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Statistics on the actual power consumption of mechanical equipment.

2017-08-28View Original

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I work in a chemical company. Since there are only a few main meters in the power distribution room, I want to calculate the actual electricity consumption of each unit. How can I do that? The calculated rated power is definitely inaccurate; the actual operating power will not reach the rated value. For example, the actual operating power of a centrifugal pump motor equipped with an external inverter, whose rated power is 110 kW. Is it the measured current at runtime multiplied by the voltage of 380? That is quite different from the rated power. Once measuring the operating current is done, how does a clamp meter manage to determine the actual current value? I would appreciate it if you could share your insights.
Reply #22017-08-28
Three-phase AC motors have two connection methods: delta and star. A clamp meter measures line current ; Star connection: Line voltage = 380, Phase voltage = Line voltage/SQRT(3), Line current = Phase current ; Triangle: Line voltage = Phase voltage = 380, Line current = SQRT(3) * Phase current ; For a three-phase symmetrical load, the active power P = 3 * phase voltage * phase current = SQRT(3) * line voltage * line current; multiplying this value by time t gives a figure that is close to the reading on the electricity meter. You are missing the square root of 3. Go back and review the electrical engineering course; it’s something that students in science and engineering fields should have learned.
Reply #32017-08-28
√3 *U*I*cosφ
Reply #42017-09-04
Is it still necessary to multiply by square root three for the compensation cabinets in substations?
Reply #52017-09-04
Is it still necessary to multiply by square root three for the compensation cabinets in substations?
Reply #62017-09-04
Is it still necessary to multiply by square root three for the compensation cabinets in substations?

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