Let's solve this V-belt drive design problem **step-by-step**. --- ## **Given Data** - **Input Power, \( P \)** = 25 hp - **Input Speed, \( N_1 \)** = 870 rpm (motor) - **Output Speed, \( N_2 \)** = 310 rpm (hammer mill) - **Linear Belt Speed, \( V \)** ≈ 400 ft/min (fpm) - **Daily Use** = 8 hours --- ## **Step 1: Calculate the Speed Ratio** \[ \text{Speed Ratio} = \frac{N_1}{N_2} = \frac{870}{310} = 2.81 \] --- ## **Step 2: Find Required Pitch Diameter of the Motor Sheave** The **linear speed** of the belt on the driving (motor) sheave: \[ V = \pi D_1 N_1 \] where: - \( V \) = linear velocity of belt (ft/min) - \( D_1 \) = pitch diameter of motor sheave (ft) - \( N_1 \) = speed of motor sheave (rpm) **Convert D₁ to inches** to match standard sheave sizes: - 1 ft = 12 inches - So, \( D_1 \) (in inches) = \( d_1 \) \[ V = \pi d_1 \frac{N_1}{12} \] \[ d_1 = \frac{12V}{\pi N_1} \] Plug in the numbers: \[ d_1 = \frac{12 \times 400}{\pi \times 870} \] \[ d_1 = \frac{48000}{2733.6} \] \[ d_1 \approx 17.57\, \text{inches} \] --- ## **Step 3: Find the Required Driven Sheave Diameter** From the speed ratio: \[ \frac{N_1}{N_2} = \frac{d_2}{d_1} \] So, \[ d_2 = d_1 \times \frac{N_1}{N_2} \] \[ d_2 = 17.57 \times \frac{870}{310} \] \[ d_2 = 17.57 \times 2.81 \] \[ d_2 \approx 49.4\, \text{inches} \] --- ## **Step 4: Review Practicality and Adjust** - **17.57 inches** for a motor sheave is quite large for standard V-belt sheaves. Often, practical sheave diameters are between 3–20 inches. - **49.4 inches** is extremely large and typically impractical. ### **Alternative: Use Multiple Reduction Stages** For such a high ratio, consider **two-stage reduction** or a gear reduction in combination. --- ## **Step 5: Summary and Recommendations** ### **Calculated Values:** - **Driving sheave (motor, d₁):** ≈ **17.57 inches** - **Driven sheave (mill, d₂):** ≈ **49.4 inches** - **Speed ratio:** ≈ **2.81** - **Linear speed:** ≈ **400 fpm** ### **Design Note:** - The calculated sheave sizes are unusually large for standard V-belt drives. In practice, use **smaller sheaves** and **two-stage reduction** if possible. - Check standard catalogues for available sheave sizes and adjust the design accordingly. --- ## **Final Answers** **Pitch Diameter of Motor Sheave:** \[ \boxed{17.6\ \text{inches}} \] **Pitch Diameter of Driven Sheave:** \[ \boxed{49.4\ \text{inches}} \] **Speed Ratio:** \[ \boxed{2.81} \] **Linear Belt Speed:** \[ \boxed{400\ \text{fpm}} \] --- **If you need calculation for number of belts, tension, or further design steps, let me know!**