved to be the same, i.e. all the diagrams will have one and the same angle \(\rho\) between the chord and the tangent.
We compute the values of the coefficients \(\dot C\) and \(\dot D\):
\[ \dot C=(1+j2)^2+2(1+j2)(-j20)=77-j36=85\angle -25.0^\circ, \]
\[ \dot D=(1+j2)+(-j20)=1-j18=18.02\angle -86.8^\circ, \]
or
\[ C=85;\quad D=18.02;\quad \gamma=-25.0^\circ;\quad \delta=-86.8^\circ. \]
Fig. 16.
For the total network current \(I_1\), from the equations we have
\[ \dot A=\dot V_1(\dot Z_0+\dot Z_2)=200(1+j2-j20)=200-j3600=3604\angle -86.8^\circ \]
\[ \dot B=\dot V_1=200=200\angle 0 \]
or
\[ A=3604;\quad B=200;\quad \alpha=-86.8^\circ;\quad \beta=0. \]
Since the variable quantity is the load resistance \(\dot K=\dot Z_{\mathrm{n}}\), the limiting values will correspond to no-load and short-...
the closing of the circuit. Denoting the corresponding angles of rotation of the vectors with respect to the voltage vector $\dot V_1$ by $\varphi_{xx}$ and $\varphi_{kz}$ and arranging the latter vector along the vertical axis, we have from equations (2) and (2a)
\[ \dot I_{\sim}=\dot I_{xx}=\frac{B}{D}=\frac{200}{18.02}=11.1\ A;\qquad \varphi_{xx}=\beta-\delta=86.8^\circ; \]
\[ \dot I_0=\dot I_{kz}=\frac{A}{C}=\frac{3604}{85.0}=42.4\ A;\qquad \varphi_{kz}=\alpha-\gamma=-61.8^\circ; \]
which indicates that the current $I_{xx}$ leads the voltage, while the current $I_{kz}$ lags. From equation (8a) we determine the chord length (on the current scale)
\[ H^2=11.1^2+42.4^2-2\cdot 11.1\cdot 42.4\cdot \cos[86.8^\circ-(-61.8^\circ)]=2724 \]
or
\[ H=52.2\ A. \]
Carrying out similar calculations for the remaining currents and voltages, we collect all the results obtained in a table:
| $N$ | $A$ | $\alpha$ | $B$ | $\beta$ | $N_{xx}$ | $\varphi_{xx}$ | $N_{kz}$ | $\varphi_{kz}$ | $H$ |
|---|---|---|---|---|---|---|---|---|---|
| $I_1$ | 3604 | $-86.8$ | 200 | 0 | 11.1 | 86.8 | 42.4 | $-61.8$ | 52.2 |
| $I_2$ | 4000 | $-90.0$ | 0 | — | 0 | — | 47.0 | $-65.0$ | 47.0 |
| $I_0$ | 448 | 63.5 | 200 | 0 | 11.1 | 86.8 | 5.27 | 88.5 | 5.85 |
| $V_0$ | 8960 | $-26.5$ | 4000 | $-90.0$ | 222 | $-3.2$ | 105.4 | $-1.5$ | 117.0 |
| $V_2$ | 0 | — | 4000 | $-90.0$ | 222 | $-3.2$ | 0 | — | 222.0 |
| $V_{n1}$ | 8080 | $-23.3$ | 448 | 63.5 | 24.8 | 150.3 | 95.0 | 1.7 | 116.8 |
| $V_{n2}$ | 8960 | $-26.5$ | 0 | — | 0 | — | 105.4 | $-1.5$ | 105.4 |
Choosing the scale of currents $M_I=2\ \frac{A}{cm}$ and of voltages $M_v=10\ \frac{V}{cm}$, we plot all the obtained values $N_{xx}$ and $N_{kz}$ on the drawing in the form of segments arranged at the corresponding angles $\varphi_{xx}$ and $\varphi_{kz}$ to the vertical axis. Such a construction is shown in Fig. 16.
To draw the arcs of the circles, we find from equations (7) the values of the angle $\rho$, taking into account that $\Psi=\varphi_n=0$ (ohmic resistance of the load)
\[ \rho=\gamma-\delta-\Psi=(-25.0)-(-86.8)-0=61.8^\circ. \]
In the diagrams constructed, the letters $a$ and $b$ mark the points corresponding to no-load operation and short circuit, and the points $c$ mark the positions of the ends of the vectors for the load resistance $Z_n=10\ cm$. According to the relations derived, the triangles $abc$ must be similar.
The construction of the diagram in Fig. 16 is very difficult to use, since, in order to find the corresponding points $c$, it is necessary to construct a series of similar triangles or equal angles. It is clear from the figure that such a construction will be very difficult when the number of values is large.
For simplification, we rotate all the circles so that the chords coincide in direction, and the points \(a\) coincide into one. It is convenient to perform the construction by intersections according to the values \(N_{xx}\), \(N_{kz}\), and \(H\), and, when the angle differences \(\varphi_{xx}-\varphi_{kz}\) are small or close to \(180^\circ\), by transferring from the normal diagram.
Such a combined diagram is shown in Fig. 17, in which the beginnings of the vectors are denoted by the same letters as the arcs of the circles. The points \(c\), constructed for a load resistance of \(10\) ohms, now lie on one straight line, so that to find them it is sufficient to draw this line through one of them and the point \(a\).
For still greater simplification, we change the scales so that the circles of all diagrams merge into one. Taking the chord length of such a diagram equal to 15 centimeters, we find the new scale for the total current
\[ M_1=\frac{H_1}{15}=\frac{52.2}{15}=3.48\ \frac{A}{cm}. \]
Fig. 17.
The lengths of the no-load and short-circuit vectors corresponding to this scale are
\[ \frac{I_{xx}}{M_1}=\frac{11.1}{3.48}=3.2\ \text{cm};\quad \frac{I_{kz}}{M_1}=\frac{42.4}{3.48}=12.2\ \text{cm}, \]
while the angle between them is
\[ \varphi_{xx}-\varphi_{kz}=148.6^\circ. \]
Having performed similar calculations for the remaining currents and voltages, we enter the obtained results in a table.
