0.2 Chapter 3:electromechanical-energy-conversion  (Page 4/4)

$W_{\text{fld}}^{\text{'}}={\int }_V{\int }_0^{H_0}\text{BdH}\text{dV}$ (3.50)

$W_{\text{fld}}^{\text{'}}={\int }_v\frac{{\mathrm{\mu H}}^2}{2}\text{dV}$ (3.51)

For permanent-magnet (hard) materials

$W_{\text{fld}}^{\text{'}}={\int }_v{\int }_{H_c}^{H_0}\text{BdH}\text{dV}$ (3.52)

• For a magnetically-linear system, the energy and coenergy (densities) are numerically equal: ${\lambda }^2/\mathrm{2L}=\frac{1}{2}{\text{Li}}^2,B^2/\mathrm{2\mu }=\frac{1}{2}{\mathrm{\mu H}}^2$ . For a nonlinear system in which $\lambda$ and i or B and H are not linearly proportional, the two functions are not even numerically equal.

$W_{\text{fld}}+W_{\text{fld}}^{\text{'}}=\mathrm{\lambda i}$ (3.53)

Figure 3.5Graphical interpretation of energy and coenergy in a singly-excited system.

• Consider the relay in Fig. 3.3. Assume the relay armature is at position x so that the device operating at point a in Fig. 3.6. Note that

$f_{\text{fld}}=-\frac{\partial W_{\text{fld}}(\lambda ,x)}{\partial x}{\mid }_{\lambda }\simeq \underset{\mathrm{\Delta x}\to 0}{\text{lim}}\frac{-{\mathrm{\Delta W}}_{\text{fld}}}{\mathrm{\Delta x}}{\mid }_{\lambda }\text{and}{f}_{\text{fld}}=\frac{\partial W_{\text{fld}}^{\text{'}}(i,x)}{\partial x}{\mid }_i\simeq \underset{\mathrm{\Delta x}\to 0}{\text{lim}}\frac{\Delta W_{\text{fld}}^{\text{'}}}{\mathrm{\Delta x}}{\mid }_i$

Figure 3.6Effect of $\Delta$ x on the energy and coenergy of a singly-excited device:

(a) change of energy with $\lambda$ held constant; (b) change of coenergy with i held constant.

• The force acts in a direction to decrease the magnetic field stored energy at constant flux or to increase the coenergy at constant current.

• In a singly-excited device, the force acts to increase the inductance by pulling on members so as to reduce the reluctance of the magnetic path linking the winding.

§3.6 Multiply-Excited Magnetic Field Systems

• Many electromechanical devices have multiple electrical terminals.

• Measurement systems: torque proportional to two electric signals; power as the product of voltage and current.
• Energy conversion devices: multiply-excited magnetic field system.
• A simple system with two electrical terminals and one mechanical terminal: Fig.3.7.
  • Three independent variables: $\left\{\theta ,{\lambda }_1,{\lambda }_2\right\}$ , $\left\{\theta ,i_1,i_2\right\}$ , $\left\{\theta ,{\lambda }_1,i_2\right\}$ , or $\left\{\theta ,i_1,{\lambda }_2\right\}$

${\text{dW}}_{\text{fld}}({\lambda }_1,{\lambda }_2,\theta )=i_1{\mathrm{d\lambda }}_1+i_2{\mathrm{d\lambda }}_2-T_{\text{fld}}\mathrm{d\theta }$ (3.54)

Figure 3.7Multiply-excited magnetic energy storage system.

$i_1=\frac{\partial W_{\text{fld}}({\lambda }_1,{\lambda }_2,\theta )}{\partial {\lambda }_1}{\mid }_{{\lambda }_2,\theta }$ (3.55)

$i_2=\frac{\partial W_{\text{fld}}({\lambda }_1,{\lambda }_2,\theta )}{\partial {\lambda }_2}{\mid }_{{\lambda }_1,\theta }$ (3.56)

$T_{\text{fld}}=-\frac{\partial W_{\text{fld}}({\lambda }_1,{\lambda }_2,\theta )}{\partial \theta }{\mid }_{{\lambda }_1,{\lambda }_2}$ (3.57)

To find $W_{\text{fld}}$ , use the path of integration in Fig. 3.14.

$W_{\text{fld}}({\lambda }_{1_0},{\lambda }_{2_0},{\theta }_0)={\int }_0^{{\lambda }_{2_0}}{i}_2({\lambda }_1=\mathrm{0,}{\lambda }_2,\theta ={\theta }_0){\mathrm{d\lambda }}_2+{\int }_0^{{\lambda }_{1_0}}{i}_1({\lambda }_1,{\lambda }_2={\lambda }_{2_0},\theta ={\theta }_0){\mathrm{d\lambda }}_1$ (3.58)

Figure 3.8Integration path to obtain $W_{\text{fld}}({\lambda }_{1_0},{\lambda }_{2_0},{\theta }_0)$ .

