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--- electrical_engineering_2:polyphase_networks [2024/06/18 01:06] – mexleadmin
+++ electrical_engineering_2:polyphase_networks [2025/06/24 00:40] (aktuell) – mexleadmin
@@ Zeile 362: / Zeile 362: @@
 ==== 7.2.2 Three-Phase System ====
-See also: [[https://de.mathworks.com/videos/series/what-is-3-phase-power.html|MATHWORKS Onramp Video: What is 3-phase power?]]
+See also: __ BROKEN-LINK:[[https://de.mathworks.com/videos/series/what-is-3-phase-power.html|MATHWORKS Onramp Video: What is 3-phase power?]]LINK-BROKEN__
 The most commonly used polyphase system is the three-phase system. The three-phase system has advantages over a DC system or single-phase AC system:
   * Simple three-phase machines can be used for generation.
-  * Rotary field machines (e.g. synchronous motors or induction motors) can also be simply connected to a load, converting the electrical energy into mechanical energy.
+  * 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.
   * 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 ===
-  * 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>). \\ <WRAP>
+  * 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>). \\ <WRAP>
 <imgcaption imageNo10 | Motor Terminal></imgcaption> {{drawio>Motorterminal.svg}} </WRAP>
-  * The typical **winding connections**  in a three-phase generator are called **Delta connection** (for ring connection) and **Wye connection** (for star connection). This winding connection can simply be changed by reconnecting the motor terminal. In <imgref imageNo11> the two types of winding connections are shown. For the Wye connection, is often the star configuration shown, and for the Delta connection the ring configuration. For the Wye connection, it is also possible to have the star point on a separate terminal. <WRAP>
+  * The typical **winding connections**  in a three-phase generator are called **Delta connection** (for ring connection) and **Wye connection** (for star connection). This winding connection can simply be changed by reconnecting the motor terminal. In <imgref imageNo11> the two types of winding connections are shown. For the Wye connection, it is often the star configuration shown, and for the Delta connection the ring configuration. For the Wye connection, having the star point on a separate terminal is also possible. <WRAP>
 <imgcaption imageNo11 | Motor Terminal Setup for the two Connections></imgcaption> {{drawio>MotorterminalConnections.svg}} </WRAP>
   * The **phase voltages**  are given by: <WRAP>
@@ Zeile 540: / Zeile 541: @@
 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: $I_{\rm N} =0$
   - 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 **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>
@@ 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 $\varphi=+60°$.
-. Draw the equivalent circuits based on a series and on a parallel circuit. \\
+. Draw the equivalent circuits based on a series and a parallel circuit of two components. \\
 #@HiddenBegin_HTML~71111,Result~@#
@@ 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 $\cos \varphi = 0.30$
-  - 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 series circuit. Calculate the ohmic and inductive impedance and the value of the inductivity.
@@ Zeile 1255: / Zeile 1256: @@
 </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
+#@HiddenBegin_HTML~7221,Solution~@#
+  * The **string voltage** \( U_\rm s \) is always the voltage measured on a single coil ($230 ~\rm V$).
+  * The **string current** \( I_\rm s \) is always the current measured on a single coil ($10 ~\rm A$).
