Ch8 Steady Incompressible Flow in Pressure Conduits (PartB)

8/11/2019 Ch8 Steady Incompressible Flow in Pressure Conduits (PartB)

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Chapter 8 Steady Incompressible Flowin Pressure Conduits (Part B)


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We will be looking here at the flow of real

fluid in pipes real meaning a fluid thatlooses energy due to friction as it

interacts with the pipe wall as it flows.

Head loss (Review)


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Re

8.320

d=

vd=Re

Viscous sublayer


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8.8 Head loss in Turbulent Flow

When flow is turbulent, the viscous dissipation effects cannot

be derived explicitly as in laminar flow, but the following

relation is still valid.

g

V

d

l

fhf 2

2

=l = pipe length

d= pipe diameterV = pipe velocity

f = friction factor


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8.8.1 Nikuradses Experiment

Nikuradse made a great contribution to the theory of pipe

flow by differentiating between rough and smooth pipes.

A rough pipe is one where the mean height of roughness is

greater than the thickness of the laminar sub-layer. Nikuradse

artificially roughened pipe by coating them with sand. Hedefined a relative roughness value ks/d (mean height of

roughness over pipe diameter) and produced graphs of

against Re for a range of relative roughness 1/30 to 1/1014.


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d /2ks

Figure : Regions on plot of Nikuradess data


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d /2ks


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d /2ks


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d /2ks

Figure : Regions on plot of Nikuradess data


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8

lg )

l

g

1

0

0

)

15

30.6

60

126

252

507

k

s

d /2ks

Laminar flow Re < 2000 (lg Re = 3.30) fRe

64 /Re


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8

lg )

l

g

1

0

0

)

15

30.6

60

126

252

507

d /2ks

Transition from laminar to turbulent: 2300< Re < 4000

(3.3 < lgRe < 3.6)

Pipe flow normally lies outside this region.


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8

lg )

l

g

1

0

0

)

15

30.6

60

126

252

507

d /2ks

Smooth turbulentThe limiting line of turbulent flow.

All value of relative roughness tend toward this as Re decreases.


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0.0

0.2

0.4

0.6

0.8

1.0

1.2

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8

lg )

l

g

1

0

0

)

15

30.6

60

126

252

507

d /2ks

Transitional turbulent

The region which ss

varies with

both Re and relative roughness. Most pipes lie in this region.

)(Re,d

kf s=


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0. 0

0. 2

0. 4

0. 6

0. 8

1. 0

1. 2

2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6 5.8

lg )

l

g

1

0

0

)

15

30.6

60

126

252

507

d /2ks

Rough turbulent. remains constant for a given relative

roughness. It is independent of Re.


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Nikuradses Experiments

In general, friction factor

Function of Re androughness

Laminar region

Independent of roughness

Turbulent region

Smooth pipe curve

All curves coincide @

~Re=2300

Rough pipe zone All rough pipe curves

flatten out and

become independent

of Re

Re

64=

( )Blausius

Re4/1

k=

Rough

Smooth

Laminar Transition Turbulent

Blausius OK for smooth pipe

)k

(Re, s

DF=

Re

64=

274.5

log

25.0

+

= e9.010 Re7.3 D


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Regions on plot of Nikuradess data


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Turbulent flow in a circular pipe may be classifiedas: smooth pipe region, rough pipe region and

transition region.


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8.8.2 Moody chart

The Moody chart is a graphical method to find the friction factor

in pipes.

Colebrook and White proposed the following general equationafter studying flow in real pipes:

)Re

51.2

7.3lg(21

fd

e

f +=

The values of friction factor obtained from the equation are

plotted on a Moody diagram, which shows a family of curvesfor f plotted against the relative roughness and Reynolds

number.

Colebrook-White Equation


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A good approximate equation for the turbulent region

of the Moody chart is given by Haalands equation:

Haalands equation is valid for turbulent

flow (Re > 2300)


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Pipe roughness

pipe material pipe roughness e (mm)

glass, drawn brass, copper 0.0015

commercial steel or wrought iron 0.045

asphalted cast iron 0.12galvanized iron 0.15

cast iron 0.26

concrete 0.18-0.6

rivet steel 0.9-9.0

corrugated metal 45

PVC 0.12


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8.24


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Q ti


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Question

If the flow is in turbulent transition region, the Frictional

factor f of the industrial pipes _____ with the increase of theReynolds number

A increases B reduces C keeps constant.

Q ti


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Question

There are two pipes, one transports oil and the other transports

water. If diameter d,length l and roughness coefficient of thetwo pipes are all the same, kinematic viscosityoil is bigger

than water, the Reynolds Numbers are equal, then the

Frictional Loss______

A. hfo=hfw B. hfo>hfw

C. hfo


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8.9 Single Pipe Flowssolution basics

2

4

D

Q

V =

VDVD==Re

g

V

D

L

fhL 2

2

=

Re

6464

== DVf

+

=

Re

9.6

7.3log8.1

1 11.1De

f

Four simultaneous equations:

continuity energy loss

Reynolds

number

)/(Re, Deff =

For ColebrookWhite )Re51.2

7.3

/lg(2

1

fd

De

f+=

Haaland


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3 types of pipe flow problems:1. Head loss (find h

LgivenD, Q or V)

2. Discharge (find Q givenD and hL

)

3. Sizing problem (FindD given Q and hL)


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Example ( Laminar flow):

Water, 20oC flows through a 0.6 cm tube, 30 m long, at aflow rate of 0.34 liters/min. If the pipe discharges to the

atmosphere, determine the supply pressure if the tube is

inclined 10o

above the horizontal in the flow direction.

