RELATIVE EQUILIBRIUM

I. Rectilinear Acceleration (Moving Vessel)

A. Horizontal Motion

v = constant
a = 0
C

  

W = Mg
The forces
acting on a N  W = Mg
REF = Ma particle of

liquid in
REF = Ma
N equilibrium
Force Polygon
Comments:
If the vessel moves horizontally with constant velocity, the fluid surface remains
horizontal.
If the vessel moves horizontally with constant acceleration the fluid surface will
form an incline surface making an angle  with respect to the horizontal. The
horizontal surface of the fluid seem to rotate about an axis perpendicular to the
paper running through the point of intersection of the central axis C and the
horizontal surface of the fluid (marked by a red cross). In this condition, the fluid
retains its original volume. Meaning, there is no spillage. The highest point of
The water surface coincides with the mouth of the container at the rear portion
and also giving you the maximum  with no spillage. a
When the horizontal surface no longer
appear to rotate about the designated
intersection with the central axis then,
there is already spillage happening. The
value of  is greater.
Applying the conditions of static equilibrium:

REF Ma
tan  = =
W Mg

a
tan  =
g

Therefore: the surface and all planes of equal hydrostatic pressure must be
inclined at this angle  with the horizontal.
Example
A vessel 2 m by 3m by 2 m (into the paper/screen is moving at a constant
acceleration to the right carrying water 1.2 m deep

a) What is the maximum acceleration it can have without spillage of the water?

a
tan  = Eqn 1
g

tan  = 0.8 m Eqn 2

1.5 m
 b) What is the condition if its is subjected
Equate 1 & 2
to a 4.8 m/s2 acceleration?

𝒂 𝟎.𝟖 Solving for 

=
𝒈 𝟏.𝟓 tan  = a = 4.8 m/s2
g 9.81 m/s2
𝟎.𝟖×𝟗.𝟖𝟏
a =  = 28.97O
𝟏.𝟓

a = 5.232 m/s2

Note: The max  it can have without

spillage is that  with the computed
acceleration?

tan  = 0.8 m
1.5 m

 = 31.19O
c) What is the acceleration for this Equate 1 & 3
condition? Is there spillage?
𝒂 𝟎.𝟖
=
𝒈 𝟏.𝟓

𝟐.𝟎×𝟗.𝟖𝟏
a =
𝟑.𝟎

 a = 6.54 m/s2

Original volume of water

Note: The inclined fluid surface is not Vol. orig. = 1.2m X 3m X 2m = 7.2 m3
Passing now through the red cross inter-
Section. Voltriangular = ½ X 3m X 2m X 2m
a = 6.0 m3 (Vol. retained)
tan  = Eqn 1
g
So there is spillage
tan  = 2.0 m Eqn 3 Vol. spilled = Vol orig – Vol retained
3.0 m
= 7.2 – 6.0
 = 37.43O = 1.2 m3
d) What is the condition if its is subjected tan  = 2 m Eqn 5
to a 7.0 m/s2 acceleration? x

Solving for 
Equate 4 & 5
tan  = a = 7.0 m/s2 Eqn 4
g 9.81 m/s2 7.0 m/s2
2m
=
x 9.81 m/s2
 = 39.45O
X = 2.8 m

Volume of water retained

Volret. = ½(x)(2m)(2m)
= ½(2.8m)(2m)(2m)
Volret = 5.6 m3
Example:
An open rectangular tank mounted on a truck is 5 m long, 2 m wide and
2.5 m high is filled with water to a depth of 2 m.
a) What maximum horizontal acceleration can be imposed on the tank without
spilling any water?
b) Determine the accelerating force on the liquid mass.
c) If the acceleration is increased to 6 m/s2, how much water is spilled out?

Solution:
a)
a a

5m 5m
2.5 m 2.5 m 2.5 m 2.5 m

0.5 m 0.5 m 

2.5 m 2.5 m
2m 2m 1.5 m
Since there is no liquid spilled, the two triangular wedges shown must be equal

0.5 Equate 1 & 2

tan  = 2.5 = 0.2 eqn 1
a
g = 0.2
a
tan  = g eqn 2 a = 0.2(9.81)
a = 1.962 m/s2
horizontal acceleration

b) Accelerating Force, F = Ma
F =  (Vol) a

= 1000 (2x2x5)(1.962)

F = 39,240 = 39.24 kN

OR
F = FREAR WALL - FFRONT WALL

F = 9.81(2.5/2)[2.5(2)] - 9.81(1.5/2)[1.5(2)]

