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electrical_engineering_2:polyphase_networks [2025/06/24 00:27] – mexleadminelectrical_engineering_2:polyphase_networks [2025/06/24 00:40] (aktuell) – mexleadmin
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 ==== 7.2.2 Three-Phase System ==== ==== 7.2.2 Three-Phase System ====
  
-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 __ +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: 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:
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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. \\ 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 $6 ~\rm kW$. +1. 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 +1.a Calculate the resistor value of a single string in the heater. \\ 
-    - Calculate the RMS values of the string currents and phase currents. +1.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~@# 
 + 
 +2. The heater with the same resistors as in 1. is now configured in a wye configuration. \\ 
 +2.a Calculate the RMS values of the string currents and phase currents. \\ 
 +2.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@# #@TaskEnd_HTML@#