Fluid Flow Concepts
Flow Classification,
Ideal Fluid Flow
? Frictionless (zero viscosity)
? Incompressible
? Can be solved mathematically
Real Fluid Flow
? Shear stress develops where there is velocity gradient
? Fluid in contact with wall has zero velocity
Pipe Wall
Ideal Fluid,
Uniform velocity
Zero shear stress
Real Fluid,
Non-uniform velocity Shear stress
exists
Fluid close to wall has zero velocity
Flow Classification
Laminar Flow
? Occurs at low Reynolds number
? Fluid moves along smooth layers
? Motion governed by Newton’s law of viscosity
Turbulent Flow
? Occurs at high Reynolds number
? Fluid moves along irregular,fluctuating and random
paths
Flow Classification
Rotational (vortex) flow
? Fluid undergoes net rotation about some axis
Irrotational flow
? Fluid has no net rotation,only linear translation
? All ideal fluid flows are irrotational flows
Variation of fluid properties (velocity,pressure,
density,temperature etc)
Temporal Variation,
? Steady flow –Properties at a point not changing with flow
? Unsteady flow – changing with time
Spatial Variation,
? Uniform flow – Properties at an instant do not change with space
? Non-uniform flow – changing from point to point
Four possible flow types,
? Steady uniform flow
? Steady non-uniform flow
? Unsteady uniform flow
? Unsteady non-uniform flow
Flow Classification
One-dimensional flow
? Flow properties are function of time (t) and one space
coordinate (e.g,x)
Two-dimensional flow
? Flow properties are function of time (t) and two space
coordinates (e.g,x,y)
Three-dimensional flow
? Flow properties are function of time (t) and three space
coordinates (x,y,z)
x
x
y
Engineering Simplification
Many engineering problems are simplified as one-
dimensional problems
Pipe Flow,
Open Channel,
Vm = Q/A
v Vm
Actual Assumed 1-D
Actual Assumed 1-D
V Vm
AQV
AVVdA
m
A
m
/?
??
Engineering Simplification
Axi-Symmetric Flow through Circular Pipe
Elemental discharge through elemental area,
?Q = v, ?A = v, 2?r?r
Total Q through pipe section given by integration,
Mean velocity Vm = Q/A = Q/(?R2)
r
?r R
?A = 2?r?r
v A =?r2
Vm
Q
? ???
A R
r v d rv d AQ
0
2 ?
Fluid kinematics
? Under the ‘continuum’ hypothesis,a fluid body is
considered to be made up of infinitesimal fluid ‘particles’
tightly packed together and interact with each other
? Each fluid ‘particle’ contains numerous molecules
? Fluid motion is described in terms of velocity and
acceleration of fluid ‘particles’,and not individual
molecules
Fluid Kinematics
Study of the motion of fluid (position,velocity and acceleration)
without consideration of forces producing the motion
Fluid Dynamics
Analyses of fluid motion in relation to forces producing the motion
Fluid kinematics
System
? A definite mass of matter which distinguishes it from its
surrounding matter
? Has constant mass
? System boundary moves
? Used in Bernoulli’s equation
Control Volume (C.V.)
? Definite region in space enclosed by control surfaces,fixed
relative to observer
? Control Volume boundary is fixed
? Mass can flow in or out of C.V,
? Used in Continuity and Momentum equations
Flow Analyses
Eulerian Method
Fluid motion and properties (pressure,density,velocity
etc) are described as functions of space and time
Lagrangian Method
Follows the motion of individual fluid particles
determines how the fluid properties of the particles
change with time
Flow kinematics
Streamline,
Imaginary line through fluid such that at an instant,
velocity of ever particle on the line is tangent to it
Stream-tube
Imaginary tube formed by all streamlines passing
through a closed curve,No fluid can enter or leave a
stream tube except through its ends
Used in continuity equation
Continuity Equation
For a fixed,non-deforming control volume,
conservation of mass requires the time rate of
change of mass inside the control volume,plus
net rate of mass flow through the control surface
must be zero
For steady flow,time rate of change of mass in
control volume is 0
Where = mass flow rate in and out of control
volume in kg/s
0??
?
? ??
CS
n
CV
dAvdV
t
??
0? ?? ??? ino u t
CS
n mmdAv ???
m?
Continuity Equation
For uniform normal flow across a control surface
of area A,
= ? AV = ? Q
Mass inflow into CV = Mass outflow from CV
In terms of mean density and velocity,
?1 A1 V1 = ?2 A2 V2
For incompressible fluid,? is constant
A1 V1 = A2 V2 = Q
Q = volume flow rate along stream tube
Control volume
Stream tube
?2,A2,V2 ?1 A1 V
1
m?
Continuity Equation
For steady flow,conservation of mass requires,
For incompressible fluid,? constant,
? Qin = ? Qout
A1V1 + A2V2 = A3V3 + A4V4
V is uniform and perpendicular to A
A1,V1
A2,V2
A3,V3
A4,V4
? ?? inout mm ??
