On the Calculation of Losses in the End Region of High-Power Turbogenerators
Abstract Generated abstract
This paper develops a calculation method for additional losses caused by end-region magnetic fields in high-power turbogenerators, especially in the stator end packets and teeth. Assuming constant anisotropic magnetic permeability, it solves a plane magnetic-field problem by Fourier separation, derives expressions for flux penetration and induction components, and relates the resulting losses to steel anisotropy, packet geometry, and finite back dimensions. The analysis indicates that end-field losses may be concentrated near the surface under unsaturated conditions but become more uniformly distributed with saturation, and it proposes a practical loss formula for end teeth. Experimental heating data for a TVV-165 turbogenerator are used for comparison, supporting the conclusion that these losses can significantly affect operation under underexcited or asynchronous conditions.
Full Text
ON THE CALCULATION OF LOSSES IN THE END REGION OF HIGH-POWER TURBOGENERATORS
N. S. SIUNOV, V. I. MANGILEV
(Presented by the scientific seminar of the Departments of Electrical Machines and General Electrical Engineering)
A considerable number of works have been devoted to questions of calculating and investigating additional losses in the region of the end parts of windings [1, 2, 3].
In modern synchronous machines there is a significant increase in the intensity of the fields in the end region, associated with a substantial increase in the linear loading in high-power turbogenerators.
The fields in the end region of machines are produced by the end parts of the stator and rotor windings. These fields, reaching the lateral surface of the pressure plates, shields, and the end packets of the stator core, cause additional losses in them.
The end flux, penetrating into the end packet from the end space in the region of the teeth and the stator yoke, then branches in the radial direction (in the teeth) and in the circumferential direction (in the yoke).
The study of the propagation of the flux is complicated by the fact that the magnetic permeability of the steel depends on the magnitude of the induction.
In the present article a particular problem is considered for the case of constant magnetic permeability.
In Fig. 1 a section of a magnetic core with faces into which the magnetic flux enters is presented. We assume that the flux enters the magnetic core over the segment \(h\) through the plane \(X=0\). For the teeth of the end packet, the plane \(y=0\) coincides with the generatrix of the stator bore, and the \(X\)-axis is directed along the axis of the machine.
We assume that the magnetic permeability along the \(X\) and \(y\) axes is constant, i.e. \(\mu_x=\text{const}\) and \(\mu_y=\text{const}\), with \(\mu_x\ne\mu_y\).
Let us consider the problem in which the flux enters through one face over the segment \(0\div h\) and exits over the segment \(h—2h\).
The end flux \(B_s\) may be specified according to any law.
From Fig. 1 the boundary conditions may be represented under the following assumptions:
a) the main flux through the air gap is absent;
b) the entire flux enters and exits only through the plane \(X=0\);
c) the induction \(B_s\) is distributed according to the law of an even function.
The boundary conditions on the planes are:
\[ \begin{aligned} y&=0, & B_y&=0;\\ y&=2h, & B_y&=0;\\ X&=0, & B_x&=B_s;\\ X&=b, & B_x&=0. \end{aligned} \]
Fig. 1
The induction \(B_x\) through the plane \(X=b\) will be equal to zero for a sufficiently large package width. The equation of the magnetic field is written in the form:
\[ \frac{dB_x}{dx}+\frac{dB_y}{dy}=0 . \tag{1} \]
Solving the problem for the case when there are no currents in the region under consideration, we obtain
\[ \mu_x \frac{d^2 U_m}{dx^2}+\mu_y \frac{d^2 U_m}{dy^2}=0 , \tag{2} \]
where
\[ U_m \]
is the scalar potential of the magnetic field.
We shall find the solution of equation (2) by the Fourier method.
Then
\[ U_m = X(x)\cdot Y(y). \tag{3} \]
where
\[ X(x) \]
depends only on \(x\),
\[ Y(y) \]
depends only on \(y\).
