CENGR 3260 - HYDRAULICS
RELATIVE EQUILIBRIUM
OF LIQUIDS
ENGR. JHOREENE A. JULIAN
Instructor
Department of Civil Engineering, CLSU
CENTRAL LUZON
STATE UNIVERSITY
RECTILINEAR TRANSLATION(MOVING VESSEL)
Horizontal Motion
a
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RECTILINEAR TRANSLATION(MOVING VESSEL)
Horizontal Motion
W = Mg
REF = Ma
θ
W = Mg
θ
N
θ
N
REF = Ma
tan θ =
REF
W
Ma
tan θ =
Mg
𝐚
𝐭𝐚𝐧 𝛉 = 𝐠
a = acceleration
g = gravitational acceleration
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RECTILINEAR TRANSLATION(MOVING VESSEL)
Inclined Motion
𝛂
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RECTILINEAR TRANSLATION(MOVING VESSEL)
av
Inclined Motion
a
α
REFv = Mav
Mav
REFh = Mah
W = Mg
θ
θ
ah
N
W = Mg
θ
Mah
N
𝛂
Ma
tan θ =
tan θ = Mg + hMa
v
Mah
M(g + av)
𝐚
𝐭𝐚𝐧 𝛉 = 𝐠 ±𝐡𝐚
𝐍𝐨𝐭𝐞: Use + sign for upward
motion and − sign
for downward motion
𝐯
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RECTILINEAR TRANSLATION(MOVING VESSEL)
Vertical Motion
𝛾
𝑀 = 𝜌𝑉 = 𝑔 𝑉
෍ Fv = 0
REF = Ma
𝑉𝑜𝑙𝑢𝑚𝑒, 𝑉 = 𝐴ℎ
F = Ma + γV
𝐹 = 𝑃𝐴
γ
a
W = γV
h
F = g Va + γV
γ
PA = g Ah (a) + γ Ah
h
a
Area = A
P = γh 1 + g
𝐏 = 𝛄h 𝟏 ±
F = PA
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𝐚
𝐠
𝐍𝐨𝐭𝐞: Use + sign for
upward motion and
− sign for downward motion
+a acceleration
−a deceleration
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3
SAMPLE PROBLEM NO. 1
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 is the 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
5.0 m
0.5 m
2.5 m
2.0 m
a. Maximum horizontal acceleration
a
5.0 m
tan θ =
0.5 m
2.5 m
tan θ = 0. 2
0.5 m
θ
0.5 m
2.5 m
2.0 m
1.5 m
tan θ =
a
g
0.2 =
a
9.81 m/s²
𝐚 = 𝟏. 𝟗𝟔𝟐 𝐦/𝐬²
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Solution:
b. Accelerating force
F = Ma
ρ=
M
V
M = ρV
M = 1000 kg/m³ 5.0 m 2.0 m 2.0 m
M = 20,000 kg
F = 20,000 kg 1.962 m/s²
F = 𝟑𝟗, 𝟐𝟒𝟎 𝐍
Solution:
c. When a = 6 m/s²
5.0 m
a = 6 m/s²
x = 4.087 m
2.5 m
θ
tan θ =
tan θ =
a
g
6 m/s²
9.81 m/s²
tan 31.45° =
2.5 m
31.45°
2.5 m
x
x = 4.087 m < 5.0 m
x
θ = 31.45°
5
Solution:
Vspilled = Vtotal − Vleft
Vtotal = 5.0 m 2.0 m 2.0 m
Vtotal = 20.0 m³
1
4.0875 m 2.5 m 2.0 m
2
Vleft = 10.22 m³
Vleft =
2.5 m
31.45°
4.0875
Vspilled = 20.0 m³ − 10.22 m³
𝐕𝐬𝐩𝐢𝐥𝐥𝐞𝐝 = 𝟗. 𝟕𝟖 𝐦³
SAMPLE PROBLEM NO. 2
A vessel containing oil is accelerated on a plane inclined 15° with
the horizontal at 1.2 m/s². Determine the inclination of the oil
surface when the motion is (a) upwards, and (b) downwards.
