UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
S.G. = 0.652
PROPERTIES OF FLUIDS:
1. Unit Weight
Problem 2:
Aluminium Ammonium Sulfate has a mass of 600kg and a
volume of 0.40m3 .
a. Compute its mass density.
Weight (W)
𝜸 = Volume (V)
2. Mass density:
Mass (m)
𝝆 = Volume (V)
Mass (m)
b. Compute its specific weight.
Weight (W)
4. Specific Gravity
𝜸 = Volume (V) =
specific gravity of fluid
S.G. = specific gravity of water
= unit weight of water
density of fluid
= density of water
= 14.715 kN/m3
specific gravity of fluid
dy
Where:
𝜏 (tau) = shear stress
𝜇 (mu) = dynamic or absolute viscosity
𝑑𝑉
= time rate of strain or velocity gradient
𝑑𝑦
6. Kinematic Viscosity:
μ
ʋ=ρ
7. Bulk Modulus of Elasticity
∆P
EB = - ∆V
V
Where:
EB = bulk modulus of elasticity
∆P = change in unit pressure
∆V
= volume change per unit volume
V
8. Surface Tension
F Pd
σ=L= 4
Where:
σ = surface tension in N/m
F = elastic force transverse to a length L
9. Capillarity
4σ cos θ
γd
Where:
h = height of capillarity rise or depression
σ = surface tension
γ = specific weight of liquid
d = diameter of tube
For an ideal gas:
P
𝝆 = RT
Where:
P = absolute pressure of gas in Pa
R = gas constant
= 287 J/kg-OK
= 1716 lb-ft/slug-OR
T = absolute temperature in OKelvin
= OK = OC + 273
= OR = OF + 460
Problem 1
If 4.6m3 of fluid weighs 3000 kg.
a. calculate its mass density in kg/m3.
1500
Common Fluids and their Specific Gravity
(if not given)
Water………………………………….. 1.0
Sea Water…………………………….. 1.03
Oil…………………………………...… 0.8
Mercury……………………………..… 13.6
Glycerine……………………………… 1.26
Problem 3:
A certain fluid has a unit weight of 133.416kN/m3.
a. Compute its mass density.
𝜸 = 133.416 kN/m3 x 1000 = 133,416 kN/m3
133,416
𝝆 = 9.81 = 13, 600
b. Compute its specific gravity.
specific gravity of fluid
13,600
S.G. = specific gravity of water = 1000 = 13.6
c. What is the most probable liquid that is under concern?
a. Water
b. Glycerine
c. Mercury (Answer)
d. Ammonia
Problem 4:
A liquid which is compressed in a cylinder has a volume of
1000 cu.cm. at 2 MPa and a volume of 990 cu.cm at 2.5Mpa
a. Compute the bulk modulus of elasticity
a. 20MPa
b. 30MPa
c. 40MPa
d. 50MPa
b. Compute the coefficient of compressibility.
a. 0.01
b. 0.02
c. 0.03
d. 0.04
c. Compute the percentage of volume decreased
a. 1%
b. 2%
c. 3%
d.4%
SOLUTION
∆P
(2 - 2.5)
a.
EB = - ∆V = - (1000 - 990) = 50 MPa
V
b.
c.
1000
1
1
Cc = 𝐸 = 50 = 0.02
𝐵
∆V
V
=
(1000 - 990)
1000
= 1%
3000 kg
𝝆 = Volume (V)= 4.6𝑚3 = 652.174 kg/m3
b. calculate its unit weight in N/m3.
Weight (W)
0.4𝑚 3
S.G. = specific gravity of water = 1000 = 1.5
5. Shear Stress of Fluids:
dV
𝝉=𝝁
Mass (m)
5.886 kN
W = mg
W = 600 ( 9.81)
W = 5886 N = 5.886 kN
c. Compute its specific gravity.
unit weight of fluid
h=
600 kg
𝝆 = Volume (V)= 0.4 𝑚3 = 1,500 kg/m3
3. Specific Volume:
1
Vs = ρ
29,430 N
𝜸 = Volume (V) = 4.6𝑚3
c. calculate its specific gravity.
specific gravity of fluid
652.174
S.G. = specific gravity of water = 1000
Problem 5:
A volume of one cu.m. of water is subjected to a pressure
increase of 14MPa.
a. Compute the change in volume if it has a bulk modulus of
elasticity of 2200MPa in cu.m.
