Three-Phase system problem

  • #1
208
1

Homework Statement


In the given AC circuit, electromotive forces form a symmetric and direct three-phase system.
Angular frequency [itex]\omega[/itex], magnitudes [itex]U_{AB},U_{BC},U_{CA}=U[/itex],
resistance [itex]R[/itex], inductance [itex]L[/itex] and coefficient of inductive coupling [itex]k\neq 1[/itex]
are given.
Find capacitance [itex]C[/itex] such that the three-phase generator develops only the active power and find that active power.

2. The attempt at a solution
[itex][/itex]
Total active power is given by [tex]P=3\cdot\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}[/tex]
where [itex]\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}[/itex] is the active power of one phase, and
[itex]\frac{U}{\sqrt 3}[/itex] is the voltage of one phase.

Generator needs to develop only the active power, so the reactive power must be zero,
[tex]Q=\mathfrak{I}(\underline{U_{AB}}{\underline{I_1}}^{*}+\underline{U_{BC}}{\underline{I_2}}^{*}+\underline{U_{CA}}{\underline{I_3}}^{*})=0[/tex]

[tex]L_{12}=+kL\Rightarrow \underline{U_{AB}}=j\omega L\underline{I_1}+j\omega kL\underline{I_2},\underline{U_{BC}}=j\omega L\underline{I_2}+j\omega kL\underline{I_1},\underline{U_{CA}}=\frac{\underline{I_3}}{j\omega C}[/tex]
[tex]\Rightarrow \underline{I_1}=\frac{\underline{U_{AB}}-k\underline{U_{BC}}}{j\omega L(1-k^2)},\underline{I_2}=\frac{\underline{U_{BC}}-k\underline{U_{AB}}}{j\omega L(1-k^2)}, \underline{I_3}=j\omega C\underline{U_{CA}}[/tex]

First, we find [itex]\underline{U_{AB}}{\underline{I_1}}^{*}[/itex].
[tex]\underline{I_1}=\frac{\underline{U_{AB}}-k\underline{U_{BC}}}{j\omega L(1-k^2)}\cdot \frac{-j\omega L(1-k^2)}{-j\omega L(1-k^2)}=\frac{-j\omega L(1-k^2)\underline{U_{AB}}+j\omega L(1-k^2)k\underline{U_{BC}}}{{\omega}^2L^2(1-k^2)^2}=j\frac{k\underline{U_{BC}}-\underline{U_{AB}}}{\omega L(1-k^2)}[/tex]
[tex]\Rightarrow \underline{{I_1}^{*}}=-j\frac{k\underline{U_{BC}}-\underline{U_{AB}}}{\omega L(1-k^2)}[/tex]

[tex]\underline{U_{AB}}{\underline{I_1}}^{*}=-j\frac{kU^2-U^2}{\omega L(1-k^2)}=j\frac{U^2}{\omega L(1+k)}[/tex]

We get [itex]\underline{U_{BC}}{\underline{I_2}}^{*}=\underline{U_{AB}}{\underline{I_1}}^{*}[/itex]

[tex]\underline{U_{CA}}{\underline{I_3}}^{*}=-j\omega CU^2[/tex]

Now, [tex]\underline{U_{AB}}{\underline{I_1}}^{*}+\underline{U_{BC}}{\underline{I_2}}^{*}+\underline{U_{CA}}{\underline{I_3}}^{*}=j\frac{U^2(2-{\omega}^2LC(1+k))}{\omega L(1+k)}[/tex]

[tex]Q=\frac{U^2(2-{\omega}^2LC(1+k))}{\omega L(1+k)}[/tex]
If [itex]Q=0[/itex], then [tex]C=\frac{2}{{\omega}^2L(1+k)}[/tex]

In my book, it says that the solution is [tex]Q=U^2\left(\frac{2+k}{\omega L(1-k^2)}-\omega C\right)=0\Rightarrow C=\frac{2+k}{{\omega}^2L(1-k^2)}[/tex]

Is it a mistake in the book, or in my solution?
 

Attachments

  • circuit.PNG
    circuit.PNG
    1.7 KB · Views: 159

Answers and Replies

  • #2
18,416
8,233
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

Related Threads on Three-Phase system problem

  • Last Post
Replies
11
Views
842
  • Last Post
Replies
11
Views
910
  • Last Post
Replies
1
Views
609
Replies
2
Views
783
  • Last Post
Replies
2
Views
717
  • Last Post
Replies
1
Views
748
  • Last Post
Replies
2
Views
3K
Replies
0
Views
2K
  • Last Post
Replies
13
Views
2K
Replies
0
Views
9K
Top