Calculating loads on 120/208V system

I'm a system engineer responsible for specifying power quantities for my systems. There are a lot of ways that are accepted in my industry to do this that I know to be incorrect, so I'd like to hear from someone with more specific expertise.

The systems are comprised of a combination of 120V resistive loads, and 208V resistive and inductive loads, although the devices that have inductive loads actually operate at various voltages, their power supplies accept 208VAC. Most of the power available is 208/120VAC

60hz, usually in 400A increments.

What a lot of people do is add up the load P, divide by operating voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at

208V, it would look something like

((5000/110)+(2500/203))/3 = 20A needed.

I can't imagine that's correct. What I have done based on advice and reading is divide the load P by 1.73, a relevant power factor and the supply voltage, so this:

(5000+2500)/(1.73*0.9*203) = 24A needed

But I can imagine that being wrong, also. What is really the way to do this?

Thanks in advance.

Reply to
MammalianFish
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As an electrician, I thought we always added all the loads in watts to find the total watts. Then we used the P=pf*1.73*I*E for 3-phase. However, for single phase the 1.73 is removed. But for 208/120 volts this sounds like a three phase system. I=P/(pf*1.73*E) where I is the curent in each phase.

Which looks like what you did.

I have written some JavaScript programs on this at:

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Reply to
Gerald Newton

1) Add the power: 5000+2500 = 7500 watts. 2) Divide by 3 (legs): 7500/3=2500 3) Divide by phase-neutral voltage: 2500/120 = 20.8A 4) Divide by power factor, unless power in 1) is VA not watts.

Round up according to NEC/local rules and future expansion.

This assumes the loads are balanced, but in your case they are not, since neither 10 nor 5 are evenly divisible by 3. Of the 10 500 watt 120V loads, you'll have to load the phases 4-3-3 for example.

Reply to
Michael Moroney

Just out of curiosity, why doesn't the operating voltage matter? It would seem that 208V loads would ultimately draw less amperage than

120V loads. No?

Thanks, Moo

Reply to
MammalianFish

120*sqrt(3)=208 208 three-phase is three 120V legs. He's accounted for that in 2) and 3).
Reply to
krw

208 divided by 1.732 = 120 volts

So I= 2500/120 equals I =2500/(208/1.732) or I=2500*1.732/208 I = 20.8 amperes

or I = 7500/(208*1.732) I = 20.8 amperes

Then 7500/(208*1.732) is identical to 2500*1.732/208 Why? Multiply both sides by 1.732 and you get 7500/208 ==7500/208 since

1.732*1.732 = 3. They are identities. So why all the multiplying by three, just use the standard three phase equation. Of course this assumes a balanced load, for unbalanced loads the neutral will carry the unbalance.
Reply to
Gerald Newton

That makes a lot of sense, thank you.

So when the system is out of phase and there is voltage in the neutral, is the copper in the neutral system electrified in the same way as the phases, or is it different somehow? Meaning, would I measure 120V from the Neutral to Ground?

Reply to
MammalianFish

"electrified" where did you come up with this? Somehow, I think you must be pulling a leg, so to speak. Of course you measure 120 volts to the neutral. 208 volts from phase to phase and 120 volts from each phase to the neutral. The current changes with the load, not the voltage. This assumes what we electricians work with - steady state conditions and not an engineering analysis that would require Thevenin's Theorem. It also does not include voltage drop.

Reply to
Gerald Newton

Really? What state/province did you get your licence in? An odd sort of question to ask on a news group.

Bill

Reply to
Bill Shymanski

The neutral is normally connected to ground at some point, usually your local sub-station. You should NOT be able to measure 120V between neutral and ground unless you have a dangerous fault.

Depending on the resistance of the neutral cable between your measurement point and the connection to ground, and the current flowing through it, it will develop a potential to earth but this should not normally be more than a few hundreds of millivolts.

Reply to
Stuart

The supply was not specified, but for 208 volts and 120 volts on the same system we encounter this in the field about 99 percent of the time on a three phase 4-wire 208/120 volt solidly grounded wye system where the neutral is connected at XO at the transformer and is also connected through a grounding electrode conductor to earth. This is a very common system that any licensed journeyman finds familar. We then measure 120 volts from each of the three phases to the neutral and 208 volts from phase to phase. This is a very common configuration and I cannot undrstand why there is any confusion with this.

Reply to
Gerald Newton

No, neutral is the "grounded conductor". It will remain close (IR drop, where I is the imbalance in the legs) to ground. The neutral current is the vector sum of the current in the three phases. In a balanced system it is zero so no current flows in the neutral, thus zero voltage across the neutral conductor.

Reply to
krw

Indeed

Reply to
Stuart

Sorry to rebut your claim, but I just returned from the IAEI NW Section meeting where the New Definition of neutral in the 2008 NEC was discussed. The NEC correlating committee assigned a specific task group to examine the necessity of such a definition in the Code. A definition of a neutral conductor was developed as a result of work by the NEC TCC. The new definition given in Article 100 is: "Neutral conductor is defined as the conductor connected to the neutral point of a system that is intended to carry current under normal conditions." The next point made on page 32 of the Analysis of Change Book states, "Grounded conductors are not always neutral conductors." The illustration shows a grounding electrode conductor as an example.

