On Nov 9, 8:00*pm, Tony Miklos wrote:
On 11/9/2011 6:58 PM, wrote:





On Wed, 09 Nov 2011 14:54:40 -0600, *wrote:

On 11/9/2011 5:44 AM, wrote:
...

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.

...

(I hope it's not over 100 lbs, because my tractor loader cant handle
over 1000 lbs of lifting weight).

As others said, it'll add another 200 lb+/-

But, you're forgetting about what the breakout force will be; the
snubbers and that it's been sitting in place for however long means the
initial friction forces will likely be close to that of half the weight
or perhaps even more. *I don't think you have any chance w/ your front
loader in a direct lift; not to mention the hassle of trying to deal
with such a rube goldberg setup.

I was just informed that the drop pipe is actually inch and a quarter,
not inch and a half. That takes off a few pounbut I know the initial
breaking loose the pitless and all that is gonna be tough.

If your bucket lifts 1000lbs you have two (or three) other things to
consider. *The pivot action of the bucket tilting up is probably well
over 1000lbs. of force. *That should be enough to break it loose. *Also,
you could remove the heavy bucket and put something else across the arms
to use to lift with. *Now seriously, if I were going to attempt
something like this I would be buying the proper tool to both hold the
pipe in between lifts, and the proper tool to attach to lift it with.
Don't know if it was mentioned, but if the well isn't dry you also have
the buoyancy of the water so the whole thing won't weigh as much.

Another very important thing to keep in mind. *While working, if that
thing slips and lets loose, the wire is going to be moving fast down the
well. *If your foot or hand were to get tangled in it, you could loose
it. *If it happened to jump around and get around your chest, we could
loose you!- Hide quoted text -

- Show quoted text -

Good ponts, when doing a project lke this, always have a second and
maybe even a third person standing by, out of the danger area but with
a celllphone to call for help if needed.

On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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?

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

You could always pump out the water.

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

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.
;-)

On 10 Nov 2011 01:51:33 GMT, Han wrote:

" wrote in
news
I retired five years ago, but didn't like it so went back to work. I
was retired earlier this year but I'm still too young to do it
permanently, so...

I can see that, but I'm 67 now, and I had my fill of certain aspects of my
position, so I will not likely unretire grin.

I'm 59, so... I could actually retire now, or retire later and have a lot
more money to spend. ;-)

Metric is easier if the inputs and output are in metric. Imperial is
easier if the units are imperial, as in this case. Converting is just
asking for trouble. Ask NASA. ;-)

Grew up in Hlland, so metric is more natural, really, although imperial has
grown on me the last 42 years ...

I'm an engineer and use both equally. I have a neat little applet on my
desktop to do the weird conversions, when necessary (ConvCalc).

What I like about metric is the powers of 10 rather than funny factors
(12"/ft, whatever cubic inches/gal, 11 sq feet/m^2, whatever). also, in my
educatgion it was really important to get the powers of 10 (orders of
magnitude) straight. Sometimes I am good at that, too grin.

When I was in college, slide rules were the rule. ;-) Powers of 10 mattered,
since I could do most calculations in my head, with just a few flicks of the
fingers. Since I switched to calculators 38 years ago, that skill has
atrophied to where I'm pretty much dependent on them; entering a few more
numbers doesn't slow me down. ...and I'm not old enough that remembering them
is a problem. ;-)

zzzzzzzzzz wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

zzzzzzzzzz wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.



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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.


You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)

how does the OP know for certain the pump is bad? it could be a lower
water table. a leaking or rusted out line or other issue?

On Wed, 09 Nov 2011 19:03:56 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

zzzzzzzzzz wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

I don't buy it!

You don't buy simple physics?

The PIPE is experiencing buoyancy forces.

Exactly.

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.

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.

As you raise the pipe, it will take more force because of the reduced
buoyancy.

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

Still, the change is unrelated to what's in the pipe.

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

It's all about the displacement of the PIPE.

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).


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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

The system weight will include the weight of the included 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.

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.

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.

The more I think about it, I think we're in heated agreement.

Which you've lost. ;-)

I calculated the maximum weight of the water if the pipe were full.

...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.

What matters is the actual volume of water in the pipe.

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.

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.

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

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.

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

On Wed, 9 Nov 2011 20:11:15 -0800 (PST), bob haller
wrote:

how does the OP know for certain the pump is bad? it could be a lower
water table. a leaking or rusted out line or other issue?