Continuity Equation
Summary,
Apply continuity equation to a Control Volume,
? Qin =? Qout
? AinVin =? AoutVout
Assumptions,
? ? is constant
? Flow is steady
? V is uniform across A
? V is perpendicular to A
Example
Find Q,VA,VC?
Apply continuity equation to CV1,
AA VA = AB VB
? 0.12/4 VA = ? 0.0052/4 x 3
VA = 0.0075 m/s
Apply continuity equation to CV2,
AC VC = AB VB
2 ? (0.15) x 0.005 x VC = ? 0.0052/4 x 3
VC = 0.0125 m/s
Q
VA
VC
0.1 m diam
0.005 m diam
Circular plate,0.3 m diam
VB=3 m/s
0.005 m thick
CV1
CV2
Continuity Equation
For steady flow,conservation of mass,
For incompressible fluid,? constant,
? Qin = ? Qout
A1V1 + A2V2 = A3V3 + A4V4
V is uniform and perpendicular to A
A1,V1
A2,V2
A3,V3
A4,V4
? ?? inout mm ??
Bernoulli Equation
Free body diagram of a fluid particle along a streamline
For steady flow (?v/ ?t =0),apply Newton’s 2nd Law of
Motion to free body,
? ?Fs = ?M as = ? ?V v (?v / ?s)
The following equation of motion can be derived,
- ?g sin ? - ?p / ?s = ? v ?v / ?s
Bernoulli Equation
- ?g sin ? - ?p/?s = ? v ?v/ ?s
As sin ? = dz/ds,
- ?g dz/ds - ?p/?s = 1/2 ? ?v2/?s
Deleting ?s,
?g dz + dp + 1/2 ? ?(v2) = 0
Euler’s equation
Integrating along a streamline,
?gz + p + 1/2 ?(v2) = constant
Bernoulli equation,
? Along a streamline
? Steady flow
? Frictionless
? Incompressible
Bernoulli Equation
Energy per unit volume (J/m3)
p + ?(v2)/2 + ?gz = constant
Pressure Kinetic Potential Total mechanical
energy
Energy per unit mass (J/kg)
p/ ? + (v2)/2 + gz = constant
Pressure Kinetic Potential Total M.E,
Energy per unit weight (J/N or m)
p/ ?g + v2/(2g) + z = constant
Pressure Kinetic Potential Total head
Bernoulli Equation
Apply Bernoulli’s equation to two points along a
streamline,
P1/(?g)+V12/(2g)+z1 = P2/(?g)+V22/(2g) +z2
If there is energy loss in motion of fluid from point 1 to
point 2,
P1/(?g) + V12/(2g) + z1 =
P2/(?g) + V22/(2g) + z2 + hL
hL = head loss due to friction from 1 to 2
Datum
1
2
V1
V2
z1
z2 P1
P2
Example
Apply continuity equation between Sections A & B,
AAVA = ABVB
VB = (0.15/0.1)2 4 = 9 m/s
Apply Bernoulli’s equation between points A & B,
PA/?g + VA2/2g + ZA = PB/?g + VB2/2g + ZB
80000/9810 + 16/(2g) = PB/9810 + 81/(2g)
PB = 47.5 kPa
B
0.1m diam
P=? kPa
A
0.15m diam
V=4m/s
P=80 kPa
10 m
Example
Apply continuity equation between Sections A & B,
AAVA = ABVB
VB = (0.15/0.1)2 4 = 9 m/s
Apply Bernoulli’s equation between points A & B,
PA/?g + VA2/2g + ZA = PB/?g + VB2/2g + ZB
80000/9810 + 16/(2g) + 10 = PB/9810 + 81/(2g)
PB = 145.6 kPa
B
0.1m
dia
m
P=
kP
a
A 0.15m
dia
m
V=
4m
/s
P=
80
kP
a 10 m
Example
Compute Q from orifice
Apply Bernoulli’s equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
0 + 0 + 3 = 0 + V22/2g + 0
V2 = (2 g 3)0.5 = 7.672 m/s
Q = A2 V2 = ? 0.12/4 x V2 = 0.0603 m3/s
3m
?
0.1 m diam
Q=?
1
2
Example
Compute required D for Q to be doubled
Apply Bernoulli equation to 1 and 3,
P1/?g + V12/2g + Z1 = P3/?g + V32/2g + Z3
V3 = (2 g 3)0.5 = 7.672 m/s
Q3 = A3 V3 = ? D2/4 x 7.672 = 2x0.0603 m3/s
D = 0.141 m
A2V2 = A3V3
V2 = 15.36 m/s
Apply Bernoulli equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
P2 = -88.5 kPa
3m
?
0.1 m diam
Q1=2Q
D=?
