0.3 Chapter 4: introduction to rotating machines  (Page 9/9)

${\lambda }_s=L_{\text{ss}}+L_{\text{sr}}({\theta }_{\text{me}})i_r=L_{\text{ss}}{i}_s+L_{\text{sr}}\text{cos}({\theta }_{\text{me}})i_r$ (4.55)

${\lambda }_r=L_{\text{sr}}({\theta }_{\text{me}})i_s+L_{\text{rr}}{i}_r=L_{\text{sr}}\text{cos}({\theta }_{\text{me}})i_s+L_{\text{rr}}{i}_r$ (4.56)

In matrix notation

$\left[\begin{array}{c}{\lambda }_s\\ {\lambda }_r\end{array}\right]=\left[\begin{array}{cc}{L}_{\text{ss}}& L_{\text{sr}}({\theta }_{\text{me}})\\ L_{\text{sr}}({\theta }_{\text{me}})& L_{\text{rr}}\end{array}\right]\left[\begin{array}{c}{i}_s\\ i_r\end{array}\right]$ (4.57)

The terminal voltages $v_s$ and $v_r$ are

$v_s=R_s{i}_s+\frac{{\mathrm{d\lambda }}_s}{\text{dt}}$ (4.58)

$v_r=R_r{i}_r+\frac{{\mathrm{d\lambda }}_r}{\text{dt}}$ (4.59)

where $R_s$ and $R_r$ are the resistances of the stator and rotor windings respectively.

• When the rotor is revolving, ${\theta }_{\text{me}}$ must be treated as a variable. Differentiation of Eqs.4.56 and 4.57 and substitution of the results in Eqs.4.59 and 4.60 then give

$v_s=R_s{i}_s+L_{\text{ss}}\frac{{\text{di}}_s}{\text{dt}}+L_{\text{sr}}\text{cos}({\theta }_{\text{me}})\frac{{\text{di}}_r}{\text{dt}}-L_{\text{sr}}{i}_r\text{sin}({\theta }_{\text{me}})\frac{{\mathrm{d\theta }}_{\text{me}}}{\text{dt}}$ (4.60)

$v_r=R_r{i}_r+L_{\text{rr}}\frac{{\text{di}}_r}{\text{dt}}+L_{\text{sr}}\text{cos}({\theta }_{\text{me}})\frac{{\text{di}}_r}{\text{dt}}-L_{\text{sr}}{i}_s\text{sin}({\theta }_{\text{me}})\frac{{\mathrm{d\theta }}_{\text{me}}}{\text{dt}}$ (4.61)

$\frac{{\mathrm{d\theta }}_{\text{me}}}{\text{dt}}={\omega }_{\text{me}}=\left[\frac{\text{poles}}{2}\right]{\omega }_m$ (4.62)

is the instantaneous speed in electrical radians per second. In a two-pole machine, θme and ωme are equal to the instantaneous shaft angle θm and the shaft speed ωm respectively. In a multipole machine, they are related by Eqs.4.54 and 4.46. The second and third terms on the fight-hand sides of Eqs.4.61 and 4.62 are L(di/dt) induced voltages like those induced in stationary coupled circuits such as the windings of transformers. The fourth terms are caused by mechanical motion and are proportional to the instantaneous speed. These are the speed voltage terms which correspond to the interchange of power between the electric and mechanical systems.

• The electromechanical torque can be found from the coenergy.

$\begin{array}{}{W}_{\text{fld}}^{\text{'}}=\frac{1}{2}{L}_{\text{ss}}{i}_s^2+\frac{1}{2}{L}_{\text{rr}}{i}_r^2+L_{\text{sr}}{i}_s{i}_r\text{cos}{\theta }_{\text{me}}\\ =\frac{1}{2}{L}_{\text{ss}}{i}_s^2+\frac{1}{2}{L}_{\text{rr}}{i}_r^2+L_{\text{sr}}{i}_s{i}_r\text{cos}\left[(\frac{\text{poles}}{2}){\theta }_m\right]\end{array}$ (4.63)

Note that the coenergy of Eq.4.63 has been expressed specifically in terms of the shaft angle ${\theta }_m$ because the torque expression requires that the torque be obtained from the derivative of the coenergy with respect to the spatial angle ${\theta }_m$ and not with respect to the electrical angle ${\theta }_{\text{me}}$ . Thus,

$\begin{array}{}T=\frac{\partial W_{\text{fld}}^{\text{'}}(i_s,i_r,{\theta }_m)}{\partial {\theta }_m}{\mid }_{i_s,i_r}=-\left[\frac{\text{poles}}{2}\right]L_{\text{sr}}{i}_s{i}_r\text{sin}\left[\frac{\text{poles}}{2}{\theta }_m\right]\\ =-\left[\frac{\text{poles}}{2}\right]L_{\text{sr}}{i}_s{i}_r\text{sin}{\theta }_{\text{me}}\end{array}$ (4.64)

where T is the electromechanical torque acting to accelerate the rotor (i.e., a positive torque acts to increase ${\theta }_m$ ). The negative sign in Eq.4.64 means that the electromechanical torque acts in the direction to bring the magnetic fields of the stator and rotor into alignment.

4.7.2 Magnetic Field Viewpoint

• In the discussion of Section 4.7.1 the characteristics of a rotating machine as viewed from its electric and mechanical terminals have been expressed in terms of its winding inductances. This viewpoint gives little insight into the physical phenomena which occur within the machine. In this section, we will explore an alternative formulation in terms of the interacting magnetic fields.

