CE140 Abcdef

MAPUA INSTITUTE OF TECHNOLOGY
SCHOOL OF CIVIL, ENVIRONEMENTAL AND GEOLOGICAL ENGINEERING

INTRAMUROS, MANILA
CE140-1P
FLUID
MECHANICS
ENGR. KEVIN PAOLO V. ROBLES

2009106139
CE140-1P FLUID MECHANICS
GRADING SYSTEM
Quizzes 45 %
Final Examination 20 %
Exercises 05 %
Laboratory Reports 20 %
Notebook 05 %
Project 05 %
Total 100 %
Passing Rate is strictly 70. 00 %
CONTACT INFORMATION:
JOB DESCRIPTION : Part Time Faculty

Mapua Institute of Technology
Engineer II
Department of Public Works and Highways
Unified Project Management Office
Roads Management Cluster I
PJHL-PMO Building, DPWH NCR Compound, 2nd Street,
Port Area, Manila
FACEBOOK : Kevs Robles (facebook.com/kevinpaolorobles)
INSTAGRAM/TWITTER : @itskevmosby
FACEBOOK GROUP : CE140-1P 1st Term20162017
EMAIL ADDRESS : kpvroblesmapua@yahoo.com
HOUSE RULES :
1. Strictly no cheating.
2. No taking pictures of lectures during discussions. Do it after the instructor is finished
writing on the board.
3. No unreasonable absences.
4. Always follow the format for quizzes/ exercises / lab reports.
2
FORMAT for QUIZZES/EXERCISES/HOMEWORKS
0.5in x 0.5in margin
FORMAT of LABORATORY REPORT
Margin : left x right x top x bottom
0.75 in x 0.5 in x 0.5 in x 0.5 in
Front page to be provided
CONTENTS OF LABREPORT:
I. INTRODUCTION
II. OBJECTIVES
III. SKETCH OF APPARATUS
IV. LABORATORY PROCEDURE
V. RESULTS
VI. SAMPLE COMPUTATION
3
VII. CONCLUSION
VIII. APPLICATION TO ENGINEERING
IX. REFERENCES
TOPICS TO BE DISCUSSED
1. Fluid Properties
2. Pressure Measurements
3. Principle of Manometry
4. Thin Walled Cylinders
5. Hydrostatic Forces
6. Buoyancy
7. Relative Equilibrium
8. Bernoulli’s Energy Equation
4
INTRODUCTION TO FLUID MECHANICS
FLUID STATICS is the branch of fluid mechanics that studies fluids at rest. It embraces
the study of the conditions under which fluids are at rest in stable equilibrium
FLUID DYNAMICS, in physics, is a sub discipline of fluid mechanics that deals with fluid
flow—the natural science of fluids(liquids and gases) in motion
5
FLUID PROPERTIES
A. MASS DENSITY - mass per unit of volume
mass m kg g slugs
ρ= = = != =
volume ∀ m cm! ft !
m = ρ∀
kg slugs
ρ = 1000 = 1.94
m! ft !
B. SPECIFIC WEIGHT ("! ) (UNIT WEIGHT / WEIGHT DENSITY ) - is the weight per
unit of volume
W mg N lb
γ" = = = ρg = # = #
V V m ft
$ '(
g = 9.81 γ = 9.81 (metric )
%! & $"
)* +,
g = 32.2 %! γ& = 62.4 )*" (English)
C. SPECIFIC VOLUME - is the volume per unit of weight

1 m#
V% = =
ρ kg
D. SPECIFIC GRAVITY (:! ) - is the ratio of specific weight of liquid in question to that of
water
γ" ρ"
S" = =
γ-! . ρ-! .
S-! . = 1
E. BULK MODULUS OF ELASTICTY - is the incremental change in volume when the

pressure is changed by an incremental amount.
∆F
=>?@AA ∆G
</ = =
=>?BCD G
0
H= (coefficient of compressibility)
1#
F. VISCOSITY - is the property of fluid which determines its resistance to shearing stress.
It is sometimes called coefficient of viscosity, or dynamic viscosity.
I = JK
6
L
I=
MG
MN
Where L = shear stress

