Liza anna jj309 fluid mechanics (buku kerja

2013
POLITEKNIK SULTAN
AZLAN SHAH
liza_anna
[JJ309 FLUID MECHANICS]
[Type the abstract of the document here. The abstract is typically a short summary of the contents of
the document. Type the abstract of the document here. The abstract is typically a short summary of the
contents of the document.]

JJ309: Fluid Mechanics
STUDENT WORKBOOK
JJ308-FLUID MECHANICS
LIZA ANNA BINTI MAT JUSOH
JABATAN KEJURUTERAAN MEKANIKAL
POLITEKNIK SULTAN AZLAN SHAH

Hakcipta © Liza Anna binti Mat Jusoh
Cetakan Pertama 2012
Cetakan Kedua 2013
Tiada bahagian daripada terbitan ini boleh diterbitkan semula, disimpan untuk pengeluaran atau ditukarkan ke
dalam sebarang bentuk atau dengan sebarang alat juga pun, sama ada dengan cara elektronik, gambar serta
rakaman dan sebagainya tanpa kebenaran bertulis daripada penulis terlebih dahulu.
All right reserved. No part of this publication may be reproduced or transmitted in any form or by any means,
electronic or mechanical including photocopy, recording, or any information storage and retrieval system, without
permission in writing from the writer.
E-mel: liza_anna@psas.edu.my
Dicetak oleh/Printed by

Background
This workbook published with a goal to ease reference, strengthen understanding, and
increase achievement. So that this objective filled, planning and this book manipulation
are made with very latest curriculum and past year final examination paper that. This
workbook contain inquisition technique with various difficulty level.
Objectives
This module focuses on the following objectives, typically found in syllabus content:
a. Explain clearly the characteristics of fluid
b. Solve problems correctly related to fluid properties, fluid statics and fluid
dynamics
c. Explain the theory of fluid mechanics related to engineering field .

CONTENTS
Unit 1: Fluid Properties
Covers characteristics of fluid, pressure gauge measurement,
physical properties of fluid, viscosity and compressibility
Unit 2: Fluid Statics
Introduces the relationship between pressure and depth, analyze
pressure head, buoyancy and pressure.
Unit 3: Fluid Dynamic
This topics covers flow, discharge. Mass flow rate in pipe,
continuity equation, Bernoulli equation and measurement of fluid
motion
Unit 4: 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 5: Nozzle
This topic covers types and shapes of nozzles, critical pressure
ratio, changes in pressure, temperature, maximum mass flow
and cross- sectional area

Unit:
Kilo Hekto Deko unit desi centi mili
1 liter = 1000m3
I kg x 9.81 = 1N
1 bar = 105
N/m @ Pa

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
- The pressure below the atmospheric pressure ( 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 the mass per unit volume.
b. Specific weight,  is defined as the weight per unit volume.

c. Specific gravity or relative density, s is the ratio of the weight of the substance to
the weight of an equal volume of water at 4 ºC.
d. Specific volume, v is defined as the reciprocal of mass density. It is used to mean
volume per unit mass.
e. Viscosity
A fluid at rest cannot resist shearing forces but once it is in motion, shearing
forces 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 shearing
stresses.
Kinematic viscosity
This ratio is characterized by the kinematic viscosity (Greek letter nu, ν),
defined as follows:
Dynamic viscosity, μ

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 is10-3
Ns/m2
(s=0.89, ρ=890.41kg/m3
,ω=8734.92N/m3 ,
1.123x10-6
m2
/s)

c. Determine the specific volume if it mass is 500g and the volume is 400cm3
(8x10-4
m3
/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
, 0.793,7.78x103
N/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
, 5.352x10-5
m3
/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 it s 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 x10-4
m3
/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 x10-3
m3
, 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 10-3
m3
/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:
ghp 
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
(2.03x10-3
m)
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
(200.1240x103
N/m2
,18.24 x103
N/m2
) A

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, a of a small piston is 22 cm2
and the area A 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?(275.970 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, a of
a small piston is 20 cm2
while the area, A 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. (139.85x103
N)

h. A 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 (101.6kN)

m.The basic elements of a hydraulic press are shown in Figure i. The plunger has an
area of 3cm
2
, and a force, F
1
, can be applied to the plunger through a lever
mechanism having a mechanical advantage of 8 to 1. If the large piston has an
area of 150 cm
2
, what load, F
2
, can be raised by a force of 30 N applied to the
lever? Neglect the hydrostatic pressure variation. (12 kN)
Figure i
Solution
F2 = 12 kN

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
( 17929.9N)

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)
Fig. b

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

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)
12cm
water
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)
Fig. e
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
)
Figure g

