electrical energy costs
- Thread starter DougMc
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cadpoint
Senior Member
- Location
- Durham, NC
- Occupation
- Electrician - but not by NC Law.
Welcome to the Forum !
Could it be as easy as filling in the blanks, yes.
Is that the exact equation what's needed to proof for that requirement that's based on the question, to gain the corrrect answer ?
I don't know, I'm just talking out loud!
You might also "search" the Forum for earlier topic of the same nature and
use variables of the subject or theme, to see whats found!
The host of the Forum at the home page has a sub-page devoted to calculations...
Could it be as easy as filling in the blanks, yes.
Is that the exact equation what's needed to proof for that requirement that's based on the question, to gain the corrrect answer ?
I don't know, I'm just talking out loud!
You might also "search" the Forum for earlier topic of the same nature and
use variables of the subject or theme, to see whats found!
The host of the Forum at the home page has a sub-page devoted to calculations...
Rick Christopherson
Senior Member
The first thing I noticed is that you are missing the term for the motor's efficiency, which is due to hysteresis losses in the core and windage of the spinning rotor. The estimate that I use for smaller motors is an efficiency of about 0.8, but this would likely increase with larger motors, and you might find it on the motor nameplate.
P = I * V * eff * pf * 1000 = KW
KWhr = KW * time
Cost = KWhr * rate($ / KWhr) = $
P = I * V * eff * pf * 1000 = KW
KWhr = KW * time
Cost = KWhr * rate($ / KWhr) = $
Rick Christopherson
Senior Member
Oops. I am so accustomed to using this when converting hp to current draw that I think I made a mistake by including the efficiency in this equation. I think that term comes in only when dealing with the shaft output power.
Another Edit: Also, if your amperage is per-phase, then don't forget the squareroot-3 term.
Another Edit: Also, if your amperage is per-phase, then don't forget the squareroot-3 term.
Last edited:
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
Eff values are better now for new energy efficient motors, but if you don't know, 0.8 is a "safe" bet in that if you use it, you will end up estimating a bit high as opposed to a bit low. But large motors, i.e. above NEMA frames (roughly over 250HP), are typically 90-95% efficient, so I use 90% most of the time. The best thing is to get the efficiency values from the motor nameplates or from the suppliers if it isn't there.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
D'oh! I just got caught doing the same thing...
You're right of course. From a standpoint of determining consumption, the efficiency is accouted for because although it isn't getting to the load, it still is being supplied by the power system.
weressl
Esteemed Member
Energy efficiency has nothing to do with calculating the cost of electricity for an existing system.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
You equation is right to find the kWh, assuming that the current is constant for the whole time of operation. Actually, the current (and pf) will change if the load changes. The current, voltage, and pf also have to be where the energy meter is. Using values at the motor will not count the cost of losses in the electrical system.
The cost is seldom a straight $/kWh. Usually there is a demand charge, a tiered rate, a power factor penalty or something else to make it more complicated. If you want to accurately know how much adding a motor will change the power bill, you have to know the rate schedule and apply it both with and without the new motor.
The cost is seldom a straight $/kWh. Usually there is a demand charge, a tiered rate, a power factor penalty or something else to make it more complicated. If you want to accurately know how much adding a motor will change the power bill, you have to know the rate schedule and apply it both with and without the new motor.
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