Can you increase P in water via air pressure? 18

[Mechbob1]

Please help settle a discussion…
As per the picture, can you increase the pressure in the water by increasing the pressure in the air above? It’s a sealed container.
If yes/no, why?
I thought you couldn’t, due to P equaling rho x g x h. But someone else reckons you can, and I’m second guessing myself.
Thanks.

 

[LittleInch]

bimr,

I think you're over thinking this based on a poorly worded question.

It is quite clear that what the issue was that mechbob1 forgot that the pressure of the water, not "in" the water will increase relative to the outside fixed atmospheric pressure if you add more pressure to the air above the water.

Also water is not incompressible. It is virtually incompressible, but not 100%. That's what bulk modulus is.

So in a perfectly rigid container you can add more volume and the pressure of the water increases. The amount of water is small for a large rise in pressure, but it is not zero.

So if you pumped 20,000 psig into a perfectly rigid full container of water it would store some energy and volume. A small amount (way less than 1%) for sure but not zero.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[EngrPaper]

bimr, I'm afraid I dont follow your logic at all.

If you have quiescent air and water in the same vessel, and they contact each other at a free surface, then they have the same pressure at that surface. So if the air increases in pressure, then the pressure at that surface increases (same contact area, equalization of forces). Thus the water pressure increases by the same amount the air does. The water just has the additional hydrostatic head as you go down that we normally neglect in air.

 

[bimr]

LittleInch, I agree with your comments. Perhaps 20,000 psig is an exaggeration based on from TiCl4 (Chemical) post, but no one is talking about a theoretical issue here.

The compressibility of water is a function of pressure and temperature. At 0 °C, at the limit of zero pressure, the compressibility is 5.1×10−10 Pa−1. At the zero-pressure limit, the compressibility reaches a minimum of 4.4×10−10 Pa−1 around 45 °C before increasing again with increasing temperature. As the pressure is increased, the compressibility decreases, being 3.9×10−10 Pa−1 at 0 °C and 100 megapascals (1,000 bar).[40]

The bulk modulus of water is about 2.2 GPa.[41] The low compressibility of non-gasses, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume.[41]

The bulk modulus of water ice ranges from 11.3 GPa at 0 K up to 8.6 GPa at 273 K.[42] The large change in the compressibility of ice as a function of temperature is the result of its relatively large thermal expansion coefficient compared to other common solids.

Link

 

[TiCl4]

Bimr, you haven’t addressed my question. If P1 is 1 atm (gauge) and a standard pressure gauge at the bottom of P2 is reading 1.1 atm, what will the gauge read when you increase P1 to 2 atm (gauge)?

Also, are you also arguing that my equation given above for pressure at the bottom of P2 is incorrect?

 

[bimr]

If you have quiescent air and water in the same vessel, and they contact each other at a free surface, then they have the same pressure at that surface. So if the air increases in pressure, then the pressure at that surface increases (same contact area, equalization of forces). Thus the water pressure increases by the same amount the air does. The water just has the additional hydrostatic head as you go down that we normally neglect in air.

Would you neglect static water pressure at the base of a dam? I think not.

The tank is supposed to be sealed. Yes, the pressure at the bottom of the container increases, but it is from the air not the water. Water is incompressible at the stated conditions.

A pump for example does not create pressure, it only creates flow (kinetic energy). Pressure is a measurement of the resistance to flow. Figure 2.

Link

 

[IRstuff]

https://www.grc.nasa.gov/

http://www/k-12/WindTunnel/Activiti...ontainer is open,total pressure on an object.

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[bimr]

mechbob1 forgot that the pressure of the water, not "in" the water will increase relative to the outside fixed atmospheric pressure if you add more pressure to the air above the water.

It is the pressure on the water, not "of" the water nor in the water.

 

[EngrPaper]

No it's definitely pressure of the water. Because if that water is at 212F in atmosphere, it'll boil but if it's 212F at 1000 psi it definitely won't boil. The pressure of the water, whether caused by a blanket of air or by flowing through a pump, matters.

EDIT: Also, I said neglect the static head of the air, not the water. After I specifically said you take the water head into account.

 

[TiCl4]

Bimr,

Would you then agree to the following statement?

The pressure, as measured by a pressure gauge, placed anywhere in the P2 water phase will measure P1(gauge) plus rho*g*h, where “h” is the submerged depth of the pressure gauge?

It seems like you are saying an increase in P1 will not change the the fact that the additional pressure added by the water column is only rho*g*h. If that is so, then we agree.

This discussion may stem from not clearly defining P2. You seem to understand it as the pressure increase exerted by the water phase (i.e. rho*g*h), while I, and most others, I suspect, regard it as the total pressure as measured by a pressure gauge (P1 + rho*g*h)

 

[bimr]

Forget the static water pressure for the moment.

Forget about the sealed container.

Forget about the water as it has nothing to do with answering the post.

Assume P1 = 0 psig. Assume P2 = 100 psig.

Yes, the pressure at P2 has increased in an empty container as it is due to the compressed air, not the water.

