010a (PPT) Flow through pipes.pdf .

Flow through pipes
Dr. Vijay G. S.
Professor
Dept. of Mech. & Ind. Engg.
MIT, Manipal
email: vijay.gs@manipal.edu
Mob.: 9980032104

• A pipe is a closed conduit carrying a fluid under pressure.
• Fluid motion in a pipe is subjected to a certain resistance.
• Such a resistance is assumed to be due to friction.
• Mainly due to the viscous property of the fluid.
• Fluid flow in pipes is of considerable importance in many of the processes,
like
• Animals and plants circulation systems.
• In our homes.
• City water.
• Irrigation system.
• Sewer water system, etc.
Introduction:

(a)
(b)
Fig.1. Flow through pipes. (a) From pump (b) From Tank

• To describe any of these flows, conservation of mass and conservation of
momentum equations are the most general forms could be used to
describe the dynamic system.
• The key issue is the relation between flow rate and pressure drop.
Introduction - contd…

Losses in Pipe Flow
• Losses in pipe flow can be two types:
(a) Major Loss
 Head loss due to friction
(b) Minor Loss
 Entry Loss
 Exit loss
 Sudden expansion loss
 Sudden contraction loss
 Losses due to bends & pipe fittings.

Losses in Pipe Flow – contd…
• When the ratio of the length of the pipeline, to the diameter, exceeds 2000:1,
pipe system energy losses are predominantly the result of pipe friction.
• The energy losses resulting from pipe accessories are termed “minor” losses
and are usually neglected in the calculation of pipe system energy losses.
• In short lengths of pipe, however, these minor losses can become major
sources of energy loss.
• Energy loss is usually called power loss or head loss.
• Head loss is the measure of the reduction in the total head (sum of elevation
head, velocity head and pressure head) of the fluid as it moves through a
fluid system.
• Head loss includes friction loss and local losses (minor losses)
• The Darcy-Weisbach equation is used to express energy loss caused by pipe
friction

• As per Reynold’s experiment, in case of laminar flow, the frictional loss hf
was found to be proportional to the velocity of flow.
• In case of turbulent flow, it was found that the frictional loss hf was
found to be approximately proportional to the square of the velocity of
flow.
Or more exactly,
f
h V

2
approximately
f
h V

where, n = 1.75 to 2
n
f
h V


As per William Froude, the frictional resistance for turbulent flow is:
1. Proportional to V n, where n varies from 1.5 to 2.0
2. Proportional to the density  of the fluid
3. Proportional to the area of surface of contact, Awet
4. Independent of pressure
5. Dependent on the nature of surface of contact
Frictional resistance  V n    Awet

Major losses in pipe flow
• The major losses in pipe flow is due to the frictional loss.
• There is a drop in pressure due to this frictional loss in the direction of
fluid flow.
• These losses are calculated by using
(a) The skin friction coefficient equation and the Moody’s chart, OR
(b) The Darcy-Weisbach equation

Major losses in pipe flow – contd…
Consider a fluid flow in a pipe of diameter ‘D’ as shown in the Fig.
The fluid has a density ‘’ and dynamic viscosity ‘’.
Consider a control volume having length ‘L’ with in the fluid flow domain (shown as the shaded
rectangle in figure).
Let w be the shear stress at the wall
p* be the piezo metric pressure (If the pipe is horizontal, p* = p = static pressure head)
S be the wetted perimeter
A be the area of cross section of the pipe
Derivation of Darcy’s equation
& relationship between Skin
friction coefficient Cf and the
Dracy’s friction factor f

Writing the force balance equation
  w
p* A p* p* A SL

  
w
p* A SL

 
wSL
p*
A

 
The fluid can flow through pipes with any cross-sectional shape. Hence to generalize the equation consider
‘hydraulic diameter’ defined as
4
h
A
D
S

Substituting for S/A = 4/Dh in equation (1) we get,
4
4
w w
w
h h
SL L
p* L
A D D
 

 
   
 
 
