Motor protection relay not reading current when PFCs are connected

[Ingenieur]

BTW, Injunear...

Your estimate of PF & Eff'y values for the 4.16 kV (by OP) motor is very dated.

And, R-O-T means Rule of Thumb

Phil

it's generalized
and when plugged in 52.5 vs the data sheet 52.8 pretty close

I used pf/eff, 0.9/0.9 but use 0.82, seems close for a wide hp/v range
actual 0.83/0.95 total 0.79
close enough

 

[Ingenieur]

Assume
60 kvar required C correction
53 A fla
4000 v
assume no load S of 15% or 55 kva ~8 A
No load power factor of 0.5

P 28 kw
Q 48 kvar
add the 60 kvar C
P is the same 28
Q is now 12 leading
new S 30 kva
Current is now 4.3 A

drops very low, in fact if no load is <15% of full load or C > 60 kvar or no load pf > 0.5 the current may 'disappear' lol

basically a tuned LC circuit
motor Ind +j var
Cap -j var
sum cancels reducing S leaving a small residual P

 

[Jraef]

NullSig,
Why is the No Load current with PFCs connected something you are concerned about? Was this just curiosity?

 

[nullsig]

Oh boy, lots of responses. Okay, makes sense. The PFCs are causing the current to switch from lagging to leading and the current magnitude is dropping as a result. We also determined that the relay/CTs do not pick up until the 4A mark (we aren't sure if it's the CT or relay). So if the current is less than 4A (which it sounds like it very well could be) then it won't be read.

However, we have dozens of motors with PFC and there's no issue reading current under no-load operation on those. I believe we have determined the CTs need to be upstream of the PFC connection, not downstream.


The caps are 75 kvar. We operate at no-load for a three hour run-in test that we perform on all new and reconditioned motors to ensure there are no issues before it is put into service.

 

[nullsig]

Thanks for the help guys. I've requested the no-load and full-load efficiency/PF from the manufacturer to run some firm calculations myself... you'd think that info would be included on a datasheet for a brand new motor, but I guess Siemens doesn't agree...

 

[nullsig]

Got the data.

FLA: 50.3
No load amps: 10.4
No load PF: .112

No load S = 4000 * 10 * 1.73 = 69 kVA
No load P = S * pf = 69 * .112 = 7.7 kW
Q = cos-1(.112) = 7.7 kVAR

Now if you add the 75kVAR caps...

Qcap = 7.7kvar - 75kvar = -67.3 kvar
Scap = sqrt[ (67.3 kvar)^2 + (7.7 kw)^2 ] = 68 kVa
Acap = ~9.8A

The CT/relay should still be picking up 9.8A.... looks like this might confirm the CT placement theory...

 

[Jraef]

You'll find that most motor mfrs don't provide no-load amps readily, because they are fearful of people misusing that as a selection criteria when in reality, it's relatively meaningless out of context. The only reason I ever want to know the NLA on a motor is for comparison purposes in a maintenance schedule, ie if the NLA is 4.2A now, and 5 years from now it has crept up to 5.6A at the same line voltage, that tells me the bearings are wearing out. For that purpose, I'm going to measure it when installed, because line voltage is relevant, so getting the info from the mfr is pointless anyway.

 

[Bugman1400]

Most likely the CT ratio is 600/5 (120/1). So, 4A, primary translates to 0.033A, secondary, which is extremely low and probably can't be measured by the relay. Also, if the no-load motor current is 10.4A with most of that produced by reactive load, I assume the ~10A of capacitive current offsets the reactive load to almost zero primary current.

 

[Jraef]

Most likely the CT ratio is 600/5 (120/1). So, 4A, primary translates to 0.033A, secondary, which is extremely low and probably can't be measured by the relay. Also, if the no-load motor current is 10.4A with most of that produced by reactive load, I assume the ~10A of capacitive current offsets the reactive load to almost zero primary current.

You too have missed the fact that this is 4160V, not 460V. So it would not have 600:5 CTs in it.

 

[Ingenieur]

Got the data.

