internal incompressible viscous flow

INTERNAL INCOMPRESSIBLE INTERNAL INCOMPRESSIBLE VISCOUS FLOWVISCOUS FLOW

Nazaruddin SinagaNazaruddin Sinaga

Laboratorium Efisiensi dan Konservasi Energi

Universitas Diponegoro11

Outline

2

Flow Measurements

3

4

Osborne Reynolds Experiment

5

6

Laminar and Turbulent FlowLaminar and Turbulent Flow

7

Entrance LengthEntrance Length

8

Developing FlowDeveloping Flow

9

10

11

12

13

14

15

16

17

1

2

5

18

19

20

21

22

23

24

25

26

27

Types of FlowTypes of Flow• The physical nature of fluid flow can be categorized into three

types, i.e. laminar, transition and turbulent flow. Reynolds Number (Re) can be used to characterize these flow.

(6.3)

where = density = dynamic viscosity = kinematic viscosity ( = /)V = mean velocity D = pipe diameter

In general, flow in commercial pipes have been found to conform to the

following condition:Laminar Flow: Re <2000Transitional Flow : 2000 < Re <4000Turbulent Flow : Re >4000

28

VDVD

Re

Laminar FlowLaminar Flow

Viscous shears dominate in this type of flow and the fluid appears to be moving in discreet layers. The shear stress is governed by Newton’s law of viscosity

In general the shear stress is almost impossible to measure. But for laminar flow it is possible to calculate the theoretical value for a given velocity, fluid and the appropriate geometrical shape.

29

dy

du

30

Pressure Loss During A Laminar Flow In Pressure Loss During A Laminar Flow In A PipeA Pipe

- In reality, because fluids are viscous, energy is lost by flowing fluids due to friction which must be taken into account.

• - The effect of friction shows itself as a pressure (or head) loss. In a pipe with a real fluid flowing, the shear stress at the wall retard the flow.

• - The shear stress will vary with velocity of flow and hence with Re. Many experiments have been done with various fluids measuring the pressure loss at various Reynolds numbers.

• - Figure below shows a typical velocity distribution in a pipe flow. It can be seen the velocity increases from zero at the wall to a maximum in the mainstream of the flow.

A typical velocity distribution in a pipe flow

31

In laminar flow the paths of individual particles of fluid do not cross, so the flow may be considered as a series of concentric cylinders sliding over each other. Lets consider a cylinder of fluid with a length L,

radius r, flowing steadily in the center of pipe.

Cylindrical of fluid flowing steadily in a pipe

The fluid is in equilibrium, shearing forces equal the pressure forces. Shearing force = Pressure force

(6.5)

Taking the direction of measurement r (measured from the center of pipe), rather than the use of y (measured from the pipe wall), the above equation can be written as;

(6.6)

Equatting (6.5) with (6.6) will give:

2

r

L

P

rPPArL2 2

dr

du r

2

r

L

P

dr

dudr

du

2

r

L

P

In an integral form this gives an expression for velocity, with the values of r = 0 (at the pipe center) to r = R (at the pipe wall)

(6.7)

where P = change in pressure L = length of pipe

R = pipe radius r = distance measured from the center of pipe

The maximum velocity is at the center of the pipe, i.e. when r = 0.

It can be shown that the mean velocity is half the maximum velocity, i.e. V=umax/2

Rr0r rdr

2

1

L

Pu

L

P

4

rRu

22

r

L

P

4

Ru

2

max

Shear stress and velocity distribution in pipe for laminar flow

The discharge may be found using the Hagen-Poiseuille equation, which is given by the following;

(6.8)

The Hagen-Poiseuille expresses the discharge Q in terms of the pressure gradient , diameter of pipe, and viscosity of the fluid.