(1)
(2)
(3)

The head lost hf due to pressure drop of p* over a length L can be expressed as
4 w
f
h
L
p*
h
g gD

 

 
Let us consider the skin friction coefficient
(4)
𝐶𝑓 =
𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙
𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝐻𝑒𝑎𝑑
=
𝜏𝑤
1
2
𝜌𝑉2
where ‘V’ is the reference velocity. When the fluid is flowing through a pipe, then the reference velocity is
equal to the average velocity.
2
1
2
w f
C V
 
 
Cf is called as Fanning friction factor or skin friction coefficient
(5)

Now substituting for 𝜏𝑤 from (5) in (4) we get
2
2 2
1
4
4 4
4 2
2 2
f
f f
w
f
h h h h
C V L
C LV C L
L V
h
gD gD D g D g


 
 
   
 
     
  
(6)
Equation (6) expresses the head lost due to friction hf in terms of the skin friction coefficient Cf.
But as per Darcy, the shear stress due to frictional resistance is expressed as w  V n 
Taking n = 2 and  is constant for incompressible flow,
w  V 2
 w = F × V 2
where, F is the constant of proportionality.
(7)

Substituting for w from (7) in (4), we get
 
2
4
4 w
f
h h
F V L
L
p*
h
g gD gD

  


  
But as per Darcy’s experiments, he found that
4
2
F f
g g

 where f is known as the “Darcy’s friction factor”
(9)
(8)
Substituting (9) in (8) we get
 
2 2
4
2
f
h h
F V L
p* f L V
h
g gD D g
 
  

    
 
(10)
Now, comparing (6) and (10) we get the relationship between the ‘Skin friction coefficient’Cf and the Dracy’s
friction factor f .
i.e., f = 4Cf
(11)

For circular pipes
2
4
4
4
h
D
A
D D
S D



   (12)
2 2
4
2 2
f
f
C L
p* f L V V
h
g D g D g


   
 
  
   
   
(13)
By rearranging equation (13), we can write an expression for friction factor f as:
2
2
p* f L V
g D g

 
 
  
    
2 2
1
2
2
D p* p*
f
L
L V V
D
 
 
 
 
  
 
 
   
2
2
f
p* f L V
h
g D g

 
 
   
    
2 2
2
2
f f
h g D h
f
LV V
L
D g
 
 
(14a)
(14b)
OR

(1) For laminar flow (Re < 2000), 𝐶𝑓 =
16
𝑅𝑒
• where Re is the Reynolds number and can be calculated by 𝑅𝑒 =
𝜌𝑉𝐷
𝜇
• Therefore, Darcy friction coefficient, 𝑓 = 4𝐶𝑓 =
64
𝑅𝑒
(2) When the flow is turbulent (Re > 4000), f will be a function of Reynolds
number Re and the roughness factor , i.e., f =  (Re, )
• where Re is the Reynolds number and  is the roughness factor

• Moody documented these relations in the form of a chart, popularly known
as the Moody’s chart.
• It gives the relationship between the Reynolds number Re (on the horizontal
axis), friction factor f (left vertical axis) and the relative roughness ( / D) (on
the right vertical axis).
Moody’s chart

Moody’s chart

• We can see that for laminar flow the curve is linear (Re < 2000). It means
that one can determine the laminar flow friction coefficient from the
equation f = 64/Re
• For turbulent flow (Re > 4000), for a particular value of Reynold’s number Re
and the relative roughness ( / D), friction factor f can be obtained from the
Moody’s chart.
• Each curve on the chart corresponds to a particular value of the relative
roughness ( / D).
• The intersection of the vertical through Re and a ( / D) curve gives the
corresponding value of the friction factor f.
• Further it can be observed that the curves become horizontal towards the
right, beyond the rightmost dashed curve. i.e., with increased values of
Reynold’s number, the curves tend to become horizontal.

