How many units are consumed in KW/h ?

45 KWH

mechanical:

1 HP = 0.746 KW

60 HP = 60*0.746 KW

Electrical Power

= mech power/ motor η

But that depends on the motor loading. It is highly unlikely (unless it is homework) that 60HP motor is loaded at 60HP

Units consumed in KWH (not KW/H) = KW * no of hours (1 for you)

= KW calculated as above.

- You must know motor loading,

- Efficiency at that load.

Don't forget the utility's premium for starting and running it!

It depends upon how much of the motor's 60hp capacity is utilised.

Say it is fully loaded. 1hp ≈ 746W, which comes out at ~45kW, and there will be a bit more to account for motor losses, say 10%, which comes out at approaching 50kW. So the maximum it could be is around 50kW = 50kWh per hour = 50 "units" per hour, though it does rather depend upon how well loaded it is: if it is spinning freely and not doing much work then it will be far less than this though one might want to switch it off to save a bit on the tariff, particularly if there is a low power factor penalty on it.

At 60 HP, (60 x 0.75) / Power Factor would be the no of units consumed per hour at full load.

I.e., if the power factor you can maintain at abot 0.9,

60 x 0.75

--------

0.9

= 50 KWH power consumption,

i.e. 50 units per hour would be consumed.

Regards,

Ashok Shukla

Delhi

9810145598

Hello Ashok - what you wrote would be correct if you substitute efficiency for power factor.

Changing power factor does not change the no of units (kWh) consumed. Depending on the tariff, it might change the amount you pay the electric supplier, but that's another story.

On the other hand, efficiency does affect it. I don't think any other poster has mentioned it, but a 60 HP motor produces 60 HP shaft power (at rated load). The input electrical power is greater. Of course, as others have said, it may not be operating at rated load.

Cheers.........Codey