Reference book of fluid mechanics

Fluid Mechanics work book

By Mohamad Pazlin bin Saion

CONTENTS

Unit 1: Fluid Properties
Unit 2:
Unit 3: Covers characteristics of fluid, pressure gauge measurement,
Unit 4: physical properties of fluid, viscosity and compressibility

Fluid Statics

Introduces the relationship between pressure and depth, analyze
pressure head, buoyancy and pressure.

Fluid Dynamic

This topics covers flow, discharge. Mass flow rate in pipe,
continuity equation, Bernoulli equation and measurement of fluid
motion

Energy Loss In Pipelines

This topic cover velocity profile in circular pipe. Type of head loss
in pipelines, flow characteristics, head loss equation for rate and
pipelines problems

Unit Convertion: Unit multiply
×106 = 10 ×10 ×10×10×10×10
UNIT
Mega
Kilo
Desi
Centi
Mili
Mikro
Nano
Piko

1 liter = 1000m3
1 m3 =0.001 liter
1 kg x 9.81 = 1N
1 bar = 105 N/m @ Pa

To convert between small value with
smaller unit to a bigger unit. The

answer should be in small value. For
instance: 1centimeter into meter is
equal to 0.01m…thus 0.01 is smaller
compared to 1



SYMBOL PHYSICAL BASIC UNIT RELATED FORMULA
QUANTITY
F Newton (N) F= P/A
W Force Newton (N) di mana P=tekanan;A=luas
ρ Weight W= m×g
υ Density (Rho) kg/m3 di mana m=jisim;g= pecutan graviti
Specific velocity m3/kg ρ= m/v
J Energy (Joule) di mana m=jisim dan v=isipadu
J υ = v/m
Patm Atmospheric di mana v=isipadu dan m=jisim
pressure Pa @ N/m2
di mana N=newton,
m=meter,s=second
Pabsolute=Pgauge+ Patm

=101.3kPa

t Masa (second) s
N/m3
Specific weight weight,W ω = ρ×g di mana ρ=ketumpatan
volume,V dan

 subs tan ce g=pecutan graviti
Specific gravity  water
s or
 subs tan ce
 water No unit

Specific volume volume,V

 mass, m
or
1 m3/kg

density, 

Formula Pa=Pg+ Patm
can be developed by

making acronyms. For
Example: I need to treat
my Personal assistance
thus I have to go to
ATM machine.



PROPERTIES FORMULA SI UNIT
Mass density kg/m3
Specific weight  = mass, m N/m3
Specific gravity volume,V
No unit
Specific volume  = weight,W
volume,V m3/kg

s =  subs tance or s = subs tan ce
 water water

 = volume,V or  = 1
mass, m 

Unit 1: Fluid and Properties

1. Define Fluid

2. Compare the characteristics between liquid, gas and solid
a. Liquid
b. Gas
c. Solid

3. Define of pressure
a. Atmospheric
b. Gauge
c. Absolute
d. Vacuum

4. Example problem of pressure
a. What is the pressure gauge of air in the cylinder if the atmospheric gauge is
101.3 kN/m2 and absolute pressure is 460 kN/m2.

(358.7kN/m2)

b. A Bourdon pressure gauge attached to a boiler located at sea level shows a
reading pressure of 7 bar. If atmospheric pressure is 1.013 bar, what is the
absolute pressure in that boiler (in kN/m2) ?
(801 kN/m2)

5. Fluid properties

a. Mass density, ρ is defined as ______________________________

b. Specific weight,  is defined as _____________________________
c. Specific gravity or relative density, s is _______________________

d. Specific volume, v is defined as the__________________________

e. Viscosity
A fluid at rest cannot resist ________ but once it is in motion, ________ are set
up between layers of fluid moving at different velocities. The viscosity of the fluid
determines the ability of the fluid in resisting these ___________.

f. Fluid compressibility is defined as a measure of the relative volume change of a fluid or
solid as a response to a pressure change.

6. Example problems of fluid properties
a. If 5.6 m3 of oil and weights 46000 N ,determine :
i. Mass density ,ρ in unit kg/m3
ii. Specific weight ,ω

iii. Specific gravity of oil , S

(837.33kg/m3,8214.3N/m3,0.837)

b. Get the relative density , density , specific weight and kinematik viscosity
of an oil which are 7.3 m3 in volume , 6500 kg in mass and dynamic
viscosity is10-3Ns/m2

(s=0.89, ρ=890.41kg/m3,ω=8734.92N/m3 ,1.123x10-6m2/s)

c. Determine the specific volume if it mass is 500g and the volume is 400cm3
(8x10-4m3/kg)

d. Given specific weight of fluid is 6.54 N/litter and its mass is 830 g.

