- 1 -

CHAPTER 3 VISCOUS FLOW IN PIPES 3.1 Introduction

This chapter will deal almost exclusively with pipes or conduits that are flowing full. A free surface will not exist in the pipe itself, and the Froude number, which is important to open channel flows, will accordingly have no significance. The Reynolds number will be the significant dimensionless parameter, but its importance will also diminish at high Reynolds numbers as the flow becomes independent of viscous effects. Flow in a circular pipe is undoubtedly the most common closed fluid flow. It is encountered in the veins and arteries in a body, in a city’s water system, in a farmer’s irrigation system, in the piping systems transporting fluids in a factory and in the ink jet of a computer’s printer. For a sufficiently low Reynolds numbers (Re < 2000 in a pipe) a laminar flow results, and at sufficiently high Reynolds number a turbulent flow occurs.

For all flows involved in this Chapter, we assume that the pipe is completely filled with the fluid being transported as is shown in Figure 3.1a Thus, we will not consider a concrete pipe through which rain water flows without completely filling the pipe, as is shown in Figure 3.1b. Such flows, called open-channel flow will be treated in next chapters. The difference between the open channel flow and the pipe flow is in the fundamental mechanism that drives the flow. For open channel flow, gravity alone is the driving force such that the water flows down a hill like in streams and rivers. For pipe flow, gravity may be important (the pipe need not be horizontal), but the main driving force is likely to be a pressure gradient along the pipe. If the pipe is not full, it is

not possible to maintain this pressure difference 21 pp .

Figure 3.1 (a) pipe flow and; (b) open channel flow

3.2 Laminar or Turbulent flow

The flow of a fluid in a pipe may be laminar flow or it may be turbulent flow. Osborne Reynolds (1842-1912), a British scientist and mathematician, was the first to distinguish the difference between these two classifications of flow by using a simple apparatus as shown in Figure 3.2

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Figure 3.2 (a) Experiment to illustrate type of flow and; (b) typical dye streaks

If water runs through a pipe of diameter D with an average velocity v, the following characteristics are observed by injecting a dye as shown in Figure 3.2. For “small enough flowrates” the dye streakline will remain as a well – defined line as it flows along; for a somewhat larger “intermediate flowrate” the dye streakline fluctuates in time and space. On the other hand, for “large enough flowrates” the dye streakline almost immediately becomes blurred and spreads across the entire pipe in a random fashion. These three characteristics are denoted as laminar, transitional and turbulent flow, respectively.

The curves shown in Figure 3.3 represents the x-component of the velocity as a function of time at a point A in the flow.

Figure 3.3 Time dependence of fluid velocity at a point

Question 3-1

The 2-cm diameter pipe is used to transport water at 20oC. What is the maximum average velocity that may exist in the pipe for which laminar flow is guaranteed?

Solution 3-1

The Reynolds Number ranges for which laminar, transitional or turbulent pipe flows are obtained cannot be precisely given. The actual transition from laminar, transitional or turbulent flow may take place at various Reynolds numbers, depending on how much the flow is “distributed” by vibrations of the pipe, roughness of the entrance region, and the like. In general, for engineering purposes the following values are appropriate.

The flow in a round pipe is laminar if Reynolds number is less than approximately 2100

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The flow in a round pipe is turbulent if Reynolds number is greater than approximately 4000

The flow in a round pipe is in transition if Reynolds number is ranging between 2100 and 4000

The kinematic viscosity of water at 20oC is sec/10 26 m . Using a Reynolds number of 2000 so that a laminar flow is guaranteed, we can simply find that,

UD

Re (1)

sec)/10(

)02.0(2000

26 m

mU

sec/1.0 mU

This average velocity is quite small. Such small velocities are not usually encountered in actual situations. Hence laminar flow is seldom of engineering interest except for specialized topics such as lubrication. Most internal flows are turbulent flows and thus the study of turbulence gains much attention.

3.3 Entrance Region and Fully Developed Flow

Any fluid flowing in a pipe had to enter the pipe at some location. The region of flow near where the fluid enters the pipe is termed the entrance region and is illustrated in the Figure 3.4. As it is shown in the figure, the fluid typically enters the pipe with a nearly uniform velocity profile. As the fluid moves through the pipe, viscous effects cause it to stick to the pipe wall. Thus, a boundary layer in which viscous effects are important is produced along the pipe wall such that the initial velocity profile changes with distance along the pipe, until the fluid reaches the end of the entrance length beyond which the velocity profile does not vary with distance x.

Figure 3.4 Entrance region of a laminar flow in a pipe or a wide rectangular channel

Question 3-2

A laminar flow is to occur in an experimental facility with 20oC water flowing through a 4cm diameter pipe. Calculate the average velocity and the entrance length if the Reynolds number is 8000?

Solution 3-2

The shape of the velocity profile in the pipe depends on whether the flow is laminar or turbulent, as does the length of the entrance region, LE. As with many other properties of pipe flow, the dimensionless entrance length, LE/D, correlates quite well with Reynolds number. Typical entrance lengths are given by

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Re06.0D

LE for laminar flow (2)

and

6/1(Re)4.4D

LE for turbulent flow (3)

Once the fluid reaches the end of the entrance region, the flow is simpler to describe because the velocity is a function of only the distance from the pipe centreline, r and independent of x.