MG/MN = velocity gradient
Unit : (N*s)/m2 = Pa * s
1 poise = 0.1Pa*s
G. KINEMATIC VISCOSITY (P = QR) − is the ratio of the dynamic viscosity to that of

density
µ
ν=
ρ
H. VAPOR PRESSURE- the pressure exerted by vapor in a closed space
I. SURFACE ENERGY - is the energy per unit area caused by relative forces of cohesion
and adhesion
Example no. 01
A reservoir of glycerin has a mass of 1200kg and a volume of 0.952 m3.
Find:
a. Weight
b. Unit Weight
c. Mass Density
d. Specific Gravity
Example no. 02
A liquid is compressed in a cylinder if it has a volume of 1000 cu.cm (1liter) at 2 Mpa and has
a volume of 990 cu.cm at 2.5 MPa.
a. Compute for the bulk modulus of elasticity

b. Compute the percentage of volume decreased
c. Compute the coefficient of compressibility
Example no. 03
A gas having a volume opf 40 liters has a pressure of 0.24 Mpa at 24 deg. Celcius. If the gas
constant R is equal to 212 M.N/kg.K, compute
a. density of gas
b. mass of the gas
c. weight of the gas
7
PRINCIPLES OF HYDROSTATICS
1. The Liquid at rest cannot resist shearing stress.
2. The total force is always normal to the plane where it acts. (Perpendicular).
3. Pressure exists at every point.
4. At any point in a liquid at rest. The pressure is equal in all directions (Pascal’s Law).
5. The Pressure of all points lying on a plane parallel to the liquid surface is equal.
6. The Pressure varies linearly with depth.
!"#$% ! (!
PRESSURE = = =
&#%' & (&
" = #$
PRESSURE HEAD - Represents the height of a column of homogeneous fluid that will
produce an intensity of pressure
#
!=
$
8
PRESSURE MEASUREMENTS
#!"# = #!$% + #&!&'

Where :
F234 = Absolute Pressure
F256 = Atmospheric Pressure
F7278 = Gage Pressure
ABSOLUTE PRESSURE - measured from absolute zero perfect vacuum
GAGE (GAUGE) PRESSURE - measured from the atmosphere
!!"# = #
F256 = 1B>J = 1VB? = 101.3212FB = 760JJYZ = 10.94JY9 [
PRESSURE ACTING PERPENDICULAR TO THE SURFACE
Example no. 04
Find the mass density of helium at a temperature of 4 deg. Celcius and a pressure of 194 kPa
gauge if atmospheric pressure is 101.92 kPa (R = 2079 J/(kg*K))
9
Example no. 05
The reading of an automobile fuel gage is proportional to the gage pressure at the bottom of
the tank. The tank is 32cm deep and is contaminated with 3cm of water.
1. If the tank is full of gasoline what should be the reading at the gauge in Pa?
Use \724:;<=8 = 6670]/J# and \2<> = 11.8]/J#
2. How many centimeters of air remains at the top when the gage indicates full?
Example no. 06
A closed compartment shown contains water 3 m. deep at the left side and 5m. deep on the
right side. If the pressure at point A is 98 kPa abs on the closed compartment shown.
1. What is the absolute pressure at point B considering the specific weight of air.
2. What is the absolute pressure at point B neglecting the specific weight of air.
Example no. 07
For the open tank with piezometers attached on the side, contains two different liquids
1. Find the elevation of the liquid at piezometer A.
2. Find the pressure at the bottom of the tank
3. Find the elevation of the liquid in piezometer B.
10
PRESSURE MEASUREMENTS
MANOMETER - a tube bent in a form a “U” containing a fluid of known specific gravity. The
difference in elevations of the liquid surfaces under pressure indicates the difference in
pressure at the two ends. Basically, there are two types of manometers.
1. An OPEN MANOMETER has one end, open to atmospheric pressure and is capable
of measuring the cause pressure in a vessel.
2. A DIFFERENTIAL MANOMETER connects each end to a different pressure vessel

