On the rotor of an asynchronous motor, the windings are designed in copper.
The length of the winding wire is $40~\rm{m}$.
The diameter is $0.4~\rm{mm}$.
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 $T = 20~°\rm{C}$.
Solution
\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*}
Final result
\begin{align*}
R_{20~°\rm{C}} &= 5.666 ~\Omega \rightarrow 5.7 ~\Omega \\
\end{align*}
2. what is the increase in resistance $\Delta R$ between $20~\rm °C$ and $90~\rm °C$ for one winding?
Solution
\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*}
Final result
\begin{align*}
\Delta R &= 1.56 ~\Omega \rightarrow 1.6 ~\Omega \\
\end{align*}