UD

Re

sec)/10(

)04.0(8000

26 m

mU

sec/2.0 mU

and since the flow in the pipe is turbulent flow,

6/1(Re)4.4D

LE

6/1)8000(4.404.0

EL

.787.0)8000(4.404.0 6/1 mLE

3.4 Pressure change and shear stresses

Fully developed steady flow in a constant diameter pipe is driven by pressure forces. For horizontal pipe flow, gravity has no effect except for a hydrostatic pressure variation across the

pipe, D , that is usually negligible. It is the pressure difference, 21 ppp between one

section of the horizontal pipe and another which forces the fluid through the pipe. Viscous effects provide the restraining force that exactly balances the pressure force, thereby allowing the fluid to flow through the pipe with no acceleration. If viscous effects were absent in such flows, the pressure would be constant throughout the pipe, except for the hydrostatic variation.

The need for the pressure drop can be viewed from two different standpoints. In terms of a force balance, the pressure is needed to overcome the viscous forces generated. In terms of an energy balance, the work done by the pressure force is needed to overcome the viscous dissipation of energy throughout the fluid.

Figure 3.5: Element of fluid in a pipe

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If gravitational effects are neglected, the pressure is constant across any vertical cross-section of the pipe, although it varies along the pipe from one section to the next. Thus, if the pressure is

1pp at section (1), it is ppp 12 at section (2). We can foreseen the fact that the pressure

decreases in the direction of flow so that, 0p .

The driving force due to pressure (F = Pressure x Area) can then be written as

= Pressure force at 1 - pressure force at 2

))(())(())(())(())(( 221112111 ApApApAppAp

Since the pipe area at section (1) and at section (2) are same, A1=A2=A, the equation reduces to,

))(())(())(( 11 ApApAp

))(())(( 2rpAp

The retarding force is that due to the shear stress by the walls and can be written as

= shear stress × area over which it acts

= τw × area of pipe wall

rlw 2)(

As the flow is in equilibrium, the driving forces must be equal to the retarding forces.

rlrp w 2)()( 2

which can be simplified to give

rl

p w2

(4)

the above equation represents the basic balance in forces needed to drive each fluid particle along the pipe with constant velocity. It is clear that at 0r (the centerline of the pipe) there is

no shear stress ( 0 ). At 2Dr (at the pipe wall) the shear stress is maximum, denoted w ,

the wall shear stress. The pressure drop and wall shear stress are related by

D

lp w4 (5)

it is important to note that small shear stress can produce a large pressure differences in the

closed conduits if the pipe is relatively long ( 1Dl ).

The shear stress dependence for turbulent flow is very complex. However, for laminar flow of a Newtonian fluid, the shear stress is simply proportional to the velocity gradient

dy

du

dy

du (6)

the negative sign is included to give 0 with 0drdu (the velocity decreases from the pipe

centerline to the pipe wall). There are two governing laws for fully developed laminar flow of a Newtonian fluid within a horizontal pipe. The one is Newton’s second law of motion and the other is the definition of a Newtonian fluid. By combining these two equations we can obtain,

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rl

p

dr

du

2 (7)

which can be integrated to give the velocity profile as

222 21

21

16)(

D

rV

D

r

l

pDru c

(8)

where Vc is the centerline velocity. An alternative expression can be written by using the relationship between the wall shear stress and the pressure gradient to give

2

14

)(R

rDru w

(9)

where 2DR is the pipe radius. This velocity profile plotted in Figure 3.5 is parabolic in the

radial coordinate, r, has a maximum velocity, Vc, at the pipe centerline, and a minimum velocity at the pipe wall. The volume flowrate through the pipe can be obtained by integrating the velocity profile across the pipe.

rdrR

rVrdrruudAQ

R

c

Rr

r

0

2

0

122)(

or

2

2cVR

Q

(10)

By definition, the average velocity is the flowrate divided by the cross-sectional area and can be attained as the one-half of the maximum velocity.

l

pD

R

VR

R

QVV cc

3222

2

2

2

2

l

pDQ

128

4 (11)

Figure 3.5 Shear stress distributions within the fluid in a pipe and typical velocity profiles.

The above results confirm the following properties of laminar pipe flow. For a horizontal pipe the flowrate is:

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a. Directly proportional to the pressure drop

b. Inversely proportional to the viscosity

c. Inversely proportional to the pipe length

d. Proportional to the pipe diameter to the fourth power

If we introduce the friction factor f, a quantity of substantial interest in pipe flow, a dimensionless wall shear, defined by

2)8/1( Vf

(12)

we see that the change in pressure per unit weight in pipe can be written as

g

V

D

Lfh

pL 2

2

(13)

where hL is the head loss with dimension of length. This equation is often referred to as the Darcy-Weisbach equation, named after Henri P.G. Darcy (1803-1858) and Julius Weisbach (1806- 1871). Combining above equations:

Re

64f (14)

for laminar flow in a pipe. Substituting this back into above equation we see that

2

32

D

Lfh

pL

(15)

the head loss is directly proportional to the average velocity (and hence the discharge also) to the first power, a result that generally is applied to developed, laminar flows in conduits, of shape other than circular.

