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. \( \therefore \)
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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 \( 1.29 \text{ hours} \).
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