and is capable of measuring the difference in pressure between two vessels.
Example no. 08
Calculate the difference in pressure between points A and B in
a. KPa
b. Meters of Water
11
Example no. 09
A 1cm-dia U-tube contains mercury (SHg = 13.6)as shown. If 10 cubic meter of water is
poured into the right hand leg. What would be the free surface heights after it has died
down.?
HYDRAULIC JACK
In the figure shown, the bigger piston or plunger is in the same elevation as the smaller piston
or plunger. By principle of surface of equal to the pressure under the smaller piston that is
PA = Pa
The forces causing these pressures may include the weights of the pistons as the case may
be, the above equation becomes,
$ + '$ ) + '!
=
( *
If the weights of the bigger and smaller pistons which are ^? and ^2 respectively are not
considered it becomes therefore:
$ )
=
( *
If the pistons are not in the same elevation as the figure below. The principle of manometry
is applied.
From the figure shown, the following equation holds,
12
!% − ,& - = !$ ./ !$ + ,& - = !%
If the weights of the pistons are considered, the following equation holds:
) + '! $ + '$
− ,'!( 0& - =
* (
But if the weights of the pistons are not considered, it becomes:
) $
− ,'!( 0& - =
* (
Example no. 10
In the given figure, calculate the magnitude of the force FR that is required to keep the system
in equilibrium. Consider the weights of the bigger and smaller pistons to be 20kN and 10kN
respectively, consider the intervening passages filled with oil of specific 0.80
Example no. 11
13
Calculate the total weight acting on the piston if the gage reads 250 kPag
14
HYDROSTATIC FORCE
3)*
1( 2 =
$' = ,& -
1
(y
_@ = Hydrostatic Force
ℎa = depth of liquid above the center of gravity of the submerged area
\A = unit weight of liquid
bB7 = Moment of Intertia about the center of gravity
@ = eccentricity
Recall : MOMENT OF INERTIA OF PLANE SURFACES
15
Example no. 01
A given plane below is submerged in water as shown. Calculate the total hysdrostatic force
acting on it . Specify its point of application.
16
Example no. 02
1. Compute the depth of the center of pressure for a vertical triangular gate having a
height h and a base width b and submerged in a liquid with its vertex at the liquid
surface
2. If the vertex is 1m below the surface, determine the location of center of pressure
from the surface if b = 4m and height = 3m
3. Compute also the total hydrostatic force on one side of the gate
Example no. 03
Gate AB 5m Wide into the board is hinged at point A. The Gate supporting liquids as shown
has a force Fb applied at point B to keep the gate from opening. Neglecting the Weight of the
gate, calculate the required magnitude of Fb
PRESSURE PRISM METHOD
17
Example no. 04
A rectangular gate of dimension 1 m by 4 m is held in place

by a stop block at B. This block exerts a horizontal force of
40 kN and a vertical force of 0 kN. The gate is pin-connected
at A, and the weight of the gate is 2 kN. Find the depth h of
the water.
18
Example no. 05
As shown, a round viewing window of diameter D = 0.8 m is situated in a large tank of

seawater (S = 1.03). The top of the window is 1.2 m below the water surface, and the window
is angled at 60° with respect to the horizontal. Find the hydrostatic force acting on the window
and locate the corresponding CP.
Example no. 06
A rectangular gate is hinged at the water line, as shown. The gate is 4 ft high and 10 ft wide.
The specific weight of water is 62.4 lbf/ft3. Find the necessary force (in lbf) applied at the
bottom of the gate to keep it closed.
HYDROSTATIC FORCE ON CURVE SURFACES
This section describes how to calculate forces on surfaces that have curvature. These
results are important for the design of components such as tanks, pipes, and curved gates.
Consider the curved surface AB in Fig. 3.17a. The goal is to represent the pressure
distribution with a resultant force that passes through the center of pressure. One approach
is to integrate the pressure force along the curved surface and find the equivalent force.
However, it is easier to sum forces for the free body shown in the upper part of Fig. 3.17b.
The lower sketch in Fig. 3.17b shows how the force acting on the curved surface relates to
the force F acting on the free body. Using the FBD and summing forces in the horizontal
direction shows that
19
The line of action for the force FAC is through the center of pressure for side AC, as
discussed in the previous section, and designated as ycp.
Example no. 07
Surface AB is a circular arc with a radius of 2 m and a width of 1 m into the paper. The
distance EB is 4 m. The fluid above surface AB is water, and atmospheric pressure prevails on
the free surface of the water and on the bottom side of surface AB. Find the magnitude and
line of action of the hydrostatic force acting on surface AB
Example no. 08
The profile of a dam is an arc of a circle having a radius of 30m and subtending an angle of
60deg at the center of curvature which lies in the water surface.
a. Determine the load on the dam
b. Position of the line of action
c. Depth of center of pressure
20
Example no. 09
A hemispherical Cap AC is located on the side of the tank which is under a pressure of 8KPa.
Calculate the total hydrostatic force acting on the cap AC and specify its line of action
21
BUOYANCY
The same principles used to compute hydrostatic forces on surfaces can be applied to the net
pressure force on a completely submerged or floating body. The results are the two laws of
buoyancy discovered by Archimedes in the third century B.C.:
1. A body immersed in a fluid experiences a vertical buoyant force equal to the weight of the
fluid it displaces.
2. A floating body displaces its own weight in the fluid in which it floats.
These two laws are easily derived by referring to Fig. 2.16. In Fig. 2.16a, the body lies between
an upper curved surface 1 and a lower curved surface 2. From Eq. (2.45) for vertical force,
the body experiences a net upward force
4$ = , 5
Also the BUOYANT FORCE is equal to the difference between the weight of the object in
air and the weight of the object in liquid. The weight of the object in the liquid is sometimes
called apparent weight.
That is,
4$ = '+ −'!
Example no. 01
A block of wood when placed in water projects 6cm above the water surface. When placed
in a liquid of specific gravity 1.60 it projects 10cm above the surface of that liquid. What is
the specific gravity of the wood.
Example no. 02
A 60 cm cube weighing 2000N is lowered in a tank containing 1.80 m of water below 1.80 m
of oil of specific gravity 0.80. if the sides of the cube remain vertical:
22
a) How much of the cube protrudes above the oil-water interface?

