Module-5 Flow Through Pipes

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Module-5

Flow Through Pipes

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98795 10743

Fluid Mechanics & Hydraulics (2140611)

10-02-2020

Prof. Mehul Pujara

2Module-5 Flow Through Pipes Darshan Institute of Engineering & Technology, Rajkot

Fluid:

• The term fluid refers to both aliquid and a gas; it is generallydefined as a state of matter inwhich the material flows freelyunder the action of a shearstress.


CONTINUITY EQUATION:

It follows “Principle of conservationof mass”

“Mass neither be created nor bedestroyed, total mass of systemremains constant.”

Mass offluidenteringunit time

Mass offluidmovingout perunit time

Rate ofincreaseof massper unittime

- =



Considering flow though the stream tube.

Mass of fluid entering into stream tube perunit time is

= Density * Discharge

= Density * Area * velocity

= 𝜌 𝐴 𝑣 (1)

Mass of fluid coming out from stream tubeper unit time is

= 𝜌 𝐴 𝑣 +𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠ds (2)

Net mass of fluid reaming in stream tubeper unit time is

= -𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠ds (3)

𝜌 𝐴 𝑣

𝜌 𝐴 𝑣 +𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠ds


The mass of fluid in stream tube is

= density * volume of fluid in tube

= 𝜌 𝐴 ds (4)

Rate of increase of mass of fluid with timein stream tube is

=𝜕 (𝜌 𝐴𝑑𝑠 )

𝜕𝑡

=𝜕 (𝜌 𝐴 )

𝜕𝑡𝑑𝑠 (5)

Net mass remained in the stream tube per unittime is equal to rate of increase of mass withthe time.

Equation (3) = Equation (5)

-𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠𝑑𝑠 =

𝜕 (𝜌 𝐴 )

𝜕𝑡𝑑𝑠


-𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠= 𝜕 (𝜌 𝐴 )

𝜕𝑡

𝜕 (𝜌 𝐴 )

𝜕𝑡+ 𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠= 0 (6)

Equation (6) is known as continuityequation for one dimensional flow, it isapplicable to all cases of flow.


For Steady flow𝜕 (𝜌 𝐴 )

𝜕𝑡= 0

Therefore𝜕 (𝜌 𝐴 𝑣)

𝜕𝑠= 0

𝜌 𝐴 𝑣 = Constant

At section 1 and 2 we can write continuityequation

𝜌1 𝐴1 𝑣1 = 𝜌2 𝐴2 𝑣2

In case of incompressible flow

𝜌1 = 𝜌2

𝐴1 𝑣1 = 𝐴2 𝑣2


1

2


Hydrodynamics

Aerodynamics

Electromagnetism

Quantum mechanics

Flow of fluid through pipe, ducts or tubes

Rivers

Process plants

Power plants

Dairies

Logistics in general

CONTINUITY EQUATION APPLICATION:

1

2


Consider Steady flow

Incompressible flow

Momentum at section 1-1

= 𝜌1 𝑄 𝑣𝑥1

Momentum at section 2-2

= 𝜌2 𝑄 𝑣𝑥2

Change of momentum

= (𝜌2 𝑄 𝑣𝑥2) – (𝜌1 𝑄 𝑣𝑥1)

For incompressible flow

𝜌1 =𝜌2

Change of momentum = 𝜌 𝑄 (𝑣𝑥2- 𝑣𝑥1)

MOMENTUM EQUATION:

• By impulse momentum principle…F = 𝜌 𝑄 (𝑣𝑥2- 𝑣𝑥1)


Now, components of velocity

𝑣𝑥1 = 𝑣1 cosθ1 , 𝑣𝑥2 = 𝑣2 cosθ2

𝑣𝑦1 = 𝑣1 sinθ1 , 𝑣𝑦2 = 𝑣2 sinθ2

Component of force along x-axisand y-axis

Fx= ρ Q (𝑣2 cosθ2 - 𝑣1 cosθ1 )

Fy= ρ Q (𝑣2 sinθ2 - 𝑣1 sinθ1 )

Above equations represent the forceexerted by the pipe bend on fluidmass.