Question 3-3

An oil with a viscosity of =0.4 Ns/m2 and density = 900 kg/m3 flows in a pipe of diameter

D=0.02m. What pressure drop, 21 ppp is needed to produce a flowrate of 2x10-5 m3/s if the

pipe is horizontal with x1=0 and x2=10m.

Solution 3-3

The question mentions that the Reynolds number is less than 2100.The average velocity in the pipe can be calculated as:

A

QV

0637.0)01.0(

1022

5

V m/sec.

The Reynolds number is;

VD

Re in which

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87.2sec)/1044.4(

)02.0()0637.0(Re

24

m

m

Hence, the flow is laminar. Therefore the pressure difference can be deducted by the following

equation with mxxl 1012 .

4

128

D

Qlp

kPamNp 4.20/204001002.5

101024

)02.0(

)102)(10)(4.0(128 27

5

4

5

Question 3-4

A small diameter horizontal tube is connected to a supply reservoir as shown in the following Figure. If 6600 mm3 is captured at the outlet in 10 seconds estimate the viscosity of the water.

Solution 3-4

The tube is very small, so we expect viscous effects to limit the velocity to a small value. Using Bernoulli’s equation from the surface to the entrance to the tube, and neglecting the velocity head, we have, letting 0 be a point on the surface:

tubetubetube

kkk z

g

vPz

g

vP

22

2

tan

2tantan

where we have used gage pressure with 0op and thus 0tan kp and the velocity head in the

reservoir is accepted as negligible. The equation becomes

00200 tubeP

tubeP2

kPamNPtube 6.19/1960098002 2

at the exit of the tube the pressure is zero; hence,

mkPal

p/3.16

2.1

19600

The average velocity is found to be

smA

QV /840.0

)001.0(4

10/)106600(

2

9

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Check to make sure the velocity head is negligible: (v2/2g)=0.036m compared with pressure head. So the assumption of negligible velocity head is valid. The viscosity of the fluid can be found by:

l

pDV

32

2

where

Vl

pD

32

2

242

/1006.6)840.0(32

001.0)16300(mNs

we should check the Reynolds number to determine if our assumption of a laminar flow was acceptable. It is

VD

Re in which

6.138sec)/1006.6(

)001.0(sec)/84.0)(1000(Re

24

m

mm

Hence, the flow is laminar. Therefore the pressure difference can be deducted by the following

equation with mxxl 1012 .This is obviously a laminar flow since Re<2100, so the

calculations are valid providing the entrance length is not too long. It is

Re06.0D

LE

mDLE 083.0138606.0001.0Re06.0

this is approximately 8% of the total length, a sufficiently small quantity, hence calculations are assumed reliable. 3.5 Turbulent Flow in a pipe The study of turbulent flow in a pipe is of substantial interest in actual flows since most flows encountered in practical applications are turbulent flows in pipes. In a turbulent flow all three velocity components are nonzero. If we measure the components as a function of time, the graphs similar to those shown in Figure 3.6 would result for a flow in a pipe where u, v and w are in the x-, r- and θ-directions. However, there is seldom interest to the civil engineers in the details of the randomly fluctuating velocity components; hence, here, only the notation of a time-averaged quantity will be introduced. The velocity components (u,v,w) are written as

'uuu (16) 'vvv (17)

'www (18) where a bar over a quantity denotes a time average and a prime denotes the fluctuating part. Using the component u as an example, the time average is defined as,

T

dttuT

u0

)(1

(19)

Where T is a time increment large enough to eliminate all time dependence from u . As the flow

in the pipe becomes developed it is observed that the horizontal velocity parameter u would be

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nonzero and v =0, w =0 will be attained.

Figure 3.6 Velocity components in a turbulent pipe flow

All materials are rough when viewed with sufficient magnification, although glass and plastic assumed to be smooth with e=0 (sometimes mentioned as ks or δ). As noted in the preceding sections shear stresses are significant only near the wall in the viscous wall layer with thickness δv. If the thickness δv is sufficiently large, it submerges the wall roughness elements so that they have negligible effect on the flow; it is as if the wall were smooth. Such a condition is often referred to as being hydraulically smooth. If the viscous wall layer is relatively thin, the roughness elements protrude out of this layer and the wall is rough. The relative roughness e/D and the Reynolds number can be used to determine if a pipe is smooth or rough. (See Figure 3.7)

Figure 3.7: (a) A smooth wall and (b) a rough wall

Question 3-5

Water at 20oC flows in a 10 cm diameter pipe at an average velocity of 1.6 m/sec. If the friction factor, f of the pipe is 0.018, calculate the wall shear stress and friction (shear) velocity of the pipe.

