Calculating Motor Operating Costs

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GeorgeB

ElectroHydraulics engineer (retired)
Location
Greenville SC
Occupation
Retired
then why does V*I*PF*1.732 not equal 56kW. Equations should be interchangeable.
The issue I see, as pointed out by others but perhaps not in a way you follow, is that INPUT power (whether kW or HP) is not the same as OUTPUT power (same kW/HP issue) I've added some comments in your portion below indicated by {}
Single Phase{ or DC is you allow that PF with DC==1}

kW = I x E x PF/1000{if a motor or any LOAD, this will be INPUT power}

HP = I x E x Eff x PF/746{if a motor or any LOAD, this will be OUTPUT power}

HP{output} = kW{input} x Eff/746

Three Phase

KW{input} = I x E x 1.73 PF/1000

HP{output} = I x E x Eff x 1.73 x PF/746

HP{output} = kW{input} x Eff/746
{I can accurately state also HP{output} = HP{input} x Eff and kW{output} = KW{input} x Eff}


Whether Single phase or three phase HP = kW x EFF/746{not really true; what is true is that OUTPUT = INPUT x EFF

I suppose the differences in our assumed numbers were due to Efficiency and Power factor. If i used a power factor of .67 i would arrive at the same answer of 56 kW.{the difference is in the lack of identifying input and output; POWER=POWER, and all that goes in comes out somewhere, usually as heat.}

I suppose from now on i will just use the 1 hp = .746 kW for my calcs. {That is true.}
 

Finite10

Senior Member
Location
Great NW
Actually not so. kW is power. Not a rate of power, just power.


Again, not so.
kWh is a measure of energy. It isn't a rate. For example, if you use 1 kWh it could be a steady 1 kW for one hour or 60 kW for one minute.

Lets see where I'm heading for the ditch here.:confused: Regarding 'energy' I learned that;

1 Watt = 1 Joule per second
W=J/s
A 'second' is a unit of time, which means a 'rate' of work

Work = (Force)(distance) <-- physics, energy
Power is the 'rate' of work done by some force.
Power = work done over time by some force acting on a body
P = Work/time

*whenever you see a unit of time -that means work is being done at a certain 'rate'.

Apple to apples; Since W=J/s therefore a Watt is a rate by definition. W or KW, both are a rate.

Since a Watt is known to be a rate, the question is how much energy is used over a period of time. In this case, time is 1 hour.

Rate over a certain time period
Rate per time
Rate/time = KW/H
commonly written as KWH

edit to add; People call it a power bill, but it's actually an energy bill. I think that may be the underpinning of your statement.
 
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Besoeker

Senior Member
Location
UK
Lets see where I'm heading for the ditch here.:confused: Regarding 'energy' I learned that;

1 Watt = 1 Joule per second
W=J/s
A 'second' is a unit of time, which means a 'rate' of work
Yes, power, measured in Watts, is a rate of doing work.
Not a rate of power as you stated:
KW is a rate of power,

Power = work done over time by some force acting on a body
Not so.
Power is a rate. An instantaneous value. The time for which it prevails is a measure of energy.

*whenever you see a unit of time -that means work is being done at a certain 'rate'.
Not so. Think about 1 kWh. It is a measure of energy. Though it has a unit of time it gives you no information whatsoever about rate. Could be one Watt for a thousand hours or a thousand Watts for one hour. Rate doesn't come into it.
 

Finite10

Senior Member
Location
Great NW
Actually not so. kW is power. Not a rate of power, just power.


Again, not so.
kWh is a measure of energy. It isn't a rate. For example, if you use 1 kWh it could be a steady 1 kW for one hour or 60 kW for one minute.

Ah well, we may have to agree to disagree - but I'm open to learn! I'm still trying to grasp it all.

My post:
KW is a rate of power,
Your correction:
Actually not so. kW is power. Not a rate of power, just power.