Using these data, we construct, by intersections, the new diagram shown in Fig. 18. It is easy to see that the beginnings of the vectors for \(I_2\) and \(V_{n2}\) coincide with point \(a\) of the diagram, and for the vector \(V_2\)—with point \(b\). The beginnings of the vectors \(I_1\) and
| $N$ | $M_N$ | $\dfrac{N_{xx}}{M_N}$ | $\dfrac{N_{kz}}{M_N}$ | $\varphi_{xx} - \varphi_{kz}$ |
|---|---|---|---|---|
| $I_1$ | 3.48 | 3.2 | 12.2 | 148.6 |
| $I_2$ | 3.13 | 0 | 15.0 | — |
| $I_0$ | 0.39 | 28.5 | 13.5 | -1.7 |
| $V_0$ | 7.80 | 28.5 | 13.5 | -1.7 |
| $V_2$ | 14.80 | 15.0 | 0 | — |
| $V_{n1}$ | 7.78 | 3.2 | 12.2 | 148.6 |
| $V_{n2}$ | 7.03 | 0 | 15.0 | — |
$V_{n1}$ will be at a single point lying below the chord $ab$, while for the vectors $I_0$ and $V_0$ it will be above the chord (the angle $\varphi_{xx} - \varphi_{kz}$ is negative).
Choosing on the chord a segment $ad$ equal to 5 centimeters, we draw from point $d$ a straight line parallel to the tangent $aa_1$. The segment $de$ of this straight line, obtained from its intersection with the straight line $ac$, will give the load resistance on the scale determined by equation (10)
\[ M_z = \frac{C}{D \cdot ad} = \frac{85}{18.02 \cdot 5} = 0.945 \ \frac{\mathrm{ohm}}{\mathrm{cm}} . \]
For a resistance $Z_n = 10 \ \mathrm{ohm}$ this segment will be equal to $de = 10 : 0.945 = 10.6 \ \mathrm{cm}$. Laying it off and drawing the straight line $ae$, we find the point $c$, and consequently all the currents and voltages corresponding to this resistance (constructed in Fig. 18).
Fig. 18.
Let us find the powers of the separate sections of the circuit; for convenience, we shall put all the constructions in a separate Fig. 19. The total power supplied from the network to the given circuit will be equal to
\[ P_2 = V_1 I_1 \cos \varphi_1 \]
and will be represented by the segment \(cd_1\), drawn parallel to the vector \(\dot V_1\) up to the straight line drawn from the origin of the current vectors \(\dot I_1\) perpendicular to the vector \(\dot V_1\). The scale of this power will be, by equation (25), equal to
\[ M_1 = 3{,}48,\ 200 = 696 \frac{W}{cm}. \]
The power consumed by the load will be equal to
\[ P_2 = V_2 I_2 \cos \varphi_2. \]
Fig. 19.
Since in this case \(V_2 = I_2 Z_h\), this power will be represented by the segment \(cd_2\) from the point \(c\) to the chord. If this segment is taken perpendicular to the chord, then the scale is obtained, by equation (24), as
\[ M_2 = 15{,}14{,}8. 3{,}13 \cdot \frac{1}{0{,}88} = 791 \frac{W}{cm}; \]
but if the direction is chosen parallel to the tangent \(aa_1\), then by equation (24a)
\[ M_{21} = 15.14{,}8.3{,}13 \cdot \frac{1}{0{,}88} = 696 \frac{w}{cm} \]
The latter scale is more convenient, since it is the same as the scale for \(P_1\). The power lost in the resistance \(Z_2\) will be equal to
\[ P_3 = I_2^2 \cdot R_2. \]
Since the equation for the current \(I_2\) has coefficient \(B=0\), this power will be determined by the segment \(cd_3\), taken from point \(c\) to the tangent \(aa^1\) and drawn perpendicular to it. We find the scale from equation (22)
\[ M_3=1\cdot \frac{3.13^2\cdot 15}{0.88}=167\ \frac{W}{cm}. \]
The power at the branching point
\[ P_0=P_2+P_3=V_0' I_2' + V_0'' I_2'' \]
will be determined by the segment \(cd_0\) to the corresponding straight line found from the general equations. For the present case we have
\[ A_1=4000;\quad \alpha_1=-90^\circ;\quad B_1=0; \]
\[ A_2=8960;\quad \alpha_2=-26.5^\circ;\quad B_2=4000;\quad \beta_2=-90^\circ . \]
From equations (15) we obtain
\[ R=\frac{4000\cdot 4000}{85\cdot 18.02}\cdot \cos[(-90^\circ)-(-90^\circ)-0]=10430, \]
\[ S=0, \]
\[ T=\frac{4000\cdot 8960}{85^2}\cdot \cos[(-90^\circ)-(-26.5^\circ)]=2220. \]
From equation (16) we obtain
\[ \tan\eta=\frac{2220\cdot 0.88}{10430-2220\cdot 0.472}=0.208;\quad \eta=11.7^\circ . \]
If the direction of the segments \(cd_0\) is taken parallel to the tangent \(aa^1\), then from equations (17) and (19) we find the scale
\[ M_0'=\frac{2220\cdot \sin(61.6^\circ+11.7^\circ)} {15\cdot \sin 11.7^\circ}=696\ \frac{W}{cm}; \]
i.e., again the same as the scale for \(P_1\).
Since from equation (18) \(F=S:M_0^1=0\), the power line will pass through point \(a\) at an angle of \(11.7^\circ\) to the chord \(ab\), being shifted from it clockwise.