Pressure drop throughout the length of pipe can then be calculated by (6.9)

128

D

L

PQ

4

L

P

dx

dP

D

LV

D

LV

VDRV

R

L

R

LQP

2Re

64

2

6488 222

44

Fully Developed Laminar FlowFully Developed Laminar Flowin a Pipein a Pipe

• Velocity Distribution

Shear Stress Distribution

35

Fully Developed Laminar FlowFully Developed Laminar Flowin a Pipein a Pipe

• Volume Flow Rate

Flow Rate as a Function of Pressure Drop

36

Fully Developed Laminar FlowFully Developed Laminar Flowin a Pipein a Pipe

• Average Velocity

Maximum Velocity

37

38

39

Turbulent FlowTurbulent Flow This is the most commonly occurring flow in engineering

practice in which fluid particles move erratically causing instantaneous fluctuations in the velocity components.

These fluctuations cause additional shear stresses. In this type of flow both viscous and turbulent shear stresses exists.

Thus, the shear stress in turbulent flow is a combination of laminar and turbulent shear stresses, and can be written as:

where = dynamic viscosity = eddy viscosity which is not a fluid property but

depends upon turbulence condition of flow.

dy

dUturlam

40

41

Velocity Profile for Fully Developed Velocity Profile for Fully Developed Pipe FlowPipe Flow

42

The velocity at any point in the cross-section will be proportional to the one-seventh power of the distance from the wall, which can be expressed as:

(6.10)

where Uy is the velocity at a distance y from the wall, UCL is the velocity at the centerline of pipe, and R is the radius of pipe. This equation is known as the Prandtl one-seventh law.

Figure below shows the velocity profile for turbulent flow in a pipe. The shape of the profile is said to be logarithmic.

7/1

CL

y

R

y

U

U

Velocity profile for turbulent flow

43

44

For smooth pipe:

(6.11a)

For rough pipe:

(6.11b)

In the above equations, U represents the velocity at a distance yfrom the pipe wall, U* is the shear velocity =

y is the distance form the pipe wall, k is the surface roughness and

is the kinematic viscosity of the fluid.

5.5log75.5 *10

*

yU

U

U

5.8k

ylog75.5

U

U10

*

Turbulent Velocity Profiles in Fully Developed Pipe Flow

45

46

47

48

Head LossesHead Losses

Head LossesHead Losses

49

The momentum balance in the flow direction is thus given by

50

51

• Head Loss

53

54

Friction FactorFriction Factor

55

56

57

Nikuradse’s ExperimentsNikuradse’s Experiments

• In general, friction factor

– Function of Re and roughness

• 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

64f

Blausius

Re 4/1k

f Rough

Smooth

Laminar Transition Turbulent

Blausius OK for smooth pipe

)(Re,D

eFf

Re

64f

2

9.010Re

74.5

7.3log

25.0

D

ef

59

60

Moody DiagramMoody Diagram

61

Calculation of Minor Head LossCalculation of Minor Head Loss

• Minor Losses– Examples: Inlets and Exits; Enlargements and Contractions;

Pipe Bends; Valves and Fittings

62

Pipe EntrancePipe Entrance

• Developing flow– Includes boundary layer

and core, – viscous effects grow inward

from the wall• Fully developed flow

– Shape of velocity profile is same at all points along pipe

flowTurbulent 4.4Re

flowLaminar Re06.01/6D

LeeL

Entrance length LeFully developed flow region

Region of linear pressure drop

Entrance pressure drop

Pressure

x

Entrance Loss in a PipeEntrance Loss in a Pipe

• In addition to frictional losses, there are minor losses due to – Entrances or exits– Expansions or contractions– Bends, elbows, tees, and

other fittings– Valves

• Losses generally determined by experiment and then corellated with pipe flow characteristics

• Loss coefficients are generally given as the ratio of head loss to velocity head

Abrupt inlet, K ~ 0.5

g

VKh

g

V

hK L

L

2or

2

2

2

K – loss coefficentK ~ 0.1 for well-rounded inlet (high Re)K ~ 1.0 abrupt pipe outletK ~ 0.5 abrupt pipe inlet

