Basics of Fluid Mechanics and Pressure

[Tanya Sharma]

Hello

I am having difficulty in comprehending the basics of fluid mechanics . I have few questions which I would like to put one at a time , so as not to create any confusion .

In the following attachment , a container with a liquid rests on a surface .I would like to understand what are the forces acting on a liquid molecule(marked in red) at the surface.

1.Atmospheric Pressure P₀ acting vertically downwards(Purple)
2.Pressure from the surrounding liquid P_sur acting vertically upwards(black)
3.Weight of the particle W acting vertically downwards(green)

Is it correct to say P_sur = P₀ + W ?

But pressure at the surface due to surrounding liquid P_sur should be equal to P₀ . Does that mean we neglect W ?

Edit :The red dot represents a liquid molecule at the surface .

 

[boneh3ad]

Well, you are neglecting two forces here which throw of your simple expression up there: buoyancy (due to the displacement of water by the particle) and surface tension (due to the deflection of the water surface). Depending on the size of the particle, these would likely be small, but so would the pressure from the water.

Ultimately, the gauge pressure from the water would be incredibly tiny for a small particle since only a tiny bit of the particle would dip below the surface and get any kind of pressure on it to support the weight. This is due to the fact that the absolute hydrostatic pressure is p_{\text{atm}} + \rho g h, and in this case the depth of the particle, h, ranges from zero to incredibly tiny.

 

[Tanya Sharma]

Well, you are neglecting two forces here which throw of your simple expression up there:

Okay...that does increase the complexity of the analysis . One thing I want to clear is that by particle I am talking about a liquid molecule at the surface of the liquid .I am sorry if I may have caused some confusion.

buoyancy (due to the displacement of water by the particle)

Are you sure buoyancy comes into picture when we are analysing forces on a liquid molecule at the surface ? I think it applies only to external object submerged in water .

and surface tension (due to the deflection of the water surface).

Is the force due to surface tension acting vertically downwards ?

Depending on the size of the particle, these would likely be small, but so would the pressure from the water.

Again,I am talking about a liquid molecule at the surface .

Ultimately, the gauge pressure from the water would be incredibly tiny for a small particle since only a tiny bit of the particle would dip below the surface and get any kind of pressure on it to support the weight. This is due to the fact that the absolute hydrostatic pressure is p_{\text{atm}} + \rho g h, and in this case the depth of the particle, h, ranges from zero to incredibly tiny.

Right

 

[voko]

The problem with analysing molecules is that they don't behave as some stationary objects. At the molecular level, there is no pressure to speak of, there are random walk and collisions. No buoyant force either, because it depends on pressure and density, and "density" of a molecule is something that is very tricky to define.

Usually fluid dynamics deals with "parcels" of matter, which are tiny but still contain huge numbers of molecules, and the notions like pressure and density are applicable.

 

[Tanya Sharma]

Okay...so if I need to write a force equation then I should consider a column of water of say height 'h' and cross section area 'A' instead of a single liquid molecule . Is it correct ?

 

[voko]

Yes, that is how it is typically done. Surface tension, however, will need to be accounted for differently, if you want to do that at all.

 

[Tanya Sharma]

Thanks...

Another doubt I have is regarding the pressure at the surface of liquid .

Is there a single pressure acting at the air liquid interface or are there different pressures on either side of the surface(air on top and liquid beneath the surface) ?

 

[voko]

If there is surface tension, the pressures must be equal. Whether that is a "single pressure" is not a very meaningful question, because microscopically pressure is really just motion of molecules, and at the interface you will find molecules of both fluids.

If the surface tension is present, there is a pressure difference.

 

[Tanya Sharma]

If there is surface tension, the pressures must be equal. Whether that is a "single pressure" is not a very meaningful question, because microscopically pressure is really just motion of molecules, and at the interface you will find molecules of both fluids.

If the surface tension is present, there is a pressure difference.

I think you meant "if there is no surface tension" in the first line .