F = 39.24 kN
c) When: a = 6 m/s2

a 6
tan  = g = Voriginal = (2)(2)(5) = 20 m3
9.81

 = 31.45o Vspilled = Vorig - Vleft

= 20 - 10.22
a = 6 m/s2

Vspilled = 9.78 m3
5m
x

2.5 m

x = 2.5 cot 31.45o = 4.0875 < 5 m

1
Volleft = 2 (4.0875)(2.5)(2) = 10.22 m3
B. Inclined Motion

ay Considering a mass of fluid accelerated upwards at

a an inclination of  with the horizontal:

ax ax = a cos 
ay = a sin 

REFy = May


W = Mg

• REFx = Max

 N 

In the force polygon:

Max
tan  =
 Mg + May
N
Mg + May
ax
tan  =
g + ay
Max
 ax
tan  =
g  ay

Use (+) if the acceleration is upward

Use (-) if the acceleration is downwards

Problem:
A vessel containing oil is accelerated on a plane inclined 12o with the horizon-
tal at 1.5 m/s2. Determine the inclination of the oil surface when the acceleration
is:
(a) upwards
(b) downwards
ay
a
ax
 = 12 o




 = 12 o
Solution:

ax
tan  =
g  ay

ax = a cos  a) when the acceleration is upwards:

= 1.5 cos 12o 1.467
tan  =
= 1.467 m/s2 9.81 + 0.311
 = 8.247o
ay = a sin 
b) when the acceleration is down-
= 1.5 sin 12o wards:
= 0.311 m/s2
1.467
tan  =
9.81 - 0.311
 = 8.779o
C. Vertical Motion
REF = Ma
Considering a mass of fluid accelerated
upwards or downwards with an acceleration
p=0
of a as shown in the figure.
W = Mg

Forces acting:
h
Weight of the liquid above the point = Mg dA
The inertia force = Ma p
The pressure force, F = pdA
F = pdA

Fv = 0
F = Ma + Mg

W  Vol
M= =
g g
  (h)(dA)
M = g

 (h)(dA)
pdA = g (a + g)

 (h) a
p = g (g + a) =  h( 1 + )
g

a
p =  h( 1  g )

Use (+) if the acceleration is upward

Use (-) if the acceleration is downwards

Problem:
A vessel 3 m in diameter containing 2.5 m of water is being raised. Find
(a) the pressure at the bottom of the vessel in kPa when the velocity is constant
(b) the pressure at the bottom of the vessel when it is accelerating 0.5 m/s2
upwards

Solution:

For vertical motion:

a
p =  h( 1  g )

h = 2.5 m

a) when the velocity is constant, a = 0, then

p =  h = 9.81(2.5)
= 24.525 kPa (pressure at the bottom)

b) when a = 0.6 m/s2 (use “+” for upward acceleration)

0.5
p = 9.81(2.5)( 1 + )
9.81
= 25.775 kPa
Problem: (test yourself)
An open tank containing oil (sp.gr. = 0.82) is accelerated vertically at 7.5 m/s2.
Determine the pressure 4 m below the surface if the motion is
a) upward with a positive acceleration,
b) upward with a negative acceleration,
c) downward with a positive acceleration
d) downward with a negative acceleration.
 Cylindrical Vessels with Free Liquid Surfaces
I. Rotation (Rotating Vessel)
 
r

h/2
h
h/2 h
H
H D
D

For no spillage
 There is spillage

2r 2
h= H
h
2g D

There is spillage and an imaginary volume ( paraboloid)

 The mechanics of rotating fluids is an
 important part of the analysis of
r numerous scientific and engineering
h/2
problems.
h
h/2 Fluids contained in open or closed
H
D cylindrical vessels can be subjected to
rotation at constant angular speed in
rpm (revolution per minute). Rotation
will be about the centroidal axis of the
For no spillage
cylinder .

When the cylinder is subjected to rotation the fluid

will form a hollow that has its surface forming like a
parabola. Since it is a volume, we call it a paraboloid.
h
The height of the paraboloid inside the cylinder should
Vertex be based from its vertex. Height “h” is the designa-
of the
paraboloid
tion of the paraboloid’s height.

Centroidal axis
of the cylinder
During rotation when the highest point of the fluid’s surface (highest
point of the paraboloid coincides with the mouth or brim of the
cylinder then with no spillage at all, it is that condition that the
height h is bisected by the fluid’s original horizontal surface while at
rest. In the figure, it is designated as h/2.