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)
Figure h
b
a
water
merkury

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. Figure f (47.77kN/m2
)
Figure i

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)
Figure j

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
N/m
3
). 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, p
A
− p
B
, is −5 kPa. Determine
the differential reading, h. (0.46 mm )
Fig.o

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)
Figure p

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

s. A piston having a cross-sectional area of 0.07 m
2
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
1
= 60 mm and h = 100 mm, what is the value of the applied force, P,
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
Define Buoyancy Force
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
R2
R 1 = R2
ρ1 g1 v1 = ρ2 g2 v2
R1

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
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. Un- uniform flow
the velocity and other hydrodynamic parameters do not change from point to
point at any instant of time.

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.
vAQ .
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 .
vAm 



Example questions
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
Another example in the use of the continuity principle is to determine the
velocities in pipes coming from a junction.
Total discharge into the junction = Total discharge out of the junction
Q1 = Q2 + Q3
A1v1 = A2v2 + A3v3
RP
SYSTEM
P R
QRQP
1
3
2

Example:
a. A pipe is split into 2 pipes which are BC and BD The following information is
given:
diameter pipe AB at A = 0.45 m
diameter pipe AB at B = 0.3 m
diameter pipe BC = 0.2 m
diameter pipe BD = 0.15 m
Calculate:
a) discharge at section A if vA = 2 m/s
b) 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)
Fig. b

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
B
E
D
A

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)
Fig. e

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)
Au = 31.415x10-3
m2
Av = 49.09x10-3
m2
QT = Qu + Qv
Qu = 0.25 m3
/s
Qv = 0.4136 m3
/s
Vv = 8.430 m/s
Vu =7.96m/s
Rajah f
T
=

g. One pipe branching to 2 pipe TU and TV as in fig g.
Following information known:
Diameter pipe ST in part S = 0.45m
Diameter pipe ST in part T = 0.3 m (in ST is acute from part S to part T)
Diameter pipe TU = 0.2 m
Diameter pipe TV = 0.5 m
Determine :
i) Discharge for S if Vs = 2 m / s
ii) velocity in part T and part V, if velocity in U = 4 m/s ( 0.318m3
/s, 4.5m/s, .98m/s)
Figure g

h. Oil flow in a pipe 20 mm diameter as figure h . The pipe divide two branches is
10 mm diameter with velocity 0.3 m/s and another is 15 mm dia meter with
velocity 0.6 m/s . Calculate QP, QR,VS, ( 2.355 x10-5
m3
/s, 1.06 x10-4
m3
/s,
0.41m/s)
Figure h

i. The raw oil flowed through a pipe with a diameter of 40 mm and entered a pipe a
velocity of both pipes and the density of raw oil if the mass flow rate is 3.23 kg/s.
(v1=2.984m/s, v2=7.46m/s,861.3kg/m3
)

Energy of a flowing fluid
a. Potential energy
Potential energy per unit weight = z
b. Pressure energy (Pressure Head)
pressure energy per unit weight =

p
=
g
p

c. Kinetic energy
Kinetic energy per unit weight =
g
v
2
2
Total energy per unit weight =
g
vp
z
2
2


Bernoulli’s Theorem,
Total energy per unit weight at section 1 = Total energy per unit weight at section 2
g
vp
z
g
vp
z
22
22
2
2
11
1 

The limits of Bernoulli’s Equation
Bernoulli’s Eqution is the most important and useful equation in fluid mechanics. It may
be written,

2
2
2
1
1
2
1
1
22
p
g
v
z
p
g
v
z 
Bernoulli’s Equation has some restrictions in its applicability, they are :
 the flow is steady
 the density is constant (which also means the fluid is compressible)
 friction losses are negligible

 the equation relates the state at two points along a single streamline (not
conditions on two different streamlines).
Application of Bernoulli equation
a. Water flows through a pipe 36 m from the sea level as shown in figure a.
Pressure in the pipe is 410 kN/m2
and the velocity is 4.8 m/s. Calculate total
energy of every weight of unit water above the sea level. (78.96J)
b. A pipe measure 15 m length, supplying water to a house that located on a hill,
5.5 m above sea level . Diameter of the pipe is 30 cm . If the water velocity is 2
m/s, calculate the total energy . The water pressure is 5000 Pascal .(6.21m)
36
m
Figure a
m

c.
Figure b
A bent pipe labeled MN measures 5 m and 3 m respectively above the datum
line. The diameter M and N are both 20 cm and 5 cm. The water pressure is 5
kg/cm2
. If the velocity at M is 1 m/s, determine the pressure at N in kg/cm2
.
d. Ventury meter is flow meter device. Sketch and main part of horizontal ventury
meter.
5 m 5 m
3 m