Source of my interpretation, Applied Physics -Schaum's 4th ed.
pg 76 "'Power' is the rate at which work is done..."
pg 78 "'Energy' is that property something has which enables it to do work."
Power has units of watts, 1W = 1 Joule per second of time period. (A rate, by definition)

_________________
My post;
KWH is a rate of power use over time
Your corrction:
Again, not so.

kWh is a measure of energy. It isn't a rate. For example, if you use 1 kWh it could be a steady 1 kW for one hour or 60 kW for one minute.

If I assume Applied Physics text is correct;
"Power is the rate at which work is done by a force".
And it is done at a certain 'watts' rate over the period of time which is 1 hour. That's a rate of power use -over 1 hour of time, as I understand it.

Wikipedia; "power is the rate at which work is performed or energy is converted"
Watts has units of Joules per second (a rate).

Your example;
"if you use 1 kWh it could be a steady 1 kW for one hour or 60 kW for one minute"
That is changing the time period over which different wattage rates are at work.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
110116-2238 EST

Finite10:

You are getting things confused by not clearly reading the definitions to which you refer.

Power is the rate of doing work, change of energy, and is an instantaneous value. If work is done at a steady rate, then power is constant. If the torque applied to a shaft is constant and the speed of the shaft is constant, then the power is constant.

In calculus you would describe power as
p = d(e)/dt
where
p = the instantaneous power
e = the value of energy in the system at the present instant of time
t = instantaneous time
If the magnitude of energy in a system is constant, a mass has been raised 1 foot and remains at that position, then there is no power transfer after it has been raised and p = 0, but there was some power used to raise the mass. If it was done fast then the power would be high, and if slow then power would be low.

If you apply a constant force of 1 # to this mass during a period of one second and in that period raise the mass 550 ft, then the rate of during this work is 550 ft-#/second and that equals 1 HP.

A non-zero rate of change of power, I do not believe it has a specific name, would result from accelerating a mass.


If a KWH meter is at 10,000 and is not rotating, then there is no power transfer. If over a 1 hour period it changes from 10,000 to 11,000 then there was a change of energy of 1 KWH. If that changed at a uniform rate over that 1 hour, then the power would have been a constant 1 KW. But there could have been a 10 KW load for 1/10 hour and the meter change over that hour would still have been 1 KWH. But the instantaneous power would have had two different levels, 0 and 10 KW. In normal applications there are much more complex power vs time curves for a 10 KWH change in energy.

Your statement:
My post:
KW is a rate of power,
Your correction:
Actually not so. kW is power. Not a rate of power, just power.

Source of my interpretation, Applied Physics -Schaum's 4th ed.
pg 76 "'Power' is the rate at which work is done..."
pg 78 "'Energy' is that property something has which enables it to do work."
Power has units of watts, 1W = 1 Joule per second of time period. (A rate, by definition)
pg 76 "'Power' is the rate at which work is done..."
Your page 76 statement is correct.


1W = 1 Joule per second of time period. (A rate, by definition)
This statement is also correct because a Joule is a unit of energy, not power.
You need to read your reference clearly.

Read the rest of your comments and see if you can clarify your thoughts on this.

.
 

Besoeker

Senior Member
Location
UK
Ah well, we may have to agree to disagree - but I'm open to learn! I'm still trying to grasp it all.

My post:
KW is a rate of power,
Your correction:
Actually not so. kW is power. Not a rate of power, just power.

Source of my interpretation, Applied Physics -Schaum's 4th ed.
pg 76 "'Power' is the rate at which work is done..."
Yes. The rate at which work is done.
Don't confuse work with power.


If I assume Applied Physics text is correct;
"Power is the rate at which work is done by a force".
And it is done at a certain 'watts' rate over the period of time which is 1 hour. That's a rate of power use -over 1 hour of time, as I understand it.
The Applied Physics text you quote is correct, of course.
An analogy. Drive along in your car. Your speed is analogous to power and the distance you travel the energy. Speed is a rate. An instantaneous measure.


