ONLINE
STORE
"ASK THE BACH MAN"
FORUM
PARTS, SERVICE,
& INFORMATION
CATALOGS AND
BROCHURES

Welcome, Guest. Please login or register.
Did you miss your activation email?
November 21, 2017, 04:39:40 AM
Home Help Search Login Register
News: Please read the Forum Code of Conduct   >>Click Here <<
+  Bachmann Message Board
|-+  Discussion Boards
| |-+  General Discussion
| | |-+  Doorbell wire
« previous next »
Pages: 1 [2] Print
Author Topic: Doorbell wire  (Read 7822 times)
Seasaltchap

View Profile
« Reply #15 on: February 11, 2007, 11:04:40 AM »

Dick,

I think you are playing with semantics.

Why do you take it out on me, when Jim Banner has also previously told you that Amps drive up the heating effect to blow the wiring with:-

"What happens if we drive too much current through a wire?  The insulation melts off the wire, rendering it useless or the wire itself melts.  Typical insulations used in wiring chassis will fail at around 11 amps.  The wire itself will melt in 5 seconds at about 68 amps."

Wattage is directly related to Horse Power.

When you write:-

"Wattage does not describe the amount of work a unit will do, only the amount of power the unit can use if operating properly."

I think you create a conundrum of - when it is working properly that is the work it is doing.

Regards

Logged

Phoenix AZ: OO enthusiast modelling GWR 1895-1939, Box Station Wiltshire; S&DJR Writhington Colliery, Nr. Radstock.

Interested in making friends on the site with similar interests.
Seasaltchap

View Profile
« Reply #16 on: February 11, 2007, 11:22:38 AM »


I also meant to edit in their place:-

From observaltion many years ago, it was seen that one good horse could lift 550 lbs through 1 foot in 1 second; hence the measure for 1 horse power.

Regards, Aye.


Logged

Phoenix AZ: OO enthusiast modelling GWR 1895-1939, Box Station Wiltshire; S&DJR Writhington Colliery, Nr. Radstock.

Interested in making friends on the site with similar interests.
r.cprmier

View Profile
« Reply #17 on: February 11, 2007, 02:06:00 PM »

Doorbell wire will work OK for short distances; I wouldn't touch phone (#24) wire on a bet; not for power applications, anyway.

I speak from my experience as an electrician when I say that you are going to-far and away-be better off using a larger size wire, such as #12THHN-or at least a #14- and stranded at that.   Do yourself a favour, and do not pinch pennies on your electrical work; you will kick yourself in the dupa abundantly if you do, unless you have an extremely small layout; and by that, I mean a shelf on a bookcase!

As an addendum; if you are unfamiliar with how electricity works, I suggest you pick up a book on its basics and read it.  This stuff isn't rocket science, but it can cause untold grief to those who approach it with no more than a gleam in their eye...

Rich
Logged

Rich

NEW YORK NEW HAVEN & HARTFORD RR. CO.
-GONE, BUT NOT FORGOTTEN!
Terry Toenges


View Profile
« Reply #18 on: February 11, 2007, 03:56:32 PM »

I'd guess that, if you are in doubt, twist a couple of them together before you run them to put your mind a little more at ease.
Logged

Feel like a fourfouro.
r.cprmier

View Profile
« Reply #19 on: February 11, 2007, 05:37:32 PM »

I think I should qualify what I said-after having read it.  #12THHN would be used for your power buss.  Phone wire could be used for controlling a pilot device- telling it what to do.  If you are running a typical DCC system, you are running it at about 16 volts, and that #12 stranded will be good for that.  Even DC; if you are using a block system at twelve VDC, you will still be better off.

I am using #18 to run my turnout machines (Tortoise); but the machines have two three-wire configurations that can be used for controlling block-related devices.

Rich
Logged

Rich

NEW YORK NEW HAVEN & HARTFORD RR. CO.
-GONE, BUT NOT FORGOTTEN!
Jim Banner

Enjoying electric model railroading since 1950.


View Profile WWW
« Reply #20 on: February 11, 2007, 07:43:23 PM »

Rich, may I suggest doing the calculations.

