In 1 minute, the conveyor lifts a load weighing 190 kg to a height of 15 m.Visit the efficiency of the conveyor

In 1 minute, the conveyor lifts a load weighing 190 kg to a height of 15 m.Visit the efficiency of the conveyor if the current in the winding of its electric motor is 3A, and the voltage in the network is 380 V.

To find the efficiency of the used conveyor, we apply the formula: η = Ap / Qz = m * g * h / (U * I * t).

Variables: m – weight of the cargo (m = 190 kg); g – acceleration of gravity (g = 9.81 m / s2); h – lifting height (h = 15 m); U – mains voltage (U = 380 V); I is the current in the motor winding of the used conveyor (I = 3 A); t is the operating time of the conveyor (t = 1 min = 60 s).

Calculation: η = m * g * h / (U * I * t) = 190 * 9.81 * 15 / (380 * 3 * 60) = 0.40875 ≈ 41%.

Answer: The used conveyor has an efficiency of 41%.