• In a magnetically-linear system,

${\lambda }_1=L_{\text{11}}{i}_1+L_{\text{12}}{i}_2$ (3.59)

${\lambda }_2=L_{\text{21}}{i}_1+L_{\text{22}}{i}_2$ (3.60)

$L_{\text{12}}=L_{\text{21}}$ (3.61)

Note that $L_{\text{ij}}=L_{\text{ij}}(\theta )$

$i_1=\frac{L_{\text{22}}{\lambda }_1-L_{\text{12}}{\lambda }_2}{D}$ (3.62)

$i_2=\frac{-L_{\text{21}}{\lambda }_1+L_{\text{11}}{\lambda }_2}{D}$ (3.63)

$D=L_{\text{11}}{L}_{\text{22}}-L_{\text{12}}{L}_{\text{21}}$ (3.64)

The energy for this linear system is

$\begin{array}{}{W}_{\text{fld}}({\lambda }_{1_0},{\lambda }_{2_0},{\theta }_0)={\int }_0^{{\lambda }_{2_0}}\frac{L_{\text{11}}({\theta }_0){\lambda }_2}{D({\theta }_0)}{\mathrm{d\lambda }}_2+{\int }_{}^{}\frac{(L_{\text{22}}({\theta }_0){\lambda }_1-L_{\text{12}}({\theta }_0){\lambda }_{2_0})}{D({\theta }_0)}\\ =\frac{1}{\mathrm{2D}({\theta }_0)}{L}_{\text{11}}({\theta }_0){\lambda }_{2_0}^2+\frac{1}{\mathrm{2D}({\theta }_0)}{L}_{\text{22}}({\theta }_0){\lambda }_{1_0}^2-\frac{L_{\text{12}}({\theta }_0)}{D({\theta }_0)}{\lambda }_{1_0}{\lambda }_{2_0}\end{array}$ (3.65)

• Coenergy function for a system with two windings can be defined as:

$W_{\text{fld}}^{\text{'}}(i_1,i_2,\theta )={\lambda }_1{i}_1+{\lambda }_2{i}_2-W_{\text{fld}}$ (3.66)

$$ $d{W}_{\text{fld}}^{\text{'}}(i_1,i_2,\theta )={\lambda }_1{\text{di}}_1+{\lambda }_2{\text{di}}_2+T_{\text{fld}}\mathrm{d\theta }$ (3.67)

${\lambda }_1=\frac{\partial W_{\text{fld}}(i_1,i_2,\theta )}{\partial i_1}{\mid }_{i_2,\theta }$ (3.68)

${\lambda }_2=\frac{\partial W_{\text{fld}}^{\text{'}}(i_1,i_2,\theta )}{\partial i_2}{\mid }_{i_1,i_2}$ (3.69)

$$ $T_{\text{fld}}=\frac{\partial W_{\text{fld}}^{\text{'}}(i_1,i_2,\theta )}{\partial \theta }{\mid }_{i_1,i_2}$ (3.70)

$W_{\text{fld}}^{\text{'}}(i_1,i_2,{\theta }_0)={\int }_0^{i_{2_0}}{\lambda }_2(i_1=\mathrm{0,}{i}_2,\theta ={\theta }_0){\text{di}}_2+{\int }_0^{{\lambda }_{1_0}}{\lambda }_1(i_1,i_2=i_{2_0},\theta ={\theta }_0){\text{di}}_1$ (3.71)

• For the linear system:

$W^{\text{'}}(i_1,i_2,{\theta }_0)=\frac{1}{2}{L}_{\text{11}}(\theta )i_1^2+\frac{1}{2}{L}_{\text{22}}(\theta )i_2^2+L_{\text{12}}(\theta )i_1{i}_2$ (3.72)

$T_{\text{fld}}=\frac{\partial W_{\text{fld}}^{\text{'}}(i_1,i_2,{\theta }_0)}{\partial \theta }{\mid }_{i_1,i_2}=\frac{i_1^2}{2}\frac{{\text{dL}}_{\text{11}}(\theta )}{\mathrm{d\theta }}+\frac{i_2^2}{2}\frac{{\text{dL}}_{\text{22}}(\theta )}{\mathrm{d\theta }}+i_1{i}_2\frac{{\text{dL}}_{\text{12}}(\theta )}{\mathrm{d\theta }}$ (3.73)

• Note that (3.70) is simpler than (3.57). That is, the coenergy function is a relatively simple function of displacement.
• The use of a coenergy function of the terminal currents simplifies the determination of torque or force.
• Systems with more than two electrical terminals are handled in analogous fashion.
  • System with linear displacement:

$W_{\text{fld}}({\lambda }_{1_0},{\lambda }_{2_0},x_0)={\int }_0^{{\lambda }_{2_0}}{i}_2({\lambda }_1=\mathrm{0,}{\lambda }_2,x=x_0){\mathrm{d\lambda }}_2+{\int }_0^{{\lambda }_{1_0}}{i}_1({\lambda }_1,{\lambda }_2={\lambda }_{2_0},x=x_0){\mathrm{d\lambda }}_1$ (3.74)

$W_{\text{fld}}^{\text{'}}(i_{1_0},i_{2_0},x_0)={\int }_0^{{\lambda }_{2_0}}{\lambda }_2(i_1,i_2,x=x_0){\text{di}}_2+{\int }_0^{{\lambda }_{1_0}}{\lambda }_1(i_1,i_2=i_{2_0},x=x_0){\text{di}}_1$ 

$f_{\text{fld}}=-\frac{\partial W_{\text{fld}}({\lambda }_1,{\lambda }_2,x)}{\partial x}{\mid }_{{\lambda }_1,{\lambda }_2}$ (3.76)

$f_{\text{fld}}=-\frac{\partial W_{\text{fld}}^{\text{'}}(i_1,i_2,x)}{\partial x}{\mid }_{i_1,i_2}$ (3.77)

For a magnetically-linear system,

$W_{\text{fld}}^{\text{'}}(i_1,i_2,x)=\frac{1}{2}{L}_{\text{11}}(x)i_1^2+L_{\text{22}}(x)i_2^2+L{}_{\text{12}}(x)i_1{i}_2$ (3.78)

$f_{\text{fld}}=\frac{i_1^2}{2}\frac{{\text{dL}}_{\text{11}}(x)}{\text{dx}}+\frac{i_2^2}{2}\frac{{\text{dL}}_{\text{22}}(x)}{\text{dx}}+i_1{i}_2\frac{{\text{dL}}_{\text{12}}(x)}{\text{dx}}$ (3.79)