 

[Tanya Sharma]

In the attached figure the container is accelerating upwards with acceleration 'a' .

The forces acting on a column of height 'h' and cross section 'A' are

1) Force due to atmospheric pressure P₀A acting vertically downwards (purple)
2) Force due to liquid beneath the column PA acting vertically upwards(red)
3) Weight ρhAg acting vertically downwards (pink)
4) pseudo force ρhAa acting vertically downwards (black)

Doing ∑F = 0

P₀A+ρhAg+ρhAa = PA

or P = P₀+ρh(g+a)

Is this the correct expression for the pressure at a point in the liquid in a container accelerating upwards ?

 

[voko]

Things get tricky here, because now air will have dynamic pressure in addition to the static pressure. Ignoring that, your equation is correct.

It follows immediately from Einstein's equivalence principle, by the way.

 

[Tanya Sharma]

Things get tricky here, because now air will have dynamic pressure in addition to the static pressure. Ignoring that, your equation is correct.

And if that is a closed container with vacuum at top of liquid then P=ρh(g+a) . Right ?

It follows immediately from Einstein's equivalence principle, by the way.

What does this principle tell us ?

 

[voko]

And if that is a closed container with vacuum at top of liquid then P=ρh(g+a) . Right ?

Right.

What does this principle tell us ?

That acceleration cannot be distinguished from gravitation in small volumes. So if you have "real" gravity $\vec g$ and "real" acceleration $\vec a$, you cannot really say whether you have "pure" gravity $\vec g - \vec a$, or "pure" acceleration $\vec a - \vec g$.

 

[Chestermiller]

Just a few comments:

1. To elaborate on what Voko said earlier, when you are dealing with concepts like pressure and density, you need to be thinking at the "continuum" level, not the molecular level. Once you do that, all your difficulties will vanish.

2. The analysis you did on the accelerated column was right on target. Nice job. As far as what was said about the added pressure required to push the air above, this is the same kind of thing as worrying about air drag on a solid. Not very important from your current perspective.

3. As far as surface tension is concerned, if the surface is flat, it does not come into play in the force balance perpendicular to the surface. Only if the surface is curved is there a difference in pressure across the interface. The surface tension always acts tangent to the surface.

Hope this helps.

Chet

 

[Tanya Sharma]

Hi Chet

Thanks for chipping in .

I would like to discuss the case of accelerating fluid in horizontal direction.

I have attached the figure .

If I set up my coordinate axis with + y-axis upwards and + x-axis towards right .θ is the angle which the liquid surface makes with the horizontal .Then,

PAcosθ-P₀Acosθ-ρhAg = 0

PAsinθ-P₀Asinθ-ρhAa = 0

Hence,tanθ = a/g

If I set up my coordinate axis with + y-axis upwards perpendicular to the surface and + x-axis downwards towards right .Then,

PA - P₀A-ρhAg_eff = 0

where g_eff = g - a and magnitude of g_eff = √(a²+g²)

Is my analysis correct ?

 

[voko]

Yes, your analysis and its result are correct. C.f. #13.

 

[Chestermiller]

I think this analysis is correct, but it is not the way I would have done it. I'll present the method I would have used, just to provide a different perspective.

At a given time, the pressure is going to be a function of both x and y, so p = p(x,y), and