Same but opposite force applied byfluid on pipe bend so equation is

Fx= ρ Q (𝑣1 cosθ1 - 𝑣2 cosθ2 )

Fy= ρ Q (𝑣1 sinθ1 - 𝑣2 sinθ2 )

MOMENTUM EQUATION:

The resultant force exerted by the fluid is

F = 𝐹𝑥2 + 𝐹𝑦2

The direction of resultant force is

Θ = tan-1 (FyFx

)


LOSS OF ENERGY IN PIPE:


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

hf = loss of head due to friction

Darcy-Weisback equation for head loss due to friction:

L1

1

2

2

P1 P2F1

F1

d

P2 = pressure intensity at section 2-2v2 = velocity of flow at section 2-2


Applying Bernoulli’s equation at section 1-1 and 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


Now,

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

F1 = f’ * πdL * v2

= f’ * P * L * v2


1

1

2

2

P1 P2F1

F1

d

The force acting on the fluid between the sections1. Pressure force at section 1-1 = p1 * A2. Pressure force at section 2-2 = p2 * A3. 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


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


1

1

2

2

P1 P2F1

F1

d

So,

hf = f’ * 4 * L * v2 / d * ρ g

hf = 4 ∗f’ ∗ L ∗ v2

ρ g d

Putting f’/ ρ = f/2 , where f is co-efficient of friction.

hf = 4 ∗f ∗ L ∗ v2

2 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.


hf = f ∗ L ∗ v2

2 g d


In previous derivation we have seen that

hf = f’ * P * L * v2 / A * ρ g

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

V2 = ρ g

f’* m *

hf

L

V = ρ gf’

* m ∗hf

L

V = C 𝑚 ∗ 𝑖

Chezy’s formula for head loss due to friction:

C = Chezy’s Constant, hf

L= i, i = loss of head per unit length


Example:


Losses due to

1. Sudden contraction

2. Sudden enlargement

3. Bend in pipe

4. At entrance of pipe

5. An obstruction

6. Various pipe fitting

7. At exit of pipe

MINOR LOSSES:


Flow separation is takes place due to suddenchange in diameter of pipe, it results inturbulent eddies formation as shown in figure.

The loss of head takes place due to eddiesformation.

Let,

P1 = pressure intensity 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

1. Loss of head due to sudden enlargement:

P1 A1 P2 A2


Applying Bernoulli’s equation to section 1-1 and2-2

p1/ρg + v12/2g + Z1 = p2/ρg + v2

2/2g + Z2 + he

Pipe is horizontal so Z1 = Z2 rewriting the equ.

p1/ρg + v12/2g = p2/ρg + v2

2/2g + he

he = (p1/ρg - p2/ρg) + (v12/2g - v2

2/2g )


P1 A1 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

Equation is simplified in the form of ….Fx = p1A2 – p2A2

= (p1– p2) A2

Momentum of liquid at section 1-1 = mass * velocity

= 𝜌1 𝐴1 𝑣1 * 𝑣1= 𝜌1 𝐴1 𝑣1

2

Similarly momentum at section 2-2 = 𝜌2 𝐴2 𝑣22

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

2 )

From continuity equation 𝑨𝟏 𝒗𝟏 = 𝑨𝟐 𝒗𝟐So 𝑨𝟏 = 𝑨𝟐 𝒗𝟐 / 𝒗𝟏

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

2 )

= 𝜌2 𝐴2 𝑣22 - 𝜌1 𝐴2 𝑣1𝑣2



P1 A1 P2 A2

Change of momentum = 𝜌2 𝐴2 𝑣22 - 𝜌1 𝐴2 𝑣1𝑣2

= 𝜌2 𝐴2 (𝑣22 - 𝑣1𝑣2 )

Now net force acting on the fluid is equal to change of momentum,

Fx = 𝜌2 𝐴2 (𝑣22 - 𝑣1𝑣2 )

(p1– p2) A2 = 𝜌2 𝐴2 (𝑣22 - 𝑣1𝑣2 )

∴ (p1– p2)/ 𝜌2 = (𝑣22 - 𝑣1𝑣2 )

Fluid is incompressible so (𝜌2 = 𝜌1 = 𝜌 )∴ (p1– p2)/ 𝜌 = (𝑣2

2 - 𝑣1𝑣2 )

Dividing both the sides with g,

(p1– p2)/ 𝜌 g = (𝑣22 - 𝑣1𝑣2 )/g

Remember……he = (p1/ρg - p2/ρg) + (v1

2/2g - v22/2g )

he = (𝑣22 - 𝑣1𝑣2 )/g + (v1

2/2g - v22/2g )



P1 A1 P2 A2

he = (𝑣22 - 𝑣1𝑣2 )/g + (v1

2/2g - v22/2g )

=𝑣1

2 −2𝑣1𝑣2 + 𝑣22

2g

∴ he =(𝑣1−𝑣 2)

2

2g

The above equation is used for calculation ofhead loss due to sudden enlargement.


