Home Menu

Menu


Home Repair (alt.home.repair) For all homeowners and DIYers with many experienced tradesmen. Solve your toughest home fix-it problems.

DIYbanter » Do - it - Yourself » Home Repair > Wiring electric baseboard
Reply
Page 1 of 6 1 23456
 
LinkBack Thread Tools Search this Thread Display Modes
  #1   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 216
Default Wiring electric baseboard
Hi,
I'm wiring four new electric baseboards in my basement.
All four are 220Volt units: 500W, 500W, 1000W, 1500W, for a total of 3500W. From the main panel to the farthest unit is less than 100ft.
If I understood the rules, putting all of this on one circuit will require a 20Amp breaker (1.25*3500W/220V), and the wire can be #12AWG.
Did I get that right, and am I missing any other design factor?

Many thanks, especially if you can reference the NEC so I know I'm doing it right.
Theodore
Reply With QuoteReply With Quote
  #2   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Wed, 25 May 2016 21:59:39 -0700 (PDT), wrote:

Hi,
I'm wiring four new electric baseboards in my basement.
All four are 220Volt units: 500W, 500W, 1000W, 1500W, for a total of 3500W. From the main panel to the farthest unit is less than 100ft.
If I understood the rules, putting all of this on one circuit will require a 20Amp breaker (1.25*3500W/220V), and the wire can be #12AWG.
Did I get that right, and am I missing any other design factor?

Many thanks, especially if you can reference the NEC so I know I'm doing it right.
Theodore

OK first things first, heaters are rated at 240v not 220 so your total
load is 14.6a.

OK lets look at the code (Fixed electric space heating art 424)
**************
424.3 Branch Circuits.
(A) Branch-Circuit Requirements. Individual branch circuits shall be
permitted to supply any size fixed electric space-heating equipment.
Branch circuits supplying two or more outlets for fixed electric
space-heating equipment shall be rated 15, 20, 25, or 30 amperes.
*************

OK there, 20a is a listed size

*******************
424.3(B) Branch-Circuit Sizing. Fixed electric space-heating equipment
shall be considered continuous load.
*******************

"Continuous load" means you size the circuit to 125% of total load
(14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker


Modern space heating equipment should meet the "disconnecting means"
requirement with a switch marked "off" that disconnects both hot
wires. (424.19(C))

I think you are good to go.
Reply With QuoteReply With Quote
  #3   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 216
Default Wiring electric baseboard
Fantastic. Thank you soooo much for the detailed reply. Very much appreciated.
Reply With QuoteReply With Quote
  #5   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 216
Default Wiring electric baseboard
Why the hell wood anybuddy want electric baseboards?
Are you nuts?

No, no, no... it's only for backup. Definitely NOT the primary heat source for the rooms.
Reply With QuoteReply With Quote
  #14   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Sat, 28 May 2016 20:04:33 -0000 (UTC), Diesel
wrote:


Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

"Continuous load" means you size the circuit to 125% of total load
(14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker

You're going over the 80% trade standard load (16amps on a 20amp
circuit) at that point,

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so the
80% has been accounted for.
Reply With QuoteReply With Quote
  #15   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard

Sat, 28 May 2016 23:17:05 GMT in alt.home.repair, wrote:

On Sat, 28 May 2016 20:04:33 -0000 (UTC), Diesel
wrote:


Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

"Continuous load" means you size the circuit to 125% of total
load (14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker

You're going over the 80% trade standard load (16amps on a 20amp
circuit) at that point,

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so
the 80% has been accounted for.

The load is 14.6A under perfect conditions of a wire 100ft or less
in total length, yes. However, if the last heater happens to be a
bit further away from the panel than 100ft as the OP suggested,
there's a real risk of voltage drop to the last heater with
the others running because the 12/2 wire just can't provide the full
load that far out to the last device. A voltage drop will increase
the amount of amps required by that heater to do it's job. The 12/2
isn't going to appreciate that and neither will the breaker. It will
tolerate being over 20amps until it heats up enough to trip.


It's why I suggested 10/2. So, the OP won't even have to worry about
problems down the road if their estimates are off. It certainly
won't hurt the situation.

http://handymanwire.com/ubbthreads/u...ngest_distance

100feet is the normal usual maximum spec'd
The code does not rule on lenghts and voltage drop.. --only
suggests!

The end of run outlet is where the starvation is the worst and any
inductive load will suffer the worst at that point so you must
consider this as you overall design limit at 100feet.

A 20amp circuit fully loaded at this outlet will drop 5.6% where 5%
is the RECOMMENDED maximum and we must install 10gage wiring if we
still insist on a 20amp rated circuit. The other choice is to derate
this long run to a 15 amp circuit on this 12g wire and you could
possibly expect to be at 5% drop at 140-150feet. Tthe full 15 amp
dribbles off at around 120 feet!

