Query on a 3-phase motor

Ok..the 1 hp is OUTPUT power. The inverter seems small for the motor load. There may be 746v watts to the horsepower, but you have to send in more than that to account for various inefficiencies. A rough guess would be 1000 watts IN for 1 hp out and that's not even taking into account the initial inrush current, which is probably three hundred plus percent of normal.

The motor seems to be connected in Star (or 'Y') configuration, with the mid point being brought out. There is no power advantage to be had by trying to change it to Delta. There may well be a problem, also. If the windings are designed to be run in Star ONLY and then are changed to Delta, you will now be applying almost double (square root of 3) the voltage across each one of them.

mike

Thanks for confirming my thought that the motor is wired star with the star-centre connected to neutral. Although the inverter documentation states that it will drive a 1hp motor you point is well taken, and since I'm a hobbyist I doubt I will ever actually use 1 hp.

There is no power advantage to be had by

I don't understand this statement. As it stands, each phase is sent through 2 windings - from the input to the star centre is 1 winding, and from the centre to the output is the next. This is how the motor was designed to run on a 415v 3-phase supply

- each winding 'sees' only about 230v. _If_ I can separate the star centre and convert it to delta, then each phase will go through 1 winding. But since I'm then using a 3-phase supply at 230v, each winding gets exactly the same current as before, and the motor will run as intended. Is this logic faulty?

Exactly. If you connect it in a delta, each winding would get the full voltage..but..in a Y connection, the voltage across each winding is equal to the (line voltage) / (root three). The line voltage is thus root three times greater than phase voltage

Note that the Y connection phase current is equal to the line amperage. In a delta connection, the winding sees FULL line voltage, but only has (line current) / (root three) phase current, or put another way, the line current is root three times that of the phase current.

So, if power in watts = line voltage * line current * (sqrt 3) * power factor * efficiency, there is nothing to be gained by either connection. You are merely multiplying a different commodity by the root three, but the answer will be the same.

Another way to calculate power is to sum the three phases. Then we get

power in watts = 3 * (phase voltage) * (phase current) * (power factor) * (efficiency)

That will give exactly the same result as the previous formula.

Nope. the current will be divided into the two windings attached to a given line. Each phase winding current will be root three divided into line current.

Well, not really, but it's easy to forget that the other lines are still busy while we're looking at just one of them.

This will help if I've muddied things up too much.

I'd be careful about applying full line voltage to one phase. Is this a dual voltage motor as noted on the nameplate? The manufacturer may have saved money by making it suitable for Y only. The higher voltage per phase winding in a delta needs a more expensive coating on the windings.

mike

Some very wrong stuff.

The power output in the motor does not change IF it's dual voltage and converted from delta to a higher voltage in the wye configuration. A method of reducing startup current (with a fixed line voltage) is called the wye-delta starting.

sorry for the error.

mike