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

March 24, 2021 | education

| **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%.

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