(Created page with givens and basis equations) |
(Wrote expansion of electrical energy) |
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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): | 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: <math>v_1(0) = 10 \, \textrm{V}</math> | * Initial primary winding voltage: <math>v_1(0) = 10 \, \textrm{V}</math> | ||
| − | * Secondary winding current (initially | + | * Secondary winding current (initially open circuited): <math>i_2(t) = 0, \; 0 \leq t < 1 \, \textrm{s}</math> |
| − | * Primary winding current (becomes | + | * Primary winding current (becomes open circuited): <math>i_1(t) = |
\begin{cases} 10 \left(1 - e^{-t}\right) \, \textrm{A} \; &0 \leq t < 1 \, \textrm{s} \\ | \begin{cases} 10 \left(1 - e^{-t}\right) \, \textrm{A} \; &0 \leq t < 1 \, \textrm{s} \\ | ||
0 \; &t \geq 1 \, \textrm{s} | 0 \; &t \geq 1 \, \textrm{s} | ||
\end{cases}</math> | \end{cases}</math> | ||
** For convenience, it is given that <math>i_1 \left(t = 1^- \, \textrm{s}\right) = 6.32 \, \textrm{A}</math> | ** For convenience, it is given that <math>i_1 \left(t = 1^- \, \textrm{s}\right) = 6.32 \, \textrm{A}</math> | ||
| − | * Secondary winding voltage (becomes | + | * Secondary winding voltage (becomes short circuited): <math>v_2(t) = 0, \; t \geq 1 \, \textrm{s}</math> |
The electrical energy provided to the machine is denoted <math>W_E</math>. | The electrical energy provided to the machine is denoted <math>W_E</math>. | ||
| Line 34: | Line 34: | ||
<math>\begin{equation} | <math>\begin{equation} | ||
W_E = \sum_{k = 1}^{N} \int v_k i_k \, dt | W_E = \sum_{k = 1}^{N} \int v_k i_k \, dt | ||
| − | |||
\end{equation}</math> | \end{equation}</math> | ||
| − | From KVL, the voltage equations are found (<math>p</math> denotes the Heaviside operator). | + | From KVL, the voltage equations needed for the energy calculation are found (<math>p</math> denotes the Heaviside operator). |
<math>\begin{align} | <math>\begin{align} | ||
v_1 &= r_1 i_1 + p\lambda_1 \\ | v_1 &= r_1 i_1 + p\lambda_1 \\ | ||
v_2 &= r_2 i_2 + p\lambda_2 | v_2 &= r_2 i_2 + p\lambda_2 | ||
| − | |||
\end{align}</math> | \end{align}</math> | ||
| Line 51: | Line 49: | ||
\lambda_1 &= L_{\ell 1} i_1 + L_{m1} \left(i_1 + \frac{N_2}{N_1} i_2\right) \\ | \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) | \lambda_2 &= L_{\ell 2} i_2 + L_{m2} \left(i_2 + \frac{N_1}{N_2} i_1\right) | ||
| − | \ | + | \end{align}</math> |
| + | |||
| + | The electrical energy supplied to this transformer may be found by combining the voltage equations and flux linkage equations to the first equation for <math>W_E</math>. The leakage terms are omitted since <math>L_{\ell 1} = L_{\ell 2} = 0</math>. <math>\tau</math> is used as a dummy variable for indefinite integrals. | ||
| + | |||
| + | <math>\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}</math> | \end{align}</math> | ||
Revision as of 15:18, 10 January 2018
Answers and Discussions for
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} $