The definition of Neutral Point has been revised and is now defined as the common point on a wye-connection in a polyphase system, or midpoint on a single-phase, 3-wire system, or mid-point of a single- phase portion of a 3-phase delta system, or a midpoint of a 3-wire direct current system.

A grounded conductor is still defined as a system or circuit conductor that is intentionally grounded.

So a neutral may be a grounded conductor but a grounded conductor is not always a neutral but is always grounded. Nowhere in the definition of a neutral conductor or neutral point does it say the neutral or neutral point must be grounded.

Now this certainly means more confusion in the field. But it looks as if a ungrounded system can have both a neutral point and a neutral conductor that are not grounded, and a separate grounded conductor that is not a neutral.

Reply to
Gerald Newton

It's not a claim. You, and the NEC, are welcome to redefine terms [*], then show how brilliant you are. The electrons don't know any differently.

Ok, so why would you not ground the neutral point in a Wye? ...sounds dangerous.

Again...

Isn't that their intention? ;-)/2

Again, YOY would you float the neutral? Sounds dangerous.

Reply to
krw

How about the high impedance grounded system. We have a neutral that is not grounded in a system that is commonly used and considered safe.

Reply to
Gerald Newton

How about it? How does it work? Why don't you let us in on the theory behind it, making it safe? Why? What keeps neutral at a reasonable voltage, say, if one of the legs drops? Call me skeptical.

Reply to
krw

On Sep 23, 7:30 am, krw wrote:

The IEEE Green book has ample diagrams of high impedance grounding in the last part of the book. The theory is simple; we install impedance in the ground return path between XO and ground. For 3-phase 480/277 volt wye systems this is usually a 55 ohm resistor but can be an inductor. The resistor allows charging current to return, but limits the ground fault current to about 5 amperes (277/55). A ct is place on the ungrounded neutral that runs from XO to one end of the resistor. The ct sends a signal to a relay set to energize at 3 to 5 amperes or at some value just above the charging current which is roughly 1 ampere for each system 1000 KVA . The relay sends a signal to the control room. The first ground fault does not take the system down but allows maintenance persons time to troubleshoot and repair. However, I have seen such ground faults allow to exist for months at the Valdez Marine Terminal on the Trans Alaska pipeline. While the ground fault existed the other phase conductors measure 480 volts to ground instead of 277 volts. In Valdez we experienced a second ground fault in this system on a separate phase which then gave us a phase to phase fault. The

400 ampere breaker did not trip because the second ground fault was some distance from the first ground fault and this terminal has one of the greatest engineering blunders of the century. Fluor engineers permitted almost all 480 volt, 5 kv, and 13.8 kv cables to be installed in the 14 miles of cable tray in 1975 when the cables DID NOT HAVE EQUIPMENT GROUNDING CONDUCTORS INSTALLED IN THEM. Equipment grounding is done through an elaborate system of metal bonding and grounding electrode bonding. Believe me this is true. Almost 20 percent of the nation's oil has been going through this terminal for about 28 years and the electrical system has a major grounding anomaly. That second ground fault I am referring to took place near a crude tank in a Class 1, Division 1 hazardous location! It really shook some people up to see that phase to phase arc and not see a circuit breaker trip. The reasons the cables do not have equipment grounding conductors vary, but one is after the error was discovered and the cables were on site, if the error had been corrected Arco would have had to file for bankruptcy. The oil had to flowing in the pipeline by a certain date and that was more critical than reordering the correct cables. Please excuse me for being off topic here but then these threads go all over the place. The bottom line is there are systems where the neutral is not grounded and where there is a grounded conductor that is supposed to be safe. These are primarily found in industrial locations where a ground fault detection system has been installed.
Reply to
Gerald Newton

floating neutrals?

perhaps xformer isolated and within a separated enclosure to prevent faulted grounding? this seems inherently dangerous a definition.

ground is generally a physical hard connection somewhere to "earth/terra firma" the great return loopback to the nuclear plant.

neutral is, i thought it was anyway, a localised pathway among the power circuits at the devices/installation themselves, rather than a "faux ground"

where standards diverge we open many problematic situations for dangerous installs and site specific alterations. this requires again specialized maintenance and education of technicians and subsequent add-ons later.

basic standard and un-ambiguous design eliminates potentially dangerous confusion and death.

Reply to
HapticZ

A high impedance grounded neutral is not an ungrounded or "floating" neutral. It is a neutral purposely grounded through a resistor or inductor. This way the charging current (or leakage current) can return, but ground fault current is limited to several amperes. This is primarily used for a wye wound secondary. For a 480/277 volt 3- phase system this does not permit the 277 volts to be used for power or lighting. If 277 volts is required for power a separate isolation

480/480/277 volt delta/wye transformer is used with a solidly grounded neutral. For ungrounded delta systems a zig zag transformer is used to allow charging current to return and to prevent high voltages on the system in case of an arcing ground fault in the 480 volt system. These are common configurations used in the oil industry and mine mills in Alaska. Traditionally, I used to think of the neutral as the grounded conductor that carried the unbalanced current in a 3-phase 4-wire system. However, the new definition broadens the definition to where we now can call the grounded white conductor in a 240/120 volt single phase residential wiring system the neutral conductor.
Reply to
Gerald Newton

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