Rated Current Calculations

Hello Abo,

Are you certain you have the spec right? The supply voltage seems odd to me! Is it a 3 phase system? What does it say on the spec plate of the unit! Can you post a picture of the said plate?

Happy 2008

From the details i feel its 2 pole machine suitable for 60Hz Supply.As pointed out Voltage mentioned is incorrect.Alos effiecy at half load is more than full load which is not the case for most motors unless specially designed .The formula is

Rated kw÷√3×Voltage×Full load Eff×Full load PF

with given figs if u calculate current is approx 188amp assuming it is 4.16KV

As long as you are talking about a motor, its rated power is a mechanical one, thus the current will be the the mechanical power divided by the rated efficiency, rated power factor, square root of 3 (supposing a three phase system) and rated voltage, that is 189 (A) rather than the 195 (A) present.

Welcome,

You need to clarify your question a bit. The data is good, but you already have full load current which could be taken as rated. Or are you talking max continuous current? Or starting current?

You did not specify service factor. Typically 1 to as much as 1.5, which would tell you how much current you can draw safely under continuous conditions.

The rated current is 195 Amps. i.e. the full load current.

It is possible that the motor is capable of higher than that, i.e. the full load current may be the full load current driving the fan at full fan load, but you can use that as the rated current.

Best regards,
Mark Empson

http://www.LMPhotonics.com | http://www.LMPForum.com

The rated current mentioned in the name plate is the current drawn by the motor when delivering full out put.The actual current will be less depending upon the loading as well the margin considered while arriving at the rated power for motor.Normally minimum 10% over fan BHP will be considered so the fan power required will be 1100kw and current drawn will be this KW divided by root 3 pf and Efficiency at 90% load and the voltage.

Since you have as you say "v. good knowladge about electrical engneering, could you please explain to a layman:

1. How can efficiency @ 50% rated voltage be better than @100% rated voltage?
2. What is exactly "rated current cal"?

Beside, if you are asking about FLA, isn't it already in your specs? enlighten me please.

Wangito.

PD: What you obviously lacking is the knowledge about using the spell checker.

It is normal for the peak efficiency of an induction motor to be at less than 100% load. Most machines have a peak efficiency at around 75% load, so 50% is not impossible.

Best regards,

Mark Empson
http://www.LMPhotonics.com | http://www.LMPForum.com

Seems to calculate. I = P/1.73/U/cos*/efficiency. therefor calculated @ 50% data: I = 1200000/1.73/4160/.91/.962 = 190.47A regards.