ECE PhD QE ES 2007 Problem2 - Rhea

Question 2

The following information about the circuit model of the transformer may be obtained from the figure:

• Winding resistances: $ r_1 = r_2 = 1 \, \Omega $
• Winding leakage inductances: $ L_{\ell 1} = L_{\ell 2} = 0 $
• Magnetizing inductances: $ L_{m1} = L_{m2} = 1 \, \textrm{H} $
• Winding turns: $ N_1 = N_2 = 10 $

The following information about the circuit excitation and state of the transformer is given in the text (note that it is easier to avoid referred quantities because of how the information is given):

• Initial primary winding voltage: $ v_1(0) = 10 \, \textrm{V} $
• Secondary winding current (initially open circuited): $ i_2(t) = 0, \; 0 \leq t < 1 \, \textrm{s} $
• Primary winding current (becomes open circuited): $ i_1(t) = \begin{cases} 10 \left(1 - e^{-t}\right) \, \textrm{A} \; &0 \leq t < 1 \, \textrm{s} \\ 0 \; &t \geq 1 \, \textrm{s} \end{cases} $
  • For convenience, it is given that $ i_1 \left(t = 1^- \, \textrm{s}\right) = 6.32 \, \textrm{A} $
• Secondary winding voltage (becomes short circuited): $ v_2(t) = 0, \; t \geq 1 \, \textrm{s} $

The electrical energy provided to the machine is denoted $ W_E $.

$ \begin{equation} W_E = \sum_{k = 1}^{N} \int v_k i_k \, dt \end{equation} $

From KVL, the voltage equations needed for the energy calculation are found ($ p $ denotes the Heaviside operator).

$ \begin{align} v_1 &= r_1 i_1 + p\lambda_1 \\ v_2 &= r_2 i_2 + p\lambda_2 \end{align} $

The flux linkage equations for a transformer are used. Recall that the magnetic flux linked to winding 1 due to a current in winding 2 is $ \Phi_{1,2} = L_{m1} \frac{N_2}{N_1} i_2 $.

$ \begin{align} \lambda_1 &= L_{\ell 1} i_1 + L_{m1} \left(i_1 + \frac{N_2}{N_1} i_2\right) \\ \lambda_2 &= L_{\ell 2} i_2 + L_{m2} \left(i_2 + \frac{N_1}{N_2} i_1\right) \end{align} $

The electrical energy supplied to this transformer may be found by combining the voltage equations and flux linkage equations to the first equation for $ W_E $. The leakage terms are omitted since $ L_{\ell 1} = L_{\ell 2} = 0 $. $ \tau $ is used as a dummy variable for indefinite integrals.

$ \begin{align} W_E &= \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} r_1 \cancelto{0}{i_1^2} \, d\tau \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} L_{m1} \cancelto{0}{i_1} \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 0}^{1 \, \textrm{s}} \frac{N_2}{N_1} L_{m1} i_1 \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} \frac{N_2}{N_1} L_{m1} \cancelto{0}{i_1} \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} r_2 \cancelto{0}{i_2^2} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau \\ &{}+ \int_{\tau = 0}^{1 \, \textrm{s}} L_{m2} \cancelto{0}{i_2} \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} L_{m2} i_2 \frac{di_2}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 0}^{1 \, \textrm{s}} \frac{N_1}{N_2} L_{m2} \cancelto{0}{i_2} \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} + \int_{\tau = 1 \, \textrm{s}}^{t} \frac{N_1}{N_2} L_{m2} i_2 \frac{di_1}{\cancel{d\tau}} \, \cancel{d\tau} \\ \end{align} $

Discussion

Back to ES-1, August 2007