3 Phase Heater Control

P

Paul O'Keeffe

I have six heater elements each rated at 2.5kW, 110V and wish to connect these across a 3 phases, 380V supply. If I connect 2 of the heater elements in series per phase and connect the phases in Delta (Wye) formation what will my current and voltage per phase be ? Will the currents and voltages change if I connect the elements in star formation ?

As I come from an electronics backround and have little 3 phase experience all assistance would be appreciated.

Paul

J

Jeffrey W. Eggenberger

Do not connect anything rated for 110V across 380V or you WILL have a fire.

Get a Three-phase transformer to transform your 380V source down to 110V at the rating that you need. I would suggest that you get in contact with a knowledgeable electrician in your area and not count on your "electronics background".

Can you make it work? Probably. If I start a fire will the insurance company pay even though it was caused by my ignorance of local electrical codes?
Probably not. Most electricians are used not only because they are "electrically savvy", but also because they are trained to know, understand,
and implement local electrical codes.

Jeffrey W. Eggenberger
Electrician: Industrial, Commercial, and Residential

R

R A Peterson

I think you will end up frying the heaters if you connect them in delta configuration. If you connect them in delta as described above you will have 380V across each pair of elements, or 190V across each element. Instead of 2.5KW per heater, it will put out 7.5KW (remember P=V^2/R).

If you were to put them in wye (star) configuration, each phase to neutral in a 380V system is 220V, so each heater would see 110V. So in this configuration you'd get the original 2.5KW per heater.

T

Tony Firth

Paul,
Three Phase 101:
Where 3 phase supply is 380 VAC i.e. the voltage between each phase line to any other phase line is 380 VAC, (European system), the voltage between each of the phase lines and neutral is 380/sqrt3 = 380/1.732 = 220 VAC. This indicates that the 3 phase supply is coming from a "Wye", (star), connected supply transformer and the
neutral is coming from the star node of the transformer. This is referred to as 380 VAC 3 Phase, 4 wire. (+ Ground of course).
So you connect your heaters in series as you suggested and connect each pair to a Phase wire and to Neutral. i.e. you have connected your load in "Wye" fashion. Do NOT connect your load in Delta fashion! Why not? Some basic math: Power = V2/R i.e. for twice the voltage we get 2 squared times the power so that your 2 heaters in series that should have 110 x 2 = 220 volts across them will now have 380 volts and the power will be (2.5kW + 2.5kW) x 380/220 x 380/220 = 14.9kW which, if it doesn't fry your heaters is way more heat than your design calls for. Cross check: Power = V2/R, therefore R = (110 x 110)/2500 = 4.84 ohms. 2 heaters in series = 9.68 ohms. Power when delta connected with 380 volts = (380 x 380)/9.68 = 14.9 kW as before.
But Wye connected, each heater pair gives (220 x 220)/9.68 = 5 kW which is what your design calls for.
The current in each heater = V/R = 220/9.68 = 22.7 A.
Hope this helps,
Tony Firth,EE,
Quester Technology Inc.,Fremont,CA.

C

Crystal Majercik

Hello Paul,

After working for Emerson Electric's Chromalox division, and producing their 760 page catalog of industrial electric resistance heaters and related heating products, you may want to visit this catalog on-line at www.chromaloxheating.com.

Also, you may post this question to the webmaster who is also the companies trainer.

Best of luck to you,

Crystal Majercik
Integrated Industrial Technologies

PhilCorso

Responding to Paul O'keeffe's query:

First identify the currents and voltages:

Subscripts y, d, s, and r, denote value in Wye, Delta, Supply, and Heater element (resistive), respectively. Furthermore the line-to-midpoint voltage is Vn = Vs/SQRT(3), say 220 v, for
simplicity. Then,

For one heating element, Pr = 2,500 W, and , and Vr = 110 V, so that,
· Ir = Pr/Vr = 2,500/110 = 22.73 A.
· Rr = (Vr)^2/Pr = 4.84 Ohms.

Wye connection relationships,
· Iy = Vn/(2xRr) = [Vs/SQRT(3)]/(2xRr) = 22.73 A.
· Py = Vn x Iy = 5,000 W.
* Is = Iy.
· Ps = SQRT(3)xVs x Is = 15,000 W.

For the Delta Connection
· Id = Vs / (2xRr) = 39.26 A.
· Pd = Vs x Id = approx 15,000 W.
* Is = SQRT(3) x Id = 68 A.
· Ps = SQRT(3) x Vs x Is = approx 45,000 W.

In conclusion, it is quite obvious that the Delta configuration will greatly exceed the heater element ratings.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

PhilCorso

As a result of the recent to and fro discourse on whether a 'slug' is added to the motor's MoI or the system's, I must respond to the 3-Phase Heater thread... the intent being to make it more Technically Correct. Call it more of that "PEing contest mentality" I addressed many months ago.

When a 3-phase load is connected in a wye (or star) arrangement, the common point is called the "midpoint." It is called the "neutral" if, and only if, it is connected to the neutral of a 3-phase, 4-wire supply system.

If anyone is interested in why 3-phase is the most commonly used multi-phase system, and not 2-phase, 4-phase, 5-phase, etc, i'll be happy to elaborate.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

R

Robert R. Stephens Pennzoil Products

At 110 volts and 2.5 kW the heaters are 4.89 ohms each. If you connect two heaters in series and connect each series pair in a wye connection V line to neutral is 219 volts. The current per leg will be 22.6 amps and the watts per wye leg will be 4.95 kW. Each individual element will dissipate about 2.5 kW and the voltage drop across each heater element will be 110 volts.
Total power dissipation (or heat generated depending on your view) with wye configuration is 14.85 kW.

