Energy cost calculation of motor

The power factor [PF] number is missing.

Is there a tariff penalty for connecting low PF loads to the supply?

Is there a tariff penalty for high starting currents?

Mention the energy cost in $ per unit of usage.

You need the efficiency, load factor and operating hours. The equation is:

kWh=0.746*HP*LF/Eff*Hours

For a rough estimate, you could assume the efficiency and load factor are equivalent. But if this is a realistic assumption will depend on the use of hte motor.

IF you are located in the U.S. contact your energy utility and ask how they bill usage.

The above equation is what I use, although xcel energy charges a "demand charge"

which often doubles the bill when using large motors.

The "demand charge" is "The demand charge covers the cost of maintaining extra generating capacity so we can meet your peak demand needs at any time during the month. Your bill is based on a lower rate per kWh and the additional demand charge for the number of kilowatts your company used at its highest 15-minute period of usage during the month."

Sorry Guys for the late reply. I truly thank you all for the answers so far.

I have come to know that this particular motor runs anywhere between 62-68 A at any given time. Now taking 65 A as a standard load, can we figure out Cost?

For Example:

P = 1.732 X V X I X Pf

To find out Pf, we can use HP Vs RPM graph based on Load percentage. For our case, Load % would be 65/76 which is 0.85. Now there are graphs for 3/4 Load which I used. Knowing HP and RPM, I took an educated guess of 0.91 from the graph.

From that, I was able to calculate to tal power consumed @ 65 A.

Is this right way to find out Pf? And are these calculations valid?

A induction motor star conected 4.5kw motor applied voltage415.

no load riding 1) R/ph 2.9. 2) 4.65 amp 3) 565 w 4) 49.50 hz

full load riding. 1 . 10.02 amp

5.434 kw

full load calculation is as follows:

motor kw = (horsepower * .746) divided by % efficiency

for example:

a 5hp motor rated at 80% efficiency

5hp*.746 / .8 = 4.66kw * cost per kWh = cost per hour to run

for your example motor, assuming 94% efficiency (typ for a motor this large)

600 * .746 / .94 = 476.17kw

476.17kw @ $.10/KwH = $ 47.617 per hour