Difference between revisions of "ECE PhD QE PE1 2012 Problem1" - Rhea

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Problem 1

Electromechanical Energy Balance

The electromechanical energy conversion process for a device with $ J $ electrical inputs and single mechanical input follows the energy balance equations given below. It is assumed that there are no radiation energy transfers and that the inputs have frequency content sufficiently low that lumped parameter models are valid.

$ \begin{align} W_{E,j} &= W_{e,j} + W_{e\ell,j} + W_{eS,j} \\ W_M &= W_m + W_{m\ell} + W_{mS} \\ W_f &= \sum_{j=1}^J W_{e,j} + W_m - W_{f\ell} \end{align} $

Each energy term has the following associations.

• $ W_{E,j} $ is the total energy supplied to the $ j^{\text{th}} $ electrical system.
  • $ W_{e,j} $ is the energy transferred from the $ j^{\text{th}} $ electrical system to the coupling field.
  • $ W_{e\ell,j} $ is the energy loss (dissipated as heat) of the $ j^{\text{th}} $ electrical system.
  • $ W_{eS,j} $ is the energy stored (electric field or magnetic field) uniquely in the $ j^{\text{th}} $ electrical system.
• $ W_M $ is the total energy supplied to the mechanical system.
  • $ W_m $ is the energy transferred from the mechanical system to the coupling field.
  • $ W_{m\ell} $ is the energy loss (dissipated as heat) of the mechanical system.
  • $ W_{mS} $ is the energy stored (kinetic or elastic) uniquely in the mechanical system.
• $ W_f $ is the energy stored in the coupling field.
  • $ W_{f\ell} $ is the energy loss (such as core losses or dielectric losses) within the coupling field ($ W_{f\ell} = 0 $ is equivalent to the coupling field being conservative).

Since electromagnetic force $ f_e $ is defined positive in the same direction that applied force $ f $ and displacement $ x $ are defined positive, it can be said that $ W_m = - \int f_e \, dx $ since the work done by the electromagnetic force on the mechanical system is the line integral of the force with the differential displacement. (Negation arises from the way $ W_m $ is defined.)

If the voltage drop in the $ j^{\text{th}} $ electrical system associated with the coupling field is $ e_{f,j} $, then it can be said that $ W_{e,j} = \int e_{f,j} i_j \, dt $ since the energy in a circuit is the product of the voltage and current waveforms integrated over time. The energy stored in the coupling field may now be related to circuit quantities and the free body diagram. The coupling field is assumed to be conservative.

$ \begin{equation} W_f = \int \sum_{j=1}^J e_{f,j} i_j \, dt - \int f_e \, dx \end{equation} $

Coupling Field Energy

In order to calculate the field energy of the conservative coupling field as a function of flux linkages and mechanical position, it is helpful to fix the mechanical position of the device so that $ dx = 0 $ and hence $ W_m = 0 $. Thus, the equation for the coupling field reduces to $ W_f = \int \sum_{j=1}^J e_{f,j} i_j \, dt $. For a magnetic coupling field, $ e_{f,j} = \frac{d\lambda_j}{dt} $ and so a more useful equation for the field energy results.

$ \begin{equation} W_f = \int \sum_{j=1}^J i_j \, d\lambda_j \end{equation} $

For a conservative coupling field, field energy and electrical system currents may be expressed as functions of the $ J $ flux linkages and mechanical position: $ W_f = W_f(\vec{\lambda}, x) $ and $ i = i(\vec{\lambda}, x) $ where the vector $ \vec{\lambda} = \begin{bmatrix} \lambda_1 & \lambda_2 & \cdots & \lambda_J \end{bmatrix}^\mathrm{T} $ is used as shorthand. The summation and integration may be interchanged because both are linear operations.

$ \begin{equation} W_f(\vec{\lambda}, x) = \sum_{j=1}^J \int_{\lambda_j^{'} = 0}^{\lambda_j} i_j(\vec{\lambda}^{'}, x) \, d\lambda_j^{'} \end{equation} $

Electromagnetic Force Calculation

With an equation for magnetic coupling field energy in possession, the mechanical position of the device is allowed to shift again $ dx \neq 0 $, necessitating a return to the full equation of coupling field before looking at the total derivative.

$ \begin{align} W_f &= \int \sum_{j=1}^J e_{f,j} i_j \, dt - \int f_e \, dx \\ W_f(\vec{\lambda}, x) &= \sum_{j=1}^J \int_{\lambda_j^{'} = 0}^{\lambda_j} i_j(\vec{\lambda}^{'}, x) \, d\lambda_j^{'} - \int_{x^{'} = 0}^x f_e(\vec{\lambda}, x^{'}) \, dx^{'} \\ dW_f(\vec{\lambda}, x) &= \sum_{j=1}^J i_j(\vec{\lambda}, x) \, d\lambda_j - f_e(\vec{\lambda}, x) \, dx \\ f_e(\vec{\lambda}, x) \, dx &= \sum_{j=1}^J i_j(\vec{\lambda}, x) \, d\lambda_j - dW_f(\vec{\lambda}, x) \end{align} $

The total derivative of field energy $ dW_f(\vec{\lambda}, x) = \sum_{j=1}^J \left(\frac{\partial W_f(\vec{\lambda}, x)}{\partial \lambda_j} \, d\lambda_j\right) + \frac{\partial W_f(\vec{\lambda}, x)}{\partial x} \, dx $ is needed for substitution.

$ \begin{equation} f_e(\vec{\lambda}, x) \, dx = \sum_{j=1}^J i_j(\vec{\lambda}, x) \, d\lambda_j - \sum_{j=1}^J \frac{\partial W_f(\vec{\lambda}, x)}{\partial \lambda_j} \, d\lambda_j - \frac{\partial W_f(\vec{\lambda}, x)}{\partial x} \, dx \end{equation} $

By equating the coefficients of $ dx $, one obtains the final equation for electromagnetic force.

$ \begin{equation} \boxed{f_e(\vec{\lambda}, x) = -\frac{\partial W_f(\vec{\lambda}, x)}{\partial x}} \end{equation} $

Discussion

Back to PE-1, August 2012