UNIT Properties of Fluids

Lesson 1: Fundamental Properties of Fluids

Learning Outcomes:

 Identify the Fundamental Properties of Fluids

 Solve Related Problems

Pretest:

Simple Recall from Physics and Chemistry:

Fill in the blanks. Write the correct answer on the space provided for:

__________ 1. It is the absolute quantity of matter in a substance or body

that does not change with the change of gravity.
__________ 2. It is the force of gravity on the body that could be
determined by a spring scale. This quantity changes with
the of gravity.
__________ 3. The instrument used to measure atmospheric pressure.
__________ 4. A property that is defined as the force per unit area.
__________ 5. What is the standard atmospheric pressure at the surface of
The earth , near sea level expressed in kPa?
__________ 6. The measure of the hotness and coldness of a body or a
substance.
__________ 7. A substance that is homogeneous in composition and
homogeneous and invariable in chemical aggression
__________ 8. It is the product of the component of a force in the direction
of motion and the distance through which the point of
application of the force moves during its action.
__________ 9. It is the capacity of a system or a substance to do an effect.
__________ 10. It is the energy of a body because of its velocity.

Basic Concepts

Mechanics is a physical science that deals with the action of fluids at rest
or in motion and with applications and devices in engineering using fluids.

Fluid Mechanics can be subdivided in 2 major areas:

1. Fluid Statics is the branch of mechanics that deals with bodies at rest.
2. Fluid Dynamics is the branch that deals with bodies in motion.
 Fluid is a substance that is capable of flowing. It deforms continuously
under the action of a shear force. It is a substance in liquid/gas phase. It offers
resistance to a change of shape. Although liquids and gases share some
common characteristics, a liquid fluid is a visible substance that conforms to its
containing vessel and has a free surface, while a gas fluid is a compressible
substance and completely occupies its containing vessel. A solid can resist an
applied shear by deforming its shape whereas a fluid deforms continuously
under the influence of shear stress no matter how small is its shape. In solid,
stress is proportional to strain, but in fluids, stress is proportional to strain rate.

Fig. 1 : Illustration of solid and fluid deformation

Referring to Fig 1, the shear modulus of solid and coefficient of viscosity for fluid
can be defined by (S) and (µ)

F F
Shear Stress A Shear Stress A
S = = μ = Shear Strain Rate μ = Δμ
Shear Strain Δx
h h

The shear force (F) is acting on the certain cross-sectional area (A), (h) is the
height of the solid block/height between two adjacent layer of the fluid
element, (∆x) is the elongation of the solid block, and (∆µ) is the velocity
gradient between two adjacent layers of the fluid.

Fluids are divided into 2 categories:

1. Ideal fluids (Ideal Gas) Assumed to have no viscosity
 (no resistance to shear)
 Incompressible
 Have uniform velocity
when flowing
 No friction between moving
layers of fluids
 No eddy current or turbulence.
 2. Real Fluids

 Exhibit infinite viscosities

 Non-uniform velocity distribution when flowing
 Compressible
 Experience friction and turbulence in flow

Dimension and Unit

A dimension is the measure by which a physical variable is expressed

quantitatively.

A unit is a particular way of attaching a number to the quantitative

dimension. Thus length is a dimension associated with variables such as
distance, displacement, deflection, height, width. Meters, centimeters, inches
and that such are numerical units for expressing length.

Properties of Fluids

Any characteristic of a system is called a property. It may either be

intensive (mass independent) or extensive (depends on the size of system). The
state of a system is described by its properties.

Though each fluid is different from others in terms of composition and

specific qualities, there are some properties which every fluid shares.

Density of a fluid is defined as the ratio of mass of the fluid to its

volume. Density of gases is dependent on pressure and temperature, while the
density of liquid remains constant. It is expressed in three different ways; Mass
density, specific weight and relative density /specific gravity.

Mass Density (ρ) is mass per unit For an Ideal Gas:

volume
P
ρ =
mass of fluid RT
ρ = volume

M
ρ = V Where:
P = Absolute pressure of gas
 Water: 1000 kg/m3; R = Gas constant
 Mercury: 13546 kg/m3; T = Absolute Temperature
 Air: 1.23 kg/m3;
 Absolute temperature is the temperature of a fluid in reference to
absolute zero, expressed in Rankine for the English System and Kelvin in Metric
System.
Temperature in Kelvin; K = oC + 273
Temperature in Rankine; R = oF + 460

Specific Weight (γ) or unit weight is defined as the weight possessed

by a unit volume of fluid or the force of gravity per unit volume of a substance.
Specific Weight is dependent on acceleration due to gravity as it changes from
place to place. Acceleration is due to gravity. The specific weight of water is
9.81x100 N.m-3

We have:

Weight mass x g
γ = Volume γ = Volume γ =ρxg

Specific Volume (vs) is the volume occupied by a unit mass of the fluid.
It is the reciprocal of density.

Volume 1
vs = Mass vs = Density

V 1
vs = M vs = ρ

Specific Gravity is a dimensionless ratio of density of the fluid to the

density of a standard fluid (reference density of water). It is also defined as the
ratio of the specific weight of a certain fluid to that of the specific weight of
water at standard temperature.

Where:
γ = Specific weight of a certain fluid
ρfluid γ w = specific weight of water at
S =
ρstandard fluid standard condition
= 62.4 lbs/ft
Viscosityγ is the property of a fluid
= which
9.81 kN/mdetermines
3 the amount of its
SG =
resistance toγwshearing forces. It is the =resistance
1 kg/liter of a fluid to motion. Viscous
force is that force of resistance offered by a layer
= 1000kg/m 3 of fluid for the motion of

another layer over it. In case of ρliquids, viscosity

= density is due to cohesive force
of a fluid
between the molecules of adjacent layers of liquid. In case of gases, viscosity is
ρw= layers.
the molecular activity between adjacent
density of water at standard condition
A perfect fluid has no viscosity.
 μdv where:
τ = dy τ = shear stress lb/ft2 or Pa
μ = absolute viscosity lb.sec/ft2
τ y = distance between plates/layers
μ = dv v = velocity in ft/sec or m/sec
dy

Surface Tension is the membrane of skin that seems to form on the

free surface of a fluid due to the intermolecular cohesive forces. When a liquid
and gas or two immiscible liquids are in contact, an unbalanced force is
developed at the interface stretched over the entire fluid mass; water droplets
from rain or dew hanging from branches or leaves of plants, water dripping from
leaky faucet falls as spherical droplets, a soap bubble released into the air forms
a spherical shape, waters beads up into small drops on flower petals. In these
observances liquid droplets behave like small spherical balloons filled with the
liquid and the surface of the liquid acts like a stretched elastic membrane under
tension. The membrane that seems to form on the free surface of a fluid is due
to intermolecular cohesive forces. The magnitude of this force per unit length is
called surface tension. Surface tension also causes bubbles and droplets to take
on a spherical shape since any other shape would have more surface area per
unit volume.