20amp circuit loaded at 18amp is 5%

16amp will result in 4.5% missing at that far outlet.

The loss will be accumulative along the run as other 'stuff' is
plugged in.

Most equipment that uses electricity is designed to function well at
plus or minus 10%,whetner it is a resitive load like light bulb or
an inductive load like a lighting ballast or motor.

This all becomes even more crucial when the power company is
experiencing voltage sags in the summer high loads and the volatage
that arrives at the pole down the street is already considerably
LOW!!!---and if you designed your circuit lengths at the extra long
end ---you are causing motor loads to labor,and lights to be dim!

Go with the far outlet at 100feet MAX...and you will be with the
majority!!!

from the OPS original post:
From the main panel to the farthest unit is less than 100ft.

So, the last heater on the circuit, depending on how far away it
actually is from the panel and if the others are all running or not,
could exceed the rated capacity for the wire and/or the breaker. IE:
it's too far and the wire is too skinny to get the needed amps to
the device.

I understand there is a cost difference going with 10/2 and some
individuals consider that to be more important than the total amount
of amps the circuit might pull when fully energized (all heaters on
- high) So, for short distance, and if the OP is very close with his
figure on distance to last device on the circuit, your suggestion
will save him money and work fine. If he's wrong or a bit off and
it's a bit further, he might have to break up the heaters into
seperate fed circuits to keep the amps low enough. Which will use
additional wire and another breaker or two in his panel, assuming
there's room for that.

At this point, whatever cost savings the OP had by not running 10/2
have disappeared.

Why even have to be concerned with it? Use the thicker wire, you
KNOW each device will get all the amps it needs, without any
possibility of an overload condition.

Like I said though, whatever choice the OP makes, it looks fine from
here.


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #16   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Sun, 29 May 2016 03:47:25 -0000 (UTC), Diesel
wrote:


Sat, 28 May 2016 23:17:05 GMT in alt.home.repair, wrote:

On Sat, 28 May 2016 20:04:33 -0000 (UTC), Diesel
wrote:


Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

"Continuous load" means you size the circuit to 125% of total
load (14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker

You're going over the 80% trade standard load (16amps on a 20amp
circuit) at that point,

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so
the 80% has been accounted for.

The load is 14.6A under perfect conditions of a wire 100ft or less
in total length, yes. However, if the last heater happens to be a
bit further away from the panel than 100ft as the OP suggested,
there's a real risk of voltage drop to the last heater with
the others running because the 12/2 wire just can't provide the full
load that far out to the last device. A voltage drop will increase
the amount of amps required by that heater to do it's job. The 12/2
isn't going to appreciate that and neither will the breaker. It will
tolerate being over 20amps until it heats up enough to trip.

The heater is not going to decrease it's resistance and draw more
current to compensate for the voltage drop. It will just be a slightly
smaller heater. Heaters are rated at 120, 208 or 240v. If the voltage
is lower, the heater will just have a lower output.
An example is an oven element. At 240v it is rated 3600w, at 208 it is
only 2700w
Reply With QuoteReply With Quote
  #17   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 15,279
Default Wiring electric baseboard
On Saturday, May 28, 2016 at 11:51:10 PM UTC-4, Diesel wrote:


The load is 14.6A under perfect conditions of a wire 100ft or less
in total length, yes. However, if the last heater happens to be a
bit further away from the panel than 100ft as the OP suggested,
there's a real risk of voltage drop to the last heater with
the others running because the 12/2 wire just can't provide the full
load that far out to the last device. A voltage drop will increase
the amount of amps required by that heater to do it's job. The 12/2
isn't going to appreciate that and neither will the breaker. It will
tolerate being over 20amps until it heats up enough to trip.


Like Gfre said, a resistance heater isn't going to increase it's
current draw because there is some additional resistance in the
circuit. It's like saying a light bulb will burn just as bright
with a rheostat in the circuit to dim it, because it's going to
pull additional current to compensate for the lower voltage.
Put additional resistance in the circuit and the total current
goes down, not up. And the difference here, the resistance in
that wire is negligible.



It's why I suggested 10/2. So, the OP won't even have to worry about
problems down the road if their estimates are off. It certainly
won't hurt the situation.


There won't be problems if he follows the code. I can see using
the heavier gauge wire if he thinks he might need to go to a larger
load in the future, ie adding a heater or using larger ones.


http://handymanwire.com/ubbthreads/u...ngest_distance

100feet is the normal usual maximum spec'd
The code does not rule on lenghts and voltage drop.. --only
suggests!


