Flow Through Pipes

Lecture slides by

Sachin Kansal

NATIONAL INSTITUTE OF TECHNOLOGY

KURUKSHETRA

2

Objectives

• Understand the various types of losses in Pipe flow

• To know concept of Total Energy Line and Hydraulic

Grade Line

• Understand the head loss in combination of pipes

• To know the fluid power transmitted by the pipe and

nozzles

• Understand the concept of water hammer

3

4

Introduction

• Since the fluid in a pipe is in motion, it has to

overcome the frictional resistance between the

adjacent fluid layers and that between the fluid

layer and pipe walls. As fluid flows from one point

to another, there is a loss of head due to friction,

which is regarded as a Major loss .

• Moreover, there are some other type of losses will

also be there considered as Minor loss

• Thus there is a drop in energy gradient line. In a

fully developed pipe flow, the pressure drops

linearly along the length of the pipe. Therefore the

pressure gradient along the flow remains constant.

LOSS OF ENERGY IN PIPE Flow

5

LOSS OF ENERGY IN PIPE Flow

6

• The loss of head or energy due to friction in a pipe is

known as major loss.

• Loss of energy due to change of velocity of the flowing

fluid in magnitude or direction is called minor loss of

energy. The minor loss of energy (or head) includes the

following cases :

1. Loss of head due to sudden enlargement

2. Loss of head due to sudden contraction

3. Loss of head at the entrance of a pipe

4. Loss of head at the exit of a pipe

5. Loss of head due to an obstruction in a pipe

6. Loss of head due to bend in the pipe

7. Loss of head in various pipe fittings

Consider uniform horizontal pipe with steady flow.

Let 1-1 and 2-2 are two section

of pipe.

Let,

P1 = pressure intensity at section

1-1 v1 = velocity of flow at

section 1-1

L = length of pipe between

sections d = diameter of the

pipe

f’ = frictional resistance per unit wetted

are per unit velocity

h = loss of head due to friction

Darcy-Weisback equation for head loss

due to friction: L1

1

2

2

P1 P2F1

F1

d

7

P2 = pressure intensity at section

2-2 v2 = velocity of flow at

section 2-2

Applying Bernoulli’s equation at section 1-1and

2-2

p1/ρg + v12/2g + Z1 = p2/ρg + v2

2/2g + Z2 +

hf

But pipe is horizontal and diameter same at both the section

so Z1 = Z2 and v1 =v2 p1/ρg =

p2/ρg + hf

Therefore,

p1/ρg - p2/ρg = hf or (p1 - p2) = ρ

g hf

1

1

2

2

P1 P2F1

F1

d

8

Now,

Frictional resistance = frictional resistance per unit wetted area per unit velocity * wetted area * (velocity)2

F1 = f’ * πdL * v2

= f’ * P * L * v2

The force acting on the fluid between the

sections

1

1

2

2

P1 P2F1

F1

d

1. Pressure force at section 1-1 = p1 * A

2. Pressure force at section 2-2 = p2 * A

3. Frictional force as shown in the figure = F1

Resolving all the forces in horizontal direction, we

have

p1 * A - p2 * A - F1 = 0, or ( p1 - p2 ) * A = F1

Putting F1 = f’ * P * L * v2 and (p1 - p2) = ρ

g hf

9

Rewriting the equation

ρ g hf A = f’ * P * L * v2

hf = f’ * P * L * v2 / A * ρ g

Now, P/A = wetted perimeter / area =4/d

So,

hf = f’ * 4 * L * v2 / d * ρ g

1

1

2

2

P1 P2F1

F1

d

fh = 24 ∗f’ ∗ L ∗ v

ρ g d

Putting f’/ ρ = f/2 ,

where f is co-efficient of

friction.

fh =2

10

4 ∗f ∗ L ∗ v2 g d

Co-efficient of friction is given by

f = 16/Re for laminar flow

f = 0.0791/(Re)1/4 for turbulent flow (Re≥4000 but

≤105)

f = 0.0008 + 0.5525/0.257 * Re (Re ≥ 105 but ≤ 107)

