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Power mains question: wire gauge

D

DraconisExtinctor

You implied that it could be used, AlwaysWrong.
Click to expand...

NO! I DID NO SUCH THING! EVER! You fucking RETARD!
You're a liar, AlwaysWrong. ...but you know that as well as everyone
else.
Click to expand...

The person who mentioned PVC is who fucked up. Let's go take a look at
who that was... Oh, that's right... it was YOU, you fucking
RETARD-RW.

YOU mentioned incorrect media. To which I immediately responded is the
WRONG choice.

YOU can squirm all you want, but the truth will not change.
 
K

[email protected]

Of course you project you inadequacies on others. It's always been
your MO.

Oh, I had no trouble with that,
Click to expand...

You're a liar.
I was just referring to the fact that
you didn't mention where in the string the "#" is supposed to be,
which made your statement ambiguous.
Click to expand...

You're a liar. But that's nothing new.
 
D

DarkMatter

I don't need you to tell me you're stupid. Like DimBulb, you
advertise the fact constantly.
Click to expand...

The blind shall not see...

This is why, despite the only pre-requisite being a pre-death
repentance, most sinners will *still* not ever be saved.

Despite being told that you were acting like an adolescent worm, you
continued to do so, and enhanced your pathetic wormage even further.

You illustrate for the intelligent among us, just what the flaws in the
human race are.

Not with your diatribes, but with your attempts at making a diatribe.
It borders on funny, considering that you tout yourself as intelligent
and civil.

Your senility onset is showing again.

Must be that curse I put on you about 8 years ago, when you started
your baby bullshit back then, over in a.e.e. I hope the pain settles in
soon.

Maybe you can figure out why an Easy-Bake-Oven works better with a 100W
heating coil than it does with a 100 W light bulb.
 
J

josephkk

It is not. If 700 amps is available, there's a huge margin for the
startup current for a dinky motor like this one. And plenty of current
to blow a breaker if needed.
Click to expand...
No John Larkin, you are full of baloney on this one. I do the wire gauge
for motor starts all the time. I(sc) is NOT used.
Exactly. It needs 15, we have 700. We don't need #10 wire.
Click to expand...
Incorrect setup. What is of concern is the drop during motor start. Not
motor run and not wiring fault conditions. The goal is to keep it below
5% including all other losses to the service transformer. Generally a 2%
to 3% drop during normal operation, and no more than 8% during motor
start, is allowed in the building, all the way to the main service lugs.

Thus #14 is a bit wimpy but #12 is fine. No real advantage going to #10.
 
J

josephkk

Actually, that's what I plan to do. Put the compressor in the garage
and pipe the air downstairs. 4" oughtta do it. ;-)

I just have to figure out how much work it's going to be (how things
line up).
Click to expand...

4" really? How big it that compressor, 500 kW? Does it go over 3000'?
Not that i am all that expert but that is the size of compressor and the
run length i saw last time i saw one plumbed at 4". And that one was done
by an expert.

?-)
 
J

josephkk

Conduit is mentioned in the leadoff post by OP. The cost of putting in
conduit almost completely swamps the wire gauge issue.

?-)
 
J

josephkk

#12 is normal stuff, sold by the pallet at HD. #10 is "special" and
quite expensive because of it. Anything outside the norm has a
premium attached to it. For example, 12-3 has only 33% more copper
than 12-2 but costs almost double ($120 vs. $70 per 250' roll). And,
10-2 is $135 per roll, so yes, *DOUBLE*.

When *was* the last time you bought this stuff? You really need to
get out more and bitch less.
Click to expand...


Small wonder you get raped on #10. You don't know where to buy stuff.

?-)
 
K

[email protected]

4" really?
Click to expand...

Of course not. Michael is being silly with his suggestion of running
#6 wire for the compressor. I thought I'd join in. For anything I've
done, 1/2" Type-L copper is fine. That's probably what I'll use again
but it's getting rather pricey.
 
J

John S

No John Larkin, you are full of baloney on this one. I do the wire gauge
for motor starts all the time. I(sc) is NOT used.
Incorrect setup. What is of concern is the drop during motor start. Not
motor run and not wiring fault conditions. The goal is to keep it below
5% including all other losses to the service transformer. Generally a 2%
to 3% drop during normal operation, and no more than 8% during motor
start, is allowed in the building, all the way to the main service lugs.
Click to expand...

