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Voltage Drop

jar546

jar546

CBO
Joined
Oct 16, 2009
Messages
12,980
Location
Not where I really want to be
OK so a lot has changed in the code since I first picked up an NEC in 1987. We no longer have the hard and fast rules of voltage drop. I would, however like to do something for educational purposes. Let's try to solve a problem and see who can come up with the answer based on today's codes of the 2008/2011 NEC. Here is the scenario, let's see who still has this in their memory banks. Show your work please.

You have to run a feeder to a compressor across the warehouse from your nearest, usable panelboard. The compressor has a 10HP 240 single phase motor. The distance to the compressor is 200' and the conduit will be run inside the web of the metal roof trusses. Your boss wants you to use up the THHN that he has stockpiled. Conduit size not relevant unless you want to make it relevant.

Let's see how this goes and discuss it. Please show your work so this can be educational. I wonder if we will all have the same answer.

What size copper wire THHN will you pull for this job?
 
Are you implying the the roof truss conduit routing makes ambient temp derating a player?

Also can't see nameplate so with pressure switch control can I rule out the last column of T 430.22(E) as a player? ( no continuous RATING)
 
I'll take a swing at it

Without a nameplate I will start with 10HP that 430.248 says is 50a. Multiply that times 1.15 per 440.12(A)(1) ... 57.5a.

That gets you #6 but you are 200' out so you are dropping 9.82 volts (based on 50a). That is about 4%, not within the "design" criteria in the FPN (<3%)

Bump that up to a #4 you only drop 6.16v and that is 2.57%, under the maximum recommendation in the FPN so I would use #4
 
chris kennedy said:
Are you implying the the roof truss conduit routing makes ambient temp derating a player?Also can't see nameplate so with pressure switch control can I rule out the last column of T 430.22(E) as a player? ( no continuous RATING)
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Sorry Chris, I thought I replied but there was a problem with it posting from the work computer.

Attic temp is not an issue.

Not continuous.

Thanks for swinging at it George. Let's see what others come up with on this one.
 
T430.248 10HP 230V 50A.

50Ax1.25=62.5A

T310.16 #6 AWG use 75 C can carry 65A

For 200' #6 AWG should be safe to carry 3% VD.
 
If this is an air compressor using a motor per 430 you use a design current of 125% of FLA (62.5a) #6 is still going to work but the VD stays the same and I still end up with a #4 if I want to stay under 3%
 
OK, I made this one up and I am going to take a stab at it since I did not do it ahead of time. Here is my thinking and explanation. Your mileage may vary.

First things first lets look at the motor. I said 10HP single phase but did not specify continuous duty until asked. Since I did not post a nameplate (that would have been nice) we really don't know much, therefore I am going to stick with 125% of FLA based on 430.22(A) to play it safe. Again basic information and KISS (keep it simple stu__). But that does not help us until we know the Amperage (I) of the motor. We all know that that it takes 746 watts to = 1 hp but instead of doing the math, I just looked at table 430.248 and determined it to be 50A. There are other factors involved when doing the math but I'll let the NEC determine it for me.

So now I know that I have a 50A design amperage draw 200' from the source nearest panelboard. Once I apply my 125% rating for the conductors based on 430.122(A), I now get 62.5 amps.

Now on to the voltage drop or as I like to refer to it. Vd which does not stand for venereal disease. Vd=IR

We know I=62.5 but we don't know R (resistance) so we have to figure that out first.

We first have to decide what size wire we are going to start with. That takes us to Table 310.16 and I already said we are using THHN. Our target load is 62.5 and the nearest size for that is #6 which is rated for 65 amps based on 75 degree terminals.

*Post Drift- The Vd is based off of what is now an FPN with the NEC and FPNs are not enforceable, however there may be a go around as the motor manufacturers have minimum voltage ratings for supply that they want you to meet, but I digress.

Now that we know we are going to use a #6, we can go to Chapter 9 and find Table 8 which gives us a resistance per 1,000 ft of wire. If you look closely at Table 8, you will see that #6 copper has a resistance of .510 ohms per 1,000'.

Now we have what we need to continue. But an important factor comes into play. We just can't calculate 200' because current travels in both directions so we will double the length of the wire (many forget to do this)

If we take the fact that there are .510 ohms per 1,000' and multiply that by .4 (.4 represents 400' of the 1,000') we then get .204 ohms of resistance in the 400' of wire.

Now we have what we need, remember Vd=IR so 62.5 x .204 = 12.75 volts or Vd=12.75

12.75/240= 5.3% voltage drop.

From a code standpoint without the FPN taken into consideration, this is a compliant installation when using a #6 (maybe)

An inspector would like to see a #4, and electrician wants to install a #6 (more than likely). There are 2 different ways of looking at this

Again, your mileage may vary and the nameplate rating of the motor and its exact use may change this answer.
 
What is funny is that Southwire has a voltage drop calculator and even when I put it down to 50A and limited 3%, it still wants us to use a #4

Oh and by the way

NEC RECOMMENDATIONSThe National Electrical Code contains six Fine Print Notes to alert the Code user that equipment can have improved efficiency of operation if conductor voltage drop is taken into consideration.1. Branch Circuits – This FPN recommends that branch circuit conductors be sized to prevent a maximum voltage drop of 3%. The maximum total voltage drop for a combination of both branch circuit and feeder should not exceed 5%. [210-19(a) FPN No. 4], Figure 2.2. Feeders – This FPN recommends that feeder conductors be sized to prevent a maximum voltage drop of 3%. The maximum total voltage drop for a combination of both branch circuit and feeder should not exceed 5%. [215-2(d) FPN No. 2], Figure 2
 
Thank you jar546 for your easy VD calculation. I'm using VD= 2 x one way lengh x resistantance x current / 1000, it also have same answer but yours is better.
 