Some times equation can be written as

f is called friction factor (f = 4f), which is dimensionless

quantity.

fh =2

11

f ∗ L ∗ v

2 g d

In previous derivation we have seen that hf= f’ * P * L * v2 / A * ρ g

Let P/A = wetted perimeter / area = 1/m

f’V2 =

ρ g* m *

hf

L

V =f’

ρ g *

m

∗hfL

V = C 𝑚 ∗ 𝑖

Chezy’s formula for head loss due to friction:

C = Chezy’s Constant,Lhf = i, i = loss of head per unit

length

12

Losses dueto

1. Suddencontraction

2. Sudden

enlargement

3. Bend inpipe

4. At entrance ofpipe

5. An obstruction

6. Various pipefitting

7. At exit of pipe

MINOR LOSSES:

13

Flow separation is takes place due tosudden change in diameter of pipe, itresults in turbulent eddies formation asshown in figure.

The loss of head takes place due to eddiesformation.

Let,

P1 = pressure intensity at section 1-1

v1 = velocity of flow at section 1-1

A1 = area of pipe at section 1-1

P2 ,v2 , A2= corresponding value at section 2-2

P’ = pressure intensity of liquid eddies on the area(A2 – A1 )

he = loss of head due to sudden enlargement

Loss of head due to sudden enlargement:

P1 A1

14

P2 A2

Applying Bernoulli’s equation to section 1-1 and 2-2p1/ρg + v1

2/2g + Z1 = p2/ρg + v22/2g + Z2 + he

Pipe is horizontal so Z1= Z2 rewriting the

equ.p1/ρg + v1

2 /2g = p2/ρg + v22 /2g+ he

he =(p1/ρg - p2/ρg) + (v 2/2g -v 2/2g )1 2

P1 A1

15

P2 A2

Consider the control volume of liquid between sections 1-1

and 2-2.

Force acting on the liquid in the direction of flow is given by,

Fx = p1A1 +p’(A2 – A1) – p2A2

Experimentally it is found that p’ = p1

So equation becomes…

Fx = p1A1 +p1(A2 – A1) – p2A2

P1 A1 P2 A2

1 1= 𝜌1 𝐴 𝑣2

Equation is simplified in the form of ….

Fx = p1A2 – p2A2

= (p1– p2) A2

Momentum of liquid at section1-1 = mass * velocity= 𝜌1 𝐴1 𝑣1 * 𝑣1

Similarly momentum at section 2-2 = 𝜌2 𝐴2 𝑣22

∴ Change of momentum = (𝜌2 𝐴2 𝑣22 -𝜌1 𝐴1 𝑣1

2 )

From continuity equation 𝑨𝟏 𝒗𝟏= 𝑨𝟐𝒗𝟐So 𝑨𝟏 = 𝑨𝟐 𝒗𝟐 /𝒗𝟏

2 2 1 1 1Change of momentum =(𝜌2𝐴 𝑣 2 - 𝜌 𝐴 𝑣2 )

= 𝜌2 𝐴2 𝑣22 - 𝜌1 𝐴2 𝑣1𝑣2

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P1 A1 P2 A2

∴ (p1– p2)/ 𝜌2 = (𝑣22 - 𝑣1 𝑣2 )

Fluid is incompressible so, (𝜌2 = 𝜌1 = 𝜌 )

∴ (p1– p2)/ 𝜌 = (𝑣22 - 𝑣1 𝑣2 )

Dividing both the sides with g,

(p1– p2)/ 𝜌g = (𝑣22 -𝑣1 𝑣2)/g

Change of momentum = 𝜌2 𝐴2 𝑣22 - 𝜌1 𝐴2 𝑣1 𝑣2

= 𝜌2 𝐴2 (𝑣22 - 𝑣1 𝑣2 )

