4 4 Pressure Distribution in Ro


Pressure Distribution in Rotating Flows http://edugen.wiley.com/edugen/courses/crs2436/crowe9771/crowe9771...
4.4 Pressure Distribution in Rotating Flows
Situations in which a fluid rotates as a solid body are found in many engineering applications. One common
application is the centrifugal separator. The centripetal accelerations resulting from rotating a fluid separate the
heavier elements from the lighter elements as the heavier elements move toward the outside and the lighter
elements are displaced toward the center. A milk separator operates in this fashion, as does a cyclone separator
for removing particulates from an air stream.
To learn how pressure varies in a rotating, incompressible flow, apply Euler's equation in the direction normal to
the streamlines and outward from the center of rotation. In this case the fluid elements rotate as the spokes of a
wheel, so the direction ! in Euler's equation, Eq. (4.8), is replaced by r giving
(4.9)
where the partial derivative has been replaced by an ordinary derivative since the flow is steady and a function
only of the radius r. From Eq. (4.5), the acceleration in the radial direction (away from the center of curvature) is
and Euler's equation becomes
(4.10)
For a liquid rotating as a rigid body,
Substituting this velocity distribution into Euler's equation results in,
(4.11)
Integrating Eq. (4.11) with respect to r gives
(4.12)
or
(4.13a)
This equation can also be written as
(4.13b)
These equivalent equations describe the pressure variation in rotating flow.
The equation for pressure variation in a rotating flow is used in Example 4.4 to predict the surface profile of a
liquid in a rotating tank.
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Pressure Distribution in Rotating Flows http://edugen.wiley.com/edugen/courses/crs2436/crowe9771/crowe9771...
EXAMPLE 4.4 SURFACE PROFILE OF ROTATI G
LIQUID
A cylindrical tank of liquid shown in the figure is rotating as a solid body at a rate of 4 rad/s. The
tank diameter is 0.5 m. The line AA depicts the liquid surface before rotation, and the line A2 A2 shows
the surface profile after rotation has been established. Find the elevation difference between the liquid
at the center and the wall during rotation.
Sketch:
Problem Definition
Situation: Liquid rotating in a cylindrical tank.
Find: Elevation difference (in meters) between liquid at center and at the wall.
Assumptions: Fluid is incompressible.
Plan
Pressure at liquid surface is constant (atmospheric).
1. Apply equation for pressure variation in rotating flow, Eq. (4.13a), between points 1 and 2.
2. Evaluate elevation difference.
Solution
1. Equation (4.13a) applied between points 1 and 2.
The pressure at both points is atmospheric, so p1 = p2 and the pressure terms cancel out. At
point 1, r1 = 0, and at point 2, r = r2. The equation reduces to
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Pressure Distribution in Rotating Flows http://edugen.wiley.com/edugen/courses/crs2436/crowe9771/crowe9771...
2. Evaluation of elevation difference:
Review
Notice that the surface profile is parabolic.
Example 4.5 illustrates the application of the equation for pressure variation in rotating flows to a rotating
manometer.
EXAMPLE 4.5 ROTATI G MA OMETER TUBE
When the U-tube is not rotated, the water stands in the tube as shown. If the tube is rotated about the
eccentric axis at a rate of 8 rad/s, what are the new levels of water in the tube?
Problem Definition
Situation: Manometer tube is rotated around an eccentric axis.
Find: Levels of water in each leg.
Assumptions: Liquid is incompressible.
Sketch:
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Pressure Distribution in Rotating Flows http://edugen.wiley.com/edugen/courses/crs2436/crowe9771/crowe9771...
Plan
The total length of the liquid in the manometer must be the same before and after rotation, namely
90 cm. Assume, to start with, that liquid remains in the bottom leg. The pressure at the top of the
liquid in each leg is atmospheric.
1. Apply the equation for pressure variation in rotating flows, Eq. (4.13a), to evaluate difference
in elevation in each leg.
2. Using constraint of total liquid length, find the level in each leg.
Solution
1. Application of Eq. (4.13a) between top of leg on left (1) and on right (2):
2. The sum of the heights in each leg is 36 cm.
Solution for the leg heights:
Review
If the result was a negative height in one leg, it would mean that one end of the liquid column would
be in the horizontal leg, and the problem would have to be reworked to reflect this configuration.
Copyright © 2009 John Wiley & Sons, Inc. All rights reserved.
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