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I need to know how much electricity a machine uses in kW please.

On the machine information plate it states: 460 volts, 60 Hz, 161 amps, and it is a 3 phase.

I took the readings for the volts and amps: Phase 1- 271 volts & 37.68 Amps Phase 2- 271 volts & 37.62 Amps Phase 3- 273 volts & 38 Amps

I cannot figure out the Power Factor to apply to the formula I am using: kW= (Volts x Amps x Power Factor x Sqrt 3) / 1,000

I know Power Factor = Real Power/ Apparent Power, however I cannot figure out the real power. I believe the apparent power is (460 x 161 x Sqrt 3) / 1,000 = 128.27 kVA?

Could I just average the voltage readings and then average the amperage readings for the 3 phases and then do (Avg volt x Avg amp x Sqrt 3) / 1,000 to get the true/real power without using the Power Factor?

Are my assumptions correct? Can someone please help me calculate the electrical consumption please! Thank you

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  • \$\begingroup\$ The "proper" way is to measure V and I and relative phase angle OR compute RMS power by calculating instantaneous power many times across a mains cycle and doing an RMS calculation. HOWEVER - there is equipment available which does all the hard work for you. Someone mentioned a watt meter - which is what you want. And/but you can but ICs which do the whole task of development kits which can be persuaded to do it either all 3 phases at once or a phase at a time. A typical energy meter IC inputs Vin/1000 and the voltage across a current shunt (which can just be two points on a feed cable ... \$\endgroup\$ – Russell McMahon Dec 5 '14 at 14:44
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Your assumptions are not correct. The machine plate states the maximum voltage and current at full load. But in your measures the machine is far from full load.

You are measuring voltage from phase to neutral. But in the formula $$P=\sqrt 3 U_L I \cdot PF$$ you must use the line voltage (from phase to phase).

If you're using the phase to neutral voltage, the formula becomes $$P=3U_fI·PF$$

The PF must be calculated by $$PF=cos \theta $$ where angle theta is the phase difference between voltage and current. You would need a current probe, a x100 voltage probe, and an oscilloscope for this measure.

You might instead use a wattmeter to perform this measure.

If none of this instruments is available, we could follow another approach. It is not something standard as far as I know, but a way it could be done.

Let's do $$S=3U_fI$$ to calculate the apparent power (letter S), in your case it is more or less 30 kVA. You can draw it as a circle of radius 30 (using some scale, for example 1 cm=1 kVA=1 kW=1 kvar). The x-axis is the active power (real power in W) while the y-axis is the reactive power (in var). This is circle blue in the figure. The working point is somewhere in the circle, but we don't know where yet.

Power diagrama

We will substract some reactive power from the system by placing 3 capacitors as in the next schematics:

schematic

simulate this circuit – Schematic created using CircuitLab

The capacitors modify the total Q (reactive power) but not the total P (active or real power) of the whole system seen by the mains. The total Q will be now: $$Q_{new}=Q_{old}-3·U_f^2·2\pi C·f$$

If C=100 uF, and f=50 Hz (frequency of the electrical network which depends on the country), then new Q will be $$Q_{new}=Q_{old}-7 kvar$$

Therefore, the new working point of the whole system (capacitors + machine) is inside the circle red in the figure. It has the same radius as before but it is shifted 7 kvar down in the y-axis.

Finally, you will have measure the new current with your tools and calculate the new apparent power. $$S'=3U_f I$$

It will have changed, and now it is for example 26 kVA. Whatever radius value it has, you draw it and this will be circle green.

Finally you must look at the intersection between circle green and cicle red (point A). If you look at the x-axis coordinates of this point, you get the real (or active) power.

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  • \$\begingroup\$ I have 2 tools at my disposal, a Fluke 115 True RMS Multimeter & a Fluke 325 True RMS Clamp Meter. If you could please guide me through the steps needed to calculate the most accurate kW (true power) with the resources at my disposal, I will be eternally grateful. Thank you! \$\endgroup\$ – Berk Dec 4 '14 at 0:01
  • \$\begingroup\$ No problem. I've added in the answer a possible way of do it with your available tools. You only need to buy three capacitors with proper rating! \$\endgroup\$ – Roger C. Dec 4 '14 at 16:31
  • \$\begingroup\$ @RogerC. I like how you and I are approaching the reactive measurement different and somewhat complementary ways. You add VAR (or subtract) and I suggest adding pure watts mechanical load. What fun :-). \$\endgroup\$ – Russell McMahon Dec 5 '14 at 14:39
  • \$\begingroup\$ @RussellMcMahon, completely agree that EE is fun. Not everybody sees that way though! \$\endgroup\$ – Roger C. Dec 5 '14 at 15:15
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Best guess with data so far is
VA input = Uf x (I1 + I2 + I3)
~= 113A x 272V = 30.736 kVA.

This is the upper limit for PF=1.
PF is likely to be in the 0.8 - 0.9 range giving a guestimated power of around 24 to 28 kW. Meters available do not allow you to determine phase angle which is needed for PF calculation from electrical data only.

If you know the mechanical power output you can determine power factor from
PF = Watt_out / VA in
If you do not know the load power but are able to incrementally alter power then you may be able to determine the PF for the delta load change. eg if you can add 1 kW of friction load and the electrical input rises by say 1.2 kVA then PF for the incremental amount = 1/1.2 = 0.83. This would be very rough as delta PF is probably not the same as total PF.

What is your load?
Can you measure load power?
Can you add measure delta load?

For interest, where are you located?

As a guide only from here this table relates motor sizes, loads and typical power factors.

enter image description here

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  • \$\begingroup\$ Torrance, California. I will look into the motor when I get a chance \$\endgroup\$ – Berk Dec 4 '14 at 18:30
  • \$\begingroup\$ Baldor JMM3713T 15HP 3450RPM 3PH 60HZ 215JM 3744M TEFC mrosupply.com/motors/ac...tric-motors/… This is the motor on the parts washer. What is the next step to figure out the kW (True Power)? \$\endgroup\$ – Berk Dec 4 '14 at 22:44
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The "proper" way is to measure V and I and relative phase angle
OR compute RMS power by calculating instantaneous power many times across a mains cycle and doing an RMS calculation.

HOWEVER - there is equipment available which does all the hard work for you. Someone mentioned a watt meter - which is what you want.
And/but if you want to do this ultra cheaply and with not too much effort you can buy ICs which do the whole task or development kits which can be persuaded to do it either all 3 phases at once or a phase at a time.

A typical energy meter IC inputs Vin/1000 and the voltage across a current shunt (which can just be two points on a feed cable. One IC which is very easy to use for 1 phase at a time is the Microchip MCP39F501. The development kit here gives and idea of the circuitry. There is a cheaper more useful one available from Microchip. By changing the voltage divider and using a lower R current shunt you could do your task. Dev kit is about $US120 fro uChip AFAIR BUT the IC is under $5 1 off and a suitably keen person using the dev kit and available free software could get all you want. Software allows an optically isolated USB interface to a PC where you can see eg V I VA reactive Watts and a bit more.

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