Static pressure inside the glass of water

  • #1
adjurovich
97
16
If we had a perfectly isolated glass of water, the total pressure inside will be equal to the sum of hydrostatic and static pressure. In this case, would the static pressure be totally dependent on the temperature of water and the volume of glass?

Also, consider that the glass is completely filled with water so there is no any additional air above it.
 
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  • #2
What is the difference between static pressure and hydrostatic pressure in this system? What is your definition of these terms?
 
  • #3
The equation of state of liquid water is approximated by $$\rho=\rho_0e^{(-\alpha(T-T_0)+\frac{P-P_0}{B})}$$where ##\alpha## is the coefficient of thermal expansion and B is the bulk modulus; ##T_0=293 K## and ##P_0=1 atm##. So the hydrostatic equation reads $$\frac{dP}{dh}=\rho g=\rho_0ge^{(-\alpha(T-T_0)+\frac{P-P_0}{B})}$$where h is the depth. From this relationship, what do you get for P as a function of depth below the surface h?
 
  • #4
Chestermiller said:
What is the difference between static pressure and hydrostatic pressure in this system? What is your definition of these terms?
Hydrostatic pressure is simply pressure exerted by the weight of fluid molecules (so it’s usually ignored in gases). However, static pressure as I understand is simply some additional pressure due to motion of molecules of fluid. In gases, we observe only static pressure since hydrostatic pressure is usually small enough so it can be neglected. If we were observing a system where we have cylinder container with piston, the gas would contract until its static pressure balances atmospheric pressure + pressure exerted by the weight of the piston. If we applied the similar logic on liquids, and put the water inside the piston, the water should experience much weaker contractions because its molecular forces are much stronger, but it would still balance atmospheric + pressure due to the weight of the piston. If we were saying that static and hydrostatic pressure are the same, what could balance the piston at the top of cylinder since there can be no hydrostatic pressure at depth of 0 meters (surface). Using the same logic, which pressure are we talking about when we talk about gases? Hydrostatic is certainly not the one.

Also, if possible, please avoid further math because I’m still a high school student and it won’t be of much help since I am not yet at that level of education
 
  • #5
If I understand you correctly, the hydrostatic pressure must match the static pressure at all depths in the glass.
 
  • #6
Chestermiller said:
If I understand you correctly, the hydrostatic pressure must match the static pressure at all depths in the glass.
No, I didn’t say that. I said they are two independent and very much different pressures. I said static pressure is basically pressure caused by Brownian motion. When some glass of water is left open to atmosphere, the pressure of water inside it at certain depth ##h## will be:

##p = p_0 + \rho g h##

If we somehow ignored the hydrostatic pressure (just hypothetically, we should never):

##p = p_0##

Where ##p## is the total pressure. And static pressure ##P = p_0##
 
  • #7
adjurovich said:
No, I didn’t say that. I said they are two independent and very much different pressures. I said static pressure is basically pressure caused by Brownian motion. When some glass of water is left open to atmosphere, the pressure of water inside it at certain depth ##h## will be:

##p = p_0 + \rho g h##

If we somehow ignored the hydrostatic pressure (just hypothetically, we should never):

##p = p_0##

Where ##p## is the total pressure. And static pressure ##P = p_0##
Your equation for the hydrostatic pressure is incorrect if you are allowing the water to have a pressure which varies with specific volume (i.e,. your 'static pressure relationship). That is, if the water is considered compressible.
 
  • #8
Chestermiller said:
Your equation for the hydrostatic pressure is incorrect if you are allowing the water to have a pressure which varies with specific volume (i.e,. your 'static pressure relationship). That is, if the water is considered compressible.
We consider water to be incompressible ideal fluid here. The hydrostatic pressure as I stated in equation varies with depth only. Also I didn’t specify exact equation for static pressure of water because I haven’t heard about it yet?