a. 0.0062
b. 0.0064
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
c. 0.0066
d. 0.0068
b Compute the percentage of volume decreased.
a. 0.62%
b. 0.64%
c. 0.66%
d. 0.68%
c. Compute the coefficient of compressibility
a. 0.000454
b. 0.000323
c. 0.000282
d.0.000898
SOLUTION:
∆P
5.a
EB = - ∆V
V
Solution 8:
Gay Lussac’s Law states that, for a given gas at constant
volume, the ratio of pressure to Absolute temperature is
constant.
p1
𝑇1
p
= 𝑇2
2
101.325
p
(25+273.15)
14
2,200 = - ∆V
2
= (320+273.15)
P2 = 201.579 kPa
1
5.b
5.c
∆V = 0.0064 m3
PROBLEM 9
∆V
At -230C, a sample of gas occupies a volume of 0.05 m3
under a pressure of 80 kPa. Find the pressure of the gas
when it is heated to 1100C and expanded to a volume of
0.1m3.
𝑉
=
0.0064
1
1
𝑥 100 = 0.64%
1
Cc = 𝐸 = 2200 = 0.00045
𝐵
Problem 6:
The radius of the test tube is 1mm. The surface tension of
water is equal to 0.0728N/m.
a. Find the capillary rise in the tube in mm.
b. If wetting angle is 80°, determine the surface tension
Solution 9:
Combined Gas Law
𝑃1 𝑉1
𝑇1
SOLUTION:
6.a
R = 1 mm; d = 2 mm = 0.002 m
σ = 0.0728 N/m
θ = 0 (Water)
h=
𝑃 𝑉
= 2𝑇 2
80 x 0.05
(273−23)
2
𝑃 (0.1)
2
= (273+110)
P2 = 61.28 kPa
PROBLEM 10
4σ cos θ
γd
4(0.0728) cos 0
h = 9810 (0.002)
h = 0.01484 m = 14.842 mm
6.b
Find the volume occupied by 2 g mol of an ideal
gas at Standard Absolute Temperature and
atmospheric Pressure
Solution 10:
h=
4σ cos θ
In terms of gram mole, the ideal gas equation can
be re written as:
γd
4σ cos 80
0.01482 = 9810 (0.002)
σ = 0.4192 N/mm
PROBLEM 7
pV = nRT
P – Pressure in kPa
A 200 mL gas sample is at 250C and atmospheric
pressure. It is heated to 1210C under constant pressure.
Find the new volume of the gas.
V – Volume occupied by 1g mol of an ideal gas in
liters
SOLUTION 7
T – Absolute Temperature (oK)
This problem can be solved using the concept of Charles’
Law; For a given gas at constant pressure, the ratio of
volume to absolute temperature is constant.
𝑉1
𝑇1
R – Universal gas constant, 8.314 J / (g mol)( oK)
n – number of gram mole
101.325 (Vmol) = 2(8.314)(273)
𝑉
= 𝑇2
2
Vmol = 44.8 liters
T = 273.15 + oC
200mL
(25+273.15)
𝑉
2
= (121+273.15)
V2 = 264.4 mL
PROBLEM 8
PRESSURE
SITUATION 1
A cylindrical tank shown below has a height of 5 meter
filled with water to a depth of 3 m and oil (SG = 0.82) up
to a depth of 2 m.
450 ml of unknow gas sample is at 250C and experiences
atmospheric pressure (Patm = 101.325kPa). It is heated
to 3200C while the volume remains the same. Find the
new pressure of the gas.
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
1. Calculate for the pressure created by the fluids at the
bottom of the cylindrical tank.
find the pressure at the bottom if the tank contains water
and the gauge at the top reads 76 kPa.
a. 115.24 kPa
b. 102.53 kPa
c. 77.28 kPa
d. 95.62 kPa
a. 49.050 kPa
b. 42.518 kPa
c. 12.060 kPa
d. 99.121 kPa
SOLUTION 4:
2. Calculate for the force to be carried by the tank at the
bottom if it has a diameter of 2 m.
a. 37.888 kN
b. 143.0 kN
c. 154.095 kN
d. 311. 398 kN
3. What is the equivalent height of water if oil is
converted?
a. 1.2 m
b. 1.4 m
c. 1.6 m
d. 1.8 m
SOLUTION 1:
Remember:
The pressure experienced by one point at a certain
elevation is equal to the summation of the pressure
(including the hydrostatic pressure) above that point.