Solution 3-5

In the wall region the characteristic velocity and the length are the shear velocity uτ is defined by

/ou (20)

and the wall shear is calculated from

fVo2

8

1 (21)

Therefore the wall shear stress occurring on the walls of pipe is calculated as:

Pa8.5018.06.110008

1 2

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and the friction velocity is

076.01000

8.5 m/sec

Question 3-6

The 4-cm diameter smooth, horizontal pipe as shown in the Figure transports 0.004 m3/sec of water at 20oC. Calculate (a) the wall shear, (b) the pressure drop over a 10-m length, f of the pipe is 0.018.

Solution 3-6

The average flow velocity in the pipe can be obtained by using the discharge relation, AVQ

sec/18.302.0

004.02

mA

QV

As it was same in the previous question, the wall shear stress occurring on the walls of pipe can be calculated as:

Pao 23018.018.310008

1 2

As it was defined before, the pressure drop and wall shear stress are related by

D

lp o4

2300004.0

23104

p Pa

3.6 Losses in Developed Pipe Flow

Perhaps the most calculated quantity in pipe flow is the head loss. If the head loss is known in a developed flow, the pressure change can be calculated; The head loss that results from the wall shear in a developed flow is related to the friction factor by the Darcy-Weisbach equation, namely,

g

V

D

Lfh

pL 2

2

(22)

Consequently, if the friction factor is known, we can find the head loss and then the pressure drop. The friction factor f, depends on the various quantities that affect the flow, written as,

),,,,( eDVff (23)

where the average wall roughness height e accounts for the influence of the wall roughness elements. A dimensional analysis provides us with

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),(D

eVDff

(24)

where e/D is the relative roughness. Experimental data that relate the friction factor to the Reynolds number have been obtained for fully developed pipe flow over a wide range of wall roughnesses. The results of this data are presented in Figure 3.8, which is commonly referred to as the Moody diagram, named after Lewis F. Moody (1880-1953). There are several features of the Moody diagram that should be noted.

For a given wall roughness, measured by the relative roughness e/D, there is a sufficiently large value of Re above which the friction factor is constant, thereby defining the completely turbulent regime. The average roughness element size e is substantially greater than the viscous wall layer thickness δv, so that viscous effects are not significant.

For the smaller relative roughness e/D values it is observed that, as Re decreases, the friction factor increases in the transition zone and eventually becomes the same as that of a smooth pipe. The roughness elements become submerged in the viscous wall layer so that they produce little effect on the main flow.

For Reynolds numbers less than 2000, the friction factor of laminar flow is shown. The critical zone couples the turbulent flow to the laminar flow and may represent an oscillatory flow that alternately exists between turbulent and laminar flow.

The following empirical equations represent the Moody diagram for Re>4000:

Smooth pipe flow:

8.0Reln86.01

ff

(25)

Transition zone:

fD

e

f Re

51.2

7.3ln86.0

1 (26)

Completely turbulent zone:

D

e

f 7.3ln86.0

1 (27)

The transition zone equation that couples the smooth pipe equation to the completely turbulent regime equation is known as the Colebrook Equation. Smooth pipe flow is the Colebrook equation with e=0, and completely turbulent zone equation is the Colebrook equation with Re=∞

Three categories of problems can be identified for developed turbulent flow in a pipe of length L.

Catagory Known Unknown

1 Q,D,e,ν hL

2 hL,D,e,ν Q

3 Q,e,ν,hL D

A category 1 problem is straightforward and requires no iteration procedure when using the

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Moody diagram. Category 2 and 3 problems are more like problems encountered in engineering design situations and require an iterative trial and-error process when using the Moody diagram. Each of these types will be illustrated with following examples.

Figure 3.8: Moody diagram (Note that if e/D or ks/D=0.01 and Re=104, the f reads f=0.043

Question 3-7

Water at 20oC is transported for 500m in a 4 cm-diameter wrought iron (commercial steel) horizontal pipe with a flow rate of 3L/sec. Calculate the pressure drop over the 500-m length of pipe, using Moody diagram.

Solution 3-7

The average flow velocity in the pipe can be obtained by using the discharge relation, AVQ

sec/38.202.0

1032

3

mA

QV

The Reynolds number is 4

61052.9

10007.1

100438.2Re

UD

Obtaining e or ks from the Moody diagram we have, using D=4/100

00115.040

046.0

mm

mm

D

e

The friction factor is read from the Moody diagram to be 023.0f

The head loss is than calculated as

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g

V

D

Lfh

pL 2

2

mhL 8381.92

38.2

04.0

500023.0

2

The pressure drop can also be calculated as

839810 Lhp =814230N/m2 or

23.814p kPa

Question 3-8

A pressure drop of 700kPa is measured over a 300-m length of horizontal, 10-cm diameter commercial pipe that transports oil (S=0.9, ν=10-5 m2/s). Calculate the flow rate using Moody diagram.

