Motor Energy Consuption Formula

[Sofistioelevib]

Hello to everyone,
this post came from another post but i want to threat in different because different object.

The question is EASY

I have VOLTAGE, CURRENT and V/I ANGLE

I want to calculate the real energy consumption in a year.

What formula i need? please give me a description of every parameters on the formula

 

[Sofistioelevib]

Another consideration.

How low PF impact in energy consuption? Low power on motor lower PF so, it is equal consumption of other motor with same KW but with high PF?

Regards

 

[waross]

Volts times Amps times the cosine of the phase angle divided by 1000 = Kilo-Watts.
Watts times 8760 Hours (24 hours x 365 Days) = Kilo-Watt Hours consumption in a year.
That's the easy answer.
OR for three phase:
Volts times Amps times the cosine of the phase angle times 1.73 (square root of 3) divided by 1000 = Kilo-Watts.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

Lower PF will not directly cause more consumption, however, indirectly, the lower PF causes more current for the same work done and this higher current will cause greater losses in the motor and in the supply conductors and transformers.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[jraef]

But this is also important; where did your data come from and what is the purpose of this calculation? If the motor current is from a NAME PLATE and you are only looking for a WORST CASE GROSS ESTIMATE of consumption, i.e. the motor would be running fully loaded at all times, then that's all you can get with this information.

The ACTUAL consumption is always going to be based on the connected load, which is always deemed to be LESS than the full load capability of the motor. So the only way to know the ACTUAL consumed power is to MEASURE it when the load is connected.


" We are all here on earth to help others; what on earth the others are here for I don't know." -- W. H. Auden

 

[edison123]

Motor power rating = What the motor is designed to deliver

Motor energy consumption = What the motor actually delivered on hourly/weekly/monthly/yearly basis

Muthu

http://www.edison.co.in

 

[Sofistioelevib]

@jraef
nameplate data of motors is 88A and 0,82cosphi
some motor load is 82A with 0,8 cosphi>>>no problem
one motor load is 60A with 0,6 cosphi>>>no process problem
because all of this motor are FAN and probably the 60A load could be due to incorrect pitch angle of the blades,
Suppose the reduced capacity flow have no impact for the process
my dubt is...
How can 60A cosphi 0,6 can impact on total consuption? is better to run the motor in this condition (so low current absorption but hig cosphi) or could be better to rise load to motor in order to correct LOW PF but increasing current absorption?

 

[edison123]

You can't simply raise the load to correct low pf. Google how to improve motor pf.

Muthu

http://www.edison.co.in

 

[waross]

Increased load means increased Kilo-Watt-Hours means increased cost every month.
Power Factor is the ratio between apparent power and real power.
Increasing the load will increase the apparent power, the real power and to a lessor extent increase the KVARs.
Overall the ratio will increase.
But this is a wasteful way to improve the power factor.
When driving on the highway, you can slow down for a reduced speed zone by riding the brakes without taking you foot off of the throttle
It works but it wastes fuel and brake linings.
Are you being charged a penalty for low PF?
If you are not paying a PF penalty, you have no problem.
Move on to something else.
If you are being charged a penalty, then add some PF correction capacitors to the motor.
The payback time for PF correction is often less than one year.

Bill
--------------------
"Why not the best?"
Jimmy Carter