Therefore:
Pbottom = ∑ (γh)
= 0.82 (9.81 kN/m3) (2 m)
+ 1 ( 9.81 kN/m3) ( 3 m )
= 45.518 kPa
SOLUTION 2:
F=PxA
π
F = 45.518 kPa x 4 (2 m)2
F = 143.0 kN
SOLUTION 3:
Take note that the pressure created by one fluid should
be equiuvalent to the other, in this case, you can use:
PA = PB
SA γA HA = SB γA HB
Pbottom = Ptop + ∑ (γh)
= 76 kPa + ( 9.81 kN/m3) ( 4 m )
= 115.24 kPa
SITUATION 2
An open tank is inclined at 70o from the horizontal
contains 3 types of fluid with specific gravity as shown
above.
5. Find the pressure at the interface of oil and water
a. 55.144 kPa
b. 15.696 kPa
c. 20.981 kPa
d. 36.115 kPa
6. Find the pressure at the interface of water and sea
water.
a. 99.261 kPa
b. 45.126 kPa
c. 57.191 kPa
d. 60.953 kPa
7. Find the pressure at the bottom
a. 88.302 kPa
b. 70.331 kPa
c. 90.782 kPa
d. 65.335 kPa
Cancelling same value of unit weight of water,
SA HA = SB HB
SOLUTION 5:
0.8 x 2 = 1 x HB
The inclination of the tank does not affect the pressure at
a certain elevation
HB = 1.6 m
PROBLEM 4
Two pressure gauges are attached at the top and at the
bottom side of the tank. If the tank has a height of 4 m,
Poil – water
= ∑(γh)
= 0.8 (9.81 kN/m3) (2 m)
+ 1 ( 9.81 kN/m3) ( 3 m )
= 15.696 kPa
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
SOLUTION 6:
Pwater- sea water = 0.8 (9.81) (2 m) + 1 ( 9.81) ( 3 m )
= 45.126 kPa
c. Given that the elevation goes higher, at the
same time, the pressure goes higher.
d. Vaccum pressure is the difference between
the atmospheric pressure and absolute pressure and
gives a negative value
SOLUTION 7:
Pbottom = 0.8 (9.81) (2 m)
+ 1 ( 9.81) (3 m )
+ 1.03 ( 9.81) (3 m)
= 65.335 kPa
ALTERNATIVE SOLUTION 6:
You can also use the pressure at the interface of water
and oil (15.696 kPa) then add the hydrostatic pressure
created by the water to up to the interface of water and
sea water
Pwater- sea water = 15.696 kPa + 1 9.81) ( 3 m )
= 45.126 kPa
ALTERNATIVE SOLUTION 7:
Pbottom = 45.126 kPa +
1.03 ( 9.81) (3 m)
= 65.335 kPa
SITUATION 3
Two pressure gauge was established at the top and at
the bottom of the building, the gauge at the top reads 723
mmHg while at the bottom reads 760 mmHg. The unit
weight of the air is 0.01235 kN/m3
8. What is the approximate height of the building?
a. 300 m
b. 400 m
c. 500 m
d. 600 m
9. If each floor measures 5 m, what is the top floor of the
building?
a. 77th
b. 78th
c. 80th
d. 81st
SOLUTION 8:
“mmHg” is a unit of pressure that means __ (blank) mm
height of column of Mercury (SG = 13.6)
Pbottom = Ptop + ∑ (γh)
γbottomh = γtoph + γairh
SOLUTION 10:
A is correct, it is known as Pascal’s Law
B is correct, the direction acts perpendicularly at
all times
C is wrong, as the elevation goes higher the
magnitude of the pressure decreases.
D is correct, a negative value of pressure is a
vaccum pressure and creates suction.
FORCES ON PLANE SURFACES
PROBLEM 1
Determine the hydrostatic force acting on one face of the
plane having a rectangular cross section width of 3m and
a height of 2m with the top flushed with water.
Solution 1:
F = γ h̅ A
𝑚
F = (9.81 kN/m3) (22 ) (3m x 2m)
F = 58.86 kN
SITUATION
An rectangular gate on one face of a prism container is
hinged at the bottom as shown retains a oil (SG = 0.81).
