# Weird question of the day



## backstay (Feb 3, 2011)

HP = V x I x Eff/746


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## wildleg (Apr 12, 2009)

glen1971 said:


> ....
> 
> Is there a calculation to determine the hp being produced at a given speed? What would it be used for on a system that has been in use for 30 years and running with no issues?


the specs for the replacement system that is being designed


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## MDShunk (Jan 7, 2007)

746 watts per horsepower or amps / 1.5 for a 480 motor both get you in a pretty close ballpark for management types that will probably do nothing with that number anyhow.


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## JRaef (Mar 23, 2009)

There is a problem with the way the question is asked, but there is a way to answer what I THINK was meant.

Problem: "...how many horsepower a motor is using..." in and of itself is not something you can CALCULATE, it's something that must be MEASURED. That's because it is the LOAD that determines that the motor uses, not the motor that determines what the load uses. So what the load USES is what the load uses, regardless of frequency.

Now, what they may have wanted to KNOW, is how the change in frequency changes the CAPACITY of the motor loading. That one is easier. Load capacity of the motor changes directly with the frequency. So if a motor is rated to be capable of delivering 30HP at 60Hz, if you turn the frequency down to 30Hz, the motor capacity becomes 15HP; 50% speed = 50% HP. Why? because simply, the term "Horsepower" is just a shorthand expression of so much torque and such and such a speed. HP = Tq (in ft-lbs) x RPM / 5250. So when we say a motor is capable of delivering "30HP" , and it is 4 pole (1750RPM), that really means that the motor will deliver 90ft-lbs of torque at 1750RPM; 30 x 5250/1750. What a VFD is really doing is allowing the motor to deliver the SAME torque even as the speed is lowered by maintaining the correct V/Hz ratio that the motor was designed for. So if the motor is now delivering 90ft-lbs of torque at 875RPM (30Hz), then the HP is 90 x 875/5250 = 15HP. 

Whether or not the motor USES all 15HP is not something that you can determine up front, unless you have a lot of mechanical information and a good computer software program.

But honestly, that's pretty much irrelevant; why do they care? If what they want to do is know if the VFD can save ENERGY, that is a COMPLETELY different exercise, having little to do with only the speed. I'll warn you however, it is a common salesman's myth. If that's what they want to know, there are a LOT more questions that need to be asked, starting with "What kind of machine are we discussing?" If it is anything other than a centrifugal pump or fan, then the answer will be NO.


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## glen1971 (Oct 10, 2012)

Thanks Jraef.. I was kinda thinking in that ball park too, but wasn't sure if I had missed something or not... 
I think it's a waste of time, as they won't be changing the fan blade size, shiv size or motor size any time soon... The system keeps up with the change in temps and load, with lots of room to spare..

Wildleg... The only changes that I can forsee coming to this site would be:
- getting rid of the PLC5 and going to a Control Logix
- upgrading the original Moeller MCC to Allen Bradley. This was already done on a sister site and will probably happen during a 2017 turnaround.

There are some mechanical upgrades coming, but they shouldn't involve much on the I & E side...


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## NC Plc (Mar 24, 2014)

Seriously JRaef, the amount of knowledge you have is staggering and impressive.


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## AK_sparky (Aug 13, 2013)

JRaef said:


> If what they want to do is know if the VFD can save ENERGY, that is a COMPLETELY different exercise, having little to do with only the speed.
> "What kind of machine are we discussing?" *If it is anything other than a centrifugal pump or fan, then the answer will be NO.*


Why?


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## JRaef (Mar 23, 2009)

AK_sparky said:


> Why?


A VFD can only save energy that is being wasted. AC motors only use the energy the load needs, plus a little overhead to make the motor become a motor. That overhead only represents a few percentage points, and can't chage much. So the only way for a VFD to save any appreciable energy is to reduce losses somewhere in a system. If you have something like a conveyor, reducing the speed reduces the power consumed, but also reduces the amount of work performed. Energy is power consumed over time, so if you use less power, but it takes longer to do a task, the net energy remains the same.

The exception is on a centrifugal load like a pump or fan where flow needs to be varied, because in whatever means you use to vary that flow, there are losses involved in it. But if you vary the flow by changing the motor speed with a VFD, there are far fewer losses than any other means, so the net difference is the energy that is saved.

