basic differential equations in fluid mechanics
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OverviewOverview
Continuity EquationNavier-Stokes Equation
(a bit of vector notation...)Examples (all laminar flow)
Flow between stationary parallel horizontal plates
Flow between inclined parallel platesPipe flow (Hagen Poiseuille)
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Why Differential Equations? Why Differential Equations?
A droplet of waterCloudsWall jetHurricane
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Conservation of Mass in Differential Equation Form
Conservation of Mass in Differential Equation Form
v x z
x
y
z
FHG
IKJy
y
t
y x z
Mass flux into differential volume
Mass flux out of differential volume
Rate of change of mass in differential volume
vvy
y
FHG
IKJ x z
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Continuity EquationContinuity Equation
vy
vy t
Mass flux out of differential volume
vvy
y vy
yvy y
y x z
FHG
IKJ
2 Higher order term
out in Rate of mass decrease
vy t
0 1-d continuity equation
vvy
y vy
y x z v x zt
y x z
FHG
IKJ
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u, v, w are velocities in x, y, and z directions
Continuity EquationContinuity Equation
0t
V
t
u
x
v
y
w
z
af af a f0 3-d continuity equation
If density is constant...
ux
vy
wz
0
Vector notation
0 Vor in vector notation
True everywhere! (contrast with CV equations!)
divergence
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Continuity IllustratedContinuity Illustrated
ux
vy
wz
0
x
y
What must be happening?
0v
y
0u
x
\< >
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Shear
GravityPressure
Navier-Stokes EquationsNavier-Stokes Equations
momentum
Derived by Claude-Louis-Marie Navier in 1827
General Equation of Fluid MotionBased on conservation of ___________ with forces…
__________________________________________________
U.S. National Academy of Sciences has made the full solution of the Navier-Stokes Equations a top priority
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If _________ then _____
Navier-Stokes Equation
Inertial forces [N/m3], a is Lagrangian acceleration
Pressure gradient (not due to change in elevation)
Shear stress gradient
Navier-Stokes EquationsNavier-Stokes Equations
V a F2p a g V
a
p g
2 Vdu
dx
Is acceleration zero when V/ t = 0?
V
2p a g V
g is constanta is a function of t, x, y, z
NO!
0p g 0V
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Lagrangian acceleration
Notation: Total DerivativeEulerian Perspective
Notation: Total DerivativeEulerian Perspective
u v wt x y z
V V V V
a
( , , , )D
t x y z u v wDt t x y z
( , , , )D
t x y z u v wDt t x y z
V V V V V
t
V
a V V
kjizyx
()()()
()
() () ()() u v w
x y z
V
( , , , )D dt dx dy dz
t x y zDt t dt x dt y dt z dt
Total derivative (chain rule)
Material or substantial derivative
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Application of Navier-Stokes Equations
Application of Navier-Stokes Equations
The equations are nonlinear partial differential equations
No full analytical solution existsThe equations can be solved for several
simple flow conditionsNumerical solutions to Navier-Stokes
equations are increasingly being used to describe complex flows.
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0 p g
Navier-Stokes Equations: A Simple Case
Navier-Stokes Equations: A Simple Case
No acceleration and no velocity gradients
p gy C
p g
x y z
p p pg g g
x y z
0 0p p p
gx y z
xyz could have any orientation
Let y be vertical upward
g
For constant
2p a g V
Component of g in the x,y,z direction
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Infinite Horizontal Plates: Laminar Flow
Infinite Horizontal Plates: Laminar Flow
2p a g V
20 p g V
Derive the equation for the laminar, steady, uniform flow between infinite horizontal parallel plates.
2
20
p u
x y
0p
gy
0 0
2 2 2
2 2 20 y
p v v vg
y x y z
2 2 2
2 2 20 z
p w w wg
z x y z
2 2 2
2 2 20 x
p u u ug
x x y z
y
x
Hydrostatic in y0v =
0w =
x
y
z
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Infinite Horizontal Plates: Laminar Flow
Infinite Horizontal Plates: Laminar Flow
2
20
p u
x y
2
2
dp d u
dx dy
2
2
dp d udy dy
dx dy
dp duy A
dx dy
dp duy A dy dy
dx dy
2
2
y dpAy B u
dx
Pressure gradient in x balanced by shear gradient in y
du
dy
d
dy
No a so forces must balance!