dp=\frac{\partial p}{∂x}dx+\frac{\partial p}{∂y}dy

I'm going to take a tiny control volume of fluid in the middle of figure with sides Δx and Δy, and width Δz into the paper, and I'm going to do horizontal and vertical force balances on the fluid in the control volume:
[p(x,y)-p(x+Δx,y)]ΔyΔz=ρ(ΔxΔyΔz)a
[p(x,y)-p(x,y+Δy)]ΔyΔz=ρ(ΔxΔyΔz)g
If we divide both sides of this equation by the volume (ΔxΔyΔz) and take the limit as the increments become small, we obtain:
-\frac{∂p}{∂x}=ρa
-\frac{∂p}{∂y}=ρg
The surfaces of constant p are characterized by dp = 0, or
\frac{\partial p}{∂x}dx+\frac{\partial p}{∂y}dy=0
If we substitute the results of the force balance into this equation, we get:
adx+gdy=0
or, equivalently,
\left(\frac{∂y}{∂x}\right)_p=-\frac{a}{g}
Since the free surface is a surface of constant p, this is the slope of the interface.
If we want to find the pressure at a distance h perpendicular to the interface, we take Δx=-hsinθ and Δy=-hcosθ so that:
P=P_0-hsinθ\frac{∂p}{∂x}-hcosθ\frac{∂p}{∂y}=P_0+ρahsinθ+ρghcosθ
But sinθ=\frac{a}{\sqrt{a^2+g^2}} and cosθ=\frac{g}{\sqrt{a^2+g^2}}
So,
P=P_0+ρ\sqrt{a^2+g^2} h

This is the same as your result.

Chet

 

[Tanya Sharma]

Excellent analysis...Thank you very much for enhancing my knowledge .

Why does the liquid surface of accelerating fluid forms an angle with the horizontal and becomes perpendicular to g_eff ?

My thoughts - Because if it weren't perpendicular then there would be a component of g_eff parallel as well as perpendicular to surface.

There can be no net force perpendicular to the surface (otherwise water would move in that direction and change the surface shape)

There can be no net force parallel to the surface (otherwise layers of water would slip against each other).I think fluids have a no slip condition .But if layers are undergoing same acceleration parallel to surfaces ,why should they slip ?

Not sure . What are Your views ?

 

[voko]

Tanya, your reasoning about the orientation of the surface is correct. Liquids can withstand compression, but they cannot resist any shear. So if they can have any stable configuration at all, this is the configuration where the net force is everywhere perpendicular to the surface.

 

[Chestermiller]

Excellent analysis...Thank you very much for enhancing my knowledge .

Why does the liquid surface of accelerating fluid forms an angle with the horizontal and becomes perpendicular to g_eff ?

In order to accelerate the fluid to the right, you need a higher pressure on the left than on the right (at a given value of y). So to get the higher pressure on the left, you need more of a fluid column to the left than to the right. That translates into a tilted surface. The pressure gradient vector is perpendicular to surfaces of constant pressure. The interface is one such surface. So the maximum directional pressure derivative is perpendicular to the interface.

My thoughts - Because if it weren't perpendicular then there would be a component of g_eff parallel as well as perpendicular to surface.

There can be no net force perpendicular to the surface (otherwise water would move in that direction and change the surface shape)

There can be no net force parallel to the surface (otherwise layers of water would slip against each other).I think fluids have a no slip condition .But if layers are undergoing same acceleration parallel to surfaces ,why should they slip ?

Not sure . What are Your views ?

The stress vector at the interface must be continuous across the interface. If you do a force balance on a tiny element of area of interface, you can see this right away because the mass involved is zero. The stress on the air side of the interface is P₀, and is perpendicular to the interface. There is no component tangent to the interface (shear stress). So, on the fluid side of the interface, the normal stress must be P₀ and the shear stress must be zero.

Incidentally, the no-slip boundary condition means that the velocity vector is continuous at an interface (even if the interface is solid). A fluid can be shearing at a stationary solid boundary and still satisfy the no-slip boundary condition (zero velocity, but velocity gradient normal to boundary). In our problem, there is no shearing at the interface because the shear stress imposed by the air is zero.

 

[Tanya Sharma]

In order to accelerate the fluid to the right, you need a higher pressure on the left than on the right (at a given value of y). So to get the higher pressure on the left, you need more of a fluid column to the left than to the right. That translates into a tilted surface. The pressure gradient vector is perpendicular to surfaces of constant pressure. The interface is one such surface. So the maximum directional pressure derivative is perpendicular to the interface.

Very Nice...

