Motor kW Reading 1

[JiJiP]

Greetings fellow seniors, I am a new mechanical guy who just got into the line in a utility plant and I am a lost one with electrical stuff, so I was wondering I can get some insight and understanding from the brains.

So part of my job was to get the real input(absorbed)motor kW reading from motors of a machine and log them down, however I am not sure what I am asked to do is correct ?

So the motor has a name plate of 160kW with a motor eff of 95% at full load with a power factor of 0.85.

The VFD reading says in a given time averages shows me 114.5kW, 388.0V, 203.0A

I thought I was supposed to divided 114.5kW with 0.96 to get actual motor absorbed power 120.53kW. But instead I was told to take 114.5kW to multiply by 0.96 to get 108.78kW ?!

My plant technician senior shows me that when they use a fluke power analyzer, they will too take the active power reading and multiply by motor efficiency to get the real absorbed power of motor ? Is this correct ?

 

[LittleInch]

What exactly are you trying to measure?

If it's simply electrical power consumed then you have it already - 114.5 kW

If you want shaft power into the machine then yes you multiply by the efficiency as there will be losses within the motor.

"Absorbed power" can mean different things to different people.

And that's before the sparkies get excited about kVA.... and reactive power.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[JiJiP]

Let me try rephrasing, i will like to know the power consumed at the motor shaft, therefore i guess it call shaft power ?

 

[JiJiP]

"If you mean power FROM the electric motor into the driven equipment (aka shaft power) then multiply by efficiency."

This statement is correct to me then, so it multiply be efficiency ! thanks tons !!!

 

[che12345]

"...I am trying to get and determine the power input to the motor".
I have the following opinion for your consideration.
1. The VFD shows 114.5kW, 388.0V, 203.0A. This is the VFD Output Power to the Motor = the Input Power of the Motor. Ignoring the small I[sup]2[/sup] R losses on the cable between the VFD and the Motor.
2. Power supply by the utility (electric bill you are paying) = the [VFD Output Power + VFD losses], e.g. 114.5kW/0.96 =120.53kW.
3. Power to the Load = [Input Power to the Motor - Motor losses] e.g. 114.5kW x Motor efficiency = < 114.5kW.
Che Kuan Yau (Singapore)

 

[LittleInch]

che,

Where did you get 0.96 for the efficiency of the VFD? I commonly use 8% losses on top of the supplied power to the motor.

The 95% he refers to is the motor efficiency. Nothing there about the VFD losses....

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[waross]

So part of my job was to get the real input(absorbed)motor kW reading from motors of a machine and log them down

This is what you pay for and may be what is desired.
Note: For total power used on behalf of the motor you must add the VFD losses.

So the motor has a name plate of 160kW with a motor eff of 95% at full load with a power factor of 0.85.

As the load is reduced the efficiency improves.
Peak efficiency will typically be around 2/3 to 3/4 of full load
As the load drops, the PF drops.

KW / PF = KVA, Nothing to get excited about.
Reactive power (KVAR) = √(KVA[sup]2[/sup] - KW[sup]2[/sup])= KVAR A few more clicks but still seldom prompts excitement.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[JiJiP]

Yea so my understanding is, we need to know the absorbed power or the shaft power whichever we are going to call or name it.

So it should be lower than the kW that we are paying for. We need to find out what is the real power after the losses at the motor shaft after motor efficiency and VFDS losses. So i guess LittleInch reply was more to what I was looking for, thanks everyone though !

 

[jghrist]

So, if you have a VFD efficiency of 0.96 and a measured power into the VFD of 114.5 kW, the power out of the VFD into the motor is 114.5*0.96 = 109.9 kW. If the motor has a 0.95 efficiency, then the power into the shaft is 109.9*0.95 = 104.4 kW.

 

[Gr8blu]

SUPPLY -> CABLE 1 -> TRANSFORMER -> CABLE 2 -> DRIVE -> CABLE 3 -> MOTOR -> GEARING -> PROCESS

Sounds like our OP is tasked with an efficiency study (electrical power paid for vs mechanical power used at the process). Remember that no conversion is 100% efficient - so as we move from left to right in the chain above, we get progressively lower SYSTEM efficiency because the SYSTEM efficiency is the product of all the conversion efficiencies that take place along the way.

To know how much power actually goes to the process, we need to either measure the torque and speed at the gearbox output/process input, or start from the other end. Note that making the measurements at the process end is not a trivial task, which is why most folks approximate the answer by starting at the electrical (supply) end.

OP measured a real power value at the drive input = 114.5 kW. This is now a known point in the system. Cable losses are going to be minimal for the drive-to-motor run (a couple of hundred watts at most and we're talking a through power rating in the hundreds of kilowatts). So we can say that the power into the motor is going to be (roughly) what was measured, multiplied by the drive "efficiency". If we assume drive EFF = 0.96, we have 114.5 * 0.96 = 109.9 kW into the motor. But motor is actually rated 160 kW, so we're running at only a bit over two-thirds load (68.6%). That means the motor efficiency at that load point is going to be a bit different from the full load EFF - maybe up by 0.5% or so.

So now we have power into the gear (if there is one, otherwise it will be power to the process) that is 109.9 * 0.955 = 104.9 kW. If there was a gear, it's efficiency gets multiplied in the same way, resulting in an even lower value.