I'm curious about this. Doesn't a motor draw about 6-10 times FLA during
start? If so, wouldn't that cause a 18-30% drop during start if you have
a 3% drop normally? Or, looking at it the other way, if you size for 8%
starting drop, your normal operating drop would be only 1.3%.
 
S

SoothSayer

I'm curious about this. Doesn't a motor draw about 6-10 times FLA during
start? If so, wouldn't that cause a 18-30% drop during start if you have
a 3% drop normally? Or, looking at it the other way, if you size for 8%
starting drop, your normal operating drop would be only 1.3%.
Click to expand...

---
From the OP's example we have a 240VAC 1/2HP motor with a startup
current of 15A being fed through 200 feet of #14 AWG wire.

1HP is 746 watts, so if his motor delivers 1/2HP at the shaft and it's
80% efficient it'll be taking about 467 watts from the mains under
full load.

200 feet of #14 solid copper looks like about half an ohm, so we have,
for the startup case:

15A--->
240AC>---[0.25R]---+ <------+
| |
[16R] 233V
| |
240AC>---[0.25R]---+ <------+

Because of the wire resistance, instead of 15A in the circuit we'll
have 14.55A, and that will drop 233V across the motor.

233V is 97% of 240V, so there's only a 3% loss in the wire.

In the full load case we have this:

1.94A--->
240AC>---[0.25R]---+ <------+
| |
[123R] 239V
| |
240AC>---[0.25R]---+ <------+

so the loss is less than half a percent.
Click to expand...


That is right on the mark.

It was always fun to un a 100 foot extension cord to a construction
work area and run a device like a circular saw or such. Then try it with
two 100 foot cords together for 200 feet. It (the difference) is very
noticeable.

Most of those cords are #14 at those lengths (or they'd better be), but
at the price difference for raw wire and a buried run, I'd say that
dropping to #12 or eve the #10 run was not an improper suggestion,
considering it was going to be a permanent install, and the device being
powered may not be, and wiring a service outlet is what we are doing.

So, yes, as you illustrate, #14 is sufficient, even good, considering
what I have seen houses and such wired with.

At 200 feet away, and not knowing if the device will be the only item
being powered, I would at least examine the cost tables for the job using
#10, #12 and #14 Thhn, no conduit.

As far as the conduit goes though, if one gets small diameter PVC, and
all the fittings, no 200 foot pull needs to be done. The segments and
fittings can all be fed down the wire bundle and build the conduit in
place right on the wires, then bury the finished run.
 
B

bud--

I'm curious about this. Doesn't a motor draw about 6-10 times FLA during
start? If so, wouldn't that cause a 18-30% drop during start if you have
a 3% drop normally? Or, looking at it the other way, if you size for 8%
starting drop, your normal operating drop would be only 1.3%.
Click to expand...

---
From the OP's example we have a 240VAC 1/2HP motor with a startup
current of 15A being fed through 200 feet of #14 AWG wire.

1HP is 746 watts, so if his motor delivers 1/2HP at the shaft and it's
80% efficient it'll be taking about 467 watts from the mains under
full load.

200 feet of #14 solid copper looks like about half an ohm, so we have,
for the startup case:

15A--->
240AC>---[0.25R]---+<------+
| |
[16R] 233V
| |
240AC>---[0.25R]---+<------+

Because of the wire resistance, instead of 15A in the circuit we'll
have 14.55A, and that will drop 233V across the motor.

233V is 97% of 240V, so there's only a 3% loss in the wire.

In the full load case we have this:

1.94A--->
240AC>---[0.25R]---+<------+
| |
[123R] 239V
| |
240AC>---[0.25R]---+<------+

so the loss is less than half a percent.
Click to expand...

The NEC gives the full load current of a 1/2 HP 230V motor as 4.9 A.
Power factor adds to the motor current. (In most cases this will be
high, but is used to determine wire size.)

The NEC gives the locked rotor amps as 29.4 (which is 6x the running amps).

The NEC gives the resistance of 200 ft of #12 as 0.62 ohms (about the
same as you used).

The running voltage drop is 4.9V = 3%.
The starting (locked rotor) voltage drop is 18.2V = 7.6%.

#14 sounds reasonable. Can't remember if the motor was actually 1/2 HP.
 