A random internet posting is your cited source of information?



The end of run outlet is where the starvation is the worst and any
inductive load will suffer the worst at that point so you must
consider this as you overall design limit at 100feet.


There are no inductive loads.



A 20amp circuit fully loaded at this outlet will drop 5.6% where 5%
is the RECOMMENDED maximum and we must install 10gage wiring if we
still insist on a 20amp rated circuit. The other choice is to derate
this long run to a 15 amp circuit on this 12g wire and you could
possibly expect to be at 5% drop at 140-150feet. Tthe full 15 amp
dribbles off at around 120 feet!


You don't show the calculations that is based on, and IDK how you
came up with it. The resistance of 12 gauge wire is .0016 ohms per foot.
100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V .
There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V
drop from 240V is just 2.7%, not 5.6%.




20amp circuit loaded at 18amp is 5%

16amp will result in 4.5% missing at that far outlet.


IKD where any of that is coming from, the relevant calcs are
above.


The loss will be accumulative along the run as other 'stuff' is
plugged in.


What plugged in stuff? He's wiring a dedicated circuit for baseboard
heaters, there are no receptacles.



Most equipment that uses electricity is designed to function well at
plus or minus 10%,whetner it is a resitive load like light bulb or
an inductive load like a lighting ballast or motor.


And he's within 2.7%.


This all becomes even more crucial when the power company is
experiencing voltage sags in the summer high loads and the volatage
that arrives at the pole down the street is already considerably
LOW!!!---and if you designed your circuit lengths at the extra long
end ---you are causing motor loads to labor,and lights to be dim!


There is no motor load. If there were, we'd use the appropriate
NEC to size for that.



Go with the far outlet at 100feet MAX...and you will be with the
majority!!!

from the OPS original post:
From the main panel to the farthest unit is less than 100ft.

So, the last heater on the circuit, depending on how far away it
actually is from the panel and if the others are all running or not,
could exceed the rated capacity for the wire and/or the breaker. IE:
it's too far and the wire is too skinny to get the needed amps to
the device.


The worse case is with all heaters on and that is what we
sized for, 12 gauge is sufficient.



I understand there is a cost difference going with 10/2 and some
individuals consider that to be more important than the total amount
of amps the circuit might pull when fully energized (all heaters on
- high)

Good grief. We have used the rated capacity of the heaters. And there
is already margin in the NEC, they don't say do the calcs, then upsize
it more. You can if you want to, but there is no operational or safety
issue here.


So, for short distance, and if the OP is very close with his
figure on distance to last device on the circuit, your suggestion
will save him money and work fine. If he's wrong or a bit off and
it's a bit further, he might have to break up the heaters into
seperate fed circuits to keep the amps low enough. Which will use
additional wire and another breaker or two in his panel, assuming
there's room for that.


Nonsense as the NEC and above calcs show.


At this point, whatever cost savings the OP had by not running 10/2
have disappeared.

Why even have to be concerned with it?

That's right, just use 12 gauge which NEC says is perfectly fine.
You're the one throwing all kinds of FUD in and using incorrect
calculations.

Reply With QuoteReply With Quote
  #18   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard
trader_4
Sun, 29
May 2016 14:45:19 GMT in alt.home.repair, wrote:


Like Gfre said, a resistance heater isn't going to increase it's
current draw because there is some additional resistance in the
circuit. It's like saying a light bulb will burn just as bright
with a rheostat in the circuit to dim it, because it's going to
pull additional current to compensate for the lower voltage.
Put additional resistance in the circuit and the total current
goes down, not up. And the difference here, the resistance in
that wire is negligible.

That isn't what I'm saying at all. I'm saying that the further out
you go with 12/2 (or any wire really, it just depends on the
thickness of the wire, the power you started with, and how far you're
actually going), the less available voltage you'll have at the end of
the run.

The less volts you have, the more amps you'll need to make up the
difference to get the desired wattage.

Also, the 10/2 isn't going to be near capacity at any time, either.
even if he has them all going on high at the same time. The 12/2
however, will be depending on the amount of heaters going and their
setting(s).

going by Ugly:

7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft

240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the line
yet.

100ft on 12/2 according to my 2014 edition of Ugly's electrical
reference is a net loss of 7.9 volts at the end of the run, with no
load present, yet. At 125 ft out, the loss increases to 9.8 volts.
And, obviously gets worse from there.

OTH, the 10/2 wire loses 4.97 volts at 100ft and 6.21 volts at 125ft
out. I'd rather get as many volts to the device (heaters) as is
realistically possible. The closer I can get to their expected input
voltage, the less amps they'll require to do their jobs. The more
heat I'll get (which is obviously the point here) and the less power
I'll use doing it. A win win win.