1
2
3
Example
Find V1,V2 and P
Apply Bernoulli equation to 4 and 1,
P1/?g + V12/2g + Z1 = P4/?g + V42/2g + Z4
V1 = (2 g 10)0.5 = 14.01 m/s
Apply Bernoulli equation to 4 and 2,
P2/?g + V22/2g + Z2 = P4/?g + V42/2g + Z4
V2 = (2 g 10)0.5 = 14.01 m/s
Apply continuity equation,
V3 = 2 V1 = 2 V2 = 28.01 m/s
Apply Bernoulli equation to 4 and 3,
P3 = -294.18 kPa
?
10m
V1=? V2=?
P=? 1 2 3
4
Example
Compute Q and PB
Apply Bernoulli equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
V2 = (2 g 4)0.5 = 8.859 m/s
Continuity between B and 2,
ABVB = A2V2
VB = 2.215 m/s
Apply Bernoulli equation to 1 and B,
P1/?g + V12/2g + Z1 = PB/?g + VB2/2g + ZB
PB= -22.07 kPa
?
4m
2m
0.02m diam
0.04 m diam
Q=?
PB=?
1
2
B
Example
What V1 is needed to suck oil up h m?
If P1/?g > -5m water,what is max permissible h?
P1 = - ?ogh (1)
Continuity equation between 1 & 2,
V2 = (D1/D2)2 V1 (2)
Bernoulli equation between 1 & 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2 (3)
Substitute (1) & (2) into (3) and solve for V1,
V1 = [2g?ohD24 / ?(D24 - D14)]0.5
P1 = - ?ogh = - 5 ?wg
h = (?w/ ?o) 5
D1 D2
h
Atmosphere V2 V1
Oil,?o=800 kg/m3
Water
2
1
Example
Apply Bernoulli equation to 3 & 2,
P3/?g + V32/2g + Z3 = P2/?g + V22/2g + Z2
+ 2 V12/2g + 0.1 V22/2g (1)
From continuity equation,
A1V1 = A2V2
V1 = (0.05/0.1)2 V2 = 0.25 V2 (2)
Substitute (2) into (1),
4 = (1 + 0.125 + 0.1) V22/2g
V2 = 8.004 m/s
Q = A2 V2
?
V2 V1
0.1 m
diam
0.05 m
diam
Head loss
2V12/2g Head loss 0.1V22/2g
4m
P=?
3
2
Stagnation Point
Static pressure at (3) P3 = ?g h4-3
Static pressure at (1) P1 = ?g h
At point 2,V2 = 0,stagnation point
From Bernoulli’s equation,
P2 = P1 + ? ? V12
P2 – P1 = ? ? V12 = dynamic pressure at the
stagnation point
Pitot Static Tube
For measurement of fluid velocity,
Measures the difference between static (P1) and
stagnation (P2) pressures,
P2 – P1 = ? ? V2
V = [2 (P2 – P1) / ?]1/2
Kinetic energy correction factor ?
Applies when flow velocity V is not uniform across
flow area
For flow along a circular pipe,
Total kinetic energy flow rate from integration of
actual velocity profile
V dA
Actual velocity
distribution
Assumed
uniform
distribution
Vm
Kinetic energy correction factor ?
For laminar flow in pipe,? ? 1.5
For turbulent flow in pipe,? ? 1.1
dAVV
dAAV
fa c t o rc o rr e c t i o nA p p l y
Vm
ekT o t a l
dAekT o t a l
dA
Qmd
V d AQ
m
m
m
3
A
A
33
3
m
2
m
A
3
3
2
)/( ( 1 / A )
V )2/1( )2/1(
AV )2/1()2/1(
Von b a s e d f l o w of,,
V ( 1 / 2 ),,
V )2/1(
V md ( 1 / 2 ) d ( k, e, )
V d A
?
?
?
?
?
??
?
?
?
???
??
?
???
?
?
?
?
??
?
?
?
Energy and Hydraulic Grade Line
Energy or Total Head Line (EL or THL)
Plot of total head (energy per unit weight,
J/N) (z+P/?g +V2/2g) along a stream line
For steady,inviscid and incompressible flow,
THL remains constant along a streamline
Hydraulic Grade Line (HGL)
Plot of piezometric head (z+P/?g) along a
stream line
Distance of V2/2g below THL
Energy and Hydraulic Grade Line
Power in fluid flow
Power = energy per unit time
=? g Q H (watts or J/s)
Q = flow rate in m3/s
H = total head in m
Quiz
Important Notice,
The first compulsory Quiz will be conducted
during your normal tutorial class in the week
from,
covering tutorials 1 to 4,
Students who miss the quiz without approved leave will
be awarded zero mark
The CA component from two quizzes could make up to
30 % of final mark for the subject
Linear Momentum Equation
For Computation of hydro-dynamic forces
When gate is closed,force on gate is due to
pressure, hydrostatic thrust
When gate is partially open,force on gate is due to
hydrostatic pressure and momentum change,
hydro-dynamic force
?
FS
Gate
closed
?
