Download Relative Equilibrium In the preceding chapter, it was mentioned that

Relative Equilibrium
In the preceding chapter, it was mentioned that when a fluid is at rest, no layer of the
fluid moves relative to an adjacent layer; no shear stresses exist. Similar condition of
absence of shear stress in a liquid arises when a liquid in a container is subjected to a
constant acceleration. Refer to Fig., Let a vessel containing liquid be accelerated at a
constant rate. After some time, the liquid attains to an equilibrium condition; distance
between any two liquid particles remains constant and thus, no shear stresses exist. Since
the liquid moves as if it were a solid, the fluid is said to be at relative equilibrium.
Uniform Linear Acceleration
Consider a liquid filled container moves with a uniform linear acceleration a as in fig. In
order to determine the variation of pressure in the liquid, an infinitesimal liquid element
of size Δx, Δy and Δz is considered. After the flow becomes at relative equilibrium with
new free surface, OO, no shear stress acts on sides of the element. Only normal forces act
on its surface.
As at equilibrium stage, the net force acting on the fluid element per unit of its volume is
equal to the product of the density of the fluid and acceleration.
F  P  ˆj   a
Where w is the specific weight of the fluid.
Expressing in component form
 P
P ˆ P ˆ  ˆ
  iˆ 
j
k   j   axiˆ  a y ˆj  az kˆ

x

y

z




Since P is a function of position (x, y and z), its total differential is
P
P
P
dP 
dx 
dy 
dz
x
y
z
Substituting from EQ() gives
P
P
P
   ax
   ay   
   az
x
y
z
dP    ax dx    a y    dy     az  dz
This is a general equation for a fluid in a container subjected to uniform acceleration.
Case-1 Liquid containers subjected to constant horizontal acceleration
As shown in fig., an open container partly filled with a fluid moves a constant horizontal
acceleration α. The vector components of the acceleration are ax   , a y  0 , az  0 in
the x, y and z directions, respectively.
EQ (), describing the variation of pressure, may be applied with the given acceleration
components:
dP    dx   dy
For an incompressible liquid ( ), integration of () yields
P    x   y  c
Where C is the constant of integration. If the origin of the flow domain is at the free
surface, the pressure at that location is the standard atmospheric pressure; thus P ( ) =
Patm. Imposing this condition into EQ () gives
x  0, y  0, P  Patm
Then, the pressure at any point P (x,y,z) may be obtained by substituting C value into
Eq() and is
P    x   y  0
Solving Eq () for y gives
 x  P  P0 
y



Then, the slope of constant pressure may be computed as:
dy
  1
tan  
 , 
dx
g  g
This equation indicates that the lines of constant pressure have the slope () and are
parallel to the free surface.
Case-II: Liquid container subjected to uniform vertical acceleration
Let an open container filled with a liquid of specific weight w be moved with a constant
vertical acceleration, α. As similar to the previous case, the vector components of the
acceleration are ax  0, a y   , az  0 respectively. By substituting the acceleration
components into (), the pressure at any point P in the liquid is
 
P   dP        dy   1   y  c
 g
where C is the constant of integration. Note that the pressure is independent of the x and z
coordinates. To evaluate C, A condition that the pressure at a point () indicated in Fig is
equal to the standard atmospheric pressure may be considered. Thus C is equal to
 
y  y0 , P  Patm, C  Pa   1   y0
 g
 
Then P  Pa   1    y0  y 
 g
if h  y  y0 ,
 
P  Pa   1   h
 g
Rearranging Eq () for the depth y,
y
Pa  P

 1  g

 y0
When the tank is accelerated downward at the rate of acceleration of g, acceleration due
to gravity; ( α = -g ), Eq() shows the pressure at any point in the liquid is equal to the
standard atmospheric pressure. In other words, any point in a liquid of a container falling
freely in space experiences the same pressure which is equal to that of the surrounding
atmospheric pressure.
Case-3 Liquid container subjected to uniform acceleration
Uniform rotation about a vertical axis
A liquid in a open container, when rotated about a vertical axis with angular velocity ω,
moves a solid after some time intervals. Every particle of the liquid has the same angular
velocity; no relative motion between any two layers of the liquid occurs. Thus, no shear
stresses exist in the liquid. Since the fluid being rotated with constant angular velocity,
the only acceleration, on the account of centrifugal effects, acts on the liquid radially
inward from the outer periphery of the container towards the axis of rotation.
Let an infinitesimal fluid element in a container rotating with constant angular velocity,
w, about the axis of rotation y, be positioned at radial distance of r from the axis. By
choosing a coordinate system with the unit vector i in the r direction and j in the vertical
upward direction (the y axis), Eq () may again be applied in this case also.
P  ˆj   a
In which a is the acceleration directed radially inward due to constant angular velocity ω.
In vector form,
a  iˆ 2
Expending EQ ()
P ˆ P ˆ P
iˆ
j
k
  ˆj  iˆ 2

y
z
Then,
P  2 P
P
  ,
 
0
 g
y
z
It may be noted in EQ () that the pressure is a function of y and r only. The total
differential dp is
dP 
P
P
d 
dy

y
Substituting
P
P
and
into EQ () and integrating EQ () for an incompressible liquid
r
y
yields

