Canging from 15 amps to 20 amps and beyond

[Rounder]

My shop runs on two 15 amp circuit breakers. So far, I have only flipped the switches a couple of times. I know that I really need more for the BT3100 to work better, so here is the question. To increase the breakers from 15 to 20 amps, do I need to have new wires run?

I know that to go up to 30 amps, I will ahve to have heavier wire run, but that can wait till the new shop is being built.

[LarryG]

quote:Originally posted by Rounder

To increase the breakers from 15 to 20 amps, do I need to have new wires run?

That depends on what the existing wire gauge is. A 15A circuit requires 14ga (minimum) wire, a 20A circuit requires 12ga wire.

I never use 14ga since the heavier 12ga is only slightly more expensive and is not bad to work with (10ga and up starts getting to be a bear), and I have known of electrical contractors who do likewise. But if you live in a typical subdivision home in which every effort was made to cut construction costs, you might well have 14ga.

[gonesailing]

What LarryG says with the additional comment that changing the circuit breakers will not effect how the BT3100/3000 runs. The difference is the wire gauge. The bigger the wire, the less voltage drop, the better things run. That means to improve operation you have to rewire. If it is a long run from the box to the saw, I might upgrade to 10ga wire and leave the circuit breakers at 20 amps.

Mike

[rbfunk]

You might have to upgrade the outlets themselves from 15 to 20 amps to be code compliant.
Bob

[LarryG]

Mike is right. If the wire is already 12ga or bigger, installing 20A breakers should reduce the number of times they trip, but whatever voltage drop currently exists will remain unchanged.

This thread tickled my curiosity bone so I called one of our electrical consultants and had him give me a Cliff Notes course in wire size and voltage drop. Turns out it's one of those questions the answer to which requires consulting a table, but basically the longer the run and the greater the load in amps, the greater the voltage drop for a given wire size. Some specific examples for a constant 15A load:

14ga wire, 50' run: 4.0V drop
14ga wire, 100' run: 9.2V drop

12ga wire, 50' run: 2.5V drop
12ga wire, 100' run: 5.75V drop

10ga wire, 50' run: 1.6V drop
10ga wire, 100' run: 3.6V drop

Those values are rounded off slightly, but they're close. Note that doubling the length of the wire run more than doubles the voltage drop; IOW, it's not linear. There is however a direct relationship between the load and the distance; that is, for any size wire, a 10A load would have the same voltage drop over 75' as a 15A load over 50'.

Our consultant told me that the voltage should ideally be within 3% of what it should be; for 120V, that would be 3.6V.

Other factors to consider: the voltage supplied by your local utility can and will vary at times, sometimes considerably; also, if your circuit starts at a subpanel, you also have to take into account the voltage drop between the main panel and the subpanel.

Fascinating stuff, electricity.

[Rounder]

Well, here is what I can tell you about the wiring. It is about 50' to the shop from the box. The breakers are at the bottom of the box and look like they were added in. As my house was built in 1892 and the carriage house (shop) was built without permits way back when, I ahve no clue what guage they used. All I know is that inside the shop, the wire is the three strand that is covered in white rubber (plastic?). Half the shop works on one light switch, the rest on the other.

Like i said, because of termite and hurricane damage it has to be torn down at some point, but other projects have take priority. So I was looking for a quick fix till I can do it right. It really hasn't been a big problem, but i wanted my saw to work at optimum speed.

Thanks for all the answers.

[Tom Miller]

quote:Originally posted by LarryG

...Some specific examples for a constant 15A load:

14ga wire, 50' run: 4.0V drop
14ga wire, 100' run: 9.2V drop

12ga wire, 50' run: 2.5V drop
12ga wire, 100' run: 5.75V drop

10ga wire, 50' run: 1.6V drop
10ga wire, 100' run: 3.6V drop

Note that doubling the length of the wire run more than doubles the voltage drop; IOW, it's not linear.

I Googled up some tables myself, and found some contradictions, which could lead to over a factor of 2 difference in these numbers. After spending a full 60 seconds researching it, it appears at least one table was taking into account a "return trip" circuit in their resistance per linear foot numbers. Not sure what's up with that!