The power lost in resistance \(Z_1\) will be equal to
\[ P_4=I_1^2R=I_1R_1\cdot I_1. \]
Since for the present case \(R_1=1\) ohm, and
\[ A_1=A_2=3604;\quad \alpha_1=\alpha_2=-86.8^\circ; \]
\[ B_1=B_2=200;\quad \beta_1=\beta_2=0, \]
then from equations (15) we obtain
\[ R=2\cdot \frac{3604\cdot 200}{85\cdot 18.02}\cdot \cos(-86.8^\circ)=52.8, \]
\[ S=\frac{200\cdot 200}{18.02^2}=123, \]
\[ T=\frac{3604\cdot 3604}{85^2}=1800. \]
From equation (16) we have
\[ \tan \eta = \frac{(1800-123)\sin 61.8^\circ} {52.8-(1800-123)\cos 61.8^\circ} = -1.725;\quad \eta=-60^\circ, \]
i.e., the power line will be shifted relative to the chord by an angle of \(60^\circ\) counterclockwise. From equation (17) we find the scale
\[ M_4=\frac{1800-123}{15\cdot \sin 60^\circ}=129\,\frac{W}{cm}, \]
and from equation (18), the distance from point \(a\) to the power line \(P_4\)
\[ F=\frac{123}{129}=0.95\,cm. \]
Laying off this distance and drawing a straight line, we obtain the power \(P_4\) in the form of the segment \(cd_4\), perpendicular to this straight line.^1)
To find the efficiency of the entire circuit, let us find point \(k\), the intersection of the power lines \(P_1\) and \(P_2\) (i.e., of the horizontal axis and the extension of chord \(ab\)); and on the power line \(P_2\) take point \(e\) at a distance of \(5\,cm\) from point \(k\). Drawing the straight line \(ef\) parallel to the power line \(P_1\) (i.e., to the horizontal axis), we find its intersection with the straight line drawn through points \(k\) and \(c\). The segment \(eg\) thus obtained will give the value of the efficiency \(\eta=P_2:P_1\) on a scale equal, according to equation (26), to
\[ M_\eta=\frac{1}{5}\cdot \frac{791}{696}\cdot 100=22.7\,\frac{\%}{cm}. \]
Having the diagram constructed in Figs. 18 and 19 (usually drawn in one figure), it is very easy, from any one of the specified values of one of the quantities, to find all the values of the remaining quantities of the given circuit.
2. REVERSAL OF PHASES OF A THREE-PHASE CIRCUIT
As a second example, let us consider the three-phase circuit shown in Fig. 20, which has in each of the phases an identical inductive coil. An additional capacitance is inserted in one of the phases. Under certain conditions, a change in the order of alternation of the phase currents and voltages on the coils is obtained, or what is called phase reversal.
Let us construct a pie diagram for this circuit, taking the capacitance resistance as the variable quantity. We shall take the phase voltages of the current source to be equal to \(120\,V\), and the coils to have an inductance of \(0.14\,H\) and an ohmic resistance of \(20\,ohm\) each. The frequency of the circuit is \(50\) periods.
The phase impedances are equal to
\[ \dot Z_2=\dot Z_3=\dot Z=R+jL\omega= \]
\[ =20+j\,0.14\cdot 314=20+j44; \]
\[ \dot Z_1=\dot Z+\dot Z_k, \]
Fig. 20.
^1) This straight line will be a semipolar of the circle. For the graphical construction, see below.
where $\dot Z_k$ is the variable resistance of the capacitor (capacitance).
The symmetrical resistances of the circuit phases are equal to
\[ \dot Z_a=\dot Z+\frac{1}{3}\dot Z_k, \]
\[ \dot Z_b=\dot Z_e=-\frac{1}{3}\dot Z_k. \]
Since there is no neutral wire, the component of the current of the first phase for the positive sequence will have the expression
\[ \dot I_I=\dot V\,\frac{\dot Z_a}{\dot Z_a^2-\dot Z_b\dot Z_e} =\dot V\,\frac{3\dot Z+\dot Z_k}{3\dot Z^2+2\dot Z\dot Z_k}, \]
and the component of the negative sequence
\[ \dot I_{II}=\dot V\,\frac{-\dot Z_e}{\dot Z_a^2-\dot Z_b\dot Z_e} =\dot V\,\frac{-\dot Z_k}{3\dot Z^2+2\dot Z\dot Z_k}. \]
For both components the denominator has identical expressions, so their diagrams will have the same angles $\rho$. The coefficients of the denominator will be equal to
\[ \dot C=3\dot Z^2=3(20+j44)^2=-4608+j5280=7000\angle 131^\circ, \]
\[ \dot D=2\dot Z=2(20+j44)=40+j88=96.5\angle 65.5^\circ \]
or
\[ C=7000;\quad D=96.5;\quad \gamma=131^\circ;\quad \delta=65.5^\circ. \]
The coefficients of the numerator of the positive-sequence current are equal to
\[ \dot A=\dot V\cdot 3\dot Z=120\cdot 3(20+j44)=7200+j15800=17400\angle 65.5^\circ, \]
\[ \dot B=\dot V=120=120\angle 0 \]
or
\[ A=17400;\quad B=120;\quad \alpha=65.5^\circ;\quad \beta=0, \]
and for the negative sequence
\[ A'=0;\quad \dot B=-\dot V=-200=200\angle 180^\circ;\quad B=200;\quad \beta=180^\circ. \]
The current components for the limiting values of $\dot Z_k$ for the positive sequence will be
\[ \dot I_0=\frac{A}{C}=2.49A;\quad \dot I\infty=\frac{B}{D}=1.245A;\quad \varphi_0=\alpha-\gamma=-65.5^\circ;\quad \varphi\infty=\beta-\delta=65.5^\circ \]
and for the negative sequence
\[ \dot I'_0=\frac{A'}{C}=0;\quad \dot I'\infty=\frac{B'}{D}=1.245A;\quad \varphi'\infty=\beta'-\delta=180-65.5=114.5^\circ. \]
Choosing the current scale $M_I=0.5\ \dfrac{A}{\text{cm}}$, we plot these currents in Fig. 21, as well as the corresponding ones for the second and third phases (they will be rotated by angles of $120^\circ$ and $240^\circ$).
To find the summary diagrams it is sufficient to add the corresponding vectors for the limiting values and, on their differences, construct the circles of the diagrams. Performing the addition, we find the currents in the phases and the corresponding chords (on the current scale).
| Phase | \(I_0\) | \(I_\infty\) | \(H\) | \(\varphi_\infty-\varphi_0\) |
|---|---|---|---|---|
| First . . | 2.49 | 0 | 2.49 | — |
| Second . . | 2.49 | 2.16 | 1.245 | \(+30^\circ\) |
| Third . . | 2.49 | 2.16 | 1.245 | \(-30^\circ\) |
The angle \(\rho\) between the chord and the tangent is determined by the equation
\[ \rho=\gamma-\delta-\psi=131-65.5+90=155.5^\circ, \]
since for a capacitance
\[ \psi=\varphi=-90^\circ . \]
Fig. 21.