P   dP    dy    2 d 
g
P
Then,
  2 2
g
2
 y  C
Where C is the constant of integration. At the origin of the coordinate system, as shown
in fig, the pressure is equal to Patm because of its location on the free surface.
Thus, C = Patm. The pressure at any point P is
P  P0  
 2 2
2g
y
In order to determine the position of the free surface (P = Patm), the depth of the free
surface is
y
 2 2
2g
Eq () shows that the depth of the free surface varies as the square of the radius. Therefore,
the surfaces of equal pressure are paraboloids of revolution.
If a cylindrical container filled with a liquid is rotated about its vertical axis at the rate of
constant angular velocity, then the liquid from its vertex will rise along the wall of the
container. The depth of the liquid, calculated from EQ (), is
 r0 2 
1  2 r0 2
2 2g
Where ro is the radius of the container. It may be noted that the volume of a paraboloids
of revolution is equal to half the volume of the circumscribing cylinder. Thus, the
volume of the liquid above the horizontal plane through the vertex is
1  2 r0 2
2 2g
Therefore, it may be concluded that the liquid rises along the wall the same amount as the
center drops. This relationship is used for locating the vertex when ω, r and depth before
rotation are given.
Closed cylindrical container filled with liquid rotating about its geometrical axis
Let a cylindrical tank completely filled with a liquid be closed at the top. When the tank
is rotated about its geometrical axis as shown in fig, an imaginary surface of constant
pressure, as similar to the free surface of an open rotating tank, is formed. The imaginary
surface, thus has a shape of paraboloid of revolution in accordance with Eq(). The
pressure head at any point in the fluid is equal to the vertical distance of the point from
this imaginary surface.
Solved Examples
1) A cubical tank of 3 m length is filled with water of 2 m depth. If the tank is
accelerated horizontally at constant rate of 5 m2/sec, a) calculate the angle of the
free surface to the horizontal, b) the maximum pressure on the bottom and c) the
total pressure on the bottom.
2) The tank shown in Fig., is filled with an oil of specific gravity 0.8. When the tank is
accelerated in the direction of its length at the rate of 6.0 m2/s, how many litres of
the oil is spilled?
3) A circular tank of 3m depth and 2.0 m diameter is completely filled with water.
When the tank is pulled on a horizontal plane at constant acceleration, half of the
water spells out. What is the acceleration?
4) A closed tube completely filled with a liquid of specific weight
1000 is
accelerated to the right at the rate of 5 m2/s. Draw the imaginary free surface and
determine the pressure at point A.
5) An industrial container of length 10 m is to be designed to carry liquid on an
inclined slope of 450 at the maximum rate of 10 m2/s. What is the minimum
allowance depth to be kept so that there will be no spill of the liquid?
6) An open cylindrical tank of radius r and height h is filled with oil of specific gravity
0.8. At what speed must it rotate so that the center of the bottom to have zero depth
of water ?
7) The tank given in Fig. Is filled with Carbon tetrachloride of specific weight 1590
kg/m3. When the tank is rotated at angular velocity of 2 rad/sec, what are the
pressures at points A, B and C in the figure ?
8) A 10-m length tube completely filled with water is capped at its both ends. After
being placed on a horizontal, the tube is rotated vertically at 30 rad/s about one of
its end. Calculate the pressures at its both ends.
9) The parabolic-shaped gate given in Fig is used to store water. If the width of the
gate is 10 m, compute the force components on the gate form the water ?
10) Lock gates are mainly used in irrigation or navigational channel to maintain the
difference of water levels between the upstream and downstream sides. Typical
lock gates in Fig. are of 6 m high and 3 m width. When the gates are closed,
include angle between the gates is 1300. Each gate is hinged at the top and bottom
of the gate. If the water levels are 5 m and 3 m on the upstream and downstream
sides, respectively, compute the magnitude of the force on the hinges due to the
water pressure.
11) Refer to Fig., an overhead tank supplies water to a stored tank.
a)What is the pressure reading of the pressure gauge at point C ?
b) if a circular lid of radius 10 cm is attached at point D on the tank, what is the
total resultant force on the lid ?
c) Instead of the circular lid, a square lid of the same cross-sectional area is fitted
at point D, compute the total resultant force on the square lid.
12) A rectangular tank of 2 x 3 x 4 m size is attached to a pipe, given in Fig. If
water is used to fill the tank up to xx level, what is the total force on the bottom of
the tank ? Draw the pressure distribution on all the sides of the tank.
13) A concrete dam has an arrangement to close an inclined gate as shown in Fig..
If the water level reaches 10 m during a flood event, what the weight must be
placed so that the gate will remain closed? Neglecting the weight of the gate and
assuming no frictional loss on the cable system.
14) During a flood event, the density of sediment-loaded water stored at a dam
varies linearly with the depth from the free surface. If the flood level is 40 m from
the bottom of the dam, What is the total resultant force and its line of action on
the dam ? if there is only clear water on the upstream, compute the total resultant
force and its line of action on the dam ?
15) An automatic spilling gate is used to spill excess stored water suddenly. In
Fig, a typical spilling gate is shown and the gate weighs 500kg/m normal to the
paper and its center of gravity is 50 cm from the left face and 40 cm above the
lower face. The gate also is hinged at point O. Compute the water surface depth
for the gate just to start release the water.