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Current in Unbalanced Load
2

Current in Unbalanced Load

Current in Unbalanced Load

(OP)
I have been asked to monitor the current in each phase of some delta connected heaters (unity PF) using current transducers in just two phases
I can't remember the formula for calculating current in third phase.
If someone could supply this I would be most gratefull.
Too long out of school!
Thanks in advance
Roy

RE: Current in Unbalanced Load

Kirchhoff's current law.  Consider the delta connected heating elements a single node.

RE: Current in Unbalanced Load

(OP)
Davidbeach, all the examples I see all show DC with one source, for AC it must be different because of the phase angle. If it were that simple phase C would be A+B
We are trying to detect element failure. some of the elements have several elements in parallel (overall delta)
regards
Roy

RE: Current in Unbalanced Load

Hi Roy.
What David saied.
C= -(A+B) in vectorial form for the 3-ph system.
Best Regards.
Slava
 

RE: Current in Unbalanced Load

I would recommend snap-on current coils on each leg, (not phase).  Then you can see step differences in a leg when sub heaters open.

If you want to know which elements have opened you would need a CC on each or do the troubleshooting thing.

I have had to build several devices to detect heater failures in just a few power line cycles.  You know, anal retentive semiconductor makers..

Keith Cress
kcress - http://www.flaminsystems.com

RE: Current in Unbalanced Load

(OP)
Slavag, sorry, math was never my strong point
C=-(A+B)     if I assume for the moment the load is balanced @ 6A per phase
C=-(6+6),  C=-12
I know the key is in "vectorial format", I remember working with vectors but that was at least 40 years ago and I don't work with 3 phase power on a regular basis, is there not a simple formula for 3 phase @ unity PF?

Itsmoked, unfortunately the heater terminations are in Class 1 Div 2 so I don't have access to each leg.

I remember that you can measure total kW using 2 x single phase meters so I just assumed it must be a simple C=X*(A+B).
I spent some time searching www before showing my ignornce.
Bear with me please
Roy

RE: Current in Unbalanced Load

Hi Roy.
Is OK.
Vectorial in symmetrical system:
C= -(6+6 ^120deg), that means C is also 6A.
For one phase meas you can use simple formula:
P=U(phase)xI(phase)x3 ( of course we talk about sym load and PF=1)
If you have three meters, you can use formula:
Ptotal=Ua*Ia+Ub*Ib+Uc*Ic.
For three phase system you can use formula:
P=U(phase to phase)*I*1.73.
Hope that helps.
Best Regards.
Slava
 

RE: Current in Unbalanced Load

Unfortunately, with the delta connected heaters, a change in one phase resistance (say by opening one parallel element) will result in the current phase angles not being 120 degrees apart.  This means that Slava's simple formula

Quote:

C= -(6+6 ^120deg), that means C is also 6A.
will not work.

For example, say the voltages are:
VA=120V, VB=120V@-120°, VC=120V@120°
VAB=VA-VB=207.8V@30°, VBC=207.8V@-90°, VCA:207.8@150°

With all resistances equal,

RAB=RBC=RCA=60 ohm then

IAB=3.46A@30°, IBC=3.46A@-90°, ICA=3.46A@150°

IA=6A@0°, IB=6A@-120°, IC=6A@120°

If RAB doubled to 120 ohm then

IAB=1.73A@30°, IBC=3.46A@-90°, ICA=3.46A@150°

IA=4.58A@-10.89°, IB=4.58A@-109.11°, IC=6A@120°

Using Slava's equation would result in IC=4.58A
 

RE: Current in Unbalanced Load

Hi Roy.
Of course Jghris is totally right.
My ( isn't my of course)formula is very simplify.
Thanks a lot Jghrist. Star to you.
Best Regards.
Slava

RE: Current in Unbalanced Load

Connect the CTs in "A" phase and "C" phase.
A drop in a load connected from "A"to "B" will show a current drop in the "A" phase CT.
A drop in a load connected from "B"to "C" will show a current drop in the "C" phase CT.
A drop in a load connected from "C"to "A" will show a current drop in both the "A" phase CT and the "C" phase CT..