Without drawing the arcs of the circles, let us transfer the vectors directly onto the combined diagram constructed in Fig. 22. Leaving the same scale for the current of the first phase \(\left(M_1=0.5\ \frac{A}{cm}\right)\), we lay off the segment \(ab\), equal to
\[ ab=\frac{H}{M_1}=\frac{2.49}{0.5}=4.98\ \text{cm} \]
and on it, as on a chord, construct a circle with the angle obtained above.
The scale for the currents \(I_2\) and \(I_3\) will be
\[ M'_I=\frac{H'}{ab}=\frac{1.245}{4.98}=0.25\ \frac{A}{cm}; \]
the lengths of the segments of the limiting values of the currents will be
\[ \frac{I_0}{M'_I}=\frac{2.49}{0.25}=9.96\ cm;\qquad \frac{I_{\sim\infty}}{M'_I}=\frac{2.16}{0.25}=8.64\ cm. \]
Laying off these segments from the corresponding points of the chord \(ab\), we obtain the points \(O_2\) and \(O_3\), the beginnings of the vectors of the currents of the second and third phases. The beginning of the vectors of the current of the first phase coincides with the point \(a\) of the chord.
Comparing the diagrams (Figs. 21 and 22), it is easy to see that, when the chords are superposed, the vectors of the second and third phases turn out to be rotated by \(180^\circ\)
Fig. 22.
more than the vectors of the first phase. If we now reverse their direction, regarding the points \(O_2\) and \(O_3\) as their ends rather than their beginnings, then the mutual arrangement of the current vectors of all three phases in Fig. 22 will correspond exactly to the actual one.
Thus, for some point \(c\) of the diagram, the currents of the three phases will be represented by the segments \(ac\), \(cO_2\), and \(cO_3\), the first of which has a scale twice as large as that of the other two.
Since the coil resistances are constant, the voltages across them will be
\[ V=Iz=I\sqrt{20^2+44^2}=I\cdot 48.25; \]
it follows from this that they will be determined by the same segments as the currents, with the scale for the first phase equal to
\[ M_V=M_1\cdot Z=0.5\cdot 48.25=24.125\ \frac{V}{cm} \]
and for the other two
\[ M'_v=M'_1 Z=0.25\cdot 48.25=12.06\ \frac{V}{cm}. \]
The relative positions of the voltage vectors will also be correct, but with respect to the current vectors they are in fact rotated through the angles
\[ \varphi=\operatorname{arctg}\frac{44}{20}=65.5^\circ. \]
The voltage across the capacitance, as across a variable resistance, will be determined by the segment from point \(b\) of the chord to point \(c\) of the circumference. Since this voltage has the equation
\[ \dot V_k=\dot I_1 \dot Z=\dot V\frac{3\dot Z\dot Z_k}{3\dot Z+2\dot Z_k}, \]
which, for \(Z_k=\infty\), gives the value \(V_k=\frac{3}{2}V=180\,v\), the scale is obtained from the ratio
\[ M_{vk}=\frac{V_{k\sim}}{ab}=\frac{180}{4.98}=36.19\ \frac{V}{cm}. \]
Let us find the straight line for obtaining the values of the variable capacitance. Since the capacitance is equal to \(C_k=\dfrac{1}{z_k\omega}\), it is necessary to draw from an arbitrary point \(f\) of the chord \(ab\) the straight line \(ff'\), parallel to the tangent \(bb'\). Having chosen the capacitance scale \(M_c=10\ \dfrac{\mu F}{cm}\), we find the required distance from point \(b\) to point \(f\) from equation (10a)
\[ bf=\frac{D}{M_c\cdot C\cdot \omega}. \]
From the diagram obtained it is seen that the maximum values of the voltages on the coils and on the capacitor are obtained at different capacitances. Performing the constructions, we find
| Maximum for | Point of the diagram | \(U_{max}\) | \(I_{max}\) | Corresponding capacitance, \(\mu F\) |
|---|---|---|---|---|
| 1st phase . . | \(C_1\) | 289 | 6.9 | 48 |
| 2nd phase . . | \(C_2\) | 253 | 5.25 | 45 |
| 3rd phase . . | \(C_3\) | 121 | 2.5 | 170 |
| Capacitance . . | \(C_4\) | 434 | — | 40 |
On the segment \(bc_1c_4\) of the diagram for the currents and coil voltages the correct phase sequence \((1—2—3)\) is preserved; on the segment \(c_4a\), however, the sequence changes to the reverse \((3—2—1)\). This is clearly seen from the vectors constructed with solid lines for the first case, and with dashed lines
for the second. Complete overturning, in which the currents and voltages of all phases will be equal, is determined by the point \(C_0\) of a circle with infinitely large radius, which is possible only in the absence of ohmic resistance.
In conclusion, let us note that the voltage of the first phase, taking both the coil and the capacitance together, is determined by the segment \(CO_1\). The point \(O_1\) is obtained from the condition that the distances \(O_1O_2\), \(O_2O_3\), and \(O_3O_1\) be equal and equal to the line voltages of the network. The scale of all these voltages will be the same as the scale for the vectors of the second and third phases. The maximum voltage is obtained equal to 238 volts at a capacitance of \(32\mu F\) (point \(C_5\)).
III. CONSTRUCTION OF DIAGRAMS FROM EXPERIMENTAL DATA
1. DETERMINATION OF THE ANGLE \(\rho\) WHEN THE LOAD RESISTANCE IS ACCESSIBLE
If some circuit has been constructed in practice, but the resistances included in it are unknown, and perhaps even the very scheme of its connection is unknown, then a circle diagram can be constructed on the basis of measurements made for this circuit. In doing so it is necessary only to be certain that the same angles \(\rho\) will be present for all quantities of the circuit.
