Basics of Fluid Mechanics, 2014a

108 CHAPTER 4. FLUIDS STATICS
To illustrate how to work with these equations the following example is provided.
Example 4.16:
Consider the hypothetical Figure 4.25. The last layer is made of water with density
of 1000[kg/m 3 ]. The densities are ρ 1 = 500[kg/m 3 ], ρ 2 = 800[kg/m 3 ], ρ 3 =
850[kg/m 3 ], and ρ 4 = 1000[kg/m 3 ]. Calculate the forces at points a 1 and b 1 . Assume
that the layers are stables without any movement between the liquids. Also neglect all
mass transfer phenomena that may occur. The heights are: h 1 =1[m], h 2 =2[m],
h 3 =3[m],and h 4 =4[m]. The forces distances are a 1 =1.5[m], a 2 =1.75[m], and
b 1 =4.5[m]. The angle of inclination is is β =45 ◦ .
Solution
Since there are only two unknowns,
only two equations are
needed, which are (4.136) and
(4.133). The solution method of
this example is applied for cases
with less layers (for example by
setting the specific height difference
to be zero). Equation
(4.136) can be used by modifying
it, as it can be noticed that instead
of using the regular atmospheric
pressure the new “atmospheric”
pressure can be used as
ρ 1
y
β
"O"
h 1 a 2
a 1
h 4 ρ 2
h 3 h 2
F 1
ρ 3 b 2
ρ 4
b 1
ρ 4
F 2
l
Fig. -4.25. The effects of multi layers density on static
forces.
P atmos
′
= P atmos + ρ 1 gh 1
The distance for the center for each area is at the middle of each of the “small”
rectangular. The geometries of each areas are
x c1 = a 2+ h2 ( )
sin β
2
A 1 = l
h2
sin β − a 2
x c2 = h 2+h 3
2 sin β
A 2 = l
sin β (h 3 − h 2 )
x c3 = h 3+h 4
2 sin β
A 3 = l
sin β (h 4 − h 3 )
I x ′ x ′ 1 = l „
h2
sin β −a2 « 3
36
+(x c1 ) 2 A 1
I x ′ x ′ 2 = l(h3−h2)3
36 sin β +(x c2) 2 A 2
I x ′ x ′ 3 = l(h4−h3)3
36 sin β +(x c3) 2 A 3
After inserting the values, the following equations are obtained
Thus, the first equation is
F 1 + F 2 = P atmos
′
A total
{ }} {
l(b 2 − a 2 )+g sin β
3∑
ρ i+1 x ci A i
i=1