Current in Unbalanced Load 2

[roydm]

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

 

[davidbeach]

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

 

[roydm]

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

 

[slavag]

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

 

[itsmoked]

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

 

[roydm]

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

http://www before

showing my ignornce.
Bear with me please
Roy

 

[slavag]

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

 

[jghrist]

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

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

 

[slavag]

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

 

[waross]

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

 

[roydm]

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

 

[slavag]

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

 

[jghrist]

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.

 

[waross]

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

 

[tomad]

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

 

[slavag]

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

 

[roydm]

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

 

[waross]

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

 

[roydm]

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

 

[davidbeach]

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.