Solution 3-8

The relative roughness is

00046.0100

046.0

mm

mm

D

e

Assuming that the flow is completely turbulent (Re is not needed), the Moody diagram reads to be

0165.0f

The head loss is than calculated as

g

V

D

Lfh

pL 2

2

4.799.09800

700000

oilL

ph

m note that

water

oilS

The velocity can be calculated from the head loss equation of Darcy Weisbach

g

V

D

LfhL 2

2

fL

gDhV L2

sec/61.53000165.0

4.791.081.925.0

mV

This provides us with a Reynolds number of

45

1061.510

1.061.5Re

UD

using this Reynolds number and e/D=0.00046, the Moody diagram gives the friction factor as

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023.0f

This corrects the original value for f. the velocity is recalculated to be

sec/75.4300023.0

4.791.081.925.0

mV

This provides us with a Reynolds number of

45

1075.410

1.075.4Re

UD

From the Moody diagram f=0.023 appears to be satisfactory. Thus the flow rate is

037.005.075.4 2 AVQ m3/sec

Question 3-9

Drawn tubing of what diameter should be selected to transport 0.002 m3/sec of 20oC water over a 400-m length so that the head loss does not exceed 30-m? Use Moody diagram to solve the question

Solution 3-9

In this problem we do not know D. Thus, a trial and error solution is anticipated. The average is related to D by

22

00255.0

4

002.0

DDA

QV

The friction factor and D are related as follows

g

V

D

LfhL 2

2

81.92

)/00255.0(40030

22

D

Df

Therefore,

fD 65 1042.4

The Reynolds number is

DD

DUD 2550

10

00255.0Re

62

Now, let us simply guess a value for f and check with the relations above and the Moody diagram. The first guess is f=0.03 and the correction is listed in the following table. Note: the second guess is the value for f found from the calculations of the first guess.

f D(m) Re e/D f(from Moody)

0.03 0.0421 6.06x104 0.000036 0.02 0.02 0.0388 6.57x104 0.000039 0.02

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The value of f=0.02 is acceptable, yielding a diameter of 3.88 cm. Since this diameter would undoubtedly not be a standard a diameter of

4D cm Would be the tube size selected. This tube would have a head loss less than the limit of 30-m imposed in the problem statement. 3.6 Losses in Developed noncircular conduits. Good approximations can be made for the head loss in circuits with noncircular cross sections by using the hydraulic radius R, defined by:

P

AR (28)

Where A is the cross-sectional area and P is the wetted perimeter where the fluid is in contact with the solid boundary.

Question 3-10

Air at standard conditions (ν=1.5x10-5) is to be transported through a smooth, horizontal, 30 cm x 20 cm rectangular duct whose length is 500-m. The flow rate is 0.24 m3/sec. Calculate the pressure drop.

Solution 3-10

For a circular pipe flowing full, the hydraulic radius is;

220

0

20 r

r

rR

Hence we simply replace the radius with double R and use the Moody diagram with

)4(

ReRUUD

And relative roughness as

R

e

D

e

4

The head loss therefore becomes,

g

V

R

Lf

g

V

D

LfhL 242

22

The pressure drop in the duct can be calculated by

g

V

D

Lfh

pL 2

2

Therefore it is necessary to calculate the head loss in the duct which is directly related with the friction coefficient of Darcy. On the other hand, since the duct is rectangular it is necessary to find the hydraulic radius instead of D. The hydraulic radius is

- 17 -

06.02)2.03.0(

3.02.0

P

AR m

The average velocity is

42.03.0

24.0

A

QV m/sec

This gives a Reynolds number of

45

104.6105.1

06.044)4(Re

RU

Using the smooth pipe curve of the Moody diagram, there results 0196.0f

Hence,

3.3381.92

4

06.04

5000196.0

24

22

g

V

R

LfhL m

Therefore, the pressure drop is

4023.338.923.1 Lghp Pa

The Darcy-Weisbach equation is valid for the laminar or turbulent pipe flow of any

Newtonian fluid. A number of additional equations are also available for evaluating the head loss in a pipe due to friction. These tend to be more empirical in nature and limited to the turbulent flow of water. This limitation is rarely restrictive in hydraulic engineering. Their main advantage is generally their ease of application. Although they are less accurate than the Darcy-Weisbach equation, many hydraulic engineering applications do not require or justify the higher accuracy anyway. One such empirical equation is the Hazen-Williams equation:

54.063.032.1 SRCV H (29)

or 54.063.085.0 SRCV H (30)

For use with U.S. customary or SI units, respectively. In either case, CH is a discharge coefficient with typical values given in the following table

Hazen Williams coefficient for pipes Pipe description or material CH Extremely smooth and straight 140 Asbestos-cement 140 New, unlined Steel 140-150 New, riveted Steel 110 Old, riveted Steel 95 Very smooth Steel 130 New,unlined Cast iron 130 5-year-old Cast iron 120 10-year-old Cast iron 110 20-year-old Cast iron 100 40-year-old Cast iron 65-80 Smooth wood or masonary 120 Vitrified clay 110 Ordinary brick 100 Old pipe in bad condition 60-80

- 18 -

R is the hydraulic radius (equal to D/4 in a full pipe), and S is the slope of the total head or energy grade line, that is, the ratio of the head loss to the pipe length. There are both nomograph and tabular solutions of the Hazen-Williams equation that provide rapid solutions to head loss/discharge problems. 3.7 Minor Losses in Pipe Flow We now know how to calculate the losses due to a developed flow in a pipe. Pipe systems do, however, include valves, elbows, enlargements, contractions, inlets, outlets, bends etc. that cause additional losses. Each of these devices causes a change in the magnitude and/or the direction of the velocity vectors and hence results in a loss. In general, if the flow is gradually accelerated by a device, the losses are very small; relatively large losses are associated with sudden enlargements or contractions. A minor loss is expressed in terms of a loss coefficient K, defined by

g

vKhm 2

2

(31)

the values of K have been determined experimentally for the various fittings and geometry changes of interest in piping systems. One exception is the sudden expansion from Area A1 to area A2 for which the loss can be calculated as

2

2

11

A

Ahm (32)

Question 3-11

If the flow rate through a 10-cm diameter wrought iron pipe is 0.04m3/sec, find the difference in elevation H of the two reservoirs.

