Difference between revisions of "ECE PhD QE ES 2007 Problem2" - Rhea

Latest revision as of 12:49, 16 January 2018

Problem 2

Setting Up the Problem

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 ($ \textit{p} $ denotes the Heaviside operator).

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

Method 1: Exhaustive Calculation

The electrical energy supplied to this transformer for $ t \geq 1 \, \textrm{s} $ 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 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} $

All the zero terms can be removed, and time integrals with inductance are replaced with current integrals using the currents at the time bounds.

$ \begin{align} W_E &= \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{i_1 = 10 \, \textrm{A}}^{6.32 \, \textrm{A}} L_{m1} i_1 \, di_1 + \int_{i_2 = 0}^{0} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \, di_2 \\ &{} + \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau + \int_{i_2 = 0}^{i_2(t), \, t \geq 1 \, \textrm{s}} L_{m2} i_2 \, di_2 + \int_{i_1 = 0}^{0} \cancelto{1}{\frac{N_1}{N_2}} L_{m2} i_2 \, di_1 \end{align} $

Any integration between identical bounds will have a zero result. Because the secondary winding is short circuited $ v_2(t) = 0, \; t \geq 1 \, \textrm{s} $, it is further known that $ \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau + \int_{i_2 = 0}^{i_2(t), \, t \geq 1 \, \textrm{s}} L_{m2} i_2 \, di_2 = 0 $ as a consequence of the voltage equations. The electrical energy waveform must account for the period $ 0 \leq t < 1 \, \textrm{s} $ and revert to time integrals using $ \frac{di_1}{dt} = +10 e^{-t} \, \frac{\textrm{A}}{\textrm{s}} $.

$ \begin{equation} W_E(t) = \begin{cases} \int_{\tau = 0}^{t} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{t} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases} \end{equation} $

$ \begin{equation} \boxed{W_E(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $

Method 2: Fast Calculation

The electrical energy supplied to this transformer for $ t \geq 1 \, \textrm{s} $ may be found for each time period and each winding. $ \tau $ is used as a dummy variable for integrals.

$ \begin{equation} W_E = \int_{\tau = 0}^{1 \, \textrm{s}} v_1 i_1 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} v_2 \cancelto{0}{i_2} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} v_1 \cancelto{0}{i_1} \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} \cancelto{0}{v_2} i_2 \, d\tau \end{equation} $

By removing the zero terms, it can be shown that nonzero electrical energy is only supplied to the transformer in winding 1 during the first second of time. An answer is readily obtained using the equation for $ v_1 $ (including $ \lambda_1 $), substituting the given equation for $ i_1(t) $, and removing the leakage terms since $ L_{\ell 1} = L_{\ell 2} = 0 $.

$ \begin{equation} W_E(t) = \begin{cases} \int_{\tau = 0}^{t} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{t} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau + \int_{\tau = 0}^{t} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \cancelto{0}{\frac{di_2}{d\tau}} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} r_1 i_1^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} L_{m1} i_1 \frac{di_1}{d\tau} \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} \cancelto{1}{\frac{N_2}{N_1}} L_{m1} i_1 \cancelto{0}{\frac{di_2}{d\tau}} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases} \end{equation} $

Since winding 2 is open circuited during the first second of time, the time derivative of the winding 2 current is also zero. The final answer is obtained by removing terms, finding $ \frac{di_1}{dt} = +10 e^{-t} \, \frac{\textrm{A}}{\textrm{s}} $, and substituting parameters.

$ \begin{equation} \boxed{W_E(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $

Remaining Energy Calculations

The electrical energy balance can determine the meaning of each component, where $ W_{e\ell} $ is electrical energy losses, $ W_{eS} $ is electrical energy storage, and $ W_{e} $ is electrical energy supplied to the coupling field.

$ \begin{equation} W_E = W_{e\ell} + W_{eS} + W_{e} \end{equation} $

In most cases, energy terms with resistance $ r $ and current squared are electrical energy losses, those with leakage inductance $ L_{\ell} $ or discrete winding inductors are electrical energy storage, and those with magnetizing inductance $ L_{m} $ are electrical energy supplied to the coupling field. The coupling field energy comes from electrical and mechanical sources in general. Since all transformer components are stationary, there is no mechanical energy supplied to the coupling field $ W_{m} $.

$ \begin{equation} W_f = W_e + \cancelto{0}{W_m} - \cancelto{0}{W_{f\ell}} \end{equation} $

Since no terms were identified as electrical energy storage, the only remaining term in $ W_E(t) $ for $ 0 \leq t < 1 $ must belong to $ W_e(t) $ by the electrical energy balance, and thus $ W_f(t) $ assuming no coupling field losses.

