Motor protection relay not reading current when PFCs are connected

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nullsig

Member
Location
Louisville KY
We have a 400HP 4160V motor protected by a multilin 469 relay. There are PFCs attached downstream of the CTs. When the motor is running unloaded (~10A) the relay does not read any current. Current is read if leads to the PFC are lifted. The CTs/relay were tested with a primary injection and are functional.

Any ideas why the presence of PFCs would affect current readout so significantly?
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
NullSig...

What does a clamp-on meter show upstream and downstream of the PFC location?

Regards, Phil Corso

PS: A similar situation is shown in PtonSpark's Post "Overcorrection" ~ Sep 2015!

S-M-C, is the PFC to "Correct" the 400Hp motor PF?
 
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nullsig

Member
Location
Louisville KY
Phil Corso said:
NullSig...

What does a clamp-on meter show upstream and downstream of the PFC location?

Regards, Phil Corso

PS: A similar situation is shown in PtonSpark's Post "Overcorrection" ~ Sep 2015!

S-M-C, is the PFC to "Correct" the 400Hp motor PF?
Click to expand...

We do not have the procedural ability to do this... it would require interacting with exposed 4160V parts in a confined motor starter. It would never be green lit and I would not ask the electricians to do this.

Yes. it is connect to the correct motor :cool:



Phil Corso said:
NullSig...

Is the Amp value noted (~10A) obtained via soft- or hard-ware?

Phil
Click to expand...

The relay reports 10A when the capacitors are disconnected and it's accuracy has been verified via primary injection testing.
 
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ATSman

ATSman

ATSman
Location
San Francisco Bay Area
Occupation
Electrical Engineer/ Electrical Testing & Controls
nullsig said:
We do not have the procedural ability to do this... it would require interacting with exposed 4160V parts in a confined motor starter. It would never be green lit and I would not ask the electricians to do this.

Yes. it is connect to the correct motor :cool:





The relay reports 10A when the capacitors are disconnected and it's accuracy has been verified via primary injection testing.
Click to expand...

When you run the motor loaded with the caps connected what does the relay read?
 
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GoldDigger

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Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
nullsig said:
We have a 400HP 4160V motor protected by a multilin 469 relay. There are PFCs attached downstream of the CTs. When the motor is running unloaded (~10A) the relay does not read any current. Current is read if leads to the PFC are lifted. The CTs/relay were tested with a primary injection and are functional.

Any ideas why the presence of PFCs would affect current readout so significantly?
Click to expand...
Look at it this way:

Totally unloaded motors tend to have a relatively low power factor, with high efficiency motors having the lowest.
If you have a motor which has an unloaded PF of .1, then the resistive component of the current will be 1/10 of the total current.
If the total current, including the inductive part, is 10A then the resistive component, which is all that you will see once the PFCs have cancelled out the inductive portion, will only be 1A.

The relay's meter reading may not be accurate for very low current values, or it may just read out a current of .08A as 0 rather than 1.
Any digital readout will typically be specified as +/- one unit in the least significant digit it displays. That is one reason that it was suggested that you use an amp clamp which might be more sensitive.

A motor running a full rated load will have a much higher PF, almost always above .6 for newer motors, and in some cases as high as .9.

PFCs which are sized for the unloaded condition will normally also be in the ball park for PF correction when loaded, but a design decision has to be made as to which load condition to optimize for.
 
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Ingenieur

Senior Member
Location
Earth
Off the cuff calc fla 450 A
hard to believe no load is 10 A (if 10 with tsc, with no tsc perhaps 30-50)
still 0.1 to 0.5 at ct at a device that reads in the 100's
too low for the ct to respond
probably a 500:5 ct

reads 10 A with pfc's online using a meter? Or is the 10 with no pfc?

with pfc off line what is read?
meter
relay
very important info

for the pfc you 'lift leads'?
or have a fused disconnect switch?

very confusing post
is the 10 with or without the tsc?
what is i for the opposite state?
 
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P

Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
NullSig...

The answer:

An Induction motor's No-Load Amps (NLA) consists of two components: 1) a very small "Real"component (some stator-cu loss, core-iron loss, rotor-cu loss, machinery-train bearing-friction, and windage loss; and 2) a very large "reactive" component (excitation or magnetization requirement)!

Typically, Full-Load Amps, FLA, for this machine is approximately 55A! NLA, could be between 20-30%! You measured 10A, or about 18%)... mostly inductive, and it is "lagging"! The PFC current, PFA, is capacitive, therefore "leading"!

Simply, PFA negates NLA!

Phil
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Gentlepeople...

1) PFC for motor should not exceed some 80% of a motor's NLA! I refer you to previous posts!

2) PFC for MV motors is a great waste of money! I'm betting this is for a Cost-Plus project!

Phil
 
I

Ingenieur

Senior Member
Location
Earth
Phil Corso said:
InJunEar...

My R-O-T is probably older than you... FLA = Hp /[ Sqrt(3)*kV ] !

Phil
Click to expand...

power in w = hp x 746
divide by pf x eff ~ 0.80 -0.85
result is in va
va / ( v sqrt 3) = i

you are assuming 1 hp = 1 kva
or 0.746 (kw/hp) kw/0.82 (kw/kva ~ pf x eff) = 0.90 kva/hp

people actually get dumber as they get older ;)

the motor is probably rated at 4000 v

What is rot?
some kind of old guy slam?
when I went to school I used a slide rule
 
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Ingenieur

Senior Member
Location
Earth
How much C kvar must be used to raise pf from 0.83 tp 0.90?
assume 4000 V
fla 53 A

ph ang = arc cos 0.83 = 33.9
existing
S 367.2 kva (1.732 x 4000 x 53)
P 304.8 kw S x pf
Q 204.8 kvar sin 33.9 x S

new
ph ang arc cos 0.9 = 25.8
P 304.8 kw (constant load)
S 338.7 kva P/pf
Q 147.4 kvar sin 25.8 x S

add 204.8 - 147.4 = 57.4 call it 60 kvar
 
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Phil Corso

Senior Member
Location
Boca Raton, Fl, USA
Here's an easier way:

If the original apparent load, kVAo, the original power factor, PFo, and the desired power factor, PFr, are known, then the formula for capacitive reactive power, neglecting capacitor loss, is:
kVAc = K x P, where
K = tan [ Acos ( PFo ) ] - tan [ Acos ( PFr ) ], and,
P = kVAo x PFo, in kW
Regards, Phil Corso
 
I

Ingenieur

Senior Member
Location
Earth
Phil Corso said:
Here's an easier way:

If the original apparent load, kVAo, the original power factor, PFo, and the desired power factor, PFr, are known, then the formula for capacitive reactive power, neglecting capacitor loss, is:
kVAc = K x P, where
K = tan [ Acos ( PFo ) ] - tan [ Acos ( PFr ) ], and,
P = kVAo x PFo, in kW
Regards, Phil Corso
Click to expand...

exact same thing
trig based on power triangle
I rather derive than memorize
as you get older you'll see why lol memory fades
 
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