| Beide Seiten der vorigen Revision
Vorhergehende Überarbeitung
Nächste Überarbeitung
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Vorhergehende Überarbeitung
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electrical_engineering_2:polyphase_networks [2024/06/18 01:02]
mexleadmin |
electrical_engineering_2:polyphase_networks [2024/06/18 03:20] (aktuell)
mexleadmin [Excercises] |
| </panel> | </panel> |
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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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| </panel> | </panel> |
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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> | </panel> |
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| |
| <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> | <callout title="Voltages - Currents - True Power - Apparent and Reactive Power"> |
| 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 $\varphi=+60°$. |
| |
| 1. Draw the equivalent circuits based on a series and on a parallel circuit. \\ | 1. Draw the equivalent circuits based on a series and a parallel circuit. \\ |
| |
| #@HiddenBegin_HTML~71111,Result~@# | #@HiddenBegin_HTML~71111,Result~@# |
| A magnetic coil shows at a frequency of $f=50.0 {~\rm Hz}$ the voltage of $U=115{~\rm V}$ and the current $I=2.60{~\rm A}$ with a power factor of $\cos \varphi = 0.30$ | A magnetic coil shows at a frequency of $f=50.0 {~\rm Hz}$ the voltage of $U=115{~\rm V}$ and the current $I=2.60{~\rm A}$ with a power factor of $\cos \varphi = 0.30$ |
| |
| - Calculate the real power, the reactive power, and the apparent power . | - Calculate the real power, the reactive power, and the apparent power. |
| - Draw the equivalent parallel circuit. Calculate the active and reactive part of the current. | - Draw the equivalent parallel circuit. Calculate the active and reactive part of the current. |
| - Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity. | - Draw the equivalent series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity. |
| |
| </WRAP></WRAP></panel> | </WRAP></WRAP></panel> |
| | |
| | #@TaskTitle_HTML@#7.2.2 Motor on 3-Phase System I#@TaskText_HTML@# |
| | |
| | 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@# |
| | |
| | #@TaskTitle_HTML@#7.2.3 Heater on 3-Phase System#@TaskText_HTML@# |
| | |
| | A three-phase heater with given resistors is connected to the $230~\rm V$/$400~\rm V$ three-phase system. The heater shows purely ohmic behavior and can be configured in wye or delta configuration. \\ |
| | |
| | - The heater is configured in a delta configuration and provides a constant heating power of $6 ~\rm kW$. |
| | - Calculate the resistor value of a single string in the heater |
| | - Calculate the RMS values of the string currents and phase currents. |
| | - The heater with the same resistors as in 1. is now configured in a wye configuration. |
| | - Calculate the RMS values of the string currents and phase currents. |
| | - Compare the heating power in delta configuration (1.) and wye configuration (2.) |
| | #@TaskEnd_HTML@# |
| | |
| | #@TaskTitle_HTML@#7.2.4 Motor on 3-Phase System II#@TaskText_HTML@# |
| | |
| | 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. |
| | |
| | #@TaskEnd_HTML@# |
| | |
| | |
| | #@TaskTitle_HTML@#7.2.5 Motor on 3-Phase System III#@TaskText_HTML@# |
| | |
| | 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. |
| | |
| | #@TaskEnd_HTML@# |
| |
| ==== Related Links ==== | ==== Related Links ==== |