EI s
a
a
velocity components
y
F E
acceleration
a date f ffx.y.tt
dd u
P
dt
ddIf w
a.edu
EiDEiiEyfIEiEiDE
iEiEtEay
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a
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a fFt ffI
a
adf.tvoj iwdfi.IE
Tl1ocala
aE
ax2tag2 convective accln
accln
f x
y z t a f V t V flu v w t
a a
ix I II o
a
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Types of flows
je time
aviable
and unsteady flows
Steady flows
inviscid
Viscous and V
and external flows
Internal flows
and incompressible
Compressible
Nonuniform flows
Eintifainmunstantand
dimensional flowT
andthree
One two
Streamline a
stream tube
c
line v
Time
Streamline
line is curve that is tangent
A stream a
to instantaneous velocity vector
da I
dr dei't dy'T't
Streamline
dy IV
da
e
TI up vJ wk
a ZD plane
From similar triangles on
1 Eor
Streamtuby
lines
It is a bunch of stream
0
oTy Stream
lines
Streaklined
It is locus of fluid particles passing
through a common
point sequentially
of Streak line
0 o_O
Pathling
fluid particle
by
a
Path followed
o
path line
t at t seat
Timeline
Line
joining
the fluid particles at
time instant
a
particular
O
s
I
att att tat
Note
state conditions
Under steady
streamline
the same
streak line are
timeline
Massflow ral m
dye
la fav
II
mi fav
for
Rotation and Vorticity
Rotation we Omega
Vorticity S zeta or r omega
2x angular velocity
Vorticity
hors 2 5
Vorticity is also curl V
a txt i j e
or
to curl V
S
u v w
r or s
Ey Fa ti CE I
I
C8
Fy
off i foot of I
a
E
of
N z
so he
r 0
r
Ey
a 8
w
wz
Fhif.IT sg the motion offluid particles
are
considered
Gravity forces
Pressure
Tension
Compressibilly
F ma
Gravity force Euler's wavier's
Pressure force egg stokes
Viscous force
Turbulent force
compressibilityforIpetynoldsen
d
Surface tension force
Continuityequation
is based on
Continuity equation law
principle of cat fass
ta
t É
neeO
1
Rate flow right face in the x direction
of mass through
series That is
is given byTaylor's expansion
feudA so 810
t
Jaya
xx Jadz
higher order term
Neglecting
I lot 8,10 E dyde
At
rate mass flow through leftface
Similarly of
x direction
is
in the
series
dyda by Taylor's
JIF flu di neglecting higher
order terms
in the direction is
Net rate of mass flow
0M dy da
to HE Yu
0 07
dadydz
direction is
in the g
Similarly rate of
mass flow
doe dy da
flyer
in the L direction is
Similarly rate of
mass flow
dxdyd2
0µW
Net change in mass flow rate is
given by
foldxdgdri fydbdxdydrtfzfewjdxdgdz
f.FI fly wYdxdyda
But f is a function of ffx y z t and using
conservation of mass principle
9 wyd.caydz o
fz J t ff
so 03 o
IgM FEW
compressible
This isContinuity equation for
fluids gases flow
f is constant
For incompressible flows
If Fat
o
constant
steady flow conditions
For is
0
912 0
streamlines
s alongstreamline
to streamer
Formal
a
7,41
a
as
17
V
Is
an YE
and
Ema Eg normaltostreamline
F ma or Newton's second law along a
streamline
normal to stream line
Fema
Ima along a
streamline
da
pdp
Streamline
is
to streamline
ttrormal
dz
10
Assumptions
Fluid is incompressible steady in rotational
1
non viscous and continous
Only gravity and pressure forces
2 are
considered
As per Fema
da dinsino m digs
pda ptdp
Here ma f f e f dads
