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electrical_engineering_2:polyphase_networks [2023/09/19 23:52] mexleadmin |
electrical_engineering_2:polyphase_networks [2024/06/18 02:38] mexleadmin [Excercises] |
Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </callout> | Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </callout> |
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Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \sin \varphi = S \cdot \sin \varphi $ and the reactive power $Q = S \cdot \cos \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02>). | Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \cos \varphi = S \cdot \cos \varphi $ and the reactive power $Q = S \cdot \sin \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02>). |
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<WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </WRAP> | <WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </WRAP> |
The phase currents are given as the currents through a single line: $I_1$, $I_2$, $I_3$. \\ | The phase currents are given as the currents through a single line: $I_1$, $I_2$, $I_3$. \\ |
The potential of the star point is called **neutral** $\rm N$ </WRAP> | The potential of the star point is called **neutral** $\rm N$ </WRAP> |
* **Star-voltages** $U_\rm Y$ (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: //Sternspannungen//): the voltages of the lines can be also measured or used referring to the neutral potential. | * **Star Voltages** $U_\rm Y$ (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: //Sternspannungen//): the voltages of the lines can be also measured or used referring to the neutral potential. |
<WRAP> <imgcaption imageNo13 | Example of an Three-Phase System></imgcaption> {{drawio>ExampleThreePhaseSystem.svg}} </WRAP> | <WRAP> <imgcaption imageNo13 | Example of an Three-Phase System></imgcaption> {{drawio>ExampleThreePhaseSystem.svg}} </WRAP> |
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The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages. | The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages. |
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==== 7.2.3 Load and Power in Three-Phase Systems ==== | ==== 7.2.3 Load and Power in Three-Phase Systems ==== |
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<callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
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In the case of a symmetric load, the situation and the formulas get much simpler: | In the case of a symmetric load, the situation and the formulas get much simpler: |
- The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$. | - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are related by: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$ (equal to the asymmetric load). |
- For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. \\ Since the phase currents have the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on the neutral line: $I_{\rm N} =0$ | - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. \\ Since the phase currents have the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on the neutral line: $I_{\rm N} =0$ |
- The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. \\ Based on the line voltages $U_{\rm L}$, the formula is $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ | - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. \\ Based on the line voltages $U_{\rm L}$, the formula is $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ |
- The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. \\ This corresponds to three times the apparent power of a single phase. | - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. \\ This corresponds to three times the apparent power of a single phase. |
- The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$. | - The **reactive power** leads to: $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin \varphi$. |
</callout> | </callout> |
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<callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
<panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> | <panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> |
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A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=60°$. | A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=+60°$. |
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- Draw the equivalent circuits based on a series and on a parallel circuit. | 1. Draw the equivalent circuits based on a series and on a parallel circuit. \\ |
- Calculate the equivalent components for both circuits. | |
- Calculate the real power, the reactive power, and the apparent power based on the equivalent components for both circuits from 2. . | |
- Check the solutions from 3. via direct calculation based on the input in the task above. | |
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<button size="xs" type="link" collapse="Loesung_7_1_1_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_1_1_Rechnung" collapsed="true"> | |
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| #@HiddenBegin_HTML~71111,Result~@# |
