Dimensions and Physical Characteristics of Copper Tube: TYPE ACR 3/4 OUTER IS .750 INNER IS .666 CONTENT GAL .242 PER FT

Dimensions and Physical Characteristics of Copper Tube: TYPE M 3/4 OUTER IS .875 INNER IS .811 CONTENT GAL FT .269 PER FT

Dimensions and Physical Characteristics of Copper Tube: TYPE L 3/4 OUTER IS .875 INNER IS .785 CONTENT GAL FT .251 PER FT

http://www.copper.org/publications/pub_list/pdf/copper_tube_handbook.pdf

What was the ID?

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FF

FF

bill a wrote:

What is the ID?

What is the ID?

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FF

FF

I am putting a tankless gas water heater in my home. All of the copper
is in the concrete slab so I am going to run an insulated copper tube
"up and over" to the kitchen. The water heater will be a foot from the
two bathrooms and showers. the "up and over" tube will feed the kitchen
sink and dish water - nothing else.

How much water is in a 100 foot by 1/2 inch copper tube? How much water is in a 100 foot by 3/4 inch copper tube? How much water is in a 100 foot by 1/4 inch copper tube?

I am too lazy to look it up - I am wondering if any of you experts on here have the info off-hand. I am thinking of putting thinning tubing to the kitchen because less water would be in it to cool down etc.

Right now I am running two 50 gallon electric water heaters in a house for two adults - I am wasting alot of energy keeping all that water hot - and the tanks are far away from where the hot water is needed anyway running thru a cold concrete slab.

Harry

This is ReRe

1/2" = .0158 gallons per lineal foot or 1.58 gallons in 100' 3/4" = .0277 gallons per lineal foot or 2.77 gallons in 100'

How much water is in a 100 foot by 1/2 inch copper tube? How much water is in a 100 foot by 3/4 inch copper tube? How much water is in a 100 foot by 1/4 inch copper tube?

I am too lazy to look it up - I am wondering if any of you experts on here have the info off-hand. I am thinking of putting thinning tubing to the kitchen because less water would be in it to cool down etc.

Right now I am running two 50 gallon electric water heaters in a house for two adults - I am wasting alot of energy keeping all that water hot - and the tanks are far away from where the hot water is needed anyway running thru a cold concrete slab.

Harry

This is ReRe

1/2" = .0158 gallons per lineal foot or 1.58 gallons in 100' 3/4" = .0277 gallons per lineal foot or 2.77 gallons in 100'

Refigeration tubing is ID, Plumbing pipe is sized by OD.

--

Bob Pietrangelo

snipped-for-privacy@comcast.net

Bob Pietrangelo

snipped-for-privacy@comcast.net

Click to see the full signature.

Bob Pietrangelo wrote:

Bob, Check your references. You have it backwards. Refrigeration tubing (ACR) is sized by the actual OD. Plumbing & heating tubing is sized by the nominal I.D. Therefore 1/2" plumbing tubing has an actual OD of 5/8". I just checked my textbooks to verify this so as not to give dis-information.

Source: Refrigeration & Air Conditioning Technology, 3rd Edition, 1995, page 99.

Bob, Check your references. You have it backwards. Refrigeration tubing (ACR) is sized by the actual OD. Plumbing & heating tubing is sized by the nominal I.D. Therefore 1/2" plumbing tubing has an actual OD of 5/8". I just checked my textbooks to verify this so as not to give dis-information.

