# Keeping well reservoir/pipes from freezing.

### #1

Posted December 16 2011 - 09:49 AM

Now that winter is approaching, albeit a generally mild NC winter, I guess I need to insulate everything. I have a 110v electric socket out there to work with. Would R13 insulation along with an incandescent light bulb (maybe on a thermostat) be enough? My only concern is to keep the well house (5'x5'x5') above freezing. It's rare that we have sub-32degree weather for any length of time, but I'd rather be prepared. Can anyone point me in the right direction?

### #2

Posted December 16 2011 - 10:22 AM

### #3

Posted December 16 2011 - 01:51 PM

probably cheaper in the long run than just leaving a flood light on.

### #4

Posted December 16 2011 - 02:08 PM

What?

### #5

Posted December 16 2011 - 05:30 PM

I have a well head sticking out of the ground with a fake (fiberglass) rock covering it. It has never frozen.

### #6

Posted December 16 2011 - 06:08 PM

Do the math for yourself.

Q = U x Area x temp diff.

Q is BUTH or the amount of heat

U is the heat transfer rate or the reciprocal of R or more commonly known as R value

Area is the square foot area of all the walls and ceiling of your shack.

Temperature difference is the diff you anticipate between .....say 40 and the lowest temp you'll see.

If you want to use light bulbs, 100 % of the electric they use will turn into BTUH's and you'll make or more correcly put turn 100 watts of light into 341 BTUH'S as there is 3.412 BTUH's per watt.

Pretty simple really .

I can look up the U values if you want, or you can use google to try and find it, off the top of my head plywood and R-13 will have a U value of .6.

### #7

Posted December 16 2011 - 06:47 PM

### #8

Posted December 17 2011 - 12:35 AM

^ that is a pitless and it's nothing like the OP is talking about, your pump is in the ground his is above ground. I bet I have the same fake rock on mine.

Do the math for yourself.

Q = U x Area x temp diff.

Q is BUTH or the amount of heat

U is the heat transfer rate or the reciprocal of R or more commonly known as R value

Area is the square foot area of all the walls and ceiling of your shack.

Temperature difference is the diff you anticipate between .....say 40 and the lowest temp you'll see.

If you want to use light bulbs, 100 % of the electric they use will turn into BTUH's and you'll make or more correcly put turn 100 watts of light into 341 BTUH'S as there is 3.412 BTUH's per watt.

Pretty simple really .

I can look up the U values if you want, or you can use google to try and find it, off the top of my head plywood and R-13 will have a U value of .6.

So if the U value is the reciprocal of R, that would make the U value of R-13 insulation 1/13, or about 0.076. Since the shed is approx 125 sq ft.,

Q = U x Area x Temp Diff

341 = .076 x 125 x Temp Diff

341 = 9.5 x Temp Diff

35.89 = Temp Diff

So, R13 and a 100 watt light should hypothetically keep everything above a toasty 40degrees as long as the temp doesn't dip below 5degrees, correct? I'm sure some other factors come into play, since it's not a sealed environment, but that gives me a big margin of error.

Or I actually considered plan B....just draping an old electric blanket over the damn thing and plugging it in on cold nights.

### #9

Posted December 17 2011 - 07:29 AM

Cabo

### #10

Posted December 17 2011 - 07:55 AM

^ that is a pitless and it's nothing like the OP is talking about, your pump is in the ground his is above ground. I bet I have the same fake rock on mine.

Do the math for yourself.

Q = U x Area x temp diff.

Q is BUTH or the amount of heat

U is the heat transfer rate or the reciprocal of R or more commonly known as R value

Area is the square foot area of all the walls and ceiling of your shack.

Temperature difference is the diff you anticipate between .....say 40 and the lowest temp you'll see.

If you want to use light bulbs, 100 % of the electric they use will turn into BTUH's and you'll make or more correcly put turn 100 watts of light into 341 BTUH'S as there is 3.412 BTUH's per watt.

Pretty simple really .

I can look up the U values if you want, or you can use google to try and find it, off the top of my head plywood and R-13 will have a U value of .6.

So if the U value is the reciprocal of R, that would make the U value of R-13 insulation 1/13, or about 0.076. Since the shed is approx 125 sq ft.,

Q = U x Area x Temp Diff

341 = .076 x 125 x Temp Diff

341 = 9.5 x Temp Diff

35.89 = Temp Diff

So, R13 and a 100 watt light should hypothetically keep everything above a toasty 40degrees as long as the temp doesn't dip below 5degrees, correct? I'm sure some other factors come into play, since it's not a sealed environment, but that gives me a big margin of error.

Or I actually considered plan B....just draping an old electric blanket over the damn thing and plugging it in on cold nights.

Air infiltration is the only other heat loss that should take place, so a tight weather stripng on the door to the shack is a good idea. Half of your calculated loss would be a good guess for the loss due to infiltration.

A transformer and a cheap thermostat operating a relay to turn on and off a light would give you a hassle free way to insure your water works, it's how I'd do it if it were my place. But then again I have that stuff on hand so there is really no cost for me, maybe a cheap electric heater with a built in thermostat like Ted sugjested is right for you.

At the very least you know how much heat is needed now.