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AAiT Hydraulics I By Zerihun Alemayehu
Kinematics of Fluids • Ideal fluid flow
– No viscosity no shear stresses – Boundary effects ignored and uniform velocity
• Real fluid flow – Fluid properties – Kinematics and dynamics
Kinematics of fluids deal with geometry of motion (space-time relationships) for fluids in motion.
AAiT Hydraulics I By Zerihun Alemayehu
Velocity field • Two ways to describe fluid motion
–Lagrangian method –Eulerian method
AAiT Hydraulics I By Zerihun Alemayehu 4
Lagrangian Method
• Follows a single particle and describe its characteristics (velocity and acceleration)
y
x
z
a
b c to
(x,y,z) t
X = f1 (a,b,c,t)
Y = f2 (a,b,c,t)
Z = f3 (a,b,c,t)
AAiT Hydraulics I By Zerihun Alemayehu
Lagrangian method (contd.)
2
2
z
2
2
y
2
2
x
a
a
a
tz
tzw
ty
tyv
tx
txu
∂∂
=∂∂
=
∂∂
=∂∂
=
∂∂
=∂∂
=
5
• Velocity and acceleration
AAiT Hydraulics I By Zerihun Alemayehu 6
Eulerian Method • Focus on a point in space filled with fluid and
describe the characteristics of flow at various points in the flow field at any time.
Velocity u = f1(x,y,z,t) v = f2(x,y,z,t) w = f3(x,y,z,t)
𝑑𝑑𝑑𝑑
= 𝑢 𝑑,𝑦, 𝑧, 𝑑 𝑑𝑦𝑑𝑑
= 𝑣(𝑑,𝑦, 𝑧, 𝑑) 𝑑𝑧𝑑𝑑
= 𝑤(𝑑,𝑦, 𝑧, 𝑑)
The relationship between Lagrangian and Eulerian method
AAiT Hydraulics I By Zerihun Alemayehu 7
Lagrangian Vs Eulerian Lagrangian Method • difficult because it is not easy to identify a fluid
particle and trace its path • Each particle has a random path Eulerian Method • Easier and more practical because usually one is
interested in flow parameters at certain points than what happens to the individual particles
AAiT Hydraulics I By Zerihun Alemayehu
Velocity and Acceleration
tzLimw
tyLimv
txLimu
tsLimV
t
t
t
ts
∆∆
=
∆∆
=
∆∆
=
∆∆
=
→∆
→∆
→∆
→∆
0
0
0
0
8
• Velocity is the rate of change of displacement
s Vs
x
y
Δx A
B
Δy Δs
AAiT Hydraulics I By Zerihun Alemayehu
Acceleration
tVV
sVa
dtdt
tV
dtds
sV
dtdVa
ss
ss
ssss
∂∂
+∂∂
=
∂∂
+∂∂
==
9
Acceleration is rate of change of velocity. It has two components: Tangential acceleration and normal acceleration
s Vs
A
B
dtt
Vdss
VdV sss ∂
∂+
∂∂
=
Total change in velocity
Tangential acceleration
AAiT Hydraulics I By Zerihun Alemayehu
The Normal Acceleration
tV
RVa
RVV
RV
dtds
dsdV
dtdV
a
RdsVVddV
nn
nnn
n
∂∂
+=
====
==
2
2
1
θ
10
RV
tVa
sVV
tVa
nn
ss
ss
2
+∂∂
=
∂∂
+∂∂
=
A
∆s
R
D v
v+∆v
dθ
Convective component
Temporal component
v +∆v
v ∆v
dθ ∆vn
AAiT Hydraulics I By Zerihun Alemayehu 12
Pathline and streakline Pathline: the path traced by a fluid particle during a
given time interval Streakline: path showing the positions of all
particles that passed through a given point at a given time interval
AAiT Hydraulics I By Zerihun Alemayehu 13
Streamline Streamline: An imaginary line drawn through a flow field
such that the tangent to the line at any point on the line indicates the direction of the velocity vector at that instant.
Note: since the flow is tangent to the stream line there will be no flow normal to a stream line, i.e., flow between any two stream lines remains constant
AAiT Hydraulics I By Zerihun Alemayehu
Stream tube Stream tube: is a collection of streamlines drawn
around the perimeter of a small area of stream cross section. Since the perimeter is composed of streamlines, there will not be any flow across a streamtube as well.