Source: Refrigeration & Air Conditioning Technology, 3rd Edition, 1995, page 99.

wrote:

Hi - No need to argue. I just wanted to find out how much water was in a 1/2 inch by 100 foot copper tubing. Just a ball park amount. I was surprised at just how little an amount of water it is. I am not worrying about wasting water - I am trying to cut down the time it takes for hot water to get to my showers. I am surprised it hold less than a gallon of water. If I insulated that 100 feet of copper tubing - the water in it will not cool off so quickly. Presently it is running through a 3/4 copper tube that is buried in the concrete slab. It is a double whammy - since it is in concrete and not insulated - you lose heat thru the heat sink - and since it is 3/4 copper the tube holds much more water. Harry

Hi - No need to argue. I just wanted to find out how much water was in a 1/2 inch by 100 foot copper tubing. Just a ball park amount. I was surprised at just how little an amount of water it is. I am not worrying about wasting water - I am trying to cut down the time it takes for hot water to get to my showers. I am surprised it hold less than a gallon of water. If I insulated that 100 feet of copper tubing - the water in it will not cool off so quickly. Presently it is running through a 3/4 copper tube that is buried in the concrete slab. It is a double whammy - since it is in concrete and not insulated - you lose heat thru the heat sink - and since it is 3/4 copper the tube holds much more water. Harry

Let me write this down first you are probably right and I worked a hard day
today.

7/8 Refrigeration equals.........3/4 Plumbing

I just bought a few hundred refrigeration fittings today too!

My bad, thanks for the correction. My brain is right my hands are stupid today

7/8 Refrigeration equals.........3/4 Plumbing

I just bought a few hundred refrigeration fittings today too!

My bad, thanks for the correction. My brain is right my hands are stupid today

--

Bob Pietrangelo

snipped-for-privacy@comcast.net

Bob Pietrangelo

snipped-for-privacy@comcast.net

Click to see the full signature.

Dear William - Thanks for the chart. I appreciate all the work. Does that mean 100 feet of 3/4 inch holds .282 gallons? 1/2 inch holds .120 gallons? 1/4 inch holds .026 gallons? I expected 100 feet of pipe to hold so much more. Harry

Harry Everhart wrote:

No. That's CUBIC FEET. Gallons per 100 ft are: 0.9, 2.1, 0.2

No. That's CUBIC FEET. Gallons per 100 ft are: 0.9, 2.1, 0.2

On Tue, 15 Mar 2005 13:05:04 -0500, Harry Everhart

(ID/2) x 3.1416 x (pipe length)

Just guessing.....

tom

(ID/2) x 3.1416 x (pipe length)

Just guessing.....

tom

www.WorkAtHomePlans.com> wrote:

Not a good guess. The formula is A = pi*** radius SQUARED,
not pi *** radius / 2.

-- Regards, Doug Miller (alphageek at milmac dot com)

Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?

Not a good guess. The formula is A = pi

-- Regards, Doug Miller (alphageek at milmac dot com)

Nobody ever left footprints in the sands of time by sitting on his butt. And who wants to leave buttprints in the sands of time?

Regarding;

I agree with Doug: i.e. pi times, the radius raised to the power 2 (squared) So; Inside diameter divided by two = Radius of X.Section. In this case one half divided by 2 = one quarter. Area of X.Section = Radius squared, multiplied by pi. This is one quarter times one quarter times pi = 1/16 x 3.142 = 0.196 Length of 100 feet = 100 x 12 inches. Thus (1/4 x 1/4 x 3.142 x 1200) = 236 cubic inches. Multiply that 236 by 0.004 to get gallons = 0.9 gallons approx. (Not sure if that's US or Imperial gallons but "A bit less than a gallon" is close enough). Anybody else agree? BTW those 236 cubic inches will weigh approx 8.5 pounds. Those 8.5 pounds will require 8.5 BTUs for every degree Fahrenheit change of temperature. So if that 8.5 lbs comes out of the tank at temperature of, say, 160 degrees, sits in the pipe and cools down, to say 60 degrees it will lose 100 x 8.5 = 850 BTUs of heat. If electrically heated that's equivalent to about one quarter of a kilowatt hour (unit) of electricity. If your electricity costs 10 cents per kilowatt hour that's a waste (sort of) of 2 to 3 cents. Of course that heat, slight though it is, could end up helping to heat the house! BTW There is a very good site at < www.tedmongomery.com/convrsns/ > for those NOT too lazy to look it up! PS. In school we remembered circular area by "Two are(a) squared pies". i.e. Pies were normally round; not square. Even if we were!