Here is the nomenclature of the figure:

H = the height of the cylinder
D = the depth of the fluid
h = height of the paraboloid (parabola)
Dia = the diameter of the cyylinder (not in the figure)
r = radius of the circular cross section of the cylinder
 = constant angular speed (in rpm) that the cylinder is
subjected to

When there is already spillage, the bisection of height “h” by the

fluid surface when it was still at rest will no longer hold true.
The highest point of the parabola is always at the brim of the open
cylinder.
The determination of the hollow paraboloid is expressed in the form of:

2r 2
h =
2g

Where:
h = the height of the paraboloid
 = constant angular speed in rpm
r = radius of the circular cross section of the cylinder
g = gravitational acceleration

The section of the paraboloid is a

parabola and any point in the
parabola can be determine using
the SQUARED PROPERTY OF THE
PARABOLA
Volume of the paraboloid is given to be:

𝟏
V= 𝝅 𝒓𝟐 𝒉
𝟐

REMEMBER: VERTEX CAN ALWAYS BE FOUND AT THE AXIS OF ROTATION.

NOTE: For closed vessels, there can never be any liquid spilled, so that the
initial volume of liquid (before rotation) is always equal to the final volume
of the liquid (after rotation) or the initial volume of air inside is equal to the
final volume of air inside.
 2r 2
h=
2g

where:
 = constant angular speed (in radians per second)
r = farthest distance of a particle in the paraboloid
equivalent to the radius of the cylinder
h = height of the paraboloid made during rotation
 Cylindrical Vessels with Free Liquid Surfaces
Problem:
An open cylindrical vessel, 0.60 m in diameter, 1.00 m high and three-
fourths full of water, rotates about its vertical axis with a constant speed of 80
rpm, determine :
a) the depth of water at the center of the vessel
b) the total pressure on the cylindrical walls
c) the total pressure on the bottom of the vessel
d) is there some spillage? If there is, what is the volume that was spilled over
during rotation?
e) What is the maximum constant speed that can be given to the vessel
without water spilling over the sides?
f) For a constant speed of 110 rpm subjected to the cylinder what is the
volume that spilled out and the volume retained?
g) For the same dimensions and depth of water, determine the water that
spilled out if 140 rpm is subjected to the cylinder.
a) Given: Dia = 0.60 m; H = 1.0 m h = 0.322 m (computed)
D = ¾(H) = 0.75 m
𝑟𝑒𝑣 2𝜋 𝑟𝑎𝑑 1 𝑚𝑖𝑛
 = 80 × × h/2 = 0.322/2
𝑚𝑖𝑛 1 𝑟𝑒𝑣 60 𝑠𝑒𝑐
= 0.161 m
 = 8.38 rad/sec
D(a) = 0.75 – 0.01635
1 = 0.589 m
8.38 2 (𝑠𝑒𝑐2)(0.30)2 𝑚2
h=
2 9.81 𝑚/𝑠𝑒𝑐2

In the figure from the given,

the computed height of the
paraboloid is lesser than the
Maximum h for a condition
of no spillage. So, no spillage
at all.
Max h for no spill = 0.50 m
h/2 = 0.25 m
 b) @  = 80 rpm, the total Magnitude P per 1 m circumferential
Hydrostatic force on the wall Length of the wall (the triangular
distribution of pressure)

P = ½ bH
D + h/2 H = 1.0 m
P = ½ (8.94 kN/m2)(0.911 m)
P = 4.072 kN/m
p = (D + h/2)

p = (D + h/2) PT = (4.072 kN/m)(π)(Dia)

p = (9.81 kN/m3) (0.75 + 0.161)m = 7.676 kN
= 8.94 kN/m2
c) The Total Pressure Force d) Spillage? @  = 80 rpm there was/
acting @ the bottom is no spillage

e) What is the max  that can be given
for no spillage
For no spill, h/2 = 0.25 m
So h = 0.50 m
P=W
Using:

2r 2
h =
2g
The Total Pressure Force is
equal to the weight W of the 𝜔2 (0.302 )
water 0.50 =
2×9.81
P = W = Vol =  πr2D
 = 10.44 rad/sec
= (9.81)(π)(0.30)2(0.75)
 = 95.49 rpm
= 2.08 kN
 
f) @  = 110 rpm

2𝜋
(110× 60 )2 (0.30)2 h = 0.608 m
h=
2×9.81

h = 0.608 m
h/2 = 0.304 m > 0.25 m
So, there is spillage. The
horizontal water surface at rest Vol of paraboloid = ½ πr2h
no longer bisects the height of = ½ π(0.30)2(0.608)
the paraboloid.
= 0.086 m3
Vol of cylinder = πr2H Vol of water ret. = Vol cyl – Vol prbld
= π(0.30)2(1.0) = 0.283 – 0.086
= 0.283 m3 = 0.197 m3
Vol of water orig. = πr2D Vol of spill = Vol H2O orig – Vol ret .
= (π)(0.30)2(0.75) = 0.212 - 0.197
= 0.212 m3 = 0.015 m3
g) @  = 140 rpm

