credit to DR ADAM

1. FLUID MECHANICS – 1
Semester 1 2011 - 2012
Week – 8, 9 and 10
FLOW IN PIPES-
CO3
Compiled and modified
by
Sharma, Adam

2. Review
• Conservation of mass (mass balance)
• Continuity Equation
• Bernoulli Equation
• Momentum Equation

3. Objectives
• Understand Laminar and Turbulent flow in pipes
• Identify types of flow using Reynolds number
• Explain minor losses and major losses for flow in
pipes
• Determine friction factor and major losses using
moody chart and simplified Colebrook equation
• Calculate minor losses, major losses, pressure
losses and head losses
• Understand equivalent length
• Solve piping system using Bernoulli equation
considering all losses.
3

4. 1. Introduction
• There are two kind of flow
a. Internal flow ( flow in pipes)
b. External flow ( flow over bodies, drag, lift) – will
be covered in fluids mechanics 2 (BMM2543)
• Examples of internal flow
1. Water flow in pipes
2. Blood flow
3. Oil and Gas industry
4. Cooling system of a car (Radiator)
5. Air Conditioning (Chilled water system) 4

5. examples & pictures
5

6. 6

7. 2. Types of flow
• Flow in pipes has three types, Laminar,
Transition, Turbulent
• Laminar flow
― Smooth streamlines
Transition
flow (flow)
― Highly ordered motion
― Short in length
― normally appeared in
high viscosity flow and
small pipe/passage i.e.
oil in small pipe
7

8. 2. Types of flow (cont’)
• Turbulent flow
― Rough streamlines (flow)
― Highly disordered motion
― Most flow in reality is turbulent
― High momentum, thus high friction
• Transition flow
― flow from laminar does not
suddenly change to turbulent, it
will enter transition flow first
― Normally ignored in calculation
― fluctuation of laminar and
turbulent randomly 8

9. Is there other method to distinguish
these flow?
• Yes, use Reynolds number, Re
ν = kinematic viscosity, (m 2 s )
µ = dynamic viscosity, (kg m.s )
ρ = density of flowing fluid (kg/m 3 )
Vavg = Average pipe velocity
D = Internal diameter of pipe
• Re < 2300, Laminar flow
• 2300 < Re < 4000, Transition flow
• Re > 4000, Turbulent flow
9

10. Is there other method to distinguish
these flow?(cont)
• For non-circular pipes, hydraulic diameter, Dh is used
for calculating the Reynolds number.
4 × Crosssectional Area (internal) 4 Ac
Dh = =
Internal Perimeter p
• however, in practical, type of flow depend on
smoothness of pipe, vibrations and fluctuation in the
flow.
10

11. Reynolds Number (example)
• Q1: Water at 20°C flow with average velocity of
2cm/s inside a circular pipe. Determine flow
type if the pipe diameter, a) 2 cm, b) 15 cm,
and c) 30 cm From table (Cengel book, 2010) At 20°C
µ = 1.002 × 10-3 kg m.s , ρ = 998kg/m 3
ρVave D 998(0.02)(0.02)
a) Re = = = 398 • (a) Laminar flow
µ 0.001002
ρVave D 998(0.02)(0.15)
b) Re = = = 2988 • (b) Transition flow
µ 0.001002
ρVave D 998(0.02)(0.3)
c) Re = = = 5970 • (c) Turbulent flow
µ 0.001002
11

12. Reynolds Number (example) cont’
• Q2: Water at 20°C flow in a circular pipe of 3.5
cm diameter. Determine the range for the
average velocity so the flow is always
transition flow From table (Cengel book, 2010) At 20°C
µ = 1.002 × 10-3 kg m.s , ρ = 998kg/m 3
ρVave D 998(Vave ,min )(0.035)
Re min imum = 2300 = =
µ 0.001002
2300 × 0.001002
Vave ,min = = 0.066 m/s
998(0.035)
ρVave D 998(Vave ,max )(0.035)
Re maximum = 4000 = =
µ 0.001002 4000 × 0.001002
Vave ,max = = 0.115 m/s
998(0.035)
Range ∴ 0.066 m/s < Vave < 0.115 m/s
12