Span said:
Thank you jar546 for your easy VD calculation. I'm using VD= 2 x one way lengh x resistantance x current / 1000, it also have same answer but yours is better.
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Thanks, I'm just hoping that I am right!

For the record, if I was the electrician I would use a #4 and I would strongly suggest it as an inspector. Strongly because there are other factors that come into play.
 
You really have to be careful using the instructions that come with a motor. I have a booklet from a Dayton (Grainger) motor that says the conductors must hold starting current to <5%.

They say you need #8 for a 1/2 HP motor 50 feet away.
 
Since the load is 50 amp I would not worry about the 125% since the wire size for the VD will take care of that. With a 3% VD then you need a #4.

Now with a 240 V motor and the tables based on 230V we know that the motor will run a bit lower and the tables are the worse case scenario for power factor. Given that a #6thhn copper wire will have a 9.5V drop. The voltage supplied to most buildings is usually higher than 240v-- about 245V or 246v . So with that increase in voltage and a 9.5v drop I would feel very comfortable with a #6
 
Dennis said:
Since the load is 50 amp I would not worry about the 125% since the wire size for the VD will take care of that. With a 3% VD then you need a #4.Now with a 240 V motor and the tables based on 230V we know that the motor will run a bit lower and the tables are the worse case scenario for power factor. Given that a #6thhn copper wire will have a 9.5V drop. The voltage supplied to most buildings is usually higher than 240v-- about 245V or 246v . So with that increase in voltage and a 9.5v drop I would feel very comfortable with a #6
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Understood with the Vd taken care of by the wire. We are doing an exercise in Vd and that was the basic point so that we have an educational value to this thread. Thank you for your input; it looks like we all want to see a #4
 
1) I don't think about doing a VD calc until the one way length of my circuit is considerably above the supplied voltage, rule of thumb gleaned from experience.

2) Working in the field for a contractor my job in to deliver a quality, problem free installation to our customers at as low of cost as possible. In the case posted here there is no reason to take FLA at 125%. (T430 22(E)) That said if we had #4 at the shop for a single motor, sure, in a heart beat.

If I had 20 of these motors I'm running 6's and sleeping well.

3) I use the calc; VD=(2×L×R×I)÷1000 That gives me a 4.25V drop using #6 @ 50A Given the 10% built in tolerance motor manufactures design into the equipment, again I sleep well.
 
I am in total agreement with Chris as I stated that myself.

Generally we don't have to worry about a 3% or 5% VD. However lets make sure that we understand that a fire pump is a different animal.

695.7 Voltage Drop.(A) Starting. The voltage at the fire pump controller line terminals shall not drop more than 15 percent below normal (controller-rated voltage) under motor starting conditions.

Exception: This limitation shall not apply for emergency run mechanical starting. [20:9.4.2]

(B) Running. The voltage at the motor terminals shall not drop more than 5 percent below the voltage rating of the motor when the motor is operating at 115 percent of the full load

current rating of the motor.
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chris kennedy said:
1) I don't think about doing a VD calc until the one way length of my circuit is considerably above the supplied voltage, rule of thumb gleaned from experience. 2) Working in the field for a contractor my job in to deliver a quality, problem free installation to our customers at as low of cost as possible. In the case posted here there is no reason to take FLA at 125%. (T430 22(E)) That said if we had #4 at the shop for a single motor, sure, in a heart beat.

If I had 20 of these motors I'm running 6's and sleeping well.

3) I use the calc; VD=(2×L×R×I)÷1000 That gives me a 4.25V drop using #6 @ 50A Given the 10% built in tolerance motor manufactures design into the equipment, again I sleep well.
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1) I think I can agree on that one except with smaller wire like #12 where the resistance is significantly higher.

2) I have no problem with this statement and yes, if there were 20 of these motors, the cost factor would be significant at 200' runs each.

3) Your Vd is less because you did not use the 125% which I understand

The bottom line is that #6 is a code compliant installation as stated by me in my opinion.
 
Copper.org likes to point out that you are paying for that voltage drop in I2R losses and you will get the cost of the larger wire back in a year or two.
 
gfretwell said:
Copper.org likes to point out that you are paying for that voltage drop in I2R losses and you will get the cost of the larger wire back in a year or two.
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Of course they do. Try selling that concept to the electrician who does not pay the electric bill. Smaller wire is not only less expensive for a better profit but it is lighter to carry! I can just hear the sales pitch now.
 
For example a 3.66 delta in voltage drop at 50a and 15 cents a KWH electricity, you are spending $240.46 a year warming up the gophers.

3.66v x 50a = 183 watts or .183kw x .15 x 24 x 365 = 240.462
 
gfretwell said:
For example a 3.66 delta in voltage drop at 50a and 15 cents a KWH electricity, you are spending $240.46 a year warming up the gophers.3.66v x 50a = 183 watts or .183kw x .15 x 24 x 365 = 240.462
Click to expand...
If it is a continuous duty motor on 24/7 that is a perfect example of how money is wasted on the back end to save copper on the front end.
 
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