Pressure inside a droplet of Where:

liquid:
P = pressure inside the droplet (Pa)
4σ
P = d σ = surface tension (N/m)d = diameter
of the droplet (m)

1. Capillarity (Capillary action) is the name given to the behavior of the

liquid that rise or fall in a thin-bore tube inserted into the liquid which is
caused by surface tension and depends on the relative magnitudes of the
cohesion of the liquid and the adhesion of the liquid to the walls of the
containing vessel. The curved free surface of a liquid in a capillary tube is
called the meniscus. It is commonly observed that water in a glass
container curves up slightly at the edges where it touches the glass
surface; but the opposite occurs for mercury, it curves down at the
edges. Liquids rise in tubes they wet (adhesion > cohesion), liquids fall in
tubes they do not wet (cohesion > adhesion)

Capillary is important when using tubes smaller that about 3/8 inch
(9.5mm) in diameter.

Where:
4 σcosθ
h=
γd h = capillary rise or depression in
meter
For complete wetting as with water γ = unit weight in N/m3
on Clean glass, angle θ is zero:
d = diameter of the tube in m
4θ
h= γd σ = surface tension in Pa.
 2. Compressibility (β) also known as coefficient of compressibility is the
fractional change in the volume of a fluid per unit change in pressure in a
constant temperature process.

Δv / v 1
β = β =
Δp EB

Bulk Modulus of Elasticity (EB) expresses the compressibility of the

fluid. It is the ratio of the change in unit pressure to the corresponding volume
change per unit volume.

Δp
EB = Δv/ v
Where:
∆V = change in volume
V = original volume
∆p = change in pressure

3. Pressure Disturbances is the velocity or celerity of pressure which

move in waves. This is also known as acoustic or sonic velocity.

√ E B gas
Property changes in Ideal
C=
βP
For any ideal gas experiencing any process; the equation of state is
express as:
P1V 1 P2V 2
T1
= T2

When temperature is held constant: When Pressure is held constant:

P1 V1 = P2 V2 V1
=
P2V 2
T1 T2

Where:
P1 = Initial absolute pressure of gas
P2 = Final absolute pressure of gas
V1 = Initial volume of gas
V2 = Final volume of gas
T1 = Initial absolute temperature of gas
T2 = Final absolute temperature of gas
Sample Problems:

1. If 6.0 m3 of oil weighs 47,200 N, determine the following:

a. Specific weight
b. Density
c. Specific Gravity

Solution: Solution: Solution:

Specific Weight Density Specific Gravity

W γ γ oil
γ= V
ρ=
g Sp. gr =
γ water

47,200 7,866.67 7,866.67

γ= 6
ρ=
9.81 Sp. gr =
9810

γ = 7,866.67 N/m3 ρ = 801.90 kg/m3 Sp. gr = 0.802

2. A specific liquid has a unit weight of

55kN/m3

a. Compute the mass density

b. Compute the specific volume
c. Compute the specific gravity

Solution: Solution: Solution:

Mass Density Specific Volume Specific Gravity

W 1 55
ρ= Vs = Sp. Gr =
g ρ 9.81

55000 1 Sp. Gr = 5.61

ρ= Vs =
9.81 506.52

ρ = 506.52 kg/m3 Vs = 0.000178m3/kg

3. Estimate the height to which water will rise in a capillary tube of
3mm diameter. Use γ = 9810n/m and σ = 0. 0728N/m Ɵ = 90o for
water in a clear tube.
Solution:

Capillary rise h = 4 (0.0728)

9810(0.003)
h = 0.0099 m or 9.9 mm

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 2.5 MPa.
Compute:
a. The bulk modulus of elasticity
b. The percentage of volume decreased
c. The coefficient of compressibility
Solution:
a. Bulk modulus of elasticity c. Coefficient of
ΔP
Ev = compressibility
ΔV /V
2.5−2 1
Ev = β=
10/1000 E
1
Ev = 50 MPa β=
50

b. Percentage of volume decrease β = 0.02

% = 10 (100)
% =1

5. The pressure in a closed tank reads 62.45 kPa. Compute for:

a. The equivalent height in water.
b. The equivalent height in terms of oil having a specific gravity of 0.85
c. Equivalent height in terms of Mercury having a specific gravity of
13.60.

a. Height in water b. Height in Mercury

P = γ wh P = γ wh
62.45 62.45
h = 9.81 h = 9.81(13.6)
h = 6.37 m h = 0.47 m of mercury

b. Height in oil
Learning Activity 1: Multiple Choice

Directions: Choose the letter of the correct answer: write your answer on the
space provided for:
__________ 1. Which type of matter do not have a free surface?
a)Solid b) Liquid c) Gas d) Fluid

__________ 2. If a person studies about a fluid which is at rest, what is his

domain of study?
a) Fluid Mechanics b) Fluid Dynamics c)Fluid Kinematic
d) Fluid Statics

__________ 3. What do you call the force that is applied by a fluid

Perpendicular to the surface of an object per unit area ?
a) Weight Density b) Pressure c) Specific Gravity
d) Specific Weight

__________ 4. The specific gravity of a liquid has,

a) The same unit as that of weight density
b) The same unit as that of specific volume
c) The same unit as that of mass density
d) No unit
Learning Activity 2: Solve the following problems regarding Properties of
Fluids. Use a separate sheet of bond paper for your solutions.

1. Carbon tetrachloride with a mass of 500 kg is placed in a container

with a volume of 0.375 m3. Determine:
a. Density
b. Specific weight
c. weight
2. An object has a specific weight of 2.5 kN/m3. Compute the following:
a. mass density
b. mass if the volume is 0.005m3
c. specific volume

3. A liquid is compressed in a cylinder. If it has a volume of 1100 cm 3

at 2 MPa and a volume of 1090 cu.cm at 2.5 MPa. compute:
a. Bulk modulus of elasticity
b. Coefficient of compressibility

4. Estimate the capillary depression for mercury in a glass capillary

tube 2mm in diameter use σ = 0.514 N/m and Ɵ = 140o.
 Mastery Test
1. A certain liquid has a unit weight of 60 kN/m3. Compute for the
following:
a. Mass density
b. Specific volume
c. Specific gravity
2. A liquid which is compressed in a cylinder having a volume of 1010
cu.cm. at 1 MN/m2 and a volume of 990 cm3 at 2 MN/m2. Compute;
a. Change in volume
b. Change in pressure
c. The bulk modulus of elasticity
3. Determine the surface tension in a tube with 0.25 m radius and a
capillary rise of 6mm.
a. If wetting angle is 0o
b. if wetting angle Ɵ is 80o

UNIT HYDROSTATIC FORCES ON

2 SURFACES

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a

given point within the fluid due to the force of gravity. It increases in proportion
to depth measured from the surface because of the increasing weight of fluid
exerting downward force from above.