However, for the ideal gas, I’d use this formula for static pressure:

##p = \dfrac{2}{3} n_0 \overline{E_k}##

But this doesn’t apply to liquids at all, so we should ignore this one.
 
  • #9
If you are considering the liquid water as incompressible. then its specific volume is independent of pressure, and the static pressure (as you call it) is equal to the hydrostatic pressure at all depths.

The equation of state for an ideal gas is $$P=\frac{\rho RT}{M}$$where M is the molecular weight.
 
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  • #10
Chestermiller said:
If you are considering the liquid water as incompressible. then its specific volume is independent of pressure, and the static pressure (as you call it) is equal to the hydrostatic pressure at all depths.
If that’s the case, why do we use this formula for pressure of water inside some container:

##p = p_0 + \rho g h##

Where the second term is hydrostatic pressure. And the first is obviously static (atmospheric) pressure.

When dealing with problem of water flowing through small opening on a container, we setup Bernoulli’s equation this way:

##p_0 + \dfrac{1}{2} \rho v_1^2 + \rho g h_1 =
p_0 + \dfrac{1}{2} \rho v_2^2 + \rho g h_2
##

By cancelling out atmospheric pressures (in this case basically the static pressure) from both sides of equation, we are left with:

##\dfrac{1}{2} \rho v_1^2 + \rho g h_1 =
\dfrac{1}{2} \rho v_2^2 + \rho g h_2
##

Since the gravitational potential energy at the ground level is 0, and since the speed of water level decreasing is so much smaller than the speed of water flowing through the opening, we can transform the equation and thus we get the very well known Torricelli’s theorem:

## v = \sqrt{2gh}##
 
  • #11
adjurovich said:
If that’s the case, why do we use this formula for pressure of water inside some container:

##p = p_0 + \rho g h##

Where the second term is hydrostatic pressure. And the first is obviously static (atmospheric) pressure.

When dealing with problem of water flowing through small opening on a container, we setup Bernoulli’s equation this way:

##p_0 + \dfrac{1}{2} \rho v_1^2 + \rho g h_1 =
p_0 + \dfrac{1}{2} \rho v_2^2 + \rho g h_2
##

By cancelling out atmospheric pressures (in this case basically the static pressure) from both sides of equation, we are left with:

##\dfrac{1}{2} \rho v_1^2 + \rho g h_1 =
\dfrac{1}{2} \rho v_2^2 + \rho g h_2
##

Since the gravitational potential energy at the ground level is 0, and since the speed of water level decreasing is so much smaller than the speed of water flowing through the opening, we can transform the equation and thus we get the very well known Torricelli’s theorem:

## v = \sqrt{2gh}##
All the equations you have written are the incompressible case.
 
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  • #12
Chestermiller said:
All the equations you have written are the incompressible case.
So? Aren’t we talking about the incompressible case?
 
  • #13
adjurovich said:
So? Aren’t we talking about the incompressible case?
Yes, but for a compressible fluid, the equation of state says that the pressure is independent of specific volume.
 
  • #14
Chestermiller said:
Yes, but for a compressible fluid, the equation of state says that the pressure is independent of specific volume.
I’m kinda confused now. What exactly did I say wrong? Also, by specific volume, do you mean just volume of the fluid?
 
  • #15
adjurovich said:
I’m kinda confused now. What exactly did I say wrong? Also, by specific volume, do you mean just volume of the fluid?
Volume per unit mass, For an incompressible fluid,fhe equation of state is degenerate.
 
  • #16
Chestermiller said:
Volume per unit mass, For an incompressible fluid,fhe equation of state is degenerate.
How does this oppose the fact that pressure of water in pool at certain depth will be equal to the sum of static (atmospheric) and hydrostatic pressure?
 
  • #17
adjurovich said:
How does this oppose the fact that pressure of water in pool at certain depth will be equal to the sum of static (atmospheric) and hydrostatic pressure?
It doesn’t.
 
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