The stress vector at the interface must be continuous across the interface. If you do a force balance on a tiny element of area of interface, you can see this right away because the mass involved is zero. The stress on the air side of the interface is P₀, and is perpendicular to the interface. There is no component tangent to the interface (shear stress). So, on the fluid side of the interface, the normal stress must be P₀ and the shear stress must be zero.

Does this reasoning go equally good for why pressure at the liquid surface is equal to the atmospheric pressure ?

I have few more doubts

1)How do we say that the pressure at the liquid surface is equal to atmospheric pressure ?

Using P=P₀ + gh ,and putting h=0 makes sense . But is there a different qualitative explanation ? If the reasoning you have given above goes with this question also,then it is fine.

2)What exactly is pressure ?

I think of it as Force/Unit Area due to fluid molecules colliding with each other .It only pushes not pulls .Is it correct thinking ?

3)Can we treat liquids just as a rigid body ? Is there a center of mass of the fluid where we can consider all the forces to be acting ?

4)Does Surface Tension act only on curved surface .Does that mean it comes into picture only in case of tubes of small radius containing fluids . In containers with large radius are we neglecting Surface Tension or is it totally absent .

 

[Chestermiller]

Does this reasoning go equally good for why pressure at the liquid surface is equal to the atmospheric pressure ?

Yes.

I have few more doubts

1)How do we say that the pressure at the liquid surface is equal to atmospheric pressure ?

Using P=P₀ + gh ,and putting h=0 makes sense . But is there a different qualitative explanation ? If the reasoning you have given above goes with this question also,then it is fine.

The reasoning is the same. The pressure is continuous across the interface, and is equal to the pressure imposed by the air (neglecting surface tension on highly curved surfaces).

2)What exactly is pressure ?

I think of it as Force/Unit Area due to fluid molecules colliding with each other .It only pushes not pulls .Is it correct thinking ?

Yes.

3)Can we treat liquids just as a rigid body ? Is there a center of mass of the fluid where we can consider all the forces to be acting ?

Not really. Liquids can undergo huge deformations, and forces are transferred between fluid parcels. Not all parts of the fluid are moving identically. Often, we have to look at fluids by considering differential elements.

4)Does Surface Tension act only on curved surface .Does that mean it comes into picture only in case of tubes of small radius containing fluids . In containers with large radius are we neglecting Surface Tension or is it totally absent .

Surface tension is a property of the fluid, and is independent of the container. It doesn't only act on curved surfaces. But it does always act tangent to the surface. Think of it as an extendable membrane stretched over the surface, separating the gas phase from the liquid phase.

Chet

 

[Tanya Sharma]

Chet...Thanks for your valuable input.

While studying FLT , work done by an expanding gas is given by pΔV .

My very basic doubts are

1)Is 'p' pressure of the gas or external pressure of the surroundings ?If gas is expanding then surely gas pressure is more than that of the surroundings and vica versa .What is the relationship between the two pressures?

2)There are three things ,gas,container fitted with piston and surroundings .Now ,gas is exerting pressure on the piston i.e doing work on the piston .Piston is doing work on the surroundings. Then why is that it is considered that gas is doing work on the surroundings ? I mean who is doing work on whom ?

3)Is the piston considered in the arrangement always massless or it may have some mass ? How does that affect our results ?

 

[Chestermiller]

Chet...Thanks for your valuable input.

While studying FLT , work done by an expanding gas is given by pΔV .

My very basic doubts are

1)Is 'p' pressure of the gas or external pressure of the surroundings ?If gas is expanding then surely gas pressure is more than that of the surroundings and vica versa .What is the relationship between the two pressures?

2)There are three things ,gas,container fitted with piston and surroundings .Now ,gas is exerting pressure on the piston i.e doing work on the piston .Piston is doing work on the surroundings. Then why is that it is considered that gas is doing work on the surroundings ? I mean who is doing work on whom ?

3)Is the piston considered in the arrangement always massless or it may have some mass ? How does that affect our results ?

Tanya,

My Blog at my PF personal page may answer many of these questions for you. If not, might I suggest starting a new thread under the heading of first law thermo?

