Confusing nameplate data

[cfordyce]

I have 10 machines, each with a specified maximum machine load of 168kVA. Each has 126 1kW motors, 2 5kW motors and some Inverters, controls, electronics, etc.
The machines run at 380VAC. They are powered by a 600V bus through a 1000KVA 600:400 stepdown Transformer.

I have an old record of an electrician's current reading on one machine drawing 60 Amps on the primary of a smaller 600:400 step down Trans when powered separately.

60x600x1.73=62kW

10 machines on the 1kVA transformer represent a loading of
10x62kW= 620kW or 62% loading on the transformer. That makes sense.

But why does the manufacturer specify a 'max machine load' of 168kVA? That would mean that my 1000kVA transformer is 168% loaded!! Can anyone explain this?

I am installing additional machines from this company and they are specifying a max machine load of 207kVA for the new machine. It has 152 1kW motors and approx. 10kW of additional motors and inverters, controls, etc. similar to the old machines. I need to size a new transformer and bus bar, cable, etc. for the new equipment...

 

[peterb]

Your nameplate is stating the maximum machine load in kilo volt-amperes (kVA), while you are adding up kilowatts. The difference is due to the power factor of the motor loads (typically in the region of 0.8); kW = kVA*PF.
Yes, your 1000 kVA transformer will be overloaded if all of the motors on all of the machines are running simultaneously at full load (maximum machine load). This is usually not the case, so there is a diversity factor to be applied based on the actual load conditions. Get some help/information from the machine supplier or an experienced consultant in sizing the new distribution equipment.

 

[IRstuff]

A factor of 2.5 for transient and startup seems plausible.

TTFN

 

[busbar]

Is it possible that the mechanical output of the motor is mistakenly being used as the rated electrical input? Based on North American standards, a 1KW motor is ~1.34hp and approximates an NEC table 430-150 electrical load of ~2120 voltamperes.

 

[peterb]

Oops! Of course, Busbar is correct. What I MEANT to say was that:
INPUT kVA = (Output kW)/(efficiency * power factor)

 

[jbartos]

Suggestion: Sometimes, the max kVAs are indicated to reflect the largest possible load that actually may or may not materialize. The load then varies from the application-to-application.

 

[Trillian]

Just a small reminder...

VAs^2 = Watts^2 + VARs^2
Watts/VAs = cosx = Power Factor (where x is the phase angle)

VAs are used in transformer ratings to better reflect the overcurrent danger. Watts won't do this.

Remember also, that the voltage ratio on the nameplate is NOT a true reflection of the turns ratio, as this is not an ideal transformer: it has core and copper losses to account for. This will affect your current calculations as well.

Efficiency = Pout / Pout + Iprimary^2Requivalent,
where Iprimary^2Requivalent = core (fixed) and copper losses (varies with the load)

Requivalent can be found with two simple tests: Open and Short Circuit.

 

[electricpete]

I think the nameplate voltage ratio is the turns ratio for that tap.

Actual voltage ratio may differ from both of the above under load due to voltage drop across the leakage reactance. Correct me if I'm wrong.