15
AAiT Hydraulics I By Zerihun Alemayehu 16
Classification of flow
• Flow is classified based on change in velocity • Depending on temporal variations of velocity
– Steady or unsteady
• Depending on spatial variations of velocity – uniform or non-uniform
AAiT Hydraulics I By Zerihun Alemayehu 17
Classification of flow
Steady flow: average flow velocity at a particular point does not change with time
Unsteady flow
Uniform flow: velocity at any point on a stream line remains the same at any given instant
Non- uniform flow
0=∂∂
tV
0=∂∂
sV
0≠∂∂
tV
0≠∂∂
sV
AAiT Hydraulics I By Zerihun Alemayehu
Classification of flow
Steady if flow rate is constant
Uniform Non-uniform Unsteady non-uniform
AAiT Hydraulics I By Zerihun Alemayehu 19
One Dimensional flows One dimensional: When velocity components normal to the main
flow are neglected and only average conditions at a section are considered
Applicable when: • No wide variation of cross- section • Stream lines are not highly curvilinear • Velocity variation across a section is negligible
AAiT Hydraulics I By Zerihun Alemayehu 20
Two and Three Dimensional flows Two Dimensional: When real fluids are considered, the effects of
boundaries and viscosity brings about variations in velocity at different sections
Three Dimensional: Velocity vector varies in the three principal
directions x, y and z.
AAiT Hydraulics I By Zerihun Alemayehu
Discharge and Mean velocity
avgvAQ .=
22
Discharge: The volume of fluid passing a certain cross section in unit time. Measured in m3/s
If velocity distribution is assumed to be uniform, Note: A should be perpendicular to the direction of flow
A
V, t
vavg
𝑄 = 𝐴𝑉𝑎𝑎𝑎
AAiT Hydraulics I By Zerihun Alemayehu
When the actual velocity distribution is considered,
For a circular section, the
annular element
Discharge and Mean velocity
∫ ∫==
=
A
dAudQQ
dAudQ
.
.
23
∫=
=
R
drruQ
anddrrdA
0
..2
..2
π
π
r
R
dr
AAiT Hydraulics I By Zerihun Alemayehu
Continuity Equation • law of conservation of mass :In steady flow, the
mass flow per unit time passing through each section does not change, even if the pipe diameter changes.
For incompressible flow ρ1=ρ2
Q is the volumetric rate of flow called discharge, expressed in m3/s
AAiT Hydraulics I By Zerihun Alemayehu
Example 3.1 If u = 1.1 + 2.8 x + 0.65 y and v = 0.95 – 2.1 x – 2.8 y calculate the acceleration at (x, y) = (-2, 3)
AAiT Hydraulics I By Zerihun Alemayehu
Example 3.2
akjiV
on,Accelerati:Find3:Given 2tyxzt ++=
kjikjia )2()3(3 22 yxyzttxytzaaa zyx ++++=++=
222
22
2
2)(0)(2)3(0
30)()(0)3(
33)(0)(0)3(0
yxyztytyxztyttww
zwv
ywu
xwa
txyzttyxxztztvw
zvv
yvu
xva
tyxzttuw
zuv
yuu
xua
z
y
x
+=+++=∂∂
+∂∂
+∂∂
+∂∂
=
+=+++=∂∂
+∂∂
+∂∂
+∂∂
=
=+++=∂∂
+∂∂
+∂∂
+∂∂
=
2;;3 tywxzvtu ===
AAiT Hydraulics I By Zerihun Alemayehu
Examples 3.3
smd
QAQV
VAQ
/611.0)25.0(
4
03.0
422====
=
ππ
Discharge in a 2-cm pipe is 0.03 m3/s. What is the average velocity?
AAiT Hydraulics I By Zerihun Alemayehu
Example 3.4
Find: )1()(RrVrv o −=
oVV
313
131
)32
(2)32
(2
2)/1(
2
2
2
22
0
32
0
===
=
−=−=
∫ −=∫=
o
oo
o
o
o
R
o
Ro
A
VR
RVV
AQ
VV
RV
RRVR
rrV
rdrRrVVdAQ
π
π
π
ππ
π