Elbow Loss in a PipeElbow Loss in a Pipe• A piping system may have many minor losses which are all

correlated to V2/2g • Sum them up to a total system loss for pipes of the same

diameter

• Where,

m

mm

mfL KD

Lf

g

Vhhh

2

2

lossheadTotalLhlossheadFrictionalfh

mhm fittingfor lossheadMinormKm fittingfor tcoefficienlossheadMinor

Calculation of Minor Head LossCalculation of Minor Head Loss

66

67

68

69

70

71

72

73

74

75

Example 1Example 1 Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The

pipe has 150-mm diameter, 500 m long, and the surface roughness is estimated at 0.06 mm. Find the head loss and the pressure drop throughout the length of the pipe.

Solution: From Table 1.3 (for water): = 1000 kg/m3 and =1.30x10-3 N.s/m2

V = Q/A and A=R2

A = (0.15/2)2 = 0.01767 m2

V = Q/A =0.03/.0.01767 =1.7 m/sRe = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000 turbulent

flowTo find , use Moody Diagram with Re and relative roughness (k/D).

k/D = 0.06x10-3/0.15 = 4x10-4

From Moody diagram, 0.018The head loss may be computed using the Darcy-Weisbach equation.

The pressure drop along the pipe can be calculated using the relationship: ΔP=ghf = 1000 x 9.81 x 8.84ΔP = 8.67 x 104 Pa

.m84.881.9x2x15.0

7.1x500x018.0

g2

V

D

Lh

22

f

76

Example 2Example 2 Determine the energy loss that will occur as 0.06 m3/s water

flows from a 40-mm pipe diameter into a 100-mm pipe diameter through a sudden expansion.

Solution: The head loss through a sudden enlargement is given by;

Da/Db = 40/100 = 0.4From Table 6.3: K = 0.70Thus, the head loss is

g2

VKh

2a

m

smA

QV

aa /58.3

)2/04.0(

06.02

m47.081.9x2

58.3x70.0h

2

Lm

77

ExExample 3ample 3

Calculate the head added by the pump when the water system shown below carries a discharge of 0.27 m3/s. If the efficiency of the pump is 80%, calculate the power input required by the pump to maintain the flow.

Solution:Applying Bernoulli equation between section 1 and 2

(1)

P1 = P2 = Patm = 0 (atm) and V1=V2 0 Thus equation (1) reduces to:

(2)

HL1-2 = hf + hentrance + hbend + hexit

From (2):

21L

22

22

p

21

11 H

g2

Vz

g

PH

g2

Vz

g

P

21L12p HzzH

g2

V4.39

14.05.04.0

1000x015.0

g2

VH

2

2

21L

81.9x2

V4.39200230H

2

p

The velocity can be calculated using the continuity equation:

Thus, the head added by the pump: Hp = 39.3 m

Pin = 130.117 Watt ≈ 130 kW.

s/m15.2

2/4.0

27.0

A

QV

2

in

pp P

gQH

8.0

3.39x27.0x81.9x1000gQHP

p

pin

80

Pipe Flow AnalysisPipe Flow Analysis

• Pipeline system used in water distribution, industrial application and in many engineering systems may range from simple arrangement to extremely complex one.

• Problems regarding pipelines are usually tackled by the use of continuity and energy equations.

• The head loss due to friction is usually calculated using the D-W equation while the minor losses are computed using equations 6.16, 6.16(a) and 6.16(b) depending on the appropriate conditions.

81

Pipes in SeriesPipes in Series

When two or more pipes of different diameters or roughness are connected in such a way that the fluid follows a single flow path throughout the system, the system represents a series pipeline.

In a series pipeline the total energy loss is the sum of the individual minor losses and all pipe friction losses.