Solution 3-11

The energy equation written for a control volume that contains the two reservoir surfaces can be given as

2

222

1

211

22z

g

vPhhz

g

vPmL

where velocities and pressures at point 1 and point 2 are equivalent and equal to zero.

21 zhhz mL

Thus letting z1-z2=H, we have

- 19 -

Figure 3.9: Minor Loss coefficients K

- 20 -

g

v

D

Lf

g

vKKKKH exitelbowvalveentrance 22

)2(22

The average velocity, Reynolds number and relative roughness are

09.505.0

04.02

A

QV m/sec

This gives a Reynolds number of

56

1009.510

01.009.5)(Re

DU

00046.0100

046.0

mm

mm

D

e

Using the Moody diagram we find that

0173.0f

Using the loss coefficient from the given table, for an entrance, a globe valve, screwed 10-cm diameter standard elbows, and an exit there results

81.92

09.5

1.0

500173.0

81.92

09.5)1)64.0(27.55.0(

22

H

mH 6.224.112.11 The minor losses are about equal to the frictional losses as expected, since there are five minor loss elements in 500 diameters of pipe length.

3.8 Hydraulic and Energy Grade Lines When the energy equation is written in the form of

2

222

1

211

22z

g

vPhhz

g

vPmL

The terms have dimensions of length. This has led to the conventional use of the hydraulic grade line and the energy grade line. The hydraulic grade line (HGL), the dashed line in Figure 3.10, in a piping system is formed by the locus of points located a distance p/γ above the center of the pipe, or (p/γ) + z above a preselected datum; the liquid in a piezometer tube would rise to the HGL. The energy grade line (EGL), the solid line in Figure 3.10 is formed by the locus of points a distance v2/2g above the HGL, or the distance (v2/2g) + (p/γ) + z above a preselected datum; the liquid in a pitot tube would rise to the EGL. The following points are noted relating to the HGL and the EGL.

As the velocity approaches to zero, the HGL and the EGL approach each other. Thus, in a reservoir, they are identical and lie on the surface.

The EGL and HGL slope downward in the direction of the flow due to the head loss in the pipe. The greater the loss per unit length, the greater the slope; As the average velocity in the pipe increases the loss per unit length increases

A sudden change occurs in the HGL and EGL whenever a loss occurs due to a sudden geometry change as represented by the valve or the sudden enlargement.

- 21 -

A jump occurs in the HGL and the EGL whenever useful energy is added to the fluid as occurs with a pump, and a drop occurs if useful energy is extracted from the flow as occurs with a turbine.

At points where the HGL passes through the centerline of the pipe, the pressure is zero. If the pipe lies above the HGL, there is a vacuum in the pipe, a condition that is often avoided. An exception to this condition is the design of a siphon.

Figure 3.10: Hydraulic Grade Line (HGL) and Energy Grade Line (EGL) for a piping system.

Question 3-12

Water at 20oC flows between two reservoirs at the rate of 0.06m3/sec, Sketch the HGL and EGL What is the minimum diameter DB allowed to avoid the occurrence of cavitation?

Solution 3-12

The EGL and the HGL will be sketched on the figure including sudden changes at the entrance, contraction, enlargement and the exit. Note the large velocity head (the difference between the EGL and the HGL) in the smaller pipe because of the high velocity. The velocity, Reynolds number, and relative roughness in the 20-cm diameter pipe are calculated to be

91.11.0

06.02

A

QV m/sec

This gives a Reynolds number of

56

1080.310

2.091.1)(Re

DU

- 22 -

0013.0200

26.0

mm

mm

D

e

Using the Moody diagram we find that

022.0f

The velocity, Reynolds number, and relative roughness in the smaller pipe are

22

0764.0

4

06.0

BB DDA

QV

This gives a Reynolds number of

BB

B

DD

DDU 76400

10

0764.0)(Re

62

BDD

e 00026.0

The minimum possible diameter is established by recognizing that the water vapor pressure at 20oC is the minimum allowable pressure. Since the distance between the pipe and the HGL is an indication of the pressure in the pipe, we can conclude that the minimum pressure will occur at section 2. Hence the energy equation applied between section 1, the reservoir surface and section 2 gives

2

22

2222

1

211

2)

2222(

2z

g

vP

g

v

D

Lf

g

v

D

Lf

g

vK

g

vKz

g

vP B

B

BB

A

AA

Bcont

Aentrance

Where the subscript A refers to the 20-cm diameter pipe. This simplifies to

81.92

91.1

2.0

30022.0

81.92

)0764.0(25.0

81.92

91.15.0(200

9810

101000 2222BD

081.92

)0764.0(

9810

2450)

81.92

)0764.0(20 2222

BB

BB

DD

Df

Which results

54

2025.198600

BB

B Df

D

Where we assume that Kcont=0.25. This requires a trial and error solution. The following illustrates the procedure

Let DB=0.1 m

Then (e/D) =0.0026 and Re 5106.7 Therefore 026.0f

520001250098600?

