We wish to manipulate Euler's equation for this situation,
to expose the pressure. Since partial derivatives of the pressure appear,
we expect progress if we integrate both sides of the equation along a well
defined path. It turns out to be enough if we take the scalar product of both
sides with respect to an element 
, and then integrate both
sides along a convenient path.
(Since pressure is a scalar field, with a well defined value at each point in space, the result must be independent of the path that we choose.)
|
Path of integration from origin to a point with position vector
 |
We choose our path in segments so that along each segment, only one
coordinate varies. This way, the path takes over the job of holding the other
coordinates constant, and a partial derivative along the path can be written as
an ordinary derivative. We begin the path at the origin, and end it at the
location specified by , anticipating that the
definite integral will give information about the pressure difference between
the two locations.
Our path is illustrated with dotted lines in the figure which defines
and
: We move
along the x-axis till we reach the x component of . Then we move along the
z-axis till we reach 
. Finally we move along the
y-axis till we reach .
We recall the definition of ,
,
and see that the left side of Euler's equation integrates as follows:
where 
.
The integral of the 
term on the right side is
even easier:
.
The integral over the pressure gradient takes a few more lines. To make it
as easy as possible we assume that the density is constant. We factor out the
term
and integrate only the gradient:
Because the integration path holds the proper variables constant, we can integrate as though the arguments were full derivatives of the pressure. Thus
Since the pressure is a scalar field, uniquely defined at each point in space, this pressure difference is independent of path, and
where 
is the pressure at position and 
is the pressure at the
origin. Our integral of Euler's equation,
becomes
,
giving us
.
The pressure difference includes the "usual" 
term, plus a "centrifugal"
term.
DISCUSSION:
The pressure increase in a rotating paint bucket with depth is the "usual" pressure; it is the same as for a stationary bucket of water. The "centrifugal" term increases with (the square of) the angular velocity and the distance from the axis of rotation.
These latter two effects are important for centrifuges: A high speed centrifuge spins samples to achieve severe pressure gradients. The pressure may be increased either by spinning faster, or by increasing the distance of the sample from the axis of rotation.
In the case of the paint bucket, the lid on the bucket must be firmly attached, since it must "provide" high pressure near the bucket walls. The one-atmosphere pressure provided by the air is not enough to hold the lid down when the bucket is spinning fast. If the lid is loose, the high pressure water at the bucket wall will push the lid off, and escape the bucket. Our example then changes to something more like the spin cycle in a washing machine.
In fact, if we put holes in the side of the bucket, the high pressure at the walls will squirt water out through the holes. If we arrange to replace the lost water by feeding it into the center of the bucket, then the spinning bucket keeps squirting water out. We have invented the centrifugal pump. The centrifugal pump has been invented several times already, and is a quite useful tool. Sump pumps and the gasoline engine pumps used to empty ponds and ditches of water are usually centrifugal pumps.
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