CV2601/G263, Fluid Mechanics
Dr Chiew Yee Meng,Lecturer/Course Coordinator
Office, N1-1b-53
Tel, 67905256
Email, cymchiew@ntu.edu.sg
Dr Shuy Eng Ban,Lecturer
Office, N1-1a-23
Tel, 67905326
Email, cshuyeb@ntu.edu.sg
TEXT
Munson,B R,Young,D F and Okiishi,T H,Fundamentals of Fluid
Mechanics,4th Edition,John Wiley & Son,2002
REFERENCES
? Shames IH,Mechanics of Fluids,3rd Edition,McGraw-Hill,1992,
? Potter MC and Wiggert DC,Mechanics of Fluids,Prentice Hall,1991,
Course Outline
Fluid Mechanics
12 hrs lectures,5 Tutorials,1 Quiz
W e e k
No
C o n t e n t L e c T u t R e f
1 D e f i n i t i o n s o f f l u i d, F l u i d p r o p e r t i e s
F l u i d S t a t i c s, B a s i c e q u a t i o n f o r h y d r o s t a t i c
p r e s s u r e
1
2
C h a p, 1
C h a p, 2
2 P r e s s u r e m e a s u r e m e n t, M a n o m e t e r
c o m p u t a t i o n s
H y d r o s t a t i c t h r u s t o n a p l a n e s u r f a c e,
M a g n i t u d e a n d c e n t r e o f p r e s s u r e
3
4
T1
C h a p, 2
C h a p, 2
3 H y d r o s t a t i c t h r u s t o n c u r v e s u r f a c e s
B u o y a n c y, S t a b i l i t y o f f l o a t i n g b o d i e s
5
6
T2
C h a p, 2
C h a p, 2
4 B a s i c f l u i d f l o w c o n c e p t s, C l a s s i f i c a t i o n o f
f l o w, S y s t e m a n d c o n t r o l v o l u m e, C o n t i n u i t y
e q u a t i o n
E n e r g y e q u a t i o n f o r s t e a d y i n c o m p r e s s i b l e f l u i d
f l o w
7
8
T3
C h a p, 4
C h a p, 5
C h a p, 3
5
B e r n o u l l i ’ s e q u a t i o n a n d i t s a p p l i c a t i o n s,
F l o w m e a s u r i n g d e v i c e s
9
10
T4
C h a p, 3
C h a p, 5
6 M o m e n t u m e q u a t i o n f o r s t e a d y f l o w,
A p p l i c a t i o n s o f t h e m o m e n t u m e q u a t i o n, F o r c e s
o n o b j e c t s
11
12
Q1
C h a p, 5
C h a p, 5
Course Outline
Hydraulics
14 hrs lectures,5 Tutorials,1 Quiz
7 D i m e n s i o n a l a n a l y s i s
B u c k i n g h a m P i t h e o r e m
13
14
T 5
C h a p, 7
C h a p, 7
8 S i g n i f i c a n c e o f c o m m o n d i m e n s i o n a l g r o u p s
S i m i l i t u d e a n d s c a l e m o d e l s
15
16
T6
C h a p, 7
C h a p, 7
9 C o n c e p t s o f B o u n d a r y L a y e r, L a m i n a r f l o w
b e t w e e n p a r a l l e l p l a t e s
L a m i n a r a n d t u r b u l e n t f l o w s i n p i p e s
17
18
T7
C h a p, 8
C h a p, 8
10 E n e r g y c o n c e p t s i n p i p e f l o w s
D a r c y - W e i s b a c h e q u a t i o n
19
20
T8
C h a p, 8
C h a p, 8
11
M o o d y d i a g r a m
F r i c t i o n a n d m i n o r l o s s e s
21
22
T9
C h a p, 8
C h a p, 8
12 P r i n c i p l e s o f f l u i d m a c h i n e s, P e r f o r m a n c e
c h a r a c t e r i s t i c s o f p u m p s,
S i m i l a r i t y l a w s, S p e c i f i c s p e e d a n d m a c h i n e
s e l e c t i o n
23
24
Q2
C h a p, 1 2
C h a p, 1 2
13 S y s t e m c h a r a c t e r i s t i c s a n d m a t c h i n g,
C a v i t a t i o n a n d N P S H,
P a r a l l e l a n d s e r i e s o p e r a t i o n s o f p u m p s,
25
26
T 1 0
C h a p, 1 2
C h a p, 1 2
Assessment
? CA (2 quizes), Up to 30 %
? Final Examination, At least 70 %
? Do Not Skip Lectures
– Highlight Important Topics/Concepts/Equation
– Solve Additional Examples and Past Year
Questions
? 10 Tutorials,Start on Week 2
– Tutorials 1 to 4
– Quiz 1
– Tutorials 5 to 9
– Quiz 2
– Tutorial 10
Why Study Fluid Mechanics?