Now net force acting on the fluid is equal to change

of momentum,

Fx= 𝜌2 𝐴2 (𝑣22 - 𝑣1 𝑣2 )

(p1– p2) A2 = 𝜌2 𝐴2 (𝑣22 - 𝑣1 𝑣2 )

Remember……

he= (p1/ρg - p2/ρg) +(v12/2g – v2

2/2g)

he = (𝑣22 -𝑣1 𝑣2 )/g + (v 2/2g -v 2/2g )

2 217

P1 A1 P2 A2

he= 1 1 2 2𝑣 2 −2𝑣 𝑣 + 𝑣 2

2g

e∴ h

=

1

18

2(𝑣 −𝑣 )2

2g

The above equation is used for calculation of head loss due

to sudden enlargement.

2. Loss of head due to sudden contraction:

C

C

P1 A1 P2 A2

1

19

2

Figure shows flow of liquid through section1-1 and 2-2.

Flow of liquid is from large pipe 1-1 to small pipe 2-2.

Flow at section c-c is minimum so sectionc-c is called vena-contract.

Let,

Ac = area of flow at section c-c

vc = velocity of flow at section c-c

A2 = area of flow at section 2-2

v2 = velocity of flow at section 2-2

hc = loss of head due to sudden enlargementfrom section c-c to 2-2

hc = 0.5𝑣 2

2

2g25

C

C

P1 A1 P2 A2

1 2

Similar to loss of head due to suddenenlargement,

ch = c 2(𝑣 −𝑣 ) 2

2g

2g 𝑣2

= 𝑣2

2 ( 𝑣c − 1)2

𝑣2

2g𝟏

𝑪𝒄hc = 2 ( − 1)2

= 𝐾 𝑣2

2

2g 𝑪𝒄where K=(

𝟏 − 1)2

From continuity

equation So

𝑨𝒄𝒗𝒄= 𝑨𝟐𝒗𝟐𝒗𝒄/ 𝒗𝟐 = 𝑨𝟐/𝑨𝒄 =1/ Cc (Cc = 𝑨𝒄/𝑨𝟐)

∴ hc = 0.375 2

If Cc is assumed to be 0.62,K = 0.375𝑣 2

2gGenerally,

The loss of head when pipe is connectedto a large tank or reservoir.

The loss is similar to loss of head due to sudden contraction.

In general,

i enth or h =𝐾 𝑣

22

2g

The value of k is as per given

table.

3. Loss of head at the entrance of pipe:

21

The loss of head at the exit of pipeas a result of form of a free jet orit may be connected to reservoir.

General equation

oh =𝑣2

2g

v = velocity at the outlet of the

pipe.

4. Loss of head at the exit of pipe:

22

As shown in the figure if there is anyobstruction in pipe, it cause loss ofenergy.

. There is sudden enlargement of thearea of flow beyond the obstructiondue to which loss of head take place.

Loss of head due to obstruction is

2

g

𝐶𝑐(𝐴−𝑎)=

𝑣2 (

𝐴 − 1)2

Where,

Cc = coefficient of contraction

A = are of pipe

a = maximum area of obstruction

v = velocity of liquid in pipe

5. Loss of head due to an obstruction in pipe:

23

g

Due to bend in pipe, change in velocity as well formation of eddies take place.

It result in loss of energy due to bend in pipe.

It can be express as per below:

fh =𝐾 𝑣

22

2g

The value of K is different for different bend angle and R/D ratio.

So, The value of k depends on

• Angle of bend,

• Radius of curvature of bend,

• Diameter of pipe.

6. Loss of head due to bend in pipe:

24

6. Loss of head due to bend in pipe:

25

To analyze the pipe problem, the conceptof HEL or TEL is very useful.

Both are graphical representation of thelongitudinal variation in the total head atsalient points of pipe line.