2. Find the value of the hydrostatic force in kN that is
experienced on the gate
3. Find the value of the force P that is necessary to be
applied on the top to hold the gate
13.6 (9.81) (0.76m) = 13.6 (9.81) (0.723m) + 0.01235(H)
Solving for H,
H = 399.708 m ≈ 400 m
SOLUTION 9:
For this kind of questions, we will be starting to count on
the 1st floor..
0 m = 1st floor
5 m = 2nd floor
5m

+ 1 = 2nd floor
5
10 m = 3rd floor
10m

+ 1 =3rd floor
5
4. If P is limited only to 1 kN, determine the height of the
water that this container can retain measured from the
hinge
400
Floor level = 5 + 1 = 81st floor
PROBLEM 10
Which of the following statement is NOT true?
a. The pressure at all points inside a static fluid
are all equal.
b. The pressure acts perpendicular or normal to
the boundary surfaces
SOLUTION 2:
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
F = γ h̅ A
𝑚
F = 0.81 ( 9.81) (12 ) (0.8m x 1m)
F = 3.178 kN
SOLUTION 3:
Solve for the location of the center of pressure
for you to be able to take moment on the hinge.
Take note that we are considering water now (SG = 1)
The height “Z” of the water from the bottom, varies,
The h̅ also varies, and the area also varies.
Yp = y̅+ e
e = (A𝐼𝑋y̅ )
3
3
Ix = (𝑏ℎ
) = (0.8(1)
)
12
12
4
Ix = 1/15 =0.06667 m
y̅ = h̅ (since our gate is on upright position)
h̅ = 0.5 m
0.066667
e = ((0.8)(1)(0.5)
) = 0.166667 m
Yp = y̅+ e = 0.5+0.167777
Yp = 0.667 m from the oil surface
(Location of the center of pressure)
Or simply you can assume a triangular pressure
diagram acting on the gate with zero pressure at the
top, giving you the location of the center of pressure to
be 2/3 from the top and 1/3 from the hinge.
↻∑M0 = 0
F (0.33333Z) = P(1)
We will be replacing F with new value considering
water.
Since F = γ h̅ A, then
F (0.33333 z) = P(1) will become
γ h̅ A (0.33333 z) = P (1). P = 1.
We will be having the equation in terms of variable z
z
1(9.81) (2) (z x 0.8) (0.3333z) = (1) (1)
1(1)
Z3 = 1.308
z = 0.914 m
SITUATION
A rectangular gate on one face of a prism container is
hinged at the bottom as shown retains water
The vertical distance from the center of gravity to the
center of pressure was measured to be 0.3 m.
Yp = (23)(1m) = 0.667 m
5. What is the total pressure experienced by the gate
6. How far from the bottom is the total thrust acting
Now, take moment at the hinge, the moment at the
hinge is zero, because it is free to rotate at that point.
↻∑M0 = 0
F(0.333) = P (1)
(3.178 kN) (0.333) = P (1)
P = 1.058 kN
SOLUTION 3:
7. What would be the value of P enough to close the
gate if the tank is full of water?
SOLUTION 5
The vertical distance between the center of gravity and
the center of pressure is the eccentricity, e =0.3
e = (AIXy̅ )
2(y)3
( 12 )
y
(2)(y)(2 )
0.3 = (
)
y = 1.8 m
F = γ h̅ A
F = 9.81 (1.8/2) ( 1.8 x 2)
F = 31.784 kN
SOLUTION 6
z = (1/3)(1.8) = 0.6 m
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
SOLUTION 7
F = γ h̅ A
F = 9.81 (3/2) (3 x 2)
F = 88.29 kN
↻∑M0 = 0
F(1/3) (3) = P (3)
(88.29) (1/3) (3) = P (3)
P = 29.43 kN
h̅ = 1.2 + y
Z = 13 h
Z = 13 (3.6 m)
z = 1.2 m
Solving for y,
sin 300 = 𝑦𝑧
y
sin 300 = 1.2
; y = 0.6 m
h̅ = 1.2 + y
h̅ = 1.2 + 0.6
h̅ = 1.8 m
F = γ h̅ A
= 9.81 (1.8) ( 12 x 2.4 x 3.6 )
= 76.283 kN
SOLUTION 9
SITUATION
A triangular gate as shown is submerged on water.
8. Find the value of the hydrostatic force acting on one
side of the gate.
9. Determine the location of the hydrostatic force from
the water surface.