I agree with Doug: i.e. pi times, the radius raised to the power 2 (squared) So; Inside diameter divided by two = Radius of X.Section. In this case one half divided by 2 = one quarter. Area of X.Section = Radius squared, multiplied by pi. This is one quarter times one quarter times pi = 1/16 x 3.142 = 0.196 Length of 100 feet = 100 x 12 inches. Thus (1/4 x 1/4 x 3.142 x 1200) = 236 cubic inches. Multiply that 236 by 0.004 to get gallons = 0.9 gallons approx. (Not sure if that's US or Imperial gallons but "A bit less than a gallon" is close enough). Anybody else agree? BTW those 236 cubic inches will weigh approx 8.5 pounds. Those 8.5 pounds will require 8.5 BTUs for every degree Fahrenheit change of temperature. So if that 8.5 lbs comes out of the tank at temperature of, say, 160 degrees, sits in the pipe and cools down, to say 60 degrees it will lose 100 x 8.5 = 850 BTUs of heat. If electrically heated that's equivalent to about one quarter of a kilowatt hour (unit) of electricity. If your electricity costs 10 cents per kilowatt hour that's a waste (sort of) of 2 to 3 cents. Of course that heat, slight though it is, could end up helping to heat the house! BTW There is a very good site at < www.tedmongomery.com/convrsns/ > for those NOT too lazy to look it up! PS. In school we remembered circular area by "Two are(a) squared pies". i.e. Pies were normally round; not square. Even if we were!

snipped-for-privacy@milmac.com (Doug Miller) writes:

Not to nitpick, but the OP had suggested pi*radius (i.e., ID/2), not pi*radius/2.

So in terms of ID, the formula is: pi*(ID/2)*(ID/2) where pi=3.14159....

Not to nitpick, but the OP had suggested pi*radius (i.e., ID/2), not pi*radius/2.

So in terms of ID, the formula is: pi*(ID/2)*(ID/2) where pi=3.14159....

blueman wrote:

The table I published above used the formula =PI()*(B2/2)^2 in the Excel spreadsheet. The B column holds the Inside Diameter. Excel supplies the right value for pi.

The table I published above used the formula =PI()*(B2/2)^2 in the Excel spreadsheet. The B column holds the Inside Diameter. Excel supplies the right value for pi.

On Tue, 15 Mar 2005 20:40:22 GMT, snipped-for-privacy@milmac.com (Doug Miller)
wrote:

Caught it after I posted it, that's why I reposted it.

Thought the second post would 'over-write' the first. Hmmmmmm......

later,

tom

Caught it after I posted it, that's why I reposted it.

Thought the second post would 'over-write' the first. Hmmmmmm......

later,

tom

On Tue, 15 Mar 2005 13:05:04 -0500, Harry Everhart

(ID/2)^2 x 3.1416 x (pipe length)

Sorry to square the radius.

So should be radius squared times pi times length of straight tube.

hth,

tom

(ID/2)^2 x 3.1416 x (pipe length)

Sorry to square the radius.

So should be radius squared times pi times length of straight tube.

hth,

tom

Yes, we do.

Edwin Pawlowski wrote:

LOL. I was somewhat surprised to see Harry get all of the help he did get, after that remark. Now. Who's he going to get to do the work? ;o)

... This is a good group of people.

LOL. I was somewhat surprised to see Harry get all of the help he did get, after that remark. Now. Who's he going to get to do the work? ;o)

... This is a good group of people.

People can sense honestly. :-)

I really like this group - no one ripped me - and I got great answers. Now I must decide whether I want to run 100 feet of 1/4 inch - 1/2 inch - 3/4 inch to a kitchen sink and dishwasher 100 feet away from the tankless water heater. The tube will be overhead and insulated.

Opinions on that?

Harry

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