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electrical_engineering_1:aufgabe_1.7.6_mit_rechnung [2021/10/06 11:28] – [Bearbeiten - Panel] tfischerelectrical_engineering_1:aufgabe_1.7.6_mit_rechnung [2023/03/19 17:52] (aktuell) – mexleadmin
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 On the rotor of an asynchronous motor, the windings are designed in copper. On the rotor of an asynchronous motor, the windings are designed in copper.
-The length of the winding wire is 40 m. +The length of the winding wire is $40~\rm{m}$. 
-The diameter is 0.4 mm. +The diameter is $0.4~\rm{mm}$. 
-When the motor is started, it is uniformly cooled down to the ambient temperature of 20°C. +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°C. \\ +During operation the windings on the rotor have a temperature of $90~°\rm{C}$. \\ 
-$\alpha_{Cu,20°C}=0.039 \frac{1}{K}$ \\ +$\alpha_{Cu,20~°\rm{C}}=0.0039 ~\frac{1}{\rm{K}}$ \\ 
-$\beta_{Cu,20°C}=0.6 \cdot 10^{-6}  \frac{1}{K^2}$ \\ +$ \beta_{Cu,20~°\rm{C}}=0.6 \cdot 10^{-6}  ~\frac{1}{\rm{K}^2}$ \\ 
-$\rho_{Cu,20°C}=0.0178 \frac{\Omega mm^2}{m}$+$  \rho_{Cu,20~°\rm{C}}=0.0178 ~\frac{\Omega \rm{mm}^2}{\rm{m}}$
  
 Use both the linear and quadratic temperature coefficients! Use both the linear and quadratic temperature coefficients!
-1. determine the resistance of the wire for $T = 20°C$.+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*} \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~°\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°C} &= \rho_{Cu,20°C} \cdot \frac{4 \cdot l}{d^2 \cdot \pi} && \\ +R_{20~°\rm{C}} &= \rho_{Cu,20~°\rm{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} && \\+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*} \end{align*}
  
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 \begin{align*} \begin{align*}
-R_{20°C} &= 5.666 \Omega \rightarrow 5.7 \Omega \\+R_{20~°\rm{C}} &= 5.666 ~\Omega \rightarrow 5.7 ~\Omega \\
 \end{align*} \end{align*}
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-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?
  
  
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 \begin{align*} \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\\ +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°C} \cdot ( \alpha_{Cu,20°C} \cdot \Delta T + \beta_{Cu,20°C} \cdot \Delta T^2 ) \\ +\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}{K} \cdot 70K + 0.6 \cdot 10^{-6}  \frac{1}{K^2} \cdot (70K)^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*} \end{align*}
  
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 \begin{align*} \begin{align*}
-\Delta R &= 1.56 \Omega \rightarrow 1.6 \Omega \\+\Delta R &= 1.56 ~\Omega \rightarrow 1.6 ~\Omega \\
 \end{align*} \end{align*}
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