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

Figure 3.4Integration paths for Wfld

$W_{\text{fld}}({\lambda }_0,x_0)=\underset{\text{path 2a}}{\int }{\text{dW}}_{\text{fld}}+\underset{\text{path 2b}}{\int }{\text{dW}}_{\text{fld}}$ (3.17)

On path 2a, $\mathrm{d\lambda }=0$ and $f_{\text{fld}}=0$ . Thus, ${\text{dW}}_{\text{fld}}=0$ on path 2a.

On path 2b, dx  0 .

Therefore, (3.17) reduces to the integral of $\text{id}\lambda$ over path 2b.

$W_{\text{fld}}({\lambda }_0,x_0)={\int }_0^{{\lambda }_0}i(\lambda ,x_0)\mathrm{d\lambda }$ (3.18)

For a linear system in which $\lambda$ is proportional to i , (3.18) gives

$W_{\text{fld}}(\lambda ,x)={\int }_0^{\lambda }i({\lambda }^{\text{'}},x)d{\lambda }^{\text{'}}={\int }_0^{\lambda }\frac{{\lambda }^{\text{'}}}{L(x)}d{\lambda }^{\text{'}}=\frac{1}{2}\frac{{\lambda }^2}{L(x)}$ (3.19)

• V : the volume of the magnetic field

$W_{\text{fld}}={\int }_v({\int }_0^{B}H\text{.}d{B}^{\text{'}})\text{dV}$ (3.20)

If $B=\mathrm{\mu H}$ ,

$W_{\text{fld}}={\int }_v(\frac{B^2}{\mathrm{2\mu }})\text{dV}$ (3.21)

§3.4 Determination of Magnetic Force and Torque form Energy

• The magnetic stored energy $W_{\text{fld}}$ is a state function, determined uniquely by the values of the independent state variables $\lambda$ and x.

${\text{dW}}_{\text{fld}}(\lambda ,x)=\text{id}\lambda -f_{\text{fld}}\text{dx}$ (3.22)

$\text{dF}(x_1,x_2)=\frac{\partial F}{\partial x_1}{\mid }_{x_2}{\text{dx}}_1+\frac{\partial F}{\partial x_2}{\mid }_{x_1}{\text{dx}}_2$ (3.23)

${\text{dW}}_{\text{fld}}(\lambda ,x)=\frac{\partial W_{\text{fld}}}{\partial \lambda }{\mid }_x\mathrm{d\lambda }+\frac{\partial W_{\text{fld}}}{\partial x}{\mid }_{\lambda }\text{dx}$ (3.24)

Comparing (3.22) with (3.24) gives (3.25) and (3.26):

$i=\frac{\partial W_{\text{fld}}(\lambda ,x)}{\partial \lambda }{\mid }_x$ (3.25)

$f_{\text{fld}}=-\frac{\partial W_{\text{fld}}(\lambda ,x)}{\partial x}{\mid }_{\lambda }$ (3.26)

• Once we know $W_{\text{fld}}$ as a function of $\lambda$ and as a function of $\lambda$ and i( $\lambda$ , x) .
• Equation (3.26) can be used to solve for the mechanical force $f_{\text{fld}}(\lambda ,x)$ .The partial derivative is taken while holding the flux linkages $\lambda$ constant.
• For linear magnetic systems for which $\lambda =L(x)i$ , the force can be found as

$f_{\text{fld}}=-\frac{\partial }{\partial x}\frac{1}{2}\frac{{\lambda }^2}{L(x)}{\mid }_{\lambda }=\frac{{\lambda }^2}{\mathrm{2L}(x{)}^2}\frac{\text{dL}(x)}{\text{dx}}$ (3.27)

$$ $f_{\text{fld}}=\frac{i^2}{2}\frac{\text{dL}(x)}{\text{dx}}$ (3.28)

• For a system with a rotating mechanical terminal, the mechanical terminal variables become the angular displacement $\theta$ and the torque $T_{\text{fld}}$ .

${\text{dW}}_{\text{fld}}(\lambda ,\theta )=\text{id}\lambda -T_{\text{fld}}\mathrm{d\theta }$ (3.29)

$T_{\text{fld}}=-\frac{\partial W_{\text{fld}}(\lambda ,\theta )}{\partial \theta }{\mid }_{\lambda }$ (3.30)

• For linear magnetic systems for which $\lambda =L(\theta )i$ :

$W_{\text{fld}}(\lambda ,\theta )=\frac{1}{2}\frac{{\lambda }^2}{L(\theta )}$ (3.31)

$T_{\text{fld}}=-\frac{\partial }{\partial \theta }\frac{1}{2}\frac{{\lambda }^2}{L(\theta )}{\mid }_{\lambda }=\frac{1}{2}\frac{{\lambda }^2}{L(\theta {)}^2}\frac{\text{dL}(\theta )}{\mathrm{d\theta }}$ (3.32)

$T_{\text{fld}}=\frac{i^2}{2}\frac{\text{dL}(\theta )}{\mathrm{d\theta }}$ (3.33)

§3.5 Determination of Magnetic Force and Torque from Coenergy

• Recall that in §3.4, the magnetic stored energy $W_{\text{fld}}$ is a state function, determined uniquely by the values of the independent state variables $\lambda$ and x .