FD
Gate partially open
Linear Momentum Equation
Hydrodynamic force is governed by Newton’s
second law of motion,Apply law to a system of
fixed mass m kg
F = m.a = m, dV/ dt
= d (m.V)/ dt
= d M / dt
Force = rate of change of momentum of system
System, fixed mass m
Force F Velocity V
Acceleration a
Flow rate calculations
Volume flow rate Q = AV
Mass flow rate = ? A V
Momentum flow rate = V
= ? A V2
= ? Q V
= ? Q2/A
Flow area A
Uniform mean
normal velocity V
m?
m? M?
Momentum equation applied to Control Volume
F = net force on fluid in C.V,in x-direction
Rate of increase of momentum of fluid in C.V,
= F + ?AinVin2
Rate of decrease of momentum of fluid in C.V,
= ?AoutVout2
Net rate of change of momentum of fluid in C.V,
= F + ?AinVin2-?AoutVout2
= 0 for steady flow
Hence,
F = ?AoutVout2 - ?AinVin2
Vout
Aout
Vin
Ain F
Control
Volume
Mass of fluid in
C.V,= m
x
Linear Momentum Equation
In equation form,
For steady flow,
Along a given direction,
In words,
Net force on fluid inside CV
= Momentum outflow rate from CV
- Momentum inflow rate into CV
inoutn MMF ?? ??
? ?? ????
CV
n
CS
CVinF l u i d dAvvdVvtF ??
0?
?
?
?
CV
dVv
t
?
Net force acting on fluid inside C.V,
Net force acting on a body of fluid
= sum of forces acting on its surfaces
+ its self weight
On surface in contact with solid boundary,in
general force can be resolved into two
components,pressure and shear,
?
W
F3
F4
F5
F6
F7 F8
Momentum Correction Factor ?
Applies when velocity across a flow area is not
uniform,e.g,flow along circular pipe
It can be shown that total momentum flow
rate given by
M = ? ? A Vm2
Where,
V dA
Actual velocity
distribution
Assumed
uniform
distribution
Vm = Q/A
dA
V
v
A
A
m
2
1
? ??
?
?
??
?
?
??
Example
Jet impact force on hemisphere = 0.5g N
Reaction force on jet = 0.5g N
From Bernoulli equation,
Vi = Vo = V
Apply Momentum equation to CV in
Fn = -
- 0.5g = - ? Q V - ? Q V = -2 ? Q V
V = 1.533 m/s
V12 - V2 = 2 g h
V1 = Q/A = 3.26 m/s
h = 0.42 m
0.5 g N
h=?
Q=0.0016m3/s
0.025 m diam
Control volume Mass = 0.5 kg Fj =0.5g
Vo Vi
V1
outM? inM?
Example
Continuity equation 1 & 2,
A1V1 = A2V2
V2 = 9 V1
Bernoulli equation 1 & 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
Solving,
V1 = 3.873 m/s,V2 = 34.857 m/s
Apply Momentum equation in
Fn = -
F1 - Fn - F2 = ? A2 V22 - ? A1 V12
600,000 x ? x 0.32/4 - Fn - 0 = ? A2 V22 - ? A1 V12
Fn = 33.929 kN
0.3m dia 0.1m dia P=600 kPa
Fn on nozzle
Water jet V2
F2=0
V1
F1=P1A1 Fn
outM? inM?
Example
Continuity equation 1 & 2,
A1V1 = A2V2
V2 = 1.2 / 0.8 x 2.5 = 3.75 m/s
F1 = hydrostatic thrust = ?gAhc
= 1000 x 9.81 x 1.2 x 0.6 = 7063.2 N
F2 = ?gAhc = ?gx 0.8 x 0.4 = 3139.2 N
Apply Momentum equation in
Fn = -
F1 -FD - F2 = ? A2 V22 - ? A1 V12
FD = 175 N
?
FD
0.8m 1.2m
V1=2.5m/s
FD
V2
Control volume
F1 F
2
outM? inM?
Example
From Bernoulli equation,
V1 = V2 = V3 = V
Apply momentum equation along t
Fn = -
- Ft = ?CQV - ?(1-C)QV - ?QVCos?
For smooth plane,Ft = 0
C = (1 + Cos?)/2
Apply momentum equation along n
Fn = -
Fn = 0 - (-?QVSin?) = ?QVSin?
?
Q,V1
CQ,V2
(1-C)Q,V3
Smooth plate
Jet sheet
Fn
Ft
Ft
Fn
outM? inM?
outM? inM?
Example
Continuity equation between 1 & 2,
A1 V1 = A2 V2
V1 = 0.03/0.32 V2 = 0.225 m/s
Momentum equation along
Fn = -
F1 - Fg - F2 = ? A2 V22 - ? A1 V12
?g 0.32 x 0.16 - Fg - ?g 0.03 x 0.015 =
? 0.03 x 2.42 - ? 0.32 0.2252
Fg = 341.3 N
?
?
Fg
0.32 m
V2=2.4 m/s
0.03m
Fg V
1 F1
F2
Control volume
outM? inM?