Typical DCC booster produces 5 amps at 16 volts.  Required voltage at locomotive is 14 volts.  So allowable voltage loss is 2 volts. 
Question: what is the maximum run using #20 wire to keep within the 2 volt loss?  For #12 wire?

Answer: Maximum allowable wire resistance is 2 volts/5 amps = .4 ohms.
#20 copper wire has a resistance of 10 ohms per foot.  So the maximum length of wire = (.4/10) x 1000 = 40 feet, out plus back.  Maximum run = 40/2 = 20 feet.  And this assumes that all of the 5 amps is being drawn through the entire length of that one run of wire.  On a small layout (4' x 8'), you would have to have a very convoluted wiring layout to exceed 20 feet.

The resistance of #12 copper wire is 1.6 ohms/1000 feet.  So the maximum length of wire = (.4/1.6) x 1000 = 250 feet, out plus back.   Maximum run is 125 feet. 

Why would you want to use wire with that kind of capability on a small layout?

In fact, with DCC, you could not use that length even on a large layout because of the parallel capacitance and series inductance of the pair of wires.

Just for fun, lets look at #24 wire and Tortoise switch machines.  This is a low power application where the switch machine draws only 50 milliamps at 12 volts dc.  A Tortoise machine in good condition will throw a turnout at as little as 8 volts.  So what length of #24 wire could you use?  Well, a 4 volt drop at 50 milliamps is 4/.05 = 80 ohms.  That is, we can have up to 80 ohms of wire in the circuit and still have the Tortoise machine work.  The resistance of #24 copper wire is 26 ohms per 1000'.  So we could use an astounding (80/26) x 1000 = 3000 feet, out plus back, or a run of 1500 feet.  However, take note of the proviso "in good condition."  Suppose a Tortoise on the point of wearing out needs 11 volts to run (this number is a guess, I have never measured current consuption on an almost worn out Tortoise.)  Then the length would drop to about 400 feet, more or less.

These are just applications of Ohm's Law, which says the voltage across a resistance equals the current through the resistance multiplied by the resistance.  That is, Volts = Amps x Ohms.  This can be stated two other ways (if your algebra is a little rusty.)  Ohms = Volts / Amp  and Amps = Volts / Ohms. 

The numbers for the resistance of copper wires can be found at any number of websites - just Google "resistance of copper wire."  These calculations ignore the increase in resistance of copper wire when heated.  This is minor at temperatures below the melting point of the insulation.  Listed ampacity of #20 wire is around 3 amps, which would be the limit for bell wire if installed in an insulated wall.  In free air, the limit is about double that.  And as mentioned before, #20 appliance wire, which has higher temperature insulation, has a capacity in chassis work of around 11 amps.

Logged

Growing older is mandatory but growing up is optional.
ddellacca

View Profile
« Reply #21 on: February 11, 2007, 07:48:39 PM »

Stewart,

Why do you take it out on me, when Jim Banner has also previously told you that Amps drive up the heating effect to blow the wiring with:-


Take what out on you?

I merely stated what I have learned in over 50 years of electricity,
electronics, and computers.

If you think that is picking on you, so be it.

Dick
Logged
SteamGene

View Profile
« Reply #22 on: February 11, 2007, 07:55:33 PM »

Question.  My under construction layout is around three walls with two penninsulas.  One wall is 26' and the other two have 17' going to the penninsulas, which are about 9 feet each.  I'm planning on using DCC and have a mostly steam fleet, most of which have or will have sound decoders in them.  What size for the bus wires and what for the feeders.  My plan is to have the command station approximately centered on the 26' side.
Gene
Logged

Chief Brass Hat
Virginia Tidewater and Piedmont Railroad
"Only coal fired steam locomotives"
Nathan

View Profile
« Reply #23 on: February 11, 2007, 09:42:26 PM »

SteamGene,

Assumption: HO Scale Layout, one 5 Amp output used on the whole layout, normal draw when running your trains is say 3.5 Amps. This allows for the times that you may have and extra train running or are throwing turnouts and still stay under 5 amps.  If I read what you say right the longest distance from the DCC unit to the end of track is 39 feet.

