High current & warm wires

[mike33]

We just finished installing 26 High Bay Fluorescent fixtures in a warehouse. We knew we wouldn't be back for a couple days so we stayed a little late to make up all connections, land wires to breakers and install some temporary switches to see them light up. Everything looked great.

I decided to check Amps before replacing cover and found one 20A circuit was drawing 19.5 Amps. This circuit has 8 lights on it, one of which I measured with an extension cord at 2 Amps on a different day.

The wires in the junction box were noticeably warm to the touch.

Could both of these problems be caused in a loose connection somewhere?
Voltage drop?

I had no more time to troubleshoot. I will be going back soon.

 

[skeshesh]

Maybe... a bit more detail would be nice: can you provide more details about the fixture? A link to a data sheet for the fixture would help. Also length of runs/wire size used.

 

[mike33]

FBD FLUORESCENT HIGH BAY LUMINAIRE T5/HO

http://www.daybrite.com/pdfspecs/HB-10032.pdf

http://www.daybrite.com/brochures/260.pdf

Four 54w lamps each.

120volt.

12 AWG THHN.

Three phase.

5 - 20A circuits.

1" EMT from sub panel to 12x12 junction.
3/4" EMT from 12x12 junction to four switch locations (four separate runs).
Combination of existing 3/4" EMT and new 1/2" EMT from junction to fixtures.
4" boxes feeding 1 or 2 fixtures each with 12/2 MC.

How should I count the length of run?

suppose I have:

5' grounded & ungrounded from sub panel to j-box
10' ungrounded AND travelers from j-box to switch#1
15' travelers line and load from j-box to (4way) switch#2
20' travelers AND switch leg from j-box to switch#3
25' grounded & ungrounded switch leg from j-box to 4" box
5' 12/2 MC from 4" box to fixture

It seems to me that with these numbers the ungrounded would be 125' long while the neutral is only 35' long. Would I then say its a 125' run?

Is such an uneven wire length a problem?

This design of pulling everything to one main box didn't sound bad when we started, but now I'm worried. I'll have to measure when I get back there, but I think the runs are hundreds of feet long.

 

[ptonsparky]

Warm wires with a fully loaded circuit is normal. Circuits are feeding only your lights and not an exit or emergency light some place?

 

[stevebea]

We just finished installing 26 High Bay Fluorescent fixtures in a warehouse. We knew we wouldn't be back for a couple days so we stayed a little late to make up all connections, land wires to breakers and install some temporary switches to see them light up. Everything looked great.

I decided to check Amps before replacing cover and found one 20A circuit was drawing 19.5 Amps. This circuit has 8 lights on it, one of which I measured with an extension cord at 2 Amps on a different day.

The wires in the junction box were noticeably warm to the touch.

Could both of these problems be caused in a loose connection somewhere?
Voltage drop?

I had no more time to troubleshoot. I will be going back soon.

Did you check current draw on your other lighting circuits?

 

[patriot]

FBD FLUORESCENT HIGH BAY LUMINAIRE T5/HO

http://www.daybrite.com/pdfspecs/HB-10032.pdf

http://www.daybrite.com/brochures/260.pdf

Four 54w lamps each.

120volt.

12 AWG THHN.

Three phase.

5 - 20A circuits.

1" EMT from sub panel to 12x12 junction.
3/4" EMT from 12x12 junction to four switch locations (four separate runs).
Combination of existing 3/4" EMT and new 1/2" EMT from junction to fixtures.
4" boxes feeding 1 or 2 fixtures each with 12/2 MC.

How should I count the length of run?

suppose I have:

5' grounded & ungrounded from sub panel to j-box
10' ungrounded AND travelers from j-box to switch#1
15' travelers line and load from j-box to (4way) switch#2
20' travelers AND switch leg from j-box to switch#3
25' grounded & ungrounded switch leg from j-box to 4" box
5' 12/2 MC from 4" box to fixture

It seems to me that with these numbers the ungrounded would be 125' long while the neutral is only 35' long. Would I then say its a 125' run?

Is such an uneven wire length a problem?

This design of pulling everything to one main box didn't sound bad when we started, but now I'm worried. I'll have to measure when I get back there, but I think the runs are hundreds of feet long.

If your runs are hundreds of feet long, I think VD could be a factor.

 

[mike33]

I estimate the 19.5A loaded circuit has 225' of neutral wire and 885' of ungrounded wire.

I think the four circuits were:

19.5A (8) fixtures
17.6A (8) fixtures
10A (4) fixtures
7A (4) fixtures

There's definitely something wrong here

I won't be going back until after the weekend.

 

[hurk27]

I estimate the 19.5A loaded circuit has 225' of neutral wire and 885' of ungrounded wire.

I think the four circuits were:

19.5A (8) fixtures
17.6A (8) fixtures
10A (4) fixtures
7A (4) fixtures

There's definitely something wrong here

I won't be going back until after the weekend.

Yes there is, you have 1110' of circuit path, at 19.7 amps, that would equal to a 24.8% voltage drop, or your end voltage is only 90.2 volts, I'm very surprised they even light, these T-5 ballast do have a voltage regulator in them that as the voltage sags it turns up the power sort-to-speak and this in turns causes more current.

even with a # 6 you would have a 5.1% VD using the 17.6 amps the shorter run had.

Anyway to use multiwire circuits? this will bring you down to a #8 at 4% VD if you can balance the circuit?

This is why most commercial lighting is 277 volts or even 240 volts, many of these newer T-5 fixtures will operate on 120 Thu 277 volts,
Design of lighting can be costly if all things are not considered.

 

[Finite10]

Nuts... I did a Vd calc and it got lost in the nether or something and didn't post.