J

John S

On 1/2/2013 8:48 PM, josephkk wrote:
On Sat, 29 Dec 2012 11:04:23 -0800, John Larkin

On Sat, 29 Dec 2012 12:20:01 -0600, John Fields

On Fri, 28 Dec 2012 16:05:49 -0800, John Larkin

Air compressor 1/2 hp motor rated 220v (2-wire, not 3-phase) @
15A. Distance
from load panel ~100 ft (as the conduit runs).

15A can be handled by 14 gauge, but I'd normally go with 12
gauge due to
start current.

With such a distance, is it recommend to up-scale the wire to 10
ga?

Thanks.

#12 is 1.6 mohms/foot. That's 0.32 ohms total. The short-circuit
current from 220 volts is almost 700 amps. A half horse is only
around
400 watts, about 2 amps. I'd use #14.

---
The short circuit current through the wire resistance is irrelevant.

It is not. If 700 amps is available, there's a huge margin for the
startup current for a dinky motor like this one. And plenty of current
to blow a breaker if needed.

No John Larkin, you are full of baloney on this one. I do the wire
gauge
for motor starts all the time. I(sc) is NOT used.

What matters is the stall current through the motor when it's
starting
up; no doubt the 15A the OP mentioned.

Exactly. It needs 15, we have 700. We don't need #10 wire.

Incorrect setup. What is of concern is the drop during motor
start. Not
motor run and not wiring fault conditions. The goal is to keep it
below
5% including all other losses to the service transformer. Generally
a 2%
to 3% drop during normal operation, and no more than 8% during motor
start, is allowed in the building, all the way to the main service
lugs.

I'm curious about this. Doesn't a motor draw about 6-10 times FLA during
start? If so, wouldn't that cause a 18-30% drop during start if you have
a 3% drop normally? Or, looking at it the other way, if you size for 8%
starting drop, your normal operating drop would be only 1.3%.
Click to expand...

---
From the OP's example we have a 240VAC 1/2HP motor with a startup
current of 15A being fed through 200 feet of #14 AWG wire.

1HP is 746 watts, so if his motor delivers 1/2HP at the shaft and it's
80% efficient it'll be taking about 467 watts from the mains under
full load.

200 feet of #14 solid copper looks like about half an ohm, so we have,
for the startup case:

15A--->
240AC>---[0.25R]---+<------+
| |
[16R] 233V
| |
240AC>---[0.25R]---+<------+

Because of the wire resistance, instead of 15A in the circuit we'll
have 14.55A, and that will drop 233V across the motor.

233V is 97% of 240V, so there's only a 3% loss in the wire.

In the full load case we have this:

1.94A--->
240AC>---[0.25R]---+<------+
| |
[123R] 239V
| |
240AC>---[0.25R]---+<------+

so the loss is less than half a percent.
Click to expand...

The NEC gives the full load current of a 1/2 HP 230V motor as 4.9 A.
Power factor adds to the motor current. (In most cases this will be
high, but is used to determine wire size.)
Click to expand...

I believe the motor current given by the NEC includes efficiency and
power factor.

In fact, the NEC info indicates an inefficient motor with a poor power
factor.

Motor running VA is the motor volt-amps (4.9 * 230) = 1127.

The power output of a motor is nameplate horsepower times 746 = .5 * 746
= 373 Watts.

The product of motor efficiency and power factor is:
N*Pf = P/VA where N is efficiency and Pf is power factor. That product
is 373/1127 = .331

I know neither the efficiency nor power factor of the NEC's motor, but
as a quick rough "feel" for the numbers, I can take the square root of
the .331 = .575. Roughly .6 for power factor and .6 for efficiency.

If power factor was actually typical of larger HP motors, say .7, then
the efficiency is .331/.7 = .47 or about 50%.

Fractional horsepower single phase motors typically have a lower power
factor and lower efficiency than integral horsepower three-phase motors.

However, that is not my point.
The NEC gives the locked rotor amps as 29.4 (which is 6x the running amps).
Click to expand...

Yes, I was thinking somewhere between 6 and 10 times.
The NEC gives the resistance of 200 ft of #12 as 0.62 ohms (about the
same as you used).

The running voltage drop is 4.9V = 3%.
The starting (locked rotor) voltage drop is 18.2V = 7.6%.
Click to expand...

ISTM that if there is a 3% drop at FLA, then the drop will be 6 times
that at 6 times the current. That would be 18%
#14 sounds reasonable. Can't remember if the motor was actually 1/2 HP.
Click to expand...

I am not questioning the wire gauge chosen.
 
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