2𝜋
(140× 60 )2 (0.30)2
h=
2×9.81 h = 1.0 m h = 0.986 m

h = 0.986 m

Vol of paraboloid = ½ πr2h

= ½ π(0.30)2(0.986)
= 0.139 m3
Vol of water ret. = Vol cyl – Vol prbld
= 0.283 – 0.139
= 0.144 m3
Vol of spill = Vol H2O orig – Vol ret .
= 0.212 - 0.144
= 0.068 m3
h) What constant speed (in rpm) must be applied to the cylinder so
that a radial distance of 0.15 m from the central axis will no longer
have water during rotation.

()2 (0.30)2
h= 𝐄𝐪𝐧 𝟏
2×9.81

()2 (0.15)2 H = 1.00 m

y= 𝐄𝐪𝐧 𝟐 h=?
2×9.81
D = 0.75 m

y
From the figure

h= H+y 𝐄𝐪𝐧 𝟑 0.15 m

Subst. eqns 1 & 2 into 3

 = 17.049 rad/sec
( )2 (0.30)2 ( )2 (0.15)2
= 1.0 +  = 162.8 rpm
2×9.81 2×9.81
Problem:

If the mercury U-tube shown in the figure is given an

acceleration of 4.905 m/s2 towards the right, determine the
gage pressure at A in kPa.
As the U-tube moves at constant
C velocity, the fluid’s imaginary surface
will remain horizontal. If it moves at
constant acceleration, the imaginary
surface assume and inclination at an
 angle  with respect to the horizontal.
200 mm The rest of the solution here involves
y
A
• B
the principle of trigonometry and of
C Pressure determination bearing in
150 mm
mine that we always start at the
600 mm fluid’s surface when solving gage
pressure (real or imaginary).
Same principle as what we have in a vessel
moving horizontally at constant acceleration
is used here. But because this is a U-tube we Remember: Liquid always seek its
have an imaginary fluid surface between the own level.
two legs of the U-tube.
 U-Tubes Subjected to Rotation about the Vertical
Axis at Constant Speed
Problem:
If the mercury U-tube shown in the figure is rotated about
a vertical axis through the leg BC, determine the height of
mercury column in the leg DE, when the speed is 40 rpm,
and the pressure at A in kPa.

 To satisfy the condition of the problem

above, use the general formula:
E C
2r2
h=
2g

Please next slides if computed h is

h greater than 400 mm or 0.40 m
2
h
h
For pressure at A, the solver only needs
200 mm 2 the imaginary height of the imaginary
A fluid surface above point A. The square
• B properties of the parabola can be used
D
150 mm here. Ans: fluid height in leg DE = 360 mm

600 mm
 If the vertex of the parabola is still on the
vertical leg where the axis of rotation is,
 h/2 is still applicable. When the vertex
E C goes below B then h of the parabola is no
longer bisected with the horizontal surface
of the fluid when it was still at rest.
Try to find the angular speed  for the
h
2 condition on the left where the vertex of
h
the parabola is at the bottom of leg BC
h
200 mm 2
A
To satisfy the condition of the problem
•
D B above, h = 400 mm or 0.400 m. This
150 mm
will enable the solver to determine .
600 mm
Please, see next slide.

NOTE: The Squared Property of Parabola is always useful in solving problems like these. Parabolic
Curve is always symmetrical from the axis to both sides.
 2r2 still applies since this is a
h=
2g
general formula. y uses the x as its
radius in the small paraboloid (parabola)

E  but remember its part of the big parabola

and h is the total height of the big
C
parabola.
The condition on the left will only happen
200 mm + x
for a bigger value of . The legs are
h assumed to be always high that no
spillage will occur.
200 mm x Problems could be with  given and x
B
will be determined or the other way
D y
around.

r = 600 mm If  = 60 rpm what should be the height

of the mercury column in leg DE?
hDE = 6 22 mm; x = 222 mm
Problem:
A glass U-tube with vertical stems are 300 mm apart and is filled with mercury
to a depth of 150 mm in the vertical stems. It is rotated about a vertical axis
through the midpoint of the horizontal section. What angular speed  will
produce a pressure of absolute zero in the mercury at the axis?
Problem:
A glass U-tube with vertical stems are 600 mm apart is filled with mercury to a
depth of 200 mm in the vertical stems. It is rotated about a vertical axis
through its horizontal base 400 mm from one vertical stem. How fast should it
be rotated so that the difference in the mercury levels in the stems is 200 mm?