Voltage drop calculation

Status
Not open for further replies.
Hello,
I am new to this forum and I posted a question yesterday but did not get an answer. I am hoping taht some one here can help me.
I am a helper with an electrical company here in South Carolina. Two other helpers and myself are currently studying some Mike Holt DVD's to take our Journeyman exam in the next month or two. The question has to do with voltage drop and the sizing of wires.
Lets say you have a 220v, 30 amp, two hots and a ground (no neutral),50 feet from the panel, circuit that going to be on for five to six hours. This is a continuous load so it has to be multiplied by 125% which brings it up to 37.5 amps. Since there is no 37.5 amp breaker a 40 amp breaker would be used. The question that we would like to have answered is this. Do we use 30 amps or 40 amps in our calculations when figuring out the wire size? Some say that the 30 amps should be used because that is what the load is actually rated at. Some say use 40 amps to size the wires since the breaker is 40 amps and the wires must be sized to match the breaker.
What is the correct answer?
Thanks for your help in advance.
Ivan
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Electrical Contractor
First off if the circuit protection is 30 amps then the load must be less than 30. If it is a continuous load then the largest connected load on the circuit is 24 amps. 24*1.25= 30 amps. The wire must be sized to the load. So if it is a 24 amp load then you need #10 wire.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
It depends on what you are calculating. Your thread title speaks of voltage drop. The actual current flowing through the wire is 30 amps, so I would use that value to calculate VD. But I would use the 37.5 as the amount of current that the wire must be sized to handle. So you could not use #10 wire, since it can only be protected at 30 amps, and you are protecting this circuit at 40 amps. The minimum wire size is therefore #8.

Welcome to the forum.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
I see that Dennis and I interpreted the question differently.
  • I thought you were saying that the load current is 30 amps, and that you based the wire size on 125% of that value.
  • Dennis appears to have thought you were saying that the breaker is 30 amps, and he figured the maximum current you could supply with that breaker.
So please clarify: What is the amount of current that this load will draw?
 

kaichosan

Member
It depends on what you are calculating. Your thread title speaks of voltage drop. The actual current flowing through the wire is 30 amps, so I would use that value to calculate VD. But I would use the 37.5 as the amount of current that the wire must be sized to handle. So you could not use #10 wire, since it can only be protected at 30 amps, and you are protecting this circuit at 40 amps. The minimum wire size is therefore #8.

Welcome to the forum.
Charlie is right. The 40A breaker protects the wire and the wire should have the ampacity of 40A... voltage drop has to also be considered.
 
Depends on the load somewhat....

Depends on the load somewhat....

It is important to know the load being served. Assuming this is a linear resistive type load and not a motor / HVAC load, the calculations are pretty straight forward as follows (all NEC references are to the 2008 version):

Voltage: 240V, 1-phase
FLA: 30 A
Load: continuous, linear resistive type

First, determine the minimum conductor and overcurrent protection device (OCPD) based on the load. Per NEC 210.19, conductors must be sized for 125% of continuous loads. Per NEC 210.20, OCPD must be sized for 125% of continuous loads. Thus, minimum conductor to be #8 AWG and OCPD to be minimum of 40 amps.

Second, check the voltage drop with the minimum conductor size. Using NEC Table 9 for #8 AWG Cu stranded conductor.
VD = (2xLxRxI)/1000
where L = 1-way length of circuit in feet (40 feet)
R = resistance of conductor per 1000 feet (Table 9 - 0.778 ohms per 1000')
I = FLA (30 amps)
VD = 1.867 Volts
VD% = 1.867V / 240V = 0.00778 or 0.78%

Voltage drop is within allowable limits of <3%. #8 AWG wire is acceptable.

Please note there would be additional considerations if this were a motor / HVAC or specialty load.

I hope this helps.
 
