3 phase Kwatt hours

[ali68]

Hello,

Not strictly a DIY question but hoping you can help, basically I am after the formulae for working out kilowatt hours for a 3 phase load.

Bit of background, we have a vessel that plugs onto a 3 phase socket and on average uses 28 units for 14.5hrs, but if he plugs into another socket on different pier the meter on that pier registers 59 units for 14hrs, again average.

Going to clamp the 3 phases to find out what he is pulling power wise.



Also when measuring the voltage is it phase to phase or phase to neutral for the formulae?

Hopefully I might have some figures to work with tonight.

Tried googling but there seems to be different ways to do it.

Any questions and I will try to fill in the blanks.

Thank you

 

[ricicle]

If the load is not balanced and varies then it will be difficult to monitor power usage with a clamp meter.

As a basic guide 4.35 amps is approx. 1 kW @ 230v. If each phase were to pull 4.35 amps then the total load would equal 3 kw. This load for 1 hour would be 3 units ie 3 kWhs.

8.7 amps per phase = 2kw per phase. total load = 6kw. for 1 hour = 6kWhs (6 units)
8.7 amps per phase = 2kw per phase. total load = 6kw. for 0.5 hour = 3kwhs (3 units)

and so on.

 

[Paul_C]

If the load actually is balanced (i.e. same current in each phase), then you can calculate the current in each phase from:

I = P / (1.73 x E)

Where P = power in watts, and E = line voltage (i.e. 415V for a 240/415V system).

That assumes unity power factor.

 

[ali68]

Totally understand a clamp meter is not ideal as the ships load will vary

We put the suspect meter on test today in the workshop with the following results.

10.8amps load on each phase,used a hydraulic power pack as a load.
240v average phase to neutral and 440v phase to phase average.

How long should that have taken to register 1 unit on the meter?

The result we got was not what we expected but the result was the same with a new meter drawn from the store.

I am not a electrician by the way,more a mechanical/hydraulics engineer with some electrical knowledge,just trying to help out a young lad who is just out his time and his tradesman retired last month so he has kinda been thrown in at the deep end.

 

[westie101]

So total load is (10.8x240)3=7.8kW.

So in an hour that would be 7.8kWh

1kWh would be achieved in 60/7.8 = 7.7mins

(all figures to 1decimal place)

 

[Paul_C]

That's assuming a purely resistive load. I'm guessing that the hydraulic power pack referred to is basically a motor driving a hydraulic pump, so the power factor is quite likely a little less than unity, which means that the actual power being consumed is less than the measured current would suggest.

 

[ali68]

Using this formulae we came up with the answer of 4Kw.

1.732 x VOLTS x AMPS x PF=watts.

but even with two wildly contrasting answer the meter took 35 mins to register a unit as did the new one drawn from the store.

 

[westie101]

What actual figures did you use for volts & amps?

 

[ali68]

240 and 10.8 Westie, PF he used 0.9 muttered something about using this at college if the actual pf was unknown!

 

[westie101]

Using that formula you should use the ph to ph volts.
But if p-n = 240V, ph - ph =415V

1.732x415x10.8x0.9 = 7kW

So 1kWh would be in 60/7 = 8.5 mins


I'm assuming by meter you mean something like a private version of a standard electricity meter. If it is of the older rotating disc type it will have a "revs per kWh" comment on the face plate