For Civil Engineers,
? Water Supply, Dams,reservoirs,Treatment and Distribution
network systems
? Drainage and Irrigation Systems, Open Channel Hydraulics
? Environmental Hydraulics, Sewerage Systems,Pollutant
Dispersion Modelling in Air and Water
For Mechanical Engineers,
? Aerodynamics – Airfoil design,Lift and Drag,CFD
? Rotodynamic Machinery – Pumps and Turbines
? Hydraulic/Pneumatic Control Systems
? Industrial Hydraulics, Piping Systems – Gas and Oil
Industries
What is Fluid?
Three States of Matter,
? Solid
? Fluids,
? Liquid
? Gas
Solids,
? Shear Strain = function of Stress
? Body recovers when stress is removed
Fluids,
? Deform continuously when subjected to shear stress
? Rate of Shear Strain = function of Shear stress
? Body does not recover when stress is removed
Differences Between Liquids and Gases
Liquids
? Practically incompressible
? Has a finite volume at given pressure and temperature
Gases
? Highly compressible
? Always expands to fill up container
?
What is Fluid Mechanics?
Application of principles of mechanics to
fluid motion,
? Conservation Laws, Mass,Energy,Momentum
? Newton’s Laws of Motion
? Thermodynamic laws for Gases
Main Areas,
? Fluid Statics, Study of Fluid at rest
? Fluid Kinematics, Study of Fluid motion without
considering forces
? Fluid Dynamics, Study relation between motion and
forces
? Hydraulics, Application of Fluid Mechanics to practical
problems
Dimensions and Units
Fluid characteristics (properties) can be described
qualitatively in terms of certain ‘Basic Dimensions’ or
‘Primary Quantities’,
? Length,L
? Mass,M
? Time,T
? Temperature,?
‘Secondary Quantities’ can be derived in terms of the
‘primary quantities’,e.g,
? Area,L2
? Velocity LT-1
? Density,ML-3
Units are standards for quantitative measurement, m,
s,kg
Forces on a Plane in Fluid
Resultant force FR on a plane can be resolved into,
? perpendicular component FN
? tangential component FT
? Pressure P is defined as,
P = Normal force per unit area = FN / A
? Shear stress ? is defined as,
? = Tangential force per unit area = FT / A
? In Fluid statics,FT = 0,only P exists
FR
FN
FT
Plane Area A
Fluid Properties,Definitions
Density ? = mass/ volume,kg/m3
Specific Volume Vs = 1/?,m3/kg
Specific weight ? = ? g,N/m3
Specific gravity (Relative density) s = ?/?w
Density of water as function of temperature
Viscosity
Shear Deformation/Strain
F
F
y
x = u, t
u
Deformed shape after time t
Original shape at time t = 0
?
F = shear force on top surface of area A
Shear stress ?
= F / A
Shear strain ?
