Hey Martin,
The formula P=U·I·√3 is correct to calculate the so-called apparent electric power. This is the information that is needed to dimension the power supply and the cables. But if they sold you a spindle as 2.2 kW spindle based on this formula, or if that is the power that is marked on the nameplate, then they fooled you.
According to IEC 60034-1, an electric motor shall be provided with a rating plate on which shall be marked the rated output. According to IEC 60034-1, the rated output is the mechanical power available at the shaft and shall be expressed in watts (W).
For a three-phase induction motor, this calculates this way:
η = Pmech / Pelectr = Pshaft / (UN · IN · √3 · cos φ)
where
η (eta) is the efficiency,
cos φ (cosinus phi) is the power factor,
U is the rated voltage, and
I is the rated current.
So the rated mechanical output power on the nameplate of an induction motor is calculated this way:
Pshaft = UN · IN · √3 · cos φ · η
Let’s say for calculating an example let’s use my spindle. According to its datasheet, in duty type S1 (continuous running duty), the rated voltage and current are:
UN = 230 V
IN = 8 A
I don’t have power factor and efficiency at hand, but let’s use the typical values:
cos φ = 0.82
η = 85%
This spindle is sold as a 2.2 kW spindle, so for the mechanical power available at the shaft, which shall be marked on the nameplate, you calculate:
Pshaft = 230 · 8 · √3 · 0.82 · 85/100 = 2221.32 W
Okay, this is true, it says 2.2 kW on the nameplate. Now for this spindle, the so-called electrical active (real) power is:
Preal = 230 · 8 · √3 · 0.82 = 2613.31 W
And still for this spindle, the so-called electrical apparent power is:
Pappar = 230 · 8 · √3 = 3186.97 VA
Okay, you see, a spindle rated as 2.2 kW spindle in fact sucks an apparent power of 3.2 kVA. The VFDs reflect that: If you have a look at the datasheet of my 2.2 kW VFD, it says, for a typical 2.2 kW induction motor, it provides 3.8 – 4.5 kVA output power (dependent on voltage, 200-240 V) (this VFD is rated for up to 11 A output). And it is sold as 2.2 kW VFD, which means, it is for a spindle that makes 2.2 kW mechanical power available at its shaft.
So now to your spindle: So if they told you they calculate your 2.2 kW power rating by the formula P=U·I·√3, which is the formula for the so-called electrical apparent power, the current calculates this way:
I = P / ( U · √3 )
2200 / ( 220 · √3 ) = 5.77
Which corresponds approx. to the 6 A they told you.
So let’s say the power factor cos φ = 0.82 (a typical value), then the active (real) power is:
Preal = 220 · 5.77 · √3 · 0.82 = 1802.90 W
and let’s say the efficiency is η = 85 % (a typical value) then the rated (mechanically available at the shaft) power is:
Pshaft = 220 · 5.77 · √3 · 0.82 · 85/100 = 1532.47 W
So in fact, to comply to IEC 60034-1, it should have marked “1.5 kW” on the nameplate (and not “2.2 kW”)