Power Calculation Unbalanced 3 Phase

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StephenSDH

Senior Member
Location
Allentown, PA
I support a system that heats molds using 32 zones of electrical heat. Each zone is contains several single phase heater strips on a 3 phase circuit. They have had many issues with the heaters going out and not knowing until they make bad product. I had the customer add current CTs to each phase with the signal going to a PLC. I am going to energize each zone one at a time and calculate the KW to detect a heater loss.

I thought the calculation would be easy until I got into it. Heater zones are 600V 3 Phase and fed from a delta-delta transformer. Zones may be single phase, balanced, or something in between.

If it is a single phase load or a balanced 3 phase load I can calculate the KW, but the in-between I am having a hard time coming up with the formula. If anyone can point me in the right direction I'd appreciate it.

Thanks, Steve
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141122-1307 EST

Is the resistive load wired as a delta or wye?

32 resistors are not evenly divisible by 3. Thus, inherently unbalanced.

Is there any phase shift or on-off cycling of the current to the heaters for power input control?
If there is, then is this done in bulk, or individually by each three phase bank?

Are there any temperature sensors in the mold?

.
 

StephenSDH

Senior Member
Location
Allentown, PA
I'm looking for the formula to calculate power of an unbalanced 3phase delta load assuming the voltage phase to phase is 600V and measured phase currents. Looks like there is some vector calculations I need to do, but I am unable to come up with the formula.

Gar's Questions
Delta secondary and the heaters are wired phase to phase. There are around a 100 resistors and 32 zones. Around 2-4 heater strips per zone. They have many different molds depending on the product they are running. It heats the zones based on a thermocouple and pulse width modulation with a 30A contactor.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141122-1358 EST

What does pulse width modulation mean? Is this phase control of SCRs, or is it "bang-bang" control of the contactor?

If you are dealing with sine wave current, then vector math applies, and you can use an average measuring method of current adjusted to be an RMS value without using an RMS converter. For a chopped sine wave vector math does not work.

You probably don't need complex math to determine whether a resistor has failed or not. If all phases are modulated the same and not independently, then the relative relationship of each phase current to the other phases for a particular mold should allow determination of a resistor failure.

More information is needed.

.
 

StephenSDH

Senior Member
Location
Allentown, PA
PWM - PLC energizes the contactor for a period of time based on the thermocouple "BangBang".

It is an unbalanced 3 phase load with PF of 1. I don't want to average the current, because I want to detect if a heater failed or if they replaced the heater with the wrong wattage. If I average the current the calculation could be 15% off.

3 Phase, Delta 600VAC Sine Wave.

Just looking for a unbalanced 3 phase power calculation based on measured current, if anyone can point me in the right direction I would appreciate it.

Thanks, Steve
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
If you know each line current and the load is resistive simply multiply each line current by the line to neutral voltage (even if it is a delta), then add the three values to get total wattage.
You can do this because any way of producing the same load current must calculate out to the same wattage.
This is only valid for unity power factor.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141122-1554 EST

You misunderstood my comment on average measurement of an AC sinewave vs an RMS measurement of that same curve.

An average measurement of a half rectified full cycle sine wave is 0.318 times the sine wave peak. For full wave rectification the value is 0.636 of the peak. The RMS value integrated over one full cycle is 0.707 of the peak. The scaling ratio between the two is 1.112 .

A Simpson 260 meter full wave rectifies the AC signal, then averages this over about 0.1 second by the inertia of the meter movement. Then the meter face has its calibration adjusted by the 1.112 factor so that on a sine wave the meter reads the RMS value of a sine wave. Put a square wave into the meter and the reading is not the RMS value. A Fluke 27 also full wave rectifies the signal. A Fluke 87 has a chip or process to perform the RMS measurement.

I think your desire to do vector math on your three currents is the hard way to solve your problem.

Suppose your three voltages remain at exactly 600 V, which I believe is a bad assumption. But not a real problem if voltage is measured. Next assume only one resistor fails, then the current on only one supply phase remains constant. The other two phase currents will diminsh. Thus, you know where the resistor failed. It is in the delta branch opposite the supply phase that remained constant.

.
 

StephenSDH

Senior Member
Location
Allentown, PA
Here is a drawing and a sketch. The drawing doesn't show all the zones and simplifies the heaters because they are not connected balanced like it shows.
 

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winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
Unfortunately, the power factor of an _unbalanced_ three phase resistive load is not 1. This is true even though the power factor of the individual resistors that make up the load is 1.

The approach that I would take is to make a spreadsheet that tabulates the current through each resistor, identifies the phase angle, and then does the vector math to get the total current on each phase.

I wouldn't bother to try to figure out power factor and total power; I would just look to see if the current levels match the expected current levels.

-Jon
 

Phil Corso

Senior Member
Stephen, thank you for the sketch!