To do anything less is only costing me more money and time in the
future. The 12/2 is going to heat up a bit more under various
conditions than the 10/2 ever thought about doing, even if all
heaters were on at the same time, on HIGH. Over time, the 12/2 wire
will degrade due to heating/cooling cycles that the 10/2 won't have
experienced. As it degrades, it's own resistance will increase.

Another side effect of running the wire warm/possibly hot to the
touch at near full load over a period of time is that the connection
points and terminals in the panel and the heaters will also become a
little 'warmer' than they would if they'd been powered from the 10/2
line and pigtailed to it with a short 12/2 run.

A random internet posting is your cited source of information?

Uhh, no. I remembered it was 100ft or so off hand as the general rule
of thumb, but didn't remember how much voltage was lost as a result.
I've also got various NEC books and my 2014 edition of the yellow
Ugly electricians reference book. It's where the voltage drop figures
I used today came from, actually.

That's right, just use 12 gauge which NEC says is perfectly fine.
You're the one throwing all kinds of FUD in and using incorrect
calculations.

Just to clarify, the calculations I provided in this post aren't FUD,
unless you're able to dismiss the yellow Ugly electrician reference
book. I haven't seen any 240volt baseboard heater wired with a 12/2
in sometime, actually. I think the last time I actually observed that
was with a trailer. Alas, they're built with heavy consideration on
cost. IE: as cheap as you can get away with. I don't work like that,
I don't offer advice with that in mind first and foremost.

If it were me, writing only for myself, I'd spend the extra money for
the heavier gauge wire and use the 12/2 wire for pigtailing off of it
into the heater units. I've already explained why I'd run it in this
manner. No real point in doing so again.



--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #19   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Mon, 30 May 2016 02:23:31 -0000 (UTC), Diesel
wrote:


7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft

240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga


And that's due entirely to distance with no loads present on the line
yet.

With no load, there is no voltage drop at all.

If you are actually dropping 7.9v you do not have a 1500w heater
anymore. The element is going to be still around 38.4 ohms (the
element did not change, just the voltage) so at 232.1 v heater draws
6.04a and becomes a 1402.8w heater.

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

This means nothing at this point unless you are changing the element
to get 1500w at 232v.
Reply With QuoteReply With Quote
  #20   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 202
Default Wiring electric baseboard
On 05/29/2016 09:46 PM, wrote:

[snip]

This means nothing at this point unless you are changing the element
to get 1500w at 232v.

Heaters are supposed to be intelligent enough to figure out when they
need to shorten their elements :-)

Reply With QuoteReply With Quote
  #21   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 348
Default Wiring electric baseboard
On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the line
yet.

I'm often surprised by how little some people know about electricity.
Voltage drop depends on current flow. There's no power (wattage) without
current flow either.

So you have a load which is not present. How is it drawing power?

[ship]


Reply With QuoteReply With Quote
  #22   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 144
Default Wiring electric baseboard
On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

Just to clarify, the calculations I provided in this post aren't FUD,
unless you're able to dismiss the yellow Ugly electrician reference
book.

If it says you have voltage drop without load, it SHOULD be dismissed.

[snip]

Reply With QuoteReply With Quote
  #23   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard

Sat, 28 May 2016 23:17:05 GMT in alt.home.repair, wrote:

On Sat, 28 May 2016 20:04:33 -0000 (UTC), Diesel
wrote:


Thu, 26 May 2016 05:50:22 GMT in alt.home.repair, wrote:

"Continuous load" means you size the circuit to 125% of total
load (14.6*1.25=18.25a)
so 18.25a is OK on a 12 ga wire with a 20a breaker

You're going over the 80% trade standard load (16amps on a 20amp
circuit) at that point,

Excuse me but the load is 14.6a. The 18.25a IS 125% of the load so
the 80% has been accounted for.

I sincerely do apologize for having missed that in my original reply
to you. With that said, I'd still have opted for 10/2 to feed them,
and come off the feed with a 12/2 for each heater. This reduces heat
creation along the wire, extends the life of the chosen wire, ensures
the greatest possible amount of power is available to each heater on
the circuit.

Although power loss will still occur at the last one, the voltage
drop won't be as bad as it would have been with a 12/2 run from the
panel; and the feed wire won't run as hot supplying power to an
individual heater or all of them at the same time. A bit more costly,
but, a decent enough tradeoff imo, that I would personally have used
10/2.

As I said though, at the end of the day, it all looks great from
here.

--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #24   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 202
Default Wiring electric baseboard
On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the line
yet.

[snip]

Where can I get one of these magic heaters, that uses 1500 watts when it
isn't connected?