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.

On Wed, 09 Nov 2011 19:51:42 -0500, Tony Miklos
wrote:


That sounds much more reasonable. I just got a second estimate for
$2800. Better than the first one, but still seems quite high. The
pump is a 1 HP, down approx 300ft. This new estimate is everything
new, pump, pipe, fittings, control box. They said they would try to
save some of the old pipe if possible, and can put plastic pipe below
the good steel pipe, but it's hard to know what is needed. They
insist the wire be replaced because the new code requires 4 wires now
(a ground wire is added).

I must admit my well guys work amazingly cheap so I'd add about $400 to
my price for a more realistic comparison. Most likely anyone who drills
wells will be higher priced, repairs are stuff they don't like to be
bothered with.

Even if I add $400, your price would be about $1000 less than what I'm
being quoted, and you have double the depth of mine and a 1.5hp pump
v/s my 1 hp. However, you reused your pipe and wire. My estimate is
for both of them new. I asked them if they could reuse the pipe, if
they would. They said they would, or maybe just the clean stuff above
the water line. As far as the wire, by code, they must install
grounded wire now, but said that if my wire was in real good shape,
they would reuse it, but if there is even one wear spot, they wont.
(I'm sure they'll find something wrong with it).

What surprises me, is that they need to install #10 wire. I'm almost
sure what I have is #12, and I know for sure that the wire feeding the
control box from the breaker is a #12. It's a 20A dual 220v breaker.
20A is #12 wire.

On Wed, 09 Nov 2011 19:51:42 -0500, Tony Miklos
wrote:

On 11/9/2011 6:55 PM, wrote:
On Wed, 09 Nov 2011 15:24:10 -0500, Tony Miklos
wrote:

On 11/9/2011 8:40 AM, wrote:

I already called a well company. When I heard their estimate of $4000
(which will likely be $5000 or more), I'll do it myself. I'm not
worried about the tractor loader. It wont break, it just wont lift if
the weight it too much. I know I have to cut the pipe, that's what my
sawsall is for. I intend to rent a clamping device that prevents the
pipe from dropping in the well. That's the tool I have not yet found.

Wow! I had mine replaced not long ago (about 600' down, new 1.5hp
goulds pump, old wire and pipe) for a grand total of $1324.66 including tax.

I missed the beginning of the thread, what size is your pump?

That sounds much more reasonable. I just got a second estimate for
$2800. Better than the first one, but still seems quite high. The
pump is a 1 HP, down approx 300ft. This new estimate is everything
new, pump, pipe, fittings, control box. They said they would try to
save some of the old pipe if possible, and can put plastic pipe below
the good steel pipe, but it's hard to know what is needed. They
insist the wire be replaced because the new code requires 4 wires now
(a ground wire is added).

I must admit my well guys work amazingly cheap so I'd add about $400 to
my price for a more realistic comparison. Most likely anyone who drills
wells will be higher priced, repairs are stuff they don't like to be
bothered with.


Damn, I need one of these.......
http://www.youtube.com/watch?v=hB_jzeU1Xgw
One man pulls 480 feet of pipe in about 10 minutes with this machine.
Amazing!

On Nov 9, 10:03*pm, mike wrote:
wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.



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.







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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.





You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

On Nov 10, 12:54*am, wrote:
On Wed, 9 Nov 2011 20:11:15 -0800 (PST), bob haller
wrote:

how does the OP know for certain the pump is bad? it could be a lower
water table. a leaking or rusted out line or other issue?

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.

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?

wrote:
On Wed, 9 Nov 2011 20:11:15 -0800 (PST), bob haller
wrote:

how does the OP know for certain the pump is bad? it could be a lower
water table. a leaking or rusted out line or other issue?

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.


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.

On Nov 9, 6:44*am, wrote:
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

Volume in cubic inches equals Pi X .75 X.75 X length in inches. I
figured up how much water the plumbing in a house would hold. Its
amazing how little water it takes to fill the pipes.

Jimmie

Jimmie

On Nov 10, 3:05*am, "
wrote:
On Nov 9, 10:03*pm, mike wrote:





wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.



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.

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

What counts is the water displaced by the pipe below the water level..
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K

On Nov 10, 11:36*pm, Harry K wrote:
On Nov 10, 3:05*am, "
wrote:





On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

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

On Nov 11, 4:06*am, "
wrote:
On Nov 10, 11:36*pm, Harry K wrote:





On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? The water drains out the _open_ end of the pipe of course.

