Earth Currents with Submersible Pumps 3

[Marke]

I have an interesting problem as follows:
Submersible pump 185KW.
400V 50Hz controlled by a soft starter.
MEN supply. Star connected transformer secondary dedicated to this application with the neutral earthed.

12 inch bore, 10 inch pump, 125 Meters deep.

Earth cable from pump is attached to the well casing at the top of the well.

If the earth cable is not attached to the well casing, there is 15Amp AC between the well casing and the earth cable to the pump motor. If the earth is connected, 90 Amps flows between the motor earth and the well casing.
If the motor stops, no voltage and no current.

Any thoughts?

Best regards,



Mark Empson
L M Photonics Ltd

 

[itsmoked]

Hi Mark;
Check this statement.

If the earth cable is not attached to the well casing, there is 15Amp AC between the well casing and the earth cable to the pump motor. If the earth is connected, 90 Amps flows between the motor earth and the well casing.

It isn't making logical sense to me.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[Marke]

Hi Keith

Yes, you are correct. Should read 15 VAC.

Thanks,
Mark

Mark Empson
L M Photonics Ltd

 

[waross]

Have you been able to megger the motor? Is it possible that a voltage is being induced into the ground cable by two phase conductors. We normally think of a twisted cable as canceling out induction and magnetic field influence. However, the construction of some submersible pump cables is such that the ground conductor may be influenced more by the two adjacent conductors and less than the third conductor.
The other possibility is a failing pump or cable.
Does the ground cable carry back to the control panel or does it end at the surface of the casing?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[thewellguy]

Is there any other grounds from the building for example attached to the well casing? Also is there rod rods or anoid rods in close proximity to the well casing?

 

[7anoter4]

I don't know what kind of water it is about, but I feel the well is well grounded- through the water- and the grounding wire is not. So, one has to check the grounding of grounding wire and to mend it.

 

[electricpete]

Megger check makes good sense.

Another interesting piece of data would be to check the balance among phase currents in both conditions (with ground connected and without).


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[electricpete]

Also I agree with the scenario waross suggested: one conductor closer to the casing creates induced currents in the casing.

Just imagine the pump casing is grounded solidly at bottom. Now you add a solid ground at top and you have a loop where high current can flow with relatively low voltage.

If this is the case, than 15vac is not a floating measurmeent but a measure of voltage induced along the length fo the column from the grounded end (bottom) to the open end (top). That seems a little high. But I guess long runs of cable maybe can create lond induced voltages. I started trying to think about how to calculate that for simple geometry such as cable located a distance d from a plane...didn't get very far. Maybe tonight.

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[electricpete]

Actually would need to think about the combined field of all 3 conductors (based on their positions) for anything useful.

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[Marke]

This is a brand new pump and megger tests fine.
The current flows around the loop between the pump and the casing via the earth wire. If the earth from the switchboard is removed, the current still flows. It is not coming from the switchboard earth.

It appears to be an induced current rather than a leakage current.

I had three phone calls in the same week about similar problems, I have not found this before, but in one case, the earthwire is nearly burning off the crimp lug.

Best regards,

Mark Empson
L M Photonics Ltd

 

[7anoter4]

The emf induced from one wire to another parallel wire is equal to: Eig=j*Ii*Xig
Where Xig=k*ln(disti_g/re) disti_g =distance from center line of the two wires and re it is
the equivalent stranded wire radius =radius*.7788.For SI u/m k=4*pi()*frq/10^7 H/m = 6.28E-05.
Let us say conductor A is a diameter opposite to G and B and C are as close as possible to G.
The maximum distance between two conductors within a pipe of 10" could be 220 mm if the cables are of 240 sqr.mm cross section , 20.6 mm dia copper and 30 mm overall dia of cable.
The maximum Xig=k*ln(220/8.02) *length[m].If the length could be 185 m Xigmax= 0.038598 ohm.Xigmin=k*ln(30/8.02)*185= 0.015333 ohm.
If we shall take pf=0.8 and eff.=0.9 the rated current will be 642.4 A.
Then IARE=513.9 A and IAIM=385.4 A ; IBRE= 76.8 A and IBIM= -637.7A;ICRE= -590.7A ICIM=252.3A.Calculated Ere=-IAIM*Xigmax-(IBIM+ICIM)*Xigmin=-8.97 V; Eim=IARE*Xigmax+(IBRE+ICRE)*Xigmin= 11.96 V Etot=sqrt(Ere^2+Eim^2)= 14.94 V.
Could be?