Substitute expression (3) into (2)
\[ \mu_x \frac{d^2 x}{dx^2}\,Y+\mu_y \frac{d^2 Y}{dy^2}\,X=0 . \tag{3a} \]
After a number of transformations, we obtain the value of the scalar potential of the magnetic field
\[ U_m=\sum_{k=1}^{\infty} -\frac{\dfrac{1}{h}\int_0^{2h}\dfrac{B_s}{\mu_x}\cos\frac{\pi}{2}\cdot\frac{k}{h}y\,dy \left( e^{\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}(x-2b)} + e^{-\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}x} \right)} {\dfrac{\pi}{2}\cdot\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}} \left( e^{-\frac{\pi}{2}\cdot\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}-1 \right)} \times \cos\frac{\pi}{2}\cdot\frac{k}{h}y . \tag{4} \]
The distribution of induction inside the magnetic core will be determined by the equations:
\[ B_x=-\mu_x\frac{dU_m}{dx}= \]
\[ =\mu_x\sum_{k=1}^{\infty} \frac{\dfrac{1}{h}\int_0^{2h}\dfrac{B_s}{\mu_x}\cos\frac{\pi}{2}\cdot\frac{k}{h}y\,dy \left( e^{\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}(x-2b)} + e^{-\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}x} \right)} {\left( e^{-\frac{\pi}{2}\cdot\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}-1 \right)} \times \]
\[ \times \cos\frac{\pi}{2}\frac{k}{h}y , \tag{5} \]
\[ B_y=-\mu_y\frac{dU_m}{dy}= \]
\[ =\mu_y\sum_{k=1}^{\infty} -\frac{\dfrac{1}{h}\int_0^{2h}\dfrac{B_s}{\mu_x}\cos\frac{\pi}{2}\cdot\frac{k}{h}y\,dy \left( e^{\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}(x-2b)} + e^{-\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}x} \right)} {\sqrt{\dfrac{\mu_y}{\mu_x}} \left( e^{-\frac{\pi}{2}\cdot\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}-1 \right)} \times \]
\[ \times \sin\frac{\pi}{2}\cdot\frac{k}{h}y . \tag{6} \]
Let
\[ B_s=B_0\cos\frac{\pi}{2}\frac{y}{h}. \tag{7} \]
Then the integral
\[ \frac{1}{h}\int_0^{2h}\frac{B_s}{\mu_x}\cos\frac{\pi}{2}\frac{k}{h}y\,dy = \frac{B_0}{\mu_x h}\int_0^{2h}\cos\frac{\pi}{2}\frac{y}{h}\cos\frac{\pi}{2}\cdot\frac{k}{h}y\,dy \]
will be equal to zero for all values of “\(k\)” except \(k=1\) (by orthogonality), and for the case under consideration we shall have
\[ B_x=B_0 \frac{ \left[ e^{\frac{\pi}{2}\cdot\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}(x-2b)} - e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}x} \right] \cos\frac{\pi}{2}\frac{y}{h} } { \left( e^{-\frac{\pi}{2}\cdot\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b} -1 \right) }, \tag{8} \]
\[ B_y=-B_0\sqrt{\frac{\mu_y}{\mu_x}}\, \frac{ \left[ e^{\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}(x-2b)} + e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}x} \right]\cdot \sin \frac{\pi}{2}\frac{y}{h} }{ \left(e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\,2b}-1\right) }. \tag{9} \]
As can be seen, the parameters determining the distribution of induction inside the magnetic core are the ratios
\[ \frac{\mu_y}{\mu_x} \quad \text{and} \quad \frac{2b}{h}. \]
Let us consider the influence of these parameters. Let \(b\to \infty\). Setting \(b\to \infty\) in equation (4), we obtain
\[ U_m=\sum_{k=1}^{\infty} \frac{ h^{-1}\displaystyle\int_{0}^{2h}\frac{B_s}{\mu_x}\cos \frac{\pi}{2}\frac{k}{h}y\,dy \cdot e^{-\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}}\,x} \cdot \cos \frac{\pi}{2}\frac{k}{h}y }{ -\frac{\pi}{2}\frac{k}{h}\sqrt{\frac{\mu_y}{\mu_x}} }. \tag{10} \]
Then, for \(B_s=B_0\cos \frac{\pi}{2}\frac{y}{h}\), we have
\[ B_x=B_0 e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\,x} \cos \frac{\pi}{2}\cdot \frac{y}{h}, \tag{11} \]
\[ B_y=B_0 e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\,x} \cdot \sqrt{\frac{\mu_y}{\mu_x}} \sin \frac{\pi}{2}\cdot \frac{y}{h}. \tag{12} \]
The ratio \(\frac{\mu_y}{\mu_x}\) can be physically explained as follows:
\[
\frac{\mu_y}{\mu_x}=1
\]
— the case of the distribution of induction in the stator back for hot-rolled steel and of the end flux in a solid shield;
\[
\frac{\mu_y}{\mu_x}=4\div 25
\]
— the case of the distribution of induction in the stator back for cold-rolled steel with the rolling direction along the back.
At present, in synchronous machines, cold-rolled steel has found wide application for the stator core. The main losses and the principal consumption of magnetizing ampere-turns in turbogenerators are concentrated in the yoke. Therefore the stator segments are punched so that the rolling direction coincides with the direction of the magnetic flux along the back.