Total head as per Bernoulli’s equationgiven by p/𝛾 + v2/2g + Z = Constant

p/𝛾= Pressure Head,

v2/2g= Velocity Head

Z = Potential Head

HYDRAULIC GRADIENT AND TOTAL

ENERGY LINE:

26

Because of friction effects associated with fluid flow and the local

resistance arising from pipe transmissions and fittings, a part of

energy is dissipated. Evidently there is loss of head, and energy

drops in the direction of flow by an amount equal to the head loss.

Line representing the sum of pressurehead, datum head and velocity headwith respect to some reference line iscalled total energy line (TEL).

It is represented by line connecting thevalues of the total head at successivepoints along a piping system.

So, It is also defined as the line whichis obtained by joining the tops of allvertical ordinates showing the sum ofpressure head and kinetic head fromthe centre of the pipe

For ideal fluid as there are no losses,total energy line would remain parallelto the datum.

Total energy line (TEL):or Energy Gradient

Line (EGL)

27

Hydraulic grade line (HGL):

Line representing sum of pressurehead and datum head with respectto some reference line (datum) iscalled as hydraulic gradient line(HGL).

It the line obtained by connectingthe values of the piezometric head(p/𝛾 + Z) at successive points alongthe piping system.

HGL is always vertically below TELby and equal amount to the velocityhead (v2/2g).

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Two or more pipes of different diameter are connected end to

end to form a single pipe line.

Pipes can be same or different length.

Discharge through all pipes is same.

Q = Q1= Q2 =Q3

Total loss of head is sum of losses in all individual pipes and

fittings.

H = hf1 + hf2 + hf3+……….+ minor losses

For given figure

If minor losses areneglected,2 2

H =4𝑓1𝐿1𝑣1 + +

4𝑓2𝐿2𝑣2 +4𝑓3𝐿3𝑣3

2

2𝑔𝑑1 2𝑔𝑑2 2𝑔𝑑3

PIPE IN SERIES OR COMPUNDPIPE:

29

2 2

H =4𝑓1𝐿1𝑣1 +

4𝑓2𝐿2𝑣2 +4𝑓3𝐿3𝑣3

2

2𝑔𝑑1 2𝑔𝑑2 2𝑔𝑑3+ minor losses

EQUIVALENT PIPE:

30

• This is defined as the pipe of uniform diameter having loss

of head and discharge equal to the loss of head and

discharge of a compound pipe consisting of several pipes

of different lengths and diameters.

• The uniform diameter of the equivalent pipe is called

equivalent size of the pipe.

• The length of equivalent pipe is equal to sum of lengths of

the compound pipe consisting of different pipes.

• Let, L1 = length of pipe 1 and d1 =diameter of pipe 1

• L2 = length of pipe 2 and d2 =diameter of pipe 2

• L3 = length of pipe 3 and d3=diameter of pipe 3

• H = total head loss

EQUIVALENT PIPE:

loss of head

and discharge

loss of head and discharge of

compound pipe=

L1d1v1 L2d2v2 L3d3v3

Then, L= L1 + L2 +L3

Total head loss in the compound pipe neglecting minorlosses,

H=𝟐𝒈𝒅𝟏 𝟐𝒈𝒅𝟐

𝟏 𝟏 𝟏 + 𝟐 𝟐 𝟐 + 𝟑𝟑𝟑𝟒𝒇 𝑳 𝒗 𝟐 𝟒𝒇 𝑳 𝒗 𝟐 𝟒𝒇𝑳𝒗𝟐

𝟐𝒈𝒅𝟑(Assuming f = f1 = f2 = f3)

