Let's break down the given information: - **Major diameter (d):** 3 in - **Pitch (p):** 0.2 in - **Coefficient of friction for collar (\( f_c \)):** 0.16 - **Coefficient of friction for thread (\( f_t \)):** 0.12 - **Frictional radius of collar (\( r_c \)):** 2.5 in - **Input power:** 3.5 hp = 3.5 × 746 = 2611 W - **Velocity of nut:** 2.5 in/s = 0.0635 m/s We are to **calculate the overall efficiency** of the power screw for raising the load. --- ### Step 1: Output Power \[ \text{Output Power} = \text{Load} \times \text{Velocity} \] But the load (W) is not given directly. We need to use the input power and the power screw equations. --- ### Step 2: Lead and Screw Parameters For a **double-threaded** screw: - Lead (\( l \)) = Number of threads × pitch = 2 × 0.2 = 0.4 in Mean diameter (\( d_m \)) = Major diameter = 3 in (approximate since minor diameter not given) --- ### Step 3: Rotational Speed \[ \text{Lead} = \text{distance moved by nut per revolution} \] \[ \text{Nut velocity} = \text{Lead} \times N \] \[ N = \frac{\text{Nut velocity}}{\text{Lead}} = \frac{2.5 \text{ in/s}}{0.4 \text{ in/rev}} = 6.25 \text{ rev/s} = 375 \text{ rpm} \] --- ### Step 4: Power Required to Raise Load \[ \text{Input Power} = T \cdot \omega \] where \( T \) is the torque required and \( \omega \) is the angular velocity. \[ \omega = 2\pi N = 2\pi \times 6.25 = 39.27 \text{ rad/s} \] --- ### Step 5: Torque to Raise Load (including friction) The total torque required to raise the load: \[ T = T_{\text{thread}} + T_{\text{collar}} \] #### **Thread Torque:** \[ T_{\text{thread}} = W \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} \] Where \[ \tan(\lambda) = \frac{\text{Lead}}{\pi d_m} \] \[ \tan(\lambda) = \frac{0.4}{\pi \times 3} = 0.0424 \] #### **Collar Torque:** \[ T_{\text{collar}} = W f_c r_c \] --- ### Step 6: Efficiency \[ \text{Efficiency} (\eta) = \frac{\text{Output Power}}{\text{Input Power}} \] #### **Output Power:** \[ \text{Output Power} = W \times v \] #### **Input Power:** \[ \text{Input Power} = T \cdot \omega \] --- Since the actual load \( W \) cancels out in the efficiency formula, we can proceed: \[ \eta = \frac{W v}{T \omega} \] \[ T = W \left[ \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} + f_c r_c \right] \] \[ T \omega = W \left[ \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} + f_c r_c \right] \omega \] \[ \eta = \frac{v}{\left[ \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} + f_c r_c \right] \omega} \] --- ### Substitute the values: - \( v = 2.5 \) in/s = 0.0635 m/s = 2.5 in/s - \( d_m = 3 \) in - \( \tan(\lambda) = 0.0424 \) - \( f_t = 0.12 \) - \( f_c = 0.16 \) - \( r_c = 2.5 \) in - \( \omega = 39.27 \) rad/s Calculate the denominator: \[ \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} = \frac{3}{2} \cdot \frac{0.0424 + 0.12}{1 - 0.12 \times 0.0424} = 1.5 \cdot \frac{0.1624}{1 - 0.005088} = 1.5 \cdot \frac{0.1624}{0.994912} = 1.5 \cdot 0.1633 = 0.245 \] \[ f_c r_c = 0.16 \times 2.5 = 0.4 \] Total denominator: \[ 0.245 + 0.4 = 0.645 \] Now multiply by \( \omega \): \[ 0.645 \times 39.27 = 25.33 \] Now, \[ \eta = \frac{2.5}{25.33} = 0.0987 \approx 0.10 \] --- ## **Final Answer:** The overall efficiency of the power screw for raising the load is **0.10** (or **10%**). **Correct option:** **0.10**