Having this problem

[sparo]

Hi all,

Need some help on this.

Currently this existing AHU Fan motor having 11kw Rated FLA as 20.7 Ampere. It is 415V/3ph/50Hz.
However I only manage to get 6.7 Ampere. Is this correct?
The owner need me to do a power consumption comparison between the new & existing AHU.
The new AHU fan motor is also having 11kw rated FLA 22 Ampere. 415V/3ph/50Hz.

How to justified and go about this issue.


Regards

[Tesla]

Hi sparo
FLA = Full Load Amps ie the maximum current the motor was designed to handle. It is normal to find motors running with far less current than rated. Other factors to take into account is th position of the dampers which have a small effect on current of motor, other factors would be if vav's are used and are they open or closed?, filters. Then there is the torque curve of fan - with centrif fans the current reduces with increased static pressure. Last of all is the measurement, are you using a true RMS metre for accuracy. Take into calculation the power factor of motor.

[nike123]

Hi all,

Need some help on this.

Currently this existing AHU Fan motor having 11kw Rated FLA as 20.7 Ampere. It is 415V/3ph/50Hz.
However I only manage to get 6.7 Ampere. Is this correct?
The owner need me to do a power consumption comparison between the new & existing AHU.
The new AHU fan motor is also having 11kw rated FLA 22 Ampere. 415V/3ph/50Hz.

How to justified and go about this issue.



Regards

Let assume power factor of 0.86
11kW*0,86=9,46
9,46kW/3=3,15kW power by phase
Phase voltage is 415V
3150W/415V=7,6 A phase current

By calculation phase FLA is 7,6A

You got 6,7A.
To me it sounds everything is OK

That FLA stated on unit is sum of all 3 phase currents.

[sparo]

Thanks for the information really appreaciate.:D

[Peter_1]

That FLA stated on unit is sum of all 3 phase currents.
FLA stands for me the max current/phase and not the sum of the 3 phases.
If on a compressor's nameplate 12A is the FLA, then if you measure 1 phase at full load, you will measure 12A.
You set the overload protector also at 12A, otherwise your compressor will be overloaded.
And if you clamp all 3 phases at once, you will measure AMP... Strange, strange ;-)

[Peter_1]

Current of a centrifugal fan is dependent from airflow and pressure differential.
If It draws few AMP's, then suction or pulsion is somewhere restricted. You have to verify this with the fan chart specific for that fan.

[nike123]

FLA stands for me the max current/phase and not the sum of the 3 phases.
If on a compressor's nameplate 12A is the FLA, then if you measure 1 phase at full load, you will measure 12A.
You set the overload protector also at 12A, otherwise your compressor will be overloaded.
And if you clamp all 3 phases at once, you will measure AMP... Strange, strange ;-)

I agree that is strange to state on nameplate sum of phase currents but, either is information on FLA incorrect or information on power of motor is incorrect in case that is one phase current stated, because mathematics than doesn't have any sense.:confused:

[Peter_1]

FLA stated on a compressor nameplate is max current/phase, that's sure for 100%.

[nike123]

Let assume power factor of 0.86
11kW*0,86=9,46
9,46kW/3=3,15kW power by phase
Phase voltage is 415V
3150W/415V=7,6 A phase current

By calculation phase FLA is 7,6A

You got 6,7A.
To me it sounds everything is OK

That FLA stated on unit is sum of all 3 phase currents.


I made mistake in this calculation. Corect one is this:


11kW/0,86=12,8
12,8kW/3=4,3kW power by phase
Phase voltage is 415V
4300W/415V=10,4 A phase current

If power on nameplate is given for star connection, than
4300W/240V=17,9A current in windings connected in star.
Which comes close to that FLA at nameplate

To me it is not clear what power of motor is at what voltage, and what is power factor, to be able to tell what current at maximum load should be.

[Peter_1]

If power on nameplate is given for star connection, than
4300W/240V=17,9A current in windings connected in star.
Which comes close to that FLA at nameplate
Power of a motor is independent from the voltage

[Peter_1]

To me it is not clear what power of motor is at what voltage, and what is power factor, to be able to tell what current at maximum load should be.
Let's take simple figures, 1000 W nameplate, power factor 0.9.
1000 W is available (Pa)on the shaft of a 3 phase motor and you need to add P needed (Pn)1.73x line voltage x Line Curent x power factor.
The difference between Pn and Pe is heat, reason why the motor becomes warm.

A motor of 2HP, connected on a net of 3 x 220V/50 Hz, full load, efficiency = 0.77 and PF = 0.76
Line amp = (2 Hp x 736 W/HP)/(0.77 x 1.73 x 230 V x 0.76) = 6.32 A

[nike123]

Ok, but depending on how is motor windings connected, current will be (for max load)
20,7A or 20,7A/√3 (12A)


Windings of standard three-phase single speed motors can be connected either in star or
delta connection.
Star connection
A star connection is obtained by connecting W2, U2, V2 terminals to eachother and the U1,
V1, W1 terminals to the mains. The phase current and voltage are:
Iph = In ; Uph = Un / √3
where In is the line current and Un the line voltage referred to the star connection.
Delta connection
A delta connection is obtained by connecting the end of a phase to the beginning of the
next phase.
The phase current Iph and the phase voltage Uph are:
Iph = In / √3 ; Uph = Un
where In and Un are referred to the delta connection.

http://www.lafert.com/lafert/img3/32/Technical%20Catalogue%20Lafert%202008%20eng.pdf

[Electrocoolman]

Hi Sparo,

You need to open up the motor connection box and see how the motor is connected.

As Nike says, the current draw depends on whether the motor is STAR or DELTA connected.

It is possible to run the motor in STAR at a lower speed/power - in which case the current will be lower.

Note some motors are DUAL voltage rated....in which case you need to connect it according to the line voltage.

[Peter_1]

And not to forget the part winding motors and dahlander motors