Wikipedia; "power is the rate at which work is performed or energy is converted"
Watts has units of Joules per second (a rate).
Yes. Power is a rate. It is already in the unit. J/s. Saying "KW is a rate of power" doesn't make sense. It would be like J/s/s.
 

broadgage

Senior Member
Location
London, England
Is this a record for over-complicating what should be a very simple matter ?

The output power of which a motor is capable in generally stated in HP, 1 HP=746 watts regardless of voltage, number of phases or current.

The input power must be greater than that since losses occur in the motor.
The exact figure for losses may be obtained from the suppliers data, but 10% is a fair real-world estimate.

Haveing estimated the INPUT in KW, the maximum running cost is determined by very simple maths, provided of course that one knows the cost per KWH.

The actual cost may be less if the motor is not fully loaded.
 

Finite10

Senior Member
Location
Great NW
Yes. The rate at which work is done.
Don't confuse work with power.



The Applied Physics text you quote is correct, of course.
An analogy. Drive along in your car. Your speed is analogous to power and the distance you travel the energy. Speed is a rate. An instantaneous measure.


Wikipedia; "power is the rate at which work is performed or energy is converted"
Watts has units of Joules per second (a rate).
Yes. Power is a rate. It is already in the unit. J/s. Saying "KW is a rate of power" doesn't make sense. It would be like J/s/s.

Thanks for the education and clarification Besoeker. I'm a recent fan of this stuff and no expert. Still learning, but learning at least. And I thought I had a grip on this... oh well!
 

skeshesh

Senior Member
Location
Los Angeles, Ca
A good discussion to refresh the physics concepts - I find reading stuff like this is a good practice to keep theory fresh in mind.

As far as the OP's confusion though, as was mentioned in an earlier post, is that he's calculating KVAs using FLA and voltage and comparing that value with KW value which is the result of taking the machine's power factor and efficiency values into consideration.
 

Besoeker

Senior Member
Location
UK
Thanks for the education and clarification Besoeker. I'm a recent fan of this stuff and no expert. Still learning, but learning at least.
And that's a positive.
I have heard this stated other ways. The Socrates quote is as good as any:
True wisdom comes to each of us when we realize how little we understand about life, ourselves, and the world around us.
 
Quite.
Of course, if the motors were rated in kW instead of the archaic HP, that would then eliminate the need for conversion factors.......:cool:
Runs for cover........;)

If the electrical kW is nameplated, there is still a conversion necessary for/from the mechanical output side. If the mechanical kW is nameplated the specific efficiency and power factor at the operating point is still necessary to calculate the results. If you have varying load, voltage and phase imbalance it becomes an integration nightmare.

Since the motor is primarily a MECHANICAL device, the electrical parameters take second seat, so to speak.
 

topgone

Senior Member
In the calculation formula Amp IS the largest variable factor, as no efficiency, voltage or power factor changes AS much as current does along the load profile.

If and only if, the motor is a low-voltage one!
Example: 250HP, 480V, 325 AMPS, 1791RPM versus 250HP, 4160V, 31 AMPS
The amp is the smaller variable factor when you're dealing with MV motors.
 

Besoeker

Senior Member
Location
UK
If the electrical kW is nameplated, there is still a conversion necessary for/from the mechanical output side.
To convert to what though? The difference between input power and output power is just the efficiency. With both expressed in kW it's a simple division of kW by efficiency - it's the output power in kW that is given on the nameplate for European machines.


If the mechanical kW is nameplated the specific efficiency and power factor at the operating point is still necessary to calculate the results.
Yes, but that would be the same whether you expressed the output in kW or HP.
The main difference is that you would have to do an extra conversion from HP to kW. In any case the nameplate gives rated voltage, frequency, rated output power (kW), cos phi, rated speed at full load and a few other details. Probably not a lot different to NEMA.