Types of problems in Flow through pipes
1. Determining the pressure drop (or head loss) when the pipe length and
diameter are given for a specified flow rate (or velocity)
2. Determining the flow rate when the pipe length and diameter are given for a
specified pressure drop (or head loss)
3. Determining the pipe diameter when the pipe length and flow rate are given
for a specified pressure drop (or head loss)

Problem1: Water at 10°C ( = 999.7 kg/m3 and  = 1.307×10-3 Ns/m2) is
flowing steadily in a 20 cm diameter, 15 m long pipe at an average velocity of
1.2 m/s. Determine (a) the pressure drop, (b) the head loss, and (c) the
pumping power requirement to overcome this pressure drop.
Reynolds number
𝑅𝑒 =
𝜌𝑉𝐷
𝜇
=
999.7 × 1.2 × 0.2 × 10−2
1.307 × 10−3
𝑹𝒆 = 𝟏𝟖𝟑𝟓. 𝟕𝟏
which is less than 2000. Therefore, the flow is laminar
For laminar flow
𝑓 =
64
𝑅𝑒
𝑓 = 0.0349
Pressure drop,
∆𝑝 = 𝑓
𝐿
𝐷
1
2
𝜌𝑉2
∆𝑝 = 0.0349
15
0.002
999.7 × 1.22
2
∆𝑝 = 188.403𝑘𝑃𝑎
The head loss in the pipe
ℎ𝑓 =
∆𝑝
𝜌𝑔
=
188.03 × 103
999.7 × 9.81
= 19.21𝑚

Problem1: Contd…
Flow rate calculation
𝑄 = 𝐴𝐶𝑉 =
𝜋𝐷2
4
× 𝑉 = 𝟑. 𝟕𝟕 × 𝟏𝟎−𝟔𝒎𝟑/𝒔
Pumping Power
𝑃 = ∆𝑝 × 𝑄 = 188.403 × 103 × 3.77 × 10−6= 0.71W
Power input in the amount of 0.71 W is needed to overcome the frictional losses in the flow due to
viscosity.

Problem2: Water at 15°C ( = 999.1 kg/m3 and  = 1.138 ×10-3 Ns/m2) is
flowing steadily in a 30m-long and 4 cm diameter horizontal pipe made of
stainless steel at a rate of 8 lps. Determine (a) the pressure drop, (b) the head
loss, and (c) the pumping power requirement to overcome this pressure drop.
(Take the roughness of stainless steel is 0.002 mm).
Given data:
 = 999.1 kg/m3
 = 1.138 ×10-3 Ns/m2
Flow rate = 8 lps
Roughness of stainless steel = 0.002 mm
𝑄 = 𝐴𝐶𝑉
𝑉 =
𝑄
𝐴𝐶
=
0.008
𝜋 ×
0.042
4
= 6.366𝑚/𝑠
𝑅𝑒 =
𝜌𝑉𝐷
𝜇
=
999.1 × 6.366 × 0.04
1.138 × 10−3
= 2.236 × 105
Which is greater than 4000. Therefore, the flow is
turbulent.
The relative roughness of the pipe is
𝜀
𝐷
=
0.002 × 10−3
0.04
= 5 × 10−5

Problem 2: Contd…
The friction factor can be determined from the Moody chart
𝑓 = 0.015
∆𝑝 = 𝑓
𝐿
𝐷
1
2
𝜌𝑉2
∆𝑝 = 0.015
30
0.04
999.1 × 6.3662
2
= 227.75 𝑘𝑃𝑎
ℎ𝑓 =
∆𝑝
𝜌𝑔
=
227.75 × 103
999.1 × 9.81
= 𝟐𝟑. 𝟐𝟑𝒎
Power required
𝑃 = ∆𝑝 × 𝑄 = 227.75 × 103
× 0.008 = 𝟏. 𝟖𝟐𝟐 𝒌𝑾