If the variable resistance playing the role of the load is accessible for the connection of measuring instruments (an ammeter and a voltmeter), then it is best to begin the construction of the diagram with the voltage and current of this resistance. To construct this diagram it is sufficient to measure the voltage with the resistance open (no-load), the current when the terminals of this resistance are short-circuited (short circuit), and the current and voltage for one particular value of the connected resistance (load). All these measurements must be made at the network voltage (or at the current supplied from the network) for which it is intended to construct the diagram. If the measurements were made under other conditions, it is necessary to recalculate the results in proportion to the change in the quantity that is to be constant (see the example).
The load current on the combined diagram has the origin of its vectors coincident with the point \(a\) of the chord, and the load voltage—with the point \(b\). The chord of the diagram, in the corresponding scales, will be equal both to the no-load voltage \(V_{xx}\) and to the short-circuit current \(I_{kz}\), i.e.
\[ ab=\frac{V_{xx}}{M_v}=\frac{I_{kz}}{M_I}. \]
The load current and voltage in the same scales will be represented by the segments \(ac\) and \(bc\) (Fig. 8). From the triangle \(abc\) we find
\[ (ab)^2=(ac)^2+(bc)^2+2(ac)(cb)\cos\rho \]
or
\[ \left(\frac{V_{xx}}{M_v}\right)^2 = \left(\frac{V_n}{M_v}\right)^2 + \left(\frac{I_n}{M_I}\right)^2 + 2\frac{V_n I_n}{M_v M_I}\cos\rho. \]
Eliminating the scales from the two equations obtained, we find
\[ \cos\rho = \frac{1}{2} \left( \frac{V_{xx}\cdot I_{kz}}{V_n\cdot I_n} - \frac{V_{xx}I_n}{V_n I_{kz}} - \frac{V_n\cdot I_{kz}}{V_{xx}\cdot I_n} \right) \]
This equation makes it possible to compute the value of the angle \(\rho\), and the best accuracy is obtained in the case when \(\cos\rho\) is not very close to uni-
lower, while the segments \(ac\) and \(bc\) are close in size. For greater accuracy, measurements may be made not for one, but for two loads of different resistances.
If the measurements have been made for a load having phase-shift angle \(\varphi_n\), while the diagram must be constructed for a load with angle \(\varphi'\), then, taking equation (7) into account, we can find the new angle
\[ \rho'=\rho+\varphi_n-\varphi'_n . \tag{28} \]
This relation makes it possible to carry out measurements not at the load that is to appear on the diagram; consequently, one can always select the value of the computed \(\cos \rho\) to be different from unity. In case of doubt about the sign of the angle \(\rho\), this makes it possible, after making measurements for two different \(\varphi_n\), to find it from the relation of the angles obtained.
Instead of computing the angle, the diagram may be constructed by notches on the triangle \(abc\).
Having constructed the diagram, one may immediately draw on it the straight line for the resistances or conductances of the load, as was indicated in Figure 8. The scale is determined from the relation of the triangles
\[ \frac{de}{ad}=\frac{bc}{ac}; \quad \frac{Z_n}{M_z}:ad=\frac{V_n}{M_v}:\frac{I_n}{M_I} \]
or, since \(I_n Z_n=V_n\), the resistance scale is equal to
\[ M_z=\frac{M_v}{M_I ad}; \tag{29} \]
in a similar way, the conductance scale is equal to
\[ M_I=\frac{M_I}{M_v \cdot bf}. \tag{29a} \]
The scales of the load powers may be, depending on the chosen direction of the segments, determined from one of equations (24) or (24a).
2. THE LOAD RESISTANCE IS INACCESSIBLE
If the current and voltage of the variable resistance (load), for one reason or another, are inaccessible for measurement (for example, in asynchronous motors), then the construction of diagrams has to be begun with one of the other quantities. In this case it is necessary, from the instrument readings, to compute the phase-shift angles of one of the quantities of the circuit at no-load, under load, and at short circuit, with respect to a constant vector. When constructing a diagram for a constant mains voltage, it is best to choose as this quantity the current entering the given circuit from the mains. With a constant value of the incoming current (which happens very rarely), it is convenient to choose for the construction the voltage at the terminals of the circuit. The angles may be measured either by wattmeters or by any other method.
Fig. 23.
Having determined the current values \(I_{xx}\), \(I_n\), and \(I_{kz}\), and the corresponding phase-shift angles \(\varphi_{xx}\), \(\varphi_n\), and \(\varphi_{kz}\), we plot them (Fig. 23) and draw a circle through the three points \(a\), \(b\), and \(c\) obtained. In this construction, with respect to the angle \(\rho\), everything that was said above for the current and voltage of the load may be stated in the same way. Just as there, one may construct the circle for a single load and then change the angle depending on the change in the load phase-shift angle.
The analytical determination of the angle \(\rho\), the radius, and the center of the circle of the diagram is very complicated and is usually unnecessary, which is why it is not given here.
For some circuits the value of the angle \(\rho\) can be found without measuring the current and angle under load, but only from the no-load and short-circuit tests. Thus, for a simple \(T\)-shaped equivalent circuit, the angle is found from an additional reverse short-circuit test by the equation
\[ \rho=\varphi_{okz}-\varphi_n, \]
as is evident from the ratio of the coefficients \(\dot C\) and \(\dot D\) and the value of the current \(I\) at \(\dot Z_n=0\) for the example given above. If the circuit is symmetric, i.e. \(\dot Z_1=\dot Z_2\), then the angle \(\varphi_{okz}\) is equal to the angle for a normal short circuit. In this case a reverse connection of the circuit need not be performed at all.
3. CONSTRUCTION OF DIAGRAMS FOR THE REMAINING CURRENTS AND VOLTAGES
To construct the combined diagrams of the remaining currents and voltages, it is sufficient to know, for each of them, three values corresponding to no-load operation of the circuit, short circuit, and the load for which the intermediate point \(C\) was found on the diagram. The problem reduces to finding on the drawing a point such that its distances to the points \(a\), \(b\), and \(c\) of the diagram are proportional to the available values \(N_{xx}\), \(N_{kz}\), and \(N_n\):
\[ oa:ob:oc=N_{xx}:N_{kz}:N_n. \]
Finding the point \(O\) (the origin of the vectors) by calculation proves to be very complicated, and the problem is solved much more simply by graphical means.
