ASSUMING the 400 V is phase to phase, that means 400/1.732 V per phase * 65.3 * 0.88 = 13.27 kW so over 1 hour it will be 13.27 * 3600 = 47.8  MJoules.

despite the above answer is correct , i would like to let you know some basic information to help you calculate it by your self

Cosø is the power factor

400Δ and 680Y is the voltage for delta and star connection respectively

power = square root of (3) * I * V * power_factor 

hope that helps you

Hi Yasser, I had to guess that the 400V was phase to phase. It is in fact the line voltage - so the calculation goes like this:

Find the line power: that's voltage x current x power factor: 400 x 65.3 x .88 = 22.96 kW

So the energy consumed per hour is simply 22.96 x 1 = 22.96 kWh.

There are 3600 joules per watt hour, so 3.6 10^6 joules / kWh.

So the energy consumption of your motor per phase, per hour, in MJ is 22.96 x 3.6 MJ = 82.75 MJ

than is what is that 37 KW? on the motor? how we came to answer of 22.96KWh 

Can it be possible that the motor power is 37KW *Time ...so in one hour in consume 37KW

@Tony  Maine 

at first you calculate 13.27KW and second you calculate 22.96KWh which one is right

If i want to find the power of one phase and then * by 3 did it gives me total power consumed

despite the above answer is correct , i would like to let you know some basic information to help you calculate it by your self

Cosø is the power factor

400Δ and 680Y is the voltage for delta and star connection respectively

power = square root of (3) * I * V * power_factor