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Hotwire to thaw frozen fridge waterline Answered

My fridge water dispenser waterline gets frozen from time to time. Im thinking about inserting (and leaving there) a thin piece of wire (insulated, folded by half) through the line with the two ends left out, so when water freezes i will just connect a battery for a few minutes to get the wire warm and thaw the ice. Question is what battery can I use to heat about 25 inches of wire, and what gauge should I use (the thiiner the better). I dont want need too much heat as I dont want to risk melting the plastic water line.

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Downunder35m

23 days ago

Wouldn't recommend this approach...

I know so far two types of these water dispensers, if yours if another let me know...
The most basic ones use little storage tank hidden in the fridge and the cooling coils cool it whenever the fridge is cooled.
The more modern one are so called On-Demand units and have a seperate tiny compressor to cool the water while it is running through the pipes.
The first type should never freeze, if it does then the fridge or freezer compartment are set too cold (or the thing is opened too frequently).
The second type is desgined to work with incoming water at a temp higher than 10 to 12°C.
Depending on the time of the year and area you are in the water from your tap can be already much colder.
Without a temp check of the incoming water these units tend to cool the water below freezing point, especially if the flow rate is slow.
In most cases though you will still find an adjustable thermostat controlling the outcoming water or the pipe itself.
If you are lucky then simply adjusting it will fix the issue.

However, you said the water is freezing in the outgoing line.
If you are certain about this then try to locate where the cold is originating first.
In some case you have a metal pipe before it goes into some flexible tubing.
There should be no ice forming in the metal tubing as otherwise it could cause a leak over time.
For the rest it should be suffient to just add some pipe insulation from the hardware store around the waterline - or for a quick check to use bubble wrap.

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882BFTDownunder35m

Reply 21 days ago

Thanks, I think my system (GE side by side) is just a plastic water pipe running from the bottom of the door to the dispenser. No additionsl cooling besides the temperature inside the freezer going inside the door. The water out of the dispenser is roughly at ambient temp or just a little below.

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Jack A Lopez

22 days ago

The usual method for designing an electric heating element is to start with a quantity of power P (measured in watts). Then use Ohm's law (V=I*R) to find a resistance R that will dissipate that much power, when some specific voltage V is across it (since P = V^2/R), or, equivalently, when some specific current I is pushed through it (sincce P=I^2*R).

A long wire has resistance R that is proportional to its length L. R is also inversely proportional to the wire's cross-section area A (this is a measure of how thick or thin the wire is). The only other factor is bulk resistivity, rho.

R = rho*L/A = (rho/A)*L

That is a very general formula, and it can be simplified when considering a specific kind of wire (e.g. copper, nicrome alloy, etc) of a specific thickness, or gauge, which is a measure of thickness. Then (rho/A), which has units of resistance per unit length (.e.g ohm/meter), is specific to that kind of wire, and we should be able to just look that number up in a table.

Example: What is the resistance per unit length of 33 AWG copper magnet wire?

Answer: 0.6788 (ohm/meter) or 0.2069 (ohm/foot)

How much resistance in a piece of this wire 50 inches long?

Answer: R = (0.2069 ohm/foot)*(50/12 foot) = 0.862 ohm

or
R = (0.6788 ohm/meter)*(50*0.0254 meter) = 0.862 ohm

Another example: How much power would this R dissipate, if you could somehow drop about 5 volts of potential across it? How much current would this require?

Answer: I = V/R = (5.0 V)/(0.862 ohm) = 5.80 ampere

and

P = V*I = V^2/R = (5.0 V)*(5.0 V)/(0.862 ohm) = 29.00 watt

Note that power is dissipated, as heat, pretty much uniformly over the length of the wire, and over the surface area of the volume enclosing that wire.

I dunno. You said you wanted to the wire to go some length down a tube, and then make a hairpin turn and come back up the same tube; i.e. 63.5 cm (or 25 inch) down a tube, then back along the same path. I guess that would have about twice the power density as the same wire in a long straight run, 127 cm ( 50 inch) long.

For the sake of comparison, a pencil style soldering iron also dissipates about 30 watts. However its heating element is bundled up into a roughly cylindrical volume, about 3.0 cm long, with radius about 0.5 cm.

By comparison, a water (or ice) filled tube, 0.3 cm radius, 63.5 cm length.

>>> r=0.5,h=3.0,A=(2*pi*r^2 + 2*pi*r*h)
r = 0.50000
h = 3
A = 10.996

>>> r=0.3,h=63.5,A=(2*pi*r^2 + 2*pi*r*h)
r = 0.30000
h = 63.500
A = 120.26

So I would expect this long skinny tube to get a lot less hot than a soldering iron, since it is dissipating the same power through roughly 10 times the surface area.

By the way, I would not worry about the wire getting hot enough to melt the plastic (guessing HDPE), not as long as there is liquid water, or better: water plus ice, surrounding the wire. A wire inside an empty plastic tube, containing only dry air, would be another story.

Did you know latent heat is required to melt solid water ice into liquid water, or to turn liquid water into vapor water?
https://en.wikipedia.org/wiki/Latent_heat#Table_of...

And that's the reason why I am not worried about melting the plastic tube, as long as it has water or ice in it.

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882BFTJack A Lopez

Reply 21 days ago

Thank you I think it clarifies my idea. I will need to make simple adjustments as my wire is about 25 inches in total length, so 12.5 when fokded.