# Static hydraulic pressure in a pipe

Pascal pressure
My Question:

Two pipes filled with water and caped at one end, they have the same diameter, but different lengths.
One pipe is 10 feet long and the other pipe is 1,000 feet long.
At the other end of both pipes a piston applies an equal amount of pressure on the water in the pipes.

Assumptions:
The pressure within each pipe is equal at every point within the pipe.
The pressure within the 10-foot pipe is higher than the 1,000-foot pipe.

Does anyone know what the equation is that describes this relationship?
Does anyone know where tables & charts are that describe this relationship?

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At the other end of both pipes a piston applies an equal amount of pressure on the water in the pipes.
The pressure within each pipe is equal at every point within the pipe.
The pressure within the 10-foot pipe is higher than the 1,000-foot pipe.
Seems to me these three statements are mutually contradictory. If the end of the pipes have the same pressure and the pressure in each pipe is the same throughout, it seems obvious that all the pressures are the same.

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russ_watters, berkeman and mcastillo356
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At the other end of both pipes a piston applies an equal amount of pressure on the water in the pipes.
Do you mean equal force or equal pressure?

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Welcome, Pascal!

Where at and how is that piston-induced-pressure measured?

What end is capped and for which pipe?

Are both pipes perfectly horizontal?

Pascal pressure
Seems to me these three statements are mutually contradictory. If the end of the pipes have the same pressure and the pressure in each pipe is the same throughout, it seems obvious that all the pressures are the same.
The same size piston supplies pressure on both pipes. In a closed system the static pressure is the same at all points. But, the longer pipe holds a greater volume of water. My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe

Pascal pressure
Do you mean equal force or equal pressure?
The piston's acting on each pipe are equal in size and they apply the same force on the water in both pipes. We know that the pressure created within the pipes is equal at all points, however since one pipe holds a much larger volume of water, will the pressure of the water in that pipe be smaller than the pressure in the shorter pipe?

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We know that the pressure created within the pipes is equal at all points, however since one pipe holds a much larger volume of water, will the pressure of the water in that pipe be smaller than the pressure in the shorter pipe?
How are these not contradictory statements? Where do you expect to measure the pressure and find it smaller? And smaller than what?

Pascal pressure
Welcome, Pascal!

Where at and how is that piston-induced-pressure measured?

What end is capped and for which pipe?

Are both pipes perfectly horizontal?
At this point it is a thought experiment. Two separate pipes, each one is caped on one end. So actually they are two containers, that happen to be in the shape of a pipe. Good observation, yes lets say they are both horizonal. At the other end of each pipe is a piston that delivers the same amount of force on the water in each pipe.
My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe. What do you think?

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My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe. What do you think?
Yes, you keep SAYING that. We get that that is your belief, but opinion does not trump math. Now answer my question.

Where do you expect to measure the pressure and find it smaller? And smaller than what?

Mentor
So, like, you apply 10psi to two pipes with a piston and somehow one of the pipes ends up at 5psi?

Maybe a labeled sketch would help? Because, yeah, this seems like an obvious self contradiction.
My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe. What do you think?

That's not a thing.

phinds
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At this point it is a thought experiment. Two separate pipes, each one is caped on one end. So actually they are two containers, that happen to be in the shape of a pipe. Good observation, yes lets say they are both horizonal. At the other end of each pipe is a piston that delivers the same amount of force on the water in each pipe.
My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe. What do you think?
I know both pressures will be the same.
https://en.m.wikipedia.org/wiki/Pascal's_law

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russ_watters
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@Pascal pressure I think your thought process would probably lead you to believe that the pressure at the bottom of a 10 foot tall test tube would be lower than the pressure 10 feet below the surface of the Pacific Ocean, yes? That IS, after all, identical to what you are saying.

russ_watters
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I know both pressures will be the same.
Yes, I think we've established that. What we're trying to figure out is why the OP doesn't see it.

Pascal pressure
How are these not contradictory statements? Where do you expect to measure the pressure and find it smaller? And smaller than what?
It is my understanding is, general, when you have liquid under pressure in a closed vessel, like a hydraulic bottle jack, the pressure is the same through out the vessel. However in this thought experiment we have a piston applying force on the water in the vessel. There are two separate vessels. (Pipes) one vessel contains much more water than the other vessel, but the force applied to the piston is the same for both pipes, the only difference is one pipe is longer than the other one.

My assumption is; since the applied pressure in the longer pipe (1000 ft) is applied against a greater volume of water, than the pressure in the small pipe, then the pressure in the shorter pipe (10 feet) will be higher than the pressure in the longer pipe. The pressure will be the same inside of each separate system (pipe) but each pipe will have a different pressure value that will be equally distributed within it.

What do you think?

Pascal pressure
Yes, I think we've established that. What we're trying to figure out is why the OP doesn't see it.
What does OP mean?

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What does OP mean?
Original Poster / Original Post --- the person that started the thread and or the first post in the thread.

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What do you think?
I think you have confirmed the belief that I expressed in post #12.