Reply With QuoteReply With Quote
  #25   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard
Sam E
Mon, 30 May 2016 03:14:39 GMT
in alt.home.repair, wrote:

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

I'm often surprised by how little some people know about
electricity.

I sometimes feel the same way.

Voltage drop depends on current flow. There's no
power (wattage) without current flow either.

If you put 120volts AC on a 12/2 wire, and run that wire, 100 ft out,
and take a measurement at the end of the run, the available voltage
at the end of the run isn't going to match the volts going into the
wire. Some is lost during transit due to the wires resistance and the
length.

If you try this experiment using DC power, it's even worse. DC really
doesn't travel distance well.

https://en.wikipedia.org/wiki/Voltage_drop

For example, an electric space heater may have a resistance of ten
ohms, and the wires which supply it may have a resistance of 0.2
ohms, about 2% of the total circuit resistance. This means that
approximately 2% of the supplied voltage is lost in the wire itself.
Excessive voltage drop may result in unsatisfactory operation of, and
damage to, electrical and electronic equipment.

The simplest way to reduce voltage drop is to increase the diameter
of the conductor between the source and the load, which lowers the
overall resistance. In power distribution systems, a given amount of
power can be transmitted with less voltage drop if a higher voltage
is used. More sophisticated techniques use active elements to
compensate for the undesired voltage drop.


So you have a load which is not present. How is it drawing power?

You aren't losing power due to the load, You've already lost it in
route to the load via wire resistance and the distance said
electricity has to travel on the wire. Thicker wire, less resistance,
less voltage drop. Simple concept, really.

Like you wrote though, it does sometimes amaze me how little people
know/understand about electricity.


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #26   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard
notX Mon,
30 May 2016 03:19:44 GMT in alt.home.repair, wrote:

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

Just to clarify, the calculations I provided in this post aren't
FUD, unless you're able to dismiss the yellow Ugly electrician
reference book.

If it says you have voltage drop without load, it SHOULD be
dismissed.

http://www.uglys.net/

You have voltage drop due to the length and size of the wire. Because
the wire isn't a super conductor.

At the end of the day, you might have put 120 volts on the line, but,
100ft down that line, you aren't getting 120 volts. Some has been lost
on the wire, due to the wires own resistance.

Switch from AC to DC with no other changes, and the voltage drop is
more pronounced.


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #28   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Mon, 30 May 2016 05:57:44 -0000 (UTC), Diesel
wrote:

Sam E
Mon, 30 May 2016 03:14:39 GMT
in alt.home.repair, wrote:

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

I'm often surprised by how little some people know about
electricity.

I sometimes feel the same way.

Voltage drop depends on current flow. There's no
power (wattage) without current flow either.

If you put 120volts AC on a 12/2 wire, and run that wire, 100 ft out,
and take a measurement at the end of the run, the available voltage
at the end of the run isn't going to match the volts going into the
wire. Some is lost during transit due to the wires resistance and the
length.

If you try this experiment using DC power, it's even worse. DC really
doesn't travel distance well.

https://en.wikipedia.org/wiki/Voltage_drop

For example, an electric space heater may have a resistance of ten
ohms, and the wires which supply it may have a resistance of 0.2
ohms, about 2% of the total circuit resistance. This means that
approximately 2% of the supplied voltage is lost in the wire itself.
Excessive voltage drop may result in unsatisfactory operation of, and
damage to, electrical and electronic equipment.

The simplest way to reduce voltage drop is to increase the diameter
of the conductor between the source and the load, which lowers the
overall resistance. In power distribution systems, a given amount of
power can be transmitted with less voltage drop if a higher voltage
is used. More sophisticated techniques use active elements to
compensate for the undesired voltage drop.


So you have a load which is not present. How is it drawing power?

You aren't losing power due to the load, You've already lost it in
route to the load via wire resistance and the distance said
electricity has to travel on the wire. Thicker wire, less resistance,
less voltage drop. Simple concept, really.

Like you wrote though, it does sometimes amaze me how little people
know/understand about electricity.

Reply With QuoteReply With Quote
  #29   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Mon, 30 May 2016 05:57:44 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 02:46:57 GMT in alt.home.repair, wrote:

On Mon, 30 May 2016 02:23:31 -0000 (UTC), Diesel
wrote:


7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft

240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga


And that's due entirely to distance with no loads present on the
line yet.

With no load, there is no voltage drop at all.

The voltage drop is due to the wire's own resistance, and the
distance the voltage must travel. The wire we're using isn't a super
conductor. It has a certain amount of resistance to it. As a result,
some volts are no longer available to us, we spent them getting the
rest down the line. Nothing for free, you know.