Figure 4.25 Simplified two-pole machine: (a) elementary model and

(b) vector diagram of mmf waves. Torque is produced by the tendency of the rotor and stator magnetic fields to align. Note that these figures are drawn with ${\delta }_{\text{sr}}$ positive, i.e., with the rotor mmf wave leading that of the stator $F_s$ .

• As we have seen, currents in the machine windings create magnetic flux in the air gap between the stator and rotor, the flux paths being completed through the stator and rotor iron. This condition corresponds to the appearance of magnetic poles on both the stator and the rotor, centered on their respective magnetic axes, as shown in Fig.4.35a for a two-pole machine with a smooth air gap. Torque is produced by the tendency of the two component magnetic fields to line up their magnetic axes.
• We shall derive an expression for the magnetic coenergy stored in the air gap in terms of the stator and rotor mmfs and the angle ${\delta }_{\text{sr}}$ between their magnetic axes. The torque can then be found from the partial derivative of the coenergy with respect to angle ${\delta }_{\text{sr}}$ .
• The line integral of Hag across the gap then is simply $H_{\text{ag}}g$ and equals the resultant air-gap mmf $F_{\text{sr}}$ produced by the stator and rotor windings; thus

$H_{\text{ag}}g=F_{\text{sr}}$ (4.65)

• The mmf waves of the stator and rotor are spatial sine waves with ${\delta }_{\text{sr}}$ being the phase angle between their magnetic axes in electrical degrees. They can be represented by the space vectors $F_s$ and $F_r$ drawn along the magnetic axes of the stator- and rotor mmf waves respectively. The resultant mmf $F_{\text{sr}}$ acting across the air gap, also a sine wave, is their vector sum.

$F_{\text{sr}}^2=F_s^2+F_r^2+{\mathrm{2F}}_s{F}_r\text{cos}{\delta }_{\text{sr}}$ (4.66)

The resultant radial $H_{\text{ag}}$ field is a sinusoidal space wave whose peak value $H_{\text{ag},\text{peak}}$ is, from Eq.4.65,

$(H_{\text{ag}}{)}_{\text{peak}}=\frac{F_{\text{sr}}}{g}$ (4.67)

Average coenergy density = $\frac{{\mu }_o}{2}\left[\frac{(H_{\text{ag}}{)}_{\text{peak}}^2}{2}\right]=\frac{{\mu }_o}{4}{\left[\frac{F_{\text{sr}}}{g}\right]}^2$ (4.68)

$\begin{array}{}{W}_{\text{fld}}^{\text{'}}=(\text{average coenergy density})(\text{volume of air gap})\\ =\frac{{\mu }_o}{4}{\left[\frac{F_{\text{sr}}}{g}\right]}^2\mathrm{\pi D}\text{lg}=\frac{{\mu }_o\pi \text{Dl}}{\mathrm{4g}}{F}_{\text{sr}}^2\end{array}$ (4.69)

$W_{\text{fld}}^{\text{'}}=\frac{{\mu }_o\pi \text{Dl}}{\mathrm{4g}}(F_s^2+F_r^2+{\mathrm{2F}}_s{F}_r\text{cos}{\delta }_{\text{sr}})$ (4.70)

For a two-pole machine

$T=\frac{\partial W_{\text{fld}}^{\text{'}}}{\partial {\delta }_{\text{sr}}}{\mid }_{F_s,F_r}=-\left[\frac{{\mu }_o\pi \text{Dl}}{\mathrm{2g}}\right]F_s{F}_r\text{sin}{\delta }_{\text{sr}}$ (4.71)

The general expression for the torque for a multipole machine is

$T=-\left[\frac{\text{poles}}{2}\right]\left[\frac{{\mu }_o\pi \text{Dl}}{\mathrm{2g}}\right]F_s{F}_r\text{sin}{\delta }_{\text{sr}}$ (4.72)

• In this equation, ${\delta }_{\text{sr}}$ is the electrical space-phase angle between the rotor and stator mmf waves and the torque T acts in the direction to accelerate the rotor. Thus when ${\delta }_{\text{sr}}$ is positive, the torque is negative and the machine is operating as a generator.
• Similarly, a negative value of ${\delta }_{\text{sr}}$ corresponds to positive torque and, correspondingly, motor action. The torque, acting to accelerate the rotor, can then be expressed in terms of the resultant mmf wave $F_{\text{sr}}$ ; thus

$T=-\left[\frac{\text{poles}}{2}\right]\left[\frac{{\mu }_o\pi \text{Dl}}{\mathrm{2g}}\right]F_s{F}_{\text{sr}}\text{sin}{\delta }_s$ (4.73)

$T=-\left[\frac{\text{poles}}{2}\right]\left[\frac{{\mu }_o\pi \text{Dl}}{\mathrm{2g}}\right]F_r{F}_{\text{sr}}\text{sin}{\delta }_r$ (4.74)

$T=\left[\frac{\text{poles}}{2}\right]\left[\frac{\pi \text{Dl}}{2}\right]B_{\text{sr}}{F}_r\text{sin}{\delta }_r$ (4.75)

• One of the inherent limitations in the design of electromagnetic apparatus is the saturation flux density of magnetic materials. Because of saturation in the armature teeth the peak value $B_{\text{sr}}$ of the resultant flux-density wave in the air gap is limited to about 1.5 to 2.0T. The maximum permissible value of the winding current, and hence the corresponding mmf wave, is limited by the temperature rise of the winding and other design requirements. Because the resultant flux density and mmf appear explicitly in Eq.4.75, this equation is in a convenient form for design purposes and can be used to estimate the maximum torque which can be obtained from a machineof a given size.