It is known from geometry that the locus of points for which the ratio of the distances to two fixed points is constant is a circle with its center on the straight line joining these points. Denoting the ratio
\[ \frac{oa}{ob}=\frac{N_{xx}}{N_{kz}}=\alpha, \]
Fig. 24.
we find, on side \(ab\) of triangle \(abc\) (Fig. 24) and on its extension, points \(m\) and \(m'\) satisfying the condition
\[ \frac{am}{mb}=\frac{am'}{m'b}=\alpha, \]
since in this case
\[ am+mb=ab;\qquad am'+ab=m'b, \]
then, solving the equations, we obtain the required distances
\[ am=ab\,\frac{\alpha}{1+\alpha};\qquad am'=ab\,\frac{\alpha}{1-\alpha}. \tag{30} \]
If in the last equation a negative value is obtained for the distance \(am'\), this will mean that the point \(m'\) lies not beyond point \(a\), but beyond point \(b\), i.e. on the other side of the segment \(ab\). When the ratio
\[ \alpha=\frac{N_{xx}}{N_{k3}} \]
is equal to unity, the distance \(am'\) will be infinitely large.
In the same way we find the points \(n, n', p\), and \(p'\) for the sides \(bc\) and \(ca\) of the triangle, performing the calculations by the equations
\[ bn=cb\,\frac{\beta}{1+\beta};\qquad bn'=cb\,\frac{\beta}{1-\beta}; \tag{30a} \]
\[ cp=ac\,\frac{\gamma}{1+\gamma};\qquad cp'=ac\,\frac{\gamma}{1-\gamma}; \tag{30b} \]
where the ratios are
\[ \beta=\frac{N_{k3}}{N_{n}};\qquad \gamma=\frac{N_{n}}{N_{xx}}. \]
If now, on the obtained segments \(mm'\), \(nn'\), and \(pp'\), as on diameters, circles are constructed, then their point of intersection will be the required origin of the vectors for the given quantity. From the drawing we see that there will be two such points. If there is any doubt as to which of the points \(O_1\) or \(O_2\) will correspond to the origin of the vectors, it is necessary to carry out the additional construction of one more circle for another load, which will immediately determine the actual origin of the vectors by passing through it.
Having found the point of origin of the vectors, it is easy also to determine their scale from the ratio of the segments and the values themselves
\[ M_N=\frac{N_{xx}}{oa}=\frac{N_{k3}}{ob}=\frac{N_{n}}{ac}. \]
Finding the points \(m, m', n\), etc., can be done not by computation, but by means of a graphical construction. For this, through point \(b\) we draw a straight line at an arbitrary angle (Fig. 25) and on it, at any scale, lay off the segment \(bd\), corresponding to the value \(N_{k3}\), and the segments \(df\) and \(de\), corresponding to the value \(N_{xx}\). Drawing the line \(dm\), parallel to \(ea\), and the line \(dm'\), parallel to \(fa\), we find, at their intersections with the line \(ab\) and its extension, the required points \(m\) and \(m'\). A similar construction must also be carried out for the second side, pri—
Fig. 25.
than which one may use the same straight line \(be\) and the segment \(bd\), replacing only the segments \(de\) and \(df\) by the corresponding values \(N_n\).
To reduce the number of constructions, it should be remembered that the origins of the vectors for a certain constant resistance and the voltage at the terminals of that resistance coincide with one another.
4. POWER LINES
To construct the power line of some section of a circuit, it is necessary to know the value of this power for three different loads. It is best to choose these loads as the same ones for which the diagram was constructed, i.e., no load, short circuit, and one intermediate load.
Suppose that the power line is the straight line \(mn\) (Fig. 26). Drawing the straight line \(ak\), parallel to \(mn\), and extending the side \(cb\) of triangle \(abc\) until it intersects it at point \(k\), we obtain two similar triangles \(ckf\) and \(bkh\), from which we find the relations
\[ \frac{ck}{cf}=\frac{bk}{bh}; \qquad \frac{ck-bk}{cf-bh}=\frac{bk}{bh} \]
or
\[ \frac{cb}{cd-bg}=\frac{bk}{bg-am}. \]
From this relation we find the segment \(bk\) from point \(b\) to the point \(k\) of intersection of the straight lines:
Fig. 26.
\[ bk=cb\,\frac{bg-am}{cd-bg}; \]
since the segments \(am\), \(cd\), and \(bg\) must, on some scale, give the values of the powers \(P_{xx}\), \(P_n\), and \(P_{kz}\), this equation takes the form
\[ bk=cb\,\frac{P_{kz}-P_{xx}}{P_n-P_{kz}}. \tag{31} \]
From these relations follows the following rule for constructing the power line. Knowing the three powers indicated above and measuring the segment \(cb\), we calculate the value of the segment \(bk\). Laying it off on the extension of the side \(cb\), we find the point \(k\). Through this point and the point \(a\) of the diagram we draw the straight line \(ak\). If the value obtained for \(bk\) is negative, then the segment should be laid off from point \(b\) in the direction of point \(c\).
Having chosen the direction for the power segments (perpendicular to the straight line \(mn\), parallel to the tangent, or any other), we find the lengths of the segments \(cf\) and \(bh\), and from them compute the power scale
\[ M_p=\frac{P_n-P_{xx}}{cf}=\frac{P_{kz}-P_{xx}}{bh}. \tag{32} \]
On this scale, from point \(a\) in the chosen direction we lay off the segment \(am\), corresponding to the no-load power, and through point \(m\) draw the required power line \(mn\), parallel to the straight line \(ak\).
If in the construction the point \(k\) falls outside the limits of the drawing, then the same construction may be carried out not for the side \(cb\), but for the
any other side of the triangle. Equation (32) remains of the same form, but the powers entering into it change cyclically.
Instead of calculation by equation (32), the point \(k\) can be found by a construction analogous to the construction carried out in Fig. 25 for the point \(m'\).
To construct the line of powers from experiment it is sufficient to determine only one value of the power (for example, for no load), while the others can be found from the following considerations.
