Elementary Mechanics of Fluids

CE 319 F

Daene McKinney

Viscosity

Some Simple Flows• Flow between a fixed and a moving plate

Fluid in contact with the plate has the same velocity as the plate

u = x-direction component of velocity

u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )( Fluid

Some Simple Flows

• Flow through a long, straight pipeFluid in contact with the pipe wall has the same velocity as the wall

u = x-direction component of velocity

rx

R

21)(

R

rVru

VFluid

Fluid Deformation

• Flow between a fixed and a moving plate

• Force causes plate to move with velocity V and the fluid deforms continuously.

u=VMoving plate

Fixed plate

y

xu=0

Fluid

t0 t1 t2

u=V+VMoving plate

Fixed plate

y

x

u=V

Fluid

Fluid Deformation

t t+t

x

y

L

t Shear stress on the plate is proportional

to deformation rate of the fluid

V

Lt

y

L

y

V

t

y

V

Shear in Different Fluids

• Shear-stress relations for different typesof fluids

• Newtonian fluids: linear relationship

• Slope of line (coefficient of proportionality) is “viscosity”

dy

dV

dy

dV

Viscosity

• Newton’s Law of Viscosity

• Viscosity

• Units

• Water (@ 20oC)– = 1x10-3 N-s/m2

• Air (@ 20oC)– = 1.8x10-5 N-s/m2

• Kinematic viscosity

dydV /

2

2

//

/

m

sN

msm

mN

V

V+dv

dy

dV

Flow between 2 plates

u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )( Fluid Force acting

ON the plate

21

21

222111

AA

FAAF

221

1 dy

du

dy

du

Thus, slope of velocity profile is constant and velocity profile is a st. line

Force is same on topand bottom

Flow between 2 plates

u=VMoving plate

Fixed plate

y

x

V

u=0

B yB

Vyu )(

B

V

dy

du Shear stress anywherebetween plates

Shearon fluid

mB

smV

CSAEmsN o

02.0

/3

)38@30(/1.0 2

2

2

/15

)02.0

/3)(/1.0(

mN

m

smmsN

Flow between 2 plates

• 2 different coordinate systems

rx

B

21)(

B

rVru

Vy

x yByCyu )(

Example: Journal Bearing

• Given– Rotation rate, = 1500 rpm

– d = 6 cm

– l = 40 cm

– D = 6.02 cm

– SGoil = 0.88

– oil = 0.003 m2/s

• Find: Torque and Power required to turn the bearing at the indicated speed.

Example: cont.

• Assume: Linear velocity profile in oil film

2/1242/)0002.0(

)2/06.0(1500*60

2

)003.0*998*88.0(

2/)(

)2/( StressShear

mkN

dD

d

dy

dV

mN

dl

dM

2812

06.0)4.0*

2

06.0*000,124*2(

2)

22(Torque

kWsmNMP 1.44/100,441.157*281Power

Example: Rotating Disk

• Assume linear velocity profile: dV/dy=V/y=r/y

• Find shear stress