= angular deformation
= x / y = (u.t) / y
Rate of shear strain
= ? / t = (u.t)/(y.t) = u / y
= velocity gradient in direction perpendicular to u (y-
direction)
Newton’s Law of Viscosity
For most fluids,shear stress is proportional to
rate of angular strain,
? ? (?u / ?y)
= ? (?u / ?y)
= ? (u / y)
(Where velocity variation in y direction is linear)
? = property of fluid known as dynamic or absolute
viscosity of fluid,kg/(m.s) or N.s/m2
? = Kinematic viscosity = ? /?,m2/s
Law applies to Newtonian Fluids in laminar
motion
Newtonian Fluids
For Newtonian fluids,
Linear relation
between shear stress
and rate of shearing
strain passing through
origin
Non-Newtonian
fluids
Viscosity of Common Fluids
Linear Velocity Gradient
u/y = V / S V
Stationary plate
S
Moving plate
Moving
piston V
S u/y = V / S
Stationary cylinder
S
R
Rotating
piston
? rad/s
V=R?
u/y = R.? / S
Stationary cylinder
V
S
Moving plate
Moving plate
W u/y = (W+V)/S
Non- Linear Velocity Gradient
y U
Velocity gradient at distance y varies,
?u / ?y = f (y)
Boundary layer flow
Example
1 m/s
F =?
? = 75 mm
Sleeve,? = 75.5 mm
Lubricant ? = 0.19 kg/m.s
L = 150 mm
Velocity gradient u/y = 1/0.00025 = 4000 m/s/m
? = ? (U/y)
= 0.19 x 4000
= 760 N/m2
F = ?,A
= 760 x ?(0.075)0.15
=26.9 N
Piston
Newton’s Law of Viscosity
For Newtonian Fluids in laminar
motion,
? = ? (?u / ?y)
= ? (u / y)
(Where velocity variation is linear)
? = dynamic or absolute viscosity of
fluid,kg/(m.s) or N.s/m2
Example
W = ? g Vol = 1100 (9.81) 0.23 = 86.33 N
? = ? U/y = 0.08 (Vt)/0.000005
F = ? A = 640 Vt = W Sin 10o
Vt = 0.0234 m/s
0.2x0.2x0.2 m cube
? = 1100 kg/m3
10o
Vt =?
0.005 mm gap,oil of ? = 0.08 Ns/m2
W
W Sin ? F
Pressure
Pressure P = normal force per unit area,N/m2
?1 Pascal (Pa) = 1 N/m2
?1 bar = 100,000 N/m2 = 100 kPa
Relative pressures are measured,
?Gauge pressure Pgauge
= pressure above atmospheric Patm
?Absolute pressure Pabs
= Pressure above vacuum
= Pgauge + Patm
Patm ? 1 bar ? 100,000 Pa abs
Vacuum = 0 Pa abs
Compressibility of Liquid
New Pressure P = Po + ?P
New volume V = Vo - ?V
?P ? (- ?V/Vo)
= - K (?V/Vo)
= K (?? / ?o)
where ?? is the increase in density and ?o is the original density
K = Bulk modulus of elasticity of liquid N/m2
C = Compressibility C = 1/K
Original pressure Po
Increase by ?P
Original volume Vo
Decrease by ?V
Compressibility of Liquid
oo
V
V
V
dVd
V
dV
V
dV
V
m
dV
V
m
d
V
m
?
??
?
??
??????
?
?
?
?
?
??
?
2
Example
Calculate density of sea water at 200 m below
surface,given density at surface is 1025 kg/m3,
Kwater = 2.3x109 N/m2
(Ans, 1025.896 kg/m3)
Solution,
At 200 m below surface,
?P = ? g h = 1025 x 9.81 x 200
= K ? ? / ?
? ? = 0.896 kg/m3
?’ = ? + ? ? = 1025.896 kg/m
Surface Tension
Liquid surface behaves like an elastic membrane under
tension
Surface tension ? = tension force per m of membrane,N/m
Water droplet of radius R,
Increase in Pressure due to surface tension ?P = 2?/R
?