Since the motor is primarily a MECHANICAL device, the electrical parameters take second seat, so to speak.
Well, mechanical output so fair enough. Appropriate to use HP to describe the mechanical output power I suppose.

But then, so is the Watt or multiples like the kW.
In mechanical terms, one Watt is power required to move a force of one Newton through a distance of one metre in one second. Convenient numbers.
In electrical terms, it is one Amp times one Volt. Again, convenient numbers.
If you can conveniently use the same units for both mechanical and electrical power it seems perverse not to.

I suppose there might be the perception that HP is for mechanical power and kW for electrical power. But using kW for both simplifies the calculations.
 
To convert to what though? The difference between input power and output power is just the efficiency. With both expressed in kW it's a simple division of kW by efficiency - it's the output power in kW that is given on the nameplate for European machines.



Yes, but that would be the same whether you expressed the output in kW or HP.
The main difference is that you would have to do an extra conversion from HP to kW. In any case the nameplate gives rated voltage, frequency, rated output power (kW), cos phi, rated speed at full load and a few other details. Probably not a lot different to NEMA.



Well, mechanical output so fair enough. Appropriate to use HP to describe the mechanical output power I suppose.

But then, so is the Watt or multiples like the kW.
In mechanical terms, one Watt is power required to move a force of one Newton through a distance of one metre in one second. Convenient numbers.
In electrical terms, it is one Amp times one Volt. Again, convenient numbers.
If you can conveniently use the same units for both mechanical and electrical power it seems perverse not to.

I suppose there might be the perception that HP is for mechanical power and kW for electrical power. But using kW for both simplifies the calculations.

Well it is a tortorous justification either way. I don't agree though that the electrical nameplating solves anything. The MECHANICAL conversion is a simple constant with 0.746, but when you go to the electrical it becomes complex. There are 4 nameplate datas(if you lucky) and ALL of them are nominal values at full load, not what the actual results going to be at operating conditions. The only single useful data is the FLA so that you can set your thermal protection.

I find it amusing though that someone from the British Isles argues for the elimination of the HP, which was a dubious gift from them to begin with....;)

I've lived with both of them and feel totaly ambivalent about the issue. In other words, both are fine with me and - really - no-one will convince me of the merits of either over the another. (But hey the day is not over yet, so I am open to arguments.:roll:)
 

Besoeker

Senior Member
Location
UK
Well it is a tortorous justification either way. I don't agree though that the electrical nameplating solves anything. The MECHANICAL conversion is a simple constant with 0.746,
Quite. But if you use kW it's a conversion you don't need.
but when you go to the electrical it becomes complex. There are 4 nameplate datas(if you lucky) and ALL of them are nominal values at full load,
With you.

not what the actual results going to be at operating conditions.
But, in general, the motor manufacturer won't know the range of operating conditions. Nameplate data couldn't reasonably accommodate all possible operational data.

The only single useful data is the FLA so that you can set your thermal protection.
Rated speed and output power (kW) might just be a little useful too.

I find it amusing though that someone from the British Isles argues for the elimination of the HP, which was a dubious gift from them to begin with....;)
And from a Scot at that. Salt of the earth and all that. But we are prepared to move on....


I've lived with both of them and feel totaly ambivalent about the issue. In other words, both are fine with me and - really - no-one will convince me of the merits of either over the another. (But hey the day is not over yet, so I am open to arguments.:roll:)
I also have lived with both. I can do either and know basic conversion factors. As I have said before, I'm blessed and cursed with a memory for numbers.
For maybe no better reason than the simpler relationships between the units, I prefer SI units.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
All i want to know is if you generically use the formual for 1 hp = 746 watts do you need to account for phree phase situation by multiplying by 1.732.
No. The 1.73 only comes into play when using 3ph volts and amps.

Try using 1hp = 1kw and let us know how the numbers play out.
 
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