# Math issues - Amount of water in a 1½ inch pipe

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• posted on November 9, 2011, 11:44 am
I'm no good at math. I'm trying to determine the amount of water in a 1½" (one and one half inch) galvanized steel pipe, per foot. My reason for this is because I'm trying to calculate the water weight in a well pipe. Here's the issue. My well has 300 feet of pipe going to the submersible pump. I found the weight of the actual pipe, per foot. Also the weight of the pump, as well as the wire.
One foot of 1.5" diameter schedule 40 galvanized pipe weighs 2.72 pounds. The pump is about 25 lbs. The wire weighs about 21 lbs per 100 feet.
So far I have 300 x 2.72 = 816 lbs for the pipe 300 feet of wire is 63 lbs plus 25 lbs for the pump
That totals 904 lbs.
There are a few other small parts such as the fittings, foot valve (if there is one), etc. But I know the pump is actually set at 292 feet, so I can knock off a few pounds for that, but these small parts will add a little. Also, this pump is 40 years old, so it may weight a little more than the new ones. Either way, I can assume this whole thing weighs around 900lbs.
However, there is water in the pipes and that is likely a significant amount of weight added. My problem occurs here. How the heck does a person calculate the amount (or weight) of the water in one foot of 1.5" pipe? (Or in 10 feet or 100 feet)?
I found online that the weight of one gallon of water is approximately 8.35 lb. Now I only need to figure out how to determine the amount of water in the pipe.....
Any math experts out there?
(I hope it's not over 100 lbs, because my tractor loader cant handle over 1000 lbs of lifting weight).
thanks

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<%-name%>
• posted on November 9, 2011, 12:03 pm
snipped-for-privacy@myplace.com wrote:

volume is pi*(radius squared)* length. for 300 feet I get 6358 cubic inches > 27.6 gallons > 230 pounds. Somebody should check my math, 'cause I'm senile.
You could always pump out the water.

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<%-name%>
• posted on November 9, 2011, 12:58 pm

That sounds about right, just from guessing.... I'm senile too :(

Not really, there is a foot valve or built in one way valve, and since the pump is under water, it wont come out as air.

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<%-name%>
• posted on November 9, 2011, 1:43 pm
On 11/9/2011 7:58 AM, snipped-for-privacy@myplace.com wrote:

Assuming 1.5 is id, then calculation is right.

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<%-name%>
• posted on November 9, 2011, 3:26 pm

You're correct but note that the water in the pipe below the well level doesn't count.

Isn't that what the pump is for? ;-)

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<%-name%>
• posted on November 9, 2011, 3:47 pm
wrote:

According to my ACE Pocket Reference guidebook, "Weight of Water in a Pipe" is formulated as: Pounds Water = Pipe Length feet x (Pipe diameter inches(squared)) x 0.34
So based on that formula, Pipe Length feet = 300 (Pipe diameter inches (squared)) = 1.5 squared = 2.25
so the formula = 300 x 2.25 x 0.34 = 229.5 pounds
Robin

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<%-name%>
• posted on November 9, 2011, 4:09 pm

Simple formula for a cylinder X constant (weight per unit water).

...and this disagrees with what I said, how? You still have to subtract the weight of the water in the submerged pipe.

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<%-name%>
• posted on November 10, 2011, 1:51 am
snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

Please explain. Thought experiment: Assume no friction. Drain the water outa the pipe. Pull up on the pipe to raise it off the supports. Weigh the pipe. Pour a pound of water into the pipe. Weigh it again. Does the weight not increase by a pound?
What counts is the water displaced by the pipe below the water level. Contents of the pipe does not affect that.

Good point. Because of the depth, You'd need a pump that could be lowered down into the 1.5" pipe. Or maybe a lot of trips with a sponge, or a hollow cup. or a 1: plastic pipe with a check valve on the end. ;-)

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<%-name%>
• posted on November 10, 2011, 2:17 am

Certainly. All of the water is now above the well's water line. It's not displacing the equivalent well water. It's not "floating".
To be accurate, one could also subtract the weight of the water the pipe is displacing, too.
W(pipe) = Pi(Ro^2-Ri^2)*D*L Where Ro = outer radius of pipe Ri = inner radius of pipe D = Density of pipe L = Lenght of submerged section

No, what counts is the water *above* the well's water line. If the pipe were full of air it would "float" on the well's water. That is, it would be *lighter* by the equivalent of the weight of the water that would be in the pipe (minus the small weight of the air).

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<%-name%>
• posted on November 10, 2011, 3:03 am
snipped-for-privacy@att.bizzzzzzzzzzzz wrote:

I don't buy it! The PIPE is experiencing buoyancy forces. What's inside the pipe is irrelevant as long as the pipe contents doesn't change the position of the pipe relative to the water in the well. As you raise the pipe, it will take more force because of the reduced buoyancy. Still, the change is unrelated to what's in the pipe. It's all about the displacement of the PIPE.

