Fluid Mechanics Exercises

1. Fluid Mechanics – Solved Exercises
June 18, 2020
Miguel Fernandes
1 Exercises (solved and suggested)
1. A closed tank is partially filled with gasoline (at rest) whose depth is 8 ft.. Assume
that the pressure of the air in the tank is 6 Ib/in.2
.
Find the pressure at 5 ft. from the bottom of the tank. Remark: the specific weight
of gasoline is 42.5 Ib/ft.3
(the variation of the density of the liquid is negligible).
Solution:
Since we assume that the fluid is at rest (and hence its acceleration is zero) we can
use the following relation for the pressure:
dp
dz
= −γ = −42.5
where z denotes the upward direction.
It follows that:
p = 42.5h + p0
where p0 is the air pressure (at the liquid surface) and h is the difference of heights,
z − z1, corresponding to the pressures p and p1, respectively. Hence:
p = 42.5
Ib
ft3 ∗ 3 ft + 6
Ib
in.2
144 in.2
ft2 = 991.5
Ib
ft2 .
1

2. Exercise 1. Assume that the tank has a cylindrical shape whose base has radius 1
ft.. Find the surface force in z direction at the bottom of the tank.
2. Find the moment of inertia Ixc of the ellipse:
x2
a2
+
y2
b2
= 1
for some a, b ∈ R+
. Remark: take the conventional coordinate system xy (x is the
horizontal axis and y is the vertical one) centered at the centroid of the ellipse (it is
clear that we are adopting the euclidean norm of R2
).
Solution:
We start by recalling that Ixc is given by the double integral:
Ixc =
A
y2
dA
where A denotes the area enclosed by the ellipse.
We now use Green’s theorem to find:
Ixc =
A
y2
dA =
∂A
P(x, y) dx
where we set P(x, y) = −y3
/3.
But:
∂A
P(x, y) dx =
2π
0
ab3
3
sin4
(t) dt
and using trigonometric identities and integration methods we easily get:
Ixc =
2π
0
ab3
3
sin4
(t) dt =
π
4
ab3
.
Exercise 2. For the ellipse of the previous exercise find:
(a) Iyc.
(b) Ix where x is parallel to x axis so that the distance between them is 2. (Hint:
use parallel axis theorem.)
Challenge 1. Give a geometric interpretation for the moment of inertia of a certain
plane region.
Challenge 2. Consider a rotation of angle θ, where θ ∈ [0, π[, of the ellipse around
its centroid in the coordinate system considered above. Write a formula that expresses
the moment of inertia Ixc of the rotated ellipse. (Notice that Ixc = Ixc (a, b, θ).)
Assuming that a and b are fixed, find the values of θ for which the moment of inertia
Ixc of the ellipse is maximum (interpret). Conclude that the moment of inertia Ixc
of the ellipse is not invariant under rotations. Check if it is still true for Ixyc. What
if a = b? (Hint: recall the rotation matrix and find a parametrization for the rotated
ellipse.)
2

3. 3. In Figure 1, one can observe a sketch of a reser-
voir filled with water. One of its walls is in-
clined and contains a plane surface with the
shape of an ellipse with dimensions 3 ft. × 2 ft.
(semi–major axis × semi–minor axis). Find
the magnitude and location of the resultant
force exerted on the plane surface by the water
in the reservoir.
Remark: use γwater = 62.4 Ib/ft.3
.
Figure 1
Solution:
To find the magnitude of the force (which acts perpendicularly to the side wall of
the reservoir) we use the following formula:
FR = γwaterhcA
where hc is the vertical distance from the surface of the fluid to the centroid of
the ellipse and A is the area of the latter. Clearly yc = 4 + 3 = 7 ft. and using
trigonometry we find hc = 7 sin(π/6) ft. = 7/2 ft.. Hence:
FR = 62.4
Ib
ft.3
∗ 7/2 ft. ∗ 6π ft.2
= 1310.4 Ib.
It remains to find the point on which the force acts (center of pressure). Its coordi-
nates are given by:
xR =
Ixyc
ycA
+ xc and yR =
Ixc
ycA
+ yc
where Ixc is the moment of inertia of the ellipse with respect to the x axis when the
xy coordinate system is translated so that it gets centered at the centroid and Ixyc
is the product of inertia with respect to both x and y axes on the same conditions.
Given the symmetry of ellipse we find Ixyc = 0 and so xR = xc = 4 ft.. Moreover,
using the result of exercise 2. (solved), we have Ixc = π/4 ∗ 3 ∗ 23
= 6π and hence:
yR =
6π ft.4
6π ft.2
∗ 7 ft.
+ 7 ft. =
50
7
ft.
where we used the fact that yc = 4 + 3 = 7 ft..
Therefore FR acts in the major–semiaxis of the ellipse and 1/7 ft. from the centroid.
3

