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--- electrical_engineering_2:polyphase_networks [2024/11/17 10:30] – [7.2.2 Three-Phase System] 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:
@@ Zeile 852: / 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 a parallel circuit. \\
+. Draw the equivalent circuits based on a series and a parallel circuit of two components. \\
 #@HiddenBegin_HTML~71111,Result~@#
@@ Zeile 1264: / Zeile 1264: @@
   * 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 wye configuration. \\
-  - The motor shall be in delta configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power
+Write down the string voltage, phase voltage, string current, phase current, and active power
-  - Compare the results
+#@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@#
@@ Zeile 1273: / Zeile 1355: @@
 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 $6 ~\rm kW$.
+. 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
+.a Calculate the resistor value of a single string in the heater. \\
-    - Calculate the RMS values of the string currents and phase currents.
+.b 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.
+#@HiddenBegin_HTML~7231,Solution~@#
-    - Compare the heating power in delta configuration (1.) and wye configuration (2.)
+===== 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@#