Saying the same thing over and over isn't going to make it right.

Mentor
My assumption is; since the applied pressure in the longer pipe (1000 ft) is applied against a greater volume of water, than the pressure in the small pipe, then the pressure in the shorter pipe (10 feet) will be higher than the pressure in the longer pipe. The pressure will be the same inside of each separate system (pipe) but each pipe will have a different pressure value that will be equally distributed within it.

What do you think?
The water doesn't know if there's a piston at the end or not. Pascal's Law applies the same either way.

Lnewqban
Pascal pressure
Original Poster / Original Post --- the guy that started the thread.
The pipes are horizonal and each pipe is a closed system. Lets say one pipe is 10 feet long and the other pipe is 5,000 feet long. Again one end of both separate pipes is caped, the other end of both separate pipes has a piston applying pressure on to the water in the pipe. I would like to believe that the water pressure in both systems will be the same, because they have the same piston-force acting on the water in the pipes. However, intuitively I don't think so , because the force from one piston is distributed across a much larger volume, than the smaller pipe. Therefore I'm thinking the pressure in one will be different than the pressure in the other.

I would truly like to think I'm wrong.
Thank you for your insight. I need to get back to my regular job, for now. Will check back later
thank you,
Pascal Pressure

Mentor
I would like to believe that the water pressure in both systems will be the same, because they have the same piston-force acting on the water in the pipes. However, intuitively I don't think so , because the force from one piston is distributed across a much larger volume, than the smaller pipe. Therefore I'm thinking the pressure in one will be different than the pressure in the other.

I would truly like to think I'm wrong.
You can rest assured. Your intuition is wrong. Trust the law.

My assumption is; since the applied pressure in the longer pipe acts on a grater volume of water, then the pressure (Which is the same through the pipe) in that pipe will be less than the pressure in the 10 foot pipe. What do you think?
For a fixed applied pressure, the pressure will be distributed equally throughout the volume of the water in the pipe.

The analysis I have seen in this thread involves compression of the water.
In reality, the pipes will stretch much more than the water will compress.

If you increase the volume of water in each pipe by the same fixed volume, the pressure rise will be greatest in the short pipe, where the strain per unit length will be greater than in the longer pipe.

Lnewqban
Gold Member
because the force from one piston is distributed across a much larger volume, than the smaller pipe.
Pressure ##p## is a force ##F## distributed across an area ##A##; if you have the same force & area in both cases, then you get the same pressure in both cases. There is nothing about volume in the definition of pressure:
$$p = \frac{F}{A}$$

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Pressure ##p## is a force ##F## distributed across an area ##A##; if you have the same force & area in both cases, then you get the same pressure in both cases. There is nothing about volume in the definition of pressure:
$$p = \frac{F}{A}$$
And just to be clear, the area Jack is talking about is NOT the areas of the inside of the pipes, it is the area of the face head of the piston applying the pressure.

jack action and russ_watters
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...I would like to believe that the water pressure in both systems will be the same, because they have the same piston-force acting on the water in the pipes. However, intuitively I don't think so , because the force from one piston is distributed across a much larger volume, than the smaller pipe. Therefore I'm thinking the pressure in one will be different than the pressure in the other.

We could idealize the water in your pipes as a “3D deformable solid”, which molecules are willing to adopt the shape of their container, but can’t be squeezed or stretched like a spring.

Would a long rigid steel rod push less at the opposite end of the force than a short one?
Let’s reverse the direction of the force of the pistons.
Would a long steel wirerope pull less at the opposite end of the force than a short one?

In static conditions, for each of the above cases, each molecule transfers the full pushing or pulling load to the next one, no matter how long the chain of molecules is.

Pascal pressure
Thank you for all the insightful comments.
As long as the water is not compressed, then yes I totally agree, the water in the pipe acts as a solid under compression, when the water has pressure applied to it.
I believe that I read somewhere that below 3,000 psi water is essentially non compressible. Then at about 100,000 psi water compresses about 15% of its volume.
As I think about it some more, I guess the answer to the question should be fairly straight forward:
(p-1) * (v-1) = (p-2) * (v-2)

Even though I think this equation is used for pneumatics, I think it could be applied to hydraulics

Where:
(v-1) = volume of the water in the pipe at 14.7 psi [(p-1) = 14.7 psi]
(v-2) = volume of the water in the pipe at pressures above 3,000 psi minus % compression
Then pressure drop of the water in the pipe = (p-2)

I guess I should change my name from Pascal Pressure to
"Learning Pascal Pressure"
Thank you again,

Mentor
Thank you for all the insightful comments.
As long as the water is not compressed, then yes I totally agree, the water in the pipe acts as a solid under compression, when the water has pressure applied to it.
I believe that I read somewhere that below 3,000 psi water is essentially non compressible. Then at about 100,000 psi water compresses about 15% of its volume.
As I think about it some more, I guess the answer to the question should be fairly straight forward:
(p-1) * (v-1) = (p-2) * (v-2)

Even though I think this equation is used for pneumatics, I think it could be applied to hydraulics

Where:
(v-1) = volume of the water in the pipe at 14.7 psi [(p-1) = 14.7 psi]
(v-2) = volume of the water in the pipe at pressures above 3,000 psi minus % compression
Then pressure drop of the water in the pipe = (p-2)

I guess I should change my name from Pascal Pressure to
"Learning Pascal Pressure"
Thank you again,
Sorry, but basically none of that is correct. Compressibility isn't part of Pascal's Law either. Nor does it act as a solid.