2. Loss of head due to sudden contraction:

C

C

P1 A1 P2 A2

1 2

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

Flow of liquid is from large pipe 1-1 tosmall 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


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


2. Loss of head due to sudden contraction:

C

C

P1 A1 P2 A2

1 2

Similar to loss of head due to suddenenlargement,

hc =(𝑣c−𝑣

2)2

2g

=𝑣22

2g(𝑣c

𝑣2

− 1)2

hc =𝑣22

2g(𝟏

𝑪𝒄

− 1)2

=𝐾 𝑣

22

2gwhere K = (

𝟏

𝑪𝒄− 1)2

From continuity equation 𝑨𝒄 𝒗𝒄 = 𝑨𝟐 𝒗𝟐So 𝒗𝒄 / 𝒗𝟐 = 𝑨𝟐 / 𝑨𝒄 = 1/ Cc (Cc = 𝑨𝒄 / 𝑨𝟐)

If Cc is assumed to be 0.62, K = 0.375

∴ hc = 0.375 𝑣22

2gGenerally,

hc = 0.5𝑣22

2g


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

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

In general,

hi or hent = 𝐾 𝑣

22

2g

The value of k is as per given table.

3. Loss of head at the entrance of pipe:


The loss of head at the exit of pipe as aresult of form of a free jet or it may beconnected to reservoir.

General equation

ho =𝑣2

2g

v = velocity at the outlet of the pipe.

4. Loss of head at the exit of pipe:


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

. There is sudden enlargement of the areaof flow beyond the obstruction due towhich loss of head take place.

Loss of head due to obstruction is

=𝑣2

2g(

𝐴

𝐶𝑐(𝐴−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:


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:

hf = 𝐾 𝑣

22

2g

The value of K is different for deferent bend angle and r/d ratio.

6. Loss of head due to bend in pipe:


Example:


To analyze the pipe problem the concept of HEL or TEL is very useful.

Both are graphical representation of the longitudinal variation in the total head at salient points of pipe line.

Total head as per Bernoulli’s equation given by

p/𝛾 + v2/2g + Z = Constant

p/𝛾 = Pressure Head

v2/2g = Velocity Head

Z = Potential Head

HYDRAULIC GRADIENT AND TOTAL ENERGY LINE:


Line representing the sum of pressurehead, datum head and velocity head withrespect to some reference line is calledtotal energy line (TEL).

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

For ideal fluid as there are no losses, totalenergy line would remain parallel to thedatum.

Total energy line (TEL):


Hydraulic grade line (HGL):

Line representing sum of pressure headand datum head with respect to somereference line (datum) is called as hydraulicgradient line (HGL).

It the line obtained by connecting thevalues of the piezometric head (p/𝛾 + Z) atsuccessive points along the piping system.

HGL is always vertically below TEL by andequal amount to the velocity head (v2/2g).


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

H = 0.5 𝑣

12

2g+ 4𝑓

1𝐿1𝑣12

2𝑔𝑑1

+0.5 𝑣

22

2g+ 4𝑓

2𝐿2𝑣22

2𝑔𝑑2

+(𝑣2−𝑣

3)2

2g+ 4𝑓

3𝐿3𝑣3

2𝑔𝑑3

+ 𝑣32

2g

If minor losses are neglected,

H = 4𝑓

1𝐿1𝑣12

2𝑔𝑑1

+ + 4𝑓

2𝐿2𝑣22

2𝑔𝑑2

+ 4𝑓

3𝐿3𝑣3

2𝑔𝑑3

(4f = f, f is friction factor)

PIPE IN SERIES OR COMPUND PIPE:


Compound pipe is consisting of several pipes

of varying diameter and length.