Now we must find A and B… Boundary Conditions
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negative
Infinite Horizontal Plates: Boundary Conditions
Infinite Horizontal Plates: Boundary Conditions
No slip condition
u = 0 at y = 0 and y = a a
B 0
a dpdx
Aa2
20 A
a dpdx
2
2
y y a dpu
dx
2
a dpy
dx
y
du dpy A
dy dx
dpdx
let be___________
u
2
2
y dpAy B u
dx
What can we learn about ?
x
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Laminar Flow Between Parallel Plates
Laminar Flow Between Parallel Plates
2p a g V
20 p g V
2
20 x
p ug
x y
2 2 2
2 2 20 x
p u u ug
x x y z
U
a
u
y
x
No fluid particles are accelerating
Write the x-component
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Flow between Parallel Plates Flow between Parallel Plates
2
20 x
p ug
x y
2
20 x
dp d ug
dx dy
General equation describing laminar flow between parallel plates with the only velocity in the x direction
u is only a function of y
ˆxg g i
2
2 x
d u dpg
dy dx
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Flow Between Parallel Plates: Integration
Flow Between Parallel Plates: Integration
2
2 x
d u dpg
dy dx
x
du dpy g A
dy dx
2
2 x
d u dpdy g dy
dy dx
x
du dpdy y g A dy
dy dx
2
2 x
y dpu g Ay B
dx
U
a
u
y
x
t
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u = U at y = a
Boundary ConditionsBoundary Conditions
2
2 x
y dpu g Ay B
dx
Boundary condition
B 000
Boundary condition
2
2 x
a dpU g Aa
dx
2 x
U a dpA g
a dx
2
2 x
Uy y ay dpu g
a dx
u = 0 at y = 0
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Discharge per unit width!
DischargeDischarge
2
00
2
aa
x
y y ay dpq udy U g dy
a dx
3
2 12 x
Ua a dpq g
dx
2
2 x
y y ay dpu U g
a dx
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Example: Oil SkimmerExample: Oil Skimmer
An oil skimmer uses a 5 m wide x 6 m long moving belt above a fixed platform (=60º) to skim oil off of rivers (T=10 ºC). The belt travels at 3 m/s. The distance between the belt and the fixed platform is 2 mm. The belt discharges into an open container on the ship. The fluid is actually a mixture of oil and water. To simplify the analysis, assume crude oil dominates. Find the discharge and the power required to move the belt.
hl
= 1x10-2 Ns/m2
= 860 kg/m3 60ºxg
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Example: Oil SkimmerExample: Oil Skimmer
dp
dx ˆ cos(60) 0.5xg g g g i
m 0.002a m/s 3U
3
2 3
-2 2
(3 m/s)(0.002 m) (0.002 m)0.5 9.806 m/s 860 kg/m )
2 12 1x10 N s/mq
In direction of beltq = 0.0027 m2/s (per unit width)
Q = 0.0027 m2/s (5 m) = 0.0136 m3/s
3
2 12 x
Ua a dpq g
dx
0
dominates
60ºxg
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How do we get the power requirement?___________________________
What is the force acting on the belt? ___________________________
Remember the equation for shear?_____________ Evaluate at y = a.
Example: Oil Skimmer Power Requirements
Example: Oil Skimmer Power Requirements
x
du dpy g A
dy dx
2 x
U a dpA g
a dx
2 x
a dp Uy g
dx a
Power = Force x Velocity [N·m/s]
Shear force (·L · W)
=(du/dy)
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Example: Oil Skimmer Power Requirements
Example: Oil Skimmer Power Requirements
2 x
a dp Uy g
dx a
cos602
a Ug
a
cos 60x
dpg g
dx
22
23 2
3 m N s1x10
0.002 m 860 kg Ns m9.8 m/s 0.5 19.2
2 m 0.002 m m
Power LWU
sm 3
m 5m 6mN
19.2Power 2
(shear by belt on fluid)
= 3.46 kW
FV
How could you reduce the power requirement? __________Decrease
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Potential and kinetic energy
Heating the oil (thermal energy)
Example: Oil Skimmer Where did the Power Go?
Example: Oil Skimmer Where did the Power Go?
Where did the energy input from the belt go?
h = 3 m
P Qh
m 3s
m0.0136
mN
84303
3
P
W443P
Potential energy
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Velocity ProfilesVelocity Profiles
-2
-1
0
1
2
3
0 0.0005 0.001 0.0015 0.002
y (m)
u (m
/s)
oil
water
2
2 x
y y ay dpu U g
a dx
Pressure gradients and gravity have the same effect.