V

Vitor Finkel

Paul,

If you "line" voltage is 380 (between phases) its monophase voltage ( 1 phase to neutral ) should be 220V, so 2 X 110V elements in series may be properly connected to each phase to neutral.
That much is correct.

If you connect the resistances from phase to neutral, you'd beter conect your transformer in star ( providing it can still provide you with 380 volts beween phases ) and the star center to
the center of the three combined resistors, so that any variation in resistances values, or broken ( interrupted ) resistor will not unbalance seriously the aplied voltage to remaining resistors.

>As I come from an electronics backround and have little 3 phase
>experience all assistance would be appreciated.

Do learn a little bit about it. Take the practical approach.
Make yourself a phase diahram. A Star - 3 vectors originating in the center, 120 degrees apart. Rewire that to delta, moving each vector to start were the other ends, but keep each phase in the
same direction as you had previously. ( slide them, but don't turn them ) Do that in scale, and you may easily undertand how the whole 3 phase system works, regarding delta-star connections.
Measure the lenght of your vectors to determine line voltages, ( 1 phase to ground ) or phase voltages ( between 2 phases ).
I personally like the graphic method, because I can visualize without too much effort what hapens when I rewire or unbalance, or otherwise alter the basic 3 phase configuration.
On a balanced three phase system Phase voltage = line voltage X .577 ( I think that constant .577 comes from the tangent ( sin/cos ) of the 120 degrees ( phase angle ) = (SQRT 3 )/3, but I really don't recall all the trigonometric details. Anyway if you have 380 V between phases, 380 X .577 = 220 (line voltage) Remember, for all practical purposes the 3 vectors are not static, but rotate at 60 revolutions per second on a 60 Hz power line.
The angle btween the vectors remains the same, while they rotate, so they are synchronized with respect to each other.
So your static drawn vectors maintain the correct phase relation and voltage amplitudes among them.

Let me know if you tried to play around with the vetor's drawings, and what did you find. Remeber that reversing the terminals of one transformer or motor winding is equivalent to turn the vector
180 degrees, so it is easy to provide vectors 60 degrees apart, or built an hexafasic system based on a three phase transformer with a central tap on each secondary, etc.

Presently I am also an Electronician, but on a previous life I was graduated as a technician on Eletrothecnics, so I can live a bit on both universes, and talk to specialists in both fields.

In fact I enjoy it.

Let me know if I can be of any further assistance with your problem.

Vitor Finkel [email protected]
P.O. Box 16061 tel (+55) 21 285-5641
22222.970 Rio de Janeiro Brazil fax (+55) 21 205-3339

D

Dobrowolski, Jacek

Hi
If you connect them in the star connection the current will be equal to the nominal current of the heaters:
2.5 kW / 110 V = 22.7 A
In the delta connection of them current increases by factor 1.73 (square root of 3).
22.7 A * 1.73 = 39.3 A
But be careful power dissipated from each heater in this connection increases by factor 3 !

Regards

Jacek Dobrowolski

Software Eng.
Electrical and Software Design Department
Secondary Division
International Tobacco Machinery Ltd. Poland

B

Bob Welker

I wouldn't do a delta, because element crapout would be the result. However, it looks like it could be done wye. Rely on Ohm's law when starting heater calculations, and be aware of the
differences between delta and wye. I suggest going to a local electrical shop, pick up a copy of the "Ugly Book", and study it thoroughly.

The elements are 2500W, 110V rated, so the resistance of each element will be about 4.84 ohms (using Ohm's law).

There are six equally rated heaters, so we can make 3 series pairs. Since they are equally rated half of the phase voltage will drop across each
heater in the pair (if the heaters were different wattages, and therefore different resistances, this would not be the case).

In a delta configuration the phase voltage equals line voltage, so half of the supply (190 volts - about 172% of rated voltage) would be present across each element, would draw about 39 amps, and produce about 7500 watts (for a while, anyway). Clearly this isn't a good situation, and the heaters would melt out in a few minutes.

In a wye configuration phase voltage (that is, between the line terminal, and center point) is line voltage divided by the square root of three, so 380/1.73 = about 219V phase voltage (most electricians are more familiar with 480 V line, and 277 V phase voltages in a wye system). Half of this 219V will be dropped across each heater, and just happens to be about 110V. This circuit should work.

H

Hunter Farris

380 Phase to Phase would be 220 Phase to Ground. Connecting two heaters in series and powered at the 220 Phase to Ground would be sufficient with an ~22 Amps per pair of heaters. I do not see any problem with this.

Hunter Farris

F

Ficke, Wayne

Paul,
The wye configuration should not be a problem!

2.5Kw/110vac=22.73 amps
2.5Kw/(22.73amps)^2=4.84 ohms
4.84 ohms/htr * 2 htrs=9.68 ohms total/phase

Voltage/phase(wye)=Vline/(2sin(pi/n))
where sin(pin/n) is in radians and n=the number of phases.
=380vac(2sin(pi/3))=219.39 vac/phase

Iphase=Vphase/Rphase (R is used since this is a resistive load)
=219.39Vac/9.68 ohms=22.66 amps/phase

Iphase=Iline(wye)=22.66 amps

Total KW= n * Vp * Ip = 3 * 22.66 * 219.39 = 14.9 kw vs. 15 kw to 6 x 2.5kw
htrs @ 110v

Wayne Ficke II
Research Eng.

D

DWESTONE

CALL AN ELECTRICIAN
380 VOLTS WILL KILL YOU.