The system weight will include the weight of the included water.
Here's another thought experiment: Drain the pipe and set up to weigh it as before. Pour in a pound of water. Insert a bladder at the 20 foot level and blow it up to seal the opening. Position is arbitrary as long as it's above the water level on the outside of the pipe. Weigh the pipe.
Drain the pipe again. Put in the bladder at the 20 foot level to seal the pipe. Pour in a pound of water. Convince me that the weight (force to lift the system off its support) depends on where the water is in the pipe.
The more I think about it, I think we're in heated agreement.
I calculated the maximum weight of the water if the pipe were full. What matters is the actual volume of water in the pipe.
I neglected the separate opposing force of buoyancy due to the volume of external water displaced by the pipe (and the pump). As the pipe is raised out of the water, that force goes to zero. The weight of water inside the pipe is unchanged...unless there's a leak, in which case, the original amount of water in the pipe is irrelevant.

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<%-name%>
• posted on November 10, 2011, 5:50 am

Exactly.
If there was nothing in the pipe it would weigh exactly as much as the pipe alone. If there is water in the pipe it weighs as much as the pipe + water. Obviously what's inside matters.
Now, stick it in water. Same thing.

You'e only lifting the water that's above the well level. Water lower than that is neutral.

Wrong. If there were air in the pipe it would be lighter, lead, heavier.

No, the water matters, but only that above the well level. The weight of the water in the pipe below that level is canceled by the water in the well (water weighs the same as water).

Only above the well level. Below that, the weight of the water in the pipe is canceled by the water in the well. You're only lifting the water above the well level.

A pound is a pound. It's not *where* the water is. It's how much you're lifting. You're not lifting water below the well level. That's a wash.

Which you've lost. ;-)

..ad wasn't supported by any water in the well. The real weight will be reduced by the weight of the water in the pipe below the well level *PLUS* the weight of the pipe below that level minus the weight of that volume of water.

No, only the water actually being lifted. The water in the pipe below the well level is a wash, since it has the same weight as the equivalent volume of well water.

It's now above water, so it counts. That which is still below the well level doesn't.

The volume of the pipe (and water in the pipe) below the water has changed, though.

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<%-name%>
• posted on November 10, 2011, 5:54 am
wrote:

I'm the OP.
As of today, I know for certain that the pump is bad (or a wire shorted in the well). It wont start anymore and it blows the circuit breaker. I located another well driller and had him come for an estimate. He opened the control box and said the overload protector was almost hot enough to light a cigarette. I felt it, and it sure was. He said the pump may have seized up, or pump motor has bad windings, or there could be a barespot on the cable. Either way, it all has to be pulled. His estimate was better than the first one I got, but still quite high if you ask me.

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<%-name%>
• posted on November 10, 2011, 12:28 pm
On Nov 10, 12:54 am, snipped-for-privacy@myplace.com wrote:

just to play devils advocate, strictly to prep you for a really bad day:( and I hope this doesnt occur!.
that pump has been in place for 40 years.
what if you try pulling the pump and the pipe breaks? Perhaps the well has collapsed:( nation and world wide we have had a lot of earthquakes....
Are you prepared to fund a new well?
If you try pulling it yourself and things break the well driller may not be able to get up whats left of things and recommend a new well, to say 400 feet.. you might want to kick yourself wondering if a pro would of gotten it up OK?
whats the going rate for a new well in your area?

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• posted on November 10, 2011, 2:36 pm
snipped-for-privacy@myplace.com wrote:

Are you positively absolutely sure the Motor is bad? Is there any possibility of rewiring at the surface to make it run backwards? Or maybe hook up your air compressor and put some back pressure on it. Be a shame to pull it and find it was jammed up by a pebble that coulda been expelled by reversing it or forcing air backwards thru it. Probability may not be high, but the potential reward is great.
A time domain reflectometer might help determine if and where the cable is shorted. Cable TV guys and phone guys use 'em all the time to find the distance to a short. Be a shame to pull it and find the short was two feet down.
When your business is pulling pipe out of wells, everything looks like it needs the pipe pulled. As an unbiased observer, you have more diagnostic options.

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• posted on November 10, 2011, 11:05 am

An example of why the US has such low science and math scores.
Yes, how much of the pipe is submerged and subjected to bouyant forces is a factor. But so is the amount of water in the pipe that is ABOVE the water level. Unless you're going to claim that a pipe that is empty weighs the same as pipe that is full of water.
Basicly, the part of the pipe below the water has a buoyant force on it that approximately cancels out the weight of the water inside the pipe. It would exactly cancel it if the water volume inside the pipe equalled the water displaced. But it does not because you also have the volume of the metal of the pipe.
The water in the pipe above the water line adds directly to the weight that must be lifted.

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<%-name%>
• posted on November 11, 2011, 4:36 am
wrote:

Water above the well head drains out as the pipe is lifted.
Harry K

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<%-name%>
• posted on November 11, 2011, 12:06 pm

How does that happen with a check valve at the submersible pump?

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• posted on November 11, 2011, 2:56 pm

200 feet of pipe in air will never occur, the pipe must be cut into pieces as its pulled for easier handling

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• posted on November 12, 2011, 4:23 am

Exactly. If using iron pipe it is unscrewed one joint at a time, usual y 20' but 10' sections are also seen. Of couse modern wells lusually use plasttict which is passed over a wheel and stretched along the ground.
Harry K

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<%-name%>
• posted on November 11, 2011, 3:52 pm
wrote:

No, it's taken/cut apart as it's pulled.

After it's cut, sure. The pipe that's cut off no longer counts in the pull-weight calculation either.