In the absence of pressure gradients and gravity the velocity profile is ________linear
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Example: No flowExample: No flow
Find the velocity of a vertical belt that is 5 mm from a stationary surface that will result in no flow of glycerin at 20°C (m = 0.62 Ns/m2 and =1250 kg/m3)
Draw the glycerin velocity profile.What is your solution scheme?
3
2 12 y
Ua a dpq g
dy
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Laminar Flow through Circular Tubes
Laminar Flow through Circular Tubes
Different geometry, same equation development (see Munson, et al. p 327)
Apply equation of motion to cylindrical sleeve (use cylindrical coordinates)
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Max velocity when r = 0
Laminar Flow through Circular Tubes: Equations
Laminar Flow through Circular Tubes: Equations
2 2
4l x
r R dpv g
dx
2
max 4 x
R dpv g
dx
2
8 x
R dpV g
dx
4
8 x
R dpQ g
dx
Velocity distribution is paraboloid of revolution therefore _____________ _____________
Q = VA =
average velocity (V) is 1/2 vmax
VR2
R is radius of the tube
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Laminar Flow through Circular Tubes: Diagram
Laminar Flow through Circular Tubes: Diagram
Velocity
Shear (wall on fluid)2
lx
dv r dpg
dr dx
2l
x
dv r dpg
dr dx
2lghr
l
0 4lgh d
l
True for Laminar or Turbulent flow
Shear at the wall
Laminar flow
2 2
4l x
r R dpv g
dx
Next slide!
Remember the approximations of no shear, no head loss?
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cv energy equation
Relationship between head loss and pressure gradient for pipesRelationship between head loss and pressure gradient for pipes
1 21 2
1 2l
p pz z h
g g Constant cross section
2 21 1 2 2
1 1 2 21 22 2p t l
p V p Vz h z h h
g g g g
2 1 2 1lgh p p gz gz
lh p zg g
x x x
lx
h pg g
x x
lx
h dpg g
l dx
l is distance between control surfaces (length of the pipe)
In the energy equation the z axis is tangent to g
x is tangent to V
x
zg g
x
x
z
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The Hagen-Poiseuille EquationThe Hagen-Poiseuille Equation
4
128lghD
Ql
2
32lghD
Vl
Hagen-Poiseuille Laminar pipe flow equations4
8 x
R dpQ g
dx
From Navier-Stokes
4
8lhR
Q gl
lx
h dpg g
l dx
Relationship between head loss and pressure gradient
What happens if you double the pressure gradient in a horizontal tube? ____________flow doubles
V is average velocity
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Example: Laminar Flow (Team work)
Example: Laminar Flow (Team work)
Calculate the discharge of 20ºCwater through a long vertical section of 0.5 mm ID hypodermic tube. The inlet and outlet pressures are both atmospheric. You may neglect minor losses.
What is the total shear force?
What assumption did you make? (Check your assumption!)
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Example: Hypodermic Tubing Flow
Example: Hypodermic Tubing Flow
QD h
Ll
4
128
QN m m
x Ns m
9806 0 0005
128 1 10
3 4
3 2
/ .
/
c hafa fc h
Q x m s 158 10 8 3. /
VQd
4
2V m s 0 0764. /
ReVd
3
3 2
0.0764 / 0.0005 1000 /Re
1 10 /
m s m kg m
x Ns m
Re 38
Q L s 158. /
ldhl
40
ldhrl
F lshear 4
2
2shear lF r h
= weight!
2 21 1 2 2
1 1 2 21 22 2p t l
p V p Vz H z H h
g g
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SummarySummary
Navier-Stokes Equations and the Continuity Equation describe complex flow including turbulence
The Navier-Stokes Equations can be solved analytically for several simple flows
Numerical solutions are required to describe turbulent flows
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GlycerinGlycerin
3
2 12 y
Ua a dpQ g
dy
3
02 12
Ua a g
y
dpg g
dy
2
6
a gU
2 3
2
0.005 12300 /0.083 /
6 0.62 /
m N mU m s
Ns m
RVl kg m m s m
Ns m
1254 0 083 0 005
0 620 8
3
2
/ . / .
. /.
c ha fa f
y