Determine the HP output of the motor. Convert to kW.
From applied voltage, calculate the minimum current.
Add about 30% for losses.
Still a WAG.
Or:
Three phase current before phase loss times root three.
Add about 20% for losses.
Still a WAG.
Losses may be greater than my estimate.
Bill
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"Why not the best?"
Jimmy Carter
Losses WILL be much greater. When a 3 phase motor is single phased AFTER IT WAS ALREADY RUNNING*, it’s an extreme imbalance and the unused phase will still become magnetized by the others, but at the wrong phase angles and as an additional burden on those other windings. It will then create negative sequence current in the rotor bars, which then creates negative torque, fighting the normal torque created. That fight turns into heat in both the rotor and the stator, which is why 3 phase motors burn up when fed with single phase unless the load on the motor is less than 50% of its rating. That 50% is a “rule of thumb” though, I’ve never taken the time to do the math on it myself because it’s not something I needed to know. I will say however that I used to use 33% until someone in this forum did post some info on how it could be 50%. That was many years ago now, but it should still be searchable.
* If a 3 phase motor is fed single phase before it is already spinning it will not start, so any discussion of what it looks like electrically is moot beyond Locked Rotor Current.
" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden
In 3-phase system, hp (shaft) = kW = (I x E x 1.73 x pf)
Losing a leg & holding kW constant: I x E x pf. Current will go up while pf will also degrade. Of course I^2R losses will be greater since current increased. Motor overloads will trip motor and if auto reset, will turn on if LVR type, and cycle again and again, unless phase loss relay present. In short, voltage may remain same depending on plant size, but current will increase by at least 1.73 and likely double.