If a fluid is within a container then the depth of an object placed in that

fluid can be measured. The deeper the object is placed in the fluid, the more
pressure it experiences because the weight of the fluid is above it. The more
dense the fluid above it, the more pressure is exerted on the object that is
submerged, due to the weight of the fluid.
 Pretest:
Calculate the following:
1. Area of a circle with a diameter of 500 mm.
2. Area of a trapezoid with bases of 2meters and 4 meters with a depth
of 4 meters.
3. Volume of a trapezoid with bases of 2 and 4 meters, depth of 6
meters and a width of 1 meter.

Lesson 1: Forces on Plane Surfaces

Learning Outcomes:

 Derive the equation in determining total hydrostatic Pressure

 Determine the behavior and centroids of submerged plane surfaces
 Solve related problems

When a surface is submerged in a fluid, forces develop on the surface

due to the fluid The determination of these forces is important in the design of
storage tanks, ships, dams and other hydraulic structures. For fluids at rest the
force is perpendicular to the surface. Shear stress is zero and the pressure varies
linearly with depth if the fluid is incompressible. For a horizontal surface such as
the bottom of a liquid-filled tank, the magnitude of the resultant force is F = pA
where p is the uniform pressure on the bottom and A is the area of the bottom.

Fig 1 The hydrostatic force on a horizontal plane.

If atmospheric pressure acts on both sides of the bottom the resultant

force on the bottom is due to the weight of the liquid in the tank. The force F
exerted by the liquid on a plane area A is equal to the product of the specific
weight (γ) of the liquid, depth of the center of gravity of the area (h cg) and the
area A.

F= γ hcg A
 The line of action of the force passes through the
center of pressure.

Centroids

RECTANGLE TRIANGLE

d
d
b
bh3
A = bh Icg =
12 bh bh3
A = 2 Icg =
36

CIRCLE

A = πd2

O c.g. d πd 4
Icg = 64

If a submerged plane surface is inclined, the force on the plane surface is

due to the pressure P = ρgh = γh acting over the area, i.e., where ȳ is the
distance (mean value) to the centroid of the plane area; the centroid is identified
as the point C.
Figure 2: The hydrostatic force on an inclined plane area.

dp = γ y sinƟ dA Eq. 1
P = γ sinƟ ∫ ydA

But ∫ ydA = moment of area

∫ ydA = Ay

Y = distance from center of gravity of the plane surface to the water

Surface measures along the plane

P = γ sinƟ A y Eq. 2

But h = y sinƟ

P=γhA

Location of the center of pressure:

Py p = ∫ y dp
From Eq. 1 and 2

(γ sinƟ A y) yp = ∫ γ y sinƟ dA y
 y2
yp = ∫ dA Is = Moment of Inertia of the plane with respect to
Ay
the water surface
2
y
Is = ∫ Ay dA

2
Is =∫ y dA

Y Is
p=¿ ¿
Ay

Ay = Ss = statistical moment with respect

Yp = y + e to the water surface
Is = Iy + A y 2
Y
p=¿
Is
¿ Ig = moment of inertia with respect to the
Ay

center of gravity

Y Ig
p=¿
Ig + A y 2
¿ y+e = +y
Ay Ay

Ig A y2 Ig
Yp = A + y+e= S +y
y Ay s

Ig Ig
Yp = A + y e =
Ss
y

Sample Problem:

1. A vertical triangular plate whose height is 3.6 meters has its base
horizontal and vertex uppermost in the water surface.
a. Find the depth to which it must be lowered from its vertex so
Figure : that the difference in level between the center of gravity and the
C
center of pressure shall be 20 cm.
b. How
Water far is the center of pressure below
surface y the water surface?
c. What is the base width of the plate if the total hydrostatic force
h acting
CG on the plate is 76.28 kN?

20cm 3.6m
Sample Problem 2:
A square frame 3 meters x 3 meters is submerged in water vertically with
its tops 3 meters from the surface. If oil with specific gravity of 0.80
occupies the top meter; determine the following:
a. The horizontal pressure acting on the plane
b. The pressure at the top of the gate
c. The pressure at the bottom of the gate
Solution:

Oil 1 m

h Water 2 m

P 3 meters

3 meters

a. Horizontal pressure acting on the frame

P = γ hA
h = 1(0.80) + 2 + 1.5
h = 4.3 meters
P = 9.81(4.3)(3)(3)
P = 379.65 kN
Learning Activity:

1. A square plate having one of its side equal to 4 meters is immerse

in a water surface in a vertical position such that the two edges of
the square would be horizontal in order that the center of pressure
shall be 10 centimeters from the center o gravity.
a. How far below the water surface should the upper plate be
submerged?
b. What is the distance of the center of pressure from the water
surface?
c. Determine the hydrostatic force acting on the plate at this
position.

h
e 3m

2. A circular door having a diameter of 3 meters closes a horizontal

duct from a small dam. The center of the door is 5 meters below
the dam’s water level. Compute for the following:
a. The hydrostatic force acting on the door.
b. The location of the center of pressure from the centroid of the
door.
c. The force at the bottom of the gate to maintain equilibrium if it
 o

Lesson 2 : Forces Acting on Curved Surfaces

Learning Outcomes:

 Determine the behavior and centroids of submerged curved

surfaces
 Determine the forces acting on submerged curved surfaces
 Solve related problems

To determine the hydrostatic force acting on curve surfaces, break down

the force into its component forces (horizontal and vertical components).

Horizontal Component (Ph) is the normal force on the vertical

projection of the surface. The component acts through the center of pressure for
the vertical projection. The horizontal component of the hydrostatic pressure is
equal to the total pressure on the projection of that surface on a vertical plane
which is normal to the chosen axis.

water surface 4m 1m

P h = γw h A
Ph h
h/3
 P
Pv

h/3 is the distance of the hydrostatic pressure from the water surface.

Vertical Component (Pv) of the hydrostatic pressure is equal to the

weight of that volume of water or liquid extending vertically from the surface of
the free surface of liquid or the volume of liquid above the area.

