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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_2:magnetic_circuits [2023/05/09 13:59] mexleadmin |
electrical_engineering_2:magnetic_circuits [2024/05/03 16:02] (aktuell) mexleadmin [Magnetic Circuit with two Sources] |
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Zeile 1: | Zeile 1: | ||
- | ====== 5. Magnetic Circuits ====== | + | ====== 5 Magnetic Circuits ====== |
< | < | ||
Zeile 9: | Zeile 9: | ||
In this chapter, we will investigate, | In this chapter, we will investigate, | ||
- | ===== 5.1 Linear | + | ===== 5.1 Linear |
For the upcoming calculations, | For the upcoming calculations, | ||
Zeile 168: | Zeile 168: | ||
Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions: | Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions: | ||
- | - $l=35.8~\rm cm$, $d=1.9~\rm cm$ | + | - $l=35.8~\rm cm$, $d=1.90~\rm cm$ |
- | - $l=22.5~\rm cm$, $d=1.5~\rm cm$ | + | - $l=11.1~\rm cm$, $d=1.50~\rm cm$ |
- | <button size=" | + | # |
- | - $1.5\cdot 10^5 ~\rm {{1}\over{H}}$ | + | The magnetic resistance is given by: |
- | - $3.0\cdot 10^5 ~\rm {{1}\over{H}}$ | + | \begin{align*} |
+ | \ R_{\rm m} & | ||
+ | \end{align*} | ||
- | </collapse> | + | With |
+ | * the area $ A = \left({{d}\over{2}}\right)^2 \cdot \pi $ | ||
+ | * the vacuum magnetic permeability $\mu_{0}=4\pi\cdot 10^{-7} ~\rm H/m$, and | ||
+ | * the relative permeability $\mu_{\rm r}=1$. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$ | ||
+ | - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$ | ||
+ | # | ||
</ | </ | ||
Zeile 353: | Zeile 365: | ||
The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$? | The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$? | ||
- | ==== Magnetic Circuit with 2 Sources ==== | + | ==== Magnetic Circuit with two Sources ==== |
To get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08> | To get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08> | ||
Zeile 360: | Zeile 372: | ||
< | < | ||
- | The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby the windings mark the direction of the windings: \\ | + | The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby the windings mark the direction of the windings, and therefore the sign of the generated flux. \\ |
- | All the fluxes | + | All the fluxes |
- | When there is a current flowing into a non-marked | + | When there is a current flowing into a __non-marked |
To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ | To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ | ||
Zeile 393: | Zeile 405: | ||
$k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$): | $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$): | ||
- | \begin{align*} k_{21} = {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*} | + | \begin{align*} k_{21} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*} |
- | When $k_{21}=100~\%$, | + | The sign of $k_{21}$ depends on the direction of $\Phi_{21}$ relative to $\Phi_{22}$! If the directions are the same, the positive sign applies, if the directions are oposite, the minus sign applies. |
+ | |||
+ | When $k_{21}=+100~\%$, there is no flux in the middle leg but only in the second coil and in the same direction as the flux that originates from second coil. \\ | ||
+ | When $k_{21}=-100~\%$, | ||
For $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling. | For $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling. | ||
Zeile 409: | Zeile 424: | ||
& | & | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | Note, that also $M_{21}$ and $M_{12}$ can be either positiv or negative, depending on the sign of the coupling coefficients. | ||
The formula is finally: | The formula is finally: | ||
Zeile 473: | Zeile 490: | ||
* the mutual inductances are equal: $M_{12} = M_{21} = M$ | * the mutual inductances are equal: $M_{12} = M_{21} = M$ | ||
* the mutual inductance $M$ is: $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ | * the mutual inductance $M$ is: $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ | ||
- | * The resulting *total coupling* $k$ is given as \begin{align*} k = \sqrt{k_{12}\cdot k_{21}} \end{align*} | + | * The resulting |
==== Effects in the electric Circuits ==== | ==== Effects in the electric Circuits ==== | ||
