Please forgive my ignorance but I'm trying to better understand the Voltage Drop formulas displayed in the 2002 Handbook. I've managed to figure most of it out but I'm hung up on one thing. In both Examples 1 and 2, to find the line-to-line voltage drop the line-to-neutral is multiplied by the square root of 3 because both examples are 3-phase circuits. My question is how would I determine the line-to-line voltage drop for a single phase circuit?
NEC Handbook Voltage Drop Formula
- Thread starter allang
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physis
Senior Member
Re: NEC Handbook Voltage Drop Formula
I don't have the hanbook. So I'm going to guess that your three phase formula is this:
Vd = (1.732 x K x L x I) / cmil
or
Vd = 1.732 x R x L x I
If that's the case replace the 1.732 with 2
For single phase:
Vd = (2 x K x L x I) / cmil or
Vd = 2 x R x L x I
I don't have the hanbook. So I'm going to guess that your three phase formula is this:
Vd = (1.732 x K x L x I) / cmil
or
Vd = 1.732 x R x L x I
If that's the case replace the 1.732 with 2
For single phase:
Vd = (2 x K x L x I) / cmil or
Vd = 2 x R x L x I
bphgravity
Senior Member
- Location
- Florida
Re: NEC Handbook Voltage Drop Formula
K = R x cm/ 1000
VD = 2 x K x D x I/ cm
D = cm x VD PERMITTED/ 2 x K x I
I = cm x VD PERMITTED/ 2 x K x D
cm = 2 x K x D x I/ VD PERMITTED
WHEN USING ?cm? FORMULA, USE APPROX. K OF 12.9 FOR COPPER AND 21.2 FOR ALUMINUM. USE EXACT K FOR ALL OTHERS.
?2?- The length of the wire, there and back. Use 1.732 for 3 phase problems
?D?- The length of the wire, one way.
?I?- The total load in amps.
?VD PERMITTED?- The allowable percent of voltage drop over the circuit.
?cm?- The circular mil of the conductor.
?K?- The resistance of a circular mil foot. Use entire number in calculation.
?R?- The resistance of the conductor.
K = R x cm/ 1000
VD = 2 x K x D x I/ cm
D = cm x VD PERMITTED/ 2 x K x I
I = cm x VD PERMITTED/ 2 x K x D
cm = 2 x K x D x I/ VD PERMITTED
WHEN USING ?cm? FORMULA, USE APPROX. K OF 12.9 FOR COPPER AND 21.2 FOR ALUMINUM. USE EXACT K FOR ALL OTHERS.
?2?- The length of the wire, there and back. Use 1.732 for 3 phase problems
?D?- The length of the wire, one way.
?I?- The total load in amps.
?VD PERMITTED?- The allowable percent of voltage drop over the circuit.
?cm?- The circular mil of the conductor.
?K?- The resistance of a circular mil foot. Use entire number in calculation.
?R?- The resistance of the conductor.
charlie
Senior Member
- Location
- Indianapolis
Re: NEC Handbook Voltage Drop Formula
The concept of voltage drop is very basic. Regardless of the formula that you use, it is a modification of E = IR. The formula that is given in the NEC Handbook is:
Vd = 2RLI / 1000
Vd = voltage drop (in volts)
R = resistance of the conductor per 1000 feet
I = amperes
2 = both ways, going out to the load and back to the source
L = length of the run
1000 = gets rid of the 1000 feet on the conductor resistance
In order to change the formula to 3 phase, you must multiply by the Sin of 60?, i.e. (square root 3)/2. This changes the formula to:
Vd = (square root 3)RLI/1000
The concept and formulae above are based on direct current, not alternating current. If calculations must be accurate, reactance must be accounted for as well as the power factor of the load and the resistance of the circuit to alternating current. The question is, how important is the accuracy of your calculations?
[ March 03, 2005, 01:04 PM: Message edited by: charlie ]
The concept of voltage drop is very basic. Regardless of the formula that you use, it is a modification of E = IR. The formula that is given in the NEC Handbook is:
Vd = 2RLI / 1000
Vd = voltage drop (in volts)
R = resistance of the conductor per 1000 feet
I = amperes
2 = both ways, going out to the load and back to the source
L = length of the run
1000 = gets rid of the 1000 feet on the conductor resistance
In order to change the formula to 3 phase, you must multiply by the Sin of 60?, i.e. (square root 3)/2. This changes the formula to:
Vd = (square root 3)RLI/1000
The concept and formulae above are based on direct current, not alternating current. If calculations must be accurate, reactance must be accounted for as well as the power factor of the load and the resistance of the circuit to alternating current. The question is, how important is the accuracy of your calculations?
