1. The temperature rise of an electric motor is \( 50^{\circ} \mathrm{C} \) after one hour and \( 60^{\circ} \mathrm{C} \) after two hours. The motor current is 70 A. Determine the approximately its final temperature rise when it works on load cycle of 6 minutes working, 10 minutes rest with a current of 80 A. Neglect the effects on an iron losses.

1. We first assume that the motor behaves as a lumped thermal system with a first‐order response. In general, the temperature rise under a constant load can be written as \[ \Delta T(t) = T_\infty\Bigl(1-e^{-t/\tau}\Bigr) \] where \( T_\infty \) is the final (steady‐state) temperature rise and \(\tau\) is the time constant. 2. For a current of 70 A we are given: - At \(t=1\) hour, \[ T_\infty\Bigl(1-e^{-1/\tau}\Bigr)=50 \] - At \(t=2\) hours, \[ T_\infty\Bigl(1-e^{-2/\tau}\Bigr)=60 \] 3. Let \(x=e^{-1/\tau}\). Then \(e^{-2/\tau}=x^2\). The equations become: - \(T_\infty(1-x)=50\), - \(T_\infty(1-x^2)=60\). 4. Divide the second equation by the first: \[ \frac{1-x^2}{1-x}=\frac{60}{50}\,. \] Note that \[ \frac{1-x^2}{1-x}=1+x\quad (\text{for } x\neq 1). \] Thus, \[ 1+x=1.2\quad\Longrightarrow\quad x=0.2\,. \] 5. Now solve for \( T_\infty \) using \(T_\infty(1-x)=50\): \[ T_\infty(1-0.2)=50\quad\Longrightarrow\quad T_\infty\times 0.8=50\,, \] so, \[ T_\infty=\frac{50}{0.8}=62.5\quad\text{(for 70 A continuous load)}. \] 6. The copper losses, which are the main source of heating (neglecting iron losses), scale as the square of the current, so when the motor current changes from 70 A to 80 A the steady‐state temperature rise (if it were operating continuously) would change by a factor of \[ \left(\frac{80}{70}\right)^2\,. \] Therefore, for 80 A continuous load, the final temperature rise would be \[ T_{\infty}^{80}=62.5\left(\frac{80}{70}\right)^2\,. \] Calculating, \[ \left(\frac{80}{70}\right)^2=\left(\frac{8}{7}\right)^2=\frac{64}{49}\approx 1.306\,, \] so, \[ T_{\infty}^{80}\approx 62.5\times 1.306\approx 81.6^\circ\mbox{C}\,. \] 7. Now the motor is not running continuously at 80 A but instead on a load cycle: 6 minutes working and 10 minutes rest (i.e. 16 minutes per cycle). The duty cycle is \[ \text{Duty cycle}=\frac{6}{16}=0.375\,. \] Assuming that the system behaves linearly with respect to the average heating power (which is valid when the cycle period is short compared to the thermal time constant), the effective final temperature rise will be the steady‐state temperature rise at 80 A multiplied by the duty cycle: \[ \Delta T_{\text{final}}=81.6\times 0.375\approx 30.6^\circ\mbox{C}\,. \] Thus, the approximate final temperature rise under the load cycle is about \(30.6^\circ \mathrm{C}\).