FLA vs power consumption vs motor hp
FLA vs power consumption vs motor hp
(OP)
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
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





RE: FLA vs power consumption vs motor hp
RE: FLA vs power consumption vs motor hp
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
RE: FLA vs power consumption vs motor 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
RE: FLA vs power consumption vs motor hp
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.
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).
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(2B)+(2B)' ?
RE: FLA vs power consumption vs motor hp
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(2B)+(2B)' ?
RE: FLA vs power consumption vs motor 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.
RE: FLA vs power consumption vs motor hp
Best to you,
Goober Dave
Haven't see the forum policies? Do so now: Forum Policies
RE: FLA vs power consumption vs motor hp
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
RE: FLA vs power consumption vs motor hp
For service capacity you must also consider whether the motors start simultaneously or use staggered start.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: FLA vs power consumption vs motor hp
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.
RE: FLA vs power consumption vs motor hp
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.
RE: FLA vs power consumption vs motor hp
Have you tried contacting the manufacturer?
I would think that 4-pole motors with 100rpm slip are quite inefficient.
RE: FLA vs power consumption vs motor hp
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
RE: FLA vs power consumption vs motor hp
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.
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RE: FLA vs power consumption vs motor hp
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"
RE: FLA vs power consumption vs motor hp
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(2B)+(2B)' ?
RE: FLA vs power consumption vs motor hp
=====================================
(2B)+(2B)' ?
RE: FLA vs power consumption vs motor hp
"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: Eng-Tips.com Forum Policies
RE: FLA vs power consumption vs motor 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
RE: FLA vs power consumption vs motor hp
"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: Eng-Tips.com Forum Policies
RE: FLA vs power consumption vs motor hp
It also does not say 23kVA, it says "23 FLA". So even if you are at 3A per 1-1/2HP 600V motor, that is still not getting to 23A.
My point is simply that the numbers do not add up in my opinion, so somewhere, there is data in/on that cooling system that is inaccurate or open to gross misinterpretation.
"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: Eng-Tips.com Forum Policies
RE: FLA vs power consumption vs motor hp
http://www.luvata.com/Documents/Air%20Units/Air%20...
Although for a slightly different product model number, the "-4B" specifies 460V motors with 115 volt controls. On that model the total watts is 8682 Watts (1447 W/Fan). The quote sheet also lists MCA/FLA/MOPD instead of FLA/MCA/MOPD. Would a power factor of 0.5 be reasonable. Assuming 0.5 pf gets you to the listed 21.6 FLA.
RE: FLA vs power consumption vs motor hp
I was basing the KVA on this sentence from the first post.
"That would give approximately kw = 600 x 23 x 1.73 = 23.9 kw or 32 hp."
But I pretty much agree with you.
Time for us to stop guessing.
Bill
--------------------
"Why not the best?"
Jimmy Carter