This screencast will explain how pressure in a fluid with no shear stresses varies from
point to point.
What we are going to do is look at an element of fluid with surface forces.
Our surfaces forces are going to consists of pressures forces as well as our body forces
which are going to consist of weight.
Let's start with out surface forces.
We are going to designate right here in the middle of our element the pressure P.
We are going to express P on each one of the faces in terms of our P and its derivatives
which is actually doing a taylor series and neglecting the higher order terms.
Let's start by looking at the surface forces in the z direction.
I have taken the top which is right here of our elemental cube and the bottom.
The volume of our elemental cube is dy*dx*dz.
Again we are looking at this top face and we express our surface force in the z direction
as P.
We are going to start in the center of the cube and then we are going to go up.
We go dP/dz which is that change in pressure with respect to the z direction.
Since we are going half way up the dz side we multiply that by dz/2.
Because this is a force we have to multiply it by the area that it is acting on.
Now let's look at the bottom.
Here again we start with out P and because we are going in the downward direction from
the P in the center it is minus our dP/dz times dz/2 all multiples by our area.
If we put these together what we are looking at is the force that is coming in minus the
force that is coming out.
That is P minus dP/dz*dz/2 minus P plus dP/dz*dz/2.
All of that is multiplied bu this dx*dy.
This is going to be the force in the z direction.
As we look at this if we take this and do the math we end up that this is minus dP/dz
times dx*dx*dz since right here we have this P minus this P and then this is a negative
minus a positive so we add them together.
That is why we get the dz/2 plus dz/2 which is our dz times our dx*dy.
We do this same thing for every other face.
We will get the same expressions except how P varies.
dFy is equal to -dP /dy*dx*dy*dz.
Now if we write this in terms of a vector it is going to equal minus dP/dx which is
in the i direction plus dP/dy in the j direction plus dP/dz in the k direction.
This entire expression is multiplied by our dx*dy*dz.
If you notice these dP/dx, dP/dy, dP/dz terms we can actually write this as del P where
del equals d something / dx in the i direction plus d something / dy in the j direction plus
d something / dz in the k direction.
In our case our d something is P.
Now we can rewrite this vector here.
Divide it by the volume as minus del P.
Now we have to look at the body forces.
The z axis is vertical.
The weight only acts in the vertical direction so what we are left with in out body force
as minus the differential weight in the k direction which can be written as minus the
specific weight times dx*dy*dz.
Notice it is only in the k direction.
Again it is negative because the weight is in the negative z direction.
Now let's apply Newtons second law to this and again it is to the fluid element.
The sum of the forces equals our differential mass times acceleration.
We can write that as our surface force minus our weight which is only in the k direction.
When we put this all together and simplify it.
Minus del P times our differential volume minus our specific volume times our differential
volume in the k equals rho dx*dy*dz times acceleration.
Dividing through by volume we find that minus del P minus the specific gravity in the k
direction equals rho times the acceleration.
This is the equation of motion with no shear stresses.