10. If the triangular gate will be inverted (i.e. the vertex
will be at pt. A and the base will be at pt B) , calculate
the new value of the hydrostatic force acting on the
gate.
Solve for y̅;
̅
sin 300 = hy̅
sin 300 = 1.8
; y̅ = 3.6 m
y̅
Solving for Yp = y̅ + e
e = (A𝐼𝑋y̅ )
3
SOLUTION 8:
Ix = ( 𝑏ℎ
)
36
(
𝑏ℎ3
)
e = ( A36y̅ )
e=( 1
2.4 𝑥 3.63
(
)
36
( 2 x 2.4 x 3.6 ) 3.6
)
e = 0.2 m
Yp = y̅ + e
Yp = 3.6 + 0.2
Yp = 3.8 m
SOLUTION 10
F = γ h̅ A
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
𝜋
9.81( 4 𝑥 1.12 )(1.9) + 9.81 VPb = 3.625 + 110 VPb
VPb = 0.1406 m3
W = 𝛾𝑉𝑜𝑙 = 9.81 (0.1406)
W = 15.468 kN
2. BFcyl = W cyl + W pb
𝜋
9.81( 4 𝑥 1.12 )(1.9) = 3.625 + W Pb
W Pb = 14.1 kN
3. BFcyl = W cyl + W pb
𝜋
9.81( 4 𝑥 1.12 )(2.4) = 3.625 + W Pb
W Pb = 18.75 kN
F = γ h̅ A
h̅ = 1.2 + y
Z = 23 h
Z = 23 (3.6 m)
z = 2.4 m
Solving for y,
sin 300 = 𝑦𝑧
y
sin 300 = 2.4
; y = 1.2 m
h̅ = 1.2 + 1.2
h̅ = 2.4 m
F = γ h̅ A
= 9.81 (2.4) ( 12 x 2.4 x 3.6 )
= 101.710 kN
BUOYANCY
ARCHIMEDES PRINCIPLE
“Any body completely or partially submerged in
a fluid (gas or liquid) at rest is acted upon by an
upward, or buoyant, force the magnitude of which is
equal to the weight of the fluid displaced by the body.”
BF = 𝜸Liquid x Vsubmerged
DAMS:
Overturning Moment (OM)
OM = Ʃ moment about the toe due the active forces
other than the weight of the dam
Resisting Moment (RM)
RM = Ʃ of counter clockwise moment about the toe due
to weight of dam
Factor of safety against overturning:
RM
F.S.OT = OM
Factor of safety against sliding:
μRy
F.S. sliding =
ƩFx
Where:
μ = coefficient of soil friction
ƩMTOE = 0
Note: moments are due to active and reactive forces.
SITUATION
The section of a concrete gravity dam has 2m top width
and 4m bottom width and 8n high. The depth of water at
the upstream side is 6m. Neglect hydrostatic uplift and
use unit weight of concrete equal to 23.5 kN/m3.
Coefficient of friction between the base of the dam and
the foundation is 0.6
1. Determine the factor of safety against sliding.
a. 2.40
b. 1.581
c. 1.437
d. 1.916
2. Determine the factor of safety against overturning.
a. 4.88
b. 3.904
c. 5.856
d. 5.33
SITUATION
A hollow cylindrical tank with 1.10 m diameter has
height of 2.4 m. It weighs 3.625 kN in air.
1. Determine the weight of lead attached to the outside
bottom so that the tank will be submerged vertically 1.9
m in fresh water. Density of lead is 110 kN/m3
a. 12.5 kN
b. 15.5 kN
c. 18.6 kN
d. 21.5 kN
2. Determine the weight of the lead loaded inside so
that the tank will be submerged vertically 1.9 m in fresh
water.
a. 14.1 kN
b. 9.81 kN
c. 10.6 kN
d. 16.2 kN
3. Determine the maximum weight of the lead loaded
inside so that the top of the cylinder will be flush with
the water surface.