Pipelines in series

82

Referring to Figure 6.11, the Bernoulli equation can be written between points 1 and 2 as follows;

(6.18)

where P/g = pressure headz = elevation headV2/2g = velocity headHL1-2 = total energy lost between point

1 and 2

Realizing that P1=P2=Patm, and V1=V2, then equation (6.14) reduces to

z1-z2 = HL1-2

21L

22

22

21

11 H

g2

Vz

g

P

g2

Vz

g

P

83

Or we can say that the different of reservoir water level is equivalent to the total head losses in the system.

The total head losses are a combination of the all the friction losses and the sum of the individual minor losses.

HL1-2 = hfa + hfb + hentrance + hvalve + hexpansion + hexit.

Since the same discharge passes through all the pipes, the continuity equation can be written as; Q1 = Q2

84

Pipes in ParallelPipes in Parallel

• A combination of two or more pipes connected between two points so that the discharge divides at the first junction and rejoins at the next is known as pipes in parallel. Here the head loss between the two junctions is the same for all pipes.

Figure 6.12 Pipelines in parallel

Applying the continuity equation to the system;

Q1 = Qa + Qb = Q2 (6.19)

The energy equation between point 1 and 2 can be written as;

The head losses throughout the system are given by;

HL1-2=hLa = hLb (6.20)

Equations (6.19) and (6.20) are the governing relationships for parallel pipe line systems. The system automatically adjusts the flow in each branch until the total system flow satisfies these equations.

L

22

22

21

11 H

g2

Vz

g

P

g2

Vz

g

P

86

Pipe NetworkPipe Network A water distribution system consists of complex

interconnected pipes, service reservoirs and/or pumps, which deliver water from the treatment plant to the consumer. Water demand is highly variable, whereas supply

is normally constant. Thus, the distribution system must include storage elements, and must be capable of flexible operation. Pipe network analysis involves the determination

of the pipe flow rates and pressure heads at the outflows points of the network. The flow rate and pressure heads must satisfy the continuity and energy equations.

87

Pipe NetworkPipe Network The earliest systematic method of network

analysis (Hardy-Cross Method) is known as the head balance or closed loop method. This method is applicable to system in which pipes form closed loops. The outflows from the system are generally assumed to occur at the nodes junction. For a given pipe system with known outflows, the

Hardy-Cross method is an iterative procedure based on initially iterated flows in the pipes. At each junction these flows must satisfy the continuity criterion, i.e. the algebraic sum of the flow rates in the pipe meeting at a junction, together with any external flows is zero.

88

• Assigning clockwise flows and their associated head losses are positive, the procedure is as follows:Assume values of Q to satisfy Q = 0.Calculate HL from Q using HL = K1Q2 . If HL = 0, then the solution is correct.

89

If HL 0, then apply a correction factor, Q, to all Q and repeat from step (2). For practical purposes, the calculation is usually terminated when HL < 0.01 m or Q < 1 L/s. A reasonably efficient value of Q for rapid convergence is given by;

(6.21)

QH2

HQ

L

L

90

Example 4Example 4• A pipe 6-cm in diameter, 1000m long and with =

0.018 is connected in parallel between two points M and N with another pipe 8-cm in diameter, 800-m long and having = 0.020. A total discharge of 20 L/s enters the parallel pipe through division at A and rejoins at B. Estimate the discharge in each of the pipe.

91

Solution:Continuity: Q = Q1 + Q2

(1)

Pipes in parallel: hf1 = hf2

Substitute (2) into (1)

0.8165V2 + 1.778 V2 = 7.074

V2 = 2.73 m/s

22

221

2 V)08.0(4

V)06.0(4

02.0

074.7V778.1V 21

)2(V8165.0V

V08.0

800x020.0V

06.0

1000x018.0

gD2

VL

gD2

VL

21

22

21

2

222

21

211

1

92

Q2 = 0.0137 m3/s

From (2):V1 = 0.8165 V2 = 0.8165x2.73 = 2.23 m/s

Q1 = 0.0063 m3/s

Recheck the answer:Q1+ Q2 = Q

0.0063 + 0.0137 = 0.020 (same as given Q OK!)

