Basic Voltage Drop Question

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Do You Like Ham

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I have a basic voltage drop question.


Part way through the following url:


http://ecmweb.com/code-basics/dont-let-voltage-drop-get-your-system-down


Is this example:


Let's do a sample calculation using the Ohm's Law method. What is the voltage drop of two 12 AWG THHN conductors that supply a 16A, 120V load located 100 feet from the power supply?


VD = I * R
I = 16A
R = 2 ohms/1000 ft (Chapter 9 Table 9)


CD = 16A * 2ohms/1000 ft * 200 ft = 6.4 V


Here, they doubled the length of the circuit when using the length.


Now, refer to:
http://www.eaton.com/ecm/groups/public/@pub/@electrical/documents/content/tb08104003e.pdf


Page 1.3-23, in the first column/big paragraph of text states:
"The circuit length is from the beginning point to the end point of the circuit regardless of the number of conductors"


Under the next column, Calculations section, #1, it states: "Circuit length is the distance from the point of origin to the load end of the circuit".


Compute the voltage drop for the same example above, using the details in the Eaton document:


Step 1:
Amps = 16A
Circuit length = 100 feet
Amps * Length = 1600 Amp Feet


Step 2:
Divide by 100
1600 / 100 = 16


Step 3:
0.3410 Voltage Drop - volts per Ampere per 100 feet (Table 1.3-13, on the next page).
(assume 1.0 PF, Steel conduit)



Multiply by VD value from table
0.3410 * 16 = 5.456


Multiply by 'Correction Factor'
Correction Factor = 0.577 (under the first column of text on the same page, under #2).


5.456 * 0.577 = 3.15 V


So, the question is, why does the Eaton approach result in a voltage drop about half that of the conventional wisdom that :
for single-pole line-to-neutral, one needs to use the 'back-and-forth' (2x) the circuit length? Is there something I'm overlooking in the Eaton documentation?
 
Your first example is calculating the drop of a two wire circuit.

The multipliers in the table for the second example says at top of the table it is for three phase phase to phase loads. I would guess they have a 1.73 factor applied in there plus your calculation also was assuming steel conduit and the first example did not factor type of raceway at all. (although I don't think it makes too significant of a difference on a run that short or that low of current.)
 
kwired said:
Your first example is calculating the drop of a two wire circuit.
The multipliers in the table for the second example says at top of the table it is for three phase phase to phase loads.
Click to expand...

The second example also states to use the correction factor 0.577 for:
Three-phase, four-wire, phase-to-neutral
Single-phase, three-wire, phase-to-neutral

Which I did... however, I think you're saying that the VD on the 1- or 3-phase, phase-to-neutral is not the same as that for a two wire circuit.

How would one use the Eaton data to determine the VD on a two wire circuit? Is it simply a matter of taking the phase-to-neutral case and multiplying by two (to consider the both-ways-length)?
 
Do You Like Ham said:
The second example also states to use the correction factor 0.577 for:
Three-phase, four-wire, phase-to-neutral
Single-phase, three-wire, phase-to-neutral

Which I did... however, I think you're saying that the VD on the 1- or 3-phase, phase-to-neutral is not the same as that for a two wire circuit.

How would one use the Eaton data to determine the VD on a two wire circuit? Is it simply a matter of taking the phase-to-neutral case and multiplying by two (to consider the both-ways-length)?
Click to expand...

You applied the wrong correction factor (0.577) for a 2 wire circuit.

From the ecm example, you have a 2 wire circuit and got a voltage drop of 6.4V.

From the eaton example, you should have applied a factor of 1.155 for a 2 wire circuit.

.................5.456 * 1.155 = 6.3V

You get essentially the same answer.
 
If the second example is using the methods for a three phase system then they are assuming that the system is a balanced 4 wire wye and that there are is no neutral current. Could also be a delta that has no neutral at all. This would account for the voltage drop being halved in the second case.
 
Do You Like Ham said:
The second example also states to use the correction factor 0.577 for:
Three-phase, four-wire, phase-to-neutral
Single-phase, three-wire, phase-to-neutral

Which I did... however, I think you're saying that the VD on the 1- or 3-phase, phase-to-neutral is not the same as that for a two wire circuit.

How would one use the Eaton data to determine the VD on a two wire circuit? Is it simply a matter of taking the phase-to-neutral case and multiplying by two (to consider the both-ways-length)?
Click to expand...

I did not look through that entire publication, but the table used did not have correct factors for 2 wire circuit and the title of the table did indicate this. I believe this publication was intended primarily for determining voltage drop for three phase motor circuits.

david luchini said:
You applied the wrong correction factor (0.577) for a 2 wire circuit.

From the ecm example, you have a 2 wire circuit and got a voltage drop of 6.4V.

From the eaton example, you should have applied a factor of 1.155 for a 2 wire circuit.

.................5.456 * 1.155 = 6.3V

You get essentially the same answer.
Click to expand...
 
kwired said:
I did not look through that entire publication, but the table used did not have correct factors for 2 wire circuit and the title of the table did indicate this. I believe this publication was intended primarily for determining voltage drop for three phase motor circuits.
Click to expand...

I didn't read the entire publication either, but the OP specifically directs us to page 1.3-23. It says right on that page "Tables are based on the following conditions:"

"2. Voltage Drops are phase-to-phase, for three phase, three wire or three phase, four wire...For other circuits multiply voltage drop given in tables by the following correction factors:"

"Single phase, two wire x 1.155"

The publication tells you the correction factor for a two wire circuit, and the correction factor produces the same result as the Ohm's Law method in the OP.
 
Thanks...

david luchini said:
I didn't read the entire publication either, but the OP specifically directs us to page 1.3-23. It says right on that page "Tables are based on the following conditions:"
"2. Voltage Drops are phase-to-phase, for three phase, three wire or three phase, four wire...For other circuits multiply voltage drop given in tables by the following correction factors:"

"Single phase, two wire x 1.155"


The publication tells you the correction factor for a two wire circuit, and the correction factor produces the same result as the Ohm's Law method in the OP.
Click to expand...

I saw the three phase three wire part at the top of the page, but somehow missed or it didn't sink in about the "other circuits".
 
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