a. 20.50 kN
b. 22.35 kN
c. 16.85 kN
d. 18.75 kN
SOLUTION 1
𝐹 = 𝛾ℎ̅𝐴
=9.81(3)(6x1)
F = 176.58 kN
F = 176.58 kN
y = 1/3 (6) = 2m
W1 = 𝛾𝑐 𝑉1
=23.5(2(8)(1))
W1 = 376 kN
W2 = 𝛾𝑐 𝑉2
=23.5(1/2(2)(8)(1))
W2 = 188 kN
X1 = 4 - 1/2 (2)
X2 = (2/3)(2) = 1.333m
Rx = F = 176.58 kN
Ry = W1 + W2 = 376 + 188
Ry = 564 kN
𝜇𝑅𝑦 (0.6)(564)
𝐹𝑆𝑠 =
=
= 1.916
𝑅𝑥
176.58
SOLUTION:
1. BFcyl + BFpb = W cyl + W pb
SOLUTION 2
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
RM = 376(3) + 188(1.333)
RM = 1378.604 kN-m
OM = F x y
= 176.58 (2)
OM = 353.16 kn=m
𝑅𝑀 1378.604
𝐹𝑆𝑜 =
=
= 3.904
𝑂𝑀
353.16
(𝒂𝒏𝒔𝒘𝒆𝒓)
SITUATION
An open cylindrical tank, 2.5m in diameter and 5m
high contains water to a depth of 3.75m. it is rotated
about its own vertical axis with a constant angular
speed 𝜔.
12. If 𝜔 = 2.5𝑟𝑎𝑑/𝑠𝑒𝑐. how much water is spilled
out?
a. 0
b. 1
c. 0.5
d. 1.25
ROTATING VESSELS
SOLUTION 12
𝜔2 𝑟 2
ℎ=
2𝑔
𝜔 = 2.5 rad/sec
2
(2.5) (1.25)2
ℎ=
2(9.81)
h = 0.50
h/2 = 0.25 m < (5 - 3.75)
therefore no liquid spilled (answer)
1
Volume = 2 πR2h
h=
V2
2g
=
ω2r2
2g
PROBLEM 10
An open cylindrical vessel having a height equal to
its diameter is half-filled with water and revolved
about its own vertical axis with a constant angular
speed of 150 rpm. Find its minimum diameter so
that there can be no liquid spilled.
a. 397.58mm b. 361.44mm
c. 271.15mm d. 318.07mm
SOLUTION 10:
H = D; r = D/2
𝜔 = 150 𝑟𝑝𝑚 𝑥 𝜋/30
𝜔 = 5𝜋 rad/sec
2
(5𝜋) (𝐷/2)2
𝐷=
2(9.81)
D = 0.318m or 318.07mm (answer)
13. What maximum value of angular velocity in rpm
can be imposed without spilling any liquid?
a. 53.50
b. 66.88
c. 60.80
d. 45.60
SOLUTION 13
h/2 = 5 - 3.75 ; h = 2.5
𝜔2 𝑟 2
ℎ=
2𝑔
2
(𝜔) (1.25)2
2.5 =
2(9.81)
𝜔 = 5.60 rad/sec x 30/ 𝜋
𝜔 = 53.50 𝑟𝑝𝑚 (answer)
14. If 𝜔 = 6 𝑟𝑎𝑑/𝑠𝑒𝑐, how much water is spilled out?
a. 1.03
b. 0.91
c. 0.77
d. 1.14
SOLUTION 14:
𝜔 = 6 𝑟𝑎𝑑/𝑠𝑒𝑐
2
(6) (1.25)2
ℎ=
PROBLEM 11
An open cylindrical tank 1.5m in diameter and 2.2m
high is full of water when rotated about its vertical
axis at 25 rpm, what would be the slope of the water
surface at the rim of the tank?
a. 0.655
b. 0.524
c. 0.786
d. 1.048
SOLUTION 11
Slope = 𝑡𝑎𝑛𝜃
𝜔2 𝑥
𝑆𝑙𝑜𝑝𝑒 =
𝑔
𝜔 = 25 𝑟𝑝𝑚 𝑥 𝜋/30
5
𝜔 = 𝜋 rad/sec
6
5
6
2(9.81)
h = 2.87 m
some liquid spilled but the vortex of the
paraboloid is inside the tank since h<3.75
Vspilled = Vair(final) – Vair(initial)
1
𝑉𝑓𝑖𝑛𝑎𝑙 = 𝜋(1.25)2 (2.87)
2
Vfinal = 7.044
𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝜋(1.252 )(1.25)
Vfinal = 6.14
Vspilled = 7.044 - 6.14
Vspilled = 0.91m3 (answer)
DYNAMIC EQUILIBRIUM
A. Horizontal Acceleration:
2
( 𝜋) (0.75)
𝑆𝑙𝑜𝑝𝑒 =
9.81
Slope = 0.524 (answer)
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
a
g
a
Tan 𝜃 = g
P2 = P1 + 𝜸h(1 ± )
Case 2: Open Tank
a
P2 = 𝜸h(1 ± )
g
Case 1: Open Tank
Where:
+ = for upward acceleration
- = for downward acceleration
PA = 𝜸hA
PB = 𝜸hB
Case 2: Closed Tank With Air
PA = P1 + 𝜕hA
PB = P1 + 𝜕hB
SITUATION
A vessel 2.5m in diameter containing 1.6m of water
is being raised.
