How to Calculate Electrical Consumption If Amps Measured Are Not Equal?

Add them together, then divide by 3.

what are you calculating,kw or kva .to cater for people who are unsure there are instruments which does all the doubtfull calculations to avoid disasters, some are known as digital power meters.

We have a "triple your money back" guarantee if our answers are not correct.

Are the "amps during startup" different than the amps running? (just nod 'yes') You realise of course that basing your kWh calculation on startup current will give you different results than you may be expecting, yes?

Counting beans is different than counting AMPS.

On the other hand, maybe it's an ecectrical resistance furnace.

At one time people thought that polio and then consumption would follow the path to extinction that small pox did. We still have a chance to eradicate polio if some mullahs would listen. [I know, that's more of a pipe dream but hope springs eternal. The CIA were silly fools, too.] Unfortunately, drug resistant strains will keep consumption as a scourge to humanity.

As for your energy use to run a furnace, there really is no way to tell with the information you provide. The starting current will always be a bad metric because thermal equilibrium and synchronous motor speeds have yet to be achieved. Similarly there is no information about what gets heated in this furnace or how much [tonnes, milligrams] of this mystery material gets heated. We also know nothing about the furnace heating source. If the electric energy just pumps fuel and air to heat the furnace then that's one scenario. An induction or resistance furnace are two completely different scenarios.

at least it isn't the dreaded tuberculosis

Im guessing you're in Norge or Poland if you're on 380V?

Go for LPG

Go For LPG...

Personally I would look at the question as three separate single phases, take the current, voltage and pf of each phase to calculate the power of each phase - then add the results to get the total power.

If you look at the math, an average current does the exact same thing to get a single answer. Your technique gives the power taken by each load as an intermediate step.

I'm mildly surprised that nobody has asked if this is a star or delta wiring.

are you in the electricity department assigned to calculate line losses due to theft??

I assume 380 is your phase to neutral (or ground) voltage. Total power is:

P=380 x sum of phase currents, i.e. P=380*(120+130+90)

If 380 is your phase to phase voltage, divide by sqrt(3), P=(380/1.732)*(120+130+90)

You called this start up current. For steady state power, use the steady state currents.

You didnt purchase the furnace yet. Then, how did you get this starting current or given by whome?....Why dont you get the kVA or kW of the equipment it self

I agree in principle with Rixter except that P it is actually S [VA] and not kW.

Since the O.P. said 130x380x1.73x.95 = 81, 2 KVA it is obviously 380 it is the line-to-line voltage [this could be 400 V the new rated voltage in Europe].

Then the consumed power is (120+ 130+ 92)*380/sqrt(3)*0.95/1000 [if 0.95 it is general p.f.]=71.28 kW. 71.28 kW*1 hour=71.28 kWh. In 18 hours the required energy will be 71.28*18= 1283 kWh.