Let DB=0.09 m

- 23 -

Then (e/D) =0.0029 and Re 5104.8 Therefore 027.0f

910001900098600?

We can easily observe that 0.1m is too large and 0.09 m is too small. In fact, the value of 0.09 m is only slightly too small. Consequently, to be safe we must select the next larger pipe size, of 0.1meter diameter. If there were a pipe size of 9.5-cm diameter, that could be selected. Assuming that such a size is not available we could select DB=10cm 3.9 Pipeline Systems Pipes in series One type of pipe system has already been considered in Example 3-11 and 3-12, namely pipes in series. Inspection of that example would indicate the two basic premises associated with pipes in series: first, continuity requires that the discharge is the same in all pipes, and second, the total head loss for the system equals the sum of the individual head losses of the respective pipes. These principles can be generalized for any number of pipes that are connected in series. It should be noted that if minor losses are included, they must be referenced to the appropriate pipe size and velocity head. Similarly, the head loss due to friction must be evaluated separately for each pipe size. When the Darcy-Weisbach equation is used in the analysis of any type of pipe system, the resistance coefficients are frequently treated as constants.

Figure 3.11: Three pipes in series system.

Therefore, according to above definition in the pipes connected in series; The total head loss is the sum of head losses in each pipe:

LcLbLatotalL hhhh )( The flow rate, in each of the pipes connected is the same.

cbatotal QQQQ

Pipes in Parallel A second pipe system is composed of a number of pipes connected in parallel (Fig. 3-12). Although any number of pipes could be so connected, the development of the concepts will be based on just three pipes. In passing, note that the pipe sections upstream and downstream of the parallel pipes could be considered to be in series with the parallel pipe system of our immediate interest, and once the parallel pipe system is analyzed the remaining system could be treated as a number of pipes in series. Returning to the three parallel pipes, continuity can be applied with the result

- 24 -

cbatotal QQQQ

Since each of the pipes must share common values of piezometric head at the junctions, it follows that the head loss must be identical in each of the parallel pipes. This gives a second relationship that may be expressed as

LcLbLatotalL hhhh )(

Figure 3.12: Three pipes in parallel system.

Branching Pipe System Another type of pipe system may be introducing by considering the three interconnected reservoirs of Fig. 3-13. Subsequently, we will generalize the analysis to include additional reservoirs and/or additional junctions. For the present, consider the problem of estimating the discharge in each of the pipe sections.

Figure 3.13: Three-branch reservoir system. During the solution of branch systems the necessary numbers of equations are obtained by utilizing energy and continuity concepts. However, it is difficult to incorporate the velocity head at the junction, a problem that is circumvented by assuming that it is a small quantity relative to the other energy terms and may justifiably be ignored. It is usual to make the same assumption with respect to possible minor losses. A second problem is that the flow direction in pipe DB is not known at the outset. Our procedure will be to formulate the energy equation so as to determine the actual flow direction in pipe DB and then proceed on the basis of the outcome of this first calculation. If the piezometric head (since the velocity head is neglected) at the junction is assumed equal to the elevation of reservoir surface B, then no flow would occur in pipe DB, and the continuity equation becomes QAD=QDC. In addition, the following energy equations can be written (neglecting the velocity heads):

- 25 -

ADLD

DA hP

zz

DBLD

DB hP

zz

DCLD

DC hP

zz

Where the head losses are defined using the Darcy-Weisbach equation.

g

V

D

Lfh AD

AD

ADADLAD 2

2

g

V

D

Lfh DB

DB

DBDBLDB 2

2

g

V

D

Lfh DC

DC

DCDCLDC 2

2

the continuity equation is

DCDBAD QQQ

or

DCDCDBDBADAD VAVAVA

By substituting the headloss expressions respectively into equations, the three energy equations have four unknowns PD/γ, VAD, VDB and VDC. The continuity equation provides the fourth equation to solve for the four unknowns.

Question 3-13

Water flows at a rate of 0.030-m3/sec from reservoir 1 to reservoir 2 through three pipes connected in series (f=0.025) as shown in the following figure. Neglecting minor losses, determine the difference in water surface elevation.

Solution 3-13

Write down the energy equation from reservoir 1 to reservoir 2.

