Lecture 4Steady Flow in Pipes (1)

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~𝑓_𝑛(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~𝑓_𝑛 ^𝑑

- Colebrook Eq. for 𝑒

commercial pipes

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 2

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}

^{Pdl - g + d g}

2 æ èç ö

ø÷ Adl dz

dl = ( V + dV )

^A ( ^{r + d r} ) ^{- V}

^{A r}

Pressure force

Shear force

Gravitational force

h h

A A

P R

R  P

 

  



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

V²

2g

èç

ö

ø÷ + dz = -

t

g

d p

g

V²

2g

èç

ö

ø÷ = -

t

g

 

2 2

0 2 1

1 1 2 2

1 2

2 _n 2 _{n h}

p p

z z l l

V V

g g R





 

 



 

hL

 The drop in the energy line is called head loss.

 In incompressible flow

 For pipe flow,

18

1 2

2 2

1 1 2 2

1 2

2 _n 2 _n ^L

p p

g z g h

V V z

 

   

     

   

   

 

1 2

0 2 1 0

L

h h

l l

h R

l R

 

 



  

1 2 1 2

0 2

L h L

h R h R

l l

 



2

2 2

h

A R

R R R

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

a

2g

g

a

2g

 Derivation of Darcy-Weisbach equation using Dimensional analysis

Find head loss equation for pipe flow In smooth pipe, problem parameters are

– Head loss, h_L – Pipe length, l – Pipe diameter, d – Density, 

– Viscosity, m – Gravity, g – Velocity, V

20





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 M

M L t f V d L

t L Lt

L M L

M L t f V d g L

t L t

a b c d Vd

a b c d V

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

, ,

hL l V

d d gd

f Vd m

 

  

 

   

   

0

3 3 3

3

0 4

0 0

0 0

1 3

4

0, , 0

0, , 0,

: , , ,

: , , ,

c b

c a

d

a

d L

b

L

L M

M L t f V d l L L

t L

L M

M L t f V d h L L

t L

a b d c l

d

a b d c h

d













   

      

   

   

     



 

   

    

2 2

'

2 2

L

l V l V

h d

Vd

f g f

d g

 m

 

   

 

' '

Vd (Re)

f 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, R_h=R/2)

23 2

L

2

f l V h d

 g

1 2

0 L

h

h l

R





 

t

= f r ^V

8

(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

^/ r

0

* V 8f

v







*

8 V

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 =

2g

D

L h_L =

2

9.81 4.5m / s

( )

0.150 m

30 m 5.33 m

0.026

8

4.5m / s 0.026

8

0.26m / s

2 L

2

f l V h d

 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 ^h

2l æ

èç ö

ø÷ r = m ^dv

dy = - m ^dv dr t

= g ^h

2l æ

èç ö

ø÷ R (at the wall) dv

dr = - 1

m t = - 1 m

g ^h

2l æ

èç ö

ø÷ r = - 1 m

t

R r = - t

^r m ^R

2 0

v r 2 c

R

 m

 

    

 

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

Then,

At the center of pipe

Then

28 2

0

R

2

0=- τ +c

μR

 

 

 





v =

2 R - r R

 m

2 0

v =

( 0)

2 when r

R

 R

m ^

2

v =

1 - r

v R

 

 

 

Paraboloid

→ Hagen-Poiseuille flow (A)

(4.7)

Apply the friction velocity into (A)

When y is small (near the wall), 2^nd term is negligible, then velocity profile has a linear relationship with distance from the wall.

29

   

 

2

2 2 2 2

* *

*

2 2 2

*

*

v = - v = -

v

v = - = R-y )

v 2

2 2

(where r

R R

v v

r r

v y

y R

R R

 





 

 

 

* 0

v = τ /ρ

*

*

v

v ν

 v y

=kinematic viscosity (m /s)

 m

 

(4.9) (4.8)

From Eq. A

We can get flow rate

Since

30

   

R 0 2 2 0

0

v

-

Q= 2 rdr =

R r 4 R

R r dr

 

 m ^ m

 

L 0

τ γh R

= 2l

4

2

4

2

2

8 ,

32 128

8

L L

L L

h d h

Q Q AV R

l l

R h R

d h

V

V l l

    

m m

 

m m

 





 





v =

2 R -

R 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

32

L

h lV

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

Re f Vd m

  

2 L

2

f l V h d

 g

2

32

L

h lV

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 h

Q l

d h

d l Q

Q l h g m

 





m



 





34

For water at 20°C;

→ Laminar flow