Basics of Fluid Mechanics, 2014a

96 CHAPTER 4. FLUIDS STATICS
Calculation of
the correction
factor
dA
ω
Rotation
center
S
constant pressure line
L
xL
Fig. -4.17. Schematic angular angle to explain example 4.11.
It is first assumed that the height is uniform at the tube (see for the open question on
this assumption). The pressure at the interface at the two sides of the tube is same.
Thus, equation (4.89) represents the pressure line. Taking the “left” wing of U tube
change in z direction
{ }} {
z l − z 0 =
The same can be said for the other side
change in r direction
{ }} {
ω 2 L 2
2 g
z r − z 0 = ω2 x 2 L 2
2 g
Thus subtracting the two equations above from each each other results in
(
z r − z l = Lω2 1 − x 2)
2 g
It can be noticed that this kind equipment can be used to find the gravity.
End Solution
Example 4.12:
Assume that the diameter of the U tube is R t . What will be the correction factor if the
curvature in the liquid in the tube is taken in to account. How would you suggest to
define the height in the tube?
Solution
In Figure 4.17 shows the infinitesimal area used in these calculations. The distance of
the infinitesimal area from the rotation center is ?. The height of the infinitesimal area
is ?. Notice that the curvature in the two sides are different from each other. The
volume above the lower point is ? which is only a function of the geometry.
End Solution
Example 4.13:
In the U tube in example 4.11 is rotating with upper part height of l. At what rotating