Seoul National University Lecture 4 Steady Flow in Pipes (1)ocw.snu.ac.kr/sites/default/files/NOTE/HD Ch4-1-LC4_0.pdf · 2020. 10. 7. · Seoul National University Text Ch. 9 Flow

Seoul National University

Lecture 4

Steady Flow in Pipes (1)


Text Ch. 9 Flow in Pipes

Steady flow

9.1 Fundamental equations

9.2 Laminar flow

9.3 Turbulent flow – Smooth pipes

9.4 Turbulent flow – Rough pipes

9.5 Classification of smoothness and roughness

9.6 Pipe friction factors

9.7 Pipe friction in noncircular pipes

9.8 Pipe fiction – Empirical formulation

9.9 Local losses in pipelines

9.10 Pipeline problems – Single pipes

9.11 Pipeline problems – Multiple pipes

2

LC 4

LC 6

LC 5

LC 7

LC 8


3

Laminar flow

- Shear stress- Velocity profile- Head loss- Friction factor

Turbulent flow – smooth pipe

- Velocity profile- Friction factor

𝑓~𝑓𝑛(𝑅𝑒,𝑑𝑣

𝑑)

Pipe friction

- Blasius-Stanton diagram- Moody diagram for

commercial pipes- Empirical formula

Local losses

- Enlargement & contraction- Entrances- Bends, elbows, valves

Pipe problems – single pipe

- Work-energy equation- Continuity equation- Calculation of head loss,

flow rate, pipe diameter

Pipe problems – pipe network

- Three reservoir problem- Pipe networks- Hardy Cross method

Fundamental eq.

- Energy equation- Darcy-Weisbach

equation

Turbulent flow – rough pipe

- Velocity profile

- 𝑓~𝑓𝑛𝑑

𝑒

- Colebrook Eq. for commercial pipes

Outline of Pipe Flow


Contents

4.0 Applications

4.1 Fundamentals Equations

4.2 Laminar Flow

Review the shear stress and head loss

Understand laminar flows and friction relating problems.

Objectives


Water supply system

5

4.0 Applications


Water system

6


7

Water pipe Water pipe in Libya

Water pipe in Libya (L=1,872 km; d =4 m; Q=4,000,000 ton/d)


8

Chemical pipe Gas/Oil pipe


Diffuser system Wastewater discharge from STP

Heated water discharge from power plants

Cooled water discharge from LNG terminals

Brine discharge from desalination plants

9


Heated water discharge from power plants

10


Heated water diffuser

11


Cooled water diffuser

12


RO desalination plant (Tampa Bay, US)

13


Pipe flow

14


Pipe flow vs Open-channel flow

15


4.1 Fundamentals Equations

Newton’s 2nd law of motion → Momentum eq.

In a pipe flow (Ch. 7; p. 260), apply momentum eq.

– where P is wetted perimeter

16

pA- p+ dp( )A-t oPdl - g +dg

2

æ

èçö

ø÷Adl

dz

dl= V + dV( )

2A r + dr( ) -V 2Ar

Pressure

force

Shear

force

Gravitational

force

h

h

A AP R

R P

out in

F Q v Q v


Dividing by specific weight and neglecting small terms yields

– For incompressible fluids

Integrating from 1 to 2 to yield

17

dp

g+ d

V 2

2gn

æ

èçö

ø÷+ dz = -

t 0dl

g Rh

dp

g+V 2

2gn+ z

æ

èçö

ø÷= -

t 0dl

g Rh

2 2 0 2 11 1 2 21 2

2 2n n h

p pz z

l lV V

g Rg

Lh


The drop in the energy line is called head loss.

In incompressible flow

For pipe flow,

18

1 2

2 2

1 1 2 21 2

2 2L

n n

p pz

g gh

Vz

V

1 2

0 2 1 0

L

h h

l lh

R

l

R

1 2 1 2

02

L h Lh R h R

l l

2

2 2h

A R

R

RR

P

(4.1)

(4.2)


Work-energy equation

Energy correction factor can be ignored

– In turbulent flow (~1), in the most engineering problem

– In laminar flow, when energy correction factor is large,

but the velocity heads are usually negligible

– In most case, velocity head is very small compared to

other terms

19

z1 +p1

g+a1

V12

2gn= z2 +

p2

g+a2

V22

2gn+ hL (4.3)


Derivation of Darcy-Weisbach equation

using Dimensional analysis

Find head loss equation for pipe flow

In smooth pipe, problem parameters are

– Head loss, hL

– Pipe length, l

– Pipe diameter, d

– Density,

– Viscosity, m

– Gravity, g

– Velocity, V

20

, , , , , , 0Lf h d l V g m


1. V, d, and do not combine, choose as a repeating variable; k=3

2. In this case, n=7, n-k=4

21

0

3

0

2 2

0 0

1 1

1

0 0

2

2

2

3

: , , ,

: ,

1, 1

, ,

2, 1, 0, 1

a d

a d

c

b

c

b

L M MM L t f V d L

t L Lt

L M LM L t f V d g L

t L t

Vda b c d

Va b c d

gd

m

m

2 2

3 3 4

1 1

4

, , , ; , , ,

, , , ; , , , L

f V d f V d g

f V d l f V d h

m

Apply Buckingham P theory


22

2

, ,Lh l V

d d gd

Vdf

m

0

3 3 3

3

0

4

0 0

0 0

1 3

4

0, , 0

0, , 0,

: , , ,

: , , ,

c

b

c

ad

ad

L

b

L

L MM L t f V d l L L

t L

L MM L t f V d h L L

t L

la b d c

d

ha b d c

d

2 2'

