Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

Register Log in
  • Congratulations IDS on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

FLA vs power consumption vs motor hp

Status
Not open for further replies.
mech4321

mech4321

Mechanical
Jun 8, 2009
14
Hi,

We have to evaluate the power consumption at steady load of a fluid cooler which have at 600V an FLA of 23A. That would give approximately kw = 600 x 23 x 1.73 = 23.9 kw or 32 hp. However, the fluid cooler has only a total of motor of 6 x 1.5 hp = 9 hp (plus a small amount for the controls). Where does come this big differance? How can we evaluate the consumption if this value is not available from the manufacturer.

Thanks
Alex
 
Replies continue below

Recommended for you

Sort by date Sort by votes
A power meter would be the most accurate. Your 9 hp is shaft power out and small motors are not particularly efficient, so more electrical power in than shaft power out, perhaps by a whole bunch. Then you completely left out power factor, not all of those amps are producing real power, some (a lot at that size perhaps) is providing excitation for the motor.
 
Upvote 0 Downvote
I understand the concept of PF and Efficiency. I was wondering why the differance is so big even with these values applied.
So I can use the FLA to calculate the consumption at steady full load?
I heard about the running load amp which could be 50% of the FLA? Is this value applicable to electrical consumption?
thanks for your help
Alex
 
Upvote 0 Downvote
Your apparent power is 32 HP.
Your nameplate power is 9 HP.
That is a ratio of 9:32 or 0.28. Taking the root of 0.28 we get 0.53
Assume a power factor of 0.53 and an efficiency of 0.53
Your nameplate power is 9 HP. 9 HP/0.53 = 17 HP.
Use 17 HP as your consumption.
Unless you have a power factor meter to find the actual power factor or a Watt-meter to find the actual consumption that's about the best you can do.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Upvote 0 Downvote
at 600V an FLA of 23A. That would give approximately kw = 600 x 23 x 1.73 = 23.9 kw or 32 hp.
Click to expand...
To convert amps and volts to a real electrical power value, you need to consider power factor. And to convert to output mech power you'd also need to consider efficiency. You said you understand, but you left out at least power factor.

MORE IMPORTANTLY, the power computed from nameplate represents rated conditions but tells you nothing about the power drawn by the motor, that is determined by the load. As was mentioned, measuring power would be a good way to determine it.

However, the fluid cooler has only a total of motor of 6 x 1.5 hp = 9 hp (plus a small amount for the controls).
Click to expand...
How did you determine this hp? Does it represent a shaft horsepower or a thermal power? And if it is based on a nameplate condition of the cooler, again it may not represent operating condition (actual load may depend on thermal conditions).


=====================================
(2B)+(2B)' ?
 
Upvote 0 Downvote
power drawn by the motor, that is determined by the load.
Click to expand...
to be more specific, motor shaft (output) power depends on the the load. Motor input power for a given shaft load also depends on motor efficiency.

=====================================
(2B)+(2B)' ?
 
Upvote 0 Downvote
The power factor and efficiency where not applied to simplify my calculation (assuming that these value where not high enough to get that big differance (32 and 9 hp)).
Here is the spec attached of the equipment. Can somebody tell me a way to estimate the consumption at steady load. The equipement is not installed yet and I can't measure the power.
 
 http://files.engineering.com/getfile.aspx?folder=a15dda36-0a63-4b80-9061-4166bd21ea6a&file=server_room_fluid_cooler.pdf
Upvote 0 Downvote
Motors of that size typically have a very low power factor.
Motors of that size typically have very low efficiency.
It is reasonable to assume that the efficiency is 53% and the power factor is 0.53.
If you don't like my WAG, you may measure the consumption (may not be an option), call the supplier, or make your own WAG. (Wild A$$ Guess)


Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Upvote 0 Downvote
PS. If you are sizing services, consumption is moot. For service capacity you must use the KVA per motor (23.9 KVA / 6 motors = 4 KVA per motor).
For service capacity you must also consider whether the motors start simultaneously or use staggered start.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Upvote 0 Downvote
I thought you meant actual full load amps. Name plate/code full load amps are a whole different animal and have nothing to do with anything other than sizing the supply to the motors. Nothing about expected usage.

Actual running amps of a motor will probably never reach listed FLA unless something has gone very wrong. And actual running amps will vary as the load varies. You'll need process loading information to determine the actual load on the motors and then you'll need to know the actual motor efficiency and power factor at that loading. But it won't matter, really, because the supply will be sized for the FLA and all will be well. Actual consumption will be determined by the load and conveyed to you in the monthly utility bill. If you care to know actual consumption in more detail you will need metering on the equipment that can provide you that information. You'll never calculate it out in advance.
 
Upvote 0 Downvote
Alex,

I think first up you need to speak with your mechanical counter part and get him to advise you further on this matter.The advice on this is forum free and as such may be worth what you pay for.