For $ t \geq 1 \, \textrm{s} $, energy may still move around in winding 2 by exchanging energy between the coupling field and electrical losses $ \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau $, even though the winding is short circuited. There is no place for coupling field energy to go in winding 1 when $ t \geq 1 \, \textrm{s} $ because the winding is open circuited (every term in the voltage equation will be multiplied by $ i_1 = 0 $ when creating energy terms). The energy balance for $ t \geq 1 \, \textrm{s} $ becomes $ \cancelto{0}{W_E} = W_{e\ell} + \cancelto{0}{W_{eS}} + W_{e} $, implying that electrical energy supplied to the coupling field $ \int_{\tau = 1 \, \textrm{s}}^{t} L_{m2} i_2 \frac{di_2}{d\tau} \, d\tau $ is opposite (meaning energy is sourced from the coupling field) the electrical energy lost as heat.

A detour is needed to find $ i_2(t), \; t \geq 1 \, \textrm{s} $. The energy stored in the coupling field cannot change instantaneously, thus relations may be drawn as follows.

$ \begin{align} W_f(t = 1^- \, \textrm{s}) &= W_f(t = 1^+ \, \textrm{s}) \\ \frac{1}{2} L_{m1} i_1^2(t = 1^+ \, \textrm{s}) &= \frac{1}{2} L_{m2} i_2^2(t = 1^+ \, \textrm{s}) \\ i_2(t = 1^+ \, \textrm{s}) &= + \cancelto{1}{\frac{L_{m1}}{L_{m2}}} i_1(t = 1^+ \, \textrm{s}) \\ i_2(t = 1^+ \, \textrm{s}) &= +6.32 \, \textrm{A} \end{align} $

The positive sign arises from the winding direction (influence magnetic flux linkage in magnetic equivalent circuit), current reference direction, and intuition about the flow of current when winding 2 is short circuited. The current waveform is obtained by solving the equation for $ v_2 $ (including $ \lambda_2 $) using a first-order ordinary differential equation solution.

$ \begin{align} \cancelto{0}{v_2} &= r_2 i_2 + \cancelto{0}{L_{\ell 2}} \frac{di_2}{dt} + L_{m2} \left(\frac{di_2}{dt} + \frac{N_1}{N_2} \cancelto{0}{\frac{di_1}{dt}}\right), \; t \geq 1 \, \textrm{s} \\ i_2(t), \; t \geq 1 \, \textrm{s} &= i_2(t = \infty) + \left[i_2(t = 1^+ \, \textrm{s}) - i_2(t = \infty)\right] e^{-\frac{(t - 1 \, \textrm{s})}{\tau_2}} \\ i_2(t), \; t \geq 1 \, \textrm{s} &= 6.32 \, \textrm{A} \left(e^{-\frac{(t - 1 \, \textrm{s})}{1 \, \textrm{s}}}\right) \end{align} $

In solution for $ i_2(t) $, $ i_2(t = \infty) = 0 $ is the current in winding 2 after infinite time (no sources are present to sustain it) and $ \tau_2 = \frac{L_{m2}}{r_2} = \frac{1 \, \textrm{H}}{1 \, \Omega} = 1 \, \textrm{s} $ is the time constant for current in winding 2. The field energy may finally be written using $ \int_{\tau = 1 \, \textrm{s}}^{t} L_{m2} i_2 \frac{di_2}{d\tau} \, d\tau $ and $ \frac{di_2}{dt} = -6.32 e^{-(t - 1)} \, \frac{\textrm{A}}{\textrm{s}} $.

$ \begin{equation} \boxed{W_f(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right) e^{-\tau} \, d\tau - \int_{\tau = 1 \, \textrm{s}}^{t} (6.32)^2 \, \textrm{W} \left(e^{-(\tau - 1)}\right)^2 \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $

With knowledge of $ \int_{\tau = 1 \, \textrm{s}}^{t} r_2 i_2^2 \, d\tau $ and $ i_2(t) $, $ W_{e\ell}(t) $ follows suit.

$ \begin{equation} \boxed{W_{e\ell}(t) = \begin{cases} \int_{\tau = 0}^{t} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau, \; &0 \leq t < 1 \, \textrm{s} \\ \int_{\tau = 0}^{1 \, \textrm{s}} 100 \, \textrm{W} \left(1 - e^{-\tau}\right)^2 \, d\tau + \int_{\tau = 1 \, \textrm{s}}^{t} (6.32)^2 \, \textrm{W} \left(e^{-(\tau - 1)}\right)^2 \, d\tau, \; &t \geq 1 \, \textrm{s} \end{cases}} \end{equation} $

Discussion

Back to ES-1, August 2007