If you add compressed air to the container half filled with water, the air is on top of the water and that causes the pressure to increase on the bottom. From the compressed air not the water.

 

[georgeverghese]

To get an increase in head space pressure by 0.1bar to simulate an increase in pressure by 1m of water column somewhere in the liquid space, the mechanical design pressure of the container must be higher than 0.1bar + max static head, else this container may burst.

 

[bimr]

pressure increase exerted by the water phase

There is no increase in pressure exerted by the water phase. The water is static, not moving. Water only has static pressure unless the water is moving, in which case the water has a kinetic energy.

Pump pressure, however, is a measure of resistance to flow. Without flow, there is no pressure. In piping (at the same elevation), water pressure is a measurement of the resistance to flow.

 

[TiCl4]

Bimr,

First, please define P2. Pressure as read by a pressure gauge at the top of the water is not the same as pressure read at the bottom of the water. Therefore, you need to define exactly WHERE you are measuring pressure. That is why I refer to it as an increase in pressure from the water column. If you move a pressure gauge from the top of the water level to the bottom of the water level, the pressure reading will increase. If you do not understand this, please review basic fluid mechanics.

Also, why do you keep referring to pump pressure? We aren't talking about pumps.

Lastly, you haven't answered my question. Until you answer this question, I will assume you are just trolling at this point.

Do you agree with the following statement? The pressure, as measured by a pressure gauge, placed anywhere in the P2 water phase will measure P1(gauge) plus rho*g*h, where “h” is the submerged depth of the pressure gauge?

 

[bimr]

Assume you had a tank with air pressure on the tank. Now open a valve on the tank.

The air pressure in the tank will decrease as the flow decreases out of the valve. Will the air flow stay constant until the tank is empty? No.

 

[bimr]

TiCl4 (Chemical),

Your question has already been answered above in the response to EngrPaper:

The tank is supposed to be sealed. Yes, the pressure at the bottom of the container increases, but it is from the air not the water. Water is incompressible at the stated conditions.

IF you increase the pressure from P1 to P2, it doesn't matter whether the tank has water in it or not. The pressures are still P1 and P2.

I keep repeating pump pressure because nobody seems to realize that pump pressure results from resistance to flow (kinetic energy) not water pressure. Just as a pump exerts no pressure on the pumped fluid, the sealed container without the air pressure doesn't exert any pressure on the water.

The air pressure is "on" the water not in the water, nor of the water.

 

[TiCl4]

Okay, you still haven't answered that simple yes/no question with either a yes or no. I'll end my comments on this thread here.

 

[bimr]

TiCl4 (Chemical)

Regarding:

"Do you agree with the following statement? The pressure, as measured by a pressure gauge, placed anywhere in the P2 water phase will measure P1(gauge) plus rho*g*h, where “h” is the submerged depth of the pressure gauge?"

Can't answer that question. Define "P2 water phase"? Is the container closed or open?

I have already stated that water can have a static pressure on depth if the water has atmosphere on top. When the system is motionless, i.e. the fluid is not being circulated, the water static pressure is the same whether the air pressure is increased from a point P1 to a point P2.

 

[EngrPaper]

And bimr, what we are saying is that you are missing an entire component of static pressure: the pressure on the liquid surface. Let's talk in psia, so we can clearly see. If the atmosphere was 20psia, the pressure at any submerged point in water would be 20 psia +rho(g)(h). So for 5 ft down, we are looking at 20psia+5ft/(2.31ft/psi)=22.16psia. Pardon the unit conversion, I just have that one in my head. By the same math, if the atmosphere was 10 psia, the pressure at 5 feet submerged would be 12.16 psia. That is the actual pressure. The pressure at any point INSIDE the liquid depends on the surface pressure, whether in a pressure vessel or at different altitudes in a freely open tank.

 

[bimr]

And bimr, what we are saying is that you are missing an entire component of static pressure: the pressure on the liquid surface. Let's talk in psia, so we can clearly see. If the atmosphere was 20psia, the pressure at any submerged point in water would be 20 psia +rho(g)(h). So for 5 ft down, we are looking at 20psia+5ft/(2.31ft/psi)=22.16psia. Pardon the unit conversion, I just have that one in my head. By the same math, if the atmosphere was 10 psia, the pressure at 5 feet submerged would be 12.16 psia. That is the actual pressure. The pressure at any point INSIDE the liquid depends on the surface pressure, whether in a pressure vessel or at different altitudes in a freely open tank.

Atmospheric pressure can be neglected in most examples involving water exposed to the atmosphere because the water static pressure is significantly higher than atmospheric pressure.

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.

Pascal's law

 

[Compositepro]

The OP is talking about static pressure. I cannot conceive how anyone can think anything else, in this case, or how anything bmir is saying has any relevance to the original question. This is not the first time and I don't expect him to see the light. But this behavior is detrimental to the integrity of eng-tips, and thus cannot go unchallenged. I am flabbergasted that such a simple question produces a thread like this, on a site like this. Many of us became engineers because we want to be in a world where there are right and wrong answers