# Basics of Fluid Mechanics and Pressure

1. ### 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 Psur acting vertically upwards(black)
3.Weight of the particle W acting vertically downwards(green)

Is it correct to say Psur = P0 + W ?

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

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

#### Attached Files:

• ###### fluid.PNG
File size:
4 KB
Views:
101
Last edited: Feb 19, 2014

1,938
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.

3. ### Tanya Sharma

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.

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 .

Is the force due to surface tension acting vertically downwards ?

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

Right

Last edited: Feb 19, 2014
4. ### 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.

5. ### 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 ?

Last edited: Feb 19, 2014
6. ### 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.

7. ### 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) ?

8. ### 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.

9. ### Tanya Sharma

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

10. ### 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 ?

File size:
6.4 KB
Views:
74
11. ### 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.

12. ### Tanya Sharma

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

What does this principle tell us ?

13. ### voko

Right.

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##.

### Staff: Mentor

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

1 person likes this.
15. ### 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-ρhAgeff = 0

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

Is my analysis correct ?

File size:
16.3 KB
Views:
61
16. ### voko

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

### Staff: Mentor

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

18. ### 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 geff ?

My thoughts - Because if it weren't perpendicular then there would be a component of geff 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 ?

Last edited: Feb 20, 2014
19. ### 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.

### Staff: Mentor

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.
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.

1 person likes this.