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I am an electrical technician working in a hospital. In the laundry room, all the cloth washers contain three-phase motors which are balanced loads connected to a 140 A circuit breaker. The consumption does not exceed 30 A in each phase. The hospital bought a new single-phase electric cloth dryer (single phase load 1PH+N, rating current 100 A, voltage 240 V). By adding this single-phase load to the circuit, obviously the circuit will become unbalanced.

Any suggestions to connect this dryer to the circuit without any problems? How can we solve the unbalance problem, knowing that no other loads will be connected to the circuit besides the washers and the dryer?

ocrdu
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user137684
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  • @Uwe are you sure? Intuitively this seems too good to be true. Thinking it through, if you connect the secondaries in series then the vector sim of the three would be zero, although if you reverse one winding then the phase offset goes from 120 to 60 degrees, so I can see that working but the unbalance would be reduced rather than eliminated. Am I understanding that correctly? – Frog Dec 21 '22 at 19:27
  • As it is a "big" load, no "picture" of the specs ... – Antonio51 Dec 21 '22 at 19:38
  • @Frog Imagine a equilateral triangle. With two parallel shifts you get the series connection of the three voltages. With the proper polarity you get a vector sum of 240 V if each voltage of the triangle is 120 V. – Uwe Dec 21 '22 at 20:49
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    (A 24 kVA appliance has been purchased before knowing how to connect it? Or whether indeed it is single phase: I find that hard to imagine.) – greybeard Dec 21 '22 at 20:56
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    A well designed dryer should have three heaters with equal resistance. For a three phase grid the heaters should be connected in a triangle, for a single phase grid all three heaters in parallel. Simple and flexible. – Uwe Dec 21 '22 at 21:31
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    (Wait - a kWatts heat pump would be 3-phase, does single phase imply *resistive heating*? How out-of-date.) – greybeard Dec 22 '22 at 08:48
  • what are the consequences of adding this load directly to the three phase circuit, What problems are most likely to occur? – user137684 Dec 22 '22 at 15:32
  • (You still don't *explicitly* state line voltages, just `single phase load […] voltage 240 V`. Directly connecting this load, is that 240 V split phase, 208 V line voltage connected to two phases (beware PF), or 240 V line voltage?) *Electrically*, I'd expect nothing drastic *assuming* the latter and the wiring matching the circuit breaker. One of the most unwanted consequences of an unbalanced load is (a contribution to) a *voltage unbalance* and under-voltage. Loads reacting ungraciously to an unbalanced supply include induction motors: available torque goes down, resistive losses go up. – greybeard Dec 24 '22 at 16:34
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    @greybeard No each winding should contribute 1/3 of power. We want a symmetric balanced load.We need 240 V and 100 A, the load is a pure resistor, so voltage and current should be in phase. The current of 100 A flowing through the load is also flowing through each winding. Total power is 24 kW and each winding delivers 8 kW. In one winding current and voltage will be in phase, so we need for 8 kW and 100 A 80 V. For the two other phases the shift between voltage and current should be 60 °, cos(60) is 0.5, we need 160 V to get 8 kW with 100 A. – Uwe Dec 25 '22 at 13:15
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    The series connection of the three windings will deliver 100 A and 240 V. 80 V from from the first phase and 160 V from the series connection of the two other phases with the 60° shift between voltage and current. Unfortunately the real power load will be symmetric to the three phases, but there is a lot of reactive power on two phases. – Uwe Dec 25 '22 at 13:23
  • @greybeard No, it is not your bad, it is my bad. I wrote Each secondary winding is rated for 100 A and 120 V, but this was wrong. – Uwe Dec 26 '22 at 07:14
  • Three phase network line to line voltage 415V, phase to neutral 240V the load connected to phase and neutral. – user137684 Dec 26 '22 at 17:54
  • `Three phase network line to line voltage 415V, phase to neutral 240V` Don't comment comments asking for complementary information or clarification: Edit your post. This is not chat. – greybeard Dec 28 '22 at 05:57

1 Answers1

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Connecting 100 A load in a 140 A 3 phase circuit could overload the neutral wire, in case the cross section of this wire is half section of active wires.

To minimize unbalance in 3 phases caused by a high single phase load, a 2 windings transformer can be used, vector group V-V inverted. See diagram.

enter image description here

Another but worse option is to use a one phase transformer, more easy to find, primary 415 V - sec. 240 V. Primary current will be reduce 1.73 ratio (58 A) and loaded in 2 phases.

Rated power must be 25 kVA or higher, also new breaker is necessary, so no cheap solutions.

Bravale
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  • Shouldn't it be enough to have a single single phase 8 kVA 1:1 transformer to have one phase earth-free (dang - no field of any expertise of mine, and I don't know the English technical terms) and connect that in proper phase to one of the other live potentials? – greybeard Dec 21 '22 at 21:03
  • @greybeard to get a symmetric balanced load three such transformers would be needed. If some asymmetry is possible, a single transformer may do. Especially if the single phase load is not constant. – Uwe Dec 21 '22 at 21:26
  • A bit of a learning curve: [single single phase transformer](https://tinyurl.com/2jy5pboy) – greybeard Dec 22 '22 at 00:25
  • @greybeard Excellent simulation! – Uwe Dec 22 '22 at 06:50
  • (Ah, wait. Current got to be severely (60°) out of phase with "the transformer phase" and one of the others, each supplying, of course, just one fourth of the power with the remaining one providing twice that or half the total. Meh.) – greybeard Dec 22 '22 at 11:39
  • @greybeard: as I understood, the load is the 416.7 mOhm resistor. New voltage generated in transformer is same as phase T,(but floating) and it is added to phase S. That means you get same voltage than phase R, but inverted. The load is connected between VR and new generated voltage, so the voltage applied is 2 times higher than phase voltage. Op stated phase voltage 240 V. – Bravale Dec 22 '22 at 12:31
  • @greybeard I am a little stuck with the 3 phases. I'm a little "surprised" about the representation of the 3 phases in this animation. The 3 phases seem a little "tight" to me. but hey, maybe I saw it wrong. https://i.stack.imgur.com/YTmnU.png – Antonio51 Dec 24 '22 at 18:04
  • Sorry. It is in your animation above ... The 3 phases do seem not "regular" to me. But it seems it is an "error" of "Falstad" (?) ... I just verified 0, 120 and -120 degrees are ok. But these 3 phases seem to be as this https://i.stack.imgur.com/N3il4.png at 60 degrees (?). – Antonio51 Dec 24 '22 at 20:00
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    (@Antonio51: Ah. The "crowded"/60° look comes from graphing R' in addition to R, S and T.) – greybeard Dec 24 '22 at 20:10
  • Op stated *load* voltage 240 V, added *line* voltage 6 days later [in a comment](https://electronics.stackexchange.com/questions/647219/solving-the-problem-of-connecting-a-high-current-single-phase-load-to-a-three-p#comment1719066_647219). – greybeard Dec 28 '22 at 06:00