b) What is the total hydrostatic force acting on the side of the cube?
Example no. 03
A closed empty steel cylindrical tank 2m in diameter 3m high is to be used as a buoy in fresh
water. What would be the size of the concrete cube if the tank would be submerged 2m into
the water? The steel tank weighs 500kg. Use Sconc.=2.4
Example no. 04
A concrete cube 10.0 inch on each side is to be held in equilibrium under water by attaching
a lightweight foam buoy to it. (In theory, the attached foam buoy and concrete cube, when
placed under water, will neither rise nor sink) If the specific weight of concrete and foam are
150 lb/ft3 and 5.0 lb/ft3, respectively, what minimum volume of foam is required.
Example no. 05
The homogeneous 12-cm cube in Fig. 2.116 is balanced by a 2-kg mass on the beam scale
when the cube is immersed in 20°C water. What is the specific gravity of the cube?
23
Example no. 06
An object weighs 200N in air and 190N in oil of specific gravity 0.75. find the volume and
specific gravity of the object?
24
RELATIVE EQUILIBRIUM
- liquid mass in Rigid Body Motion (the fluid will move as a rigid mass with each
particle having the same acceleration)
LINEAR ACCELERATION
1. VERTICAL ACCELERATION
Upward Acceleration Downward Acceleration
2. HORIZONTAL ACCELERATION
- liquid surface becomes inclined
tan θ = ax/g = y/x

where :
tan θ = slope the final liquid surface makes when accelerated or
decelerated
ax = super imposed acceleration or deceleration
g = acceleration due to gravity
25
HORIZONTAL ACCELERATION
A. OPEN VESSEL
casdasdasdasdasdas
CASE 1 CASE 2
CASE 3 CASE 4
CASE 5
26
B. CLOSED VESSEL
CASE 1 CASE 2
CASE 3
CASE 4
CASE 5 27
3. INCLINED ACCELERATION
EXAMPLE
1. An open vessel 3m wide, 4m long and 2.50m deep contains 2m of water. It is then
accelerated to the right along its width.
a.) Calculate the acceleration just about water to spill.

b.) At this rate, what is the force causing the acceleration if the mass of the vessel is
neglected? If the mass of the vessel is 500kg?
2. Consider the vessel below accelerated at 6.00 m/s^2. Determine the volume of water spilled
at that acceleration.
3. Consider the above vessel closed with air space subjected to a pressure of 10kpag. If the
same vessel is accelerated at a rate of 10 m/s^2, determine the following:
a.) force acting on the left side c.) force acting on the top surface
b.) force acting on the right side d.) force acting at the bottom
28
ROTATIONAL ACCELERATION (ROTATIONAL VORTEX
- liquid surface becomes inclined
A. OPEN VESSEL
CASE 1 CASE 2
CASE 3 CASE 4
CASE 5
B. CLOSED VESSEL
CASE 1 CASE 2
CASE 3 CASE 4
CASE 5
30
EXAMPLE
1. An open cylindrical 60cm in diameter and 90cm high is 2/3 full of water. If the tank is rotated
about its vertical axis at the center.
a. What is the maximum speed it can have without spilling?
b. What speed it must have in order that the depth of water at the center is zero?
c. What speed must it have so that there is no water within 15cm from axis of
rotation? At this speed, how much volume is spilled?
2. Consider the same tank closed with air space subjected to a pressure of 15kpag. If it is
rotated about the same axis at a rate of 12.50 rad/sec, determine the pressure at bottom
of the tank, center and side.
31
THIN WALLED CYLINDERS
A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which
resist bursting, developed across longitudinal and transverse sections
A. CIRCUMFERENTIAL (TANGENTIAL) STRESS, >),-)
!?
>),-) =
@A
B. LONGITUDINAL STRESS, >.+/*
!?
>.+/* =
BA
Where :
P = internal pressure caused by the fluid