If, for no load, the measurements gave the current \(I_{xx}\), the voltage \(V_{xx}\), and the power \(P_{xx}\), then the phase-displacement angle is determined from the relation
\[ \cos \varphi_{xx}=\frac{P_{xx}}{V_{xx}I_{xx}}. \]
Suppose that, on the superposed diagram, the current vector \(I_{xx}\) and the voltage \(V_{xx}\) turned out to be located at an angle \(\alpha_{xx}\). From the relation of these
Fig. 27.
angles we conclude that the difference of the rotations of the diagrams when they were superposed was equal to \(\beta=\varphi_{xx}-\alpha_{xx}\). Measuring now, for any point of the diagram, the angle between current and voltage, we find the true phase-displacement angle from the equation \(\varphi=\alpha+\beta\). If there is doubt about the signs of the angles, it is better to make such a determination for two powers, and from them then determine the third power. From the angle found, the power is calculated by the usual equation \(P=VI\cos\varphi\).
In a number of cases the construction of lines of powers is greatly simplified if one takes into account the features indicated above for the particular cases of products of vectors. When one of them is constant, the line of powers passes perpendicular to it through the origin of the second vectors (total wattless power), or parallel to it (reactive power). Joule losses (\(I^2R\) or \(V^2g\)), under the condition that the vector is reduced to zero for one of the loads, are determined by the tangent to the circle of the diagram.
If the vector does not take a value equal to zero, then the line of Joule losses coincides with the semipolar of the circle. We shall prove this and find the scale of the resulting segments.
If the pole \(O\) of the circle of the diagram (the origin of the vectors) lies outside it (Fig. 27), then the polar is the straight line drawn through
the points \(A'\) and \(A''\) of tangency of the lines \(OA'\) and \(OA''\). Drawing the line \(OF\) through the pole and the center of the circle, we divide the resulting segment \(OA\) into two equal parts. The line \(BB'\), drawn through the point \(B\) obtained in this division and situated parallel to the polar, is called the semipolar.
From the similarity of the triangles \(OA'A\) and \(OFA'\) we have
\[ \frac{OA}{OA'}=\frac{OA'}{OF}; \qquad (OA')^2=OA \cdot OF . \]
At the same time, for the tangents and secants we may write
\[ (OA')^2=OG \cdot OH; \]
equating the expressions obtained, we find
\[ OA=\frac{OG \cdot OH}{OF}; \]
denoting the radius of the circle by \(r\) and the distance from the pole to the center of the circle by \(d\), we find the distance from the pole \(O\) to the polar \(AA'\)
\[ OA=\frac{(d-r)(d+r)}{d}=\frac{d^2-r^2}{d} \]
and to the semipolar \(BB'\)
\[ OB=BA=\frac{OA}{2}=\frac{d^2-r^2}{2d}. \]
If we take on the circle some point \(c\), having coordinates \(x\) and \(y\), then its distance to the semipolar will be equal to
\[ CD=X-OB=\frac{2dx-d^2+r^2}{2d}; \tag{33} \]
the distance of the same point to the pole (i.e., the length of the vector) will be
\[ oc=\sqrt{x^2+y^2}; \]
from the equation of the circle we have
\[ y^2+(x-d)^2=r^2, \]
whence
\[ (OC)^2=y^2+x^2=2xd-d^2+r^2, \tag{34} \]
comparing equations (33) and (34), we obtain
\[ CD=\frac{(OC)^2}{2d}; \]
i.e., the segment \(CD\) is proportional to the square of the segment \(OC\).
If the diagram is constructed for the current \(I^2\) on the scale \(M_I\) and it is required to find the power determined by the equation \(P=I^2R\), then, constructing the semipolar, we obtain the required power in the form of the segments \(CD\) on the scale
\[ M_p=2dM_I^2R. \tag{35} \]
If the diagram is constructed for voltage, and the power is determined by the equation \(P=V^2g\), then the scale will be equal to
\[ M_p=2dM_v^2g. \tag{35a} \]
When the pole lies inside the circle, the polar will pass outside it. To find it, from the pole \(A\) (Fig. 27) we erect the perpendicular \(A'A''\) to the line \(AF\) joining the pole with the center of the circle. At the points obtained, \(A'\) and \(A''\), we draw tangents and find the point \(O\) of their intersection. The line \(O'O''\), drawn through this point parallel to the line \(A'A''\), will be the polar. In the same way as in the preceding case, we find the polupolar as the line passing through the midpoint of the segment \(OA\) (the line drawn through the pole and the polar will be mutually conjugate and interchange places).
Introducing the distance \(d_1\) from the new pole \(A\) to the center of the circle \(F\), we have, from the similarity of the triangles \(OA'F\) and \(A'AF\),
\[ \frac{FA}{FA'}=\frac{FA'}{OF};\quad OF=\frac{(FA')^2}{FA};\quad d=\frac{r^2}{d_1}, \]
which, on the basis of what was obtained above, gives
\[ OA=\frac{d^2-r^2}{d}=\frac{r^2-d_1^2}{d_1} \]
or the distance to the polupolar
\[ AB=OB=\frac{r^2-d_1^2}{2d}. \]
Transferring the origin of coordinates to the pole \(A\), we have for the segment \(CD\) the expression
\[ CD=AB+X_1=\frac{2x_1d_1-d_1^2+r^2}{2d_1}. \]
Since the equation of the circle in the new coordinates will be
\[ y^2+(x_1-d_1)^2=r^2, \]
the square of the length of the segment \(AC\) will be equal to
\[ (AC)^2=x_1^2+y^2=2x_1d_1-d_1^2+r^2 \]
or, comparing the equations obtained,
\[ CD=\frac{(AC)^2}{2d_1}, \]
i.e., in this case as well the scales obtained from equations (35) will be valid, but with the distance \(d\) replaced by \(d_1\).
When the pole lies on the circle itself, both the polar and the polupolar coincide with the tangent drawn through the pole. This fully corresponds to the case indicated above of the vector being reduced to zero. In this case \(d=r\), and the scale will be equal to
\[ M_p=2rM_iR, \tag{35в} \]
since the diameter of the circle and the chord are related by the equation
\[ 2r=\frac{H}{\sin p}, \]
then the expression obtained for the scale is consistent with the earlier derived equations (22) and (23).
