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Basics of Fluid Mechanics, 2014a

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142 CHAPTER 4. FLUIDS STATICS<br />

Fig. -4.50. The cross section <strong>of</strong> the interface. The purple color represents the maximum heavy<br />

liquid raising area. The yellow color represents the maximum lighter liquid that are “going<br />

down.”<br />

The maximum possible radius <strong>of</strong> the depression depends on the geometry <strong>of</strong> the<br />

container. For the cylindrical geometry, the maximum depression radius is about half<br />

for the container radius (see Figure 4.50). This radius is limited because the lighter<br />

liquid has to enter at the same time into the heavier liquid zone. Since the “exchange”<br />

volumes <strong>of</strong> these two process are the same, the specific radius is limited. Thus, it can<br />

be written that the minimum radius is<br />

√<br />

r mintube =2<br />

2 πσ<br />

g (ρ L − ρ G )<br />

(4.194)<br />

The actual radius will be much larger. The heavier liquid can stay on top <strong>of</strong><br />

the lighter liquid without being turned upside down when the radius is smaller than<br />

the equation 4.194. This analysis introduces a new dimensional number that will be<br />

discussed in a greater length in the Dimensionless chapter. In equation (4.194) the<br />

angle was assumed to be 90 degrees. However, this angle is never can be obtained.<br />

The actual value <strong>of</strong> this angle is about π/4 to π/3 and in only extreme cases the<br />

angle exceed this value (considering dynamics). In Figure 4.50, it was shown that the<br />

depression and the raised area are the same. The actual area <strong>of</strong> the depression is only<br />

a fraction <strong>of</strong> the interfacial cross section and is a function. For example,the depression<br />

is larger for square area. These two scenarios should be inserting into equation 4.168<br />

by introducing experimental coefficient.<br />

Example 4.33:<br />

Estimate the minimum radius to insert liquid aluminum into represent tube at temperature<br />

<strong>of</strong> 600[K]. Assume that the surface tension is 400[mN/m]. The density <strong>of</strong> the<br />

aluminum is 2400kg/m 3 .<br />

Solution<br />

The depression radius is assume to be significantly smaller and thus equation (4.193)

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