That aside,the catalogue you have loaded up is that of a heat exchanger with six fans.As the document shows, the maximum power it can consume is 6x1.5= 9 HP.So under steady state conditions that is the maximum power it can draw.But I would go as far as saying it will never hit 9 HP as fan motors are selected for the fan starting torque.The running torque will be less than the starting torque.So your steady state power consumption is quite likely to be in the 5~7 Hp range.
 
Upvote 0 Downvote
Why do you want to know the power consumption?

Have you tried contacting the manufacturer?

I would think that 4-pole motors with 100rpm slip are quite inefficient.
 
Upvote 0 Downvote
I am more familiar with the Canadian Electrical Code than with the NEC but that information may not meet the minimum code requirement for installation in Canada. It is common for refrigeration equipment to show the size of the branch circuit conductors and protection, as Davidbeach suggests BUT NOT IN THAT FORMAT.
There is not enough information given for us to give you an accurate value for consumption.
You may also have issues with the AHJ if the actual nameplate on the cooler does not provide more information and evidence of approvals. This may be addressed with a field approval inspection. There will be an added inspection cost and the AHJ may require changes to the equipment and possibly to the nameplate.
Been there, done that, got the tee shirt.

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Upvote 0 Downvote
However correct all these answers are, I do think the OP has a point. If indeed this cooling unit has 6 x 1.5HP motors at 600VAC, the FLAs of each motor should be only 9A in aggregate. How and why they come up with the 23 FLA for the unit(according to their drawing), is a mystery. 14A in control power, as measured on the 600V side of the transformer (which would mean 73A on the 115V side)? Seems like a lot.

So either there is more to this cooling unit than just the 6 x 1.5HP fan motors, or something is amiss in their documentation and the OP has good reason to be questioning it. Still, it has nothing to do with power consumption so nothing that was said so far is wrong. I just see this issue as very odd. I'd vote for either woefully incomplete data and the cooling unit also includes something like pumps that they fail to mention, or a poor job of documentation, i.e. they used a drawing for another cooling system that had larger motors but forgot to update the MCA/FLA/MOPD data cell.

"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376
 
Upvote 0 Downvote
Hi,
I would look for a name plate on the motor and not the cooler to see the kw rating of each motor that might solve the mystery.

"its not what you know it what you can apply"
 
Upvote 0 Downvote
However correct all these answers are, I do think the OP has a point. If indeed this cooling unit has 6 x 1.5HP motors at 600VAC, the FLAs of each motor should be only 9A in aggregate. How and why they come up with the 23 FLA for the unit(according to their drawing), is a mystery
Click to expand...
See waross' first post for example pf and efficiency values that sound reasonable for these small motors. Nothing you don't already know I'm sure.

=====================================
(2B)+(2B)' ?
 
Upvote 0 Downvote
Perhaps more relevant, it's 9hp, not 9A.

=====================================
(2B)+(2B)' ?
 
Upvote 0 Downvote
At 600VAC, HP = FLA for the most part. I have not seen a new 3 phase motor with that low of eff and PF at 75% load or higher, in fact here in the US you cannot buy one that bad any longer without special dispensation as retrofit for existing or certifying that there is no more efficient version available for some special machine. These are just fans, I seriously doubt they would qualify. No way that anything that bad could be used in any facility attempting LEED certification as well, which lately is the new big deal in facility design.



"Dear future generations: Please accept our apologies. We were rolling drunk on petroleum."
— Kilgore Trout (via Kurt Vonnegut)
For the best use of Eng-Tips, please click here -> faq731-376
 
Upvote 0 Downvote
That's 9 x 1.5 HP. I have used 1 amp per HP at 600 Volts for more years than I can remember, but at 5 HP and higher. At 1.5 HP the rules of thumb don't hold up too well. Three phase motors may be more efficient than single phase motors, but remember that these are slower, higher slip motors than the ones that will run 1 Amp per HP.
As a sanity check I looked up 1.5 HP motors on the Baldor site.
I found two, one at 65% PF and 85.5% efficiency. That is about 56% Output kW over input KVA. The other motor was slightly better.
But these were 1170 RPM motors and the motors on the cooler are 1100 RPM, so possibly the current per motor is 23/9 = 2.56 amps.
New WAG based on 65% PF. KVA = 23.9
kW = KVA x PF = 23.9 KVA x 65% PF = 15.5 kW

Bill
--------------------
"Why not the best?"
Jimmy Carter
 
Upvote 0 Downvote
Status
Not open for further replies.

Similar threads

S
  • Question
Motor rpm vs Torque
Replies
4
Views
2K
zlatkodo
zlatkodo
unclebob
  • Question
500 HP, 440V motor
2
Replies
22
Views
5K
waross
waross
Its_Ali
  • Locked
  • Question
Constant Torque vs Constant Power VFD
Replies
1
Views
384
FacEngrPE
FacEngrPE
NickParker
  • Locked
  • Question
Class F motor Vs Class F/B motor - constructional difference 4
Replies
8
Views
462
edison123
edison123
CuriousElectron
  • Question
Eutectic Alloy vs E100 Solid State Overload
Replies
6
Views
3K
RRaghunath
RRaghunath

Part and Inventory Search

Sponsor

Back
Top Bottom