I take it then you are looking for the ability to calculate current-draw for balanced and unbalanced load combinations. If so, will the number of elements per phase be the variable? For example four, 4-kW, elements supplied via phases A-B, three via B-C, and via on C-A, or something like this?

Regards, Phil
 

Smart $

Esteemed Member
Location
Ohio
Unfortunately, the power factor of an _unbalanced_ three phase resistive load is not 1. This is true even though the power factor of the individual resistors that make up the load is 1.

The approach that I would take is to make a spreadsheet that tabulates the current through each resistor, identifies the phase angle, and then does the vector math to get the total current on each phase.

I wouldn't bother to try to figure out power factor and total power; I would just look to see if the current levels match the expected current levels.

-Jon
I agree. Rather than having the PLC calculate power, just work with the values the PLC receives. Less complexity, less potential for error.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
141122-1823 EST

The approximate currents are 10.1, 5.77, and 7.22 for your sketch 2. Your other currents are OK.

The calculations are:

3 kW 3000/600 = 5 A
4 kW 4000/600 = 6.6667 A
1 kW 1000/600 = 1.6667 A

5*0.866 = 4.330 A
6.6667 * 0.866 = 5.773 A
1.6667 * 0.866 = 1.443 A

Phase line A --- 4.330 + 5.773 = 10.1 A
Phase line B --- 5.773 + 1.443 = 7.22 A
Phase line C --- 4.330 + 1.443 = 5.77 A

As a check

600/1.732 = 346.4 V

346.4*(10.1 + 7.22 + 5.77) = 346.4*23.09 = 7998.4 W

.
 
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Smart $

Esteemed Member
Location
Ohio
141122-1823 EST

The approximate currents are 10.1, 5.77, and 7.22 for your sketch 2. Your other currents are OK.

The calculations are:

3 kW 3000/600 = 5 A
4 kW 4000/600 = 6.6667 A
1 kW 1000/600 = 1.6667 A

5*0.866 = 4.330 A
6.6667 * 0.866 = 5.773 A
1.6667 * 0.866 = 1.443 A

Phase line A --- 4.330 + 5.773 = 10.1 A
Phase line B --- 5.773 + 1.443 = 7.22 A
Phase line C --- 4.330 + 1.443 = 5.77 A

As a check

600/1.732 = 346.4 V

346.4*(10.1 + 7.22 + 5.77) = 346.4*23.09 = 7998.4 W

.
As you said, approximate. Multiplying by 1/2 the square root of 3 is an assumption that both phase currents are equal in magnitude.

Doing the vector math, the results are:

IA = 7.64A @ -130.89?
IB = 10.14A @ 85.28?
IC = 6.01A @ -46.10?

Where C-A is reference 0?. I also took the top line to be A, then B and C below, respectively.
 

Smart $

Esteemed Member
Location
Ohio
Winnie,

Why is the power-factor of an unbalanced array of resistive elements not unity?

Regards, Phil Corso
Not Jon, but...

Because the unequal magnitudes vector add to an angle which is not the bisecting angle (which is what occurs when adding two 60? difference phase vectors of equal magnitude)

Vector: a@-30? + b@+30? = 1.732*(a or b)@0?... only when a = b.
 
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GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
Winnie,

Why is the power-factor of an unbalanced array of resistive elements not unity?

Regards, Phil Corso
Think of a single line to line resistive load.
The voltage across the load is not in phase with either line to neutral voltage.
Therefore resistive current flowing through that load will not be in phase with either line to neutral voltage either.
There is no way for the current to change phase as it turns the corner in the circuit diagram.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
As noted in the posts that follow, the equations posted by gar in post #14 are approximations which assume a balanced load.

Smart$ beat me to doing the vector calculations. I have a spreadsheet that does the vector addition, so I decided to see how quickly the error builds.

I think, for the original poster's purposes, that gar's calculations are probably accurate enough. They are correct for a balanced load.

If you take (for example) 30KW and divide it as 8KW, 10KW and 12KW, and calculate currents using gar's approximation then the largest error is 0.7%

If you divide it as 5KW, 10KW and 15KW, then the largest error is 4%

If you divide it as 0KW, 10KW and 20KW, then the largest error is 13.4%

-Jon
 

StephenSDH

Senior Member
Location
Allentown, PA
Thanks guys.

I need to work the formula backwards. Based on the numbers Smart$ calculated, the knowns are

IA = 7.64A, IB = 10.14A, IC = 6.01A (using current cts)
V = 600VAC (assumed)

and I need to find KW. I'm having a hard time doing the math.

There is error from the CTs, 4-20ma signal, assuming 600VAC, I hate to stack on 13% because of my inability to calculate power. They have engineered drawings that show the Wattage for each zone. It would be nice to get within 5%, so I can alarm if it is out of spec.

If someone has a spreadsheet that can calculate unbalanced power I'd appreciate it. I'm going to try and use excel to come up with a trend line approximation.

Thanks again,

Steve
 
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