[ March 03, 2005, 01:04 PM: Message edited by: charlie ]
Re: NEC Handbook Voltage Drop Formula
Sorry about that.. I didn't realize you guys might not have the handbook. The 2002 NEC Handbook formula for AC Voltage drop is as follows:
Zc = (R x cos A) / (XL x sin A)
where:
Zc = impedance (from Table 9 in the NEC)
R = ohm per 1000ft (Also from Table 9)
A = angle representing the power factor which is figured by getting the arc-cosine of the powerfactor.
This formula gives us the ohms to neutral
You take that value and find the line-to-neutral voltage drop with the following formula:
Voltage drop (L-N) =
Circular length / 1000ft x circuit load
(I've seen mentioned elsewhere in this forum that the term 'Circular Length' might be a mis-print and that you're supposed to use the length of the circuit one way)
Then you take that value and figure the line-to-line voltage drop with the following formula:
Voltage Drop (L-L) =
Voltage Drop (L-N) x SqRt of 3
This is the basis of my question.. What do I do in this step for a single phase circuit? I know there are easier ways to figure Voltage drop but my boss is insisting I use this method.. even though he can't answer my question about the single phase formula.
Sorry about that.. I didn't realize you guys might not have the handbook. The 2002 NEC Handbook formula for AC Voltage drop is as follows:
Zc = (R x cos A) / (XL x sin A)
where:
Zc = impedance (from Table 9 in the NEC)
R = ohm per 1000ft (Also from Table 9)
A = angle representing the power factor which is figured by getting the arc-cosine of the powerfactor.
This formula gives us the ohms to neutral
You take that value and find the line-to-neutral voltage drop with the following formula:
Voltage drop (L-N) =
Circular length / 1000ft x circuit load
(I've seen mentioned elsewhere in this forum that the term 'Circular Length' might be a mis-print and that you're supposed to use the length of the circuit one way)
Then you take that value and figure the line-to-line voltage drop with the following formula:
Voltage Drop (L-L) =
Voltage Drop (L-N) x SqRt of 3
This is the basis of my question.. What do I do in this step for a single phase circuit? I know there are easier ways to figure Voltage drop but my boss is insisting I use this method.. even though he can't answer my question about the single phase formula.
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: NEC Handbook Voltage Drop Formula
allang
Consider the following
In a balanced 3 phase circuit you consider the distance one way when calculating line to neutral drop, because there is no current in the neutral. When you do a single phase calculation, you must multiply the distance by 2 because the current is coming back 180 degrees out of phase from the current going out.
For example .1 ohm/1000' resistance 100 amps of load, 200' from the source to the load, 277/480 system.
Three phase line-neutral drop is .1/1000 x 100 x 200 = 2 volts drop or (2/277)x100 = .72%
Single phase line-line drop is .1/1000 x 100 x 200 x 2 = 4 volts drop or (4/480)x100 = .833%
Jim T
allang
Consider the following
In a balanced 3 phase circuit you consider the distance one way when calculating line to neutral drop, because there is no current in the neutral. When you do a single phase calculation, you must multiply the distance by 2 because the current is coming back 180 degrees out of phase from the current going out.
For example .1 ohm/1000' resistance 100 amps of load, 200' from the source to the load, 277/480 system.
Three phase line-neutral drop is .1/1000 x 100 x 200 = 2 volts drop or (2/277)x100 = .72%
Single phase line-line drop is .1/1000 x 100 x 200 x 2 = 4 volts drop or (4/480)x100 = .833%
Jim T
Re: NEC Handbook Voltage Drop Formula
Ok so the General consensus seems to be that instead of the sqrt of 3 you multiply the value by 2. This brings up entirely new questions. (sorry to be such a pain but my brain still hasn't reconciled all these options)
There are 2 types of single phase circuits, for example in a typical house you have a 120/240V service so you could have a Single Phase circuit thats 120V which would always be L-N and you also have the 2P single phase circuit or 240V which would always be L-L. You tell me that in a balanced 3phase load you use the length one-way but in a single phase situation you use the length of the run x 2.
