EXAMPLE: Water Flow in a Pipe
P1 > P 2
Velocity profile is parabolic (we will learn why it is parabolic
later, but since friction comes from walls the shape is intuitive)
The pressure drops linearly along the pipe.
Does the water slow down as it flows from one end to the other?
Only component of velocity is in the x-direction.
~v = vx~i
vy = vz = 0
Incompressible Continuity:
∂vx ∂vy ∂vz
+
+
=0
∂x
∂y
∂z
∴
∂vx
∂x
= 0 and the water does not slow down.
1
EXAMPLE: Flow Through a Tank
V = constant (always full)
Integral Mass Balance:
R
S
(~v · ~n)dA = 0
v 1 A 1 = v 2 A2 ≡ Q
Constant volumetric flow rate Q.
EXAMPLE: Simple Shear Flow
vy = vz = 0 vx = vx (y)
~ · v ⇒ ∂vx + ∂vy + ∂vz = 0
∇
∂x
∂y
∂z
satisfied identically
2
NAVIER-STOKES EQUATIONS (p. 1)
(in the limit of slow flows with high viscosity)
Reynolds Number:
Re ≡
ρvD
η
(1-62)
ρ = density
η = viscosity
v = typical velocity scale
D = typical length scale
For Re 1 have laminar flow (no turbulence)
∂~v
~ + ρ~g + η∇2~v
= −∇P
∂t
Vector equation (thus really three equations)
ρ
The full Navier-Stokes equations have other nasty inertial terms that are
important for low viscosity, high speed flows that have turbulence (airplane
wing).
3
NAVIER-STOKES EQUATIONS (p. 2)
ρ
∂~v
~ + ρ~g + η∇2~v
= −∇P
∂t
∂~v
= acceleration
∂t
ρ=
ρ
∂~v
force
=
∂t
unit volume
mass
unit volume
(F~ = m~a) Newton’s 2nd Law
Navier-Stokes equations are a force balance per unit volume
What accelerates the fluid?
~ = Pressure Gradient
−∇P
ρ~g = Gravity
η∇2~v = Flow (fluid accelerates in direction of increasing velocity gradient.
Increasing ∇~v ⇒ ∇2~v > 0
4
GENERAL FLUID MECHANICS
SOLUTIONS
Navier-Stokes equations + Continuity + Boundary Conditions
Four coupled differential equations!
Always look for ways to simplify the problem!
EXAMPLE: 2D Source Flow
Injection Molding a Plate
1. Independent of time
2. 2-D ⇒ vz = 0
3. Symmetry ⇒ Polar Coordinates
4. Symmetry ⇒ vθ = 0
Continuity equation
~ · ~v =
∇
1 d
(rvr )
r dr
=0
rvr = constant
constant
r
Already know the way velocity varies with position, and have not used
the Navier-Stokes equations!
vr =
5
EXAMPLE: Poiseuille Flow between Parallel Plates
(important for injection molding) (P. 1)
Independent of time
vy = vz = 0
Cartesian coordinates
Continuity:
∂vx
= 0 vx = vx (y)
∂x
Navier-Stokes equation:
−
∂ 2 vx
∂P
+µ 2 =0
∂x
∂y
∂P
∂P
=
=0
∂y
∂z
P = P (x) vx = vx (y)
∂P
∂ 2 vx
=µ 2
∂x
∂y
How can f (x) = h(y)? Each must be constant!
∂P
= C1 P = C1 x + C2
∂x
B.C.
x = 0 P = P1 ⇒ C2 = P1
x = L P = P2 ⇒ C1 = −∆P/L
x
P = P1 − ∆P
L
6
where : ∆P ≡ P1 − P2
EXAMPLE: Poiseuille Flow between Parallel Plates
(important for injection molding) (P. 2)
µ
∂ 2 vx
= C1 = −∆P/L
∂y 2
∂ 2 vx
∆P
=−
2
∂y
µL
∂vx
∆P
=−
y + C3
∂y
µL
vx = −
B.C. NO SLIP
∆P 2
y + C3 y + C4
2µL
top plate y = d/2 vx = 0
bottom plate y = −d/2 vx = 0
0=
−∆P 2
d
d + C3 + C4
8µL
2
0=
−∆P 2
d
d − C3 + C4
8µL
2
∴ C3 = 0
∆P
vx =
2µl
d2
− y2
4
C4 =
∆P d2
8µL
Parabolic velocity profile
7
EXAMPLE: Poiseuille Flow between Parallel Plates
(important for injection molding) (P. 3)
Where is the velocity largest?
