# Step-by-Step Solutions Let's solve each problem step by step, with clear explanations. --- ## **Problem 1** A motor is specified as a ¼ HP DC motor. The supply voltage is 48V. ### **a. What is the maximum current this motor can draw to not exceed ¼ HP?** #### **Step 1: Convert HP to Watts** 1 HP = 746 W ¼ HP = \( \frac{1}{4} \times 746 = 186.5 \) W #### **Step 2: Use Power Formula** \[ P = V \times I \] Where: - \( P \) = Power (W) - \( V \) = Voltage (V) - \( I \) = Current (A) #### **Step 3: Solve for I** \[ I = \frac{P}{V} = \frac{186.5}{48} \approx 3.89\ \text{A} \] **Final Answer:** **Maximum current = 3.89 A** --- ### **b. If the no-load speed is 300 RPM, estimate the speed at which the power consumption is maximized.** For a DC motor, maximum mechanical power output occurs at **half the no-load speed**. \[ \text{Speed for max power} = \frac{300}{2} = 150\ \text{RPM} \] **Final Answer:** **Speed at max power = 150 RPM** --- ### **c. The torque of the motor at stall is 23. N·m and at 280 RPM is 1.5 N·m. What torque at maximum power?** For a DC motor, **maximum power occurs at half the stall torque and half the no-load speed**. - Stall torque (\(T_{stall}\)) = 23. N·m \[ T_{max\ power} = \frac{T_{stall}}{2} = \frac{23.}{2} = 11.5\ \text{N·m} \] **Final Answer:** **Torque at maximum power = 11.5 N·m** --- ## **Problem 2** Motor: - No-load speed = 15,000 RPM - Torque at 14,000 RPM = .01 N·m - Stall torque = .015 N·m - Gearbox efficiency = 90% - Desired output peak power at 375 RPM --- ### **a. Gear ratio for peak power at 375 RPM** Peak mechanical power occurs at **half the no-load speed**: \[ \text{Peak power speed} = \frac{15,000}{2} = 7,500\ \text{RPM} \] Required output speed = 375 RPM Gear ratio \( G \): \[ G = \frac{\text{Motor speed at peak power}}{\text{Output speed}} = \frac{7,500}{375} = 20 \] **Final Answer:** **Gear ratio = 20:1** --- ### **b. Mechanical power delivered at the output shaft at peak power** Power at peak point (motor side): \[ \text{At peak power:} \quad T = \frac{T_{stall}}{2} = \frac{.015}{2} = .0075\ \text{N·m} \] \[ P_{motor} = 2\pi N T \] Where \( N \) is in **revolutions per second** (\( N = 7,500/60 = 125 \) rev/sec): \[ P_{motor} = 2\pi \cdot 125 \cdot .0075 \approx 5.89\ \text{W} \] Considering 90% gearbox efficiency: \[ P_{output} = .9 \times P_{motor} = .9 \times 5.89 \approx 5.30\ \text{W} \] **Final Answer:** **Mechanical power at output = 5.3 W** --- ### **c. Output torque at 375 RPM (output of gearbox, peak power point)** \[ P_{out} = 2\pi N_{out} T_{out} \] Where \( N_{out} = 375/60 = 6.25 \) rev/sec \[ T_{out} = \frac{P_{out}}{2\pi N_{out}} = \frac{5.3}{2\pi \times 6.25} \approx .135\ \text{N·m} \] **Final Answer:** **Output torque at 375 RPM = .135 N·m** --- ### **d. New stall torque at output shaft** Output shaft torque is increased by gear ratio (minus efficiency): \[ T_{stall, out} = T_{stall, motor} \times G \times \text{efficiency} = .015 \times 20 \times .9 = .27\ \text{N·m} \] **Final Answer:** **Output stall torque = .27 N·m** --- ## **Problem 3** This requires you to find a datasheet for a DC motor and fill in the values: **Example:** Suppose you pick a common hobby DC motor (e.g., RS-775). - Stall torque: .53 N·m - Maximum speed: 18,000 RPM - Maximum power produced: 150 W *(These are example values; you should fill in with real values from your chosen datasheet.)* --- # **Summary Table** | Part | Answer | |------|--------| | 1a | 3.89 A | | 1b | 150 RPM | | 1c | 11.5 N·m | | 2a | 20:1 | | 2b | 5.3 W | | 2c | .135 N·m | | 2d | .27 N·m | | 3 | Refer to datasheet | --- **Let me know if you need more explanation or want help with a specific datasheet!**