EXAMPLE: Water Flow in a Pipe

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.

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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

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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).

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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

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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 =

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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

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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

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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

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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.

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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 )

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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

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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.

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NON-NEWTONIAN EFFECTS

EXAMPLE: Poiseuille Flow in a Circular Pipe Newtonian Velocity Profile: vz =

 ∆P R2  1 − (r/R)2 4µL

Shear Rate: γ˙ = −

∂vz ∆P r = ∂r 2µL

Apparent Viscosity: where γ˙ is higher

Real Velocity Profile: Lower ηa increases vz non-parabolic

Viscosity is lower