Flow Classification,
Ideal Fluid Flow
? Frictionless (zero viscosity)
? Incompressible
? Can be solved mathematically
Real Fluid Flow
? Shear stress develops where there is velocity gradient
? Fluid in contact with wall has zero velocity
Pipe Wall
Ideal Fluid,
Uniform velocity
Zero shear stress
Real Fluid,
Non-uniform velocity Shear stress
exists
Fluid close to wall has zero velocity
Flow Classification
Laminar Flow
? Occurs at low Reynolds number
? Fluid moves along smooth layers
? Motion governed by Newton’s law of viscosity
Turbulent Flow
? Occurs at high Reynolds number
? Fluid moves along irregular,fluctuating and random
paths
Flow Classification
Rotational (vortex) flow
? Fluid undergoes net rotation about some axis
Irrotational flow
? Fluid has no net rotation,only linear translation
? All ideal fluid flows are irrotational flows
Variation of fluid properties (velocity,pressure,
density,temperature etc)
Temporal Variation,
? Steady flow –Properties at a point not changing with flow
? Unsteady flow – changing with time
Spatial Variation,
? Uniform flow – Properties at an instant do not change with space
? Non-uniform flow – changing from point to point
Four possible flow types,
? Steady uniform flow
? Steady non-uniform flow
? Unsteady uniform flow
? Unsteady non-uniform flow
Flow Classification
One-dimensional flow
? Flow properties are function of time (t) and one space
coordinate (e.g,x)
Two-dimensional flow
? Flow properties are function of time (t) and two space
coordinates (e.g,x,y)
Three-dimensional flow
? Flow properties are function of time (t) and three space
coordinates (x,y,z)
x
x
y
Engineering Simplification
Many engineering problems are simplified as one-
dimensional problems
Pipe Flow,
Open Channel,
Vm = Q/A
v Vm
Actual Assumed 1-D
Actual Assumed 1-D
V Vm
AQV
AVVdA
m
A
m
/?
??
Engineering Simplification
Axi-Symmetric Flow through Circular Pipe
Elemental discharge through elemental area,
?Q = v, ?A = v, 2?r?r
Total Q through pipe section given by integration,
Mean velocity Vm = Q/A = Q/(?R2)
r
?r R
?A = 2?r?r
v A =?r2
Vm
Q
? ???
A R
r v d rv d AQ
0
2 ?
Fluid kinematics
? Under the ‘continuum’ hypothesis,a fluid body is
considered to be made up of infinitesimal fluid ‘particles’
tightly packed together and interact with each other
? Each fluid ‘particle’ contains numerous molecules
? Fluid motion is described in terms of velocity and
acceleration of fluid ‘particles’,and not individual
molecules
Fluid Kinematics
Study of the motion of fluid (position,velocity and acceleration)
without consideration of forces producing the motion
Fluid Dynamics
Analyses of fluid motion in relation to forces producing the motion
Fluid kinematics
System
? A definite mass of matter which distinguishes it from its
surrounding matter
? Has constant mass
? System boundary moves
? Used in Bernoulli’s equation
Control Volume (C.V.)
? Definite region in space enclosed by control surfaces,fixed
relative to observer
? Control Volume boundary is fixed
? Mass can flow in or out of C.V,
? Used in Continuity and Momentum equations
Flow Analyses
Eulerian Method
Fluid motion and properties (pressure,density,velocity
etc) are described as functions of space and time
Lagrangian Method
Follows the motion of individual fluid particles
determines how the fluid properties of the particles
change with time
Flow kinematics
Streamline,
Imaginary line through fluid such that at an instant,
velocity of ever particle on the line is tangent to it
Stream-tube
Imaginary tube formed by all streamlines passing
through a closed curve,No fluid can enter or leave a
stream tube except through its ends
Used in continuity equation
Continuity Equation
For a fixed,non-deforming control volume,
conservation of mass requires the time rate of
change of mass inside the control volume,plus
net rate of mass flow through the control surface
must be zero
For steady flow,time rate of change of mass in
control volume is 0
Where = mass flow rate in and out of control
volume in kg/s
0??
?
? ??
CS
n
CV
dAvdV
t
??
0? ?? ??? ino u t
CS
n mmdAv ???
m?
Continuity Equation
For uniform normal flow across a control surface
of area A,
= ? AV = ? Q
Mass inflow into CV = Mass outflow from CV
In terms of mean density and velocity,
?1 A1 V1 = ?2 A2 V2
For incompressible fluid,? is constant
A1 V1 = A2 V2 = Q
Q = volume flow rate along stream tube
Control volume
Stream tube
?2,A2,V2 ?1 A1 V
1
m?
Continuity Equation
For steady flow,conservation of mass requires,
For incompressible fluid,? constant,
? Qin = ? Qout
A1V1 + A2V2 = A3V3 + A4V4
V is uniform and perpendicular to A
A1,V1
A2,V2
A3,V3
A4,V4
? ?? inout mm ??