D = outside diameter of the tank / container
t = thickness of the tank / container
32
Example no. 01
A cylindrical steel pressure vessel 400 mm in diameter with a wall thickness of 20 mm, is
subjected to an internal pressure of 4.5 MN/m2
a) Calculate the tangential and longitudinal stresses in the steel.
b) To what value may the internal pressure be increased if the stress in the steel is limited
to 120 MN/m2?
c) If the internal pressure were increased until the vessel burst, sketch the type of
fracture that would occur.
Example no. 02
The wall thickness of a 4-ft-diameter spherical tank is 5/16 in. Calculate the allowable internal
pressure if the stress is limited to 8000 psi.
Example no. 03
Calculate the minimum wall thickness for a cylindrical vessel that is to carry a gas at a pressure
of 1400 psi. The diameter of the vessel is 2 ft, and the stress is limited to 12ksi.
Example no. 04
A cylindrical pressure vessel is fabricated from steel plating that has a thickness of 20mm. The
diameter of the pressure vessel is 450 mm and its length is 2.0 m. Determine the maximum
internal pressure that can be applied if the longitudinal stress is limited to140 MPa, and the
circumferential stress is limited to 60 MPa
33
BOYLE’S LAW
If the temperature of a given weight of gas remains constant, the Absolute Pressure
of the Gas is Inversely Proportional w/ the Volume of the same Gas,
According to this law, the pressure exerted by a gas held at a constant temperature
varies inversely with the volume of the gas. For example, if the volume is halved, the pressure
is doubled; and if the volume is doubled, the pressure is halved. The reason for this effect is
that a gas is made up of loosely spaced molecules moving at random. If a gas is compressed
in a container, these molecules are pushed together; thus, the gas occupies less volume. The
molecules, having less space in which to move, hit the walls of the container more frequently
and thus exert an increased pressure.
Hence:
D
!C
5
Mathematically:
D
!=E×
5
Where C is the Proportionality Constant. That is
!0 50 = !1 51 = E
F0 = F< = Initial Absolute Pressure

G0 = G< = Initial Volume
F9 = FC = Final Absolute Pressure
G9 = GC = Final Volume
Example no. 01
A 3m high cylindrical tank having a diameter of 1.50m has a closed flat top and an open
bottom end. The tank is placed in sea water (Sw=1.03) with open end down until the top is
submerged 1m below the water surface. In the submerged position, determine the height of
the water inside the tank.
34
ENERGY EQUATION
The total energy <D is the sum of three heads. That is,
BERNOULLI’s ENERGY EQUATION
“Neglecting friction, the total head, or the total amount of energy per unit of weight, is the
same, at every point in the path of flow.”
G10 !0 G11 !1
+ + I0 = + + I1
@H ,& @H ,&
J0 = J1
CONTINUITY EQUATION
K = (5
35
(0 50 = (1 51
Where,
Q = Discharge
A = Water Area
v = Mean or Average Velocity
CORRECTION FACTORS
Example no. 01
A fluid is flowing in a pipe 8 in. in diameter with a mean velocity of 10 ft. per sec. The pressure
at the center of the pipe is 5 lb. per sq. in., and the elevation of the pipe above the assumed
datum is 15 ft. Compute the total head in feet if the fluid is (a) water, and (b) oil (sp. gr 0.80).
Example no. 02
A liquid (sp. gr 2.0) is flowing in a 2-in.pipe. The total energy at a given point is found to be
24.5 ft-lb per lb. The elevation of the pipe above the datum is 10 ft., and the pressure in the
pipe is 9.5 lb. per sq. in. Compute the velocity of flow.
Example no. 03
The fluid in the figure is water, with the surface 20 ft above the datum. The pipe is 6 in. in
diameter and the total loss of head between point 1 in the water surface and point 5 in the
jet is 10 ft. Determine the velocity in the pipe and the discharge Q.
36
FLOW MEASUREMENT
1. VENTURIMETER
Example no. 03
A horizontal venturimeter having a meter coefficient of 0.95 is attached to a pipe of diameter