10

L = 30 m D = .6 cm1

2

W t P ti


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Water Properties:

= 998 kg/m3 g = 9790 N/m3 = 1.005 E-6 m2/s

Energy Equation (neglecting)

P2 = 0 P1

g= Z2Z1+hf = L sin 10o + hf

V=Q

A

=0.34E

3m

3/ min*1min/ 60 s

0.3/100( )2

m2 = 0.2m /s

Re=V D

=

0.2*0.006

1.005E6 = 1197 laminar flow

P1 = 9790 N/m3*5.75 m = 56.34 kN/m2 (kPa)


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Solution Summary


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Solution Summary

To solve basic pipe flow frictional head loss problem, use the

following procedure:

1. Use known flow rate to determine Reynolds number.

2. Identify whether flow is laminar or turbulent.

3. Use correct expression to determine friction factor (withks/d if necessary).

4. Use definition of hf to determine friction head loss.

5. Use general energy equation to determine total pressure drop.

8.10 Minor Losses in Turbulent Flow


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w

Where K is minor loss coefficient

Transitions in pipe systems, such as bends, valves, changes in

diameter, entrances and exits, cause head losses in the system.

Head losses at transition points are called minor losses. A minorloss is usually a function of the velocity head as follows:

gVkh

MinorL

2

2

)( =

Sources of minor losses


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Sources of minor losses

Additional pressure (energy) losses

due to: Fittings, bends, orifice plates, and valves

Losses due to physics

Vena contracta Abrupt changes in flow area

Losses due to piping networks for fluid distribution


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Minor losses in piping networks


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Minor losses in piping networks

8.10.1 Sudden Enlargement


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g

)22

()()(2

22

2

112121

ggg

p

g

pzzhj

++=


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8.10.2 Sudden Contraction


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g

vKh

A

AK j

215.0

2

22=

1

2

A1

A2

v1 v2

A1

8.10.3 Gradual Expansion (Diffusor)


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( )

g

VV

Kh EE 2

2

21

=

diffusor angle ()0

0.10.20.30.40.5

0.60.70.8

0 20 40 60 80

KE

2

1

2

2

2 1

2

=

A

A

g

VKh EE

Loss due to gradual enlargement

8.10.4 Entrance Losses


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g

VKh ee

2

2

=

0.1eK

5.0eK

04.0eK

Losses can be

reduced byaccelerating the

flow gradually and

eliminating the

8.10.5 Head Loss in Valves


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Function of valve type and valveposition

The complex flow path through valvescan result in high head loss (of course,

one of the purposes of a valve is to

create head loss when it is not fullyopen)

g

VKh vv2

2

=


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Example


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In a sudden expansion pipe shown in figure, velocity is v1 and v2respectively. A middle diameter pipe is connected between the

two pipes to form a two-sudden-expansion pipe. The interactionof minor resistances is negligible. That is, superposition method

can be applied here. Determine: (1) the velocity of the middle

pipe when the total minor head loss of the pipe is the least. (2)the total minor head loss, and the comparison with the one

sudden expansion pipe.

Solution: (1) The minor head loss of the two-sudden-expansion


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pipe.

Assume the velocity of middle pipe is v, and make the totalminor head loss be the least, so

2The total minor head loss is

the minor head loss of the one sudden

expansion pipe is

the total minor head loss of the two-sudden-expansion pipe ishalf of the minor head loss of the one sudden expansion pipe.


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8.11 Branching Pipes


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Branching pipe systems, such as the one shown by Figure

8.27, can be solved using the following:

1. Q1 = Q2 + Q3.

2 The elevation of P is common to all pipes.

Branching pipe systems

8.12 Pipes in Series


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Pipes in series, as shown by Figure 8.29, can be solved as

follows:

Q = Q1 = Q2 = Q3

hL= hL1+ hL2+ hL3

Fig. Pipes in series


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The pipe elements of the pipes in series have:

A. the same head lossB. the same total head loss

C. the same hydraulic slope

D. the same discharge through them

Example


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Consider the two reservoirs shown in figure, connected by a single

pipe that changes diameter over its length. The surfaces of the two

reservoirs have a difference in level of 9m. The pipe has a diameter

of 200mm for the first 15m (from A to C) then a diameter of

250mm for the remaining 45m (from C to B).

For the entrance use kL = 0.5 and the exit kL = 1.0. The join at C is sudden.

For both pipes use = 0.04.


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Total head loss for the system H = height difference of reservoirs

and solve for Q, to give Q =

0.158 m3/s

8.13 Pipes in Parallel


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Parallel pipes, as shown by Figure 8.30, can be solved as

follows:

Q = Q1 + Q2 + Q3

hL= hL1= hL2= hL3

Pipes in parallel

Question


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From the figure, we can see that the relationship among the

head losses fromA toB in the pipes in parallel 1, 2, 3 is:

A. hfABhfl+hf2+hf3

B. hfABhfl+hf2

C. hfABhf2+hf3

D. hfAB

hfl=hf2=hf3.

Example


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Two pipes connect two reservoirs (A and B) which have a height difference

of 10m. Pipe 1 has diameter 50mm and length 100m. Pipe 2 has diameter

100mm and length 100m. Both have entry loss kL = 0.5 and exit loss kL=1.0and Darcy f of 0.008.

Calculate:

a) rate of flow for each pipe

b) the diameter D of a pipe 100m long that could replace the two pipes

and provide the same flow.


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Ch8 Steady Incompressible Flow in Pressure Conduits (PartB)

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