${\text{dW}}_{\text{fld}}(\lambda ,x)=\text{id}\lambda -f_{\text{fld}}\text{dx}$ (3.34)

${\text{dW}}_{\text{fld}}(\lambda ,x)=\frac{\partial W_{\text{fld}}}{\partial \lambda }{\mid }_x\mathrm{d\lambda }+\frac{\partial W_{\text{fld}}}{\partial x}{\mid }_{\lambda }\text{dx}$ (3.35)

$i=\frac{\partial W_{\text{fld}}(\lambda ,x)}{\partial \lambda }{\mid }_x$ (3.36)

$f_{\text{fld}}=\frac{\partial W_{\text{fld}}(\lambda ,x)}{\partial x}{\mid }_{\lambda }$ (3.37)

• Coenergy: from which the force can be obtained directly as a function of the current.The selection of energy or coenergy as the state function is purely a matter of convenience.

• The coenergy $W_{\text{fld}}^{\text{'}}(i,x)$ is defined as a function of i and x such that

$W_{\text{fld}}^{\text{'}}(i,x)=\mathrm{i\lambda }-W_{\text{fld}}(\lambda ,x)$ (3.38)

$d(\mathrm{i\lambda })=\text{id}\lambda =\lambda \text{di}$ (3.39)

$d{W}_{\text{fld}}^{\text{'}}(i,x)=d(\mathrm{i\lambda })-{\text{dW}}_{\text{fld}}(\lambda ,x)$ (3.40)

$d{W}_{\text{fld}}^{\text{'}}(i,x)=\lambda \text{di}+f_{\text{fld}}\text{dx}$ (3.41)

• From (3.37), the coenergy $W_{\text{fld}}^{\text{'}}(i,x)$ can be seen to be a state function of the two independent variables i and x .

$d{W}_{\text{fld}}^{\text{'}}(i,x)=\frac{\partial W_{\text{fld}}^{\text{'}}}{\partial i}{\mid }_x\text{di}+\frac{\partial W_{\text{fld}}^{\text{'}}}{\partial x}{\mid }_i\text{dx}$ (3.42)

$\lambda =\frac{\partial W_{\text{fld}}^{\text{'}}(i,x)}{\partial i}{\mid }_x$ (3.43)

$f_{\text{fld}}=\frac{\partial W_{\text{fld}}^{\text{'}}(i,x)}{\partial x}{\mid }_i$ (3.44)

• For any given system, (3.26) and (3.40) will give the same result.
• By analogy to (3.18) in §3.3, the coenergy can be found as (3.41)

$W_{\text{fld}}({\lambda }_0,x_0)={\int }_0^{{\lambda }_0}i(\lambda ,x_0)\mathrm{d\lambda }$ (3.42)

$W_{\text{fld}}^{\text{'}}(i,x)={\int }_{}^{}\lambda (i^{\text{'}},x)d{i}^{\text{'}}$ (3.43)

For linear magnetic systems for which $\lambda =L(x)i$ ,

$W_{\text{fld}}^{\text{'}}(i,x)=\frac{1}{2}L(x)i^2$ (3.44)

$f_{\text{fld}}=\frac{i^2}{2}\frac{\text{dL}(x)}{\text{dx}}$ (3.45)

• For a system with a rotating mechanical displacement,

$W_{\text{fld}}^{\text{'}}(i,\theta )={\int }_0^i\lambda (i^{\text{'}},\theta )d{i}^{\text{'}}$ (3.46)

$T_{\text{fld}}=\frac{\partial W_{\text{fld}}^{\text{'}}(i,\theta )}{\partial \theta }{\mid }_i$ (3.47)

If the system is magnetically linear,

$W_{\text{fld}}^{\text{'}}(i,\theta )=\frac{1}{2}L(\theta )i^2$ (3.48)

$T_{\text{fld}}=\frac{i^2}{2}\frac{\text{dL}(\theta )}{\mathrm{d\theta }}$ (3.49)

(3.47) is identical to the expression given by (3.33).

• In field-theory terms, for soft magnetic materials