Compound pipe is replaced by pipe of uniform diameter having

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 minor losses,

H = 𝟒𝒇

𝟏𝑳𝟏𝒗𝟏𝟐

𝟐𝒈𝒅𝟏

+ 𝟒𝒇

𝟐𝑳𝟐𝒗𝟐𝟐

𝟐𝒈𝒅𝟐

+ 𝟒𝒇

𝟑𝑳𝟑𝒗𝟑𝟐

𝟐𝒈𝒅𝟑

(Assuming f = f1 = f2 = f3)

Discharge Q = A1v1 = A2 v2 = A3 v3

Q = 𝝅

𝟒d1

2 v1 = 𝝅

𝟒d2

2 v2 =𝝅

𝟒d3

2 v3

v1 = 𝟒Q𝝅d1

2 , v2 = 𝟒Q𝝅d2

2 , v3 = 𝟒Q𝝅d3

2


EQUIVALENT PIPE:

L1d1v1

L2d2v2

L3d3v3

Putting the value of velocity in head loss equation

H = 𝟒𝒇

𝟏𝑳𝟏(𝟒Q𝝅d1

2)𝟐

𝟐𝒈𝒅𝟏

+ + 𝟒𝒇

𝟐𝑳𝟐(𝟒Q𝝅d2

2)𝟐

𝟐𝒈𝒅𝟐

+ 𝟒𝒇

𝟑𝑳𝟑

𝟒Q𝝅d3

2

𝟐

𝟐𝒈𝒅𝟑

H = 𝟒∗𝟏𝟔𝒇Q𝟐

𝝅𝟐∗𝟐𝒈

𝑳𝟏

𝒅𝟏𝟓 +

𝑳𝟐

𝒅𝟐𝟓 +

𝑳𝟑

𝒅𝟑𝟓

Head loss in equivalent pipe

H = 𝟒𝒇𝑳𝒗𝟐

𝟐𝒈𝒅= 𝟒∗𝟏𝟔𝒇Q𝟐

𝝅𝟐∗𝟐𝒈

𝑳

𝒅𝟓

Equating both head loss equation 𝟒∗𝟏𝟔𝒇Q𝟐

𝝅𝟐∗𝟐𝒈

𝑳𝟏

𝒅𝟏𝟓 +

𝑳𝟐

𝒅𝟐𝟓 +

𝑳𝟑

𝒅𝟑𝟓 =

𝟒∗𝟏𝟔𝒇Q𝟐

𝝅𝟐∗𝟐𝒈

𝑳

𝒅𝟓

𝑳

𝒅𝟓 = 𝑳𝟏

𝒅𝟏𝟓 +

𝑳𝟐

𝒅𝟐𝟓 +

𝑳𝟑

𝒅𝟑𝟓

The above equation is known as Dupuit’s equ.


The purpose of the parallel pipe is toincrease 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

𝟒𝒇𝟏𝑳𝟏𝒗𝟏𝟐

𝟐𝒈𝒅𝟏

= 𝟒𝒇

𝟐𝑳𝟐𝒗𝟐𝟐

𝟐𝒈𝒅𝟐

For f1 = f2

𝑳𝟏𝒗𝟏𝟐

𝒅𝟏

=𝑳𝟐𝒗𝟐𝟐

𝒅𝟐

PIPES IN PARALLELS:


The hydraulic power transmitted by a pipe 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

The maximum power transmission isobtained by

𝑑𝑃

𝑑𝑣= 0

hf = H/3

POWER TRANSMISSION THROUGH THE PIPES:


η = 𝑝𝑜𝑤𝑒𝑟 𝑑𝑒𝑙𝑖𝑣𝑒𝑟𝑒𝑑 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑜𝑓 𝑝𝑖𝑝𝑒

𝑝𝑜𝑤𝑒𝑟 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑒 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 𝑜𝑓 𝑝𝑖𝑝𝑒

= 𝜌 𝑔 𝑄 (H –hf)

𝜌 𝑔 𝑄 H

= (H –hf)

H

Under condition of maximum power the efficiency is given by

η = (H –H/3)

H= 2/3

= 66.67%

EFFICIENCY OF POWER TRANSMISSION:


Example:


References:

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, LaxmiPublications

3. Fluid Mechanics and Hydraulic Machines by R.K. Rajput, S.Chand & Co

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

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Module-5 Flow Through Pipes

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