 

[waross]

Hi Pete;
A suggestion. The cable is probably a twisted four, "A", "B", "C", and Ground. The ground will be nested between "A" and "C" and spaced from "B". The currents in "A" and "C" should equal the current in "B". Try your calcs with "B" phase current, but "A" and "C" phase spacing to the ground. I would use 300 Amps as a representative current.
Not exact but this should get us in the ball park to see if induction is reasonable.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

My thought is that if it’s triplexed twisted cable, the transposition in cable should make induced voltages negligible.

Even if it’s not triplexed cable, it has a vague resemblance to a donut CT which responds to sum of currents without much dependence on position. 7anoter4 has begun the math... I have to think about that.

Do you have details of the cable type?

Do you know the material (steel or something else) and wall thickness of the pump column?


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[electricpete]

vague resemblance to a donut CT

Hmmm. I was for a moment picturing cables routed inside pump column, but I’ll bet they are not. I guess they are just run along the outside of the pump column (wihtout any conduit)?


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[electricpete]

Inside our outside I'm not sure. But if the twist is symmetric then I I wouldn't think there should be induced voltage. By symmetric I mean A, B, C occupy a certain position at one elevation, then at some other elevation they will have switched positions (A where B was) and yet another elevation switched a second time (A where C was). Is the twisted four different than that Bill?


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[electricpete]

Also thinking some more about the relevant loop. I guess it would be ground wire running down the hole to pump casing then back up along the column (right?). In that case the position of ground wire needs to be a symmetric part of that twist pattern.

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[waross]

Four wires twisted. No induced voltage in the casing due to the transpositions that the twist develops. But, "A" and "C" phase will be in contact with the ground conductor but "B" phase will not be in contact. You have an asymmetric arrangement of conductors in regards to the ground conductor.
I believe that the voltage is being induced in the ground conductor by "A" and "C" phases.
Then again it may be a poor connection in the primary system. The well casing may be passing neutral current from an unbalanced primary. It could happen in a rural installation.
Are the other problem wells in the same area?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[electricpete]

Thanks. I understand now A, B, C not symmetric with respect to ground wire.

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[electricpete]

Trying to follow Bill's suggestion, I modeled the following problem in the free software F.E.M.M:

A, B, C, Ground conductors are each 0.6" diameter solid copper conductors. They are located in square arrangement 1.2" center-to-center distance. These conductors are centered within a 5000" long cylindrical iron pipe with I.D. 10.2" and O.D. 10.8" (*). Geometry is shown on slide 1 of attached powerpoint.

Current is applied in the three phase conductors is 3-phase 200A 50HZ. (IA = 200, IB =-100 + 173*I, IC =-100 - 173*I)

With ground conductor is perfectly shorted to the pipe at each end, the flux solution is on slide 2, circulating current in groundwire and pipe is approx 18A (slides 3 and 4) and induced voltage is approx 0.3 vac (slide 5)

With ground conductor open (not connected at each end) as per slide 6, there is no circulating current and induced voltage is 0.5vac in the groundwire and 0.1 vac in the pipe.

I would be glad to post the F.E.M.M files if anyone is interested. It is a pretty easy to use program that you can get for free. Of course, standard caveats apply to my analysis (garbage in gives garbage out... results require sanity check)... just providing a model for consideration.

(*) My thought was that if the conductors are placed symmetrically within the center of the pipe, I can get away without modeling the twist since at any given plane (elevation) of the solution, rotating everything changes nothing. If it is not the case that conductors are symmetrically located in the center of the pipe, then modeling of twist is required.... far beyond my capabilities.

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[electricpete]

Here is the FEMM data file for the scenario when the pipe and ground conductor are short circuited together at both ends.

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