Calculate the following in SI unit

i.Volume of fluid
ii.Specific volume of fluid
iii.Density of fluid

(1.245x10-3 m3,1.5x10-3/kg,666.67 kg/m3)

e. Volume and mass for oil are 9.2 m3 and 7300 kg

i. Mass density
ii. Relative density
iii. Specific weight

(793.4 kg/m3, 0.793,7.78x103N/m3)

f. If the mass and volume of air 11.5 kg and 650 cm3, calculate:
i. Mass density
ii. Specific weight
iii. Specific volume
iv. Specific gravity for the air
(17692.31 kg/m3,173558.5N/m3, 5.352x10-5m3/kg)

g. The volume of engine oil is 5.5m3 and the weight is 50 kN determine
i. Density of oil
ii. Specific weight of oil
iii. Specific volume of oil
iv. Specific gravity
(926.7kg/m3,9091N/m3,1.079x10 -3 m3/kg, 0.9267)

h. Determine the mass density, in SI unit if its mass is 450g and the volume is 9dm3.

(50kg/m3)

i. Determine the specific weight ω ( kN/m2) and specific gravity, s of fluid if the
weight is 100N and the volume is 500cm3
(20kN/m3, 2.039)

j. The volume of a stone is 1.5 x 10-4 m3. If the relative density of the stone is 2.6,
calculate:
i. The density
ii. The specific weight
iii. The specific volume
iv. The weight
v. The mass

(2600kg/m3, 25.506kg kN/m3, 3.85 x10-4m3/kg, 3.83 N, 0.39 kg)

k. Given the volume of oil is 3 liter and the weight is 20N, determine the specific
volume, relative density and specific weight of oil.
(1.471 x10-3m3, 0.68, 6670N/m3)

l. Specific gravity of a liquid is 0.85. determine
i. Mass density
ii. Specific volume

(850kg/m3,1.176 x 10-3m3/kg)

Unit 2: Fluid Static

1. 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
The formula that gives the pressure, p on an object submerged in a fluid is:

p = gh

Where,

▪  (rho) - the density of the fluid,

▪ g- the acceleration of gravity
▪ h - the height of the fluid above the object

2. Example Problems:
a. A barometer shows the reading 750mm merkury. Determine;
i. Atmosfera pressure in unit SI
ii. The head of water for that preassure
(100 KN/m2,10.2m)

i. P= ρgh
= 9810 x 13.6 x 075
= 100062N/m2

ii. 100062 = ρgh
h = 10.2m

b. What is the pressure experienced at a point on the bottom of a swimming
pool 9 meters in depth? The density of water is 1.00 x 103
kg/m3.(88.3kN/m2)

c. Assume standard atmospheric conditions. Determine the pressure in kN/m2
for the pressure below:

i. Depth 6m below under free space water.
ii. At the 9m under surface of oil with specific gravity 0.75.

(58.86kN/m2,66.0 kN/m2)

d. Find the height of water column, h which is equivalent to the pressure, p of
20 N/m2. Take into consideration specific weight of water , ω is 1000 kg/m2
x 9.81 m/s2
(2.03x10-3m)

e. A fluid in piezometer increased 1.5 m high from point A in a pipeline system.
What is the value of pressure in point A in N/m2 if the fluid is :
i. Mercury with specific gravity 13.6
ii. Salted water with specific gravity 1.24
A

(200.1240x103N/m2, 18.24 x103N/m2)

f. Find the head, h of water corresponding to an intensity of pressure, p of 340 000
N/m2. Take into consideration that the mass density, ρ of water is 100kg/m3
(h=34.65m)

g. A Bourdon pressure gauge attached to a boiler located at sea level shows a
reading pressure 10 bar. If atmospheric pressure is 1.01 bar , determine :
i. The absolute pressure in kN/m2
ii. The pressure head of water, h
(1101 KN/m2, 112.2m)

3. Pascal’s Law and Hyraulic Jack
iv. State the Pascal’s Law

4. Example :
a. A force, F of 900 N is applied to the smaller cylinder of an hydraulic jack. The
area, A1 of a small piston is 22 cm2 and the area A2 of a larger piston is 250
cm2. What load, W can be lifted on the larger piston if :
i. the pistons are at the same level.
ii. the large piston is 0.8 m below the smaller piston.