e. A venturimeter is used to measure liquid flow rate of 7500 litres perminute. The
difference in pressure across the venturimeter is equivalent to 8 m of the flowing
liquid. The pipe diameter is 19 cm. Calculate the throat diameter of the
venturimeter. Assume the coefficient of discharge for the venturimeter as
0.96.(11.14 cm)
m/s

f. A Venturi meter is 50 mm bore diameter at inlet and 10 mm bore diameter at the
throat. Oil of density 900 kg/m3
flows through it and a differential pressure head of
80 mm is produced. Given Cd = 0.92, determine the mass flow rate in kg/s
( 0.0815 kg/s)

g. A Venturi meter is 60 mm bore diameter at inlet and 20 mm bore diameter at the
throat. Water of density 1000 kg/m3
flows through it and a differential pressure head
of 150 mm is produced. Given Cd = 0.95, determine the flow rate in dm3
/s. (0.515
dm3
/s)

h. Calculate the differential pressure expected from a Venturi meter when the flow rate
is 2 dm3
/s of water. The area ratio is 4 and Cd is 0.94. The inlet cross section area .
is 900 mm2
.(41916 Pa)

i. Calculate the mass flow rate of water through a Venturi meter when the differential
pressure is 980 Pa given Cd = 0.93, the area ratio is 5 and the inlet cross section
area. is 1000 mm2
. (0.2658kg/s)

j. Calculate the flow rate of water through an orifice meter with an area ratio of 4 given
Cd is 0.62, the pipe area is 900 mm2
and the differential pressure is 586 Pa.
(0.156 dm3
/s).

j. A horizontal Venturi meter with 0.15 m in diameter at the entrance is use to
measures flow rate of oil . Specific gravity for oil is 0.9 . The difference of level in
manometer is 0.2 m. Calculate the throat diameter if velocity at the entrance is
3.65 m/s . Find the actual rate of flow , assuming a coefficient of discharge is 0.9
.(2.82m,0.099m,0.058m3
/s)

k. A meter ventury with diameter of 400 mm at the inlet and 200 mm at the throat .
It is horizontal and used to measure the water flow rate . A differential
manometer is used and shown the different level reading of 60 mm . Calculate
the real discharge . Given Cd = 0.95 .(0.119m3
/s)

l. A metre venturi that in a situation horizontal have neck diametrical 150 mm set
within water main pipe that diametrical 300 mm. Discharge coefficient this metre
venturi is 0.982 .Determine height difference mercury column in manometer
differential if flow rate is 0.142 m3
/ s (0.254m)

m. Horizontal a meter venturi have diameter 250 mm in inlet and 150 mm in neck
area. Manometer mercury connected to metre venturi show flow level difference
reading 55 mm. Determine rate coefficient if real discharge water which flowed is
0.063 m3
/ s .(0.9)

n. A metre venturi have diameter 400 mm in section enter and 200 mm in neck area.
It is prestigious horizontal and used to measure rate of flow water . Manometer
differential mercury / water used and show level difference 60 mm. Determine rate
of actual flow rate of water . Assume Cd = 0.95 .(0.1187m3
/ s)

o. A meter venturi horizontal used to measure fluid flow from a tank. Inlet and neck
venturi have diametrical 76 mm and 38 mm. 2200 kg water ran in 4 minutes.
Difference reading in mercury level in U-tube is 266 mm. Calculate coefficient of
flow rate. Mercury specific gravity13.6.(0.965)

p. Diameter for entry of meter ventury horizontal was 0.2 m and diameter in neck
area was 0.1 m. It used to measure flow rate oil that density comparison 0.8.
Mercury manometer difference / oil is using are showing reading 0.2 m, determine
i. Oil flow velocity
ii. Discharge in theory
iii. Actual discharge discharge coefficient, Cd = 0.9
(1.92m/s ,0.0642m3
/s ,57.85x10-3
m3
/s)

i. Energy Loss in Pipelines
__________________________________________________________________
i. sketch the velocity distribution diagram in the round pipe system
ii. explain the velocity distribution in the round pipe system
iii. The head loss in pipeline

a. A pipe caring 2100 litter /min of water increases suddenly from 27 mm to 38mm
in diameter. Calculate:
i. The head loss due to the sudden enlargement
ii. The difference in pressure in kN/m2
in two pipes.
( 46.716m, 387.3kN/m2
)