10 cm. A differential manometer attached to the inlet and to the throat of the meter shows a
deflection of 15 cm. Calculate the discharge in the pipe in liters pec sec.
2. PITOT TUBE
37
5+ = L@H- !1 − !0
-=
M
Example no. 04
A pitot tube in a pipe in which air flows (ᵞ = 12 N/m3) is connected to a manometer containing
water. If the water deflection is 10 cm, what is the flow velocity of pipe? Neglect losses.
3. ORIFICEMETER
Example no. 05
Oil flows through a pipe as shown in the figure below. The coefficient of discharge for the
orifice in the pipe is 0.63. what is the discharge of oil in the pipe?
38
ADDITIONAL EXAMPLES
Example no. 06
A 30 cm pipe delivers 2 cu.m. of water per minute through a venturimeter shown. d9 is 1/10
of d0 . Compute the mean velocities in section 1 and 2. if the pressure head in section 1 is 6m,
compute pressure head in 2
Example no. 07
A pitot tube having a coefficient of 0.98 is used to measure the velocity of water at the center
of a pipe, as shown below. What is the velocity?
39

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MAPUA INSTITUTE OF TECHNOLOGY

SCHOOL OF CIVIL, ENVIRONEMENTAL AND GEOLOGICAL ENGINEERING

ENGR. KEVIN PAOLO V. ROBLES

Passing Rate is strictly 70. 00 %

JOB DESCRIPTION : Part Time Faculty

FACEBOOK : Kevs Robles (facebook.com/kevinpaolorobles)

FACEBOOK GROUP : CE140-1P 1st Term20162017

EMAIL ADDRESS : kpvroblesmapua@yahoo.com

FORMAT for QUIZZES/EXERCISES/HOMEWORKS

0.5in x 0.5in margin

FORMAT of LABORATORY REPORT

Margin : left x right x top x bottom

0.75 in x 0.5 in x 0.5 in x 0.5 in

Front page to be provided

INTRODUCTION TO FLUID MECHANICS

A. MASS DENSITY - mass per unit of volume

C. SPECIFIC VOLUME - is the volume per unit of weight

E. BULK MODULUS OF ELASTICTY - is the incremental change in volume when the

Where L = shear stress

G. KINEMATIC VISCOSITY (P = QR) − is the ratio of the dynamic viscosity to that of

H. VAPOR PRESSURE- the pressure exerted by vapor in a closed space

A reservoir of glycerin has a mass of 1200kg and a volume of 0.952 m3.

a. Compute for the bulk modulus of elasticity

3. Pressure exists at every point.

6. The Pressure varies linearly with depth.

#!"# = #!$% + #&!&'

F234 = Absolute Pressure

F256 = Atmospheric Pressure

F7278 = Gage Pressure

ABSOLUTE PRESSURE - measured from absolute zero perfect vacuum

GAGE (GAUGE) PRESSURE - measured from the atmosphere

PRESSURE ACTING PERPENDICULAR TO THE SURFACE

2. A DIFFERENTIAL MANOMETER connects each end to a different pressure vessel

Calculate the difference in pressure between points A and B in

From the figure shown, the following equation holds,

Recall : MOMENT OF INERTIA OF PLANE SURFACES

PRESSURE PRISM METHOD

A rectangular gate of dimension 1 m by 4 m is held in place

As shown, a round viewing window of diameter D = 0.8 m is situated in a large tank of

HYDROSTATIC FORCE ON CURVE SURFACES

a) How much of the cube protrudes above the oil-water interface?

Upward Acceleration Downward Acceleration

tan θ = ax/g = y/x

a.) Calculate the acceleration just about water to spill.

THIN WALLED CYLINDERS

A. CIRCUMFERENTIAL (TANGENTIAL) STRESS, >),-)

B. LONGITUDINAL STRESS, >.+/*

P = internal pressure caused by the fluid

Where C is the Proportionality Constant. That is

F0 = F< = Initial Absolute Pressure

BERNOULLI’s ENERGY EQUATION

A horizontal venturimeter having a meter coefficient of 0.95 is attached to a pipe of diameter

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