Let us consider in greater detail the operating conditions of cold-rolled steel, which is characterized by two principal parameters:
a) the ratio of the magnetic permeability along the rolling direction to the magnetic permeability across the rolling direction,
\[ \frac{\mu_y}{\mu_x}, \]
b) the ratio of the specific losses along and across the rolling direction,
\[ \frac{a_y}{a_x}. \]
When considering the magnetic flux in cold-rolled steel, we make two assumptions:
- The maximum losses in the steel will occur at an angle of inclination to the rolling direction of \(90^\circ\).
- The flux along the back is distributed uniformly, i.e., saturation is absent.
The losses in the back for \(b \to \infty\) in a volume element (for the plane problem) are
\[ \frac{dP}{dS}=\frac{dP}{dx\,dy}=\alpha_x B_x^2+\alpha_y B_y^2 . \]
Hence
\[ P=\alpha_x \iint_S B_x\,dx\,dy+\alpha_y \iint_S B_y\,dx\,dy . \]
According to (7), we have
\[ P_x=\alpha_x \frac{B_0^2}{2\pi \sqrt{\dfrac{\mu_y}{\mu_x}}}\,h^2, \]
\[ P_y=\alpha_y \frac{B_0^2}{2\pi \sqrt{\dfrac{\mu_y}{\mu_x}}}\,h^2 \sqrt{\frac{\mu_y}{\mu_x}} . \]
Thus, the losses in a back of infinite height are determined by the relation
\[ P=P_x+P_y=\frac{B_0^2 h}{2\pi} \left( \frac{\alpha_x}{\sqrt{\dfrac{\mu_y}{\mu_x}}} +\alpha_y \sqrt{\frac{\mu_y}{\mu_x}} \right). \tag{13} \]
This equation has been obtained for constant magnetic permeability and an infinitely large back. In reality, the back has finite dimensions; therefore let us consider the case when the width of the back is equal to \(b\).
Then, from equations (8) and (9), after transformations we obtain
\[ P_x=\alpha_x \int_0^h \int_0^b B_x\,dx \cdot dy =\alpha_x B_0^2 \left[ \frac{ \left( e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\dfrac{\mu_y}{\mu_x}}\cdot 2b}+1 \right) }{ \left( 1-e^{-\frac{\pi}{2}\cdot\frac{1}{h}\sqrt{\dfrac{\mu_y}{\mu_x}}\cdot 2b} \right)\cdot 2\pi \sqrt{\dfrac{\mu_y}{\mu_x}} } - \frac{ e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\dfrac{\mu_y}{\mu_x}}\cdot 2b}\cdot b\cdot h }{ \left( 1-e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\dfrac{\mu_y}{\mu_x}}\cdot 2b} \right)^2 } \right], \]
\[ P_y=\alpha_y \int_0^h \int_0^b B_y\,dxdy =\alpha_y B_0^2 \left[ \frac{ \left(e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}+1\right)h^2 \sqrt{\frac{\mu_y}{\mu_x}} }{ \left(1-e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}\right)\cdot 2\pi } + \frac{\mu_y}{\mu_x} \frac{ e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}\cdot b\cdot h }{ \left(1-e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}\right)^2 } \right]. \]
As a result, the total losses are equal to:
\[ P=B_0^2 \left[ \frac{ \left(e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}+1\right)h^2 }{ \left(1-e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}\right)\cdot 2\pi \cdot \sqrt{\frac{\mu_y}{\mu_x}} } \left(\alpha_x+\frac{\mu_y}{\mu_x}\alpha_y\right) + \frac{ e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b} }{ \left(1-e^{-\frac{\pi}{2}\frac{1}{h}\sqrt{\frac{\mu_y}{\mu_x}}\cdot 2b}\right)^2 } \left(\frac{\mu_y}{\mu_x}\alpha_y-\alpha_x\right) \right]. \tag{14} \]
Let us consider the influence of the assumptions we have made on the results obtained. The total losses in cold-rolled steel were determined as
\[ \rho_x=\alpha_x\beta_x+\alpha_y\beta_y. \]
Thus, we assumed that, at any angle of inclination to the rolling direction, the losses would be smaller than at an angle of \(90^\circ\). However, the actual distribution of losses is such that their maximum is observed at angles of \(45\text{--}60^\circ\). Consequently, in reality larger losses should be expected.
Influence of saturation. As is known, saturation leads to an equalization of the induction distribution over the cross section of the back, which reduces the losses. Since the assumptions adopted act in opposite directions, the conclusions drawn may be regarded as sufficiently reliable.
Figure 2a shows the distribution of the tangential component of the induction over the periphery, and below it the corresponding ampere-turns. This pattern is presented under the condition \(\mu=\mathrm{const}\).
