Basic differential equations in fluid mechanics

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)

Why Differential Equations? Why Differential Equations?

A droplet of waterCloudsWall jetHurricane

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

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

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

Continuity IllustratedContinuity Illustrated

ux

vy

wz

0

x

y

What must be happening?

0v

y

0u

x

\< >

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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)

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

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?

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

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

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!)

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

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

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