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

RE: Current in Unbalanced Load

(OP)
Jghrist, so if Slavag's formula is wrong, do you have the correct one?
I'm puzzled why we have to figure out Volts & Resistance as well, I must be missing something.
Waross, I understand what you are saying, I just thought that given you have these 2 different current readings there would be a formula for calculating the 3rd phase. Assuming balanced Voltage and linear PF (resistive load) is it "Rocket Science"?
BTW I don't see the symbol "^" on my calculator, is that the recognized shorthand for Cosign?
As I say math was never my strong point, I did try to find the formula on the web, in some old text books and by asking a couple of engineers first.
Sorry about all my dumb questions!
Regards
Roy

RE: Current in Unbalanced Load

Roy.
No problem, isn't dumb questions, you aren't EE.
see, I used standard formulas for balanced load
No something...
Allways in the 3ph systems sum of all 3phase = 0, allways!!
that means Ia+Ib+Ic ( of course vectorial)=0
in my example ( I use Jghrist explanation)
IA=6A@0°, IB=6A@-120°, IC=6A@120°. Jghrist show it for the delta connected resistors Rab,Rac, Rcb=60Ohm

Jghrist show to you same for the unbalanced load. For this
he disconnect in the formula one parallel resistor Rab now=120Ohm, if you don't remeber for parallel connection of resistors:
Rab=(Ra+Rb)/2
Hope now it's more clear for you.
Regards.
Slava



 

RE: Current in Unbalanced Load

Quote:

Jghrist, so if Slavag's formula is wrong, do you have the correct one?
I'm puzzled why we have to figure out Volts & Resistance as well, I must be missing something.
Waross, I understand what you are saying, I just thought that given you have these 2 different current readings there would be a formula for calculating the 3rd phase. Assuming balanced Voltage and linear PF (resistive load) is it "Rocket Science"?
BTW I don't see the symbol "^" on my calculator, is that the recognized shorthand for Cosign?
Maybe there is a formula, but I haven't had time to work it out.  It isn't rocket science, just math.  As Slavag says, the basic formula is Ia+Ib+Ic=0.  The problem is that you only measure the magnitude of current with an ammeter, not the angle, so you can't do the vector math directly with the measured currents.  You might be able to use the fact of resistive loads to work out a formula.

Slavag was using the symbol ^ to mean "at an angle of".  I used the symbol @ for the same thing.  Often you will find the symbol ^ used for "to the power of" because that's how you would do it in an Excel formula.
 

RE: Current in Unbalanced Load

If you were using a watt-hour-meter connection, you would measure the amps and volts on "A-B" and "C-A", calculate the watts and just add the two readings together.
Any load on "B-C" will register on both "A-B" and "C-A". Because of the phase angle differences, 50% of "B-C" will register on "A-B" and 50% will register on "C-A".

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

RE: Current in Unbalanced Load

If you have the 3 phases A,B,C and you can measure the real power with 2 Wattmeters, measuring the currents on only 2 phases, as follows:

Install W-meter 1 with currents terminals to measure IA (current in phase A) and with voltage terminals to measure voltage UAB (line-line voltage between phases A and B). Measuring result: P1

Install W-meter 2 with currents terminals to measure IC (current in phase C) and with voltage terminals to measure voltage UBC (line-line voltage between phases B and C). Measuring result: P2.

Total power consumed by heater (in all 3 phases) P = P1 + P2 and P = sqrt(3)*U*I
Further : I=P/(sqrt(3)*U)= (IA+IB+IC)/3 since pf=1.

where U is the line-line voltage (system assumed symmetric), so any of the 3 voltages between either AB,BC or CA.
I = average current of the 3 phases IA,IB,IC (where IA,IC measured)(see above right side of equation)
P = determined from measurements

Finally: IB = sqrt(3)*P/U - IA - IC.
which is the current of the other phase, not measured.

* is the multiplication symbol
/ is the division symbol

 

RE: Current in Unbalanced Load

Again guys.
Please see OP.
according to two current transducers :
1. If it's balanced/symmetrical load, Ia=Ib=Ic.
2. If it's unbalanced load UNPOSSIBLE say what is value of Ib. Must some phase angle meas in the both phases.
Regards.
Slava

RE: Current in Unbalanced Load

(OP)
Thankyou everyone for your input, as Slavag says I was to have just two CTs.
I thought it would be easy to calculate the C phase based on the difference between A & B, however since it doesn't seem possible I will now allow for three CTs.
Roy

RE: Current in Unbalanced Load

As a practical solution to your problem, you could probably use off the shelf instrumentation. Use a watts transducer to convert the wattage to a 4-20 ma signal. Then use a 4-20 ma alarm module to alarm on a drop in wattage.
If you have a lot of individual heater elements, you may increase the sensitivity by using an elevated zero on the watts transducer.
This may sound complicated to anyone who is not familiar with industrial instrumentation, but it will be an apprentice level problem for an instrument mechanic.

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

RE: Current in Unbalanced Load

(OP)
Waross, I agree a Wattmeter would do what you say however three amperage readings will do so also. Actually I will use 3 x 2 wire current transmitters @ about $80 each. The voltage is essentially constant so we could calculate the Wattage however we are really just looking for element failure on a couple of large elements in a fluidized bed.
The application is an oil upgrading pilot plant. The process is operating at 500 C right on the ragged edge for electric elements.
Fortunately I have been a dual trades Inst/Elect journeyman for many years so I do understand.
Regards
Roy

RE: Current in Unbalanced Load

If all you are looking to do is detect an open heating element - out of three - your two CT arrangement will work.  When an element opens one (or both) of the CTs will see less current.  Current above a threshold is "good" while current below the threshold is "bad".  Depending on which (or both) CT current drops you will know which element went open.