dive mg f dads.g
since dads
Substituting
ldAds.VIg
PHA PHA dpdit ldads.gg
dpidt fgdtdz fdtv.dv
dp Pgdz IV du
Dividing by f and rearranging
14 91 7
This equation is known as Euler's equation of
motion
Integrating
is known as Bernoulli equation
This equation
assumptions considered
and is valued for
equation are
Other forms of this
By VI
922
ga
It
or
Ig Yg 22
Igt 2
or
constant
P I Iga
where
p is the static pressure
f v2 is the dynamic pressure
z
is the hydrostatic pressure
Ige
thestatic dynamic and hydrostatic
The sum of
pressures is called total pressure
The sum of static and dynamic pressure
is called stagnation pressure
i e
p
te Pstag
pressure energy
I
is
kinetic energy
Is is
energy
92 is potential
Applications of Bernoulli's expression
Venturi meter constant
Pigtrigth
Orificemeter
Pitot tube
DischargeQIFlowrate
crossection X velocity of flow
area of
an
Q
Q ax
VII
A V I 9242
discharge through the
Effusion Venturi meter
ÉÉ
7
Pipe
Consider a venturimeter fitted to a horizontal
as shown in
pipe carrying fluid water
the above figure
Let d and da be the diameter of the pipe
and respectively
and throat at section
Pi a and Pz Va az are
Similarly Vi
and area of cross sectional
pressure velocity
and respectively
at section
Applying
Bernoulli's expression between
section and we have
Pig Yg 12 Ig the
Since pipe is horizontal 21 22
I VIII
so
I
h pressure head
But
Big
Vig
so
he
Now Continuity equation OF Oz
applying
AV AzV2
V 921,1
the value of V in equation
Substituting 2
h
Wigs
I
III VII
00
2g h
V2 i
af
or
A V AzV2
Now discharge equation Q
aft
Q ava 92
or
YEA
Qth
a.gg
hQactual
Cdaigd
f
whereCd is co efficient of discharge
This is the expression for discharge
through the renturimeter 1.0
less than
The value of Cd is always
Epressionfor discharge through the
É
é
i q
Venacontracta
Consider a flow of fluid in a horizontal
pipe a
orifice meter is connected to
the pipe as shown in the above figure
Consider section and as
shown in the figure Section is
considered at the pipe where one of
the limbs of the manometer is connected
section is considered at the
vena contract where other limb
region
of the manometer connected Manomete
is
that flows in the
same
carrys Let h fluid
pipe be its reading
9 ti and Pa Vz as are the pressure
a
section at
velocity and area of cross
section and respectively
Applying Bernoulli's equation between
section and
Pigtrig 12
Pff Egg 122
But h
Pigg 2 22
so
he VEI
2g
Using continuity equation
Q air Arik but here az GAO
so where a is areaof orifice
a V CcAoV2 Cc is co efficient of
contraction
So Ccaovy
A
V in equation
the value of
Substituting
V f
he
29 Ciao
ri cia vicar
a
h 2g at
zg
2g ath VI
Iia or
VFR a
Vz
Taita
So Cc Ao Fgh Al
Q AzVz Coaouz
a aI
Rearranging
Cc Aoa Rgb
air Ciao
But co efficient of discharge is related to
ad
contraction by
Cc co efficient of
cd gate
ai f ciao
So 2 2
Aoa TBH
E aE
ao
a
i E
909 Agh
Qed
Tata
or
Cd Ao A Tzgh
Qa
Tata
This is the equation for discharge through
orificemeter
sionfor Velocityofthefow
pitottube.SE
seat velocity
stagnation pressure
static
FITE Pierometer
pitorube
I IF
EEE
too
III gtEgtH
Eg Es Esto k h
g
Ish v Eh
a Fsh I
This is the equation for velocity of flow
stream function
