electrical motor power factor & efficiency 3

[somacast]

hello,

I want to know where each figure recorded from the LCD panel of the power feeder of an AC motor driving a machinery goes.

in other words, I have the reading at one instance reads in the panel at electrical room:

KW = 700
KVAr = 800
AMP = 70

and this is on a 11KV feeder.

so what is the 700 over here in a power triangle?

how do I relate to power factor ?

how do I get my electrical efficiency? can I say eff= data sheet kw / reading kw from panel?

Thanks & Regard,

 

[electricpete]

KW = 700
KVAr = 800
KVA = sqrt(700^2+800^2) = 1,060
I = 1,060kva/(sqrt3*11kv) = 55.6A
but you said amps = 70

Unless I made an error (and that's very possible that I did) it seems your data does not add up quite right.

Setting that aside, power factor could be calculated from kw and kva. I'm going to stop short of telling you exactly how to calculate it, that is something you should be able to figure out.

You don't have enough information to calculate motor efficiency because you don't know the actual motor shaft output power. If you have motor factory data you might be able to estimate motor efficiency from that by adjusting for your operating point. Of course there is likely also an efficiency associated with the driven equipment.


=====================================
(2B)+(2B)' ?

 

[RRaghunath]

Reading kW is input power to motor and data sheet kW is shaft power. Thus, the ratio of shaft power to input power should give us efficiency. You are right there. But, it is important to confirm that the input power you are reading corresponds to the same shaft power indicated in data sheet.
Input kW divided by input kVA gives power factor.
Generally speaking, motor data sheet includes power factor and efficiency information at different load points.
If the voltage at the motor terminals is lower than rated (11kV), motor draws higher currents for the same kW. This is to be kept in mind.

 

[LionelHutz]

Does the data sheet not give the efficiency for various loads?

 

[somacast]

Thank you very much for your replies.

@LionelHutz : it does, however for some reason (or a misunderstanding of some part of the triangle) , I am not getting the number corresponding to what I am reading from the panel when manually calculating PF ...

 

[somacast]

for instance as one of the confusions :

datasheet says @100 load , the current is 92 Amps
I read in electric room 76 Amps only ...
quite weird to me ...

 

[LionelHutz]

The current won' be 92A unless you put rated load on the shaft. Try extrapolating between the 75% and 100% numbers.

 

[somacast]

hi, thanks for replying again

the 76A measured when the motor shaft is 100% loaded, i.e. reading taken when machine @ 100 load.

 

[waross]

the 76A measured when the motor shaft is 100% loaded, i.e. reading taken when machine @ 100 load.

The driven machine may be fully loaded, but the motor is probably oversized.
If the data sheet says that the full load current is 92 Amps, then at 76 Amps, the motor is not at full load.



Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[edison123]

The motor load current, as the name says, is dictated by the load, not by the motor. Only the motor no-load current is defined by the motor.

Muthu

http://www.edison.co.in

 

[LionelHutz]

You have something wrong. In order of most likely, the load isn't 100%, your meter isn't reading right or the motor datasheet isn't correct.

 

[bacon4life]

In the first post the amps appear to be about 25% higher than expected whereas in the post at 2 Jun 21 19:39 the amps appears to be about 20% lower than expected. How are you determining load on the motor?

A few other lower likelyhood possibilities:
[ul]
[li]Are voltage and current readings similar on all three phases? Unbalanced voltage can cause high currents.[/li]
[li]Is this a standard motor starter? Typical metering equipment can be wildly inaccurate when placed between a VFD and a motor.[/li]
[li]Are there high levels of harmonics? High levels of current harmonic distortion can increase the amp readings whereas many meters ignore harmonics when calculating reactive power.[/li]
[li]Is the voltage at the motor near the rated motor voltage? Both low and high voltage can cause excess current.[/li]
[/ul]

 

[somacast]

Thank you all very much,

as for @bacon4life questions
the first post was what I read that day in electrical room through the feeder panel, the other numbers I mentioned 92A is the rated current at full load in the data sheet of the motor,

as of the questions, i will try to answer:

1- here is a new full set of readings
I1 = 77A
I2= 75
I3 = 77.7 A
U = 10.8 KV
P = 1.01 MW
q= 1.01 MVAR

i trust those as I took them myself

2- it is a standard motor yes

3- none noticed , nor reported

4- it is ..

Do the readings look consistent?

thanks again

 

[electricpete]

these latest readings look a lot more self consistent than the first set.

=====================================
(2B)+(2B)' ?

 

[somacast]

ok great , thanks

so now what is the pf ?
efficiency?

thanks

 

[electricpete]

with p and q identical, the power factor angle is 45 degrees and pf is 1/sqrt2 = 0.707.

efficiency unknown, same reasons as above

=====================================
(2B)+(2B)' ?

 

[somacast]

sorry if I missed some information

the motor rated power in the data sheet is 1200 kw

so whats the efficiency here ?

 

[waross]

Rated KVA = 1753 KVA
Estimated KVAR = 1010 KVAR
kW based on KVAR estimate = 1433 kW
Estimated efficiency at full load = 84% efficiency.
Estimated load at 77 Amps. (77A/92A) = 84% load.
Judgement call;
1. 84% efficiency seems unreasonable low. I would expect 90% or better efficiency.
2. Neither the real current nor the reactive current are linear with respect to the per unit load.
3. The efficiency will be close to the efficiency at 75% load as shown by the data sheet.
4. The efficiency at 75% to 85% is typically better than the efficiency at 100% loading.

Bill
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[LionelHutz]

so whats the efficiency here ?

you didn't provide enough information. Efficiency is power out divided by power in. Power in is 1.01MW. What is the shaft or output power?

 

[somacast]

@waros : how did you get 1753 ?

and regarding point 4 , this 11kv motor has its highest efficiency & pf at 100% load as per its datasheet !!