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electrical_engineering_1:aufgabe_1.7.6_mit_rechnung [2023/02/11 23:07] – mexleadminelectrical_engineering_1:aufgabe_1.7.6_mit_rechnung [2023/03/19 17:52] (aktuell) – mexleadmin
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-<panel type="info" title="Exercise 1.6.6: Temperature-dependent resistance of a winding (written test, approx. 6% of a 60-minute written test, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 1.6.6: Temperature-dependent resistance of a winding (written test, approx. 6 % of a 60-minute written test, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-On the rotor of an asynchronous motor, the windings are designed in copper. The length of the winding wire is 40 m. The diameter is 0.4 mm. When the motor is started, it is uniformly cooled down to the ambient temperature of 20°C. During operation the windings on the rotor have a temperature of 90°C.\\ +On the rotor of an asynchronous motor, the windings are designed in copper. 
-$\alpha_{Cu,20°C}=0.0039 \frac{1}{K}$\\ +The length of the winding wire is $40~\rm{m}$. 
-$\beta_{Cu,20°C}=0.6 \cdot 10^{-6} \frac{1}{K^2}$\\ +The diameter is $0.4~\rm{mm}$. 
-$\rho_{Cu,20°C}=0.0178 \frac{\Omega mm^2}{m}$+When the motor is started, it is uniformly cooled down to the ambient temperature of $20~°\rm{C}$. 
 +During operation the windings on the rotor have a temperature of $90~°\rm{C}$. \\ 
 +$\alpha_{Cu,20~°\rm{C}}=0.0039 ~\frac{1}{\rm{K}}$ \\ 
 +$ \beta_{Cu,20~°\rm{C}}=0.6 \cdot 10^{-6}  ~\frac{1}{\rm{K}^2}$ \\ 
 +$  \rho_{Cu,20~°\rm{C}}=0.0178 ~\frac{\Omega \rm{mm}^2}{\rm{m}}$
  
-Use both the linear and quadratic temperature coefficients! 1. determine the resistance of the wire for $T = 20°C$.+Use both the linear and quadratic temperature coefficients! 
 +1. determine the resistance of the wire for $T = 20~°\rm{C}$.
  
 <button size="xs" type="link" collapse="Loesung_1_7_6_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_1_7_6_1_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_1_7_6_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_1_7_6_1_Lösungsweg" collapsed="true">
  
-\begin{align*} R_{20°C} &= \rho_{Cu,20°C} \cdot \frac{l}{A} && | \text{with } A = r^2 \cdot \pi = \frac{1}{4} d^2 \cdot \pi \\ R_{20°C} &= \rho_{Cu,20°C} \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\ R_{20°C} &= 0.0178 \frac{\Omega mm^2}{m} \cdot \frac{4 \cdot 40m}{(0,4mm)^2 \cdot \pi} && \\ \end{align*}+\begin{align*} 
 +R_{20~°\rm{C}} &= \rho_{Cu,20~°\rm{C}} \cdot \frac{l}{A} && | \text{with } A = r^2 \cdot \pi = \frac{1}{4} d^2 \cdot \pi \\ 
 +R_{20~°\rm{C}} &= \rho_{Cu,20~°\rm{C}} \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\ 
 +R_{20~°\rm{C}} &= 0.0178 ~\rm{\frac{\Omega mm^2}{m}} \cdot \frac{4 \cdot 40~\rm{m}}{(0.4~\rm{mm})^2 \cdot \pi} && \\ 
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_1_7_6_1_Endergebnis">{{icon>eye}} Final result</button><collapse id="Loesung_1_7_6_1_Endergebnis" collapsed="true"> \begin{align*} R_{20°C} &= 5.666 \Omega \rightarrow 5.7 \Omega \\ \end{align*}\\+<button size="xs" type="link" collapse="Loesung_1_7_6_1_Endergebnis">{{icon>eye}} Final result</button><collapse id="Loesung_1_7_6_1_Endergebnis" collapsed="true"> 
 +\begin{align*} 
 +R_{20~°\rm{C}} &= 5.666 ~\Omega \rightarrow 5.7 ~\Omega \\ 
 +\end{align*} 
 + \\
 </collapse> </collapse>
  
-2. what is the increase in resistance $\Delta R$ between $20°C$ and $90°C$ for one winding?+ 
 +2. what is the increase in resistance $\Delta R$ between $20~\rm °C$ and $90~\rm °C$ for one winding? 
  
 <button size="xs" type="link" collapse="Loesung_1_7_6_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_1_7_6_2_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_1_7_6_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_1_7_6_2_Lösungsweg" collapsed="true">
  
-\begin{align*} R_{90°C} &= R_{20°C} \cdot ( 1 + \alpha_{Cu,20°C} \cdot \Delta T + \beta_{Cu,20°C} \cdot \Delta T^2 ) && | \text{with } \Delta T = T_2 - T_1 = 90°C - 20°C = 70 °C = 70 K\\ \Delta R &= R_{20°C} \cdot ( \alpha_{Cu,20°C} \cdot \Delta T + \beta_{Cu,20°C} \cdot \Delta T^2 ) \\ \Delta R &= 5.666 \Omega \cdot ( 0.0039 \frac{1}{K} \cdot 70K + 0.6 \cdot 10^{-6} \frac{1}{K^2} \cdot (70K)^2 ) \\ \end{align*}+\begin{align*} 
 +R_{90\rm{°C}} &= R_{20\rm{°C}} \cdot ( 1 + \alpha_{Cu,20°\rm{C}} \cdot \Delta T + \beta_{Cu,20°\rm{C}} \cdot \Delta T^2 ) && | \text{with } \Delta T = T_2 - T_1 = 90~°\rm{C} - 20~°\rm{C} = 70~°\rm{C} = 70~\rm{K}\\ 
 +\Delta R &= R_{20°\rm{C}} \cdot ( \alpha_{Cu,20°\rm{C}} \cdot \Delta T + \beta_{Cu,20°\rm{C}} \cdot \Delta T^2 ) \\ 
 +\Delta R &= 5.666 \Omega \cdot ( 0.0039 ~\frac{1}{\rm{K}} \cdot 70~\rm{K} + 0.6 \cdot 10^{-6}  \frac{1}{\rm{K}^2} \cdot (70~\rm{K})^2 ) \\ 
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_1_7_6_2_Endergebnis">{{icon>eye}} Final result</button><collapse id="Loesung_1_7_6_2_Endergebnis" collapsed="true"> \begin{align*} \Delta R &= 1.56 \Omega \rightarrow 1.6 \Omega \\ \end{align*}\\+<button size="xs" type="link" collapse="Loesung_1_7_6_2_Endergebnis">{{icon>eye}} Final result</button><collapse id="Loesung_1_7_6_2_Endergebnis" collapsed="true"> 
 +\begin{align*} 
 +\Delta R &= 1.56 ~\Omega \rightarrow 1.6 ~\Omega \\ 
 +\end{align*} 
 + \\
 </collapse> </collapse>
  
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