• The loss of energy due to the change in velocity of the fluid flow, in
magnitude or direction, is termed as “minor losses”.
• In case of long pipes these losses are usually quite small as comparted to the
major losses, thus known as minor losses.
• However, in some short pipes these losses may exceed the magnitude of the
frictional (major) loss.
• Different types of minor head losses are due to:
Sudden enlargement he
Sudden contraction hc
Entrance of a pipe hi
Exit of a pipe ho
Bend in a pipe hb
Pipe fitting hfit
Obstruction hob
Minor losses in pipe flow:

Head loss due to sudden enlargement, he:
Minor losses –contd…
ℎ𝑒 =
𝑉1
2
2𝑔
1 −
𝐴1
𝐴2
2
Defining 𝐾 = 1 −
𝐴1
𝐴2
2
we can write
ℎ𝑒 = 𝐾
𝑉1
2
2𝑔
Special case
If A2 is very large compared to A1 (example pipe ending at the reservoir) then, K = 1
2
1
2
e
V
h
g

ℎ𝑒 =
𝑉1 − 𝑉2
2
2𝑔

Head loss due to sudden contraction, hc:
It can be shown that
2
2
2
c
V
h k
g

Where k is constant given by
2
1
1
c
k
C
 
 
 
 
where, Cc = Ac/A2 = Coefficient of contraction
If Cc is not given, take k = 0.5
ℎ𝑐 = 0.5
𝑉2
2
2𝑔

Head loss due to entrance at a pipe, hi:
It can be shown that,
2
2
0.5
2
i
V
h
g

Head loss due to exit from a pipe, ho:
2
2
o
o
V
h
g

Head loss due to bend in a pipe, hb:
2
2
b b
V
h k
g

where, kb = coefficient of bend, and V = average velocity of flow.

Head loss due to a pipe fitting, hfit:
2
2
fit fit
V
h k
g

where, kfit = coefficient of pipe fitting, and V = average velocity of flow.
2
2
ob ob
V
h k
g

2
1
( )
ob
c
A
k
C A a
 
 
 

 
( )
c
c
a
C
A a


Head loss due to an obstruction in the pipe, hob:
where,
kob is the coefficient of obstruction, V = average velocity of flow,
A = cross sectional area of pipe and a = Maximum area of obstruction
is the coefficient of contraction, ac = area at vena contracta.
where,

Hydraulic Grade Line (HGL) and Total Energy (TEL)
• It is often convenient to represent the level of mechanical energy graphically
using heights to facilitate visualization of the various terms of the Bernoulli
equation.
• We have the Bernoulli’s equation as:
𝑝
𝜌𝑔
+
𝑉2
2𝑔
+ 𝑧 = 𝐻 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
•
𝑝
𝜌𝑔
𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑
•
𝑉2
2𝑔
𝑖𝑠 𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑
• 𝑧 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑙𝑒𝑣𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑑
The line that represents the sum of the static pressure and the elevation heads,
𝑝
𝜌𝑔
+ 𝑧 is called
the hydraulic grade line.
The line that represents the total head of the fluid,
𝑝
𝜌𝑔
+
𝑉2
2𝑔
+ 𝑧 is called the total energy line.

HGL / TEL – contd…
HGL
Static pressure head + datum head =
𝑝
𝜌𝑔
+ z
TEL
Static pressure head + Kinetic head + datum head =
𝑝
𝜌𝑔
+
𝑉2
2𝑔
+ 𝑧

Problem 3: A horizontal pipe line 40 m long is connected to a water tank at
one end and discharges freely into the atmosphere at the other end. For the
first 25 m of its length from the tank, the pipe is 150 mm diameter and its
diameter is suddenly enlarged to 300 mm. The height of water level in the
tank is 8 m above the centre of the pipe. Take Cf = 0.01 for both sections of
the pipe. Considering all losses of head which occur,
(i) Determine the rate of flow
(ii) Draw the HGL and TEL.
Skin friction coefficient Cf = 0.01