Questions:
Wouldn't a 277/480 circuit where you're using only one phase and a neutral 277V require to also use the x 2 value of the length instead of the SqRt of 3?
What about 480V single phase? Where you run only 2 phase conductors from a 3-phase panel to a load. What do you do then?
What about a single phase 240V circuit where you have 2 phase conductors and no neutral? If the load is balanced between the 2 phases do you still use the x 2 for length?
I have learned that the formula I'm using is based on the IEEE Std 141 exact voltage drop formula which is in the IEEE Red book. Does anyone have more information on that particular formula?
Ok so the General consensus seems to be that instead of the sqrt of 3 you multiply the value by 2. This brings up entirely new questions. (sorry to be such a pain but my brain still hasn't reconciled all these options)
There are 2 types of single phase circuits, for example in a typical house you have a 120/240V service so you could have a Single Phase circuit thats 120V which would always be L-N and you also have the 2P single phase circuit or 240V which would always be L-L. You tell me that in a balanced 3phase load you use the length one-way but in a single phase situation you use the length of the run x 2.
Questions:
Wouldn't a 277/480 circuit where you're using only one phase and a neutral 277V require to also use the x 2 value of the length instead of the SqRt of 3?
What about 480V single phase? Where you run only 2 phase conductors from a 3-phase panel to a load. What do you do then?
What about a single phase 240V circuit where you have 2 phase conductors and no neutral? If the load is balanced between the 2 phases do you still use the x 2 for length?
I have learned that the formula I'm using is based on the IEEE Std 141 exact voltage drop formula which is in the IEEE Red book. Does anyone have more information on that particular formula?
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: NEC Handbook Voltage Drop Formula
allang
The answer to all three questions is yes. If you are dealing with one single phase circuit, two wires, you multiply the distance by 2. It doesn't matter if it is 120 volt, 277 volt, or 480 volt, as long as it is single phase.
I don't have any info on the Red Book equation, I'll leave that for someone else.
Jim T
allang
The answer to all three questions is yes. If you are dealing with one single phase circuit, two wires, you multiply the distance by 2. It doesn't matter if it is 120 volt, 277 volt, or 480 volt, as long as it is single phase.
I don't have any info on the Red Book equation, I'll leave that for someone else.
Jim T
bilbo336
Member
- Location
- Durango, CO
Re: NEC Handbook Voltage Drop Formula
The IEEE Red Book gives the general formula for voltage drop as V = IR cos(A) + IX sin(A) where A is the angle whose cosine is the load power factor. Also, note that V is defined as the line to neutral voltage drop in the circuit. The Red Book goes on to say that "The reason for using the line-to-neutral voltage is to permit the line-to-line voltage to be computed by multiplying by the following constants: Single phase X 2, three phase X 1.732" (square root of 3). NOTE: in order to use this formula you need to know angle A, the load power factor.
If you look at the values in Table 9 of the NEC you will note that the Effective Z values are for a power factor of 0.85 (A = 31.7 degrees) ONLY. At a different power factor you have to use the formula and the values in the first two columns for X and R. This is what they use in the sample calculation under Table 9 in the 2002 NEC Handbook.
I know that your boss requires you to use this formula but it is hard to get a handle on the power factor of specific loads and not very practical. In most cases, the dc formula used in the example under section 215.2 in the Handbook will work. The formula is close enough and is sufficiently conservative that it will not get you into trouble.
The IEEE Red Book gives the general formula for voltage drop as V = IR cos(A) + IX sin(A) where A is the angle whose cosine is the load power factor. Also, note that V is defined as the line to neutral voltage drop in the circuit. The Red Book goes on to say that "The reason for using the line-to-neutral voltage is to permit the line-to-line voltage to be computed by multiplying by the following constants: Single phase X 2, three phase X 1.732" (square root of 3). NOTE: in order to use this formula you need to know angle A, the load power factor.
If you look at the values in Table 9 of the NEC you will note that the Effective Z values are for a power factor of 0.85 (A = 31.7 degrees) ONLY. At a different power factor you have to use the formula and the values in the first two columns for X and R. This is what they use in the sample calculation under Table 9 in the 2002 NEC Handbook.
I know that your boss requires you to use this formula but it is hard to get a handle on the power factor of specific loads and not very practical. In most cases, the dc formula used in the example under section 215.2 in the Handbook will work. The formula is close enough and is sufficiently conservative that it will not get you into trouble.
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