To be honest, it sounds like you are trying to make up your own laws rather than learning Pascal's Law. You should just learn Pascal's Law and other associated real ones.

Pascal pressure
Which part of the compressibility of water is incorrect?

Pascal pressure
Ok take Pascal's law out of the question.
Our piston will move a distance that is related to the compressibility of the water.
So depending on what the applied pressure is, the total volume of the water in the pipe will be reduced by a percent of its original volume. If the pressure is 100,000 psi and if the water is compresses by 15% then the volume change is (v-1) * .15 = (V-2)
Is this correct?

Pascal pressure
Actually, depending on how you look at it;
Probably more accurate to say;

(v-2) = (V-1) - [(v-1) * .15]

Mentor
Which part of the compressibility of water is incorrect?
The parts where you said it acts as a solid and also that it's basically incompressible under 3000 psi (though that second one can be an assumption in some circumstances..

Mentor
If the pressure is 100,000 psi and if the water is compresses by 15% then the volume change is (v-1) * .15 = (V-2)
Is this correct?
(1-.15)

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Ok take Pascal's law out of the question.
Our piston will move a distance that is related to the compressibility of the water.
So depending on what the applied pressure is, the total volume of the water in the pipe will be reduced by a percent of its original volume. If the pressure is 100,000 psi and if the water is compresses by 15% then the volume change is (v-1) * .15 = (V-2)
Is this correct?
https://en.m.wikipedia.org/wiki/Properties_of_water#Compressibility

You could calculate how thick the walls and cap of your pipes need to be to hold those 100,000 psi.
Knowing or assuming the diameter of your piston, you could also calculate how much force it needs to be pushed with to generate that pressure.

Then, even after water compresses some under that tremendous pressure, which your pipes resist, and your piston is able to create, the measurable static pressure at the end of both pipes should be exactly the same that at the piston's face.

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Pascal pressure
The parts where you said it acts as a solid and also that it's basically incompressible under 3000 psi (though that second one can be an assumption in some circumstances..
Ok, I agree it only acts like a solid in regards to compression not tension or sheer.
And yes, at some pressure under 3,000 psi the compressibility of water is essentially negligible, as mentioned, in this case it acts similar to a steel rod under compression.

In regard to the main question; the pressure drop in the water when it is under enough pressure to be compressed; So we have our starting volume v-1 which is the volumes of water in the pipe before compressing the water, then we compress the water by 15% (100,000 psi) so (v-2) = (V-1) - [(v-1) * .15]

Oops, wait a minute Huston, we have no unknown variable with this approach, because I'm saying P-1 = 14.7 psi and P-2 =100,000 psi.

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As I think about it some more, I guess the answer to the question should be fairly straight forward:
(p-1) * (v-1) = (p-2) * (v-2)

Even though I think this equation is used for pneumatics, I think it could be applied to hydraulics

Where:
(v-1) = volume of the water in the pipe at 14.7 psi [(p-1) = 14.7 psi]
(v-2) = volume of the water in the pipe at pressures above 3,000 psi minus % compression
Then pressure drop of the water in the pipe = (p-2)
First, this equation can only be used in pneumatics IF the temperature is kept constant.

Second, it cannot be applied to hydraulics. Here is the correct equation:
$$\frac{p_1 V_1 - p_2 V_1}{V_2-V_1} = E$$
Where ##E## is the Bulk Modulus Elasticity of the liquid.

Reference: https://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

What that equation state is, if a liquid is at pressure ##p1## and gets increased to ##p_2##, its volume will go from ##V_1## to ##V_2##.

In your comparison, both cases have the same ##p_1## and ##p_2## (which is constant throughout the volume of liquid as the pressure increases from ##p_1## to ##p_2##), only the initial ##V_1## is different. Obviously, ##V_2## will be different as well but the ratio ##\frac{V_2}{V_1}## will be constant:
$$\frac{V_2}{V_1} = 1-\frac{p_2-p_1}{E}$$
$$\frac{V_2}{V_1} = 1-\frac{3000-14.7}{312000}=0.99$$
$$\frac{L_2}{L_1} = \frac{\frac{V_2}{A_2}}{\frac{V_1}{A_1}} = \frac{V_2}{V_1}\frac{A_1}{A_2} =\frac{V_2}{V_1}(1) = \frac{V_2}{V_1}$$
$$L_{2\ 1000\ ft} = 0.99 \times 1000\ ft = 990\ ft$$
$$L_{2\ 10\ ft} = 0.99 \times 10\ ft = 9.90\ ft$$