73.2x)08.0(4

VAQ 2222

93

Example 6.6 • For the square loop shown, find the discharge in all

the pipes. All pipes are 1 km long and 300 mm in diameter, with a friction factor of 0.0163. Assume that minor losses can be neglected.

94

Solution: Assume values of Q to satisfy continuity equations all at

nodes.

The head loss is calculated using; HL = K1Q2

HL = hf + hLm

But minor losses can be neglected: hLm = 0

Thus HL = hf

Head loss can be calculated using the Darcy-Weisbach equation

g2

V

D

Lh

2

f

95

First trial

Since HL > 0.01 m, then correction has to be applied.

554'K

Q'KH

Q554H

3.0x4

Qx77.2

A

Q77.2H

81.9x2

Vx

3.0

1000x0163.0H

g2

V

D

LhH

2L

2L

22

2

2

2

L

2

L

2

fL

Pipe Q (L/s) HL (m) HL/Q

AB 60 2.0 0.033

BC 40 0.886 0.0222

CD 0 0 0

AD -40 -0.886 0.0222

2.00 0.0774

96

Second trial

Since HL ≈ 0.01 m, then it is OK.Thus, the discharge in each pipe is as follows (to the nearest integer).

s/L92.120774.0x2

2

QH2

HQ

L

L

Pipe Q (L/s) HL (m) HL/Q

AB 47.08 1.23 0.0261

BC 27.08 0.407 0.015

CD -12.92 -0.092 0.007

AD -52.92 -1.555 0.0294

-0.0107 0.07775

Pipe Discharge (L/s)

AB 47

BC 27

CD -13

AD -53

Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems

• Single Path– Find p for a given L, D, and Q Use energy equation directly

– Find L for a given p, D, and Q Use energy equation directly

Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems

• Single Path (Continued)– Find Q for a given p, L, and D

1. Manually iterate energy equation and friction factor formula to find V (or Q), or

2. Directly solve, simultaneously, energy equation and friction factor formula using (for example) Excel

– Find D for a given p, L, and Q1. Manually iterate energy equation and friction factor

formula to find D, or2. Directly solve, simultaneously, energy equation and

friction factor formula using (for example) Excel

Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems• Multiple-Path Systems

Example:

Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems

• Multiple-Path Systems– Solve each branch as for single path

– Two additional rules1. The net flow out of any node (junction) is zero2. Each node has a unique pressure head (HGL)

– To complete solution of problem1. Manually iterate energy equation and friction factor for each

branch to satisfy all constraints, or2. Directly solve, simultaneously, complete set of equations using

(for example) Excel

Flow Measurement• Direct Methods

– Examples: Accumulation in a Container; Positive Displacement Flowmeter

• Restriction Flow Meters for Internal Flows– Examples: Orifice Plate; Flow Nozzle; Venturi; Laminar

Flow Element

Flow Measurement• Linear Flow Meters

– Examples: Float Meter (Rotameter); Turbine; Vortex; Electromagnetic; Magnetic; Ultrasonic

Flow Measurement

• Traversing Methods– Examples: Pitot (or Pitot Static) Tube; Laser Doppler

Anemometer

The EndThe End

Terima kasihTerima kasih

104104

Shear Stress in Pipes• Steady, uniform flow in a pipe: momentum

flux is zero and pressure distribution across pipe is hydrostatic, equilibrium exists between pressure, gravity and shear forces

D

Lhhh

ds

dhD

zp

ds

dD

sDds

dzsAsA

ds

dp

sDWAsds

dpppAF

f

s

021

0

0

0

0

44

)]([4

)(0

)(sin)(0

• Since h is constant across the cross-section of the pipe (hydrostatic), and –dh/ds>0, then the shear stress will be zero at the center (r = 0) and increase linearly to a maximum at the wall.