Thank you for your answer Mr. Alwon.Actually we selected a electric car charger with a name plate rating of 30 amps. We`asked one of the guys that installs them and he said he uses a 40 amp breaker with number 8 wires. When we pressed him on this he said that he mulitplied the name plate rating (30 amps) by 125% which makes that 37.5 amps, but since there is no breaker that size he used the next higher size (40 amps) and then used #8 wires to match the breaker. All three of us are confused since some believe that 30 amps should be used since this is the actual load it will be pulling. My personal feeling is (I may be wrong ) the 40 amps should be used to match the size of the breaker.
Ivan
 
Thanks again gentlemen. Thanks for the correction, it is 240v. The actual name plate says 30 amps. So I think what you are`saying is that we have to used the 125% increase which is 37.5 amps to get the right size wire. So I can assume that any time you have a continuous load the new amperage has to be used to size the wire.
 

Dennis Alwon

Moderator
Staff member
Location
Chapel Hill, NC
Occupation
Electrical Contractor
It depends on what the nameplate states. If the nameplate states MCA or whether that is the actual load. If 30 amps if the load then I agree 125% times 30 for continuous load and a 40 amp breaker. The wire must be sized for the load so #8 as Charlie stated. Remember #8 is good for at least 40 amps so a 40 amp breaker is fine. Also you generally can use the next size breaker if the amps fall in between sizes. 240.4(B).
 
Gentlemen,
I want to thank each and every one of you for your help. I have never had better explanations for this problem. I asked a master electrican on the job and he had me calculating three different voltage drop calculations.
Thanks again for your help.
Ivan
 
Gentlemen,
Some of the text books used for the apprentice program add to the confusion for helpers like myself. Here is a typical problem taken directly from the text book.
What size THHN copper conductors would be recommended for a 208v, 3 phase, four wire feeder with a length of 150 ft from the source to a fixed continuous load of 160 amps? They continue to explain: determine the minimum ampacity required by multiplying 160 amps by 125% to obtain 200 amps. Using NEC Table 310.16, 3/0 THHN copper conductors af 75 degree are satisfactory for the load of 200 amps. Use the voltage drop formula to determine if 3/0 is suitable to avoid unacceptable voltage drop.
Then they go on to note: the current that will be used in the calculation is the original 160 amps and not the computed 200 amps, which is only used for to size branch circuits or feeder conductor sizes and minimum overcurrent protective device sizes.
Here is where I am lost. They stated that this is for a feeder. Why after they multiply the 160 amps by 125% would the not use the computed 200 amps to size the breaker and wire size?
This is very confusing and some of these books only make matters worse.
Hoping one of you learned gentlemen can help me out of this confusion.
Thanks,
Ivan
 

kingpb

Senior Member
Gentlemen,
Some of the text books used for the apprentice program add to the confusion for helpers like myself. Here is a typical problem taken directly from the text book.
What size THHN copper conductors would be recommended for a 208v, 3 phase, four wire feeder with a length of 150 ft from the source to a fixed continuous load of 160 amps? They continue to explain: determine the minimum ampacity required by multiplying 160 amps by 125% to obtain 200 amps. Using NEC Table 310.16, 3/0 THHN copper conductors af 75 degree are satisfactory for the load of 200 amps. Use the voltage drop formula to determine if 3/0 is suitable to avoid unacceptable voltage drop.
Then they go on to note: the current that will be used in the calculation is the original 160 amps and not the computed 200 amps, which is only used for to size branch circuits or feeder conductor sizes and minimum overcurrent protective device sizes.
Here is where I am lost. They stated that this is for a feeder. Why after they multiply the 160 amps by 125% would the not use the computed 200 amps to size the breaker and wire size?
This is very confusing and some of these books only make matters worse.
Hoping one of you learned gentlemen can help me out of this confusion.
Thanks,
Ivan
The first tip off is that they say "recommended" for a "Feeder". That means it could be something different but, implies using the 3% VD recommended by the NEC fine print note.
Second, they are assuming worst case that all loads are continuous, i.e. (continuous x 125% + non-continuous x 100%)
Third, they want you to check that the wire selected will have a suitable or perceived, depending on how you look at it, VD over the 150ft.
They are saying cable will need to be sized for the 200A due to breaker being 200A. But the maximum load that can be put on the cable is 160A, so that is all you need to use for the VD calc.
 

Smart $

Esteemed Member
Location
Ohio
May as well throw in my two cents...