Consider the mass density ρ of the liquid in the jack is 103 kg/m3
(10.227 kN,10.423kN)

b. Two cylinders with pistons are connected by a pipe containing water. Their

diameters are 75 mm and 600 mm respectively and the face of the smaller piston

is 6 m above the larger. What force on the smaller piston is required to maintain

a load of 3500 kg on the larger piston? (275.91 N)

c. A diameter of big piston in hydraulic jack is three times bigger than the diameter of

small piston. The small diameter is 630 mm and is used to support a weight of 40

kN. Find the force which is needed to rise up the big piston 2 m above the small

piston. Given the specific gravity of oil is 0.85. (313.18kN)

d. A force, F = 500 N is applied to the smaller cylinder of hydraulic jack. The area A1 of

a small piston is 20 cm2 while the area, A2 of a large piston is 200 cm2. What mass

can be lifted on the larger piston. (509.68 kg)

e. A hydraulic press has a diameter ratio between the two pistons of 8:1. The diameter
of the larger piston is 600 mm and it is required to support a mass of 3500 kg. The
press is filled with a hydraulic fluid of specific gravity 0.8. Calculate the force
required on the smaller piston to provide the required force ;
i. When the two pistons are at same level
ii. When the smaller piston is 2.6 m below the larger piston.
(536.48 N, 627.92 N)

f. A hydraulic jack has diameter cylinder 5 cm and 18 cm. A force has put on small
cylinder to lift the load 1100 kg at bigger cylinder. Determine force F for lift the both
cylinders.
(832.64N)

h. An area of big piston in hydraulic jack is three times bigger than the area of small
piston. The small diameter is 630 mm and is used to support a weight of 40 KN. Find
the force which is needed to rise up the big piston 2 m above the small piston. Given
the specific gravity of oil is 0.85

(104.4kN)

m. The basic elements of a hydraulic press are shown in Figure i. The plunger has an

2

area of 3cm , and a force, F , can be applied to the plunger through a lever

mechanism having a mechanical advantage of 8 to 1. If the large piston has an

2

area of 150 cm , what load, F , can be raised by a force of 30 N applied to the

lever? Neglect the hydrostatic pressure variation. (12 kN)

Figure i

n. The diameter of plunger and ram of a hydraulic press are 30 mm and 200 mm
respectively. Find the weight lifted by the hydraulic press when the force applied
at the plunger is 400N and the difference level between plunger and ram is 0.5 m.
Given ρ fluids is 1065 kg/m3
(17934.26N)

5. Concept of manometer
i. Manometer Simple

ii. Manometer U tube

iii.Manometer Differential

6.Example
a. Assume that Patm= 101.3 kN/m2 water flow in pipe and in merkuri in manometer

a= 1m h=0.4 m. Determine the absolute pressure. As figure a (38.1kPa)

Figure a

b. A U tube manometer is used to measure the pressure of oil (s= 0.8)
flowing in a pipeline as in figure b. Its right limb is open to the atmosphere
and the left limb is connected to the pipe. The centre of the pipe is 9 cm
below the level of mercury in the right limb. If the difference of mercury
level in the two limbs is 15 cm, determine the gauge pressure of the oil in
the pipe in KPa.
(19.541 KPa)

c. Determine absolute pressure at A if Patm = 101.3 kN/m2, h1=20cm,h2= 40 cm
(45.971KPa)

water

mercury

d. For a gauge pressure in pipe is 5kN/m2, determine the specific gravity of
the liquid B in the figure given below.

(6.54)

water

12cm

Liquid B

e. Find the level of h if P1 is absolute pressure 150kN/m2, ρm= 13.6 x103 kg/m2
and in pipe is water in fig. e.

(0.401m)

500mm

h

m

f. One end of a manometer contain mercury is open to atmosphere, while the
other end of the tube is connected to pipe in which a fluid of specific gravity
0.85 is flowing. Find the gauge pressure the fluid flowing in pipe.
(26.271kN/m2)

Fig.f

g. A U tube manometer measures the pressure difference between two
points A and B in a fluid as shown in Figure d. The U tube contains
mercury. Calculate the difference in pressure at pipe A and B if h1 = 160
cm, h2 = 50 cm and h3 = 80 cm. The liquid at A and B is water ρ =
1000kg/m3 and the specific gravity of mercury is 13.6.1
(53955N/m2)

h. The figure e below shown a U tube manometer . The specific gravity of mercury is
13.6 . If the pressure difference between point B and A is 47 kN/m2 , h = 12cm
and a = 43 cm , determine the height of b .
(3.71m)
water
a
b

mercury

i. A manometer U tube is using to measure between A and B in pipe has water and in
manometer has mercury. Determine the differential pressure between pipe A and B, if
a =150 cm, b = 70 cm and c = 45 cm.
(47.77kN/m2)

j. Figure g shown U tube manometer. If the differential of pressure between X andY is
50KN/m2 , h=2m and a=0.85m determine b
(0.4719m)

k. he fig. k shows a differential manometer connected at two points A nd B. At A
air pressure is 100kN/m2. Find the absolute pressure at B

