Let's solve the problem step by step. ### Given Data - **Major diameter, \( d \) = 3 in** - **Pitch, \( p \) = 0.2 in** (double thread, so lead \( l = 2 \times p = 0.4 \) 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, \( P_{in} = 3.5 \) hp = 3.5 × 746 = 2611 W** - **Velocity of nut, \( v = 2.5 \) in/s** - **Required: Overall efficiency** --- ### Step 1: **Calculate Output Power** Output Power, \( P_{out} = \text{Load} \times \text{velocity} \) But Load is not given. We use the velocity and input power to find efficiency by another method. --- ### Step 2: **Calculate Torque Required** #### **1. Lead of the screw \( l \):** For double thread: \[ l = 2 \times p = 2 \times 0.2 = 0.4 \text{ in} \] #### **2. Mean diameter \( d_m \):** \[ d_m = d - \frac{1}{2} \times \text{pitch} = 3 - 0.1 = 2.9 \text{ in} \] \[ r_m = \frac{d_m}{2} = 1.45 \text{ in} \] #### **3. Velocity relationship:** \[ \text{Linear velocity, } v = \text{lead} \times n \] \[ n = \frac{v}{l} = \frac{2.5}{0.4} = 6.25 \text{ rev/sec} \] #### **4. Angular velocity:** \[ \omega = 2\pi n = 2\pi \times 6.25 = 39.27 \text{ rad/s} \] --- ### Step 3: **Efficiency Calculation** The overall efficiency \(\eta\) of a power screw is: \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \] #### **Output Power:** \[ P_{out} = F \cdot v \] #### **Input Power:** \[ P_{in} = T_{total} \cdot \omega \] #### **Thread and Collar Torque:** - **Thread torque:** \[ T_{thread} = F \cdot r_m \cdot \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} \] Where \(\lambda = \arctan\left(\frac{l}{\pi d_m}\right)\) - **Collar torque:** \[ T_{collar} = F \cdot f_c \cdot r_c \] - **Total torque:** \[ T_{total} = T_{thread} + T_{collar} \] --- #### **Calculate \(\lambda\):** \[ \lambda = \arctan\left(\frac{0.4}{\pi \times 2.9}\right) = \arctan\left(\frac{0.4}{9.11}\right) = \arctan(0.0439) \approx 2.52^\circ \] \[ \tan(\lambda) \approx 0.044 \] --- #### **Thread Torque:** \[ T_{thread} = F \cdot 1.45 \cdot \frac{0.044 + 0.12}{1 - 0.12 \times 0.044} \] \[ T_{thread} = F \cdot 1.45 \cdot \frac{0.164}{1 - 0.0053} = F \cdot 1.45 \cdot \frac{0.164}{0.9947} \approx F \cdot 1.45 \cdot 0.165 \] \[ T_{thread} \approx F \cdot 0.239 \] --- #### **Collar Torque:** \[ T_{collar} = F \cdot 0.16 \cdot 2.5 = F \cdot 0.4 \] --- #### **Total Torque:** \[ T_{total} = F \cdot (0.239 + 0.4) = F \cdot 0.639 \] --- ### Step 4: **Power Equations** \[ P_{in} = T_{total} \cdot \omega = F \cdot 0.639 \cdot 39.27 \] \[ P_{in} = F \cdot 25.09 \text{ (in-lbf/s)} \] \[ P_{out} = F \cdot v = F \cdot 2.5 \text{ (in-lbf/s)} \] #### **Efficiency:** \[ \eta = \frac{P_{out}}{P_{in}} = \frac{F \cdot 2.5}{F \cdot 25.09} = \frac{2.5}{25.09} = 0.0996 \approx 0.10 \] --- ### **Final Answer:** \[ \boxed{0.10} \] So, the overall efficiency of the power screw for raising the load is **0.10 (option 1)**.