The issue of calculating the power of motors
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Everyone, when calculating motor power, I use the formula 1.732 × rated voltage × rated current × power factor, but the resulting value is always a bit high. How can I calculate it more accurately?
The calculation formula is unique; incorrect calculations can only be due to inaccurate data. Generally, one should know the power in order to calculate the current, and if the current and power factor are known, then the calculated power should represent the actual power. If your data is accurate, then the results will be accurate as well.
Your calculation formula is wrong; efficiency must also be taken into account! ?
The efficiency of the motor is something that must be taken into account.
It would be right to consider efficiency again. . . .
The efficiency isn’t indicated on the motor’s nameplate – how can I find it?
According to the relevant **standard requirements, different power levels and different applications have distinct requirements.
Motor efficiency = Output power / Input power * %
For three-phase motors: Rated power = 1.732 × rated voltage × rated current × power factor × efficiency. For single-phase motors, the formula is Rated power = rated voltage × rated current × power factor × efficiency. The voltage, current, and power factor can be found on the motor’s nameplate; the power value can also be obtained there. As for efficiency, it is necessary to refer to the manufacturer’s technical data or **Xiangguang standards** You always get higher values in your calculations because you ignore efficiency; since efficiency is less than 1, your calculations end up being higher than the actual value. Power refers to the output power of the motor, while what you have calculated is only the input power of the motor (that is, the power absorbed by the motor from the power grid). The input power of a motor is greater than its output power, as energy is also consumed during operation (mainly due to iron losses and copper losses); therefore, you have not taken these losses into account. In reality, when considering a motor, what we take into account is its capacity to handle loads, that is, its output power.
Your algorithm is correct; the efficiency is the power factor divided by the rated power