y
Continuity equation
E 8
0
Stream function is also a function of
space and time
Here
Ogg
ve
Fx
Now to
continuity equation reduces
0
8,1 5 8,18 7
o
84g
The lines along which is constant
y
are streamlines
Slope at stream line
any point
on a
E
rdx
Udy
da
ydy of
dy dx o
8g
8
8 9
doe 0
along a streamline
Total change foe de y dy
of y
dy 84
doc
Ofdy
Velocitypotential O
Velocity potential is a function of
space and time
x y z t
0
9,9 ve
8g w
0g
Continuity equation
o
Ey
41 17 1 7 0
0
8
0
Line constant of
of
do 0
day E
Line constant of called equipotential
of are
lines
Line constant are called streamlines
of y
Determine the vorticity
an
expression for
the flow field described by
of
xy3i tyg
Is the flow in rotational
A 25
woe I Fy Fy
wy t
was t Ooty
No Wo 0
Wyo
Wz t o f Boyz
Wz x y
U 2 wait wy
wat
I 2 0 t Oj 334214
1 3 9212
is
Since I is not zero flow
not in rotational
6 1h
The velocity components an incompressible
of are
two dimensional velocity field given by
U
y sell x Vey 2 1 Show that
innotational and satisfies conservati
the flow is
of mass
Two dimensional
i rotational
flow wz u
o
component
0
wa I 7 0
fay
0
Wz t 2y
O so the flow
is irrotational
Wz
Conservation of mass
get
o
1 0
f 1 2017 201
0 0
So conservation of mass is satisfied
631 The
stream function for a
given
two dimensional flow field is
Determine the
µ
5 Ry 153193
corresponding velocity potential
µ
5 Ry 5343
501
u
Of 55 81
Integrating with
x
2
55 dx
Ido f 5
3
0 5
5 5 f y
Tv 94 y
foxy
Integrating with respect to y
fd0 f long dy
0 12 5 fatal
5 92 721 1
From and
Of 53 I 5 Xyz t c
The stream function for a two dimensional flow
is by calculate the velocity
given y Findaythe
a point 2,3 velocity potential
of
function 0
2
4 4
2x
U
Y
Of 28
201 2 2 4 units g
Up 3
Vez 3 29 2 3 6 units s
Resultant velocity R F FIT Fa
R 552
R 721 unit Is
Now 0
94 0 foe 2x
V 2
y 2g
Now 0 0 24
co r t X
Integrating
doe
S do fax
of C e is a function y
Differentiating t
war
y
d but V
g of ay
so
0g 28
w r t y
Integrating
Soc fay dy
C yr
Now
It c
2
10 92
Sketch the stream lines represented
it by
y y also find out the velocity and
its direction at
point 1,27
x
4 y
u
8g 8
u
Ey gylity ay
get Ey pity 2x
UG 2 2y 2 2 4 unit s
1,2 2X 2N zumiffs
R Ttv A2 564 fo 4.47 unitls
Directionand
I 3 t
269562
10
Sketch of stream lines
9 22 92
If 4 1 2,3
then I pity
2 22 92
3 It y 2
This is an equation of a circle So we
get
concentric circles with y values
2
22 92
convinced n
y
food
Waving wi
g at
Dti rests
u v
Ey 0
ax
ay
at
ax 4212 2x
20
dy 4yt 2y 06
2y
y
Local acceleration components are
Ax 20C ay 2g
are
Convective acceleration components
ax 4xt aye aye
V it Etterytt
To
6xottxo.co
V Omls
TEt2 4yEzyt
i
Ax 9
ay
age
'"
%%
The velocity along the centreline of a
nozzle
of length I is given by Fat
t is time
1
in seconds
2,7
where it is velocity in mls distance from
inlet
commencement of flow x is the from
the convective and local aceh
Calculate
to nozzle acceleration at t 6s so im and
1.6m
and total
The velocity along the centre line is
2 1
21
Local acceleration
Of
1
2 C
At 1 65 1 1fm se I m
97 24 I
181 172
0.945
Convedineacceleration
pgy
zt E x 2tx2 E Ee
4
11
and 1 1 6m
At t 6s x im
convective acceleration 44 1
1,1
2924m s
Iconrectiveacch