{{drawio>electrical_engineering_2:Sol711EquivCirc.svg}} | {{drawio>electrical_engineering_2:Sol711EquivCirc.svg}} |
| #@HiddenEnd_HTML~71111,Result~@# |
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</collapse> | 2. Calculate the equivalent components for both circuits. \\ |
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<button size="xs" type="link" collapse="Loesung_7_1_1_2_Rechnung">{{icon>eye}} Result for 2.</button><collapse id="Loesung_7_1_1_2_Rechnung" collapsed="true"> | #@HiddenBegin_HTML~71112,Solution~@# |
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The apparent impedance is: | The apparent impedance is: |
\end{align*} | \end{align*} |
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| \\ \\ |
| For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $ with $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$. \\ |
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| There are multiple ways to solve this problem. Two ways shall be shown here: |
| |
| === with the Euler representation === |
| Given the formula $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$ the following can be derived: |
| \begin{align*} |
| {{1}\over{\underline{Z}^{\phantom{A}}}} &= {{I}\over{U}}\cdot e^{-j\cdot \varphi} \\ |
| &= {{1}\over{Z}}\cdot e^{-j\cdot \varphi} &&= {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} \\ |
| &= {{1}\over{Z}}\cdot \left( \cos(\varphi) - {\rm j}\cdot \sin(\varphi) \right) &&= {{1}\over{R_p}} - {{\rm j}\over{X_{Lp}}} \\ |
| \end{align*} |
| |
| Therefore, the following can be concluded: |
| \begin{align*} |
| {{1}\over{Z}}\cdot \cos(\varphi) &= {{1}\over{R_p}} &&\rightarrow && R_p &= {{Z}\over{\cos(\varphi)}} \\ |
| - {\rm j}\cdot \sin(\varphi) &= - {{\rm j}\over{ X_{Lp}}} &&\rightarrow && X_{Lp} &= {{Z}\over{\sin(\varphi)}} \\ |
| \end{align*} |
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For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ | === with the calculated values of the series circuit === |
The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: | Another way is to use the formulas of $R_s$ and $X_{Ls}$ from before. |
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\begin{align*} | \begin{align*} |
&=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } \over{Z}} \\ | &=& { {\cos \varphi - {\rm j} \cdot \sin \varphi } \over{Z}} \\ |
\end{align*} | \end{align*} |
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| Therefore |
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Now, the real and imaginary part is analyzed individually. First the real part: | Now, the real and imaginary part is analyzed individually. First the real part: |
\begin{align*} | \begin{align*} |
{{1}\over{R_p}} &=& {{\cos \varphi}\over{Z}} \\ | {{1}\over{R_p}} &=& {{\cos \varphi}\over{Z}} \\ |
\rightarrow R_p &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} = \boldsymbol{92 ~\Omega} | \rightarrow R_p &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} |
\end{align*} | \end{align*} |
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\begin{align*} | \begin{align*} |
{{1}\over{X_{Lp}}} &= {{\sin \varphi}\over{Z}} & \\ | {{1}\over{X_{Lp}}} &= {{\sin \varphi}\over{Z}} & \\ |
\rightarrow X_{Lp} &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} = 53.1 ~\Omega \\ | \rightarrow X_{Lp} &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} \\ |
\rightarrow L_p &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} &= \boldsymbol{169 ~\rm mH} | \rightarrow L_p &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} |
\end{align*} | \end{align*} |
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</collapse> | #@HiddenEnd_HTML~71112,Solution ~@# |
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| #@HiddenBegin_HTML~71113,Result~@# |
| For the series circuit: |
| \begin{align*} |
| R_s &= {23 ~\Omega} \\ |
| L_s &= {127 ~\rm mH} \\ |
| \end{align*} |
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<button size="xs" type="link" collapse="Loesung_7_1_1_3_Endergebnis">{{icon>eye}} Result for 3. </button><collapse id="Loesung_7_1_1_3_Endergebnis" collapsed="true"> | For the parallel circuit: |
| \begin{align*} |
| R_p &= {92 ~\Omega} \\ |
| L_p &= {169 ~\rm mH} \\ |
| \end{align*} |
| #@HiddenEnd_HTML~71113,Result~@# |
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| 3. Calculate the real, reactive, and apparent power based on the equivalent components for both circuits from 2. . \\ |
| |
| #@HiddenBegin_HTML~71114,Solution~@# |
| The general formula for the apparent power is $\underline{S} = U \cdot I \cdot e^{\rm j\varphi}$. \\ By this, the following can be derived: |
| \begin{align*} |
| \underline{S} &= U \cdot I \cdot e^{\rm j\varphi} \\ |
| &= Z \cdot I^2 \cdot e^{\rm j\varphi} &&= \underline{Z} \cdot I^2 \\ |
| &= {{U^2}\over{Z}} \cdot e^{\rm j\varphi} &&= {{U^2}\over{\underline{Z}^{*\phantom{I}}}} \\ |
| \end{align*} |
| |
| These formulas are handy for both types of circuits to separate the apparent power into real part (real power) and complex part (apparent power): |
| - for **series circuit**: $\underline{S} =\underline{Z} \cdot I^2 $ with $\underline{Z} = R + {\rm j} X_L$ |
| - for **parallel circuit**: $\underline{S} ={{U^2}\over{\underline{Z}^{*\phantom{I}}}} $ with ${{1} \over {\underline{Z}^{\phantom{I}}} } = {{1}\over{R}} + {{1}\over{{\rm j} X_L}} \rightarrow {{1} \over {\underline{Z}^{*\phantom{I}}} } = {{1}\over{R}} + {{\rm j}\over{ X_L}} $ |