Voltage drop still depends on the load. If there is no load, there is
no voltage drop. If you have a wire with no load at all, there will be
full circuit voltage at both ends.


Watts is watts, man. Available voltage determines how many amps it's
going to take to get them, though. Lower voltage=more amps to do the
same job.

If you have a lower voltage, the watts will be lower. In our example,
that 1500 watt heater (at 240v) will be 1407 watts at 232v. Current
will actually be less, not more.

Reply With QuoteReply With Quote
  #30   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard
hah
Mon, 30 May 2016 05:25:23 GMT in
alt.home.repair, wrote:

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

[snip]

Where can I get one of these magic heaters, that uses 1500 watts
when it isn't connected?

Hmm? Oh.. ROFL, I see. That's what happens when you don't pay
attention to context and choose to selectively quote.

Those are load calculations based on a 1500watt element running at
two different voltages. It shows the amount of amps required to get
1500watts at those voltages.

The different voltages are directly related to length and diameter of
wire and have nothing to do with the load from the heater at this
point.

Thanks to the wire and the wire alone, some volts have already been
lost (as in they'll never reach the heater) in transit, due again, to
the wires own resistance and length. In this case, the length is the
same; 100ft. But, the wire size or gauge is not. 12ga is on the top,
10ga is on the bottom; for comparison.

Might I suggest a simple english comprehension class or two? Muggles
suffers from a similar problem. So, you're in very good company.


--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #31   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 1,131
Default Wiring electric baseboard

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.

The easiest way I know of to explain the relationship is this:

It may be useful to consider the image of water in a hose. Voltage is
equivalent to pressure, water flow is equivalent to current and the
diameter of the pipe is equivalent ot the thickness of the wire - or
resistance.





--
MID:
Hmmm. I most certainly don't understand how I can access a copy of a
zip file but then not be able to unzip it so I can watch it. That
seems VERY clever!
http://al.howardknight.net/msgid.cgi?ID=145716711400
Reply With QuoteReply With Quote
  #35   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 15,279
Default Wiring electric baseboard
On Sunday, May 29, 2016 at 10:23:36 PM UTC-4, Diesel wrote:
trader_4
Sun, 29
May 2016 14:45:19 GMT in alt.home.repair, wrote:


Like Gfre said, a resistance heater isn't going to increase it's
current draw because there is some additional resistance in the
circuit. It's like saying a light bulb will burn just as bright
with a rheostat in the circuit to dim it, because it's going to
pull additional current to compensate for the lower voltage.
Put additional resistance in the circuit and the total current
goes down, not up. And the difference here, the resistance in
that wire is negligible.

That isn't what I'm saying at all. I'm saying that the further out
you go with 12/2 (or any wire really, it just depends on the
thickness of the wire, the power you started with, and how far you're
actually going), the less available voltage you'll have at the end of
the run.

The less volts you have, the more amps you'll need to make up the
difference to get the desired wattage.


The heating element looks like a resistor. There is no magical making
up. Take a simple circuit with a voltage source V and a resistor of
value R1. The current is V/R. Now add an additional resistor in
series, R2 that represents the resistance of the wire in our system.
The current is now V/(R1+R2), which is less than the current of V/R1.

Like Gfre said, if you look at a heating element, if it's rated for
two voltages, the power output is stated lower at the lower voltage.





Also, the 10/2 isn't going to be near capacity at any time, either.
even if he has them all going on high at the same time. The 12/2
however, will be depending on the amount of heaters going and their
setting(s).

going by Ugly:

7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft


IDK what parameters they used. I worked out the
actual numbers, it's very simple and the voltage drop due to the
resistance in the wire is only 2.7%, at 20A, for 100 ft of #12.
Why no reply to the actual math? Here it is again:

The resistance of 12 gauge wire is .0016 ohms per foot.
100 ft, you have .16 ohms. At 20A, that produces a voltage drop of 3.2V .
There are two conductors so double it, 6.4V. It's a 240V circuit, 6.4V
drop from 240V is just 2.7%, not 5.6%.







240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the line
yet.


No idea what you're even doing here, no loads? yet 1500?


100ft on 12/2 according to my 2014 edition of Ugly's electrical
reference is a net loss of 7.9 volts at the end of the run, with no
load present, yet. At 125 ft out, the loss increases to 9.8 volts.
And, obviously gets worse from there.


Again, the resistance of #12 is .0016 per ft. The math has been
presented above, the voltage drop at the full 20A is just is just
6.4V, which at 240V is just 2.7%. Even using your higher number,
which is probably coming from using some high temperature, you still get
a voltage drop of only 3.2%. There is no 125 ft, he said the run is under
100 ft.