4. 4. Consider a two–dimensional velocity field of the form u = (u(x, y), v(x, y)). Show
that if H(x, y) = C, for some constant C, defines implicitly a solution of dy/dx =
v(x, y)/u(x, y), then H(x, y) is constant along the streamlines.
Solution:
We want to show that:
dH
dt
=
∂H
∂x
u +
∂H
∂y
v = 0
i.e., H(x(t), y(t)) is constant along the streamlines of the velocity field:
u = (u(x, y), v(x, y)) = ( ˙x(t), ˙y(t)).
Notice that H(x, y(x)) = C defines implicity a solution of dy/dx = v(x, y)/u(x, y)
regarding y as a function of x.
Hence we can differentiate both sides of H(x, y(x)) = C with respect to x and obtain:
∂H
∂x
+
∂H
∂y
˙y =
∂H
∂x
+
∂H
∂y
v
u
= 0.
Finally we find:
dH
dt
=
∂H
∂x
u +
∂H
∂y
v = −
∂H
∂y
v
u
u +
∂H
∂y
v = 0.
Thus H(x, y) is constant along streamlines.
Exercise 3. Interpret the previous result in terms of the shape of the streamlines.
5. A two–dimensional velocity field u =
(u, v) is given by u = xy and v = (x+1)y2
.
Describe and plot the streamlines of that
flow.
Solution:
We make:
dy
dx
=
(x + 1)y2
xy
=
(x + 1)y
x
.
Using the method of separation of vari-
ables we get y = Cxex
, with C ∈ R.
Thus the streamlines of such flow are the
curves in xy plane defined as above.
Some of those curves are represented in the
Figure 2. Figure 2
Exercise 4. In what quadrants is the flow approaching the origin? Justify.
4

5. 6. Consider a cube on a table with a
2 ft. edge. The cube is filled with
water and has a hole in the center
of one of faces, as shown in Figure
2.
(a) Use the equation of continu-
ity to compare the velocity
of the water at the top of the
reservoir and at the hole.
(b) Find the velocity of the wa-
ter that emerges from the
hole of the cube.
Figure 3
Solution: (a) In what follows, we assume that the fluid is incompressible. Choose
a coordinate system with origin at the center of the cube and parallel to the table.
Let A1 and A2 be the areas of one face of the reservoir and the hole and let v1 and
v2 be the velocities of the water in two points located at those regions, respectively.
Then we have the following relation A1v1 = A2v2. But it is clear that A2 is much
smaller than A1, say A2 = A1, where << 1. Hence v1 = v2, that is, the velocity
of the water at the top of the reservoir is 1/ times smaller than v2. (b) Applying
Bernoulli’s equation (and assuming that the fluid is steady, inviscid and incompress-
ible) one writes γz1 + ρv2
1/2 + p1 = γz2 + ρv2
2/2 + p2, where z1 and z2 are the vertical
coordinates of the points considered above and p1 and p2 the corresponding pres-
sures. Since both points are exposed to the atmosphere we have p1 = p2 and hence
the equation above gives gz1 + v2
1/2 = gz2 + v2
2/2, where g is the gravity accelera-
tion. But, according to the previous exercise, one can neglect v1 and thus we find
v2 = 2 ∗ 32.15 ft./s2 ∗ 1 ft. ≈ 8 ft./s.
7. A two–dimensional velocity field of an incompressible flow u = (u(x, y), v(x, y)) is
such that u(x, y) = x2
+ y. Find v(x, y) given that v(0, y) = x2
.
Solution:
Using the continuity equation we find:
div u =
∂u
∂x
+
∂v
∂y
= 0.
Thus, since ∂xu = 2x, we use integration to obtain:
v(x, y) = −2xy + f(x)
for some differentiable function f with respect to x.
But, using the condition v(0, y) = x2
, we easily compute f(x) as follows:
v(0, y) = f(x) = x2
.
Hence v(x, y) = −2xy + x2
.
5

6. 8. Assume that the components of velocity in a flow field F are:
F =



u = x2
+ y
v = −xy2
+ z
w = −2xz + 2xyz
.
Determine the volumetric dilatation rate.
Solution: The volumetric dilatation rate, V, is given by the divergence of the flow
field, that is:
V = div F = ux + vy + wz
where the underscript clearly means partial derivative. Hence V = 2x − 2xy − 2x +
2xy = 0, that is, F is a solenoidal flow field.
Exercise 5. Interpret physically the previous result.
9. Find the vorticity of the flow field of exercise 5.
Solution:
Exercise 6. Check whether the flow field is irrotational.
2 Answers to selected exercises ( )
Challenge 2.
Ixc =
πA
4
a2
sin2
θ + b2
cos2
θ . It is maximum for θ = π/2. Ixyc is invariant under
rotations. If a = b we lie in the case of a circle of radius a so that a more simple formula
for its moment of inertia can be written. Moreover, it is clearly invariant under rotations.
6