FLA: 50.3
No load amps: 10.4
No load PF: .112

No load S = 4000 * 10 * 1.73 = 69 kVA
No load P = S * pf = 69 * .112 = 7.7 kW
Q = cos-1(.112) = 7.7 kVAR

Now if you add the 75kVAR caps...

Qcap = 7.7kvar - 75kvar = -67.3 kvar
Scap = sqrt[ (67.3 kvar)^2 + (7.7 kw)^2 ] = 68 kVa
Acap = ~9.8A

The CT/relay should still be picking up 9.8A.... looks like this might confirm the CT placement theory...

No load
Q = sin (arc cos pf) S = 69 kva

with cap
Q = 69 - 75 = -6 kva
so S = 9.8 kva
i = 1.4 A

 

[Ingenieur]

You too have missed the fact that this is 4160V, not 460V. So it would not have 600:5 CTs in it.

it may
fla = 50
or 50/120 or 0.4 A out of the CT
perhaps a bit low
it needs to read a fault current of say 1000% or 500 A
so 500 A in = 500/600 x 5 = 4.2 out, below saturation
I would use a 500:5

 

[Phil Corso]

InjunEar...

What spec requires CT's be rated to accommodate 10 times nominal (expected) current?

Phil

 

[Bugman1400]

You too have missed the fact that this is 4160V, not 460V. So it would not have 600:5 CTs in it.

Removed.

 

[Bugman1400]

it may
fla = 50
or 50/120 or 0.4 A out of the CT
perhaps a bit low
it needs to read a fault current of say 1000% or 500 A
so 500 A in = 500/600 x 5 = 4.2 out, below saturation
I would use a 500:5

My previous statement about the 600:5 CTs was based on the 450A FLA given by Ingenieur in a previous post. As always, I should have checked the math myself. I see that a 400HP motor @ 4160V is around ~41 amps. My guess is that a CT ratio of 50:5 (10:1) or 100:5 (20:1) is used. With a CT ratio of 20:1, 4A, primary current translates to 0.2A, secondary current and is typically the minimum for an accurate relay reading. Some relays can go lower but, run the risk of radio interference and noise at that low level.

My apologies for the confusion.

 

[Ingenieur]

InjunEar...

What spec requires CT's be rated to accommodate 10 times nominal (expected) current?

Phil

Common sense
Contingent on relay type

 

[Ingenieur]

My previous statement about the 600:5 CTs was based on the 450A FLA given by Ingenieur in a previous post. As always, I should have checked the math myself. I see that a 400HP motor @ 4160V is around ~41 amps. My guess is that a CT ratio of 50:5 (10:1) or 100:5 (20:1) is used. With a CT ratio of 20:1, 4A, primary current translates to 0.2A, secondary current and is typically the minimum for an accurate relay reading. Some relays can go lower but, run the risk of radio interference and noise at that low level.

My apologies for the confusion.

I mis-calculated based on 480 v
as I and others pointed out it was estimated 50-55
the op got the name plate value 50.3

 

[Ingenieur]

At 4160 no load i <1.2 A
even with a 100:5 that is only 60 mA at the relay
if the v at no load is > 4160 even lower

 

[Phil Corso]

Come on Gents...

InjunEar... How did you get arrive at 1.2 A for NLA when NullSig provided 10.4A?

NullSig... remember, unless the PFC alters the motor's terminal-voltage, it (the PFC) has
little impact on the motor's PF! Did you read the topic/thread I referenced earlier?

Phil

 

[Ingenieur]

Come on Gents...

InjunEar... How did you get arrive at 1.2 A for NLA when NullSig provided 10.4A?

NullSig... remember, unless the PFC alters the motor's terminal-voltage, it (the PFC) has
little impact on the motor's PF! Did you read the topic/thread I referenced earlier?

Phil

calc no load S
10.4 A w/o capacitors
4160 v

then derive P + jQ using pf 0.112
S 75 kva
P 8.4 kW
Q 74.5 kvar

add 75 kvar
P still 8.4
Q is now -0.5 (almost negated)
new S 8.4 kva
new 1.2 A
new pf ~ 1 slightly leading

 

[Phil Corso]

InjunEar...

1.2A is the Line-current! The motor's NLA is still 10.4! It does not change!!!

Phil