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P = sqrt(3)*U * I * cos(f) is the real power for a symmetric load (U and I are RMS values of voltage and current of a single load), but what about generators?Do they have to have equal electromotive forces too?

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    \$\begingroup\$ I don't know what you are actually asking? You mention power consumption of a load then ask something about generator voltage. Could you be a bit more specific? \$\endgroup\$ – David777 Aug 31 '19 at 20:33
  • \$\begingroup\$ @David777 I was wondering when the mentioned formula can be used.It requires that all three loads in a three phase circuit are equal (equal impedances), but what about generators?Do they need to have same output voltages? \$\endgroup\$ – user3711671 Sep 1 '19 at 9:44
  • \$\begingroup\$ See if my answer below helps you any. \$\endgroup\$ – David777 Sep 1 '19 at 13:04
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Don't know if this will help as your question is still very vague to me, but you can let me know.

Remember the general power equation measured in Watts, (P= voltage x current), where the total power P is a product of the voltage and the current. This can be applied to find the power PRODUCED by a voltage source e.g. a generator, or the power CONSUMED by a load e.g. a heater. It does not matter if power is delivered or consumed, it is just a measurement. However this formula is generally more useful for DC circuits.

But your mentioned formula (P= 1.73 x V x I x p.f) is for calculating 3 phase AC power, hence the 1.73 constant (sqrt of 3) and the power factor (p.f). As you have sort of discovered, this equation requires all voltages, current and impedances to be reasonably balanced. If load impedances are different, the generator output voltages will be different and your formula cannot be used accurately. What you could do is work out the total power per phase by removing the 1.73 (sqrt of 3) from the equation, giving you the power per phase (P= V x I x p.f). Then adding the 3 different powers together will give you the total 3 phase power either DELIVERED from source or CONSUMED by the load (whatever you are trying to work out, remembering the power factors will be different for the genset and the load).

On a side note, the output voltage of a generator depends on the load connected. A generator with a balanced load will have identical output voltages (ignoring synchronous reactance variances of the windings). Now a generator with an unbalanced load connected, will NOT have identical output voltages as the current drawn will be different, producing different internal voltage drops. To work out power in this case, each phase must be treated as a single phase load (be careful with generator & load winding configurations as this can complicate it slightly).

As you stated in your question, these must be RMS values for your mentioned equation to stand true.

Also it may help if you do some background research on how 3 phase generators generate voltage, and how phase voltages may vary depending on phase loading. This is another topic which requires too much explanation in this question, but feel free to post another question if need be.

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Neither loads nor generators "have to have" symmetric or equal electromotive forces, but it is desirable for economic utilization of equipment for the loads and sources to be symmetric. It is also less complicated to calculate power is everything is symmetric. The formula cited in the question applies only to balanced (symmetric) systems.

If the load or source is not balanced, the two-wattmeter method can be used to measure the power. The two-wattmeter method can be employed to measure the power in a 3 phase, three wire, star or delta connected, balanced or unbalanced load. The total power absorbed by the three loads, is equal to the sum of the powers measured by the two wattmeters.

The method of symmetrical components is used to simplify the analysis of three-phase unbalanced systems. The method involves converting a three-phase unbalanced system into two sets of balanced phasors and a set of single-phase phasors. These sets of phasors are called the positive-, negative-, and zero-sequence symmetrical components.

The two-wattmeter method and the method of symmetrical components are derived and described in detail in course notes and reference material that can be found on the internet as well as in text books that can be found in libraries.

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  • \$\begingroup\$ But if loads aren't symmetric then the mentioned formula isn't correct, right?What if generators don't have equal output voltages?Does the following formula still apply? \$\endgroup\$ – user3711671 Sep 1 '19 at 9:45
  • \$\begingroup\$ No, see additions to my answer. \$\endgroup\$ – Charles Cowie Sep 1 '19 at 14:32

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