b. horizontal pipes X with cross-section 0.01 m2
, joined by a sudden
enlargement to a Y pipe with diameter 250 mm. The water velocity through the
pipe is 3 m/s. Determine :
i. The flow rate through the pipe
ii. Head loss due to a sudden enlargement
(0.147m3
/s,6.98m)
AX = 0.01 m2
DY = 250 mm = 0.25 m
VY = 3 m/s
i.Q = AV
QY = AY VY
QY = 0.049 x 3
QY = 0.147 m3
/s
= 0.049 m2
ii.
HL = QX = AX VX
Qx = Qy
= 0.147 = 0.01 Vx
= 6.98 m Vx =
= 14.7 m/s
X Y

c. A pipe with diameter 100 mm have a flow rate of water is 0.047 m3
/s have
suddenly enlargement to 259 mm diameter . Calculate :
i. The head loss of sudden enlargement .
ii. The pressure difference between the small and big diameter of pipe in
kN/m2
.
(1.319m,-4.539N/m2
)

d. A horizontal pipes diameter decrease suddenly from 15 cm to 5 cm . The flow
rate of water entrance the pipe is 0.081 m3
/s . If coefficient of contraction is
0.602, calculate pressure difference in between a pipe .(1217kN/m2
)
e. The raw oil flowed through a pipe with a diameter of 40 mm and entered a pipe a
velocity of both pipes and the density of raw oil if the mass flow rate is at 3.23 kg/s

f. Two tanks filled with water connected by serial pipe as in figure e AB pipe
has a diameter 10 cm and BC pipe 6 cm . The flow rate of water
entering the pipe is 0.007 m3
/s and coefficient of contraction is 0.62. If
energy losses because shock loss at sudden contraction and friction only,
calculate level difference the two tanks . Given f = 0.04 for both pipes .
(4.8m)
Figure e

g. A tank is connected with a pipe which has a length 100 m . The outlet channel is
open which is 10 m below the water surface of tank . The inlet channel of pipe is
sharp . Calculate the diameter of pipe if the water’s velocity in pipe is 2.5 m/s ,
given f for pipe is 0.005.(66.89mm)

h. Water transmitted from a reservoir to atmosphere through a pipe 45 m long
such as fig.f .The enter is sharp and diameter is 45 mm of long 20 m from inlet
.The pipe suddenly enlargement to 80 mm for length that remainder .with take
into account loss of column, calculate level difference between pooled water
surface and drain if rate of flow was 3.0 x 10-3
m3
/ s . If f = 0.045 for small pipe
and 0.065 for big pipe. (16.0m)
Fig. f

A tank which is connected with a pipe which has a diameter of 150 mm as shown in Figure 2. The outlet
channel of the pipe is open which is 10 m below the water surface of the tank. The inlet channel of the
pipe is sharp. Calculate the length of the pipe if the water’s velocity in pipe is 2.5 m/s. Given f = 0.01 for
the pipe. (10 markah)
Fig g
( )
10 = ( )
10 = 0.159 + 0.085 L + 0.319
10 – 0.159 – 0.319 = 0.085 L
9.522 = 0.085L
L =
L = 112 m
Pipe Ø 150 mm
10 m

i. Water from a large reservoir is discharge to atmosphere through a 50mm
diameter pipe 250m long as figure i. The entry from the reservoir is sharp and
out let is 12m below the surface level in the reservoir. Taking f= 0.01, calculate
the discharge (2.123 x10-3
m3
/s)
Fig. i
d=50mm
L=250m
H = 12m

j. Two tank have column difference 45m links by serial pipe ABC such as Figure
j under. Pipe AB diametrical 60 mm and long 50 m, while pipe BC diametrical
80 mm and long 75 m. Calculate rate of flow water which flowed through pipe.
Assume energy loss only due to friction only.
Take ƒ = 0.04 for both pipe ( 6.24 x 10-
m3
/s )
Fig. j

k. A 40 m long horizontal pipe line is line is connected to a water tank at one end
discharges freely into the atmosphere at the other end as show in figure k below.
For the first 25 m of its length from the tank, the pipe is 150mm in diameter and its
diameter and its diameter is suddenly enlarge to 300mm. The height of water level
in the tank is 8m above the center of the pipe. Considering the losses at entry is
negligible and f = 0.001 for the both of pipe, determine the rate of flow. (0.2569
m3
/s)
Fig. k

q. Water flows from a reservoir to the pipe measuring 15m length and a diameter of
40mm due to sharp inlet as shown in the figure below. The pipe is suddenly
enlarged to 70mm and a length of 25m. Given discharge is 2.8 x10-3
and
coefficient of friction for both pipe is 0.03, calculate:
h. Velocity at point 2, v2
iii. Velocity at point 3, v3
iv. Head loss due to sharp inlet, hc2
v. Head loss due to friction hf23
vi. Head loss due to sudden enlargement,hL3
vii. Head loss due to friction hf34
(2.22m/s, 0.73m/s, 0.13m, 11.3m, 0.11m,1.16m)

