module 3b: flow in pipes - unix.eng.ua.eduunix.eng.ua.edu/~rpitt/class/water resources...

1

Module 3b: Flow in Pipes

Darcy-Weisbach

Robert Pitt

University of Alabama

and

Shirley Clark

Penn State-Harrisburg

Darcy-Weisbach Equation

•Based upon theory.

•Used to estimate energy loss due to friction in pipe.

Where

hf= head loss (feet)

L = length of pipe (feet)

f= friction factor

D = internal diameter of pipe (feet)

V2/2g = velocity head (feet)

=

g

V

DLf

hf

2

2


•Darcy-Weisbach can be written for flow (substitute

V = Q/A, where A = (π/4)D2in the above

equation):

•Darcy-Weisbach can be rewritten to solve for

velocity:

=gQ

DLf

hf

2

528 π

5.0

2

×

××

=f

L

Dg

hV

f


Friction factor for Darcy-Weisbach Equation:

•Based upon the Reynolds number, NRor Re, and a

dimensionless parameter called the relative roughness,

e/dor ε/d(ε= absolute roughness; d = diameter).

•For laminar flow:

•For turbulent flow, friction factor must be read off a

Moody diagram (or off a relative roughness vs. friction

factor diagram for completely turbulent flow).

Re

64

64=

=RN

f

2

Equivalent

sand

roughness

diagram

Moody Diagram


•Example:

–A polymeric coagulant, undiluted, has an absolute viscosity

of 0.48 kg/(m-sec) [0.01 lb-sec/ft2] and a specific gravity of

1.15. This fluid is to be pumped at the rate of 3.78 L/min (1

gallon/min) through 15.25 m (50 ft) of ½-inch diameter

schedule 40 pipe (ID = 0.622 in = 15.8 mm = 0.0158 m).

What is the head loss due to friction?

•To use Darcy-Weisbach to calculate head loss, need f:

=g

V

DLf

hf

2

2


•If the flow is laminar, the friction factor can be calculated.

Otherwise, it must be looked up off the chart.

•Need to determine velocity in order to calculate Reynolds

number (determine if flow is laminar or turbulent).

•By Continuity Equation:

Q = VA

or

V = Q/A

•Substituting:

[]

[]

sec

/32.0

sec

/321

.0

m)

58

01.0)(4/

(

sec)

min/60

L)(1

/1000

m L/min)(1

(3.78

2

3

mVor

mVV

===π

3


•Calculate the Reynolds number to see if flow is

laminar or turbulent.

•Substituting:

Definition of Specific Gravity = Fluid Density/Water Density

Density of Water = ρH2O= 1000 kg/m

3µρ

VD

=Re

15.

12

Re

)sec

kg/m

0.48

(

)kg/m

000

m)(1.15)(1

158

m/sec)(0.0

0.321

(Re

3

=

−=


•HAVE LAMINAR FLOW!!

•Therefore, the friction factor for Darcy-Weisbach is

calculated as follows:

27.5

15.

1264

Re

64

===

ff

f


•Substituting into Darcy-Weisbach equation:

•Friction Slope:

mhh

ff

7.26

)]m/sec

06

m)[(2)(9.8

0.0158

(

)m/sec

m)(0.321

25

(5.27)(15.

2

2

==

%175

75.1

25.

15

7.26

Slope

Friction

==

=mm

Darcy-Weisbach

Equation

•Example:

–A 24-inch class I ductile

iron pipe (ID = 0.63 m =

24.79 in.) 90 m long

with a neat cement

lining (asphalted cast

iron) carries a flow of

water at 1.5 m3/sec (52.9

ft3/sec). What is the

friction loss in the pipe?

–From the relative

roughness diagram

(assume similar to

asphalted cast iron, read

ε/d. ε/d= 0.0002

―――――――――――――――――――――――――

―――――――――――――――――――――――――

4


•Calculate Reynolds number.

Assuming that the fluid is water: ν= 1.003 x 10-6m2/sec at

20oC.

–Find velocity of flow.

