In this lesson, we will derive the equations
for the hydrostatic pressure gradient equation.
Imagine a small fluid element of mass dm,
constant density rho, and dimensions dx, dy,
and dz.
This element is part of a fluid where no shearing
stresses exist.
That is, fluid elements are not sliding against
each other.
The small fluid element would experience a
small gravitational force dFg, and small pressure
forces on each of the six faces.
The pressure forces in the y-direction are
labeled dF1 and dF2, the pressure forces in
the z-direction are labeled dF3 and dF4, and
the pressure forces in the x-direction are
labeled dF5 and dF6.
Applying Newton’s second law in the x-direction,
dF6 minus dF5 equals the mass of the fluid
element dm times the acceleration of the fluid
element in the x-direction, ax.
If we label the center of the fluid element
as some point (x, y, z), then both dF5 and
dF6 act at a distance one-half dx away from
the center.
dF5 acts at point (x+dx/2, y, z), and dF6
acts at point (x-dx/2, y, z).
Recalling the pressure is force per area,
we can rewrite the pressure force dF5 as the
pressure at point (x+dx/2, y, z) times the
area dy dz.
The pressure force dF6 is the pressure at
point (x-dx/2, y, z) times the area dy dz.
The mass of the fluid element dm is equal
to the density of the fluid element rho, times
the volume of the fluid element dx dy dz.
Now we substitute the expressions for dF5,
dF6, and dm, into the equation of motion in
the x-direction.
Applying Newton’s second law in the y-direction,
we see that dF2 minus dF1 equals the mass
dm times the acceleration in the y-direction,
ay.
Both dF1 and dF2 act at a distance one-half
dy away from the center.
So dF1 acts at point (x, y+dy/2, z), and dF2
acts at point (x, y-dy/2, z).
We can rewrite the pressure force dF1 as the
pressure at point (x, y+dy/2, z) times the
area dx dz.
And the pressure force dF2 is the pressure
at point (x, y-dy/2, z) times the area dx
dz.
And now we substitute the expressions for
dF1, dF2, and dm into the equation of motion
in y-direction.
Applying Newton’s second law in the z-direction,
dF4 minus dF3 minus dFg equals the mass dm
times the acceleration in the z-direction,
az.
Both dF3 and dF4 act at a distance one-half
dz away from the center.
dF3 acts at point (x, y, z+dz/2), and dF4
acts at point (x, y, z-dz/2).
We can rewrite the pressure force dF3 as the
pressure at point (x, y, z+dz/2) times the
area dx dy.
The pressure force dF4 is the pressure at
point (x, y, z-dz/2) times the area dx dy.
The gravitational force is dm times the gravitational
acceleration g.
Now we substitute in the expressions for dF3,
dF4, dFg, and dm, into the equation of motion
in the z-direction.
There are three equations of motion that contain
six different pressures at locations that
are very close to a central point (x,y,z).
We will now express the six pressures as a
Taylor series expansion about that central
point.
Recall that the Taylor series expansion of
some arbitrary function f about a point is
an infinite series of terms that are calculated
from the values of the function's derivatives
at that point.
For example, a function f evaluated at point
(x+dx,y,z) is equal to the value of f evaluated
at point (x,y,z) plus partial f partial x
evaluated at point (x,y,z) times dx, plus
the second partial derivative of f with respect
to x times dx squared over 2, plus other terms
that contain higher order derivatives and
dx to higher powers.
If dx is very small, as is the case with our
fluid element, we can neglect terms that have
dx squared and higher because they are relatively
very small, and we are left with just two
terms of the Taylor series expansion.
Note that if the positive sign in front of
dx on the left side of the equation were changed
to a negative sign, the positive signs circled
in red on the right side of the equation would
change to negative signs as well.
For the equation of motion in the x-direction,
The pressure at x-dx/2 is approximately the
pressure at (x,y,z) minus partial P partial
x at point (x,y,z) times dx divided by 2.
The pressure at x+dx/2 is approximately the
pressure at (x,y,z) plus partial P partial
x at point (x,y,z) times dx divided by 2.
Plugging in these expressions into the equation
of motion, we see that the pressure at (x,y,z)
drops out.
We can also eliminate dx dy dz from both sides
of the equation and we are left with negative
partial P partial x evaluated at (x,y,z) equals
rho times ax.
For the equation of motion in the y-direction,
the pressure at y-dy/2 is approximately the
pressure at (x,y,z) minus partial P partial
y at point (x,y,z) times dy divided by 2.
The pressure at y+dy/2 is approximately the
pressure at (x,y,z) plus partial P partial
y at point (x,y,z) times dy divided by 2.
Plugging in these expressions into the equation
of motion, we see that the pressure at (x,y,z)
drops out, and we can eliminate dx dy dz from
both sides of the equation.
We are left with negative partial P partial
y evaluated at (x,y,z) equals rho times ay.
For the equation of motion in the z-direction,
the pressure at z-dz/2 is approximately the
pressure at (x,y,z) minus partial P partial
z at point (x,y,z) times dz divided by 2.
The pressure at z+dz/2 is approximately the
pressure at (x,y,z) plus partial P partial
z at point (x,y,z) times dz divided by 2.
Plugging in these expressions into the equation
of motion, we see that the pressure at (x,y,z)
drops out, and we can eliminate dx dy dz from
both sides of the equation.
We are left with negative partial P partial
z evaluated at (x,y,z) minus rho times g,
equals rho times az.
Starting with the equations of motion, we
have obtained three equations that all contain
partial derivatives of pressure evaluated
at point (x,y,z).
For simplicity, the pressure evaluated at
(x,y,z) will now be denoted by the letter
P.
We can rewrite these three scalar equations
as one vector equation.
The sum of the external force vectors acting
on the fluid element is equal to the fluid
element’s mass times the fluid element’s
acceleration vector.
Expanding the vector equation into x, y, and
z components, and comparing this equation
to the three equations of motion above, we
obtain one equation of motion in vector form.
We distribute k-hat to the two terms on the
left and bring the equation to the top of
the screen.
We can write this equation even more compactly
by recalling the definition of the gradient
of a scalar function.
The gradient of some arbitrary scalar function
f that is itself a function of x, y, and z,
is partial f partial x, i-hat, plus partial
f partial y, j-hat, plus partial f partial
z, k-hat.
We write the gradient of f as del f, where
del is the del operator.
The equation of motion can be written in vector
form as negative gradient P, minus gamma k-hat,
equals density times the acceleration vector.
This equation is valid for any fluid where
shear stresses are negligible.
Often the fluid of interest is not accelerating,
such as water in a swimming pool.
In those situations, we set the acceleration
vector to zero.
Moving the term gamma k-hat to the right side
of the equation, and equating the x-terms,
y-terms, and z-terms, we have three equations.
Partial P partial x equals zero and partial
P partial y equals zero mean the pressure
does not vary in the horizontal direction.
Partial P partial z equals negative gamma
means the pressure increases as you descend
through a fluid.
So pressure only varies in the vertical direction.
Since pressure is only a function of z, we
can rewrite the partial derivatives as full
derivatives.
dP/dz equals negative gamma is called the
hydrostatic pressure gradient.