+In a **wye** configuration:
+  * The **phase voltage** \( U_{\rm ph} = U_{\rm 12} = U_{\rm 23} = U_{\rm 31} \) is the **string voltage** \( U_\rm s \) times \( \sqrt{3} \)
+  * The **phase current** \( I_{\rm ph} \) is just the **string current** \( I_\rm s \)
+So, we can compute:
+  * **Phase voltage** \( U_{\rm ph} = \sqrt{3} \cdot U_{\rm s} = \sqrt{3} \cdot 230 ~\rm V \approx 398.37 ~\rm V \)
+  * **Active power**:
+\begin{align*}
+P &= \sqrt{3} \cdot U_{\rm s} \cdot I_{\rm ph} \cdot \cos(\phi) \\
+  &= \sqrt{3} \cdot 398.37 \cdot 10 \cdot \cos(25^\circ) \\
+  &= 1.732 \cdot 398.37 \cdot 10 \cdot 0.9063 \, \text{W} \\
+  &\approx 6254.79\, \text{W}
+\end{align*}
+#@HiddenEnd_HTML~7221,Solution ~@#
+#@HiddenBegin_HTML~72210,Result~@#
+  - string voltage \( U_{\rm s} = 230\, \text{V} \)
+  - phase voltage \( U_{\rm ph} = 398\, \text{V} \)
+  - string current \( I_{\rm s} = 10\, \text{A} \)
+  - phase current \( I_{\rm ph} = 10\, \text{A} \)
+  - and active power \( P = 6254.79\, \text{W} \)
+#@HiddenEnd_HTML~72210,Result~@#
+. The motor shall be in delta configuration. \\
+Write down the string voltage, phase voltage, string current, phase current, and active power
+#@HiddenBegin_HTML~7222,Solution~@#
+  * The **string voltage** \( U_\rm s \) is always the voltage measured on a single coil ($230 ~\rm V$).
+  * The **string current** \( I_\rm s \) is always the current measured on a single coil ($10 ~\rm A$).
+In a **delta** configuration:
+  * The **phase voltage** \( U_{\rm ph} = U_{\rm 12} = U_{\rm 23} = U_{\rm 31} \) is just the **string voltage** \( U_\rm s \) \)
+  * The **phase current** \( I_{\rm ph} \) is the **string current** \( I_\rm s \) times \( \sqrt{3}
+So, we can compute:
+  * **Phase current** \( I_{\rm ph} = \sqrt{3} \cdot I_{\rm s} = \sqrt{3} \cdot 10 ~\rm V \approx 17.32 ~\rm A \)
+  * **Active power**:
+\begin{align*}
+P &= \sqrt{3} \cdot U_{\rm s} \cdot I_{\rm ph} \cdot \cos(\phi) \\
+  &= \sqrt{3} \cdot 200 \cdot 17.32 \cdot \cos(25^\circ) \\
+  &= 1.732 \cdot 200 \cdot 17.32 \cdot 0.9063 \, \text{W} \\
+  &\approx 6254.79\, \text{W}
+\end{align*}
+#@HiddenEnd_HTML~7222,Solution ~@#
+#@HiddenBegin_HTML~72220,Result~@#
+  - string voltage \( U_{\rm s} = 230\, \text{V} \)
+  - phase voltage \( U_{\rm ph} = 230\, \text{V} \)
+  - string current \( I_{\rm s} = 10\, \text{A} \)
+  - phase current \( I_{\rm ph} = 17.32\, \text{A} \)
+  - and active power \( P = 6254.79\, \text{W} \)
+#@HiddenEnd_HTML~72220,Result~@#
+. Compare the results
+#@HiddenBegin_HTML~7223,Solution ~@#
+^ Configuration ^ String Voltage (V) ^ Phase Voltage (V) ^ Line Voltage (V) ^ String Current (A) ^ Phase Current (A) ^ Line Current (A) ^ Power (W)     |
+| Wye           | 230                 | 230                | 398.37           | 10                  | 10                | 10               | 6254.79       |
+| Delta         | 230                 | 230                | 230              | 10                  | 10                | 17.32            | 6254.79       |
+**Conclusion:**
+Both configurations deliver the same active power to the motor. However:
+  * In **wye**, voltage across each phase is lower, resulting in lower current per line.
+  * In **delta**, current per line is higher, even though voltage across each phase equals line voltage.
+#@HiddenEnd_HTML~7223,Solution ~@#
+#@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 $6 ~\rm kW$. \\
+.a Calculate the resistor value of a single string in the heater. \\
+.b Calculate the RMS values of the string currents and phase currents. \\
+#@HiddenBegin_HTML~7231,Solution~@#
+===== 1. Delta configuration =====
+The line-to-line (string) voltage in delta is the system line voltage
+\[
+U_{\mathrm s,\Delta}=U_{\mathrm{LL}}=400~\mathrm{V}.