a = 1.2 m/s²
15°
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Solution:
a = 1.2 m/s²
a. When the motion is upward
θ
a
tan θ = g +ha
v
tan θ =
15°
1.159 m/s2
9.81 m/s2 +0.31 m/s2
𝛉 = 𝟔. 𝟓𝟑𝟑°
a
tan θ = g ±ha
cos 15°=
v
av
a
15°
ah
a
ah
a
sin 15° = av
a
cos 15° = 1.2
v
sin 15° = 1.2
ah = 1.159m/s²
av = 0.31 m/s²
b. When the motion is downward
a
tan θ = g −ha
v
tan θ =
1.159 m/s2
9.81 m/s2 − 0.31 m/s2
𝛉 = 𝟔. 𝟗𝟓𝟔°
ah
ROTATION (ROTATING VESSEL)
ω
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ROTATION (ROTATING VESSEL)
ω
W = Mg
x
CF
r
y
N
θ
y
r
θ
x
y
y1
θ
W = Mg
N
CF = W/g ω2x
CF
tan θ = W
tan θ =
W/g ω2x
W
𝐭𝐚𝐧 𝛉 =
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𝛚𝟐 𝐱
𝐠
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ROTATION (ROTATING VESSEL)
tan θ =
ω 2x
g
dy
= tan
dx
ω = angular speed in rad/sec
r
θ
x
y1
dy
ω2 x
= g
dx
ω2
‫ ׬‬dy = ‫ ׬‬g
𝐲=
y
y1 = height of the paraboloid at a
distance x from the axis
y = total height of the paraboloid
g = gravitational acceleration
x dx
𝐲=
𝛚𝟐 𝐫 𝟐
𝟐𝐠
𝐍𝐨𝐭𝐞: 1 rpm = π/30 rad/sec
𝛚𝟐 𝐱 𝟐
𝟐𝐠
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Liquid Surface Condition for Open Cylinder (h > H/2)
D
y
D
D
y/2
y/2
y
H
Initial liquid level
y/2
H
Initial liquid level
y
H
Initial liquid level
y/2
𝐲
<
𝟐
h
h
h
𝐲
=
𝟐
𝐃
𝐲
>
𝟐
𝐃
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𝐃
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Liquid Surface Condition for Open Cylinder (h > H/2)
D
y
D
H
Initial liquid level
h
y
y= 𝐇
Vortex at the bottom
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H
Initial liquid level
h
y>𝐇
Vortex imaginary
below the bottom
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Liquid Surface Condition for Closed Cylinder (h > H/2)
D
y
y/2
y
D
y/2
y/2
D
y
H
Initial liquid level
H
Initial liquid level
H
Initial liquid level
y/2
𝐲
<
𝟐
h
h
h
𝐲
=
𝟐
𝐃
𝐲
>
𝟐
𝐃
𝐃
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with imaginary
paraboloid above
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Liquid Surface Condition for Closed Cylinder (h > H/2)
D
D
y
H
Initial liquid level
H
Initial liquid level
y
h
h
𝐲 > 𝐇𝟐 /𝟐𝐃
𝐲 = 𝐇𝟐 /𝟐𝐃
(with imaginary paraboloid
above and vortex just touching
the bottom)
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y2
(with imaginary paraboloid above
and imaginary vortex below the
bottom)
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SAMPLE PROBLEM NO. 3
An open cylindrical tank, 2 m in diameter and 4 m high
contains water to a depth of 3 m. It is rotated about its
own axis with the constant angular speed ω.
a. If ω = 3 rad/sec, is there any liquid spilled?
b. What maximum value of ω (in rpm) can be
imposed without spilling any liquid?
c. If ω = 8 rad/sec, how much water is spilled out ?
d. What angular speed ω (in rpm) will just zero the
depth of water at the center of the tank?
1.0 m
4.0 m
3.0 m
r = 1.0 m
Solution:
a. Liquid spilled
y=
1.0 m
ω2 r2
2g
32 12
y = 2 9.81 m/s²
y
4.0 m
3.0 m
y = 0.46
𝐲/𝟐 = 𝟎. 𝟐𝟑 < 𝟏 𝐦 ; 𝐧𝐨 𝐥𝐢𝐪𝐮𝐢𝐝 𝐬𝐩𝐢𝐥𝐥𝐞𝐝
r = 1.0 m
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Solution:
b. What maximum value of ω (in rpm) can be imposed without spilling
any liquid?
y/2 = 1 or y = 2 m
2=
y
2
= 1. m
ω2 r2
2g
ω 2 12
2 = 2 9.81 m/s²
4.0 m
3.0 m
ω = 6.26 rad/sec
rad
ω = 6.26 sec x
30
π
𝛚 = 𝟓𝟗. 𝟕𝟖 𝐫𝐩𝐦
r = 1.0 m
Solution:
c. If ω = 8 rad/sec, how much water is spilled out and to what
depth will the water stand when brought to rest?
y=
ω2 r2
2g
82 12
y = 2 9.81 m/s²
1.0 m
y = 3.26 m
y
4.0 m
3.0 m
y/2 = 1.63 m
D=1m
y/2 > D
r = 1.0 m
12
Solution:
c. If ω = 8 rad/sec, how much water is spilled out and to what
depth will the water stand when brought to rest?
r = 1.0 m
Vspilled = Vair(final) - Vair(initial)
Vair(final) = Vparaboloid
1
2
1
Vair(final) = π
2
Vair(final) = πr 2y
1.0 m
y
1m
3.26 m
2
3.26 m
2
1m
Vair(final) = 5.12 m³
4.0 m
Vair(initial) = π 1 m
3.0 m
Vair(initial) = 3.1416 m³
1.0 m
Vspilled = 5.12 m³ - 3.1416 m³
𝐕𝐬𝐩𝐢𝐥𝐥𝐞𝐝 = 𝟏. 𝟗𝟕𝟖 𝐦³
r = 1.0 m
Solution:
d. What angular speed ω (in rpm) will just zero the depth of water at the
center of the tank?
y=
ω2 r2
2g
ω 2 12
4 = 2 9.81 m/s²
1.0 m
ω = 8.86 rad/sec
rad
H = y = 4.0 m
3.0 m
ω = 8.86 sec x
30
π
𝛚 = 𝟖𝟒. 𝟔𝟏 𝐫𝐩𝐦
r = 1.0 m
13
THANK YOU!
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STATE UNIVERSITY
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