But even then, no variation in flow means no reduction of losses, so no savings. And if you are just going to permanently CHANGE flow, not VARY it, then there are better ways to change the flow than a VFD as well, such as trimming the impeller of a pump or changing the blade pitch of a fan. So VFDs only save energy in VARIABLE flow applications with those centrifugal machines.


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## AK_sparky (Aug 13, 2013)

I guess in my experience I've seen a VFD used to slow a conveyor that was really designed too fast to begin with. It was moving parts from the output of one process to the hopper of another process, where the part would wait for a long time before being needed. We put a VFD on the conveyor (for safety reasons) and the speed of the entire process remained the same, just the "wait" time in the hopper was reduced. I guess in that situation it would save energy.

Similarly we had a mixer that was essentially an agitating holding tank on the input of a process. We slowed it down because it really didn't need to mix fast, just needed to keep the mixture in motion. Again a case of "should have been designed slower to start with."


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## JRaef (Mar 23, 2009)

AK_sparky said:


> I guess in my experience I've seen a VFD used to slow a conveyor that was really designed too fast to begin with. It was moving parts from the output of one process to the hopper of another process, where the part would wait for a long time before being needed. We put a VFD on the conveyor (for safety reasons) and the speed of the entire process remained the same, just the "wait" time in the hopper was reduced. I guess in that situation it would save energy.
> 
> Similarly we had a mixer that was essentially an agitating holding tank on the input of a process. We slowed it down because it really didn't need to mix fast, just needed to keep the mixture in motion. Again a case of "should have been designed slower to start with."


In both cases, the same effect could have been achieved by changing pulley or grear ratios, if there was never a need for changing it. If it needed to change, that's a perfectly valid reason to use the VFD, and it may improve productivity for sure, but it really isn't DIRECTLY saving energy per se.

Don't get me wrong, I love VFDs and that's what I do for a living now. I just like to see people use them for the right reasons.


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## AK_sparky (Aug 13, 2013)

JRaef said:


> In both cases, the same effect could have been achieved by changing pulley or grear ratios, if there was never a need for changing it. If it needed to change, that's a perfectly valid reason to use the VFD, and it may improve productivity for sure, but it really isn't DIRECTLY saving energy per se.
> 
> Don't get me wrong, I love VFDs and that's what I do for a living now. I just like to see people use them for the right reasons.


Ya, in both cases it was really a quick and easy way to change the speed and leave flexibility for future changes. Neither of them was intended to save energy. My thought was that "theoretically" they were now using less energy without affecting productivity, due to the original over design. Ideally it would have been better to use a lower HP motor to begin with, with appropriate mechanical reduction, since variable control isn't required.


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## Safari (Jul 9, 2013)

JRaef said:


> There is a problem with the way the question is asked, but there is a way to answer what I THINK was meant.
> 
> Problem: "...how many horsepower a motor is using..." in and of itself is not something you can CALCULATE, it's something that must be MEASURED. That's because it is the LOAD that determines that the motor uses, not the motor that determines what the load uses. So what the load USES is what the load uses, regardless of frequency.
> 
> ...


Why the constant 5250?


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## JRaef (Mar 23, 2009)

nickson said:


> Why the constant 5250?


Actually that was a typo on my part, it's 5252. It has to do with the official definition of horsepower. 1HP is the work needed to lift 550lbs one foot in one second. That equates to 33,000 lb-ft/minute. To relate that to rotation, it's 2 x pi / 33,000, which is 5252, which becomes a constant that applies regardless of the circumference of the circle you are looking at.


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## Safari (Jul 9, 2013)

JRaef said:


> Actually that was a typo on my part, it's 5252. It has to do with the official definition of horsepower. 1HP is the work needed to lift 550lbs one foot in one second. That equates to 33,000 lb-ft/minute. To relate that to rotation, it's 2 x pi / 33,000, which is 5252, which becomes a constant that applies regardless of the circumference of the circle you are looking at.


Got it. 
In this part of the world we are used to Nm(newton meter) when reffering to torque. Had to Google the relation.
one pound-foot is approximately
1.355 818 newton meters.


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## JRaef (Mar 23, 2009)

There is an equivalent constant used for N-m torque calculations. 
So kW = Tq. (N-m) x RPM / 9550.


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## Safari (Jul 9, 2013)

Thanks jraef


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