RE: Current in Unbalanced Load

David is correct.  If the only load is the normally non-varying heater load, you only need to know if there is a change.  See Waross' 9/25 post for a little more detail.  The extra CT will make it easier to troubleshoot, however and would be worth the $80.
 

RE: Current in Unbalanced Load

(OP)
Davidbeach,
Yes you're right I guess I got hung up on the question of measuring 3rd phase without a transducer for a few reasons.
1/ My control system is Opto22 with two inputs per module.
2/ I have often seen the 2 Wattmeter circuit for measuring 3 phase Watts so it seemed reasonable that 2 CTs would be able to measure 3 phase current.
3/ At some point finding the formula became the challenge
As I said several times math is not my strong point so I posted the question thinking it would be a simple task for someone with an engineers skills.
I got to thinking about your "Kirchhoff's current law" and although it seems to be applied to DC if you think about a 3 phase circuit at any instant in time with the phases at 0, 120 & 240 degrees it should still apply.
Jghrist,
"Maybe there is a formula, but I haven't had time to work it out"  I searched high and low but I couldn't find one, several posters hinted at one but someone else would find a flaw.

If I think about a circuit with CTs on A & B, if both currents are equal you don't know if the load is between A-B or A-C & C-B or some combination therof
but if the A & B aren't equal then you know there's current in C but how much, not A-B obviously because of the phase angle.

It seemed like a simple question at the time, I'm sorry I wasted everyones time.
As I said I will use 3 transducers but I won't stop wondering!
Kind regards
Roy

RE: Current in Unbalanced Load

The formula is just a+b.
Each watt meter in the two watt meter method measures not only the power of the phase that it is connected to but 50% of the power in the third phase.
If you use two watt meters to measure the power draw of three 10 kW heaters in delta, each meter will indicate 15 kW.
If you are measuring a single phase circuit with a watt meter the voltage is in phase with the real current. When you add a second load in delta, the phase angle of the current changes. The current is the vector sum of both currents. If you extract the horizontal component of the current vector, you will find that it equals the sum of the in phase current plus 50% of the second phase current.
 

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

RE: Current in Unbalanced Load

Kirchhoff's Current Law applies to AC just as well as it applies to DC.  When working with AC it is imperative that you use the AC current, not just the magnitude of the AC current; the phase angle matters.

RE: Current in Unbalanced Load

I agree David and thank you.
My post was not clear. What I meant by a+b was the indicated wattage. Each watt meter will be measuring its share of the total load already due to the phase angles of the current as seen by the watt meters.
I should have said that the total watts is the sum of the indications of each of the two watt meters.
My bad.

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

RE: Current in Unbalanced Load

Suggestion:
OP mentions "current transducers." Not sure what the nature of the transducer is, but this method should work for CTs that are well matched, using single phase ammeters. Connect A phase CT polarity to the A phase meter polarity terminal. Likewise with B phase. Common the two non-polarity meter terminals together and route to the C phase meter non-polarity terminal. C phase meter polarity terminal is connected to the two CTs non-polarity terminals. As Slavig said, C= -(A+B), and summing the currents by parallel connection will include the phase relationship. All three currents are being monitored with two transducers as required by OP. No watt-meters (voltage connections required), or further math needed.

RE: Current in Unbalanced Load

Another solution may be the broken delta normally use with PTs. Any unbalance in the load will show up as a voltage across the corner resistor.
3 CTs a resistor and a voltage relay.
 

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

RE: Current in Unbalanced Load

Stevenal is totally right.
Is standard solution for ptotection in case of two CT's.
Regards.
Slava

RE: Current in Unbalanced Load

Stevenal hit on the easy answer.

If you're still not sure on this, look up a GE Multilin 369 manual in the back it should be section 7.6.7 (2-phase CT Configuration) for an example of this wiring configuration. Hey, it was the first manual I had to look up.

This likely won't work though with 4-20mA current sensors. They typically convert the measured current to an RMS output (which doesn't have the phase info any more). You'd have to install plain old CT's on the leads and then install 3 x 4-20mA current transducers connected to the CT's.
 

RE: Current in Unbalanced Load

Hi.
Please back to OP.
current transducers
next:
Opto22 with two inputs per module
it's 4-20mA for my pinion.
price is 80$,
If Roy have convential CT's, evry very simple power meter or 3-ph ampermeter are suitble.
Regards.
Slava

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