convective
Local acceleration
Total acceleration acceleration
0.945 29 24
28 295m
1Totdac4y
The velocity field in a
fluid flow medium is
given
É say i zxyjtfzyt3 tk
Find the magnitude and
directions of
a translational velocity
2rotational velocity
3 Vorticity
t 3
At 1,2 1 and at time
velocityrector at 1,31 and at t 3
1 Translational
can be written as
3 3 K
3 1 IT 2.1.2.5 2.1.2
121 45 1312 18 13ms
F
12 V 4 and we 13
So here 0
vector
Rotational velocity
i j K
L ext
If u v w
'&
we 8g Is E E E
8213193 8,429 31
it by 229 31 galaxy j
K 11 4 9 g 32yd
I zit of't y 32g
1 and t 3
At 1,2
i
I 2 3 1.2 É
4.123rad
471
s
17
15 l unit
o unit and we gumi
That is woe Wy
Vorticity
I 25
ai
2ft 814 I 8.246rad
I 25
''
In an incompressible flow the velocity vector is given
BY
Va Ext y2 i at xy j xy zxyz Gtz k
equation is satisfied
i Verify whether continuity 2 2,2 and at t 2
ii Determine the acceleration at point
z Gta k
y2 i ft toy ay ay
fact
U Get y22
3t xy
W Xy zxyz 612
Gt zoey zoey Gt
Of 0g
continuity equations
so
Yy 982 0
Xy 4
0
It Xy
0 0
continuity equation is satisfied
So
ii Acceleration a ait ay's tack
ax
I t u
get Vogt way
ax 62 Getty 2 ft Bt g 22 ay zxyz 61232g
At point 2 2,2 and at t 2
6 2 6.2.2 2.22 6 2 3.2 2.22 22 2.2
Ax
2 2.2.2 6 2 2 2 2
164unit
ax
get off tag
w
ay
ay 2 Xyz 612C
3 Got y22 y't t xyz zoey
At 12 2,2 and t 2.0
2.22 2 2 2
3 6.2.2 2.22 22 3.2
199 73
243
az dog t v
dog v
Yy
w
Iz
62 fatty 22 y 292 3 29276 222
Az
fly 2292 Gtz zoey ft
At 2 2,2 and at t 2.0
2 2 2 2 2
2.22 2 2.2 2 3.2
6.2 6.2.2
Az
6.2.2 2 2.2 6.2
2.2 2.2.2
Az 432 units
Sa a ax tag azt
43212
1641 2433
A
A TÉ43t432
192522Mt
y
13.1814in
!*'
D
!*(
Y
!*)
122
!!*
!!!
!!"
!!#
Cengel
TO 8100kPa
absolute 800 100
700 KPalgage
low velocity
1001 0 kPa
100 KPa 1100
Igi É Bsg
g
zÉ
24
Bg VE
or VE
ve
Tj F
V2 1.18 m s
!!$
Angel
Take
fair 0909kg m3
f
2
0.909
2.56mfs
!!%
A horizontal venturi meter with inlet diamet
zoom and throat diameter room is used
to measure the flow of oil of sp gravity
oil through venturimeter
0.8 The discharge of
the oil mercury
is bolts Find the reading of
manometer Take ca 0.98
differential
Q Goals
act
Cd
gÉ I 0.06m
a
TE
0.031
IF
m2
a
to.o6
t
az
tg
0.0078 M2
h 2.89m 92
heavier lighter liquid in manometer
If
heavier fluid in
pipe
h x
É D !!&
lighter fluid in pipe
he f Ey
n
1st D
29m x 13 1
2 01817
!!'
Hagen Equation
Poiseuille
flow of the viscid through a cigar
Titi t
p HE.am
Consider horizontal pipe radius R The
a
of as
right to
viscous fluid flowing from left
is
in the figure Consider a fluid element
shown
radius r sliding cylindrical fluid
in a
of
element of radius ft dry Let the length of
element be ax R be the radius of
fluid is flowing
in which fluid
pipe
!!(
on face AB
Let p be the intensity of pressure
on face CD
be the intensity of pressure
R fax acts on
shear force of exzarase
the surface of the fluid element
Now
I Fx 0
A yoga e ar ax o
game e xxx
4 7
85
T.EEd
Whitt e
I today
Here
pudgy
But here R r
y
do
dy
so
y Edr
e
From equation
e
z if
E
w r t r
Integrating
8
1 1
To find the value of constant c
Boundary conditions
At ER 4 0
Ier't
So c becomes
u
o
8 R't C
R2
C
8
substituting c in equation we get
a
fpdf.si offer
fµT
u i
Velocity distribution