Pv = γw vol Resultant Force

R2 = Ph2 + Pv2

Sample Problem 1

The 6 feet diameter cylinder weighs 5000 lbs. and is 5 feet long.
Determine:
a. The upward force due to the effect of oil on the left side.
b. The horizontal reaction at A
c. The vertical reaction at B
Figure:
E C

Oil A
Oil D
(0.80)
B
B
a) Upward force due to the effect of oil on the other side

Pv = γVol E C
Pv = (0.80)(62.4) π ¿ ¿ D
Pv = 3529 lbs. upward Ph A

Pv B
R
b) Horizontal Reaction at A
Ph = γ hA
Ph = 62.4(0.80)(3)(6)(5)
Ph = 4492.8 lbs. to the left

c) Vertical reaction at B
R +Pv = weight of cylinder
R = 5000 – 3529
R = 1471 lbs. (upward)

Sample Problem 2:

From the figure, compute for the following:

a. The horizontal component of the hydrostatic force.
b. The vertical component of the hydrostatic force.
c. The location of the vertical component horizontally from B.
 Solution to sample problem 2:

a) Horizontal component of the hydrostatic force

Ph = γ hA = 62.4(3)(6)(1) Ph = 1123 lbs.

b) Vertical component of the hydrostatic force

Pv = γVol = (62.4) ( π ¿(6 ¿¿¿ 2¿¿ 4 )(1) Pv = 1764 lbs.

c) Location of the vertical component horizontal from B

Summation of moment at C:
Pv(x) = Ph(6 – 2)
1764(x) = 1123(4)
x = 2.55 m from B

Learning Activity

A 3 feet diameter log (sp.gr.= 0.82) divides two shallow ponds as

shown.
1) Compute the net vertical reaction at point C if the log is 12 ft
long.
2) Compute the horizontal reaction at point C
3) Compute the direction of the resultant force with the
horizontal at point C.

Water surface

WP
P1 P2

F1 F2

R
Mastery Test

1. The 1.2-meter diameter cylinder is 9m long into the board

and rests in static equilibrium against a frictionless wall at
point B. Determine the following:
a. The weight of the cylinder.
b. The specific gravity of the cylinder.
c. The net horizontal force acting on the cylinder.

H2O
B

2. A vertical rectangular sluice gate at the bottom of the dam

is 0.80 m wide and 1.6 m high and exposed to water
pressure on one side corresponding to a head of 16 m
above its center. Assuming the gate on stem 20 weigh
2300 N and the coefficient of friction of gate on runner s to
be 0.25.
a. Determine the hydrostatic force acting on the depth.
b. Determine the friction force of gate on runners.
c. Find the force necessary to raise the gate.

Water surface T

0.80m

16m 2300N 1.6m

3. Gate AB rotates about an axis through B. If the width of

the gate is 4 feet, compute
a. The hydrostatic force on the left side of the gate.
b. The hydrostatic force on the right side of the gate.
c. What torque applied to the shaft through B is required
to hold the gate closed?
Water surface
 Unit
3 Relative Equilibrium of Liquids

Pretest
Discuss/describe the following:
1. Equilibrium 4. Tangential force
2. Circumference 5. Dams
3. Hydrostatic Pressure 6. Pipes

Lesson 1: Hoop Tension on Circular Pipes and Tanks

Learning Outcomes:

 Know the behavior of fluids confined inside a tank or cylinder

 Determine the forces acting on the tank or cylinder
 Solve related problems

Figure:
 Hoop Tension or Circumferential Tension (T/S) is the
circumferential force per unit areas in the pipe wall due to internal pressure. It
can be explained as the largest tensile stress in a supported pipe carrying a fluid
under pressure. Hoop stress is equal to the internal pressure times the internal
diameter of the cylinder divided by twice the wall thickness.

PD Or consider 1 unit length

S=
2t PD
T =
2
Where: Where:

S = Tensile Stress P = pressure

Where:
P = Unit pressure P = γw h
T = Thickness of wall h = height of water
D = inside diameter
T = Tensile Force

Sample Problem:

A steel pipe having a diameter of 36 inches carries water under a

head of 700 feet.
1. Compute the internal pressure in the pipe
2. Compute the wall thickness of the steel pipe need to resist the
static pressure if the allowable working stress of steel is
16,000 Psi.
3. Compute the tangential force on the steel pipe.

Solution:

1. Internal pressure in the pipe

P = γw h
62.4(700)
P =
144
P = 303.33 Psi

PD
2. Ss = 3. Tangential force on the steel
Learning Activity

1. A 45- inch diameter steel pipe ¼ inch thick carries oil with specific
gravity equal to 0.82 under a head of 350 feet of oil. Compute:
a. The pressure inside the pipe in psi.
b. The stress in the steel required to carry a pressure of 240 psi, with
an allowable stress of 16,000 psi.
c. The thickness of steel required to carry a pressure of 240 psi with
an allowable stress of 16000 psi.

P DT

2. A cylindrical container 7 ft high and 3 ft in diameter is reinforced with

two hoops one foot from each end.
a. Compute the hydrostatic pressure
b. What is the tension of hoop at the bottom?
c. When it is filled with water, what is the tension in the upper hoop?

7 ft hoops
 Lesson 2 : Dams

Learning Outcomes:

 Identify the different dam cross sections and the forces acting on it
 Determine the hydrostatic pressure and hydrostatic uplift acting on
dams
 Solve related problems

Japan Water Agency

Dam is a large man-made structure built across a stream, a river or an

estuary to retain water for human consumption, irrigation, and industrial
processes. It is also used to increase the amount of water available for
generating hydroelectric power, to reduce peak discharge of floodwater, control
river flow and measure water flow.

Dams subjected to water pressure on the upstream or downstream side

are analyzed by considering only 1-meter strip of the dam.

h
P= γw h A h= 2

h
A = h (1) P= γ h (1)
 h2
 P = γw2
 R. M. = W 1 X 1 + W 2 X 2 Righting Moment – the force that
Keeps the dam in place
Ph
 O.M. = Overturning Moment – the force
3
that tends to overturn the dam
 R x = RM – OM
 Factor of Safety against factor of safety against sliding
overturning
RM μR
 F.S. = OM F.S. = P
μR
 F.S. = R = W 1 +W 2
P
Where:
P = total hydrostatic force
D = depth of the dam
h = depth of the water surface
B = base of the dam
W1 = weight of the rectangular section of the dam
W2 = weight of the triangular section of the dam
R = resultant force
X1 = location of centroid of the rectangular section of the dam
reckoned from the toe
X2 = location of the centroid of the triangular section of the dam
reckoned from the toe
x = location of the resultant reckoned from the toe
 OM Case 2: Considering
= overturning moment Hydrostatic Uplift
γw = unit weight of water
FS = factor of safety
μ = coefficient of friction between the soil and base of the dam

Due to the presence of water on the upstream face of the dam which is
raised up to a certain height “h” the water tends to seep under the base of the
dam and flows towards the lower level on the downstream side. This seepage is
under pressure and exerts a hydrostatic uplift on the dam which reduces the
stability of the dam against sliding and overturning.

h 2
 RY = W1 + W2 – U O.M. = P ( ) + U ( B)
3 3

h
 R x = R.M. – O.M U = γw 2 B

 R.M. = W1 X1 + W2 X2 U = Uplift Force

 Case 3: Overflowing dam with hydrostatic uplift neglecting the
effect of water on the downstream side

 D = Height of the Dam

 P = γw h A
 A = D (1)
Ig
 e =
Ss

e = the eccentricity from the centroid of the dam towards the center of
pressure.