The wire must be sized to carry no less than the calculated load (in the case of your example, that is 37.5A) and, among other factors concerning raceway fill adjustments and ambient temperature correction, termination temperature limitations of 110.14(C) will/may affect the size determination. AFAIK, this must be adhered to in all cases.

The OverCurrent Protective Device is sized to protect the conductor (i.e. wire). This OCPD is usually sized to protect conductors on its load side. But sometimes this device is sized to protect conductors before it (on the line side), such as for, but not limited to, Service Conductors, Tap Conductors, or Transformer Secondary Conductors. This generalization does not apply to all. Most notably, motors feeders and circuits.
 
Last edited:
Thanks again for your help. Words cannot express my appreciation. Things are becoming clearer now thanks to you guys.
I have been in the trade for about six years and I have see things on the job that as a helper just blow my mind. I have seen master electricans full conduits with so much cables that a couple of days later it was almost impossible to pull out one cable out. I remarked to one of the experts that it should not be so difficult to remove one cable and in my opinion the conduit had too many wires in it. They just laughed at me. One can look at a conduit and see that it has too many wires in it. I have come to the conclusion that it is not that they dont know better, it has to do with money, they dont want to spend the extra money for bigger conduit.
Thanks again for your help.
Sincerely,
Ivan
 

George Stolz

Moderator
Staff member
Location
Windsor, CO NEC: 2017
Occupation
Service Manager
Ivan, don't let that attitude infect you. Reflect back on your last post, and you should see that the poisonous attitude you describe in the others is present in your words.

We're all learning, maintain that positivity and it will take you far.

Also, back on topic, remember that a voltage drop formula can give you satisfactory results from a 25A load on a 14 AWG conductor, but it is a code violation. You have to keep the formula in context.
 
Thanks George and Action Dave,
I agree with you both. I just have a final question.
When using the resistance of wires for calculating voltage drop, Table #8 in chapter #9 is used. This is for DC. Why isn't Table #9 used instead since it is for AC and that is what we are working on? The values are almost identical but if taking a test and table #9 is used then your answer is going to be wrong. What am I missing here.
Ivan
 

Volta

Senior Member
Location
Columbus, Ohio
... from the source to a fixed continuous load of 160 amps? They continue to explain: determine the minimum ampacity required by multiplying 160 amps by 125% to obtain 200 amps. Using NEC Table 310.16, 3/0 THHN copper conductors af 75 degree are satisfactory for the load of 200 amps. Use the voltage drop formula to determine if 3/0 is suitable to avoid unacceptable voltage drop.
Then they go on to note: the current that will be used in the calculation is the original 160 amps and not the computed 200 amps, which is only used for to size branch circuits or feeder conductor sizes and minimum overcurrent protective device sizes.
Here is where I am lost. They stated that this is for a feeder. Why after they multiply the 160 amps by 125% would the not use the computed 200 amps to size the breaker and wire size? ...
It helps to understand why we add 25% to continuous loads. Overcurrent devices feeding these loads are used, of course, for these long periods of time too. To dissipate the heat that is generated in them, we oversize the wire. As far as the fuse/breaker is concerned, the attached wire serves as a heatsink.

But the load hasn't changed, it is still the same 30 amps that it was three hours ago. So to determine the voltage drop the wires will experience, we only need to consider the actual amperage, not the final "required ampacity" of the conductor.
 

George Stolz

Moderator
Staff member
Location
Windsor, CO NEC: 2017
Occupation
Service Manager
Thanks George and Action Dave,
I agree with you both. I just have a final question.
When using the resistance of wires for calculating voltage drop, Table #8 in chapter #9 is used. This is for DC. Why isn't Table #9 used instead since it is for AC and that is what we are working on? The values are almost identical but if taking a test and table #9 is used then your answer is going to be wrong. What am I missing here.
Ivan
There are more than one way to get a result. There really is no 100% correct answer. On a typical test, you'd select the VD you come up with and hope it's close to their answer.

FWIW, Mike teaches using Table 9 for VD on wire over 1/0, because the values differ after that point. For a true AC VD calculation, it is a much more cumbersome formula. The payoff is limited for the extra work involved. Simply put, dealing with a DC problem instead of the true AC problem is easier and produces a similar result.
 
Status
Not open for further replies.
Top