Continuity Equation
Summary,
Apply continuity equation to a Control Volume,
? Qin =? Qout
? AinVin =? AoutVout
Assumptions,
? ? is constant
? Flow is steady
? V is uniform across A
? V is perpendicular to A
Example
Find Q,VA,VC?
Apply continuity equation to CV1,
AA VA = AB VB
? 0.12/4 VA = ? 0.0052/4 x 3
VA = 0.0075 m/s
Apply continuity equation to CV2,
AC VC = AB VB
2 ? (0.15) x 0.005 x VC = ? 0.0052/4 x 3
VC = 0.0125 m/s
Q
VA
VC
0.1 m diam
0.005 m diam
Circular plate,0.3 m diam
VB=3 m/s
0.005 m thick
CV1
CV2
Continuity Equation
For steady flow,conservation of mass,
For incompressible fluid,? constant,
? Qin = ? Qout
A1V1 + A2V2 = A3V3 + A4V4
V is uniform and perpendicular to A
A1,V1
A2,V2
A3,V3
A4,V4
? ?? inout mm ??
Bernoulli Equation
Free body diagram of a fluid particle along a streamline
For steady flow (?v/ ?t =0),apply Newton’s 2nd Law of
Motion to free body,
? ?Fs = ?M as = ? ?V v (?v / ?s)
The following equation of motion can be derived,
- ?g sin ? - ?p / ?s = ? v ?v / ?s
Bernoulli Equation
- ?g sin ? - ?p/?s = ? v ?v/ ?s
As sin ? = dz/ds,
- ?g dz/ds - ?p/?s = 1/2 ? ?v2/?s
Deleting ?s,
?g dz + dp + 1/2 ? ?(v2) = 0
Euler’s equation
Integrating along a streamline,
?gz + p + 1/2 ?(v2) = constant
Bernoulli equation,
? Along a streamline
? Steady flow
? Frictionless
? Incompressible
Bernoulli Equation
Energy per unit volume (J/m3)
p + ?(v2)/2 + ?gz = constant
Pressure Kinetic Potential Total mechanical
energy
Energy per unit mass (J/kg)
p/ ? + (v2)/2 + gz = constant
Pressure Kinetic Potential Total M.E,
Energy per unit weight (J/N or m)
p/ ?g + v2/(2g) + z = constant
Pressure Kinetic Potential Total head
Bernoulli Equation
Apply Bernoulli’s equation to two points along a
streamline,
P1/(?g)+V12/(2g)+z1 = P2/(?g)+V22/(2g) +z2
If there is energy loss in motion of fluid from point 1 to
point 2,
P1/(?g) + V12/(2g) + z1 =
P2/(?g) + V22/(2g) + z2 + hL
hL = head loss due to friction from 1 to 2
Datum
1
2
V1
V2
z1
z2 P1
P2
Example
Apply continuity equation between Sections A & B,
AAVA = ABVB
VB = (0.15/0.1)2 4 = 9 m/s
Apply Bernoulli’s equation between points A & B,
PA/?g + VA2/2g + ZA = PB/?g + VB2/2g + ZB
80000/9810 + 16/(2g) = PB/9810 + 81/(2g)
PB = 47.5 kPa
B
0.1m diam
P=? kPa
A
0.15m diam
V=4m/s
P=80 kPa
10 m
Example
Apply continuity equation between Sections A & B,
AAVA = ABVB
VB = (0.15/0.1)2 4 = 9 m/s
Apply Bernoulli’s equation between points A & B,
PA/?g + VA2/2g + ZA = PB/?g + VB2/2g + ZB
80000/9810 + 16/(2g) + 10 = PB/9810 + 81/(2g)
PB = 145.6 kPa
B
0.1m
dia
m
P=
kP
a
A 0.15m
dia
m
V=
4m
/s
P=
80
kP
a 10 m
Example
Compute Q from orifice
Apply Bernoulli’s equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
0 + 0 + 3 = 0 + V22/2g + 0
V2 = (2 g 3)0.5 = 7.672 m/s
Q = A2 V2 = ? 0.12/4 x V2 = 0.0603 m3/s
3m
?
0.1 m diam
Q=?
1
2
Example
Compute required D for Q to be doubled
Apply Bernoulli equation to 1 and 3,
P1/?g + V12/2g + Z1 = P3/?g + V32/2g + Z3
V3 = (2 g 3)0.5 = 7.672 m/s
Q3 = A3 V3 = ? D2/4 x 7.672 = 2x0.0603 m3/s
D = 0.141 m
A2V2 = A3V3
V2 = 15.36 m/s
Apply Bernoulli equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
P2 = -88.5 kPa
3m
?
0.1 m diam
Q1=2Q
D=?