Liquid jet of radius R,
Increase in Pressure due to surface tension ?P = ?/R
Capillary Rise/Depression
Cohesion,
?Attraction between liquid molecules
Adhesion
?Attraction between liquid molecules and solid
boundary
Capillary rise/depression
h = ( 4? Cos ? ) / ( ? g D)
? = Contact angle
Dr Chiew Yee Meng,Lecturer/Course Coordinator
Office, N1-1b-53
Tel, 67905256
Email, cymchiew@ntu.edu.sg
Dr Shuy Eng Ban,Lecturer
Office, N1-1a-23
Tel, 67905326
Email, cshuyeb@ntu.edu.sg
TEXT
Munson,B R,Young,D F and Okiishi,T H,Fundamentals of Fluid
Mechanics,4th Edition,John Wiley & Son,2002
REFERENCES
? Shames IH,Mechanics of Fluids,3rd Edition,McGraw-Hill,1992,
? Potter MC and Wiggert DC,Mechanics of Fluids,Prentice Hall,1991,
Course Outline
Fluid Mechanics
12 hrs lectures,5 Tutorials,1 Quiz
W e e k
No
C o n t e n t L e c T u t R e f
1 D e f i n i t i o n s o f f l u i d, F l u i d p r o p e r t i e s
F l u i d S t a t i c s, B a s i c e q u a t i o n f o r h y d r o s t a t i c
p r e s s u r e
1
2
C h a p, 1
C h a p, 2
2 P r e s s u r e m e a s u r e m e n t, M a n o m e t e r
c o m p u t a t i o n s
H y d r o s t a t i c t h r u s t o n a p l a n e s u r f a c e,
M a g n i t u d e a n d c e n t r e o f p r e s s u r e
3
4
T1
C h a p, 2
C h a p, 2
3 H y d r o s t a t i c t h r u s t o n c u r v e s u r f a c e s
B u o y a n c y, S t a b i l i t y o f f l o a t i n g b o d i e s
5
6
T2
C h a p, 2
C h a p, 2
4 B a s i c f l u i d f l o w c o n c e p t s, C l a s s i f i c a t i o n o f
f l o w, S y s t e m a n d c o n t r o l v o l u m e, C o n t i n u i t y
e q u a t i o n
E n e r g y e q u a t i o n f o r s t e a d y i n c o m p r e s s i b l e f l u i d
f l o w
7
8
T3
C h a p, 4
C h a p, 5
C h a p, 3
5
B e r n o u l l i ’ s e q u a t i o n a n d i t s a p p l i c a t i o n s,
F l o w m e a s u r i n g d e v i c e s
9
10
T4
C h a p, 3
C h a p, 5
6 M o m e n t u m e q u a t i o n f o r s t e a d y f l o w,
A p p l i c a t i o n s o f t h e m o m e n t u m e q u a t i o n, F o r c e s
o n o b j e c t s
11
12
Q1
C h a p, 5
C h a p, 5
Course Outline
Hydraulics
14 hrs lectures,5 Tutorials,1 Quiz
7 D i m e n s i o n a l a n a l y s i s
B u c k i n g h a m P i t h e o r e m
13
14
T 5
C h a p, 7
C h a p, 7
8 S i g n i f i c a n c e o f c o m m o n d i m e n s i o n a l g r o u p s
S i m i l i t u d e a n d s c a l e m o d e l s
15
16
T6
C h a p, 7
C h a p, 7
9 C o n c e p t s o f B o u n d a r y L a y e r, L a m i n a r f l o w
b e t w e e n p a r a l l e l p l a t e s
L a m i n a r a n d t u r b u l e n t f l o w s i n p i p e s
17
18
T7
C h a p, 8
C h a p, 8
10 E n e r g y c o n c e p t s i n p i p e f l o w s
D a r c y - W e i s b a c h e q u a t i o n
19
20
T8
C h a p, 8
C h a p, 8
11
M o o d y d i a g r a m
F r i c t i o n a n d m i n o r l o s s e s
21
22
T9
C h a p, 8
C h a p, 8
12 P r i n c i p l e s o f f l u i d m a c h i n e s, P e r f o r m a n c e
c h a r a c t e r i s t i c s o f p u m p s,
S i m i l a r i t y l a w s, S p e c i f i c s p e e d a n d m a c h i n e
s e l e c t i o n
23
24
Q2
C h a p, 1 2
C h a p, 1 2
13 S y s t e m c h a r a c t e r i s t i c s a n d m a t c h i n g,
C a v i t a t i o n a n d N P S H,
P a r a l l e l a n d s e r i e s o p e r a t i o n s o f p u m p s,
25
26
T 1 0
C h a p, 1 2
C h a p, 1 2
Assessment
? CA (2 quizes), Up to 30 %
? Final Examination, At least 70 %
? Do Not Skip Lectures
– Highlight Important Topics/Concepts/Equation
– Solve Additional Examples and Past Year
Questions
? 10 Tutorials,Start on Week 2
– Tutorials 1 to 4
– Quiz 1
– Tutorials 5 to 9
– Quiz 2
– Tutorial 10
Why Study Fluid Mechanics?