x
= 0 = − ∆P
y
Maximum at ∂v
∂y
µL
maximum at y = 0 centerline
What is the average velocity?
vave
R
Z
vx dA
1
A
=
vx dA
A = zd
vave = R
A A
dA
A
Z z Z d/2
Z
1 d/2 ∆P d2
1
2
vx dydz =
− y dy
=
zd 0 −d/2
d −d/2 2µL 4
d/2
∆P d2
y3
∆P d2
vave =
y−
=
2µLd 4
3 −d/2
12µL
For constant ∆P , µ, L: double d ⇒ quadruple v
8
EXAMPLE: Poiseuille Flow in an Annular Die
(important for blow molding) (P. 1)
P1 > P 2
Independent of Time
Cylindrical Coordinates
vr = vθ = 0
vz = vz (r)
Continuity:
∂vz
∂z
=0
Navier-Stokes equation:
1 ∂
∂vz
∂P
=µ
r
∂z
r ∂r
∂r
f (z) = g(r) = a constant
Split into two parts - Pressure Part:
∂P
= C1
P = C1 z + C2
∂z
B.C.
z = 0 P = P2 ⇒ C2 = P2
z = L P = P1 ⇒ C1 = ∆P/L
P = P2 + ∆P
z
L
P = P2 +
∆P
z
L
where : ∆P ≡ P1 − P2
analogous to Poiseuille flow between parallel plates.
9
EXAMPLE: Poiseuille Flow in an Annular Die
(important for blow molding) (P. 2)
1 ∂
µ
r ∂r
r
∂vz
r
∂r
=
∆P
L
∂vz
∆P 2
=
r + C3
∂r
2µL
∂vz
∆P
C3
=
r+
∂r
2µL
r
vz =
B.C. NO SLIP
at
at
∆P 2
r + C3 ln r + C4
4µL
r = Ri ,
r = R0 ,
0=
vz = 0
vz = 0
∆P 2
R + C3 ln Ri + C4
4µL i
∆P 2
R + C3 ln R0 + C4
4µL 0
∆P
0 = 4µL
(R02 − Ri2 ) + C3 ln RR0i
0=
subtract
C3 = −
∆P (R02 − Ri2 )
4µL ln(R0 /Ri )
∆P
(R02 − Ri2 ) ln R0
2
C4 = −
R0 −
4µL
ln(R0 /Ri )
10
EXAMPLE: Poiseuille Flow in an Annular Die
(important for blow molding) (P. 3)
∆P 2 (R02 − Ri2 )
(R02 − Ri2 )
2
r −
ln r − R0 +
vz =
4µL
ln(R0 /Ri )
ln(R0 /Ri )
∆P R02
r2
(R02 − Ri2 )
vz =
−1 + 2 −
ln(r/R0 )
4µL
R0 ln(R0 /Ri )
r < R0 always, so vz < 0
Leading term is parabolic in r (like the flow between plates) but this one
has a logarithmic correction.
What is the volumetric flow rate?
Z
Q=
Z
A
Q=
π∆P R04
8µL
R0
vz dA =
"
−1 +
vz 2πrdr
Ri
Ri
R0
11
4
2 2
+
(1 − (Ri /R0 ) )
ln(R0 /Ri )
#
GENERAL FEATURES OF NEWTONIAN
POISEUILLE FLOW
Parallel Plates:
∆P d3 W
Q=
12µL
Circular Tube:
Q=
Annular Tube:
Q=
π∆P R4
8µL
π∆P R04
f (Ri /R0 )
8µL
Rectangular Tube:
Q=
∆P d3 w
12µL
All have the same general form:

Q ∼ ∆P 
Q ∼ 1/µ Weak effects of pressure, viscosity and flow length

Q ∼ 1/L
Q ∼ R4 or d3 w
Strong effect of size.
In designing and injection mold, we can change the runner sizes.
12
NON-NEWTONIAN EFFECTS
EXAMPLE: Poiseuille Flow in a Circular Pipe
Newtonian Velocity Profile:
vz =
∆P R2 1 − (r/R)2
4µL
Shear Rate:
γ̇ = −
∂vz
∆P r
=
∂r
2µL
Apparent Viscosity:
where γ̇ is higher
Real Velocity Profile:
Lower ηa increases vz
non-parabolic
Viscosity is lower