1. Find the pressure at the bottom of the vessel in
kPa when the velocity is constant
a. 17.84
b. 19.62
c. 13.38
d. 15.70
SOLUTION 1
𝜌 = 𝛾ℎ
𝜌 = 9.81(1.6)
P = 15.70 kPa (answer)
Case 3: Closed Tank Without Air Space
PA = 𝜸hA’
PB = 𝜸hB’
HGL = Hydraulic Grade Line – line connecting all points
of zero gage pressure
2. Find the pressure at the bottom of the vessel
when it is accelerating 0.4m/s2 upwards.
a. 18.57
b. 20.43
c. 16.34
d. 13.93
SOLUTION 2:
𝑎
𝜌 = 𝛾ℎ(1 + )
𝑔
0.4
)
9.81
P = 16.34 kPa (answer)
𝜌 = (9.81)(1.6)(1 +
SITUATION
An open rectangular tank mounted on a truck is
4.5m long, 1.2m wide and 2m high is filled with
water to a depth of 1.5m
1. What maximum horizontal acceleration can ba
imposed on the tank without spilling any water
a. 2.48
b. 2.18
c. 1.86
d. 2.73
SITUATION
An open tank containing oil (sp. gr. = 0.82) is
accelerated vertically at 7.5m/s2. Determine the
pressure 2.5m below the surface if the motion is;
3. if the motion is upward with a positive
acceleration
a. 54.09
b. 35.49
c. 43.275
d. 49.18
SOLUTION 1
0.5
2.25
𝜃 = 12.53
𝑎
𝑡𝑎𝑛𝜃 =
𝑔
𝑎 = tan(12.53)(9.81)
a = 2.18m/s2 (answer)
𝑡𝑎𝑛𝜃 =
2. Determine the accelerating force on the liquid
mass?
a. 17.66kN
b. 22.07kN
c. 20.07kN
d. 15.05kN
SOLUTION 2
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐹𝑜𝑟𝑐𝑒, 𝐹 = 𝑀𝑎
𝑀 = 𝜌(𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑)
𝑀 = 1000(4.5𝑥1.2𝑥1.5)
M = 8100 kg
𝐹 = (8100)(2.18)
F = 17658 N or 17.66kN (answer)
B. Downward/Upward Acceleration:
Case 1: Closed Tank
SOLUTION 3
𝑎
𝜌 = 𝛾ℎ(1 + )
𝑔
𝜌 = (9.81𝑥0.82)(2.5)(1 +
7.5
)
9.81
P = 35.49 kPa (answer)
4. if the motion is upward with a negative
acceleration.
a. 6.56
b. 5.78
c. 7.22
d. 4.74
SOLUTION 4
𝜌 = (9.81𝑥0.82)(2.5)(1 +
−7.5
)
9.81
P = 4.74 kPa (answer)
5. if the motion is downward with a positive
acceleration
a. 4.74
b. 5.78
c. 7.22
d. 6.56
SOLUTION 5
__M I J D__
UNIVERSITY OF NUEVA CACERES – NAGA CITY
COLLEGE OF ENGINEERING AND ARCHITECTURE
MODULE # 1
FLUID MECHANICS
𝑎
𝜌 = 𝛾ℎ(1 + )
𝑔
𝜌 = (9.81𝑥0.82)(2.5)(1 −
7.5
)
9.81
P = 4.74 kPa (answer)
6. if the motion is downward with a negative
acceleration.
a. 54.09
b. 43.275
c. 35.49
d. 49.18
SOLUTION 6
𝑎
𝜌 = 𝛾ℎ(1 + )
𝑔
𝜌 = (9.81𝑥0.82)(2.5)(1 −
−7.5
)
9.81
P = 35.49 kPa (answer)
__M I J D__