Figure k
(84.28kPa)

l. A U-tube manometer is connected to a closed tank containing air and water as
shown in Figure h. At the closed end of the manometer the absolute air
pressure is 140kPa. Determine the reading on the pressure gage for a
differential reading of 1.5-m on the manometer. Express your answer in gauge
pressure value. Assume standard atmospheric pressure and neglect the
weight of the air columns in the manometer. (64.5 kPa)

Figure l

m. A U-tube manometer contains oil, mercury, and water as shown in Figure i.
For the column heights indicated what is the pressure differential between
pipes A and B? (-15.1kPa )

Figure m

n. A U-tube manometer is connected to a closed tank as shown in Figure j. The
air pressure in the tank is 120 Pa and the liquid in the tank is oil (γ = 12000

3

N/m ). The absolute pressure at point A is 20 kPa. Determine: (a) the depth of
oil, z, and (b) the differential reading, h, on the manometer. Patm = 101.3 kPa (z
= 1.66 m, h = 1.33 m )

Figure n

o. The inverted U-tube manometer of Figure k contains oil (SG = 0.9) and water as
shown. The pressure differential between pipes A and B, pA− pB is −5 kPa. Determine
the differential reading, h.
(0.455 mm)

o. In the figure below, fluid Q is water and fluid P is oil (specific gravity = 0.9). If
h = 69 cm and z = 23 cm, what is the difference in pressure in kN/m2 between A
and B?
(-1.579kN/m2)

p. Figure m belows shows a u-tube manometer that used to measure the pressure
difference between pipe P and pipe Q that contains water. If the fluid in u-tube is
oil with specific gravity 0f 0.9, calculate the pressure difference between these two
pipes in kN/m3. Given M =80 cm and N = 25 cm.
(1667.7 Pa)

r. For the inclined-tube manometer of Figure n, the pressure in pipe A is 8 kPa. The
fluid in both pipes A and B is water, and the gage fluid in the manometer has a
specific gravity of 2.6. What is the pressure in pipe B corresponding to the
differential reading shown?(5.51kPa )

fig.r

2

s. A piston having a cross-sectional area of 0.07 m is located in a cylinder containing

water as shown in Figure o. An open U-tube manometer is connected to the cylinder

as shown. For h = 60 mm and h = 100 mm, what is the value of the applied force, P,

1

acting on the piston? The weight of the piston is negligible (892.7 N)

Fig. s

7. Pressure Measurement
Piezometer, Barometer

Bourdon gauge
Sketch important parts of bourdon gauge

Explain mechanism of a bourdon gauge

8. Buoyancy R 1 = R2
Define Buoyancy Force ρ1 g1 v1 = ρ2 g2v2

R1

R2

Buoyancy is the upward force that an object feels from the water and when compared to
the weight of the object

Buoyant Force=Weight of Displaced Fluid

9. Example Question
a. A rectangular pontoon has a width B of 6 m, a length l of 12 m, and a draught D of

1.5 m in fresh water (density 1000 kg/m3). Calculate :

a) the weight of the pontoon
b) its draught in sea water (density 1025 kg/m3)
c) the load (in kiloNewtons) that can be supported by the pontoon in fresh

water if the maximum draught permissible is 2 m.
(1059.5kN, 1.46m, 14126kN,353.1kN)

b. 8 cm side cube weighing 4N is immersed in a liquid of relative density 0.8

contained in a rectangular tank of cross- sectional area 12cm x 12cm. If the tank

contained liquid to a height of 6.4 cm before the immersion determine the levels of

the bottom of the cube and the liquid surface. (x =0.0796m)

0.8
0.0796 m

3.0 Fluid Dynamics

TYPES OF FLOW

a. Steady flow
The cross-sectional area and velocity of the stream may vary from cross-
section, but for each cross-section they do not change with time. Example: a
wave travelling along a channel.

b. Uniform flow
The cross-sectional area and velocity of the stream of fluid are the same at
each successive cross-section. Example: flow through a pipe of uniform bore
running completely full.

c. Laminar flow
Also known as streamline or viscous flow, in which the particles of the fluid
move in an orderly manner and retain the same relative positions in
successive cross-sections.

d. Turbulent flow
Turbulent flow is a non-steady flow in which the particles of fluid move in a
disorderly manner, occupying different relative positions in successive cross-
sections.

e. Transition flow
The process of a laminar flow becoming turbulent flow.