• Head loss is due to the shear stress.

• Applicable to either laminar or turbulent flow

• Now we need a relationship for the shear stress in terms of the Re and pipe roughness

Darcy-Weisbach Equation

)(Re,

)(Re,

;Re;

,,:variablesRepeating

),(

),,,,(

20

20

20

321

214

0

D

eFV

D

eF

V

VD

e

DV

F

eDVF

V D e

ML-1T-2 ML-

3

LT-1 ML-1T-1 L L

)(Re,82

)(Re,82

)(Re,4

4

2

2

2

0

D

eFf

g

V

D

Lfh

D

eF

g

V

D

L

D

eFV

D

L

D

Lh

f

f

Darcy-Weisbach Eq. Friction factor

Laminar Flow in Pipes• Laminar flow -- Newton’s law of viscosity is valid:

2

0

20

20

2

14

44

2

2

2

r

r

ds

dhrV

ds

dhrCC

ds

dhrV

drds

dhrdV

ds

dhr

dr

dV

dr

dV

dy

dV

ds

dhr

dy

dV

• Velocity distribution in a pipe (laminar flow) is parabolic with maximum at center.

2

0max 1

r

rVV

Discharge in Laminar Flow

ds

dhD

ds

dhrQ

rr

ds

dh

rdrrrds

dhVdAQ

rrds

dhV

r

r

128

8

2

)(

4

)2()(4

)(4

4

40

0

220

2

022

0

220

0

0

ds

dhDV

A

QV

32

2

Head Loss in Laminar Flow

2

21

12212

2

2

2

32

)(32

32

32

32

D

VLh

hhh

ssD

Vhh

dsD

Vdh

DV

ds

dh

ds

dhDV

f

f

Re

64

2

2/)(Re

64

2/))((64

2/

2/32

32

2

2

2

2

2

2

2

fV

D

Lfh

VD

L

VD

L

DV

V

V

D

VL

D

VLh

f

f

EGL & HGL for Losses in a Pipe• Entrances, bends, and other flow transitions

cause the EGL to drop an amount equal to the head loss produced by the transition.

• EGL is steeper at entrance than it is downstream of there where the slope is equal the frictional head loss in the pipe.

• The HGL also drops sharply downstream of an entrance

Ex. (10.8)Given: Kerosene (S=0.94, =0.048 N-s/m2).

Horizontal 5-cm pipe. Q=2x10-3 m3/s.

Find: Pressure drop per 10 m of pipe.

Solution:

cfs

sftV

VV

VVg

VγD

μLV

g

α

g

Vα

γD

μLV

γD

μLVh

zγ

p

g

Vαhz

γ

p

g

Vα

L

L

32

5

22

2

522

222

2

22

22

2

22

22

211

21

1

10*23.14/(0.25/12)**1.6A*VQ

(laminar)129310*4

)12/25.0(*6.1*94.1*8.0Re

/60.1

01.1645.8

05.0)32/1(*4.62*8.0

10*10*4*32

2

2

05.032

2

002

325.000

32

22

Ex. (10.34)

2/62.0,/300,12 mNsmN

Given: Glycerin@ 20oC flows commercial steel pipe.

Find: h

Solution:

mγD

μLVhh

VDVD

hzγ

pz

γ

ph

zγ

phz

γ

p

zγ

p

g

Vαhz

γ

p

g

Vα

L

L

L

L

42.2)02.0(*300,12

)6.0)(1)(62.0(3232

(laminar)5.2310*1.5

02.0*6.0Re

)(

22

22

4

22

11

22

11

22

22

211

21

1

Ex. (10.43)

Given: Figure

Find: Estimate the elevation required in the upper reservoir to produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?