Chet

 

[Tanya Sharma]

Okay...I will proceed as per your suggestions.

Thank you very much for your time and patience in explaining the subject matter.

 

[Tanya Sharma]

Hello Chet

1) When a container with liquid and having a hole in the side wall is undergoing free fall , the liquid doesn't come out . Is it correct to attribute this to the fact that since there is no hydrostatic pressure in the liquid undergoing free fall the pressure on both sides of the hole will be solely atmospheric . Since there is no pressure difference across the hole the water doen't come out ?Does the reasoning make sense ?

2) Is it correct to say that under free fall ,the pressure througout the liquid is stmospheric pressure ?

 

[Chestermiller]

Hello Chet

1) When a container with liquid and having a hole in the side wall is undergoing free fall , the liquid doesn't come out . Is it correct to attribute this to the fact that since there is no hydrostatic pressure in the liquid undergoing free fall the pressure on both sides of the hole will be solely atmospheric . Since there is no pressure difference across the hole the water doen't come out ?Does the reasoning make sense ?

Yes.

2) Is it correct to say that under free fall ,the pressure througout the liquid is stmospheric pressure ?

Yes, aside from sloshing inertial effects.

Chet

 

[Tanya Sharma]

Thanks .

 

[Tanya Sharma]

Chet ,

A block of wood floats in a bucket of water placed in a lift .If the lift starts accelerating up ,will the block sink more ,sink less or stays at the same level ?

I think the block will sink more as the buoyant force has to increase ( block needs to displace more water) in order to provide net acceleration to the block in upward direction .

Is it correct ?

 

[Chestermiller]

Chet ,

A block of wood floats in a bucket of water placed in a lift .If the lift starts accelerating up ,will the block sink more ,sink less or stays at the same level ?

I think the block will sink more as the buoyant force has to increase ( block needs to displace more water) in order to provide net acceleration to the block in upward direction .

Is it correct ?

What an interesting question! In the accelerating frame of reference of the elevator, what is the pressure of the water as a function of depth?

Chet

 

[Tanya Sharma]

$P(z) = ρzg_{eff}$ ,where z is measured from the surface of water . In this case $g_{eff} = g+a$

 

[Chestermiller]

$P(z) = ρzg_{eff}$ ,where z is measured from the surface of water . In this case $g_{eff} = g+a$

Right. So, if V is the volume of displaced water, what is the buoyant force now?

Chet

 

[Tanya Sharma]

$F_b = Vρg_{eff}$ , where $ρ$ is density of water .

 

[Chestermiller]

$F_b = Vρg_{eff}$ , where $ρ$ is density of water .

So, from Newton's second law, if M is the mass of the block, what is the displaced volume of water in the accelerating elevator? How does this compare with the displaced volume if the elevator is not accelerating?

Chet

 

[Tanya Sharma]

So, from Newton's second law, if M is the mass of the block, what is the displaced volume of water in the accelerating elevator? How does this compare with the displaced volume if the elevator is not accelerating?

Chet

When the elevator is not accelerating then $Mg = V_1ρg$ .

When the elevator is accelerating up then $Mg = V_2ρ(g+a)$ .

$V_1 = V_2(1+\frac{a}{g})$ or $V_1>V_2$ . The displaced fluid when the elevator is accelerating is less i.e the block sinks less ( moves upwards ) as compared to when the elevator is at rest .

Is this what you are suggesting ?

 

[Chestermiller]

When the elevator is not accelerating then $Mg = V_1ρg$ .

When the elevator is accelerating up then $Mg = V_2ρ(g+a)$ .

$V_1 = V_2(1+\frac{a}{g})$ or $V_1>V_2$ . The displaced fluid when the elevator is accelerating is less i.e the block sinks less ( moves upwards ) as compared to when the elevator is at rest .

Is this what you are suggesting ?

No. Try that force balance again, and this time use a free body diagram on the block.

Chet

 

[Tanya Sharma]

$V_1ρg -Mg = 0$

$V_2ρ(g+a) - Mg - Ma = 0$ i.e $V_1=V_2$ ?

