The full load current for a three phase AC motor rated at 7.5 hp and using 208 volts

It varies somewhat from motor to motor. It's usually specified on the tag on the motor. I just looked at a 7.5 HP Baldor motor, and it's full load current was spec'ed at 23 amps at 208V. I checked out an Emerson motor, and it's current was 20.6 amps at 208V. That should give you a general idea of the typical current draw of such a motor.

as u know the formula of power is P=sqt3*V*I
1hp=746w
7.5hp=746x7.5
=5595w
so 5595=208xI
and I=5595/208x1.73
I=15.5A

While your calculation is correct to arrive at the true wattage of a motor:

Quote:

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Originally Posted by ibadullah

as u know the formula of power is P=sqt3*V*I
1hp=746w
7.5hp=746x7.5
=5595w
so 5595=208xI
and I=5595/208x1.73
I=15.5A

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However, a simple "unit conversion" will not take into account the mechanical efficiency, electrical efficiency, or power factor of the motor, and the actual amperage draw will be more that your calculation shows.

Calculations won't give accurate results as amp depends on number of poles or speed of the motor too

Thanku yes I know that calculation is different frm the practical

Quote:

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Originally Posted by ibadullah

thanku yes i know that calculation is different frm the practical

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please correct me if I'm wrong,
if considering the mechanical eff, electrical eff, and power factor of the motor:-

for example,

7.5HP = 7.5 x 746
= 5595W
this is the mechanical power.

electrical power = mechanical power / efficiency (if eff. = 0.92)
= 5595/0.92
= 6081.52W (at 100%)
normal run usually 90% max = 6081.52 x 0.9
= 5473.37W

current flow at rated load (5595W) = electrical power / PF (if PF = 0.83)
= 5473.37 / 0.83
= 6594.42VA

full-load amp = current flow / (sqt3 x V)
= 6594.42 / (1.732 x 208)
= 18.3A

... another old post...
Although after all the math wouldn't table 430.250 be just as suitable for most applications?

Also it depends on design,design code,code letter,material used,efficiency etc

All I am attempting to point out is that, academically you may be correct...
But from a construction point of view that level of specificity is usually not required.

thanks, and I agree with both of you guys.. (sinnadurai & hfcarson). Generally I'm use this simple formula to calculate the amps for three phase.
amps = kW / (1.732 x 415 x eff. X PF) --> where here normally using 415V system. Eff & PF referring to manufacturer details.