2. At the beginning of the loading of an electric motor its temperature is \( 30^{\circ} \mathrm{C} \) After 2 hours run on full load its temperature is \( 60^{\circ} \mathrm{C} \), and after 4 hours run it is \( 80^{\circ} \mathrm{C} \). Calculate; a. heating time constant, b. the maximum final temperature rise with full load on the motor, and c. the time it will take after start for the motor to reach \( 3 / 4 \) of its final steady. \( \therefore \)

To address this problem, let's break down each part step by step. a. To calculate the heating time constant, we can use the formula for temperature rise in a heated object: \[ T(t) = T_f + (T_0 - T_f) e^{-\frac{t}{\tau}} \] Where: - \( T(t) \) is the temperature at time \( t \) - \( T_f \) is the final steady-state temperature - \( T_0 \) is the initial temperature - \( \tau \) is the time constant Given the temperatures, we can first determine the values of \( T_f \) and \( \tau \) through the available data points. Let's summarize the temperature data: - Initial Temperature \( T_0 = 30^{\circ}C \) - After 2 hours \( T(2) = 60^{\circ}C \) - After 4 hours \( T(4) = 80^{\circ}C \) Assuming the final steady state \( T_f \) is \( 80^{\circ}C \), we can rearrange the formula to solve for \( \tau \) using the two temperature measurements. Using the first data point at 2 hours: \[ 60 = 80 + (30 - 80) e^{-\frac{2}{\tau}} \] Solving for \( e^{-\frac{2}{\tau}} \), \[ -20 = -50 e^{-\frac{2}{\tau}} \implies e^{-\frac{2}{\tau}} = \frac{2}{5} \] Taking the natural logarithm, \[ -\frac{2}{\tau} = \ln\left(\frac{2}{5}\right) \implies \tau = -\frac{2}{\ln\left(\frac{2}{5}\right)} \approx 1.915 \text{ hours} \] b. The maximum final temperature rise achieved while running on full load is equal to the difference between the final temperature \( T_f \) and the initial temperature \( T_0 \): \[ \Delta T = T_f - T_0 = 80 - 30 = 50^{\circ}C \] c. To find out how long it will take for the motor to reach \( \frac{3}{4} \) of its final steady-state temperature: \[ T(t) = T_f \cdot \left(1 - e^{-\frac{t}{\tau}}\right) \] We need to solve for \( t \) when \( T(t) = 30 + \frac{3}{4} \cdot (80 - 30) = 30 + \frac{3}{4} \cdot 50 = 62.5^{\circ}C \). Setting up the equation: \[ 62.5 = 80 - 50 e^{-\frac{t}{\tau}} \implies -17.5 = -50 e^{-\frac{t}{\tau}} \implies e^{-\frac{t}{\tau}} = \frac{17.5}{50} \implies -\frac{t}{\tau} = \ln\left(\frac{17.5}{50}\right) \] Solving for \( t \): \[ t = -\tau \cdot \ln\left(\frac{17.5}{50}\right) \approx 1.915 \cdot 0.673 = 1.29 \text{ hours} \] To summarize: - The heating time constant \( \tau \approx 1.915 \text{ hours} \) - The maximum final temperature rise \( \Delta T = 50^{\circ}C \) - The time to reach \( \frac{3}{4} \) of its steady state is approximately \( 1.29 \text{ hours} \).