Define volume flow rate and mass flow rate

Flow rate

The volume of liquid passing through a given cross-section in unit time is
called the discharge. It is measured in cubic meter per second, or similar
units and denoted by Q.

Q = A.v

Mass Flow rate

The mass of fluid passing through a given cross section in unit time is
called the mass flow rate. It is measured in kilogram per second, or similar

•

units and denoted by m .

•

m =   A v

Exercise:

a. If the diameter d = 15 cm and the mean velocity, v = 3 m/s, calculate the actual

discharge in the pipe. (0.053m3/s)

b. Oil flows through a pipe at a velocity of 1.6 m/s. The diameter of the pipe is 8 cm.
Calculate discharge and mass flow rate of oil. Take into consideration soil = 0.85.
(6.836 kg/s)

c. The raw oil flowed through a pipe with a diameter of 40 mm and entered a pipe a
diameter of 25mm. The volume flow rate is at 3.75 liter/s. Calculate the flow
velocity of both pipes and the density of raw oil if the mass flow rate is at 3.23 kg/s
(V = 7.64m/s, 861.33kg/m3)

Continuity Equation PR
QP
SYSTEM QR

Another example in the use of the continuity principle is to determine the
velocities in pipes coming from a junction.

1P 2

R
3

Total discharge into the junction = Total discharge out of the junction
Q1 = Q2 + Q3

A1v1 = A2v2 + A3v3

a. A pipe is split into 2 pipes which are BC and BD The following information is
given:

Calculate: diameter pipe AB at A = 0.45 m
diameter pipe AB at B = 0.3 m
a) diameter pipe BC = 0.2 m
b) diameter pipe BD = 0.15 m

discharge at section A if vA = 2 m/s
velocity at section B and section D if velocity at section C = 4 m/s

(0.318m3/s,4.5m/s, 11.0 m/s)

b. If a pipe at fig.b has diameter 30.48 cm and 45.72 cm at 1 and 2. Water flow
5.06 m/s at part 2. Determine:
i. Velocity at 1
ii. Flow rate at 2

(11.367m/s, 0.83m3/s)

c. Water flows through a pipe AB of diameter d1 = 50 mm, which is in series with a
pipe BC of diameter d2 = 75 mm in which the mean velocity v2 = 2 m/s. At C the
pipe forks and one branch CD is of diameter d3 such that the mean velocity v3
is
1.5 m/s. The other branch CE is of diameter d4 = 30 mm and conditions are such
that the discharge Q2 from BC divides so that Q4 = ½ Q3. Calculate the values of
Q1,v1,Q2,Q3,D3,Q4 and v4.

(Q1 = Q2= 8.836 × 10-3 m3/s ,v1 = 4.50 m/s, Q3 = 5.891 × 10-3 m3/s,

Q4 = 2.945 × 10-3 m3/s, d3 = 71 mm v4 = 4.17 m/s D
AB

E

d. Determine the value of Q1 ,Q2 ,Q3 ,d3 ,Q4 and v1 if the water flow in the pipe
as figure d
Diameter pipe 1 , d1 = 40 mm
Diameter pipe , d2 = 60 mm
Velocity in pipe 2, v2 = 2 m/s
Velocity in pipe 3, v3 = 1.5 m/s
Diameter pipe 4 ,d4 = 25 mm
Discharge in pipe 3 , Q3 = 2 times Q4

Q1 = Q2= 5.65 × 10-3 m3/s ,v1 = 4.40 m/s, Q3 = 3.77 × 10-3 m3/s, Q4 = 1.884 × 10-3 m3/s,
d3 = 56.6 mm v4 = 3.83 m/s

e. Oil flows through a pipe RS and split into two pipes , which are ST and SU as
show in Figure e . The following information as given ;
Diameter pipe, RS = 250 mm
Diameter pipe, ST = 200 mm
Specific gravity, Soil = 0.95
Calculate;
i. Discharge and mass flow rate of oil at pipe RS if velocity is 2.5 m/s
ii. Diameter pipe SU if velocity at pipe ST is 1.5 m/s and at pipe , SU
is 3 m/s.
(0.1225m3/s,116.375kg/s,178mm)

f. A pipe ST is split into two pipes TU and TV as shown fig f below. Determine :
i. Discharge of S if the velocity at U is 8 m/s.
ii. Velocity at T and V if the velocity at U is 8 m/s.

(0.66m3/s, 6.9m/s, 7.96m/s)