CBE 150A – Transport Spring Semester 2014 Friction Losses Flow through Conduits Incompressible Flow

CBE 150A – Transport Spring Semester 2014

Friction LossesFlow through Conduits

Incompressible Flow


Goals

• Calculate frictional losses for laminar and turbulent flow through circular and non-circular pipes

• Define the friction factor in terms of flow properties

• Calculate the friction factor for laminar and turbulent flow

• Define and calculate the Reynolds number for different flow situations

• Derive the Hagen-Poiseuille equation


Flow Through Circular Conduits

Consider the steady flow of a fluid of constant density in fully developed flow through a horizontal pipe and visualize a disk of fluid of radius r and length dL moving as a free body. Since the fluid posses a viscosity, a shear force opposing the flow will exist at the edge of the disk


Balances

Mass Balance

→222111 SVSV 21 VV


Balances

4

2

12

DppFw

Momentum Balance

gw FFSpSpVVm 22111122

HorizontalSSS 1212


Momentum Balance (contd)If we imagine that the fluid disk extends to the wall, Fw is just due to the shear stress τw acting over the length of the disk.

Equating and solving for p over a length of pipe L.

ww LDF )(

wD

Lp

4


Mechanical Energy Balance

fh

pzg

VW

2

ˆ2

H o r iz o n t a lW 120ˆ

p

h f


Viscous Dissipation (Frictional Loss) Equation

Combining the Momentum and MEB results:

• Applies to laminar or turbulent flow• Good for Newtonian or Non-Newtonian fluids• Only good for friction losses as result of wall shear.

Not proper for fittings, expansions, etc.

wf D

Lh

4


The Friction Factor

w is not conveniently determined so the dimensionless friction factor is introduced into the equations.

headvelocitydensity

stressshearwall

Vf w

22


Fanning Friction Factor

• Increases with length• Decreases with diameter• Only need L, D, V and f to get friction loss• Valid for both laminar and turbulent flow• Valid for Newtonian and Non-Newtonian fluids

–

24

2V

D

Lfh f


Calculation of f for Laminar Flow

wD

Lp

4

22

2rR

Ru w

x

First we need the velocity profile for laminar flow in a pipe. We’ll rely on Chapter 8 for that result.

2221 1

4 R

r

L

Rppux

Recall our earlier result: LRppw 221


Laminar FlowFind Bulk Velocity (measurable quantity).

S

dSuV S

2

02

2

R

drur

R

drru

R

S

R

w drrrRR 0

323

24maxw uR

V


Reynolds Number

forceviscous

forceinertiaVDN

Re

Osbourne Reynolds (1842-1912)


Laminar Flow

4

22

RVand

Vf ww

RVf

8

←Laminar Flow←Newtonian FluidRe

f16


Hagen-Poiseuille (Laminar Flow)

Recall again:

RL

ppw 2

21

Use: Measurement of viscosity by measuring p and q through a tube of known D and L for Laminar flow.

L

DppV

32

221

L

ppRRVSVq

82142


Turbulent FlowWhen flow is turbulent, the viscous dissipation effects cannot be derived explicitly as in laminar flow, but the following relation is still valid.

24

2V

D

Lfh f

The problem is that we can not write a closed form solution for the friction factor f. Must use correlations based on experimental data.


Friction FactorTurbulent Flow

For turbulent flow f = f( Re , k/D ) where k is the roughness of the pipe wall.

Note, roughness is not dimensionless. Here, the roughness is reported in inches.

MaterialRoughness, k

inches

Cast Iron 0.01

Galvanized Steel 0.006

Commercial SteelWrought Iron

0.0018

Drawn Tubing 0.00006


How Does k/D Affect f(Text Figure 13.1)



As and alternative to Moody Chart use Churchill’s correlation:

16

16

9.0

121

23

12

37530

27.07

1ln457.2

182

ReB

DkReA

BARef



A less accurate but sometimes useful correlation for estimates is the Colebrook equation. It is independent of velocity or flow rate, instead depending on a combined dimensionless quantity

.Re f

f

Dk

f Re

255.1

7.3log4

1


Flow Through Non-Circular Conduits

Rather than resort to deriving new correlations for the friction factor, an approximation is developed for an ‘equivalent’ diameter Deq with which to calculate the Reynolds number and the friction factor.

pHeq LSRD 44

where:• RH = hydraulic radius• S = cross-sectional area• Lp = wetted perimeter

Note: Do not use Deq to calculate cross-sectional area or for laminar flow situations.


ExamplesCircular Pipe

DRR

RDeq

2

24

2

Rectangular Ducts

WH

WH

WH

WHDeq 2

24


Example 1

Water flows horizontally at a rate of 600 gal/min through 400 feet of 5 in. diameter Schedule 40 cast-iron pipe. Find the average (bulk) velocity and the pressure drop.

400 ft.

5 in.

600 GPM


Text Appendix M


10 Minute ProblemMy father is installing a sprinkler system at his lake house. The pump pulls water from the lake through a feed line and delivers 12 GPM to the sprinkler system distribution line at a point in the front yard. For the sprinkler system to operate properly, the pressure at the branch point must be 90 psig. What horsepower pump does he buy ?

40 ft.

25 ft.

10 ft.Tubing lengths:Lake to pump suction – 50 ft.Pump to distribution line – 150 ft.

Documents

CBE 150A – Transport Spring Semester 2014 Friction Losses Flow through Conduits Incompressible Flow