Pipe Volume Math Question

Hello,

I am filling a pipe with glycol.. The pipe is 1/2" (inside diameter).
I need to know how many imperial gallons per 100 feet this pipe will hold so
I can figure out how much glycol to buy.
Thanks for any help on this. Jim

 

3.397301 Gallons

 

Ask a scientist, or try this formula.
--

Question From:
Sammy, Age 8.

My Dad and I are going to build a sail boat using PVC pipes for flotation.
Is there an easy way to caculate how many 8 foot long, eight inch diameter
pipes we will need to support our weight and keep the boat floating?


Answer:

The weight that the pipes will support is the weight of the water they can
displace minus the weight of the pipes themselves. To calculate this, you
need to know the volume of the pipes, the weight of the pipes, and the
density (weight/volume) of water.

Fresh water has a density of 1 kg/liter. Salt water has a higher density.
The volume of the pipes can be calculated from their diameter and length
by the formula

V = L pi r^2, where
L is the length of the pipe, 8 ft = 96 in.,
pi = 3.14159,
r is the radius of the pipe, 4 in.

So the volume of a pipe is

V = (96 in)(3.14)(4 in)^2
= (96 in)(3.14)(16 in^2)
= 4823 in^3

Now you need to convert cubic inches to liters. 1 cubic inch = 0.01639
liters, so

V = (4823 in^3)(0.01639 L/in^3)
= 79 L.

So, in fresh water, an 8-ft length of 8-in diameter PVC pipe can float 79
kg minus its own weight.

[attrib lost, from the 'net]

 

http://www.inter-mountain.com/Pipe_Related_Formulas.htm

 

First-form maths!

The volume of a cylinder (a pipe is a cylinder) is (22*r*r*h)/7 where r is
the radius (half the diameter) and h is the height (or length, if you
like), and 22/7 is pi. Make sure that you use the same units (either feet
or inches) for both radius and height!

1 imperial gallon is 277.4193 cubic inches or 0.160544 cubic feet.

 

It is amazing that anyone does these calculations in imperial measure now.

 

Numerator = cross-sectional area of pipe cm^2 x pipe length cm x ImpGal
Denominator = conversion units for ImpGal cm^3

Values: 0.5 " ID = 0.25" radius = .635 cm
Xsec = pi x r^2 = pi x .635^2 cm^2

100' = 30480 cm = 3 x 10^4 cm
ImpGal = 4546.09 cm^3 = 4.546 x 10^3 cm^3

Numerator = pi x .635^2 x 3.048 x 10^4 cm^3 x ImpGal
Denominator = 4.546 x 10^3 cm^3

I get 8.49 Imp Gal

It seems like 100 feet of 1/2 inch pipe could hold something less than
10 gallons of liquid.

 

but i'd purchase 3.397302 gallons , just to be on the safe side.

I think it's a school math problem though

 

Perhaps he's got these old pipes and hasn't heard that shops now have to
sell stuff in litres?

 

I think you let your decimal point slip a bit somewhere in there.

My method:

22*.25*.25*1200/7=235.7142857142857143 cubic inches

235.7142857142857143/277.4193=0.8496679420439951881 imperial gallons

 

I've changed my mind on my previous answer and reckon its 0.8493253 Gallons

 

I'm trying to figure out why anyone would be filling 100 feet of 1/2"
pipe with glycol, presumably ethylene glycol.

 

1 cm of that pipe will hold 1.27 cc, that sounds right, and a meter 127
cc and 30 meters would be 3810 cc, so I think your answer is better than
mine. I must've misplaced a decimal somewhere.

 

Aha! There it is. 100' = 3048 cm = 3 x 10^3 cm

pi x .635^2 x 3.048 x 10^3 cm^3 x ImpGal

Now it gets 0.849

 

Right there.

That sounds pretty close. ;-)

 

Then is my volume calculator borkd?
I get 235.62 cubic inch = 0.849324351255694 gallon [UK]

 

To stop it from freezing?

 

1/2" pipe is seldom 0.5 inches inside diameter..
See:
http://www.keidel.com/mech/pvf/pipe-intro.htm
Depending on your type of 1/2" pipe, the responders may have to correct
their calculations by a point or two??

 

Copy/paste from Kcalc; odd when being lazy makes it easier to get answers
to a silly degree of precision than to get a rough aproximation. Let's
say 'a bit less than 4 litres' to keep things sane )

My value for pi is approximately 0.04% on the high side - but it's better
to have a drop left over than being a drop short, and differential thermal
expansion and manufacturing variations are probably going to swamp that
margin.