To add some information here, the complete picture - Power Suplying System and Load Device - will have a fixed KVA level. The load draws "X" amount of KVA and the supply must be able to deliver at least the same "X" KVA level, or the load will not be able to draw the required True Power (Wattage) it needs to develope the required Horsepower and Torque.
Results are a decrease in Rotor speed when the work exceeds the maximum HP which the Motor can develope.
This causes most Induction Motors to stall when they slip too far behind or come within a Syncronous Frequency point.

So much for the boring babble! [Linked Image]

I ran some numbers and here are the results:

<OL TYPE=1>

[*]Using the listed 142 Amp @ 240 VAC 3 Phase figure, this equals 58.96 KVA - 58,958.4 VA to be exact,

[*]58.96 KVA on the Single Phase side relates to 245.83 Amps @ 240 VAC 1 Phase.
</OL>

Just something to consider.

P.S. here's the formulas used:

KVA:

I x E / 1000 [1 ph.] I x E x 1.73 / 1000 [3 ph.]

Amperes:

KVA x 1000 / E [1 ph.] KVA x 1000 / 1.73 x E

Scott S.E.T.

Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!