And you say this is with no load present yet. That's incorrect too.
This is with the MAX load the circuit is rated for, the full 20A.
With no load, there is no voltage drop period.




OTH, the 10/2 wire loses 4.97 volts at 100ft and 6.21 volts at 125ft
out. I'd rather get as many volts to the device (heaters) as is
realistically possible.

Then use #8, #6, etc. Same argument can be made there. It does nothing
in terms of safety or proper operation of the heater.



The closer I can get to their expected input
voltage, the less amps they'll require to do their jobs. The more
heat I'll get (which is obviously the point here) and the less power
I'll use doing it. A win win win.


You're really confused here. The heater does not magically adjust to
draw more amps. Take this to an extreme. If I put just 24V across those
heaters, following your logic instead of drawing 14A at 240V they would draw
10 x 14A = 140A? It doesn't work that way.



To do anything less is only costing me more money and time in the
future. The 12/2 is going to heat up a bit more under various
conditions than the 10/2 ever thought about doing, even if all
heaters were on at the same time, on HIGH. Over time, the 12/2 wire
will degrade due to heating/cooling cycles that the 10/2 won't have
experienced. As it degrades, it's own resistance will increase.


BS. If that was the case, no electrical inspector would approve it.
You have one, Gfre, telling you that it does meet code and citing NEC
to back it up.



Another side effect of running the wire warm/possibly hot to the
touch

More BS.


at near full load over a period of time is that the connection
points and terminals in the panel and the heaters will also become a
little 'warmer' than they would if they'd been powered from the 10/2
line and pigtailed to it with a short 12/2 run.

A random internet posting is your cited source of information?

Uhh, no. I remembered it was 100ft or so off hand as the general rule
of thumb, but didn't remember how much voltage was lost as a result.
I've also got various NEC books and my 2014 edition of the yellow
Ugly electricians reference book. It's where the voltage drop figures
I used today came from, actually.


And even using those numbers, the voltage drop at the full 20A circuit
rating is just 3.2%. At the actual load, 14.6 amps, it's just 2.3%,
so again there is no problem, no hot wires.


That's right, just use 12 gauge which NEC says is perfectly fine.
You're the one throwing all kinds of FUD in and using incorrect
calculations.

Just to clarify, the calculations I provided in this post aren't FUD,
unless you're able to dismiss the yellow Ugly electrician reference
book.

Show us where Ugly's says it's a violation of NEC, unsafe, etc to wire
the circuit on a #12.


I haven't seen any 240volt baseboard heater wired with a 12/2
in sometime, actually.


That depends entirely on what the rated capacity of the heaters is.
And if you're that familiar with heating elements you should know that
there ones that are rated for either 240V or 120V and that at the lower
voltage they put out LESS power, they don't magically adjust, pull
more current, to put out the same power they do at 240V.



I think the last time I actually observed that
was with a trailer. Alas, they're built with heavy consideration on
cost. IE: as cheap as you can get away with. I don't work like that,
I don't offer advice with that in mind first and foremost.

If it were me, writing only for myself, I'd spend the extra money for
the heavier gauge wire and use the 12/2 wire for pigtailing off of it
into the heater units. I've already explained why I'd run it in this
manner. No real point in doing so again.


Which again implies that this is destined to fail, when of course it's
not.
Reply With QuoteReply With Quote
  #36   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 14,141
Default Wiring electric baseboard
On Mon, 30 May 2016 06:53:52 -0000 (UTC), Diesel
wrote:


Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor. It takes a little
to push the electrons on it. The thinner and longer the length, the
more is lost in transit. If it were a super conductor, what you've
written would be absolutely true. As long as the wire has resistance
of it's own, we're subject to voltage drop. We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.


Bull****. If there is no current flowing, there is no voltage drop.

Volts (dropped) = Amps x Resistance. If there are no Amps, it doesn't
matter what the resistance is, volts (dropped) is still zero.

You might argue that when you actually measure the volts at the far
end, you are loading the circuit with your meter but a digital meter
has in impedance in the meg ohms so the current is still virtually
zero and you will still see the full voltage within the accuracy of
the meter.

As a sanity check, notice voltage drop charts always base the number
on the load in amps times the resistance of the wire.



Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.

The easiest way I know of to explain the relationship is this:

It may be useful to consider the image of water in a hose. Voltage is
equivalent to pressure, water flow is equivalent to current and the
diameter of the pipe is equivalent ot the thickness of the wire - or
resistance.

