# Translation (Turkish → English) **Original (Turkish):** "Etiketinde üç faz üçgen 380 volt 4,5, 10 A, Cosø .86 , 145 d/d, 50 Hz yazılı olan motorun verimini ve NEWTONMETRE cinsinden döndürme momentini bulunuz." **English:** "On the nameplate a three-phase delta (triangle) motor is marked 380 volts, 4.5 kW, 10 A, cosφ = .86, 145 rpm, 50 Hz. Find the efficiency of the motor and the torque in newton-metres." --- # Step-by-step solution **Given:** * Line voltage (V_L=380\ \text{V}) * Output (rated) power (P_{out}=4.5\ \text{kW}=450\ \text{W}) * Line current (I_L=10\ \text{A}) * Power factor (\cos\varphi=.86) * Speed (n=145\ \text{rpm}) ## 1) Input electrical power (three-phase) For a balanced three-phase system: [ P_{in} = \sqrt{3};V_L;I_L;\cos\varphi ] Compute intermediate values digit-by-digit: * (\sqrt{3}=1.7320508075688772) * (\sqrt{3}\cdot V_L = 1.7320508075688772\times 380 = 658.179306875... \approx 658.1793) * Multiply by current: (658.1793\times 10 = 6581.79306875) * Multiply by power factor: (6581.79306875\times .86 = 566.342039135091\ \text{W}) So: [ P_{in} \approx 566.34\ \text{W} ] ## 2) Efficiency [ \eta = \frac{P_{out}}{P_{in}} = \frac{450}{566.342039135091} ] Compute: [ \eta \approx .7950049606 = 79.50% ] ## 3) Torque (in N·m) Mechanical angular speed: [ \omega = \frac{2\pi n}{60} = \frac{2\pi\times 145}{60} ] Compute: * (2\pi = 6.283185307179586) * (2\pi \times 145 = 6.283185307179586\times145 = 9108.618952410363) * Divide by 60: (\omega = 9108.618952410363/60 = 151.81031587350604\ \text{rad/s}) Torque: [ T = \frac{P_{out}}{\omega} = \frac{450}{151.81031587350604} \approx 29.635748024\ \text{N·m} ] Round reasonably: [ T \approx 29.64\ \text{N·m} ] --- # Final answers * **Input power:** (P_{in} \approx 566.34\ \text{W}) * **Efficiency:** (\boxed{\eta \approx 79.50%}) * **Torque:** (\boxed{T \approx 29.64\ \text{N·m}})

# Translation (Turkish → English) **Original (Turkish):** "Etiketinde üç faz üçgen 380 volt 4,5 Kw, 10 A, Cosø 0.86 , 1450 d/d, 50 Hz yazılı olan motorun verimini ve NEWTONMETRE cinsinden döndürme momentini bulunuz." **English:** "On the nameplate a three-phase delta (triangle) motor is marked 380 volts, 4.5 kW, 10 A, cosφ = 0.86, 1450 rpm, 50 Hz. Find the efficiency of the motor and the torque in newton-metres." --- # Step-by-step solution **Given:** * Line voltage \( V_L = 380\ \text{V} \) * Output (rated) power \( P_{out} = 4.5\ \text{kW} = 4500\ \text{W} \) * Line current \( I_L = 10\ \text{A} \) * Power factor \( \cos\varphi = 0.86 \) * Speed \( n = 1450\ \text{rpm} \) ## 1) Input electrical power (three-phase) For a balanced three-phase system: \[ P_{in} = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos\varphi \] Compute intermediate values digit-by-digit: * \( \sqrt{3} \approx 1.7320508075688772 \) * \( \sqrt{3} \cdot V_L \approx 1.7320508075688772 \times 380 \approx 658.1793 \) * Multiply by current: \( 658.1793 \times 10 \approx 6581.79306875 \) * Multiply by power factor: \( 6581.79306875 \times 0.86 \approx 5660.342039135091\ \text{W} \) So: \[ P_{in} \approx 5660.34\ \text{W} \] ## 2) Efficiency \[ \eta = \frac{P_{out}}{P_{in}} = \frac{4500}{5660.342039135091} \] Compute: \[ \eta \approx 0.7950049606 = 79.50\% \] ## 3) Torque (in N·m) Mechanical angular speed: \[ \omega = \frac{2\pi n}{60} = \frac{2\pi \times 1450}{60} \] Compute: * \( 2\pi \approx 6.283185307179586 \) * \( 2\pi \times 1450 \approx 6.283185307179586 \times 1450 \approx 9108.618952410363 \) * Divide by 60: \( \omega \approx \frac{9108.618952410363}{60} \approx 151.81031587350604\ \text{rad/s} \) Torque: \[ T = \frac{P_{out}}{\omega} = \frac{4500}{151.81031587350604} \approx 29.6357480240\ \text{N·m} \] Round reasonably: \[ T \approx 29.64\ \text{N·m} \] --- # Final answers * **Input power:** \( P_{in} \approx 5660.34\ \text{W} \) * **Efficiency:** \( \boxed{\eta \approx 79.50\%} \) * **Torque:** \( \boxed{T \approx 29.64\ \text{N·m}} \)