Run several feeders from the DCC unit to the track, lets say one feeder to each main line on the 26 foot wall, one to each main line on each 17 foot wall and one to each main line on each 9 foot section.  At each yard run a feeder to the lead track, the same to each engine terminal.  Here you could use #16 wire, two single wires each feeder,  one pair #16 'zip' cord each feeder, or one pair #16 'trailer' wire each feeder.  Then a short pair of #18 or #20 wires, lets say about 3 inches long, at the end of each feeder to the track.  It would not hurt to add additional feed points off each feeder pair to each track in the engine terminal or yard.

If you will have more then 3.5 amps normal draw you will want to split the layout into sections with a 5 amp booster for each section, if you main unit is 5 Amps, say one extra booster if you feel that no section would normally see more the 3.5 amps.  Place the main booster along one 17 foot wall and the other one on the other 17 foot wall, splitting the load and running multiple feeders from each one to the areas each serves.

Lets say you want to use a system like the NCE Power Cab.  The power cab it self will not run the layout, but you could use the 3 Amp Smart Booster and several DB3 3 Amp Boosters.  Here you would want to split the layout so that one booster handles the engine terminal, one for each major operating area, and maybe one for a branch line.  Then you could put each booster at the area it serves, running the control buss from the Smart Booster to the other boosters and then running several #18 or #20 wires from each booster to several places in each section it serves.

Nathan
Logged
SteamGene

View Profile
« Reply #24 on: February 11, 2007, 10:34:50 PM »

Thanks, Nathan.  I will use Digitraxx, if , for no other reason, my club uses it and other local modelers do, too.  I had planned on using at least two boosters. 
I think you solved my wire gauge question.
Gene
Logged

Chief Brass Hat
Virginia Tidewater and Piedmont Railroad
"Only coal fired steam locomotives"
Seasaltchap

View Profile
« Reply #25 on: February 12, 2007, 10:26:52 AM »


Plainly, it is better to over engineer, rather than under engineer where the costs are marginal to use #12, and the belt n' bracers of "50 feet out, and 50 feet back" make a lot of sense to also have the protection of a ring-main for the main track supply.

Working with old balls of wiring found on shelves in the garage, I think, is courting disaster.

Regards
Logged

Phoenix AZ: OO enthusiast modelling GWR 1895-1939, Box Station Wiltshire; S&DJR Writhington Colliery, Nr. Radstock.

Interested in making friends on the site with similar interests.
Jim Banner

Enjoying electric model railroading since 1950.


View Profile WWW
« Reply #26 on: February 12, 2007, 01:36:14 PM »

I see there is still a great deal of confusion when it comes to matter electrical.  But Stewart has one thing absolutely right - It is better to over engineer than under engineer.  If you don't know the numbers or don't want to do the calculations, then err on the side of caution.  It is a lot easier and cheaper to put in a larger than necessary wire the first time than it is to tear out and replace too small a wire.

Perhaps this would be a good time to clear up some of the other confusion.  Just so that we are all on the same page, there are some definitions at the bottom of the page.

Electrical Power is the product of voltage and current.  That is,

Watts = Volts x Amps

But we also know from Ohm's Law that

Volts = Amps x Ohms

So we can calculate electrical power some other ways, for example:

Watts = Volts x Volts / Ohms

Watts = Ohms x Amps x Amps

The latter explains why a piece of wire gets hotter when you push more current through it.  The piece of wire has a constant resistance, in Ohms.  But to figure how much heat that wire produces, you must multiply its resistance by the current through it (Amps) two times .  Lets do an example.  Suppose a piece of wire has a resistance of .1 ohms and we push 1 amp through it.  The heat generated is .1 ohms x 1 amp x 1 amp = .1 Watts.  Now suppose we push 2 amps through the same piece of wire. The heat generated is .1 ohms x 2 amps x2 amps = .4 watts.  As we push more current through the wire, the wire is heated more and all else being equal, gets hotter.  But note that doubling the current increased the heating by four times, not 2 times.  Note also that without the resistance of the wire, no heating would take place no matter how high the current is.

So what about current (amps) as a measure of heating effect?  It is an exact measure of heating if the voltage is known.   When we look at one space heater versus another, we can compare them on the basis of watts or amps, it does not matter, as long as the voltage is the same.  That is a direct result of Watts = volts x amps.  But do not try to compare a 10 amps heater that is rated at 12 volts to a 10 amp heater rated at 120 volts or a 10 amp heater rated at 240 volts without also taking the voltage into consideration.  They are NOT all the same.  In fact, their heat outputs are 120, 1200 and 2400 watts respectively.