The construction of the lines of efficiency and of the circles of the power factor is carried out in the same way as was indicated for the computed diagrams.
5. EXAMPLE OF CONSTRUCTING A PIE DIAGRAM FROM EXPERIMENTAL DATA
As an example, let us consider a single-phase load of a three-phase transformer with both windings connected in star, without a neutral wire in the primary circuit. We shall construct a combined diagram for all currents and voltages and find on it the line of total power.
The experiment was carried out with a three-phase transformer with a turns ratio equal to 20. The normal values of the line voltages were 120/2400 V and of the currents 60/3 A.
When the transformer was connected for no load from the low-voltage side, and the obtained results were recalculated to a voltage of 120 V, the following values were obtained:
| current in the wires and phases | 4.15 A, |
| total power of all phases | 505 W, |
| secondary phase voltage | 1380 V. |
From the equations we find the values of the phase voltage, 69.7 V, and the phase displacement angle, 54°. We neglect the inequality of the currents and of the displacement angles in the individual phases, arising from the asymmetry of the core, since their differences will be very small.
In the short-circuit test, the low voltage was applied to the high-voltage terminals, and then a recalculation was made for the proper connection and a voltage of 120 volts. Power measurements were made in each phase with respect to the neutral of the network.
| Phases | 1st | 2nd | 3rd |
|---|---|---|---|
| Line voltage | 120 V | 120 V | 120 V |
| Phase voltage of the network | 70 V | 70 V | 70 V |
| Phase voltage of the transformer | 23 V | 108 V | 98 V |
| Current | 270 A | 138 A | 133 A |
| Power of the network phase | 4750 W | 6200 W | 8900 W |
| Angle of current displacement with respect to the phase voltage of the network | 75° | 130.5° | 18.5° |
| Secondary current | 18.5 A | 0 | 0 |
The total power of all three phases is equal to 7450 W.
Connecting an ohmic load to the secondary winding of the first phase and again recalculating to a voltage of 120 volts, we obtain the values:
| 1st | 2nd | 3rd | |
|---|---|---|---|
| Phase voltages of the transformer | 55 V | 98 V | 65 V |
| Current | 173 A | 90 A | 86 A |
| Power of the network phase | 9200 W | 800 W | 5800 W |
| Secondary phase voltage | 925 V | — | — |
| Secondary current | 11.5 A | 0 | 0 |
The total power of all three phases is equal to 14200 W.
From the no-load and short-circuit data we construct the vector diagram shown in Fig. 28. The segments 01, 02, and 03 represent the phase voltages of the network and the phase voltages of the transformer under short circuit of the first phase; the segments \(Oa\), \(Ob\), and \(Oc\) are the no-load currents, and \(OA\), \(OB\), and \(OC\) are the currents under short circuit.
We compute the angle of the circles of the diagrams from equation (27)
\[ \cos \varphi=\frac{1}{2}\left(\frac{1380\cdot18.5}{925\cdot11.5} -\frac{1380\cdot11.5}{925\cdot18.5} -\frac{925\cdot18.5}{1380\cdot11.5}\right)=0.2 \]
or \(\rho=78.5^\circ\). Connecting by straight lines (dashed) the ends of the corresponding vectors of no-load and short-circuit, we construct the arcs of the circles of the diagrams.
Changing the scales and placing the chord horizontally, we construct the combined diagram shown in Fig. 29. The scales for this diagram turned out to be equal, for primary voltages, to \(1\ \frac{V}{mm}\), for secondary voltage to \(29\ \frac{V}{mm}\); for secondary current to \(0.385\ \frac{A}{mm}\), for the primary currents of the first phase to \(5.55\ \frac{A}{mm}\), of the second phase to \(2.9\ \frac{A}{mm}\), and of the third phase to \(2.7\ \frac{A}{mm}\). The beginnings of the corresponding segments are marked on the diagram by letters.
Fig. 28.
Laying off, to scale, the load current of 11.5 amperes in the form of the segment \(ac\), we find the values of the corresponding voltage, equal from the diagram to \(930\ V\), of the primary voltages \(55\), \(99\), and \(65\) volts, and of the currents \(175\), \(90\), and \(84\) amperes. Comparing the obtained results with the experimental data given above, we see that they are very close to one another.
For finding the lines of equal powers we measure the segment \(cb=32\ mm\) and compute, by equation (31), the value of the segment
\[ bk=32\frac{7450-505}{14200-7450}=32.9\ mm; \]
laying it off on the continuation of the segment \(cb\), we draw the straight line \(ak\) and find the segment \(cf=26.5\ mm\), taking it perpendicular to the chord \(ab\).
Fig. 29.
By equation (32) we compute the power scale
\[ M_p=\frac{14200-505}{26.5}=520\ \frac{W}{mm} \]
and draw the power line parallel to \(ak\) at a distance
\[ fd=\frac{P_{xx}}{M_p}=\frac{505}{520}=0.97\ \mathrm{mm}. \]
The power will be represented by the segments \(cd\).
In the same way one can also find the powers for the individual phases.
6. CONCLUSION
The construction of superposed diagrams and the use of their properties is possible in all cases when equation (1) for all currents and voltages is obtained with one and the same denominator, or, more precisely, with one and the same ratio of the coefficients \(\dot C D\). In each individual case it is better to verify this before beginning the construction.
When there is only one variable resistance (load) in the circuit, this condition is satisfied in all cases, however complex the circuit may be. This will be true both for single-phase and for polyphase circuits.
When there are several loads, circular diagrams are in general possible only when their resistances are varied simultaneously in a definite ratio to one another. Only for certain ratios are circular diagrams possible, and only under definite conditions can they be superposed. In single-phase circuits a sufficient condition is direct proportionality between the varying resistances (even if their arguments differ); for polyphase circuits, however, symmetry of the individual phases is often also necessary. Thus, when the conductors of a three-phase circuit are asymmetrical and the load resistances of all phases are equal, several different diagrams are nevertheless obtained, which cannot be superposed into one.
Submitted to the editors in October 1937.