| \\ |
| Therefore: |
^ ^ series circuit ^ parallel circuit ^ | ^ ^ series circuit ^ parallel circuit ^ |
| active power | \begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*} | \begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W} \end{align*} | | | active power | \begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*} | \begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W} \end{align*} | |
| reactive power | \begin{align*} Q_s &= Z_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*} | \begin{align*} Q_p &= {{U_p^2}\over{Z_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var} \end{align*} | | | reactive power | \begin{align*} Q_s &= X_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*} | \begin{align*} Q_p &= {{U_p^2}\over{X_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var} \end{align*} | |
| apparent power | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + Z_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{Z_{Lp}^2}}} \\ &= 1150 {~\rm VA} \end{align*} | | | apparent power | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + X_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{X_{Lp}^2}}} \\ &= 1150 {~\rm VA} \end{align*} | |
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</collapse> | #@HiddenEnd_HTML~71114,Solution ~@# |
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| 4. Check the solutions from 3. via direct calculation based on the input in the task above. \\ |
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<button size="xs" type="link" collapse="Loesung_7_1_1_4_Endergebnis">{{icon>eye}} Result for 4. </button><collapse id="Loesung_7_1_1_4_Endergebnis" collapsed="true"> | <button size="xs" type="link" collapse="Loesung_7_1_1_4_Endergebnis">{{icon>eye}} Solution </button><collapse id="Loesung_7_1_1_4_Endergebnis" collapsed="true"> |
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active power: | active power: |
apparent power: | apparent power: |
\begin{align*} | \begin{align*} |
Q &= U \cdot I \\ | S &= U \cdot I \\ |
&= 230{~\rm V} \cdot 5{~\rm A} \\ | &= 230{~\rm V} \cdot 5{~\rm A} \\ |
&= 1150 {~\rm VA} | &= 1150 {~\rm VA} |
<button size="xs" type="link" collapse="Loesung_7_1_3_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_3_1_Rechnung" collapsed="true"> | <button size="xs" type="link" collapse="Loesung_7_1_3_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_3_1_Rechnung" collapsed="true"> |
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The active power is $P = 1.80 kW$. \\ | The active power is $P = 1.80 ~\rm kW$. \\ \\ |
The apparent power is $S = U \cdot I = 220V \cdot 20A = 4.40 kVA$. \\ | The apparent power is $S = U \cdot I = 220 ~\rm V \cdot 20 ~\rm A = 4.40 ~\rm kVA$. \\ \\ |
The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 kVA)^2 - (1.80 kW)^2} = 4.01 kVar$ \\ | The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 ~\rm kVA)^2 - (1.80 ~\rm kW)^2} = 4.01 ~\rm kVar$ \\ \\ |
The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 kW}\over{4.40 kVA}} = 0.41$. | The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 ~\rm kW}\over{4.40 ~\rm kVA}} = 0.41$. |
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</collapse> | </collapse> |
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</WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
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| #@TaskTitle_HTML@#7.2.2 Motor on 3-Phase System I#@TaskText_HTML@# |
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| A three-phase motor is connected to an artificial three-phase system and can be configured in wye or delta configuration. |
| * The voltage measured on a single coil shall always be $230 ~\rm V$. |
| * The current measured on a single coil shall always be $10 ~\rm A$. |
| * The phase shift for every string is $25°$ |
| |
| - The motor shall be in wye configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power |
| - The motor shall be in delta configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power |
| - Compare the results |
| #@TaskEnd_HTML@# |
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| #@TaskTitle_HTML@#7.2.3 Motor on 3-Phase System II#@TaskText_HTML@# |
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| A three-phase motor is connected to a three-phase system with a phase voltage of $400 ~\rm V$. The phase current is $16 ~\rm A$ and the power factor $0.9$. \\ |
| Calculate the active power, reactive power, and apparent power. |
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| #@TaskEnd_HTML@# |
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| #@TaskTitle_HTML@#7.2.4 Motor on 3-Phase System III#@TaskText_HTML@# |
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| A symmetrical and balanced three-phase motor of a production line shall be configured in a star configuration and provide a power of $17~\rm kW$ with a power factor of $0.75$. The voltage on a single string is measured to be $135 ~\rm V$. \\ |
| Calculate the string current. |
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| #@TaskEnd_HTML@# |
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==== Related Links ==== | ==== Related Links ==== |