POWER TO PROCESS = POWER FROM SOURCE * (COMPONENT EFF 1) * (COMPONENT EFF 2) * ...
Thus, neglecting cable losses, we get
PROCESS POWER = SUPPLY POWER * (EFF XFR) * (EFF DRIVE) * (EFF MTR) * (EFF GEAR)
where all the EFFs are somewhere between 0 and 1 per unit.

Converting energy to motion for more than half a century

 

[waross]

What is important is that it is done the same way that it always has been done.
It is highly unlikely that whoever these results are going to knows understands or really cares about the fine points that have been discussed here.
These reports are typically used for trend tracking.
If you start using efficiency factors that have not been previously included in the results you will through the trend off.
It may be best to ask what the information is being used for.
So, KW adjusted for efficiency.
But the motor is not at 100% load and the efficiency factor is only valid at full load. Adjusted KW? Wrong answer.
The same for VFD efficiency. Adjusted KW? Wrong answer.
Gearbox efficiency and the linearity of the gearbox efficiency with less than full load?? Unknown. Adjusted KW? Wrong answer.
Driven machine efficiency and the linearity of the driven machine efficiency with less than full load? Adjusted KW? Wrong answer.
Just do it the way that it has always been done.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Heaviside1925]

Consider, you are not the first or last person to be asked to do this and they have a method they have asked you to use to arrive at a number to pump into some excel sheet or something of the like.

I thought I was supposed to divided 114.5kW with 0.96 to get actual motor absorbed power 120.53kW. But instead I was told to take 114.5kW to multiply by 0.96 to get 108.78kW ?!

this is a valid argument only, if that is how the rest of these audits/study....whatever they are calling it have/are being performed. Is it truly the actual absorbed power? No, but it is an approximation they are using for some purpose. As waross noted, I think a valid question is, what the purpose of this information is and what is it being used for. I suspect you will find that in the use of these numbers, a high level of precision is not going to be required. Consider, the drive is saying the average power is 114.5kw, what does the voltage, amperage and power factor indicate is the power? I have seen new ME's tasked with like jobs before and I doubt you have had much formal education in AC and motor theory (maybe you have but I haven't known it to be a standard offering in most ME programs). Consider this a baptism by fire of sorts because many times "things that rotate" aka motors, get stuck in the ME's lap.

 

[che12345]

"...The VFD reading says in a given time averages shows me 114.5kW, 388.0V, 203.0A...I thought I was supposed to divided 114.5kW with 0.96 to get actual motor absorbed power 120.53kW. But instead I was told to take 114.5kW to multiply by 0.96 to get 108.78kW ?!"
I simplify my earlier opinion based on the first principle, to clear the "efficiency" confusion.
1. The VFD shows 114.5kW, 388.0V, 203.0A. This is the VFD Output Power to the Motor = the Input Power of the Motor.
2. Power supply by the utility (electric bill you are paying) = the [VFD Output Power + VFD losses] = > 114.5kW.
3. Power to the Load = [Input Power to the Motor - Motor losses] = < 114.5kW.
4. BTW: In SI units, Power is in W or kW, NOT KW or kw.
Che Kuan Yau (Singapore)

 

[waross]

Sorry for any confusion that the use of historically correct units has caused you.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[FacEngrPE]

I side with Waross and Heavyside1925 on this.
Usually energy audits have enough inconsistencies and "metering errors" that it is not worth the effort to overly refine the "estimated efficiency of energy conversion. For these studies it is necessary to develop a constant method of analyzing the data, and stick with it. Fortunately the method has been provided to you.

If your purpose needed accurate motor efficiency data, be aware on a system driven by a VFD at variable speeds, the motor efficiency will be a function of:
[ul]
[li]Input frequency[/li]
[li]Input voltage[/li]
[li]Slip[/li]
[li]Motor design[/li]
[li]other stuff[/li]
[/ul]
To get the actual efficiency map, likely requires putting the motor on a dynamometer. Sounds like the stuff of higher education research papers.
The following chart may be valid for a single tested motor, and could be used to talk about the issue. It does show the point made previously, efficiency does not change much with shaft speed, until you are operating at less than 1/2 design frequency.

 

[Parchie]

@Pandit,
Correct. If the VFD panel displayed 114.5 kW, it means the power transmitted by the VFD to the motor. Subtract the VFD losses and you'll get the actual power delivered by the VFD to the motor. To consider @FacEngrPE's idea on finding the power delivered at the motor shaft, one has to subtract the motor losses from the actual power delivered by the VFD.

 

[waross]

JiJiP(mechanical)(OP)
This thread has deteriorated. There is some questionable and inaccurate information.
The inaccuracies and contradictions mostly take the form of the right answer to the wrong question.
That is difficult to argue with.
Efficiency charts for motors are based on the assumption that The motor is fed with optimum power.
That is, power with equal phase angles, equal voltages and no harmonics.
Once you add a VFD everything changes.
Motor losses are comprised of windage losses,IR losses(heating) and magnetic hysteresis related losses.
Windage is speed related, IR is current related and hysteresis may be affected by the voltage waveform.

So what is the best career advice you can have?
The person or department that is requesting these figures will value consistency over technical arguments that they "Have been doing it wrong" for years.
Your best career path is to continue to take and report the readings as they have always been done.
Personally,know that the readings may not be 100% accurate, but they are close and they are what is wanted.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!