Reply With QuoteReply With Quote
  #37   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 15,279
Default Wiring electric baseboard
On Monday, May 30, 2016 at 1:57:48 AM UTC-4, Diesel wrote:

Mon, 30 May 2016 02:46:57 GMT in alt.home.repair, wrote:

On Mon, 30 May 2016 02:23:31 -0000 (UTC), Diesel
wrote:


7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft

240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga


And that's due entirely to distance with no loads present on the
line yet.

With no load, there is no voltage drop at all.

The voltage drop is due to the wire's own resistance, and the
distance the voltage must travel. The wire we're using isn't a super
conductor. It has a certain amount of resistance to it. As a result,
some volts are no longer available to us, we spent them getting the
rest down the line. Nothing for free, you know.


Good grief, you need to take electricity 101. Gfre is correct, with no
load there is no voltage drop, the full 240V is available and can be
measured at the end of the 100 ft of wire. The only way you get voltage
drop is with a load and current flowing. Let RL be the load connected to
a voltage source V using wires that have a resistance RW. With RL not
present the voltage at the open wires where RL would be is equal to V.
With the load RL inserted, the voltage across RL is V- RW*V/(RL+RW).
You should know this just from common experience. Measure the voltage at
a receptacle on a long circuit with no loads on it, you get the same
voltage as you do at the panel. Turn all the appliances, lights, etc on
and you get some number of volts less, due to the voltage drop.




Watts is watts, man. Available voltage determines how many amps it's
going to take to get them, though. Lower voltage=more amps to do the
same job.


Resistor is a resistor man. And a heating element looks more like a
resistor that an intelligent widget that draws more amps. Here's a
though experiment. I take a 3500W heating element that draws 14.6 A at
240V. Now I hook it up instead to a 24V source. Does it draw 146 A?


Reply With QuoteReply With Quote
  #40   Report Post  
Posted to alt.home.repair
external usenet poster
  
Posts: 15,279
Default Wiring electric baseboard
On Monday, May 30, 2016 at 2:53:56 AM UTC-4, Diesel wrote:

Mon, 30 May 2016 06:09:42 GMT in alt.home.repair, wrote:

Voltage drop still depends on the load. If there is no load, there
is no voltage drop. If you have a wire with no load at all, there
will be full circuit voltage at both ends.

No, there won't. The wire isn't a super conductor.

No one ever said it was.

It takes a little
to push the electrons on it.

There is no "pushing" with no load.


The thinner and longer the length, the
more is lost in transit.

That's true, but with no load, there is no current, no pushing, no transit.
Capiche?


If it were a super conductor, what you've
written would be absolutely true.

What we've written is true, it's electricity 101 and quite frankly
if you don't understand these simple basics, you shouldn't be here
giving advice.



As long as the wire has resistance
of it's own, we're subject to voltage drop.

Sure, as long as there is a load and current flowing. The voltage drop on
each conductor will be V = I * RW, where I is the current flowing and RW is
the resistance of the wire. Set I=0 and what do you get for V?


We have various ways in
which to minimize the voltage drop, though. Short of using a super
conductor however, we can't outright prevent it.

Watts is watts, man. Available voltage determines how many amps
it's going to take to get them, though. Lower voltage=more amps to
do the same job.

If you have a lower voltage, the watts will be lower. In our
example, that 1500 watt heater (at 240v) will be 1407 watts at
232v. Current will actually be less, not more.

Not exactly. You can have high voltage and next to no current behind
it (think of a taser) or low voltage and a considerable amount of
current behind it (think of your car battery) it's no chump and your
starter motor isn't exactly a low drain device...You can also lose a
few volts in our example, and, still be able to pull enough current
to make up the difference.


Resistors don't change to make up the difference. Go look at the specs
for some heating elements rated for dual voltages. The heater in my
spa for example, is rated at 1500W at 120V, 6000W at 240V. At lower
voltages, they pull less current, not more.


Reply With QuoteReply With Quote
Reply
Page 1 of 6 1 23456
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes
Linear Mode Linear Mode

Posting Rules
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On

Similar Threads
Thread Thread Starter Forum Replies Last Post
Hydronic elecric baseboard heater vs standard electric baseboard heater John Sevinsky Home Repair 7 March 5th 14 12:13 AM
Wiring an Emerson Electric Baseboard Tiffany[_2_] Home Repair 19 November 14th 13 11:25 PM
Wiring electric baseboard heaters RonABC Home Repair 12 February 20th 09 03:14 PM
Electric baseboard and T-stat wiring Mikepier Home Repair 2 September 28th 08 08:40 PM
wiring baseboard heater Martik Home Repair 1 January 9th 05 11:08 AM


All times are GMT +1. The time now is 10:14 PM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 DIYbanter.
The comments are property of their posters.
 

About Us

"It's about DIY & home improvement"

 

Contact Us DIY Banter Forum Home Privacy Statement