Harry K

On Nov 11, 9:38*am, Harry K wrote:
On Nov 11, 4:06*am, "
wrote:





On Nov 10, 11:36*pm, Harry K wrote:

On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course.

Harry K- Hide quoted text -

- Show quoted text -

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

On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06*am, "
wrote:
On Nov 10, 11:36*pm, Harry K wrote:





On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

On Nov 11, 9:38*am, Harry K wrote:
On Nov 11, 4:06*am, "
wrote:





On Nov 10, 11:36*pm, Harry K wrote:

On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course.

Harry K- Hide quoted text -

- Show quoted text -

the pumps check valve would prevent tjhat

" wrote in
:

When I was in college, slide rules were the rule. ;-) Powers of 10
mattered, since I could do most calculations in my head, with just a
few flicks of the fingers. Since I switched to calculators 38 years
ago, that skill has atrophied to where I'm pretty much dependent on
them; entering a few more numbers doesn't slow me down. ...and I'm
not old enough that remembering them is a problem. ;-)

Used slide rules in high school, or perhaps in undergrad a bit.
Calculators were becoming the norm back in the middle 60's. My computer is
aiding me in forgeting simple math in my head ...

--
Best regards
Han
email address is invalid

On 11 Nov 2011 17:38:20 GMT, Han wrote:

" wrote in
:

When I was in college, slide rules were the rule. ;-) Powers of 10
mattered, since I could do most calculations in my head, with just a
few flicks of the fingers. Since I switched to calculators 38 years
ago, that skill has atrophied to where I'm pretty much dependent on
them; entering a few more numbers doesn't slow me down. ...and I'm
not old enough that remembering them is a problem. ;-)

Used slide rules in high school, or perhaps in undergrad a bit.
Calculators were becoming the norm back in the middle 60's. My computer is
aiding me in forgeting simple math in my head ...

Calculators didn't replace slide rules until the mid-70s (until they got small
enough to carry). I bought my first (HP-45) in '73, my senior year.

On Nov 11, 6:56*am, bob haller wrote:
On Nov 11, 9:38*am, Harry K wrote:





On Nov 11, 4:06*am, "
wrote:

On Nov 10, 11:36*pm, Harry K wrote:

On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course.

Harry K- Hide quoted text -

- Show quoted text -

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

- Show quoted text -

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

On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:





On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

- Show quoted text -

Which was my point.

Harry K

On Nov 11, 8:51*am, bob haller wrote:
On Nov 11, 9:38*am, Harry K wrote:





On Nov 11, 4:06*am, "
wrote:

On Nov 10, 11:36*pm, Harry K wrote:

On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course.

Harry K- Hide quoted text -

- Show quoted text -

the pumps check valve would prevent tjhat- Hide quoted text -

- Show quoted text -

Boggles

Harry K

On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:

On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:





On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

- Show quoted text -

Which was my point.

But you still have to lift that section of pipe, water and all. The length of
interest (for calculating the water's weight) is the length from the well
lever to the maximum pull height. IOW, the water doesn't drain out of the
pipe as it's lifted, as suggested.

On Nov 11, 9:36*pm, "
wrote:
On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:





On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

- Show quoted text -

Which was my point.

But you still have to lift that section of pipe, water and all. *The length of
interest (for calculating the water's weight) is the length from the well
lever to the maximum pull height. *IOW, the water doesn't drain out of the
pipe as it's lifted, as suggested.- Hide quoted text -

- Show quoted text -

The weight decreases as each length of pipe is removed or each foot of
poly is brought up. You are nitpicking.

Harry K

On Nov 11, 9:56*am, bob haller wrote:
On Nov 11, 9:38*am, Harry K wrote:





On Nov 11, 4:06*am, "
wrote:

On Nov 10, 11:36*pm, Harry K wrote:

On Nov 10, 3:05*am, "
wrote:

On Nov 9, 10:03*pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? *;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course.

Harry K- Hide quoted text -

- Show quoted text -

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

- Show quoted text -

Sure it will if I have a 300ft well and start pulling pipe. The
water level could be 100ft off the bottom, giving 200ft of
water filled pipe inside the casing surrounded by air.

On Nov 12, 2:56*am, Harry K wrote:
On Nov 11, 9:36*pm, "





wrote:
On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:

On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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"..