For 12 and 14 gauge (15 A draw) and 100 ft of wire, I got 2.4V and 3.8V drops. Not too big a difference. I think you have to be near 10V drop to notice the saw having trouble -- just a guess.

Also, I think the voltage drop will be linear with wire length (i.e. double length = double voltage drop).

So, I'm on the fence concerning 12g vs. 14g making any difference.

I do agree that the circuit breaker is either tripped or not, though.

Regards,
Tom

[LarryG]

quote:Originally posted by Tom Miller
For 12 and 14 gauge (15 A draw) and 100 ft of wire, I got 2.4V and 3.8V drops.

Also, I think the voltage drop will be linear with wire length (i.e. double length = double voltage drop).

Maybe those values are for a single conductor. If you double your numbers to take into account the second "round trip" wire needed to complete the circuit (which I assume you'd have to do?), your numbers and mine are a lot closer.

The conductor material makes a difference too and, according to an offhand comment our engineer made during our conversation, so does whether the wire is encased in a magnetic (metal conduit) or non-magnetic (Romex) enclosure.

You got me on the linear thing. It SEEMS like it ought to be linear, but all I know is that our engineer did the math for the examples I cited while I had him on the phone, and a few others as well, and in every case doubling the length more than doubled the voltage drop. When I noticed that pattern I queried him about that specific point and he said, Yes, that's right (i.e., it's not linear).

[Tom Miller]

quote:Originally posted by LarryG

Maybe those values are for a single conductor. If you double your numbers to take into account the second "round trip" wire needed to complete the circuit (which I assume you'd have to do?), your numbers and mine are a lot closer.

Yeah, I think that's it. In fact, I wasn't considering how the return at the motor is not at zero volts, so I would have done well to use the "doubled" numbers.

Then, I get 4.8 and 7.2 V for 12g and 14g, respectively, which still leaves me up on the fence.

Rounder, if you see copper conductor at the panel, and copper at the receptacle, then it's most likely the same all the way (assuming no junction boxes in between, or splices [:0]) So, you can check the diameter of the wire to determine if it's 12g or 14g (this info is easily found, e.g., on the web). There's a chance it's already 12g.

Regards,
Tom

[Black wallnut]

Just a thought and it's just a guess but voltage drop only matters from source to load. Which means there is no reason that I can think of why the return path would even be considered.

[Tom Miller]

quote:Originally posted by Black wallnut

Just a thought and it's just a guess but voltage drop only matters from source to load. Which means there is no reason that I can think of why the return path would even be considered.

In my haste earlier, I neglected the return, but then realized that if current is flowing in that return, there must be a voltage drop across it, meaning more voltage "stolen" from the motor.

You have to consider the full circuit from the panel, back to the panel. In this circuit, there are 3 voltage drops. The first is the supply wire, the second is the motor, and the third is the return wire. The supply from the panel starts at 110V. If the supply wire drops 5V, then you have 105 at the supply side of the motor. If you have another 5V drop on the return line, then the return side of the motor is at 5V (instead of 0V; the return at the panel is at 0V).

That means that across the motor you have 105V - 5V = 100V.


Regards,
Tom

[LCHIEN]

quote:Originally posted by Black wallnut

Just a thought and it's just a guess but voltage drop only matters from source to load. Which means there is no reason that I can think of why the return path would even be considered.

Sorry, kirchoffs law says that the voltage drops around the loop add up to the input voltage. You lose both ways!

[mccrorey]

My old wiring was a single run of 14ga that meandered around to a lot of places, supplying the kitchen lights before hitting the single outlet in the garage (shop) and then on to the back yard lights. When I ran the shop vac and then started the planer, the shop vac RPM would drop quite a bit while the planer spun up.

I added two 30' runs of 10ga directly to new outlets in the shop, each on its own breaker, and the voltage drop problem went away. The BT3100 seems a bit more robust, too. I agree that 10ga is a bear to work around corners and through holes. Probably two dedicated 12ga wires would have done the job OK. Other than the flexibility issue, larger wire won't hurt as long as the breaker rating matches the capacity of the outlets.

I really don't get why the voltage drops mentioned above are not linear with respect to the length of the run. Can somebody point me to a reference on that?