Zeile 499: | Zeile 516: | ||
< | < | ||
- | In this case, the **mutual induction | + | In this case, the **mutual induction |
The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
Zeile 513: | Zeile 530: | ||
< | < | ||
- | In this case, the **mutual induction | + | In this case, the **mutual induction |
The formula of the shown circuitry is then: | The formula of the shown circuitry is then: | ||
\begin{align*} | \begin{align*} | ||
- | u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ | + | u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\ |
- | u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ | + | u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ |
\end{align*} | \end{align*} | ||
- | <panel type=" | + | <panel type=" |
A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. | A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. | ||
On the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$. | On the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$. | ||
- | - The coils shall pass the currents with positive polarity (see the top image in <imgref ImgEx14> | + | - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14> |
- | - The coils shall pass the currents with negative polarity (see the bottom | + | - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14> |
< | < | ||
- | <button size=" | + | # |
- | - $0.40 ~\rm mVs$ | + | The resulting flux can be derived from a superposition of the individual fluxes |
- | - $0.10 ~\rm mVs$ | + | |
- | </collapse> | + | **Step 1 - Draw an equivalent magnetic circuit** |
+ | |||
+ | Since there are no branches all of the core can be lumped to a single magnetic resistance (see <imgref ImgEx14circ> | ||
+ | < | ||
+ | |||
+ | **Step 2 - Get the absolute values of the individual fluxes** | ||
+ | |||
+ | Hopkinson' | ||
+ | It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_m$. \\ | ||
+ | It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\ | ||
+ | |||
+ | \begin{align*} | ||
+ | \theta_x | ||
+ | N_x \cdot I_x &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\ | ||
+ | \rightarrow \Phi_x | ||
+ | = {{1}\over{R_{\rm m} }} \cdot N_x \cdot I_x \\ | ||
+ | \end{align*} | ||
+ | |||
+ | With the given values we get: $R_{\rm m} = 495 {\rm {kA}\over{Vs}}$ | ||
+ | |||
+ | **Step 3 - Get the signs/ | ||
+ | |||
+ | The < | ||
+ | The fluxes have to be added regarding these directions and the given direction of the flux in question. | ||
+ | < | ||
+ | |||
+ | Therefore, the formulas are | ||
+ | \begin{align*} | ||
+ | \Phi_{\rm A} & | ||
+ | & | ||
+ | & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\ | ||
+ | \Phi_{\rm B} & | ||
+ | & | ||
+ | & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | |||
+ | # | ||
+ | - $0.10 ~\rm mVs$ | ||
+ | - $0.40 ~\rm mVs$ | ||
+ | # | ||
</ | </ | ||
Zeile 558: | Zeile 616: | ||
\boxed{W_m = {{1}\over{2}}L\cdot I^2 } | \boxed{W_m = {{1}\over{2}}L\cdot I^2 } | ||
\end{align*} | \end{align*} | ||
+ | |||
+ | |||
+ | |||
+ | |||
==== magnetic Energy of a magnetic Circuit ==== | ==== magnetic Energy of a magnetic Circuit ==== | ||
- | With this formula also the stored energy in a magnetic circuit can be calculated. | + | With this formula also the stored energy in a magnetic circuit can be calculated. For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: \begin{align*} \boxed{W_{\rm m} = {{1}\over{2}} \Psi \cdot I = {{1}\over{2}} {{\Psi^2}\over{L}}= {{1}\over{2}}{{\Phi^2 }\over{N^2 \cdot L}} = {{1}\over{2}} \Phi^2 \cdot R_{\rm m} = {{1}\over{2}}{{\theta^2 }\over{R_{\rm m}}}} \end{align*} |
- | For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: | + | |
- | \begin{align*} | + | |
- | \boxed{W_{\rm m} = {{1}\over{2}} | + | |
- | = {{1}\over{2}}{{\Psi^2 }\over{L}}} | + | |
- | \end{align*} | + | |
==== magnetic Energy of a toroid Coil ==== | ==== magnetic Energy of a toroid Coil ==== | ||
Zeile 717: | Zeile 774: | ||
The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$. | The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$. | ||
This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https:// | This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https:// | ||
+ | |||
+ | ==== Moving a Plate into an Air Gap ==== | ||
+ | |||
+ | < | ||
==== Switch Reluctance Motor ==== | ==== Switch Reluctance Motor ==== |