No voltage drop? Why?

[Cirus]

I understand voltage drop has to do with wire size, voltage and distance. I've got experience in running underground lines long distances. I have a customer that has a small shed with a 200'run of 10/2G UF running underground. It's a 120 volt circuit that feeds nothing but an air compressor. Spec plate says it pulls 15 amps on 120. Someone else ran the underground line. Voltage at the receptacle is about 122 with no load. With the compressor on voltage drops to 114 and the amperage is 16.7. Sometimes it trips the breaker which is understandable. I'm running a new #6 line to it to resolve the voltage drop. I have a compressor of the same size in my shop and I thought I would perform a test just out of curiosity. I'm had a roll of 12/2g Gomez and put a cord cap on one end and receptacle on the other and plugged the compressor in while taking a amp reading. Amps never got over 14. The romex was still in a 250' coil. My question is would I get the same reading if the romex was stretched out to 250'? Why was there no voltage drop yet there was significant drop on the 10/2 underground line?

 

[Dennis Alwon]

Different compressor have different efficiency ratings and service factor ratings. Could be a difference there.

BTW, welcome to the forum.

Also is the 10/2 uf only feeding the compressor-- no lights or anything else.

 

[MAC702]

...Also is the 10/2 uf only feeding the compressor-- no lights or anything else.

That was my first thought, too. Hard to believe there aren't other lights or receptacles in the shed, which means more than one circuit being run to this outbuilding anyway. Use the #6 to do a service panel for the building.

What else is on the spec plate? MOCP? If it normally pulls 15A, my first thought it that it should probably be on a 20A circuit, if allowed.

 

[Cirus]

My mistake. There is one light but it wasn't on. Just a light and a quad receptacle. Nothing running but the compressor. It is on sp 20 breaker.

 

[gar]

190227-1347 EST

Cirus:

See https://www.google.com/search?q=#12...ome..69i57.21398j0j8&sourceid=chrome&ie=UTF-8

#10 copper is about 1 ohm per 1000 feet and #2 about 1.6 ohms per 1000 feet.

Your first location was 200*2 = 400 ft, or 0.4*1 = about 0.4 ohms. 16.7 A measured at 0.4 ohms is about 6.7 V drop at the motor. You measured 122 dropping to 114 or a change of 8 V. Some of the actual drop is breaker drop. Sufficiently good correlation.

At your own shop you never measured voltage drop you just stated there was none. A 250 ft coil is 500 ft of wire, or about 0.5*1.6 = 0.8 ohms. Ignore inductive reactance because you essentially have a bifilar wound resistance. At 14 A and 0.8 ohms the voltage drop is about 12 V, again some breaker drop. Redo your home experiment.

.

 

[LarryFine]

I understand voltage drop has to do with wire size, voltage and distance.

Actually, it's wire size, current, and distance. Wire size and distance combine to become resistance.

If you can convert the circuit and the compressor to 240v, the current, and thus the voltage drop, will halve.

 

[ggunn]

My question is would I get the same reading if the romex was stretched out to 250'?

The orientation of the wire has very little if any effect on the voltage drop.

 

[MAC702]

...If you can convert the circuit and the compressor to 240v, the current, and thus the voltage drop, will halve.

Yes, but this shed is fed with 10/2 which means they only get one voltage choice for everything in the shed.

 

[LarryFine]

Yes, but this shed is fed with 10/2 which means they only get one voltage choice for everything in the shed.

Unless they can use the existing-feeder exception and use the EGC as a neutral.

 

[MAC702]

Unless they can use the existing-feeder exception and use the EGC as a neutral.

You can do that with an uninsulated UF EGC? I wouldn't have thought so, but I've been surprised before.

 

[readydave8]

"Amps never got over 14."

Did you check voltage under load?

 

[readydave8]

"The romex was still in a 250' coil."

Saw a charred roll #12 romex when someone was doing same thing but to run space heater, he said he saw it smoking

 

[gar]

190127-2350 EST

readydave8:

Total resistance of 250 ft length of #12 Romex is about 0.8 ohms.

A typical 1500 W 120 V space heater is about 10 A when up to temperature at 120 V. I^2*R is about 100*0.8 is 80 W spread out over the spaced coil. It will be somewhat warm, but doubt that it will be smoking. Loop voltage drop is about 8 V.

60 C = 32+9*60/5 = 140 F. Try an experiment and see how hot the coil surface temperature gets.

.

 

[LarryFine]

It will be somewhat warm, but doubt that it will be smoking.

I believe the inference is AC inductive reactance due to the coil, not DC resistance. I remember being told to watch for overheating a heavily-loaded coiled-up extension cord.

 

[GoldDigger]

I believe the inference is AC inductive reactance due to the coil, not DC resistance. I remember being told to watch for overheating a heavily-loaded coiled-up extension cord.

Inductive reactance does not cause heating in a coil of wire!
What is happening is that the heat from resistive heating is concentrated in a small space rather than being conducted to free air over the full length of the wire. The result is a far higher wire temperature. The same effect is why most extension cord reels warn you to fully extend the cord off the real before trying to use it for high current. (Some reels with larger wire and provisions for ventilation on the reel do not carry that warning.)

Sent from my XT1585 using Tapatalk

 

[Sahib]

Inductive reactance does not cause heating in a coil of wire!

Sent from my XT1585 using Tapatalk

But maybe indirectly it does. It causes increased voltage drop and as OP's is motor load the current tends to rise and so more heating with resistance in the coil together with the heating effect you described.

 

[readydave8]

190127-2350 EST

readydave8:

Total resistance of 250 ft length of #12 Romex is about 0.8 ohms.

A typical 1500 W 120 V space heater is about 10 A when up to temperature at 120 V. I^2*R is about 100*0.8 is 80 W spread out over the spaced coil. It will be somewhat warm, but doubt that it will be smoking. Loop voltage drop is about 8 V.

60 C = 32+9*60/5 = 140 F. Try an experiment and see how hot the coil surface temperature gets.

.

could have been steam (or lying carpenter) but the romex was definitely charred

at the time I wondered if it was because the romex was coiled too tight to allow whatever heat there was to dissipate

 

[Strathead]

If you are planning on correcting the installation, don't forget 225.31 requirement for a disconnect. I am with others here. If I were going to up the size to #6 I would certainly pull two hots and a ground and put at least a 30 or 40 amp MCB 6 circuit panel on the far end. Given other code requirements and the future advantages.

 

[gar]

190128-1003 EST

From the original post ---

I have a compressor of the same size in my shop and I thought I would perform a test just out of curiosity. I'm had a roll of 12/2g Gomez and put a cord cap on one end and receptacle on the other and plugged the compressor in while taking a amp reading. Amps never got over 14. The romex was still in a 250' coil. My question is would I get the same reading if the romex was stretched out to 250'? Why was there no voltage drop yet there was significant drop on the 10/2 underground line?

No measured voltages at no load and the 14 A load. And obviously the voltage measurement would need to be done at the load end.

Note: the intent of the post based on the title, "No voltage drop? Why?".

.

 

[growler]

It's a 120 volt circuit that feeds nothing but an air compressor.

If you can convert the circuit and the compressor to 240v, the current, and thus the voltage drop, will halve.

Yes, but this shed is fed with 10/2 which means they only get one voltage choice for everything in the shed.

Larry is right and if there is nothing in the shed but an air compressor why not convert to 240V. You can probably buy another air compressor cheaper than you can trench 200 ft.