Electricity consumption

Output power in HP should be in the nameplate of the motor;

Efficiency tables are also available

Efficiency=output/input

so, you get the input power i.e square root of 3*V*I*cosΦ(for three phase)

and V*I*cosΦ (for single phase)

Power factor(CosΦ) chart is also available

you know the supply voltage

So you can calculate current(I)

Consumption=kWh(W=Power(V*I*CosΦ)) multiply power with time of operation of the respective motor. And devide it by 1000

You will get unit consumption

Bye

Thank you for your help...... I'll try to calculate follow this advice. And how about the cost reduction, can we reduce the cost interms of electricity consumption?

Where can i get the efficiency table for electric?

And for air-conditiner, how we can calculate the power. It is using same or different method. What i understand, the electricity consumption will depend on the temperature for the air cond, higher temperature less electricity consumption and the other wise.

Thanks.

Regards,

CT

You could take nameplate KW readings at 80% and multiply by time (HRS) to get the consumtion of each m/c per hour.

OR take actual amperes readings than I X V X PF X 1.732 and devide by 1000 this formula will give you the KW unit than multiply by the number of hours the m/c is in use to get the KWH consumption of the particular m/c

reapeat the same for each machine.

Note : PF can be assumed at 0.8 if the actual PF is not known.

Thank you for you help............ What is 1.732 stand for? For pump, how i can get the electricity consumption if the running capacity is not full. Let say, the pump capacity is 100 ton/hr but it is running only 65 ton/hr. This matter can reduce the electricity consumption also and how to calculate it?

Thanks

CT

1.732 stands for under root of 3 often used in three phase formulas .

as i said earlier the name plate KW is the 100 % load and in good designed loadings the pumps will be loaded 80 % .

the 100 ton/hr pump is running at 65 % so either go for 65% of the nameplate KW readings or take the actual current readings and follow the formula ,the current reading will be relative to the load.

For AC it will be difficult to get the actual consumption becouse of the regulation control ,you will need to attach a KWH meter or follow the rule of thumb,that is calculate the KWH and 65 % of the total hours the AC will be expected to run..

yes the consumption will definetly vary with the environment i.e room enclouser,No. of persons,etc,etc.