–Find the Reynolds number.

()(

)(

)6

26

10

02.3

Re

sec

/10

003

.1

63.0

sec

/81.4

Re

x

mx

mm

VD

=

==

−ν

()

sec

/81.4

63.0

4

sec

/5.1

2

3

mV

m

m

AQV

=

==

π


•Reading from the Moody Diagram:

f(ε/d= 0.0002 & Re = 3.02 x 106) = 0.014

•Substituting into Darcy-Weisbach:

mhh

ff

36.2

)sec

/m 6

80.9)(2)(

m 36.0(

sec)

/m 1

8.4)(

m 09)(

014

.0(

2

2

==

Moody Diagram

―――――――――――――――――――――――――

―――――――――――――――――――――――――


•Example:

•Determine the flowrate in a 500-m section of a 1-m diameter

commercial steel pipe when there is a 2-m drop in the energy

grade line over the section.

Given:

L = 500 m

D = 1 m

Commercial steel pipe

hf= 2 m

•Want to use Darcy-Weisbach equation with flow rate.

=gQ

DLf

hf

2

528 π

5

Darcy-Weisbach

Equation

•Need to find the

friction factor.

–As a first

assumption about

the flow in the

pipe, will assume

fully turbulent

flow. Using

Moody diagram

for relative

roughness in

turbulent flow:

f= 0.0105

―――――――――――――――――――――――――

―――――――――――――――――――――――――


•Solve Darcy-Weisbach for flow rate:

sec

/1472

.2

)500

)(0105

.0(

8

)1)(

sec

/81.9(

)2(

:

888

3

2/1

52

2

2/1

52

52

2

2

52

mQ

m

mm

mQ

ng

Substituti

fLgD

hQ

fLgD

hQ

gQ

DLf

h

ff

f

=

=

==

=

π

πππ


•Must check assumption of fully turbulent flow (i.e.,

was the fselected from the figure acceptable?).

•Using the continuity equation:

()

sec

/7339

.2

0.1

4sec

/1472

.2

4

23

2

mV

mV

m

VD

VA

Q

=

=

==

π

π


•Calculate the Reynolds number. (Note that no

temperature is given for the fluid. Assume the fluid is

water and the temperature is 15oC).

At 15oC, ν= 1.139 x 10-6m2/sec

•Substituting:

6

26

10

40.2

Re

sec

/10

139

.1

)0.

1sec)(

/7339

.2(

Re

Re

x

mx

mm

VD

===

−

ν

6

Darcy-Weisbach

Equation

•In order to use the

full Moody

diagram, need the

relative

roughness. For

commercial steel,

ε= 0.000046 m.

•Calculate:

ε/D = (0.000046 m)/

(1.0 m) = 0.000046

―――――――――――――――――

Moody Diagram: Draw imaginary line interpolating at

ε/D = 0.000046 ―――――――――――――――――――――――――

―――――――――――――――――――――――――

Using the full Moody diagram, for Re = 2.40 x 106

and ε/D = 0.000046:

f= 0.0115


•Substituting back into Darcy-Weisbach for flow:

sec

/05.2

)500

)(0115

.0(

8

)1)(

sec

/81.9(

)2(

:

8

3

2/1

52

2

2/1

52 m

Q

m

mm

mQ

ng

Substituti

fLgD

hQ

f

=

=

=

π

π

Are we done?


•To check this flow rate, repeat the process.

•By continuity:

•Calculate the Reynolds number associated with this velocity:

sec

/61.2

)0.1(

4

sec

/05.2

2

3

mV

m

m

AQV

=

==

π

6

26

10

3.2

Re

sec

/10

139

.1

)0.1

sec)(

/61.2(

Re

x

mx

mm

VD

=

==

−ν

7

Moody Diagram: Draw imaginary line interpolating at

ε/D = 0.000046

―――――――――――――――――――――――――

―――――――――――――――――――――――――

Using the full Moody diagram, for Re = 2.30 x 106

and ε/D = 0.000046:

f≈0.0115


•Friction factors approximately the same between last

two iterations. Can use value from previous iteration

as Q. Q = 2.05 m3/sec

Friction factors approximately the same between

last two iterations.

