### Translation **10p B)** On the label, a three-phase delta 380 volt, 4.5 kW, 10 A, Cosφ .86, 145 rpm, 50 Hz motor is written. Find the efficiency of the motor and the torque in NEWTON-METERS. --- ## Step-by-Step Solution ### **Given Data** - Voltage (\( V \)): 380 V - Power (\( P_{out} \)): 4.5 kW = 450 W - Current (\( I \)): 10 A - Power Factor (\( \cos\varphi \)): .86 - Speed (\( n \)): 145 rev/min - Frequency (\( f \)): 50 Hz --- ### **1. Find the Motor Efficiency (\( \eta \))** #### **a. Calculate Input Power (\( P_{in} \))** For a three-phase motor: \[ P_{in} = \sqrt{3} \cdot V \cdot I \cdot \cos\varphi \] \[ P_{in} = \sqrt{3} \times 380 \times 10 \times .86 \] \[ P_{in} = 1.732 \times 380 \times 10 \times .86 \] \[ P_{in} = 1.732 \times 380 \times .86 \] \[ P_{in} = 1.732 \times 3268 \] \[ P_{in} = 5661.18 \text{ W} \] #### **b. Calculate Efficiency (\( \eta \))** \[ \eta = \frac{P_{out}}{P_{in}} \times 100 \] \[ \eta = \frac{450}{5661.18} \times 100 \] \[ \eta = .7956 \times 100 \approx 79.6 \% \] --- ### **2. Find the Torque in Newton-meters** The mechanical output power is: \[ P_{out} = 2\pi n T / 60 \] Where: - \( P_{out} \) = output power in watts - \( n \) = speed in rpm - \( T \) = torque in Nm Rearrange to solve for \( T \): \[ T = \frac{P_{out} \times 60}{2\pi n} \] Substitute values: \[ T = \frac{450 \times 60}{2\pi \times 145} \] \[ T = \frac{270000}{2\pi \times 145} \] \[ 2\pi \times 145 \approx 2 \times 3.1416 \times 145 \approx 6.2832 \times 145 \approx 911.64 \] \[ T = \frac{270000}{911.64} \approx 29.64 \text{ Nm} \] --- ## **Summary of Answers** - **Motor Efficiency (\( \eta \)):** ≈ **79.6%** - **Torque (T):** ≈ **29.6 Nm**