"Doug Miller" wrote in message news:WKHZd.10109
Regarding;
How much water is in a 100 foot by 1/2 inch copper tube?

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

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

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!