Are we done?

Yes.


Example:

•A 14-inch schedule 80 pipe (commercial steel) has

an inside diameter of 12.5 in (317.5 mm). How much

flow can this pipe carry if the allowable head loss is

3.5 m in a length of 200 m?

•Need to solve for V and A to get Q, the flow.

Darcy-Weisbach

Equation

•In order to use the

full Moody

diagram, need the

relative

roughness. For

commercial steel,

ε= 0.000046 m.

•Calculate:

ε/D = 0.0046 mm/

317.5 mm =

0.000014

―――――――――――――――――

8

Refer to Moody Diagram ―

――――――――――――――――――――――――

――――――――――――――――――――――――――――――――

For this ε/d, franges from 0.0087 to 0.04, depending on the flow

(expressed as the Reynolds number).

Assume f= 0.01 (in range near low end). If f= 0.01, then the

Reynolds number is approximately 4 x 106.

――――――――――――――――――――――――――――――――

――――――――――――――――――――――――――――――――


•Substituting this into Darcy-Weisbach:

Are we done?sec

/30.3

)]01.0)(

m 020

[(

)]m 5

317

.0)(

sec

/m 6

80.9)(2)(

m 5.3[(

][

]2

[

5.0

5.0

2

5.0

5.0

mVV

Lf

Dg

hV

f

==

×

××

=


•Now need to check the friction factor assumption:

–Calculate Reynolds number for V = 3.30 m/sec.

6

26

10

04.1

Re

sec

/10

003

.1

)3175

.0

sec(

/30.3

Re

Re

x

mx

mm

VD

===

−

ν


――――――――――――――――――――――――

――――――――――――――――――――――――――――――――

From the Moody diagram, the ffor this Reynolds number is 0.012.

――――――――――――――――――――――――――――――――

――――――――――――――――――――――――――――――――

9


•Recalculate using Darcy-Weisbach:

sec

/01.3

)]012

.0)(

m 020

[(

)]m 5

317

.0)(

sec

/m 6

80.9)(2)(

m 5.3[(

][

]2

[

5.0

5.0

2

5.0

5.0

mVV

Lf

Dg

hV

f

==

×

××

=


•Calculate the Reynolds number for V = 3.01 m/sec.

CLOSE ENOUGH, So pipe will carry:

6

26

10

95.0

Re

sec

/m

10

003

.1

)m 5

317

.0

sec)(

/m 1

0.3(

Re

Re

x

x

VD

===

−

ν

Are we done?

Yes.

()

sec

/238

.0

)3175

.0(

4sec

/01.3

3

2

mQ

mm

VA

Q

=

==

π

Darcy-Weisbach

Equation

Example:

•Determine the head

loss in a 46-cm

concrete pipe with

an average velocity

of 1.0 m/sec and a

length of 30 m.

•Calculate ε/d.

D = 46 cm = 0.46 m

ε= 0.001 ft (0.3048

m/ft) = 0.000305 mm

Therefore ε/d=

0.0003 m/0.46 m =

0.00066

―――――――――――――――――


•Calculate Reynolds number: 5

26

10

58.4

Re

sec

/m

10

003

.1

)m

46.0

sec)(

/m 1(

Re

Re

x

x

VD

===

−

ν

10


――――――――――――――――――――――――

――――――――――――――――――――――――――――――――

From the Moody diagram, the ffor this Reynolds number is 0.019.

――――――――――――――――――――――――――――――――

――――――――――――――――――――――――――――――――


•Substituting into Darcy-Weisbach:

Are we done?

Yes.

mh

m

m

mmh

g

V

DLf

h

LLL

063

.0

)sec

/806

.9(

2

sec)

/1(

46.030

)019

.0(

2

2

2

2

=

=

=

module 3b: flow in pipes - unix.eng.ua.eduunix.eng.ua.edu/~rpitt/class/water resources...

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