- 26 -

2

22

1

211

22z

g

vPhz

g

vP BL

21 zhz L where

21 zzhL

Writing the total head loss in expanded form results in:

CBA LLLL hhhh

Which results in:

21 zzhhhCBA LLL

Using the Darcy-Weisbach equation

g

V

D

Lfh A

A

AALA 2

2

g

V

D

Lfh B

B

BBLB 2

2

g

V

D

Lfh C

C

CCLC 2

2

Therefore the energy equation is turn out to be

)81.9(21000/220

2000025.0

)81.9(21000/180

1500025.0

)81.9(21000/200

1000025.0

222

21CBA VVV

zz

22221 58.1162.1037.6 CBA VVVzz

Use continuity equation to determine the velocities;

955.0)1000/200(

403.02

A

A A

QV m/sec

180.1)1000/180(

403.02

B

B A

QV m/sec

79.0)1000/220(

403.02

C

C A

QV m/sec

222

21 )790.0(58.11)180.1(62.10)955.0(37.6 zz

22.779.1481.521 zz

82.2721 zz m

Question 3-14

The three pipe system shown in Figure below has the following characteristics pipe D (cm) L (m) f

A 25 500 0.020 B 30 650 0.025 C 35 1000 0.030

Find the flowrate of water in each pipe and the pressure at point 3. Neglect minor losses

- 27 -

Solution 3-14

Write down the energy equation from 1 to 2.

2

222

1

211

22z

g

vPhz

g

vPL

802

02000022 g

vhL

Where V2 is Vc and CA LLL hhh and using the Darcy Weisbach equation, we get

g

v

D

Lf

g

v

D

Lf

g

v C

C

CC

A

A

AA

C

222120

222

Because the pipes A and B are parallel, the headloss in A is equal to the headloss in B, so the headloss for B can also be used in the above energy equation instead of for A. This energy equation has two unknowns VA and VC, so that continuity can be used as a second equation.

CBA QQQ

CCBBAA vAvAvA

Which introduces a third unknown VA. Since BA LL hh the third equation is

g

v

D

Lf

g

v

D

Lf B

B

BB

A

A

AA 22

22

)81.9(230.0

650025.0

)81.9(225.0

500020.0

22BA vv

22 760.2039.2 BA vv

AB vv 859.0

So now, we have three equations and three unknowns. Using the continuity equation, we get

CBA vvv4

)35.0(

4

)30.0(

4

)25.0( 222

CBA vvv 096.007.005.0

Substituting

AB vv 859.0

CAA vvv 096.0)859.0(07.005.0

CA vv 096.011.0

CA vv 872.0

Substituting CA vv 872.0 into energy equation

- 28 -

)81.9(235.0

100003.0

)81.9(2

)872.0(

25.0

500020.0

)81.9(2120

222CCC vvv

And solve for Vc

)81.9(271.126

)81.9(2)71.85401(120

22CC vv

31.4Cv m/sec

The flow rate is then,

414.031.4435.0 2

CCvAQ m3/sec

The pressure at 3 can be computed using the energy equation from 1 to 3 or from 3 to 2. Using

2

222

)23(3

233

22z

g

vPhz

g

vPL

Since V3=V2, the velocity head terms cancel out:

80140 )23(3 Lh

P

603

P

g

v

D

Lf C

C

CC 2

2

)81.9(2

)31.4(

35.0

1000030.060

23 P

15.81603 P

15.213 P

m

and

15.21100081.93 P

5.2074813 P Pa

48.2073 P kPa (kN/m2)

Question 3-15

The pipe system shown in Figure below connects two reservoirs that have an elevation difference of 20m. This pipe system consists of 200m of 50-cm concrete pipe (pipe A), that branches into 400-m of 20-cm pipe (pipe B) and 400-m of 40-cm pipe (pipe C) in parallel. Pipes B and C join into a single 50-cm pipe that is 500m long (pipe D). For f=0.030 in all pipes, what is the flow rate in each pipe of the system?

- 29 -

Solution 3-15

The objective is to compute the velocity in each pipe. We know that VA=VD because they are the

same diameter pipe, CB LL hh because pipes B and C are in parallel, and

CBDA QQQQ

Express CB LL hh in term of the velocities,

g

v

D

Lf

g

v

D

Lf C

C

CC

B

B

BB 22

22

Since CB ff and LB=LC

C

C

B

B

D

v

D

v 22

1004010020

22CB vv

or

22

2

1CB vv or

2C

B

vv or BC vv 2

Using CBA QQQ , we get

CBA vvv4

)10040(4

)10020(4

)10050( 222

CBA vvv 222 402050

substituting BC vv 2 yields

BBA vvv 216004002500

BA vv 065.1 or AB vv 939.0

Next convert the parallel pipes to a single equivalent DA=DD=50-cm diameter pipe, with a length LE.

g

v

D

Lf

g

v

D

Lf A

A

EB

B

B

22

22

22A

A

EB

B

B vD

Lv

D

L

22

1005010020

400A

EB v

Lv

221000 AEB vLv

882EL m

Write the energy equation from reservoir surface to reservoir surface 20 Lh m.

g

vLLLf ADEA

210050

)(20

2

81.9210050

)500882200(030.020

2

Av

033.2Av m/sec.

- 30 -

399.0)033.2(4

)10050( 2

AQ m3/sec.

also

DA vv and AD QQ

909.1)033.2(939.0 Bv m/sec

060.0)909.1(4

)10020( 2

BQ m3/sec.

BAC QQQ

339.006.0399.0 m3/sec.