Voltage Drop With A Twist

[winnie]

The circuit in question is this. I'm pulling 2 legs and a neutral from a 208v 3 phase meter stack to a 100A dwelling unit panel. The breaker feeding the unit is 100A so I'm basing my voltage drop off of 80% of the breaker which would be 80A. So would I use the multiplier of 2 or 1.732 to determine the voltage drop for the sub feed feeding the unit panel?

See my post #19 for more explanation... If the bulk of the loading on this single phase panel is 208V loads, then use a multiplier of 2. If the bulk of the loading is 120V loads, then use a multiplier of 3.46 (1.732 * 2). Reality will be somewhere between the two.

-Jon

 

[Ingenieur]

The circuit in question is this. I'm pulling 2 legs and a neutral from a 208v 3 phase meter stack to a 100A dwelling unit panel. The breaker feeding the unit is 100A so I'm basing my voltage drop off of 80% of the breaker which would be 80A. So would I use the multiplier of 2 or 1.732 to determine the voltage drop for the sub feed feeding the unit panel?

if loads are mostly 120 and balanced use the 1.732
but in reality, not huge difference

 

[gar]

180105-1203 EST

In real world electrical circuits actual circuit operation is only approximated by equations. Sometimes a mathematical model accurately represents the actual circuit operation, other times the model is not too good.

One needs to accurately define a circuit to be analyzed, and then pick the best model. You can not just grab any old equation, apply numbers to it, and expect to get a useful result.

I will propose a new question comparing single phase with with a three phase wye. Use DC voltage sources, and impedances of resistance only. The question assumes copper wire and the total weight of wire is the same (for cost reasons). The total load power is the same, and in the three phase wye circuit all four wires are of equal diameter.

Single phase:
One voltage source, 10% of input power is lost in the distribution wires, and thus 90% of input power goes to the load.

Three phase wye:
Three equal voltage sources between hot lines and neutral. Three equal loads, each is 1/3 the power of the single phase load, connected between the three hots and neutral.

Which system is more power efficient?

.

 

[gar]

180105-1525 EST

In my above question to make the DC more like an AC three phase we can assume that what amounts to the neutral wire has zero resistance. So in effect what I have done is removed 1/4 of the copper from the wires.

#/linear foot is inversely proportional to resistance per foot. So with 3/4 of the copper the resistance per three phase hot leg is 3*4/3 of the total single phase resistance. That is 4 times the total single phase resistance.

For convenience and simplicity assume the single phase current was 1 A and the load voltage 90 V. A load of 90 W.

Going to the three phase circuit the load per phase is 30 W and 1/3 A.

In the single phase circuit total voltage drop was 10 V and 10 W dissipated in the wires. And wire resistance equals 10 ohms.

In the three phase circuit the wire resistance per phase is 4*10 = 40 ohms. The power dissipation in a phase wire is 40/9 = 4.44 W for a total wire loss of 13.3 W vs 10 W in the single phase circuit for the same total load power.

This is the result of having to have copper in the neutral even though there may be no neutral current.

In a delta circuit you do not need to put copper into a neutral wire.

.

 

[Grady_G]

180105-1203 EST

In real world electrical circuits actual circuit operation is only approximated by equations. Sometimes a mathematical model accurately represents the actual circuit operation, other times the model is not too good.

One needs to accurately define a circuit to be analyzed, and then pick the best model. You can not just grab any old equation, apply numbers to it, and expect to get a useful result.

I will propose a new question comparing single phase with with a three phase wye. Use DC voltage sources, and impedances of resistance only. The question assumes copper wire and the total weight of wire is the same (for cost reasons). The total load power is the same, and in the three phase wye circuit all four wires are of equal diameter.

Single phase:
One voltage source, 10% of input power is lost in the distribution wires, and thus 90% of input power goes to the load.

Three phase wye:
Three equal voltage sources between hot lines and neutral. Three equal loads, each is 1/3 the power of the single phase load, connected between the three hots and neutral.

Which system is more power efficient?

.

The problem with your first sentence is this. Its impossible to know how much of the dwelling unit will be balanced and how much will be unbalanced. When designing the dwelling units load you use 3VA per square foot and that gives you the number of general lighting circuits required. Lets say thats 2 general lighting circuits. One on A phase and one on B phase. For all I know the homeowner might plug in a lot of devices like computers, lamps, ect. into the receptacles on A phase but might not plug anything into any of the B phase receptacles. Same with the small appliance circuits. If the refrigerator is plugged into one of the small appliance circuits it will probably be the only thing ever used. At my house my refrigerator cycles all the time on one of my small appliance circuits and we only use the other circuit occasionally if running a blender for the baby food. And in reality that receptacle we plug the blender into might be the same circuit as the refrigerator so the second small appliance circuit might never be getting used. And if the refrig is on the same phase as my entertainment system then perhaps my house is not balanced so well. The only thing we can assume is the dryer and heat strips in the air handler along with the hot water heater and the range will be balanced but every 120v single pole circuit is a crap shoot to determine the balancing when you never know how devices in a residence will be plugged in. So I kinda need a one formula fits all approach to determine voltage drop with a reasonable amount of certainty. Im already using 80% of the 100A breaker feeding the unit and I think that's really over kill because that panel will probably never see more than 40A max at any one given time.

 

[JFletcher]

With everything else being equal you get 15.47% more distance with 3 phase vs 1 phase. 2 / √3 = 1.15470053838

2 legs of a 208V 3ph system is single phase, use the 2.

OP posted this:

"If I select 3 phase I can see the multiplier is 1.732050808, and the calculator says I can run a maximum distance of 178.00 feet to stay within my 3%. If I select single phase I can see the multiplier is 2, and the calculator says I can run a maximum distance of 154.15 feet to stay within my 3%."

178.00/154.15 = 1.15471942913, which is practically 1.15470053838. 178ft/1.15470053838 = 154.1525', or within 25 ten-thousandths of an inch.

 

[JFletcher]

Missed the edit time... within 25/10,000ths of a foot... still, no one is pulling out a micrometer to measure that.