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Meaning of Full Load Amperes (FLA)

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Mark_Djorn

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Hi All,

I have a FCU rated at 1.1 kW (3 speeds) but stating 2.8 A (FLA).

Can anyone explain me what FLA is and why 3.8 A? If I make my maths: 1100 W / 230 V is 4 78 A.

Not sure what I am doing wrong. Thank you.
Mark
 
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Can you tell us what you mean by FCU, over here (UK) if often means Fused connection unit.

1.1kW and 2.8A works for 400V but it would be unusual for a single phase load to be connected phase-phase.
 
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If you look at motor current charts the FLA does state its 2.8A for a 1.1kW motor at 380Vac.

[ElectriciansForums.net] Meaning of Full Load Amperes (FLA)
 
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Mark_Djorn said:
FCU: Fan Coil Unit.

But also.... apologies... I will try to modify my thread as it's a 3phase fancoil.

Yet...

1100 / (400 * sqrt 3) should be 1.58 A. Not 2.8 A.
Not sure what I'm calculating wrong.
Thank you
Click to expand...

There is also the Power Factor and Efficiency of the motor that has to be taken into account.
 
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Mark_Djorn said:
Correct... even if I wanted to consider 0.8 for both PF and efficiency... I dont get anything grrater than 2.5A.... ?
Click to expand...

Is there no plate on the motor telling you the Efficiency and Power Factor?
 
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I think my confusion comes from the fact I don't know the definition of FLA or FLC.

Also, have a look at this picture.
I cannot figure out the maths to confirm the current given once I have voltage, power, power factor and efficiency
 

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I note that the motor is 1.5kw
the fan is 1.16kw

it is normal for the motor to be slightly higher rated than the fan as it prevents the motor from being run in a slightly overloaded state and therefore overheating.

the motor itself will have a nameplate that shows the full load current.
 
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James said:
I note that the motor is 1.5kw
the fan is 1.16kw

it is normal for the motor to be slightly higher rated than the fan as it prevents the motor from being run in a slightly overloaded state and therefore overheating.

the motor itself will have a nameplate that shows the full load current.
Click to expand...
For some reason I thought that a fancoil only had 1 motor: the fan. But ot seems that this plate differentiates between motor AND fun. Not sure to understand the difference. But if I put some numbers is.... 1.5 kW; 400V; 0.83 eff; maybe a 0.8 PF... we will get more or less the 3.42 A stated.
Also, I can see that my Single Line Diagram only reported the rating of the fan (1.1 kW) but the FLA of the motor it seems.

Question is... if not the fan, what's the motor doing in a Fancoil?
 
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The motor converts electrical energy to a rotating force (speed x torque=power)
the fan converts the rotation to a forward flow of air or other fluid.

what are you trying to figure out?
wiring, suitable motor overload etc?
 
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There is only one motor, the fan motor.
By 'Motor kW' it means the maximum rated mechanical output power of the motor, as decided by the motor manufacturer.
By 'Fan absorbed power' it means how much mechanical power the impeller absorbs from the motor shaft, as decided by the FCU manufacturer.

So the motor is capable of 1.5kW but the impeller only applies 1.16kW of load to it, leaving a margin so that the motor is not worked to its maximum. However, the full load current is specified for the motor working at 1.5kW mechanical output (P2) and it is a reasonable figure given the rated 415V supply voltage and 83% efficiency (I've taken the power factor as 0.75):
1500 / 0.83 / 0.75 / 415 / sqrt(3) = 3.35A
Which compares well to the stated 3.42A.

With the fan impeller taking only 1.16kW instead of 1.5kW the actual line current will be lower, although not by the factor 1.16/1.5 because only the in-phase current will be reduced, not the magnetising current (and the pf will be lower).
 
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