For Civil Engineers,
? Water Supply, Dams,reservoirs,Treatment and Distribution
network systems
? Drainage and Irrigation Systems, Open Channel Hydraulics
? Environmental Hydraulics, Sewerage Systems,Pollutant
Dispersion Modelling in Air and Water
For Mechanical Engineers,
? Aerodynamics – Airfoil design,Lift and Drag,CFD
? Rotodynamic Machinery – Pumps and Turbines
? Hydraulic/Pneumatic Control Systems
? Industrial Hydraulics, Piping Systems – Gas and Oil
Industries
What is Fluid?
Three States of Matter,
? Solid
? Fluids,
? Liquid
? Gas
Solids,
? Shear Strain = function of Stress
? Body recovers when stress is removed
Fluids,
? Deform continuously when subjected to shear stress
? Rate of Shear Strain = function of Shear stress
? Body does not recover when stress is removed
Differences Between Liquids and Gases
Liquids
? Practically incompressible
? Has a finite volume at given pressure and temperature
Gases
? Highly compressible
? Always expands to fill up container
?
What is Fluid Mechanics?
Application of principles of mechanics to
fluid motion,
? Conservation Laws, Mass,Energy,Momentum
? Newton’s Laws of Motion
? Thermodynamic laws for Gases
Main Areas,
? Fluid Statics, Study of Fluid at rest
? Fluid Kinematics, Study of Fluid motion without
considering forces
? Fluid Dynamics, Study relation between motion and
forces
? Hydraulics, Application of Fluid Mechanics to practical
problems
Dimensions and Units
Fluid characteristics (properties) can be described
qualitatively in terms of certain ‘Basic Dimensions’ or
‘Primary Quantities’,
? Length,L
? Mass,M
? Time,T
? Temperature,?
‘Secondary Quantities’ can be derived in terms of the
‘primary quantities’,e.g,
? Area,L2
? Velocity LT-1
? Density,ML-3
Units are standards for quantitative measurement, m,
s,kg
Forces on a Plane in Fluid
Resultant force FR on a plane can be resolved into,
? perpendicular component FN
? tangential component FT
? Pressure P is defined as,
P = Normal force per unit area = FN / A
? Shear stress ? is defined as,
? = Tangential force per unit area = FT / A
? In Fluid statics,FT = 0,only P exists
FR
FN
FT
Plane Area A
Fluid Properties,Definitions
Density ? = mass/ volume,kg/m3
Specific Volume Vs = 1/?,m3/kg
Specific weight ? = ? g,N/m3
Specific gravity (Relative density) s = ?/?w
Density of water as function of temperature
Viscosity
Shear Deformation/Strain
F
F
y
x = u, t
u
Deformed shape after time t
Original shape at time t = 0
?
F = shear force on top surface of area A
Shear stress ?
= F / A
Shear strain ?