+\]
+**(a)  Resistor of one string**
+Purely ohmic, total power in delta
+\[
+P_\Delta = 3\;\frac{U_{\mathrm s,\Delta}^{2}}{R}.
+\]
+Solve for \(R\):
+\begin{align*}
+R &= 3\frac{U_{\mathrm s,\Delta}^{2}}{P_\Delta}
+   =3\frac{(400~\mathrm{V})^{2}}{6000~\mathrm{W}}
+   =\;80~\boldsymbol{\Omega}.
+\end{align*}
+**(b)  Currents**
+String (coil) current
+\[
+I_{\mathrm s,\Delta}= \frac{U_{\mathrm s,\Delta}}{R}
+                    = \frac{400}{80}=5.00~\mathrm{A}.
+\]
+In delta the **phase current equals the string current**:
+\[
+I_{\mathrm{ph},\Delta}=I_{\mathrm s,\Delta}=5.00~\mathrm{A}.
+\]
+(The line current would be \(I_{\mathrm L,\Delta}= \sqrt3\,I_{\mathrm s,\Delta}=8.66~\mathrm{A}\), but is **not** required.)
+#@HiddenEnd_HTML~7231,Solution~@#
+#@HiddenBegin_HTML~72310,Result~@#
+  * resistor of one string \(R = 80~\Omega\)
+  * string current \(I_{\mathrm s,\Delta}=5.00~\mathrm{A}\)
+  * phase current \(I_{\mathrm{ph},\Delta}=5.00~\mathrm{A}\)
+#@HiddenEnd_HTML~72310,Result~@#
+. The heater with the same resistors as in 1. is now configured in a wye configuration. \\
+.a Calculate the RMS values of the string currents and phase currents. \\
+.b Compare the heating power in delta configuration (1.) and wye configuration (2.) \\
+#@HiddenBegin_HTML~7232,Solution~@#
+===== 2. Wye configuration (same resistors \(R=80~\Omega\)) =====
+The string (phase-to-neutral) voltage is the system phase voltage
+\[
+U_{\mathrm s,\mathrm Y}=U_{\mathrm{PN}}=230~\mathrm{V}.
+\]
+**(a)  Currents**
+String current
+\[
+I_{\mathrm s,\mathrm Y}= \frac{U_{\mathrm s,\mathrm Y}}{R}
+                      = \frac{230}{80}=2.875~\mathrm{A}.
+\]
+In wye the **phase current equals the string current**:
+\[
+I_{\mathrm{ph},\mathrm Y}=I_{\mathrm s,\mathrm Y}=2.875~\mathrm{A}.
+\]
+**(b)  Heating power in wye**
+Pure ohmic load:
+\begin{align*}
+P_{\mathrm Y} &= 3\;\frac{U_{\mathrm s,\mathrm Y}^{2}}{R}
+              = 3\;\frac{(230~\mathrm{V})^{2}}{80~\Omega}
+              \approx 1.984~\mathrm{kW}.
+\end{align*}
+---
+===== Comparison =====
+\[
+\frac{P_{\mathrm Y}}{P_\Delta} = \frac{1.984~\mathrm{kW}}{6.000~\mathrm{kW}}
+                               \approx 0.331.
+\]
+The heater delivers only about **33 %** of the delta power when re-wired in wye.
+#@HiddenEnd_HTML~7232,Solution~@#
+#@HiddenBegin_HTML~72320,Result~@#
+  * string current \(I_{\mathrm s,\mathrm Y}=2.88~\mathrm{A}\)
+  * phase current \(I_{\mathrm{ph},\mathrm Y}=2.88~\mathrm{A}\)
+  * heating power \(P_{\mathrm Y}\approx1.98~\mathrm{kW}\)
+  * comparison: \(P_{\mathrm Y} \approx 0.33\,P_\Delta\) (wye power ≈ 33 % of delta power)
+#@HiddenEnd_HTML~72320,Result~@#
+#@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@#
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