Question

Электродвигатель постоянного тока потребляет от сети мощность 33кВт суммарные потери в нем равны 3кВт. Определить КПД и полезную мощность двигателя. Помогите ЗАРАНЕЕ СПАСИБО

Answer 1

P=33*10^3 Вт      ΔP=3*10^3 Вт     η=?
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η=(P-ΔP)/P=(33-3)/33≈0.91   (91%)
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Answer 2

Calculation of Efficiency and Useful Power of a DC Motor

To determine the efficiency and useful power of a DC motor, we need to use the given information about the power consumed from the power supply and the total losses in the motor.

Given: - Power consumed from the power supply: 33 kW - Total losses in the motor: 3 kW

Efficiency of the DC Motor

The efficiency of a motor is defined as the ratio of useful power output to the power input. In this case, the useful power output is the power consumed from the power supply minus the total losses.

Let's calculate the efficiency of the DC motor:

Useful Power Output = Power consumed from the power supply - Total losses Useful Power Output = 33 kW - 3 kW Useful Power Output = 30 kW

Efficiency = (Useful Power Output / Power Input) x 100% Efficiency = (30 kW / 33 kW) x 100% Efficiency ≈ 90.91%

Therefore, the efficiency of the DC motor is approximately 90.91%.

Useful Power of the DC Motor

The useful power of a motor is the power output that is available for performing useful work. In this case, the useful power is the same as the useful power output calculated earlier.

Therefore, the useful power of the DC motor is 30 kW.

To summarize: - The efficiency of the DC motor is approximately 90.91%. - The useful power of the DC motor is 30 kW.

Please let me know if you need any further assistance!