Solution:

ftz

sftA

QV

D

LfKKK

g

V

D

LfKKKh

zhz

zγ

p

g

Vαhz

γ

p

g

Vα

Ebe

EbeL

L

L

1332.32*2

73.1275.100.14.0*25.0100

/73.121*4/

10

75.101

430*025.0;0.1;(assumed)4.0;5.0

22

0000

22

2

1

2

221

22

22

211

21

1

55

2

22

1

2

1

2

11

21

1

10*910*14.1

1*73.12Re

59.0)53.1(*4.62

35.1

2.32*2

73.12

1

300025.04.05.00.17.110133

22

2*100

22

VD

psigp

ft

g

V

D

LfKK

g

Vzz

γ

p

zγ

p

g

Vhz

zγ

p

g

Vαhz

γ

p

g

Vα

b

beb

bb

bbb

L

bbb

bL

Ex. (10.68)

ftxksftx s425 105.1;/1022.1

ftDD

g

DQ

D

g

V

D

Lfh f

06.8

984,331

2

))4//((1000*015.01

2

5

22

2

Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1 ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the diameters are expressed in inches (i.e., 10 in, 12 in, etc.).

Find: Diameter.

Solution:

Assume f = 0.015

Relative roughness: 00002.006.8

105.1 4

x

D

ks

Get better estimate of f

65

2

109.31022.1)06.8)(4/(

300Re

)4/()4/(

Re

xx

D

QD

D

Q

VD

f=0.010

.8943.7

656,221

5

inftDD

Use a 90 in pipe

Ex. (10.81)

mh

h

zγ

p

g

Vαhz

γ

p

g

Vα

g

Q

h

fLD

g

DQ

D

Lf

g

V

D

Lfh

f

f

L

f

f

45.14

109810

000,60

9810

000,300

22

8

2

))4//((

2

22

22

211

21

1

5/1

2

2

222

Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m from the main? Assume galvanized-steel pipe is to be used and that the pressure required at the factory is 60 kPa gage at a point 10 m above the main connection.

Find: Size of pipe.

Solution:

Assume f = 0.020

m

g

Q

h

fLD

f

100.081.9

)025.0(140

45.14

02.08

8

5/1

2

2

5/1

2

2

Relative roughness:

022.0

0015.0100

15.0

f

D

ks

Friction factor:

mD 102.0020.0

022.0100.0

5/1

Use 12 cm pipe

Ex. (10.83)Given: The 10-cm galvanized-steel pipe is 1000 m

long and discharges water into the atmosphere. The pipeline has an open globe valve and 4 threaded elbows; h1=3 m and h2 = 15 m.

Find: What is the discharge, and what is the pressure at A, the midpoint of the line?

Solution:

002

)41(1200

222

22

22

211

21

1

g

V

D

LfKKK

zγ

p

g

Vαhz

γ

p

g

Vα

bve

L

D = 10-cm and assume f = 0.025

46

32

2

2

1071031.1

1.0*942.0Re

/0074.0)10.0)(4/(942.0

/942.01.265

24

)1.0

1000025.09.0*4105.01(24

xx

VD

smVAQ

smV

gV

Vg

So f = 0.025

kPap

mgγ

p

g

V

D

LfK

γ

p

zγ

p

g

Vαhz

γ

p

g

Vα

A

A

bA

LAAA

A

8.90)26.9(*9810

6.9152

)942.0()

1.0

500025.09.0*2(

2)2(15

22

2

2

22

22

2

2

Near cavitation pressure, not good!

Ex. (10.95)

Given: If the deluge through the system shown is 2 cfs, what horsepower is the pump supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is smooth.

Find: Horsepower

Solution:

55

22

2

22

22

22

211

21

1

1017.41022.1

)2/1(*18.10Re

611.12

/18.10)2/1)(4/(

2

)45.01(2

6003000

22

xx

VD

ftg

V

sftA

QV

D

LfK

g

Vh

hzγ

p

g

Vαhz

γ

p

g

Vα

bp

Lp

hp

hQp

ft

h

p

p

4.24550

6.107

))2/1(

17000135.019.0*45.01(611.13060

So f = 0.0135