Discharge Q =A1v1 = A2 v2 = A3 v3

Q = 𝝅d 2 v1 1𝟒

𝟒Q𝟒

= 𝝅d 2 v2 2 𝟒

31

=𝝅d 2 v3 3

1v =𝝅d1

2 2, v =𝟒Q𝝅d2

2 3, v =𝟒Q𝝅d3

2

L1d1v1

L d v2 2 2

L d v3 3 3

Putting the value of velocity in head lossequation

H=𝟏 𝟏

1𝟐𝒈𝒅𝟏

+ +𝟐 𝟐

𝝅𝝅d d22 2

𝟒𝒇 𝑳 (𝟒Q )𝟐 𝟒𝒇𝑳(

𝟒Q )𝟐

𝟐𝒈𝒅𝟐+𝟒𝒇𝑳

𝟑 𝟑𝟒Q𝝅d3

2

𝟐

𝟐𝒈𝒅𝟑

H = 𝟒∗𝟏𝟔𝒇Q

𝟐

𝝅𝟐∗𝟐𝒈

𝑳𝟏+ 𝑳𝟐+ 𝑳𝟑

𝟏 𝟐 𝟑𝒅 𝟓 𝒅 𝟓 𝒅 𝟓

Head loss in equivalentpipe

𝟐𝒈𝒅 𝝅𝟐∗𝟐𝒈H =

𝟒𝒇𝑳𝒗2

= 𝟒∗𝟏𝟔𝒇Q 𝑳

𝟐𝒅𝟓

Equating both head loss equation

𝝅𝟐∗𝟐𝒈𝟏 𝟐 𝟑

𝒅 𝟓 𝒅 𝟓 𝒅 𝟓 𝝅𝟐∗𝟐𝒈

𝟒∗𝟏𝟔𝒇Q𝟐 𝑳𝟏 + 𝑳𝟐 + 𝑳𝟑 = 𝟒∗𝟏𝟔𝒇Q𝟐 𝑳

𝒅𝟓

𝒅𝟓𝑳

=𝑳𝟏+ 𝑳𝟐 + 𝑳𝟑

32

𝟓 𝟓 𝟓𝒅𝟏 𝒅𝟐 𝒅𝟑

The above equation is

known as Dupuit’s

equ.

The purposeof the parallelpipe is to increase the discharge of the fluid.

For parallel pipes

Q = Q1 + Q2 + Q3+…….

Head loss through each branch is same

hf1 = hf2 = hf3 =……..

As per figure

Q = Q1 + Q2 &

hf1 = hf2

𝟐𝒈𝒅𝟏

𝟏 𝟏 𝟏 = 𝟐𝟐𝟐𝟒𝒇 𝑳 𝒗 𝟐 𝟒𝒇𝑳𝒗𝟐

𝟐𝒈𝒅𝟐

For f1 =f2

𝒅𝟏

𝟏𝟏 = 𝟐 𝟐𝑳 𝒗𝟐𝑳𝒗𝟐

𝒅𝟐

PIPES IN PARALLEL:

33

The hydraulic power transmitted by apipe depends on:

1.Discharge

2.Total head

Head available at the outlet of pipe isH -hf

The power available at the outlet of the pipe is

P = 𝜌 𝑔 𝑄 (H –hf) , Watts i.e.

The maximum power transmission is obtained by 𝑑𝑃 = 0 i.e.

𝑑𝑣

hf = H/3

POWER TRANSMISSION THROUGH THE

PIPES:

34

η = 𝑝𝑜𝑤𝑒𝑟 delivered at the outlet of the pipe

𝑝𝑜𝑤𝑒𝑟 supplied at the inlet to the pipe

=𝜌 𝑔𝑄 (H –hf) 𝜌 𝑔 𝑄H

=(H –hf)

H

Under condition of maximum power the efficiency is given by

η =(H –H/3)

H= 2/3

= 66.67%

35

EFFICIENCY OF POWER TRANSMISSION:

36

Flow Through Nozzles

37

38

39

40

Water Hammer in Pipes

41

42

43

44

45

References:

46

1. Fluid Mechanics and Fluid Power Engineering by D.S. Kumar, S.K.Kataria &

Sons

2. Fluid Mechanics and Hydraulic Machines by R.K. Bansal, Laxmi Publications

3. Fluid Mechanics and Hydraulic Machines by R.K. Rajput,

S.Chand & Co

4. Fluid Mechanics; Fundamentals and Applications by John. M. Cimbala Yunus A. Cengel, McGraw-Hill Publication