Problem 3: contd…
Applying Bernoulli’s equation, between the free surface of water in the tank, and the outlet of the
pipe, assuming that the reference line passes through the centre of the pipe.
𝑝1
𝜌𝑔
+
𝑉1
2
2𝑔
+ 𝑍1 =
𝑝2
𝜌𝑔
+
𝑉2
2
2𝑔
+ 𝑍2 + ℎ𝐿
hi = loss of head at entrance = 0.5
𝑉1
2
2𝑔
hf1 = head lost due to friction in pipe 1 =
4𝐶𝑓𝐿1𝑉1
2
2𝑔𝑑1
he = loss of head due to the sudden enlargement =
(𝑉1 − 𝑉2)2
2𝑔
hf2 = head lost due to friction in pipe 2 =
4𝐶𝑓𝐿2𝑉2
2
2𝑔𝑑2
(1)

Problem 3: contd…
Plotting HGL and TEL

details given in the figure, determine the (i) discharge velocity, (2) the discharge, (3) the power lost due to
all the losses in the pipe. Take the skin friction coefficient Cf = 0.002, Cc for obstruction = 0.6. Neglect length
of bend.
Top view of pipe system
Problem on major + minor losses
Problem 4:

Hydraulic Siphon
Uses of Siphon:
• To carry water from one reservoir to another reservoir separated by a hill or ridge.
• To take out the liquid from a tank which is not having any outlet.
• To empty a channel not provided with any outlet sluice.

Applying the Bernoulli’s equation at points A and B
𝑝𝐴
𝜌𝑔
+
𝑉𝐴
2
2𝑔
+ 𝑍𝐴 =
𝑝𝐵
𝜌𝑔
+
𝑉𝐵
2
2𝑔
+ 𝑍𝐵
Since 𝑝𝐴 = 𝑝𝐵 = 𝑝𝑎𝑡𝑚 and VA = 0
𝑉𝐵
2
= 2𝑔∆𝑍
𝑉𝐵 = 2𝑔∆𝑍
Applying the Bernoulli’s equation at points A and C
𝑝𝐴
𝜌𝑔
+
𝑉𝐴
2
2𝑔
+ 𝑍𝐴 =
𝑝𝐶
𝜌𝑔
+
𝑉𝐶
2
2𝑔
+ 𝑍𝐶
𝑝𝐶
𝜌𝑔
=
𝑝𝐴
𝜌𝑔
−
𝑉𝐶
2
2𝑔
− ℎ
It can be observed that the pressure head at point C
is lesser than that at A by an amount
𝑉𝐶
2
2𝑔
+ ℎ .
Hydraulic Siphon – contd…

• In the figure the point D will have the least pressure.
• The elevation of D must be such that the pressure at D should not fall below
the vapor pressure of the liquid at the existing temperature.
• If the pressure at D falls below the vapour pressure of the liquid, then the
liquid will start to boil and the liquid circuit will break, and the liquid will stop
flowing.
• Thus, there is a limitation on the elevation of D, i.e., the length of the siphon
leg on the upstream (entry side) must be carefully decided.
• Whenever designing these type of circuits care has to be taken such that the
pressure at any point should not fall below the vapor pressure of the flowing
fluid in order to avoid the cavitation problems.
Hydraulic Siphon – contd…

Problem 5: A syphon of diameter 200 mm connects two reservoirs having a difference in
elevation of 20 m. The length of the syphon is 500 m and the summit is 3.0 m above the
water level in the upper reservoir. The length of the pipe from upper reservoir to the summit
is 100 m. Determine the discharge through the syphon and also pressure at the summit.
Neglect minor losses. The co-efficient of friction (skin friction coefficient), f = 0.005.
The minor losses are neglected and only
head loss due to friction is considered.

Problem 6: A syphon of diameter 200 mm connects two reservoirs having a difference in
elevation of 15 m. The total length of the syphon is 600 m and the summit is 4 m above the
water level in the upper reservoir. If the separation takes place at 2.8 m of water absolute,
find the maximum length of syphon from upper reservoir to the summit. Take skin friction
coefficient f = 0.004 and atmospheric pressure 10.3 m of water.

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