1
2
3
Example
Find V1,V2 and P
Apply Bernoulli equation to 4 and 1,
P1/?g + V12/2g + Z1 = P4/?g + V42/2g + Z4
V1 = (2 g 10)0.5 = 14.01 m/s
Apply Bernoulli equation to 4 and 2,
P2/?g + V22/2g + Z2 = P4/?g + V42/2g + Z4
V2 = (2 g 10)0.5 = 14.01 m/s
Apply continuity equation,
V3 = 2 V1 = 2 V2 = 28.01 m/s
Apply Bernoulli equation to 4 and 3,
P3 = -294.18 kPa
?
10m
V1=? V2=?
P=? 1 2 3
4
Example
Compute Q and PB
Apply Bernoulli equation to 1 and 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
V2 = (2 g 4)0.5 = 8.859 m/s
Continuity between B and 2,
ABVB = A2V2
VB = 2.215 m/s
Apply Bernoulli equation to 1 and B,
P1/?g + V12/2g + Z1 = PB/?g + VB2/2g + ZB
PB= -22.07 kPa
?
4m
2m
0.02m diam
0.04 m diam
Q=?
PB=?
1
2
B
Example
What V1 is needed to suck oil up h m?
If P1/?g > -5m water,what is max permissible h?
P1 = - ?ogh (1)
Continuity equation between 1 & 2,
V2 = (D1/D2)2 V1 (2)
Bernoulli equation between 1 & 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2 (3)
Substitute (1) & (2) into (3) and solve for V1,
V1 = [2g?ohD24 / ?(D24 - D14)]0.5
P1 = - ?ogh = - 5 ?wg
h = (?w/ ?o) 5
D1 D2
h
Atmosphere V2 V1
Oil,?o=800 kg/m3
Water
2
1
Example
Apply Bernoulli equation to 3 & 2,
P3/?g + V32/2g + Z3 = P2/?g + V22/2g + Z2
+ 2 V12/2g + 0.1 V22/2g (1)
From continuity equation,
A1V1 = A2V2
V1 = (0.05/0.1)2 V2 = 0.25 V2 (2)
Substitute (2) into (1),
4 = (1 + 0.125 + 0.1) V22/2g
V2 = 8.004 m/s
Q = A2 V2
?
V2 V1
0.1 m
diam
0.05 m
diam
Head loss
2V12/2g Head loss 0.1V22/2g
4m
P=?
3
2
Stagnation Point
Static pressure at (3) P3 = ?g h4-3
Static pressure at (1) P1 = ?g h
At point 2,V2 = 0,stagnation point
From Bernoulli’s equation,
P2 = P1 + ? ? V12
P2 – P1 = ? ? V12 = dynamic pressure at the
stagnation point
Pitot Static Tube
For measurement of fluid velocity,
Measures the difference between static (P1) and
stagnation (P2) pressures,
P2 – P1 = ? ? V2
V = [2 (P2 – P1) / ?]1/2
Kinetic energy correction factor ?
Applies when flow velocity V is not uniform across
flow area
For flow along a circular pipe,
Total kinetic energy flow rate from integration of
actual velocity profile
V dA
Actual velocity
distribution
Assumed
uniform
distribution
Vm
Kinetic energy correction factor ?
For laminar flow in pipe,? ? 1.5
For turbulent flow in pipe,? ? 1.1
dAVV
dAAV
fa c t o rc o rr e c t i o nA p p l y
Vm
ekT o t a l
dAekT o t a l
dA
Qmd
V d AQ
m
m
m
3
A
A
33
3
m
2
m
A
3
3
2
)/( ( 1 / A )
V )2/1( )2/1(
AV )2/1()2/1(
Von b a s e d f l o w of,,
V ( 1 / 2 ),,
V )2/1(
V md ( 1 / 2 ) d ( k, e, )
V d A
?
?
?
?
?
??
?
?
?
???
??
?
???
?
?
?
?
??
?
?
?
Energy and Hydraulic Grade Line
Energy or Total Head Line (EL or THL)
Plot of total head (energy per unit weight,
J/N) (z+P/?g +V2/2g) along a stream line
For steady,inviscid and incompressible flow,
THL remains constant along a streamline
Hydraulic Grade Line (HGL)
Plot of piezometric head (z+P/?g) along a
stream line
Distance of V2/2g below THL
Energy and Hydraulic Grade Line
Power in fluid flow
Power = energy per unit time
=? g Q H (watts or J/s)
Q = flow rate in m3/s
H = total head in m
Quiz
Important Notice,
The first compulsory Quiz will be conducted
during your normal tutorial class in the week
from,
covering tutorials 1 to 4,
Students who miss the quiz without approved leave will
be awarded zero mark
The CA component from two quizzes could make up to
30 % of final mark for the subject
Linear Momentum Equation
For Computation of hydro-dynamic forces
When gate is closed,force on gate is due to
pressure, hydrostatic thrust
When gate is partially open,force on gate is due to
hydrostatic pressure and momentum change,
hydro-dynamic force
?
FS
Gate
closed
?
