# Basics of Fluid Mechanics and Pressure

by Tanya Sharma
Tags: basics, fluid, mechanics, pressure
 Thanks P: 5,869 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.
Mentor
P: 5,375
 Quote by 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 ?
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 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 ?
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.
P: 1,344
 Quote by Chestermiller 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...

 Quote by Chestermiller 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 explaination ? 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 .
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P: 5,375
 Quote by Tanya Sharma 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 explaination ? 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
 P: 1,344 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 ?
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P: 5,375
 Quote by 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 ?
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
 P: 1,344 Okay...I will proceed as per your suggestions. Thank you very much for your time and patience in explaining the subject matter.

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