Unterschiede
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| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:polyphase_networks [2024/06/18 01:56] – [Excercises] mexleadmin | electrical_engineering_2:polyphase_networks [2024/11/17 10:30] (aktuell) – [7.2.2 Three-Phase System] mexleadmin | ||
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| Zeile 367: | Zeile 367: | ||
| * Simple three-phase machines can be used for generation. | * Simple three-phase machines can be used for generation. | ||
| - | * Rotary field machines (e.g. synchronous | + | * Rotary field machines (e.g., synchronous or induction motors) can also be simply connected to a load, converting electrical energy into mechanical energy. |
| * When a symmetrical load can be assumed, the energy flow is constant in time. | * When a symmetrical load can be assumed, the energy flow is constant in time. | ||
| * For energy transport, the voltage can be up-transformed and thus the AC current, as well as the associated power loss (= waste heat), can be reduced. | * For energy transport, the voltage can be up-transformed and thus the AC current, as well as the associated power loss (= waste heat), can be reduced. | ||
| Zeile 383: | Zeile 383: | ||
| === Three-phase generator === | === Three-phase generator === | ||
| - | * The windings of a three-phase generator are called $\rm U$, $\rm V$, $\rm W$; the winding connections are correspondingly called: $\rm U1$, $\rm U2$, $\rm V1$, $\rm V2$, $\rm W1$, $\rm W2$ (see <imgref imageNo10> | + | * The windings of a three-phase generator are called $\rm U$, $\rm V$, $\rm W$. \\ Often they are also colour-coded as red, yellow and blue - and consecutively sometimes also called $\rm R$, $\rm Y$, $\rm B$. |
| + | * The winding connections are correspondingly called: $\rm U1$, $\rm U2$, $\rm V1$, $\rm V2$, $\rm W1$, $\rm W2$ (see <imgref imageNo10> | ||
| < | < | ||
| - | * The typical **winding connections** | + | * The typical **winding connections** |
| < | < | ||
| * The **phase voltages** | * The **phase voltages** | ||
| Zeile 540: | Zeile 541: | ||
| 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 related by: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S} (equal to the asymmetric load)$. | + | - 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 $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ | ||
| Zeile 851: | Zeile 852: | ||
| 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. \\ |
| # | # | ||
| Zeile 988: | Zeile 989: | ||
| 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. | ||
| Zeile 1256: | Zeile 1257: | ||
| </ | </ | ||
| + | # | ||
| - | # | + | 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 | ||
| + | * The phase shift for every string is $25°$ | ||
| - | 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 $0.85$. \\ | + | - 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 | ||
| + | # | ||
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| + | # | ||
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| + | 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.) | ||
| + | # | ||
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| + | # | ||
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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. | Calculate the active power, reactive power, and apparent power. | ||
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| + | # | ||
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| + | # | ||
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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. | ||
| # | # | ||