= angular deformation
= x / y = (u.t) / y
Rate of shear strain
= ? / t = (u.t)/(y.t) = u / y
= velocity gradient in direction perpendicular to u (y-
direction)
Newton’s Law of Viscosity
For most fluids,shear stress is proportional to
rate of angular strain,
? ? (?u / ?y)
= ? (?u / ?y)
= ? (u / y)
(Where velocity variation in y direction is linear)
? = property of fluid known as dynamic or absolute
viscosity of fluid,kg/(m.s) or N.s/m2
? = Kinematic viscosity = ? /?,m2/s
Law applies to Newtonian Fluids in laminar
motion
Newtonian Fluids
For Newtonian fluids,
Linear relation
between shear stress
and rate of shearing
strain passing through
origin
Non-Newtonian
fluids
Viscosity of Common Fluids
Linear Velocity Gradient
u/y = V / S V
Stationary plate
S
Moving plate
Moving
piston V
S u/y = V / S
Stationary cylinder
S
R
Rotating
piston
? rad/s
V=R?
u/y = R.? / S
Stationary cylinder
V
S
Moving plate
Moving plate
W u/y = (W+V)/S
Non- Linear Velocity Gradient
y U
Velocity gradient at distance y varies,
?u / ?y = f (y)
Boundary layer flow
Example
1 m/s
F =?
? = 75 mm
Sleeve,? = 75.5 mm
Lubricant ? = 0.19 kg/m.s
L = 150 mm
Velocity gradient u/y = 1/0.00025 = 4000 m/s/m
? = ? (U/y)
= 0.19 x 4000
= 760 N/m2
F = ?,A
= 760 x ?(0.075)0.15
=26.9 N
Piston
Newton’s Law of Viscosity
For Newtonian Fluids in laminar
motion,
? = ? (?u / ?y)
= ? (u / y)
(Where velocity variation is linear)
? = dynamic or absolute viscosity of
fluid,kg/(m.s) or N.s/m2
Example
W = ? g Vol = 1100 (9.81) 0.23 = 86.33 N
? = ? U/y = 0.08 (Vt)/0.000005
F = ? A = 640 Vt = W Sin 10o
Vt = 0.0234 m/s
0.2x0.2x0.2 m cube
? = 1100 kg/m3
10o
Vt =?
0.005 mm gap,oil of ? = 0.08 Ns/m2
W
W Sin ? F
Pressure
Pressure P = normal force per unit area,N/m2
?1 Pascal (Pa) = 1 N/m2
?1 bar = 100,000 N/m2 = 100 kPa
Relative pressures are measured,
?Gauge pressure Pgauge
= pressure above atmospheric Patm
?Absolute pressure Pabs
= Pressure above vacuum
= Pgauge + Patm
Patm ? 1 bar ? 100,000 Pa abs
Vacuum = 0 Pa abs
Compressibility of Liquid
New Pressure P = Po + ?P
New volume V = Vo - ?V
?P ? (- ?V/Vo)
= - K (?V/Vo)
= K (?? / ?o)
where ?? is the increase in density and ?o is the original density
K = Bulk modulus of elasticity of liquid N/m2
C = Compressibility C = 1/K
Original pressure Po
Increase by ?P
Original volume Vo
Decrease by ?V
Compressibility of Liquid
oo
V
V
V
dVd
V
dV
V
dV
V
m
dV
V
m
d
V
m
?
??
?
??
??????
?
?
?
?
?
??
?
2
Example
Calculate density of sea water at 200 m below
surface,given density at surface is 1025 kg/m3,
Kwater = 2.3x109 N/m2
(Ans, 1025.896 kg/m3)
Solution,
At 200 m below surface,
?P = ? g h = 1025 x 9.81 x 200
= K ? ? / ?
? ? = 0.896 kg/m3
?’ = ? + ? ? = 1025.896 kg/m
Surface Tension
Liquid surface behaves like an elastic membrane under
tension
Surface tension ? = tension force per m of membrane,N/m
Water droplet of radius R,
Increase in Pressure due to surface tension ?P = 2?/R
?
Liquid jet of radius R,
Increase in Pressure due to surface tension ?P = ?/R
Capillary Rise/Depression
Cohesion,
?Attraction between liquid molecules
Adhesion
?Attraction between liquid molecules and solid
boundary
Capillary rise/depression
h = ( 4? Cos ? ) / ( ? g D)
? = Contact angle
