Calculating Energy Consumption of 3 phase motors

[georgeofthejungle02]

Hello:
I have a question about calculating the energy consumption of 575V 3 phase interpole 250HP /63HP motor in one year.
8760 hr in one year
0.0504 $/kWh
From the motor data, the efficiency is 95%/92% at 1800/900 RPM respectively.
The following formula is what I use to calculate operating costs.
cost/year = hours/year x $/kWh x [ hp x (0.7457 kW/hp) / (motor efficiency) ]
What i don't know is how to use the power factor which are 80.0% and 54.5% at 1800/900 RPM respectively into the above calculation.
Any help would be appreciated.
Thanks in advance

 

[rbulsara]

You do not need to consider power factor for this purpose (as you can see its not in your formula).

PF affects the apparent power (kva) , but real power (kW) remains the same. Energy cost is only related to kw-h.

However if there are sevaral of such motors in plant and you are also paying charges for maximum kVA, demand then pf comes in play. Pooer power factor results in higher kVA demand and somewhat increased I^2R losses.

kVA=kW/pf

 

[aolalde]

I agree with Rbulsara, the PF does not modify the kW-h consumption. As this is a dual speed motor you need to know how many hours it works on each of both speeds and apply the correspondent HP.

 

[TheApprentice]

One thought to share:
HP is calculating the output power which is equal to
Output KW = horsepower *746/(1000*efficiency)
but where you are paying the bill is the input power,
and since power factor, or
input power KW= 1.73 * voltage* amp* power factor
you need higher current to deliver the same OUTPUT HP if your power factor is low,
Any thoughts?

 

[aolalde]

The Apprentice:
Your first expression is wrong.
It should be (Output kilowatts) kWo = HP*0.746
The Efficiency is the ratio of output power to input power (kWi) and the difference are the motor losses.
Eff%=kWo/kWi * 100

Certainly the power factor (P.F.) is the ratio of the real power (kW) to the total or apparent power (kVA).

P.F. = kW/kVA or kW = kVA*P.F.

 

[TheApprentice]

Thanks aolade for correcting me,
but would you agree with the statement that if you want to say 250HP of output power while your power factor is unity, then you will need buy 250HP input power.
Where as if your power factor is 0.5, then to deliver 250HP of output power, you will need to buy 2 times the input power to compensate for the poor power factor?

 

[aolalde]

”Where as if your power factor is 0.5, then to deliver 250HP of output power, you will need to buy 2 times the input power to compensate for the poor power factor?”

NO. You pay for 250 kW/EFF plus the wiring losses (i^2R). EFF is the motor efficiency and the wiring losses are from the meter to your motor. The power factor is a function of the reactive power required to magnetize the motor.
You must differentiate between real power that is measured in kW. Reactive power (kVAR), and total or apparent power (kVA). There is a vectorial relation. kVA= kW<0 + kVAR<90
The utility charges for real power consumption (kW), but regularly limits the PF (PF>0.8) because they need to provide the reactive power (kVAR) overexciting the generator. If the PF=0.5 the utility surely will penalize the monthly bill.

 

[rbulsara]

apprentice:

If the power factor is poorer you wuill need larger size of transformers, wires etc..on the supply side as the current will be higher but further out of phase with voltage. So it will not increase the kW consuption but will require higher kVA and greater current and the supply system need to be sized accordingly. So with a pooer power factor you will have to 'buy' more kvA, which is not always charged by the utiltiies company for say residential and small businesses but will be a separate charge for for larger customers. Many a time the demand charges are only for kW demand and not kVA.

Regardless of you pay for kva or not, you need to have equipment sized for required kVA. If not you will overload them. Not necessarily using more real power (kW).

I recommend you refer to some basic textbook explaining relationship between kVA,kW and power factor and what does a power factor means.

 

[georgeofthejungle02]

THanks for all your input gentlement.
Coincedentally, I've got a excel calculation sheet from one of our vendors to estimate energy consumption of the 2 speed motors, their formulas are protected so I don't see how they are getting the answers from, but say a 250HP motor running @ 1800 RPM with 95% efficiency and a PF of 1 will be $85,000/year, if the power factor is halved, it will cost $171,792. They were trying to justify how a VFD have a pay back period in less than a yr.
This result relates to what the apprentice is saying, but as both rbulsara and aolalde mentioned, you only pay for the actual kw.
Another point is that even if the motor power factors are low, there are power factor correctors upstream to tackle this problem.

 

[jghrist]

There is no way that you can cut the operating costs in half (from $171,792 to $85,000) by increasing the power factor unless the utility charges a power factor penalty of some form (either directly or by charging for kVA instead of kW). Even charging for kVA demand instead of kW demand would only result in the demand portion of the bill to be reduced by half, not the energy portion. Losses between the meter and the motor would be reduced, but not by that much.

I would be hesitant to use a vendor furnished spreadsheet to economically evaluate the vendor's product unless I knew the basis for the calculations.