1 3
 Ig = 12 D

 Ss = Ah ; Ss = D(1) h
 R = W1 + W2 + W3

W1 and W2 are the weights of the dam while W3 is the weight water
that passes on top of the dam which helps to stabilize the dam.

 R.M. = W1 X1 + W2 X2 + W3 X
D
 O.M. = P ( −e ¿
2
 Rx = R.M. – O.M.

Case 4: Overflowing Dam considering hydrostatic uplift

P = γw h A R = W 1 + W 2 + W3 - U

γw
U = hB R.M. = W1X1 + W2X2 + W3X3
2

D 2
O.M. = P ( −e ¿ + U ( B) R x = R.M. – O.M.
2 3

Ig D3
e = Ig =
Ss 12

Ss = Ah Ss = D(1) h

Case 5: Dams with water on both upstream and downstream side with
uplift pressure

γwh 2

P1 = 2
1
R = W1 + W2 + W3 – U1 – U2

γwh 2 P 2 h2
2
Pressure distribution at the base

1. Resultant at the middle third of the base

 R = P(B)(1)
R
 P = (average)
B

2. Resultant at the middle thirds nearer the toe

B
X =
3

R = ( PB+2 0 )(B)(1)
PB
R = (twice the average pressure)
2
3. Resultant outside the
middle-thirds

P (3 x)
R =
2

2R
P = 3x

4. Resultant within the middle-thirds

R MC
 Pmin = A - I Eq. 1

R MC
 Pmax = A + I Eq. 2

B
 A = B (1) C =
2

(1) B3
 M = Re I =
12

Substitute values of A, M, C and I to equations 1 & 2

B
R ℜ B /12 R ℜ( )
Pmin = B - B 3 /12
Pmax = B + 3
2
B /12

R 6 eR R ℜ6
Pmin = B - B2
Pmax = B + B2
R R 6e
Pmin = B ¿) Pmax = B (1 + B)
B
e = 2 −x
Sample Problem 1.
A trapezoidal masonry dam with vertical upstream face is 6 meters high,
0.6 meter at the top and 3.0 meters wide at the bottom. Weight of concrete is
23.5 kN/m3.
a. Find the depth of water on the vertical upstream face.
b. Using the computed depth of water, compute the hydrostatic
uplift force if the uplift varies from full hydrostatic pressure at the
heel and zero at the toe.
c. Compute the factor of safety against sliding if the coefficient of
friction on the base is equal to 0.80

Solution:
a. Depth of water
h
P = γ w hA P = 9.81( 2 )h(1) b) Uplift Force
P = 4.905h2 γ w h = 9.81(5.31)
Solve for total weight of the dam γ w h = 52.10 kN/m2
52.10
U ¿ (3)
2
W1 = (0.60) (6)(1) (23.5) U = 78.14 kN
W1 = 84.6 kN
W2 = 6(3-0.60)(1)(23.5) c) Factor of safety against sliding
μR
W2 = 169.2 kN F.S. (sliding) =
P
R = W 1 + W2
R = 84.6 + 169.2 P = 4.905(5.31)2
R = 253.8 kN P = 138.30 kN
0.80(253.8)
R.M. = W1(2.7) + W2(1.6) F.S. (sliding) =
138.30
R.M. = 84.6(2.7) + 169.2(1.6)
R.M. = 499.15 kN-m F.S. (sliding) = 1.47

O.M. = P (h/3)
O.M. = 4.905h2(h/3)
O.M. = 1.635h3
Rx = R.M.- O.M.
(253.8) (1) = 499.15 – 1.635h3
h = 5.31 meters
Sample Problem 2:
A trapezoidal dam having a top width of 0.80 m and a bottom width of 1.4m
is 2m high. The depth of water on the vertical side of the dam is 2.4 m. Unit
weight of concrete is 23.5 kN/m 3, neglecting hydrostatic uplift, determine the
following:

a. The total hydrostatic force acting on the dam.

b. The factor of safety against overturning
c. The factor of safety against sliding if coefficient of friction
between the dam and the soil is 0.80
Solution:

a. Total hydrostatic force acting on the dam

P = ɤw h A ; h = 0.4 + 1 = 1.4 m
P = 9.81(1.4)(2)(1)
P = 27.468 kN

b. Factor of safety against overturning

The weight of the overflowing water above the dam helps keep the dam in
place. Treat the trapezoidal dam as a rectangular and triangular section.

W1 = weight of the rectangular O.M. = P y y =1-e

Section
Ig 1(2)3
W1 = 0.80(2)(1)(23.5) e= Ig = = 0.67m4
Ss 12
W1 = 37.6 kN
0.67
W2 = weight of the triangular e = Ss = Ah = 2(1)(1.4)
2.8
section e = 0.239 m SS = 2.8 m3
y = 1 – 0.239
0.60(2)(1)(23.5)
W2 = y = 0.761m
2
W2 = 14.1 kN O.M. = 27.468(0.761)
O.M. = 20.14 kN-m
W3 = weight of water that passes
Over the dam c) F.S against overturning
μR
W3 = 0.40(0.80)(1)(9.81) F.S. =
P
W3 = 3.14 kN R = 37.6 + 14.1 + 3.14
R = 54.84 kN
0.80(54.84)
R.M = (37.6)(1) + 14.1(0.40) F.S. =
27.468
+ 3.14(1)
R.M.= 46.38 kN-m F.S. = 1.60 (overturning)
RM 46.38
F.S. = = F.S. = 2.30 (sliding)
OM 20.14
Learning Activity:

Solve the following problems:

1) Water is acting on the vertical side of a trapezoidal masonry dam

1.6 m wide at the top 12 m wide at the bottom and 18 m high. If
the allowable compressive stress at the toe is 345 kPa and
neglecting hydrostatic uplift, compute the following:
a) Depth of water on the upstream vertical side of the dam.
b) The factor of safety against sliding if coefficient of friction
between the soil and the base is 0.60
c) The factor of safety against overturning.
Wt of concrete = 23.5kN/m3

2) A dam is triangular in cross section with the upstream vertical.

Water is flushed at the top. The dam is 8 m high and 6 m wide at
the bottom. The dam weighs 23.50 kN/m3. Hydrostatic uplift varies
from full at the heel to zero at the toe. Determine the following:
a) The factor of safety against overturning.
b) The factor of safety against sliding if friction between the base
and the foundation is 0.80.
c) The maximum pressure at the base of the dam.