 

[Chestermiller]

$Mg - V_1ρg = 0$

$V_2ρ(g+a) - Mg - Ma = 0$ i.e $V_1=V_2$ ?

Yes. This is what will happen after the block stops bobbing. Initially, there will be a transient, and the block will dip and then bob, but, after the transient bobbing dies out, the displaced volume will be the same as without the acceleration.

Chet

 

[Chestermiller]

On second thought, I take back what I said about the dip and bobbing. There should not be a transient even if the acceleration is applied suddenly. Sorry for the confusion.

 

[Tanya Sharma]

If we write Newton's 2nd law for the block in the ground frame (not from the accelerated frame) .

$F_{b1} - Mg = 0$

$F_{b2} - Mg = Ma$ i.e $F_{b2} =M(g+a)$

$F_{b2} > F_{b1}$ . Since the buoyant force is more in accelerating elevator case , fluid displaced should be more . This is the same reasoning I gave in post #29 .

What is the flaw in this reasoning ?

 

[Chestermiller]

If we write Newton's 2nd law for the block in the ground frame (not from the accelerated frame) .

$F_{b1} - Mg = 0$

$F_{b2} - Mg = Ma$ i.e $F_{b2} =M(g+a)$

$F_{b2} > F_{b1}$ . Since the buoyant force is more in accelerating elevator case , fluid displaced should be more . This is the same reeasoning I gave in post #29 .

What is the flaw in this reasoning ?

If the water is considered incompressible, then the buoyant force imposed by the water will respond instantly to the acceleration. Your second equation here confirms the analysis in post # 37 showing that the displacement will be the same.

Chet

 

[Delta2]

it is F_{b2}>F_{b1} but because the fluid weights more now (it is like the gravitational acceleration g is increased by a) not because there is more volume of fluid displaced.

 

[Tanya Sharma]

it is F_{b2}>F_{b1} but because the fluid weights more now (it is like the gravitational acceleration g is increased by a) not because there is more volume of fluid displaced.

You are right .

 

[Tanya Sharma]

Chet

There is an ice cube floating in a glass of water . What happens to the water level if

1) ice cube has a cavity with air inside(the cavity).

2) ice cube has a cavity with water inside(the cavity) .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?

 

[Chestermiller]

Chet

There is an ice cube floating in a glass of water . What happens to the water level if

1) ice cube has a cavity with air inside(the cavity).

2) ice cube has a cavity with water inside(the cavity) .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?

I don't quite understand the question. The base case of comparison is a solid ice cube?

Chet

 

[Tanya Sharma]

No . The comparison is with the level of water present in the glass before the ice melts .Sorry for being unclear .

What happens to the original water level if

1) an ice cube having a cavity with air inside(the cavity) , floating on the water melts.

2) ice cube having a cavity with water inside(the cavity) , floating on the water melts .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?

 

[Chestermiller]

No . The comparison is with the level of water present in the glass before the ice melts .Sorry for being unclear .

What happens to the original water level if

1) an ice cube having a cavity with air inside(the cavity) , floating on the water melts.

2) ice cube having a cavity with water inside(the cavity) , floating on the water melts .

I think that the water level goes down in the first case and remains same in the second case .

Do you agree ?

Off hand, I find it rather tricky to reason out. Why don't you just model it, and let the math do the work for you? I would do either a cubical cavity or a spherical cavity.

Chet

 

[Tanya Sharma]

I have done the maths(not much involved) . Just wanted to reassure myself by confirming the result with you :)

 

[Chestermiller]

I have done the maths(not much involved) . Just wanted to reassure myself by confirming the result with you :)

I confirm your result for item 2. For item 1, it depends on whether you take into account the slight density of the air. If you include the density of the air, then the level goes down slightly, but, if you neglect the density of the air, then I get no change.

Chet

 

[Delta2]

I vent done the math but in the first case you assume that after the ice cube melts, the air is released to the atmosphere or dissolved into the water?