With regard to the role of the "return" part of the circuit, voltage drop is usually measured between the two outlet terminals (not relative to a separate fixed ground reference) so the drop across the return is included. Calculating the drop from the wire characteristics would require including the resistance of both lines, and any drop resulting from AC effects such as inductive reactance of the straight wire and electric field interactions between the two conductors, and interactions between the conductors and any nearby fixed-voltage points (e.g. ground). The nonlinear component must be related to the AC effects, since the Ohm's Law part is linear.

[LCHIEN]

Well the engineer in me finally got around to analyzing what's been presented here.

Engineers like to use equations, it gives us a better idea of what's going on than tables which are distilled down to a specific case.

First of all, there are really two things governing the size of wire
that one should use.

The first is the thermal issues and the second is the electrical performance.

If you put through too much current down a wire you risk melting the insulation (due to the temperature rise of the wire itself) and shorting the wires together causing a possible fire.
The NEC (nat'l electric code) suggests for household wiring with 140°F limit on insulation that you should have no more than 40, 30 or 25 amps for 10, 12 and 14 ga. wire respectively WHEN its ambient temperature of 87°F or less and its a single wire.
When you have a cable of three wires then its harder for the heat to get out and you have to derate it to 30, 25 and 20Amps.

When its hot and you have three wire cable then it builds up heat faster and at 105 to 113° you have to derate to 21, 17.8 and 14Amps.

This is an interesting consequence for those of us with garage shops in the south. If your garage is at 105F then you can only run 14A with 14 ga. wire regardless of the length since the heat buildup is at each point.

This wire overheating is particularly a bad deal since the resistance goes up with termperature - it could cause a runaway spiral where the wire gets hot, the resistance goes up , the wire gets hotter and the resistance goes up some more, etc.

The other problem that affects the choice of wire gauge as I said is
the electrical performance.

The loss of voltage is the product of the load current (in amps) and the resistance of the wire (in ohms) that the current travels through. In the wired setups then the resistance is the product of resistivity in ohms per foot (actually they list it in ohms per 1000 ft) and the length. In the case of houshold or extension cable wiring the length is the two-way length to the load so you have to multiply by 2.

Specifically at 77°F, for copper wire the single wire resistivity are
14 ga 2.58 ohms/1000'
12 ga 1.62 ohms/1000'
10 ga 1.02 ohms/1000'

With a 50' and 100' branch circuit the 2-way voltage drop with 15A will be
14 ga 3.87V @50', 7.74V @100'
12 ga 2.43V, 4.86V
10 ga 1.53V, 3.06V

It has to be a linear relationship to length. That table where its not linear is totally suspect. (I think for 14 ga they listed 4 and 9.2 volts).
Note with 10 ga. wire it's only a milliohm (1/1000th or an ohm) per foot. I would not be surprised that the electrical outlet and plug would have 10-50 milliohms or as much loss as 25 feet of 2-cond cable. This could be nearly an additional volt and several watts lost at the plug (and that's why you don't want a bunch of extension cords). Ever notice how the plug gets real hot when running high current appliances (hair dryer, toaster, vacuum cleaner, electric frying pan, space heater, etc) or tools?

Also note that the resistance of copper wire goes up with temperature
like .22% per degree F, a fair amount and worth considering when its 100°F, you have about 5-6% more resistance.

So what voltage loss is acceptable?
There's no black and white - where the tool works and where it won't.
The performance will degrade more and more as the voltage is dropped.
Furthermore its a squared relationship. In a resistive load sich as a light or heater, if voltage loss is 10% then you have 90% of the voltage available which means the current will also be reduced 90% and the the net power will be 81%.

For a motor, where the power drawn is affected by the torque load applied to the motor, it will try to draw more current, if starved by 10% in voltage the current will go up 10% and possibly burn out the motor or trip breakers early when fully loaded. So in order to maximize the motor power you need to be as close to full voltage as possible.

You also have to realize that the motor is intended to work over a range of 110-120V; if you have 120V then the loss of 5 or 8 volts is not so bad but if you are at 110V then 5 or 8 volts may not be good news.

As a personal judgement (this is opinion now as opposed to previous facts), I think that shooting for a loss of 3V or so would be great, 5V is begining to get dicey. A 10 V loss is definately poor.

[Black wallnut]

Thanks Loring. Your explaination makes more sense now.