Lesson 3: Principle of Archimedes

 Pretest:

Learning Outcomes:

 Familiarize the Principle of Archimedes (Buoyancy)

 Solve related Problems

Archimedes Principle states that that when an object is immersed in a fluid, it

experiences apparent weight loss. This apparent weight loss is equal to the
weight of the fluid displaced and is called the buoyant force which acts upward
through the center of gravity or centroid of the displaced volume or also known
as the center of buoyancy.

FB = ɤ W VD

Where :

FB = Buoyant Force
VD = Volume displaced
ɤ W = unit weight of the fluid
 Positive buoyancy takes place when an object happens to be lighter than
the fluid it displaces, therefore the object will float. This is because the buoyant
force is greater than the weight of the object.

Negative Buoyancy takes place when the object happens to be denser

that the fluid displaced. The object will sink because its weight is greater than
the buoyant force.

Neutral buoyancy takes place when the weight of an object is equal to

the fluid displaced when an object is completely submerged

For a freely floating object with no external forces, the weight of the object
(acting downward) is equal to the buoyant force on the object (acting upward)

W = FB
W= ɤV D

Consider a cubical block having a dimension of 2” X 2” X 2”. Place the block

below the water surface as shown:

F1 = ɤ W h1 A F1 = ɤ W h1 (2)(2)

F2 = ɤ W h2 F2 = ɤ W h2 (2)(2)

F2 ¿ F1 = (Unbalance Force)

F2 – F 1 = ɤ W h2 (2)(2) - ɤ W h1 (2)(2)

F2 – F 1 = ɤ W (2)(2) (h2 – h1) but: h2 - h1 = 2

F2 – F 1 = ɤ W (2) (2) (2)

F2 – F 1 = ɤ W V V = volume of water displaced

F2 – F1 = Buoyant Force

FB = ɤ W V

From the 2nd figure an object having a weight W floats in a liquid. Assume
the cross-sectional area of the object as “A” and a height of h. the object has a
certain specific gravity.

W = volume of object x density of the object

Density of object = sp. gr. Of object x density of water
W = A h(sp.gr.) (ɤW)

A d (ɤW) = A h (sp.gr.)( ɤ W)
d = h (sp.gr)

W = A d(ɤW)
FB = W

An object will float at a depth equal to its height multiplied by its specific
gravity. This will hold true only for rectangular or square objects.
In addition to the force of gravity on weight, all objects submerged in a
fluid are acted on by a force FB. the buoyant force acts upward and is equal to
the weight of the fluid displaced by the object. The upward buoyant force also
acts through the center of gravity or centroid of the displaced volume which is
known as the center of buoyancy.

Sample Problem:
A piece of metal weighs 350N on air and when it is submerged
completely in water it weighs 240N.

a. Find the volume of the metal

b. Determine the specific weight of the metal
c. Determine the specific gravity of the metal
 Lesson
Submerged bodies 4: STABILITY
 are those OF FLOATING
which are totally within theBODIES
liquid.

Floating  bodies are those which are totally out of the liquid but are in contact
with it.
Learning Outcomes:
Half-submerged bodies are those which are half inside the liquid and half
outside.  Discuss the relative equilibrium of bodies at rest and in motion
 Familiarize the Law of Hydrostatics which applies to both
Stability floating
of floating
andbodies is defined
submerged as its ability to return to its neutral
bodies
position after the external
Solve relatedforce has been applied and removed. For a floating
problems
body in static equilibrium and in the absence of any other external force, the
buoyant force must balance the weight of the body. The location of the
metacenter is important in determining the stability.

Metacenter is a point in the vertical neutral axis through which the

Learning Activity: Solve the following problems:

1. A prismatic object 200mm thick by 200 mm wide by 400mm long is

weighted in water at a depth of 500mm and found to be 50N.
a. Find its weight in air
b. Determine the specific gravity
c. Determine the specific weight
2. A wooden buoy of specific gravity = 0.70 floats in a liquid with a
specific gravity 0f 0.80;
a. What is the percentage of the volume above the liquid
surface to the total volume of the buoy?
b. If the volume above the liquid surface is 0.0150 m 3, what is
the weight of the wooden buoy?
c. What load that will cause the buoy to be fully submerged?

buoyant force always acts for small angles of tilt. For a large angle of tilt, the
metacenter may move along the neutral axis. For stability to exist, the object’s
center of gravity must be below its metacenter.
 For objects totally submerged in a fluid, the metacenter is located at its
center of buoyancy.

Where:
B = center of buoyancy
G = center of gravity
M = metacenter
Ɵ = angle of tilt or angular displacement
FB = Buoyant force
a = bottom of the object

For partially submerged objects in an upright position or neutral

equilibrium, the location of the center of buoyancy is half the draft of the object.

For partially submerged objects with angular displacement, location of the

metacenter is found from the following relation:

aM = aB + MB
aM = distance from the bottom of the object to the metacenter
aB = distance from the bottom of the object to the center of buoyancy

MB = B2 ( 1+ tan2 Ɵ)
12D 2
 MG = MB – GB
MG = metacentric height

Location of the buoyancy B is at the geometric center of the displacement

volume. The magnitude and geometry of the displaced volume maybe found
from the center of buoyancy to the metacenter;

MB = IS
V
Where:
IS = moment of inertia about the largest axis of the area produced if the
object were cut at the waterline.

The angular displacement of the body in the clockwise direction causes

the wedge-shapes prism BOB’ on the right of the axis to go inside the water
while the identical wedge-shaped prism represented by AOA’ emerges out of
the water on the left of the axis. This wedge represent a gain in the buoyant
force on the right side and a corresponding loss of buoyant force on the left
side. The gain is represented by a vertical force dFB acting through the C.G. of
the prism BOB ‘ while the loss is represented by an equal and opposite force
dFB acting vertically downward through the centroid of AOA’. The couple due to
these buoyant forces dFB tends to rotate the ship in the counterclockwise
direction. The moment caused by the displacement of the center of buoyancy
from B to B1 is also in the counterclockwise direction. Thus this two couples
must be equal.

Couple due to wedges.

Consider towards the right of the axis a small strip of thickness dx at a

distance x from O. The height of strip (x ) ‫ ے‬BOB’ = (x ) (Ɵ)

‫ ے‬BOB’ = ‫ے‬AOA’ = ‫ ے‬BMB1 = Ɵ

Area of strip = (height) (thickness) = (x) (Ɵ)( dx)

If L is the length of the floating body, then

Volume strip = Area ( L)
= (x) (Ɵ)(L)(dx)
Weight of strip = (ρg) (Volume)
Weight of strip = (ρg)(x)( (Ɵ)(L)(dx)
weight of strip = ρg x ƟLdx
Similarly if a small strip of thickness dx and distance x from O towards the left
of the axis is considered the weight of the strip will be ρg x ƟLdx. The two
weights are acting in the opposite direction and hence constitute a couple.

