Let's break down the problem step by step: **Given Data:** - Motor is loaded (clutched) for \( t_1 = 10 \) min (with load) - Motor is no-load (declutched) for \( t_2 = 20 \) min (without load) - **Minimum temperature rise** (when running continuously on no-load): \( \theta_{no\_load} = 15^\circ C \) - **Heating and cooling time constants**: \( \tau = 60 \) min - **Maximum temperature rise** (when running continuously on load): To find (call it \( \theta_{load} \)) - **Minimum temperature rise during cycle**: \( \theta_{min} = 40^\circ C \) (when?) Let's clarify a bit: - The **minimum temperature rise during the cycle** is given as \( 40^\circ C \). This is the temperature at the end of the cooling (no-load) period (i.e., right before the load is applied again). - The **maximum temperature during the cycle** is the temperature at the end of the heating (loaded) period (right before the load is removed again). --- ### **Step 1: Fundamental Equations** The temperature rise of the motor follows first-order exponential heating and cooling, governed by the time constant \( \tau \). **When running at load:** \[ \frac{d\theta}{dt} = \frac{\theta_{load} - \theta}{\tau} \] **When running at no load:** \[ \frac{d\theta}{dt} = \frac{\theta_{no\_load} - \theta}{\tau} \] --- ### **Step 2: Exponential Heating and Cooling** - Let \( \theta_{min} \) = temperature at the start of the heating period (start of load) - Let \( \theta_{max} \) = temperature at the end of the heating period (end of load) - Let \( \theta_{no\_load} \) = final steady-state temperature in no-load operation = \( 15^\circ C \) - Let \( \theta_{load} \) = final steady-state temperature in full-load operation (to find) **Heating (load applied for \( t_1 \))**: \[ \theta_{max} = \theta_{load} + (\theta_{min} - \theta_{load}) e^{-t_1/\tau} \] **Cooling (no-load applied for \( t_2 \))**: \[ \theta_{min} = \theta_{no\_load} + (\theta_{max} - \theta_{no\_load}) e^{-t_2/\tau} \] --- ### **Step 3: Substitute Given Values** Given: - \( t_1 = 10 \) min - \( t_2 = 20 \) min - \( \tau = 60 \) min - \( \theta_{no\_load} = 15^\circ C \) - \( \theta_{min} = 40^\circ C \) Let’s denote: - \( a = e^{-t_1/\tau} = e^{-10/60} = e^{-1/6} \) - \( b = e^{-t_2/\tau} = e^{-20/60} = e^{-1/3} \) Calculate values: - \( a = e^{-1/6} \approx 0.8465 \) - \( b = e^{-1/3} \approx 0.7165 \) --- ### **Step 4: Set Up Equations** From above: \[ \theta_{max} = \theta_{load} + (\theta_{min} - \theta_{load}) a \] \[ \theta_{min} = \theta_{no\_load} + (\theta_{max} - \theta_{no\_load}) b \] We know \( \theta_{min} = 40 \), \( \theta_{no\_load} = 15 \), \( a = 0.8465 \), \( b = 0.7165 \). Only unknowns are \( \theta_{max} \) and \( \theta_{load} \). --- ### **Step 5: Substitute and Solve** First equation: \[ \theta_{max} = \theta_{load} + (40 - \theta_{load}) \cdot 0.8465 \] \[ \theta_{max} = \theta_{load} + 0.8465 \times 40 - 0.8465\theta_{load} \] \[ \theta_{max} = \theta_{load}(1 - 0.8465) + 33.86 \] \[ \theta_{max} = 0.1535\theta_{load} + 33.86 \] Second equation: \[ 40 = 15 + (\theta_{max} - 15) \cdot 0.7165 \] \[ 40 - 15 = (\theta_{max} - 15) \cdot 0.7165 \] \[ 25 = (\theta_{max} - 15) \cdot 0.7165 \] \[ \theta_{max} - 15 = \frac{25}{0.7165} = 34.89 \] \[ \theta_{max} = 34.89 + 15 = 49.89 \] So, \( \theta_{max} \approx 49.9^\circ C \) Now, use this to find \( \theta_{load} \): \[ \theta_{max} = 0.1535\theta_{load} + 33.86 \] \[ 49.89 = 0.1535\theta_{load} + 33.86 \] \[ 49.89 - 33.86 = 0.1535\theta_{load} \] \[ 16.03 = 0.1535\theta_{load} \] \[ \theta_{load} = \frac{16.03}{0.1535} \approx 104.5^\circ C \] --- ### **Step 6: Final Answers** **i) Maximum temperature during the duty cycle:** \[ \boxed{49.9^\circ C} \] **ii) Temperature when load is clutched continuously (steady-state at full-load):** \[ \boxed{104.5^\circ C} \] --- ## **Summary Table** | Quantity | Symbol | Value | |----------|--------|-------| | Maximum temperature in cycle | \( \theta_{max} \) | \( 49.9^\circ C \) | | Steady-state temp (continuous load) | \( \theta_{load} \) | \( 104.5^\circ C \) | --- **If you need the step-by-step working or explanation for another case, feel free to ask!**