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

On Fri, 11 Nov 2011 23:56:18 -0800 (PST), Harry K
wrote:

On Nov 11, 9:36*pm, "
wrote:
On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:





On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

- Show quoted text -

Which was my point.

But you still have to lift that section of pipe, water and all. *The length of
interest (for calculating the water's weight) is the length from the well
lever to the maximum pull height. *IOW, the water doesn't drain out of the
pipe as it's lifted, as suggested.- Hide quoted text -

- Show quoted text -

The weight decreases as each length of pipe is removed or each foot of
poly is brought up. You are nitpicking.

No, you're wrong and I'm telling you how. The weight of the water being
lifted is increases as each *section* is pulled, but is otherwise a constant
for each section, until the foot is above the well level.

BTW, it's not "poly". The OP's well pipe is iron.

On Nov 11, 11:56*pm, Harry K wrote:
On Nov 11, 9:36*pm, "





wrote:
On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:

On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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"..

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

On Nov 12, 4:07*am, "
wrote:
On Nov 12, 2:56*am, Harry K wrote:

On Nov 11, 9:36*pm, "

wrote:
On Fri, 11 Nov 2011 20:24:04 -0800 (PST), Harry K
wrote:

On Nov 11, 7:52*am, "
wrote:
On Fri, 11 Nov 2011 06:38:21 -0800 (PST), Harry K
wrote:

On Nov 11, 4:06 am, "
wrote:
On Nov 10, 11:36 pm, Harry K wrote:

On Nov 10, 3:05 am, "
wrote:

On Nov 9, 10:03 pm, mike wrote:

wrote:
On Wed, 09 Nov 2011 17:51:02 -0800, mike wrote:

wrote:
On Wed, 09 Nov 2011 04:03:19 -0800, mike wrote:

wrote:
I'm no good at math. I'm trying to determine the amount of water in a
1 " (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

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

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".

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.

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.

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.

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

What counts is the water displaced by the pipe below the water level.
Contents of the pipe does not affect that.

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).

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.

You could always pump out the water.
Isn't that what the pump is for? ;-)
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.
;-)- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

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

Harry K- Hide quoted text -

- Show quoted text -

How does that happen with a check valve at the submersible
pump?- Hide quoted text -

- Show quoted text -

Did you think a couple hundred feet of pipe stands straight up in the
air?

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

The water drains out the _open_ end of the pipe of course.

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

- Show quoted text -

Which was my point.

But you still have to lift that section of pipe, water and all. *The length of
interest (for calculating the water's weight) is the length from the well
lever to the maximum pull height. *IOW, the water doesn't drain out of the
pipe as it's lifted, as suggested.- Hide quoted text -

- Show quoted text -

The weight decreases as each length of pipe is removed or each foot of
poly is brought up. *You are nitpicking.

Harry K- Hide quoted text -

I think we need to recap here. *I posted:

"The water in the pipe above the water line adds directly to the
weight that must be lifted. "

To which you replied:

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

To which I rpelied:

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

To which you replied:

"Did you think a couple hundred feet of pipe stands straight up in the
air? *The water drains out the _open_ end of the pipe of course. "

I think you, KRW and I are all in agreement on the physics.
Where this first went wrong was with you suggesting that
draining of the water was a factor. * I was never thinking of
water remaining in the whole pipe as it's being pulled. *Only
water in the pipe that is still in the well. *That seemed to be
the focus of the thread, the weight of the assembly being
pulled out of the well and I think everyone agrees that as
a section is removed, neither the weight of that pipe nor
it's contents are a factor. *The water in the pipe inside the
well above the water line does add directly to the weight
being lifted, which was my comment that you seemed to
disagree with.

I then made the mistake of thinking you meant water
draining out of the pipe that was still in the well, hence my
comment about the check valve preventing that. *But I think
we are all in agreement that:

A - The pipe is indeed removed a section at a time and
once removed it's not a weight factor

B - The water inside the pipe, between the water line and
the top of the pipe at or above the wellhead adds to the
weight being lifted.

Yep, I was thinking the same thing last night.

As to the effect of water in the pipe _below_ the static level I
dreamed up a good mind experiment to show that it has no effect.

Rig a pipe with a valve at one end that can be closed. Submerge the
pipe with valve open 10' into water - weigh pipe. Leave pipe there
and close valve - weigh pipe. No difference.

Harry K