If M were below G the couple would be an overturning couple and the original
equilibrium would have been unstable. When M coincides with G, the body will
assume its new position without any further movement and thus will be in
neutral equilibrium. Therefore for a floating body, the stability is determined not
simply by the relative position of B and G, rather by the relative position of M
and G. the distance of metacenter above G along the line BG is known as
metacentric height GM which can be written as GM = BM – BG. Hence the
condition of stable equilibrium for a floating body can be expressed in terms of
metacentric height as follows:

GM ¿ 0 ——— M is above G –—— stable equilibrium

GM = 0 ——— M coincides with G ——– neutral equilibrium

GM¿ 0 ——— M is below G ——– unstable equilibrium

The angular displacement of a boat or ship about its longitudinal axis is known
as” rolling” while that about its transverse axis is known as “pitching”.

If a floating body carrying a liquid with a free surface undergoes angular

displacement, the liquid will also move to keep its free surface horizontal. Thus
not only does the center of buoyancy B move, but also the center of gravity G of
the floating body and its contents move in the same direction as the movement
of B. Hence the stability of the body is reduced. For this reason, liquid which has
Sample Problem:

A rectangular scow 9 meters wide 15 meters long and 3.6 meters high has a
draft in sea water 0f 2.4 meters. Its center of gravity is 2.7 meters above the
bottom of the scow.
a. Determine the initial metacentric height.
b. If the scow is tilted until one end is just submerged in water, find the
sidewise shifting of the center of buoyancy.
c. Final metacentric height MG
Solution:

a. Initial metacentric height MG

Is 15 ( 9 )3
Ɵ = 0 MB = IS =
V 12

IS = 911.25 m4 (moment of inertia about the longitudinal axis)

V = (2.4) (15) (9) = 324 m3

Is 911.25
MB = = MB = 2.81 m
V 324

GB = 2.7 – 1.2 GB = 1.5 m

MG = 2.81 – 1.5 MG = 1.31 m

b. Sidewise shifting of the center of buoyancy

1.2
Tan Ɵ = Ɵ = 14.93O
4.5

x
Sin Ɵ = x = 2.81 sin 14.93o x = 0.724 m
MB
c. Final metacentric height
B2 tan 2 θ (9)2 tan 2 14.93°
MB = (1 + ) = 12(2.4) (1 + )
12 D 2 2
to be carried in a ship is put into a number of separate compartments so as to
minimize its movement within the ship.

Learning Activity

1. A cylindrical caisson having an outside diameter of 9 meters floats in

seawater (ɤ = 1030 kg/m3) with its axis vertical and its lower end
submerged 9 meters below the water surface. If its center of gravity is
on the vertical axis and is 3.6 m above the bottom, determine the
following:
a. The value of MB
b. The true metacentric height
c. The righting couple when the caisson is tipped an angle of 8 o
2. A loaded barge has a draft of 1.8 meters in fresh water when erect.
The barge is 6 meters wide, 12 meters long and 2.4 meters high. The
center of gravity of the barge is 1.8 meters above the bottom along
the vertical axis of symmetry.
a. Determine the initial metacentric height
b. What is the maximum single weight that can be moved
transversely from the center of the unloaded barge over the side
without sinking the barge?
c. If the maximum weight obtained in “b” is doubled, at what distance
from the center will the barge be on the verge of submergence?
Kinematics is defined as that branch
Lesson of scienceofwhich
5: Kinematics deals with motion of
fluid flow
particles without considering the forces causing the motion. The velocity at any
point in a flow field at any time is the focus of the study. Once the velocity is
known, then the pressure distribution and the forces acting on the fluid can be
determined.
Learning Outcomes:

 Discuss
The different thefluid
types of flowflow:
of Ideal and Real Fluids.
 Classify flow types (Pathlines, Streamlines and flownets)
  Steady and Unsteady
Solve related Flow
problems
 Uniform and Non-uniform Flow
 Laminar and Turbulent Flow
 Compressible and Incompressible Flow
 Rotational and Irrotational Flow
 One, Two, Three-dimensional Flow

1. Steady and Unsteady Flow

Steady flow is defined as that type of flow in which the fluid characteristics
like velocity, pressure, density, etc., at a point do not change with time.

For a steady flow;

∂v ∂p ∂ρ
∂ t = 0, ∂t = 0 ∂t = 0 (x0, y0, z0)

Where (x0, y0, z0) is a fixed point in fluid field.

Unsteady flow it that type of flow in which the velocity, pressure or

density at a point changes with respect to time,

For unsteady flow;

∂v ∂p ∂ρ
≠ ≠ 0, ≠ 0
∂ t 0, ∂t ∂t (x0, y0, z0)

Where (x0, y0, z0) is a fixed point in fluid field.

2. Uniform and Non-uniform flow

Uniform flow is defined as that type of flow in which the velocity at any given
time does not change with respect to space (i.e. length of direction of flow).

Thus for uniform flow;

∂v
= 0 (t = constant)
∂s

Where:
∂ υ = change of velocity
𝜕s = length of flow in the direction s
Non-uniform flow is that type of flow in which the velocity at any given time
changes with respect to space.

Thus for non-uniform flow

∂v
≠0 (t is constant)
∂s

3. Laminar and Turbulent Flow

Laminar flow is defined as that type of flow in which the fluid
particles move along well-defined paths or stream line and all the stream
lines are straight and parallel. The particles move in laminas or layers
gliding smoothly over the adjacent layer. This type of flow is also called
stream-line flow or viscous flow.

Turbulent flow is that type of flow in which the fluid particles move in a
zigzag way. Due to the movement of fluid particles in a zigzag way, the
eddy formation takes place which are responsible for high energy loss.
For a pipe flow, the type of flow is determined by a non-dimensional
number VD called the Reynold number.

Where:
D = diameter of pipe
V = mean velocity of flow in pipe
v = Kinematic viscosity of fluid
If the Reynold is less than 2000, the flow is called laminar. If the Reynold
number is more than 4000, the flow is turbulent. If the Reynold number
lies between 2000 and 4000, the flow may be laminar or turbulent.

4. Compressible and Incompressible Flow

Compressible flow is that type of flow in which the density of the fluid
changes from point to point or in other words the density (ρ) is not
constant for the fluid, thus for compressible flow
ρ ≠ Constant
 Incompressible flow is that type of flow in which the density is constant
for the fluid flow. Liquids are generally impressible while gases are
compressible. Thus for incompressible flow;
ρ ¿ Constant
5. Rotational and Irrotational Flow

Rotational flow is that type of flow in which the fluid particles while
flowing along streamlines, also rotate about their own axis. and if the
fluid particles while flowing along stream lines do not rotate about their
own axis, then that type of flow is called irrotational flow.