r. Two huge open tanks are connected with 2 types of pipe by series. The
specification is shown in table 1. The total pressure drop, PA-PB = 1.5kPa, and
the elevation drop, ZA – ZB = 5 m. Calculate the discharge.
Pipe Length Diameter Friction
1 100m 250mm 0.01
2 200m 400mm 0.05
( 0.087m3
/s)
( ) ( ) ( )
( ) ( ) ( )
{ }
√

s. Two reservoir have a difference in level of H is 8 m and are connected by a pipe
line, which is 40mm in diameter for the first 12mm and 25mm for the remaining 5
m calculate the discharge of flow in m3
s-1
if coefficient of friction , f= 0.001 for both
pipes and coefficient of contraction, Cc =0.66
( 4.034 x 10-3
m3
s-1
)
Figure s

t. Two reservoirs are connected by a pipeline which is 150 mm in diameter for the
first 6 m and 225 mm in diameter for the remaining 15 m. The entrance and exit
are sharp and the change of section is sudden. The water surface in the upper
reservoir is 6 m above that in the lower. Tabulate the losses of head which occur
and calculate the rate of flow in m3
/s. Friction coefficient f is 0.01 for both pipes.
(0.185m3
/s)

5. Nozzle
__________________________________________________
Define Nozzle
- A device that increases the velocity of a fluid at the expense of pressure
The application of nozzles in engineering fields
i. Types and shapes of nozzles
a) Convergent Nozzle
b) Convergent – divergent nozzle
- Critical temperature ratio,
1
2
11
1












p
pc
T
Tc
- Critical pressure ratio,
 1/
1
2
1










p
pc
a Steam Turbine
b Gas Turbine
c Jet Engine
d Flow Measurement
e Rocket Propulsion
f Steam Injector
g Injector
inlet throat outlet
Inlet Outlet

a. Air at 8.6 bar and 190C expands at the rate of 4.5 kg/s through a
convergent-divergent nozzle into a space at 1.03 bar. Assuming that the inlet
velocity is negligible, calculate the throat and the exit cross-sectional areas of
the nozzle.
1 C 2
8.6 bar 1.03 bar
C1=0 C2

b. A fluid at 6.9 bar and 93o
C enters a convergent nozzle with negligible velocity,
and expands isentropic into a space at 3.6 bar. Calculate the outlet
temperature and mass flow per m2
of exit area, when the fluid is helium ( Cp =
5.24 kJ/ kg K). Assume that helium is a perfect gas, and the respective
molecular weight as 4.

c. If a convergent-divergent nozzle expands the air at the rate of 5kg/s from 8.2
bar and 2500
C at the inlet and into the space at 1.15bar. Given air = 1.4, R =
287 J/kgK and Cp= 1005 J/kgK. Assuming the inlet velocity is negligible and
the flow is isentropic.
i. Sketch and label the convergent divergent nozzle based on the
information given
Calculate:
ii. The critical pressure
iii. The critical temperature
iv. The Critical volume
v. The Cross-sectional area of the throat in mm2

BIBLIOGRAPHY
1. Cengel, Y. A. and Cimbala, J. M., (2005). Fluid Mechanics: Fundamentals and
Application. International Edition, McGraw-Hill, Singapore.
2. Douglas, J.F., Gasiorek J.M. and Swaffield, J. A., (2001). Fluid Mechanics, 4th
Ed. . Prentice Hall, Spain.
3. Finnemore E.J,(2002) .Fluid Mechanics with Engineering Application, 10th Ed
McGraw Hill, Singapore, 2002
4. Robert L Mott (2005). Applied Fluid Mechanics. 5th Ed. Prentice Hall.
i. 2005 White F. M., (2003). Fluid Mechanics, 5th Edition. McGraw Hill, USA.
ii. Soalan – soalan peperiksaan akhir Politeknik Jabatan Malaysia ( JJ309)
iii. Modul J3008 – Politeknik Malaysia
iv. http://physics.tutorvista.com

Liza anna jj309 fluid mechanics (buku kerja

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Andere mochten auch

Andere mochten auch (20)

Ähnlich wie Liza anna jj309 fluid mechanics (buku kerja

Ähnlich wie Liza anna jj309 fluid mechanics (buku kerja (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Liza anna jj309 fluid mechanics (buku kerja