2 2L

l V l Vh

d

Vd

gf f

d g

m

' '(Re)

Vdf f f

m


From experiments, using a dimensionless coefficient of proportionality,

f called the friction factor, Darcy, Weisbach and others proposed

(Darcy-Weisbach equation) in long straight, uniform pipes

From momentum equation,

Two equations can be combined (D=2R, Rh=R/2)

23

2

2L

l Vf

dh

g

1 2

0

L

h

lh

R

t o =f rV 2

8 (4.5)

(4.4)


In the previous fundamental equation relating wall shear to friction

factor, density and mean velocity, it is apparent that f is

dimensionless.

Then must have the dimension of velocity.

Friction (shear) velocity is defined as

Then we have

24

t o / r

0

*8

Vf

v

*

8V

v f

(4.6)


I.P.

Water flows in a 150mm diameter pipeline at a mean velocity of 4.5

m/s. The head lost in 30 m of this pipe is measured experimentally

and found to be 5.33 m. Calculate the friction velocity in the pipe.

~ 5.8% of mean velocity

25

f =2gn

V 2D

LhL =

2 ´ 9.81

4.5m / s( )2

0.150 m

30 m5.33m = 0.026

v* =Vf

8= 4.5m / s

0.026

8= 0.26m / s

2

2L

l Vf

dh

g


4.2 Laminar Flow

Characteristics of the laminar flow in pipe

– Symmetric distribution of shear stress and velocity

– Maximum velocity at the center of the pipe and no velocity at

the wall (no-slip condition)

– Linear shear stress distribution (Eq. 7.37)

26


For laminar flow, combine Eq. 2 and Newton’s viscosity equation

Integrating once w.r.t. r yields

27

t =g hL2l

æ

èçö

ø÷r = m

dv

dy= -m

dv

dr

t 0 =g hL2l

æ

èçö

ø÷R (at the wall)

dv

dr= -

1

mt = -

1

m

g hL2l

æ

èçö

ø÷r = -

1

m

t 0Rr = -

t 0r

mR

2

0

2v c

r

R

m


Apply the no-slip boundary condition at r=R,

Then,

At the center of pipe

Then

28

2

0 R

2

τ0=- +c

μR

2 20v =2

- rRR

m

2

0

cv = ( 0)2

when rR

R

m

2

2v = 1-c

rv

R

Paraboloid

→ Hagen-Poiseuille flow

(A)

(4.7)


Apply the friction velocity into (A)

When y is small (near the wall), 2nd term is negligible, then velocity

profile has a linear relationship with distance from the wall.

29

22 2 2 2* *

*

222*

*

vv = - = -

v

v= - = R-y )

v 2

2 2

(where r

R Rv v

r r

v yy

R

R R

* 0v = τ /ρ

*

*

v

v ν

v y

2=kinematic viscosity (m /s)

m

(4.9)

(4.8)


From Eq. A

We can get flow rate

Since

30

3

R2 20 0

0 0-v

4Q= 2 =

R

rdr R rR

drR

r

m m

L0

γhτ

R=

2l

4

2

42

2

,8

32

128

8

L L

L L

h d hQ Q AV R

l l

R h

R

d h

V

Vl l

m m

m m

2 20v = 2

-RR

r

m


(4.10)

For laminar flow, head loss varies with the first power of the velocity.(Fig. 7.3

of p. 232)

These facts of laminar flow were established experimentally by Hagen

(1839) and Poiseuille (1840). → Hagen-Poiseuille law

31

2

32L

lVh

d

m


Equating the Darcy-Weisbach equation for head loss to Eq. 4.10 yields an

expression for the friction factor

(4.11)

In laminar flow, friction factor only depends on the Reynolds number.

32

64 64

ReVdf

m

2

2L

l Vf

dh

g

2

32L

lVh

d

m


I.P. 9.3 (p.329)

A fluid flows from a large pressurized tank through a 100 m long, 4 mm

diameter tube. In a 600 sec time period, 1,300 cm3 of fluid are collected in a

measuring cup. If the head loss in the tube is 1 m, calculate the kinematic

viscosity . Check to verify that the flow is laminar.

33

[Solution]

4

4

4

128

128

128

L

L

L

d hQ

l

d h

d

l

Q

Q

l

hg

m

m


34

For water at 20°C;

→ Laminar flow

Documents

Seoul National University Lecture 4 Steady Flow in Pipes (1)ocw.snu.ac.kr/sites/default/files/NOTE/HD Ch4-1-LC4_0.pdf · 2020. 10. 7. · Seoul National University Text Ch. 9 Flow