Elementary Mechanics of Fluids

CE 319 F

Daene McKinney

Description of Motion

Fluid Motion

• Two ways to describe fluid motion– Lagrangian

• Follow particles around

– Eularian• Watch fluid pass by a

point or an entire region

– Flow pattern• Streamlines – velocity

is tangent to them

kjiVdt

dz

dt

dy

dt

dx

kjiV wvu

Flow Patterns

• Uniform flow

• Non-uniform flow

• Steady flow

• Unsteady flow

0s

V

0s

V

0t

V

0t

V

Example (4.1)

• Valve at C is opened slowly

• Classify the flow at B while valve is opened

• Classify the flow at A

HW (4.2)

Laminar vs Turbulent Flow

• Laminar • Turbulent

Flow Rate

• Volume rate of flow– Constant velocity over

cross-section

– Variable velocity

• Mass flow rate

VAQ

AVdAQ

QVdAVdAmAA

Examples

• Prob. 4.17 Discharge in a 2-cm pipe is 0.03 m3/s. What is the average velocity?

smd

Q

A

QV

VAQ

/611.0)25.0(

4

03.0

422

• Prob. 4.20 A pipe whose diameter is 8 cm transports air with a temp. of 20oC and pressure of 200 kPa abs. At 20 m/s. What is the mass flow rate?

skg

VAm

mkgRT

p

/239.0)08.0(4

*20*378.2

/378.2293*287

200000

2

3

Flow Rate

• Only x-direction component of velocity (u) contributes to flow through cross-section

AV

V

Q

or

dAQ

or

dAVudAVdAQ

A

A AAcos

Example (4.24)

• Find:

)1()(R

rVrv o

oV

V

3

13

13

1

)32

(2)32

(2

2)/1(

2

2

2

22

0

32

0

o

o

oo

o

o

R

o

R

oA

VR

RVVA

Q

V

V

RV

RRV

R

rrV

rdrRrVVdAQ

Example (4.28)

• Find: mVQ ,,

skgQm

smA

QV

smy

ydyVdAQ

/65*2.1

/51

5

/52

40

2022

35.0

0

2

5.0

0

5.0

0

HW (4.30)

Acceleration

• Acceleration = rate of change of velocity

• Components: – Normal – changing direction

– Tangential – changing speed

dt

dV

dt

dV

dt

d

tsV

tt

t

ee

Va

eV

),(

nt

nt

r

V

t

V

s

VV

r

V

dt

dt

V

s

VV

dt

dV

eea

ee

2)(

Acceleration

• Cartesian coordinates

• HW (4.43)

kjiV

wvu

Convective Local

t

wwz

wvy

wux

w

t

w

dt

dz

z

w

dt

dy

y

w

dt

dx

x

w

dt

dwa

t

vwz

vvy

vux

v

t

v

dt

dz

z

v

dt

dy

y

v

dt

dx

x

v

dt

dva

t

uwz

uvy

uux

u

t

u

dt

dz

z

u

dt

dy

y

u

dt

dx

x

u

dt

dua

z

y

x

.

.

.

kjiazyx aaa

Example (4.49)

2

2

22

1

22

1

1

/49.72*743.3

/55.2)1()5.0(

4

5.0

4

)/(

/4743.3)5.0(

4

)5.0(5.0985.0

4

/2

5.0985.0

sms

VVa

smtd

Q

t

AQ

t

Va

smd

t

tQQ

A

QV

sms

V

tt

tQQQ

C

o

L

oo

oo

HW (4.50 & 4.51)

Example

a

kjiV

on,Accelerati:Find

3:Given 2tyxzt

kjikjia

)2()3(3 22 yxyzttxytzaaa zyx

2;;3 tywxzvtu

222

22

2

2)(0)(2)3(0

30)()(0)3(

33)(0)(0)3(0

yxyztytyxztytt

wwz

wvy

wux

wa

txyzttyxxztzt

vwz

vvy

vux

va

tyxztt

uwz

uvy

uux

ua

z

y

x