Anyway, to help it's best to have the specific specs. Which model - for ballast type - looks like 360W. T5 lamps? Cut sheet shows T8 and T5HO.

I assumed 360W ...
1.723x260'x20Ax12.93/6,530 = 17.83Vd

120V - 17.83Vd = 102.17V
I=P/E ... I=360W/102.17V = 3.524A x 8 fixtures = 28.19A one circuit

1.25 x 28.19A = 35.24A

 

[Finite10]

Just to add, if you have the 240W ballast then it's 23.5A

So that's my .02
.04 now

 

[broadgage]

I would agree that voltage drop is likely a problem, owing to the lengthy circuits and use of 120 volts.
If an electronic ballast is connected to a lower voltage than nominal, then it will tend to run the lamps at full power and draw more current from the line to achieve this.

The obvious way to overcome the problem is to run thicker wire, likely very expensive though.
Cheaper alternatives might include working the fittings at 208/240 volts. Switch the lights by contactors located near the load center in order to reduce the wire lengths, the switch wires will then only carry the contactor coil current and can be almost any length.

If the fittings are 120 volt only (unlikely) then the use of MWBs and contactor switching would help very significantly.

A load of 19.5 amps on a 20 amp circuit is just about compliant IF non continous, however most lighting in commericial premises would be used continually in working hours, and therefore limited to 16 amps continous on a 20 amp circuit.

 

[PetrosA]

Two questions:

1) Did you use a TRMS meter for the current readings? If not, the 19.5A reading might be an error.

2) What voltage are you reading at various points in the circuits?

 

[G._S._Ohm]

20 A through #14 copper Romex in free air after 15 minutes finally leveled off at a rise above ambient of 2 C. That's it.

20 A should give (20/15)^2 = 1.8x the rise that 15 A should give.

If your current is sinusoidal this kind of heating is really puzzling.

 

[G._S._Ohm]

I forgot that meters have a limited crest factor, probably less than 3, so it's better to use a scope because a meter may understate your RMS value.
http://en.wikipedia.org/wiki/Crest_factor
A high spike with high crest factor every cycle can give RMS values way above average values.

 

[Finite10]

Two questions:

1) Did you use a TRMS meter for the current readings? If not, the 19.5A reading might be an error.

2) What voltage are you reading at various points in the circuits?

It would be cheap to add seperate #10 neutrals for each ckt. In the past I've put my hand on 1/2" EMT that fed lights and it was very hot to the touch.

TrueRMS

 

[G._S._Ohm]

For cables similar to #14 NMB,
temp rise in free air = 1300 x (I^2)/(D^3), where temp rise is in degrees C, I is in amps and conductor diameter D is in mils.

Crunching numbers from table 310-16 shows that this formula changes slightly with larger conductors.

This kind of problem has come up before.
If anyone has free air measurements on cable sheath temp, ambient temp, current flow and cable size, I'd like to see them. The Neher McGrath formula is supposed to cover this, but some real world measurements would be good.

 

[hurk27]

For cables similar to #14 NMB,
temp rise in free air = 1300 x (I^2)/(D^3), where temp rise is in degrees C, I is in amps and conductor diameter D is in mils.

Why are you trying to calculate for "free air" the OP says these lights are in a warehouse which would mean most likely in conduit or other non-free air wiring method?

 

[G._S._Ohm]

Why are you trying to calculate for "free air" the OP says these lights are in a warehouse which would mean most likely in conduit or other non-free air wiring method?

My bad.
If I can get the formula in this paper
http://www.electriciancalculators.com/ampacity/understanding_nm.pdf
into a spreadsheet it should then be easy to crank out answers to some level of accuracy and specific to the application and so check whether there really is a problem.

The OP might already have posted enough info to do this confirmation. I'll take "warm" to mean between 40C and 50C.

 

[hurk27]

My bad.
If I can get the formula in this paper
http://www.electriciancalculators.com/ampacity/understanding_nm.pdf
into a spreadsheet it should then be easy to crank out answers to some level of accuracy and specific to the application and so check whether there really is a problem.

The OP might already have posted enough info to do this confirmation. I'll take "warm" to mean between 40C and 50C.

Here's a good online one I used in post 8 and also explains the formula:

http://www.electrician2.com/calculators/vd_calculatoradv.htm

Remember the 19.5 amps he posted in post 7 was on a 1110' (225'+885') complete run which if you cut in half to input into the formula for a run one way will be 555'

Hope this helps.

 

[G._S._Ohm]

Here's a good online one I used in post 8 and also explains the formula:

http://www.electrician2.com/calculators/vd_calculatoradv.htm

Remember the 19.5 amps he posted in post 7 was on a 1110' (225'+885') complete run which if you cut in half to input into the formula for a run one way will be 555'

Hope this helps.

Thanks.
Here's what I've got so far on my spreadsheet in trying to decode the example at the bottom of the link I posted.

Ampacity of 3 #2 conc. stranded XHHW copper in 1" steel conduit in 40 C ambient.

0.09 > enter insulation thickness in inches
0.292 > enter conductor diameter [DC] in inches
0.382 =calc'd sum [DI] in inches

Ri = 0.012*694*[r-value]*log(DI/DC)

0.576 > enter R-value of insulation on a per inch basis
0.560 =calc'd Ri value

Rsd = n'xA'/(Ds' + B')

3.2 >enter A' value
0.19 >enter B' value

Ds' = factor x DI
3 >enter # of conductors
2.15 >enter factor from table
0.821 =calc'd Ds'

9.493 =calc'd Rsd value assuming n' = the number of conductors

Re = 9.5 x n'/(1 + 1.7Dx' (e + 0.41)


This is worse than what the IRS puts me through.