FD
Gate partially open
Linear Momentum Equation
Hydrodynamic force is governed by Newton’s
second law of motion,Apply law to a system of
fixed mass m kg
F = m.a = m, dV/ dt
= d (m.V)/ dt
= d M / dt
Force = rate of change of momentum of system
System, fixed mass m
Force F Velocity V
Acceleration a
Flow rate calculations
Volume flow rate Q = AV
Mass flow rate = ? A V
Momentum flow rate = V
= ? A V2
= ? Q V
= ? Q2/A
Flow area A
Uniform mean
normal velocity V
m?
m? M?
Momentum equation applied to Control Volume
F = net force on fluid in C.V,in x-direction
Rate of increase of momentum of fluid in C.V,
= F + ?AinVin2
Rate of decrease of momentum of fluid in C.V,
= ?AoutVout2
Net rate of change of momentum of fluid in C.V,
= F + ?AinVin2-?AoutVout2
= 0 for steady flow
Hence,
F = ?AoutVout2 - ?AinVin2
Vout
Aout
Vin
Ain F
Control
Volume
Mass of fluid in
C.V,= m
x
Linear Momentum Equation
In equation form,
For steady flow,
Along a given direction,
In words,
Net force on fluid inside CV
= Momentum outflow rate from CV
- Momentum inflow rate into CV
inoutn MMF ?? ??
? ?? ????
CV
n
CS
CVinF l u i d dAvvdVvtF ??
0?
?
?
?
CV
dVv
t
?
Net force acting on fluid inside C.V,
Net force acting on a body of fluid
= sum of forces acting on its surfaces
+ its self weight
On surface in contact with solid boundary,in
general force can be resolved into two
components,pressure and shear,
?
W
F3
F4
F5
F6
F7 F8
Momentum Correction Factor ?
Applies when velocity across a flow area is not
uniform,e.g,flow along circular pipe
It can be shown that total momentum flow
rate given by
M = ? ? A Vm2
Where,
V dA
Actual velocity
distribution
Assumed
uniform
distribution
Vm = Q/A
dA
V
v
A
A
m
2
1
? ??
?
?
??
?
?
??
Example
Jet impact force on hemisphere = 0.5g N
Reaction force on jet = 0.5g N
From Bernoulli equation,
Vi = Vo = V
Apply Momentum equation to CV in
Fn = -
- 0.5g = - ? Q V - ? Q V = -2 ? Q V
V = 1.533 m/s
V12 - V2 = 2 g h
V1 = Q/A = 3.26 m/s
h = 0.42 m
0.5 g N
h=?
Q=0.0016m3/s
0.025 m diam
Control volume Mass = 0.5 kg Fj =0.5g
Vo Vi
V1
outM? inM?
Example
Continuity equation 1 & 2,
A1V1 = A2V2
V2 = 9 V1
Bernoulli equation 1 & 2,
P1/?g + V12/2g + Z1 = P2/?g + V22/2g + Z2
Solving,
V1 = 3.873 m/s,V2 = 34.857 m/s
Apply Momentum equation in
Fn = -
F1 - Fn - F2 = ? A2 V22 - ? A1 V12
600,000 x ? x 0.32/4 - Fn - 0 = ? A2 V22 - ? A1 V12
Fn = 33.929 kN
0.3m dia 0.1m dia P=600 kPa
Fn on nozzle
Water jet V2
F2=0
V1
F1=P1A1 Fn
outM? inM?
Example
Continuity equation 1 & 2,
A1V1 = A2V2
V2 = 1.2 / 0.8 x 2.5 = 3.75 m/s
F1 = hydrostatic thrust = ?gAhc
= 1000 x 9.81 x 1.2 x 0.6 = 7063.2 N
F2 = ?gAhc = ?gx 0.8 x 0.4 = 3139.2 N
Apply Momentum equation in
Fn = -
F1 -FD - F2 = ? A2 V22 - ? A1 V12
FD = 175 N
?
FD
0.8m 1.2m
V1=2.5m/s
FD
V2
Control volume
F1 F
2
outM? inM?
Example
From Bernoulli equation,
V1 = V2 = V3 = V
Apply momentum equation along t
Fn = -
- Ft = ?CQV - ?(1-C)QV - ?QVCos?
For smooth plane,Ft = 0
C = (1 + Cos?)/2
Apply momentum equation along n
Fn = -
Fn = 0 - (-?QVSin?) = ?QVSin?
?
Q,V1
CQ,V2
(1-C)Q,V3
Smooth plate
Jet sheet
Fn
Ft
Ft
Fn
outM? inM?
outM? inM?
Example
Continuity equation between 1 & 2,
A1 V1 = A2 V2
V1 = 0.03/0.32 V2 = 0.225 m/s
Momentum equation along
Fn = -
F1 - Fg - F2 = ? A2 V22 - ? A1 V12
?g 0.32 x 0.16 - Fg - ?g 0.03 x 0.015 =
? 0.03 x 2.42 - ? 0.32 0.2252
Fg = 341.3 N
?
?
Fg
0.32 m
V2=2.4 m/s
0.03m
Fg V
1 F1
F2
Control volume
outM? inM?
