• Tkopperl
In summary: Instead of a swimming pool, imagine you had a closed tank, filled with water. You want to test the welded seams of the tank for leaks.You could wait for water to leak out, but what if you wanted to help it along by increasing the pressure of the tank?You add a standpipe to the top of the tank, say about 3 or 4 feet tall, making sure the standpipe is in communication with the liquid in the tank. Add water to the standpipe, and for every foot of water in the standpipe above the top of the tank, the pressure inside the tank increases by 62.4/144 = 0.433 psi, assuming the liquid is fresh water.
Tkopperl
Ok so I'm an engineer and I ran into a physics problem the other day and I'm not quite sure how to go about it. I am very farmiliar with hyrdostatic pressures, but in this particular case I'm not exactly sure how the laws apply.

I have attached a sketch of my problem. There is an 18-3/4" ID main pipe. Inside of this pipe is a plug with a pipe running through the plug with a 6-5/8" OD and 4" ID. There are two water columns that are isolated from each other (The blue and green). The weights of the fluid are Blue = 10 lbs/gal Green = 9.8 lbs/gal

I know that the hydrostatic pressure of the blue fluid at 5000ft will be approximately (10 lbs/gal) x (5000 ft) x (.052) = 2600psi. I know this times the surface area of the top of the plug will give me the force acting downwards on the plug.

My question is in this scenario, what would the force acting upwards on the plug be? Because the smaller pipe opens up into the larger column of fluid how are the hydrostatic pressures affected. This would be like a straw on top of a swimming pool you can't measure the hydrostatic of the bottom of the pool off of the height of the straw.

I hope I have explained this well enough. Thank you in advance for anyone that might help!

Regards,
Travis Bryant

#### Attachments

• hydrostatic.png
1.7 KB · Views: 511
How about 5002*9.8*0.052 ? (funny units, I'm sorry to say :) )

BvU said:
How about 5002*9.8*0.052 ? (funny units, I'm sorry to say :) )
Yes, but that formula only works for solid columns of fluid. When the fluid changes from the smaller pipe to the larger pipe this is no longer true, so this would be true for the pressure at 5002 ft inside the smaller pipe, but not outside of it.

same depth, same pressure. So it does work. Communicating vessels and all that. If the pressure away from the axis were different, the liquid would flow towards the lower pressure. That's what it's liquid for,

BvU said:
same depth, same pressure. So it does work. Communicating vessels and all that. If the pressure away from the axis were different, the liquid would flow towards the lower pressure. That's what it's liquid for,
Hmmm. so if I had a 100ft straw on top of a 2ft swimming pool the hydrostatic pressure in the swimming pool would be based off of the 102ft? Assuming the swimming pool had a top so the water couldn't escape.

No the straw would empty itself in the pool. If that's impossible, then: yes, the pressure in the pool would be about 4 Bar (ever heard of such sensible units ? 4 Bar = about 58 psi). The top would bulge unless very very very strong.

It's the principle of the hydraulic press , very useful!

Tkopperl said:
Hmmm. so if I had a 100ft straw on top of a 2ft swimming pool the hydrostatic pressure in the swimming pool would be based off of the 102ft? Assuming the swimming pool had a top so the water couldn't escape.
Instead of a swimming pool, imagine you had a closed tank, filled with water. You want to test the welded seams of the tank for leaks.
You could wait for water to leak out, but what if you wanted to help it along by increasing the pressure of the tank?

You add a standpipe to the top of the tank, say about 3 or 4 feet tall, making sure the standpipe is in communication with the liquid in the tank. Add water to the standpipe, and for every foot of water in the standpipe above the top of the tank, the pressure inside the tank increases by 62.4/144 = 0.433 psi, assuming the liquid is fresh water.

SteamKing said:
Instead of a swimming pool, imagine you had a closed tank, filled with water. You want to test the welded seams of the tank for leaks.
You could wait for water to leak out, but what if you wanted to help it along by increasing the pressure of the tank?

You add a standpipe to the top of the tank, say about 3 or 4 feet tall, making sure the standpipe is in communication with the liquid in the tank. Add water to the standpipe, and for every foot of water in the standpipe above the top of the tank, the pressure inside the tank increases by 62.4/144 = 0.433 psi, assuming the liquid is fresh water.
So the hydrostatic pressure at the bottom of the tank would be based off of the height of the standpipe no matter what size the standpipe is?

Tkopperl said:
So the hydrostatic pressure at the bottom of the tank would be based off of the height of the standpipe no matter what size the standpipe is?
Yep. That's Pascal's Law for you.

http://en.wikipedia.org/wiki/Pascal's_law

Hi Tkopperl. Welcome to Physics Forums.

This is obviously a production or injection well operation. Is it an injection well? Are you sure that both fluids go to the surface and are open to the atmosphere at the surface? The reason I ask is that, in a real world well setting, that would not be the case. Also, if it were a real well, the fluid pressure in the injection/production string would not simply be hydrostatic. That would mean that the pressure difference across that packing would be different.

Is this really a homework problem or is it a real world well situation?

Chet

Chestermiller said:
Hi Tkopperl. Welcome to Physics Forums.

This is obviously a production or injection well operation. Is it an injection well? Are you sure that both fluids go to the surface and are open to the atmosphere at the surface? The reason I ask is that, in a real world well setting, that would not be the case. Also, if it were a real well, the fluid pressure in the injection/production string would not simply be hydrostatic. That would mean that the pressure difference across that packing would be different.

Is this really a homework problem or is it a real world well situation?

Chet
This is a real world scenario. You are correct, this is a model of a production well in 5000ft of sea water. The plug, actually a testing tool is used to test Blow Out Preventers. Both fluid columns are open to atmosphere. I was just curious about the U-Tube effect that happens.

## 1. What is hydrostatic pressure?

Hydrostatic pressure is the force per unit area exerted by a fluid at rest due to the weight of the fluid above it.

## 2. How is hydrostatic pressure calculated?

Hydrostatic pressure is calculated by multiplying the density of the fluid by the acceleration due to gravity and the depth of the fluid.

## 3. What factors affect hydrostatic pressure?

The factors that affect hydrostatic pressure include the density of the fluid, the acceleration due to gravity, and the depth of the fluid.

## 4. How does hydrostatic pressure impact objects submerged in a fluid?

Objects submerged in a fluid experience a greater hydrostatic pressure at greater depths, which can cause them to compress or break if they are not strong enough to withstand the pressure.

## 5. What is the relationship between hydrostatic pressure and altitude?

The hydrostatic pressure decreases with altitude, as there is less fluid above at higher altitudes. This is why we experience less pressure on our bodies at higher altitudes, such as when flying in an airplane.

• Mechanical Engineering
Replies
28
Views
2K
• General Engineering
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
7K
• Mechanics
Replies
2
Views
4K
• Calculus and Beyond Homework Help
Replies
3
Views
4K
• Mechanics
Replies
1
Views
1K
• General Engineering
Replies
2
Views
2K
• General Engineering
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Mechanics
Replies
7
Views
4K