6. One-, Two- and Three-Dimensional Flows

One-dimensional flow is that type of flow in which the flow parameter

such as velocity is a function of time and one space co-ordinate only, say
x. For a steady one-dimensional flow, the velocity is a function of one-
space-coordinate only. The variation of velocities in other two mutually
perpendicular directions is assumed negligible, hence mathematically for
one-dimensional flow;

u = f(x), v = 0 and w = 0

where : u, v and w are velocity components in x, y and z directions

respectively.

Two-dimensional flow is that type of flow in which the velocity is a

function of time and two rectangular space co-ordinates say x and y. F or a
steady two-dimensional flow, the velocity is a function of two space co-ordinates
only. The variations of velocity in the third direction is negligible. Thus,
mathematically for two-dimensional flow;

u = f1 (x, y), v = f2 (x, y) and w = 0

Three-dimensional flow is that type of flow in which the velocity is a

function of time and three mutually perpendicular directions. But for a steady
three-dimensional flow, the fluid parameters are functions of three space co-
ordinates (x, y and z) only. Thus, mathematically for three-dimensional flow;

u = f1 (x, y and z), v = f2 (x, y and z) and w = f3 (x, y

and z)

Rate of Flow or Discharge (Q)

Rate of flow or discharge is defined as the quantity of a fluid flowing per

second through a section of a pipe or a channel. For an incompressible fluid (or
liquid) the rate of flow or discharge is expressed as the volume of fluid flowing
across the section per unit of time. For compressible fluids, the rate of flow is
usually expressed as the weight of fluid flowing across the section. Thus:

 For liquids the units of Q are m2/sec or liters/sec

 For gases the units of Q are kgf/ sec or N/sec

Consider a liquid flowing through a pipe in which

A = Cross-sectional area of pipe

V = Average velocity of fluid across the section

Then: Q = (A)(V)
Q = AV

Continuity Equation is the equation based on the principle of conservation of

mass. Thus, for a fluid flowing through the pipe at all the cross-section, the
quantity of fluid per second is constant.

Let: V1 = average velocity at cross-section 1-1

ρ1 = density at section 1-1
A1 = area of pipe at section 1-1
V2 = average velocity at section 2-2
ρ2 = density at section 2-2
A2 = area of pipe at section 2-2

Rate of flow at section 1-1

The rate of flow at section 1-1, = ρ1 A1 V1

The rate of flow at section 2-2 = ρ2 A2 V2 `

According to the law of conservation of mass,

Rate of flow at section 1-1 = rate of flow at section 2-2

ρ1 A1 V1 = ρ2 A2 V2
 This equation is applicable to the compressible as well as incompressible
fluids and is called continuity equation. If the fluid is incompressible, then ρ 1 =
ρ2. The continuity equation reduces to:
A1 V 1 = A2 V2

Sample Problem 1:
The diameters of a pipe at sections 1 and 2 are 10 centimeters and 12
centimeters respectively. If the velocity of water flowing through the pipe at
section 1 is 5 m/sec, determine:
a. The discharge through the pipe
b. Velocity at section 2

5 m/s 12 cm
10cm

Solution:
a. Discharge through the pipe
Q = A1 V1
πd 2 π (0.10)2
A1 = = A1 = 0.00785 m2
4 4
π (0.12)2
A2 = A2 = 0.0113 m2
4

Q = 0.00785 (5)
Q = 0.008878 m3

b. Velocity at section 2
A1 V1 = A2 V2 (0.00785) (5) = (0.0113) V2
V2 = 3.47 m/s

Sample Problem 2

Pipe 1, a 25-cm diameter pipe conveying water with a velocity of 2.5 m/s,
branches into two pipes 2 and 3 of diameters 20 cm and 15 cm respectively. If
the average velocity in the 20cm diameter pipe is 1.8 m/sec. determine:
a. The discharge in pipe 1
b. Discharge in pipe 2
c. The velocity in pipe 3

Solution:
 a. Discharge in pipe 1
Q1 = A 1 V 1
A1 = π (0.25)2
4
A1 = 0.0491 m2
Q1 = (0.0491)(2.5)
Q1 = 0.12275 m3/s

b. Discharge in pipe 2
Q2 = A2 V2
A2 = π (0.20)2
4

Q2 = 0.03142 m3/s

c. Velocity in pipe 3
Q3 = A3 V3
Q1 = Q 2 + Q 3
0.12275 = 0.03142 + Q3
Q3 = 0.09133 m3/s
Substitute values:
0.09133 = π (0.15)2 V3
4
V3 = 5.17 m/s
U
Unit
Learning4Activity:
Momentum Equations of Fluid Flow
 `

Pretest:

Lesson 1: Conservation of Momentum and Momentum Equations

Learning Outcomes:

 Discuss the principle of momentum, and conservation of momentum

 Solve related problems
Learning Activity

Lesson 2: Impact on Blades

Learning outcomes:

 Determine the forces developed by moving fluids as it strikes

stationary and moving flat plates/blades
 Solve related problems

Learning Activity:

Mastery Test

Unit 5

Bernoulli/Navier-Stokes Equation
Pretest

Lesson 1: Equation of Motion Theories

Learning Outcomes:

 Discuss the Bernoulli’s Theorem on total energy head at any point

in a stream or every point in a path of flow and its effect using a
pump and a turbine.
 Solve related problems

Learning activity

Lesson 2: Trajectories

Learning outcomes:

 Determine and solve problems on fluid flows from nozzles and

pipes and conveying fluids on elevated reservoirs
 Solve related problems
 Learning Activity

Lesson 3: Measuring devices, pipe flow

Learning Outcomes:

 Determine the rate of flow and volume of fluids using different

measuring devices (Venturi meter, pipes, orifices and weirs)
 Solve related problems

Learning activity:

Mastery Test

References:

MATHalino Engineering Mathematics

ANSWERS

Unit 1:

Learning Activity 1
1) c 2) b 3) b 4) d 5) c 6) a 7) a 8) b 9) c 10) d
Learning Activity 2
1) a. 1,333.3 kg/m3 b. 13.08 kN/m c. 4.905 kN
 2) a. 254.84 kg/m3 b. 1.27 kg c. 0.0039 m3/kg
3) a. 55 MPa b. 0.02
4) 5.9 mm
5) A. 3.81 kg/m3 b. 0.152 kg c. 1.49 N

Mastery Test
1) a. 6,117.21 kg/m3 b. 0.0001635m3/kg c. 6.117
2) a. 20 cu.m. b. 1 MPa c. 50 MPa
3) a. 7.3575 N/m b. 42.37 N/m

Learning Activity

UNIT 3

BUOYANCY
1. a. 206.96 N b. 1.32 c. 12.94 kN/m3
2. a. 12.5% b. 0.824 c. 0.118 kN

STABILITY OF FLOATING OBJECTS

1. a. GB = 0.90 M b. MG = 1.46m c. M =
119,716 kg-m
2. a. MG = 0.77m b. W = 6,641 kg c. 2.76 m