Since the field in the region under consideration is irrotational, the integral of the magnetic-field strength around a closed contour is equal to zero. Accordingly, the area bounded by the magnetic-field-strength curve must be divided by a straight line drawn through zero into two equal parts. However, as can be seen from the figure, the left part is larger than the right. Consequently, for an isotropic medium the induction distribution with allowance for saturation must differ from that shown. Since the voltage drop due to the transverse flux is small, the practical induction distribution approaches a uniform one.
Figure 2b shows the limiting case of the distribution of the magnetic flux in an anisotropic medium. The case of penetration of the end flux into the outermost tooth is considered. The width of the stack is taken as \(0.25\) of the slot height. As can be seen from the figure, the flux decays so rapidly that the field-strength distribution is located entirely in the left part. Although the resistance across the stack increases greatly, the distribution of induc-
Fig. 2
…tion under the condition \(\mu=\mathrm{const}\) is not justified. Thus, rapid attenuation of the magnetic flux in the axial direction of the machine can be assumed for the case of unsaturated steel. In reality, the distribution of induction should have a more uniform character.
The equivalent depth of penetration of the end flux can be estimated as follows.
For \(\mu=\mathrm{const}\), for \(b=0.25\) of the slot height, the losses from the transverse flux as a function of the ratio \(\sqrt{\mu_y/\mu_x}\) are given in Table 1.
Losses for a uniform distribution of induction are taken as unity.
Table 1
| \(\sqrt{\dfrac{\mu_y}{\mu_x}}\) | 1 | 2 | 5 | 9 | 13 | 17 | 20 |
|---|---|---|---|---|---|---|---|
| \(P_x\) | \(\dfrac{1}{6.25}\) | \(\dfrac{2}{6.6}\) | \(\dfrac{1}{8.95}\) | \(\dfrac{1}{14.2}\) | \(\dfrac{1}{20.6}\) | \(\dfrac{1}{27}\) | \(\dfrac{1}{32}\) |
Thus, with constant magnetic permeability, the losses from the end flux are concentrated in a narrow equivalent layer. Saturation of the steel leads to a more uniform distribution of induction and to an increase in the depth of penetration of the magnetic flux. In what follows we shall take the penetration depth to be equal to \(1/3\).
In this case, the formula for calculating the losses in the teeth of the end packet can be represented in the form
\[ P_z = 0.14\left(\frac{f}{50}\right)^2 \left(\frac{B}{10000}\right)^2 b_z \cdot b_p \cdot h_y \cdot z \cdot 0.93 \cdot 10^{-3}, \quad (\mathrm{kW}), \]
where
\(f\) — frequency, Hz;
\(B\) — induction, in gausses;
\(b_p\) — packet width, cm;
$b_z$ — average tooth width, cm;
$h$ — tooth height, cm;
$\gamma$ — specific electrical conductivity, m/ohm, mm²;
$Z$ — number of teeth.
This formula does not take into account losses associated with longitudinal fluxes, which, as a rule, amount to 20–25% of the total losses.
Fig. 3.
The greatest losses in the end packets and elements of the end region should be expected when turbogenerators operate under abnormal conditions (underexcited operation, asynchronous operation), since in these conditions the mutual-induction fluxes of the rotor and stator add [4, 5]. This will be especially pronounced in turbogenerators with forced cooling, which have high current densities and linear loadings. Figure 3 shows the heating curves of the elements of the end region for the TVV-165 turbogenerator as a function of $\cos\varphi$, determined experimentally; for comparison, a curve obtained by the method presented is also given.
Conclusions
-
The induction, and consequently the magnetic flux, in the stator steel depend on the magnetic properties of the steel and on the geometrical dimensions.
-
Losses in the teeth of the end packets due to end fields substantially increase the total losses in the end teeth. Moreover, in a number of cases they limit the load under asynchronous conditions and under underexcitation.
References
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A. I. Voldek. Fundamentals of the method for calculating the magnetic fields of the end parts of windings of electrical machines. Elektrichestvo, 1963, No. 1.
-
Ya. B. Danilevich, E. G. Kasharskii. Additional losses in electrical machines. Moscow–Leningrad, Gosenergoizdat, 1963.
-
E. G. Kasharskii. On the calculation of additional losses in the stator winding of a turbogenerator. Collection of works on problems of electromechanics, Issue VI. Publishing House of the Academy of Sciences of the USSR, 1961.
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G. M. Khutoretskii, V. I. Kosachevskii. Operation of turbogenerators under underexcitation conditions. Elektrotekhnika, 1968, No. 2.
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V. I. Kosachevskii, P. V. Oshurkov, G. M. Khutoretskii. Experimental investigation of the magnetic field in the end region of a turbogenerator. In: Investigation of electromagnetic fields, parameters, and losses in high-power electrical machines. Nauka, Moscow–Leningrad, 1966.