Motor rating is a farce.  Current ratings, and horsepower ratings calculated from current ratings, are based on stall current, not operating current.  You go out and buy a router.  Chance are you buy the one with the most 'power'.  I have one rated 3 horse power.  Wow!  But let's do a little figuring.  3 horse power at 746 watts per horse power is 2238 watts.  Plugged into a 120 volt outlet, 2238 watts is 18.65 amps.  Way too much for a 15 amp circuit.  What it means is if I jam the bit and bring the router to a grinding halt, it will draw 18.65 amps and produce 2238 watts of HEAT.  With no motion, there is no mechanical horse power at all.  If my router really produced 3 horse power, it would probably rip my arms off and instead of saying WOW, I would be on my head screaming MOM.

Another fallacy to put to rest is that all motors draw more current as the voltage drops.  I hear this over and over, but unless you have some fancy constant speed control circuit driving the motor, this just does not happen.  This fallacy probably arises from the fact that furnace, air conditioner, freezer and similar motors do burn out during brownouts.  But these motors all have starting windings that draw power, and lots of it, when you first turn the motor on.  As soon as the motor is up to speed, the starting winding is shut off and the motor runs on a much lighter running winding.  You may have noticed the lights dim for a second when one of these motors kicks in.  That is when the starting winding is drawing huge amounts of power to get the motor up to speed.  Now is you lower the voltage to one of that type of motor, such as during a brown out, then the motor slows down.  If it slows down enough, the starting winding kicks back it.  And then the motor is drawing more current trying to get its speed back up.  If it cannot get up to speed, it runs with the starting winding engaged until it burns out.  Our dc motors in our model locomotives have no such starting windings.  If we turn down the voltage, the current decreases, unless you physically change something else, like the grade or the load.  There is one exception - if you run on DCC and are using back emf, which is a fancy constant speed control circuit.

One last comment about wattage ratings.  The last two microwave ovens I bought have both input and output power ratings.  The newest is 1000 watts output (to the cooking chamber,) 1200 watts input (from the power line.)  The much older one was rated 700 watts out, 1350 watts in.  Maybe this is a Canadian thing to stop false advertising - do you remember them selling "90 watt stereo amplifiers" when the only 90 watt rating on the whole thing was the amount of power it drew from the power line?

Now let's get out there, turn on our trains, and play nice!

ENERGY - the capacity to do work.  It can be potential energy (mass at a height, the energy stored in your neighbour's car parked up on his driveway,) kinetic energy (mass in motion, kinetic energy it what crumples your fenders when your neighbour's car rolls down his driveway and coasts into your car,) chemical energy (released by chemical reactions,) and thermal energy (heat.) Measured in joules, kilowatt hours, calories, foot-pounds per second, pound-feet, etc.

WORK - expenditure of energy, such as combining fuel and oxygen (chemical energy) to blast a rocket into space (potential energy as the rocket is raised off the ground plus kinetic energy of its motion.  As we do work, we convert one form of energy into another and ultimately into heat.

POWER - work or energy expended per unit of time.  Measured in joules/second, watts, calories per minute, horse power, foot-pounds per second2, pound-feet/second, etc.

VOLTAGE - electrical pressure, measured in volts

CURRENT - electrical flow rate, measured in amperes (amps)

RESISTANCE -  impeding electrical flow, measured in ohms.

Logged

Growing older is mandatory but growing up is optional.
ddellacca

View Profile
« Reply #27 on: February 12, 2007, 05:38:56 PM »

Jim,

Your latest post may just be the shortest post I have seen for quite some
time which actually explains your subject pretty much correctly.
The really complete explanations have been known to require a couple of hundred pages and a course at an educational institution.
Notice I did not state "college". College, university, trade school, high school, military, are all appropriate.

Thanks for doing it well,

Dick
Logged
Pages: 1 [2] Print 
« previous next »
Jump to:  
Powered by SMF | SMF © 2015, Simple Machines Valid XHTML 1.0! Valid CSS!