13. Reynolds Number (example) cont’
• Q3: Air at 35°C flow inside rectangular pipe of
2cmx5cm. Determine the maximum flow
velocity for the pipe before the flow enter
transition region From table (Cengel book, 2010),
Air At 35°C, ν = 1.655 ×10-5 m 2 s
4 Ac 4(0.02)(0.05)
Dh = = = 0.029 m
p 2(0.02) + 2(0.05)
Vave D Vave ,max (0.029)
Re max imum = 2300 = =
ν 1.655 × 10-5
2300 ×1.655 ×10 −5
Vave ,max = = 1.313m/s
0.029
13

14. 4. Entrance Region
• Between the entrance and fully developed flow
• Uniform velocity profile at entrance
• because of no slip boundary condition, friction at the
wall reduce the velocity of flow near the wall
• to conserve mass, velocity at the center increase
(compensate velocity decreased near the wall)
14

15. 4. Entrance Region (cont’)
• As the fluid move deeper in the pipe, the velocity
near the wall decreased further and velocity at center
increase (developing velocity profile)
• Both (up & down @ left & right) velocity profile
increase till it merge with the other side
• fully developed flow is when the velocity profile
stop to develop as it flows deeper inside the pipe
15

16. 4. Entry Region (cont’)
• hydrodynamic entry length, Lh is evaluated from
pipe entrance to where wall shear stress achieve 2%
of fully developed value or approximately
Lh ,la min ar ≅ 0.05ReD Lh ,turbulent ≅ 10 D shorter length for
turbulent pipe flow
16

17. 4. Entry Region (example)
• Q4: Determine the hydrodynamic entry region for
Q1 (a & c).
Lh ,la min ar ≅ 0.05ReD Lh ,turbulent ≅ 10 D
ρVave D 998(0.02)(0.02)
a) Re = = = 398 • (a) Laminar flow
µ 0.001002
Lh ,la min ar ≅ 0.05ReD = 0.05 × 398 × 0.02 = 0.398 m
ρVave D 998(0.02)(0.3)
c) Re = = = 5970 • (c) Turbulent flow
µ 0.001002
Lh ,turbulent ≅ 10 D = 10 × 0.02 = 0.2 m
17

18. 4.1. Turbulence velocity profile VS Laminar
velocity profile (fully developed)
• Velocity profile based on
analysis • Velocity profile is based
on analysis and empirical
• Consist of 1 layer
• Consist of 4 layer
• Small velocity gradient
• High velocity gradient
• The average velocity in fully
developed laminar pipe flow
is ½ of the maximum velocity
u max = 2Vave 18

19. 5. Losses in piping system
• There are two type of losses which is major
losses and minor losses. Losses are mainly due
to friction and obstruction
• Total losses, hL is major losses + minor losses
5.1 Major Losses (pg345)
• Major losses, hL, major , is also known pressure
losses or head losses
• The major losses is solely depend on the pipe,
nothing else
19

20. 5.1 Major Losses (cont’)
• Pressure drop across pipe can be formulated as
dP P − P2
= 1
dx L
8µLVave 32 µLVave
∆P = P − P2 =
1 2
=
R D2
L ρVave
2
Pressure loss (Smooth or Rough) ∆PL = f
D 2
∆PL 2
L Vave
major head loss, hL ,major = = f
ρg D 2g
• f, Darcy friction factor. There is another friction factor,
Fanning friction factor, but will not be covered
• for fully developed laminar flow, friction factor is obtained
from combining Darcy equation with Pressure drop. 20

21. 5.1 Major Losses (cont’)
32 µLVave L ρVave
2
∆P = ∆PL ⇒ 2
= f
D D 2
32µ ρVave ρVave 32µ
= f f =
D 2 2 D
ρVave D
f = 32 × 2
µ
f Re = 64
64
f = (friction factor for fully developed laminar flow)
Re
• from the eq. ,friction factor in Laminar flow is
independent of roughness
• The head loss represents the additional height that the
fluid needs to be raised by a pump in order to overcome
the frictional losses in the pipe
21

22. 5.1 Major Losses (cont’)
• For fully developed turbulent flow pressure losses or
head losses, the equation is more complex
• Unlike laminar flow, friction factor, f , for turbulent flow is
depend on internal pipe roughness and Reynolds
number base on Colebrook equation.
1 ε D 2.51 
= -2.0 log  +  (turbulent flow)
f  3.7 Re f 
 
• Whereas, ε is roughness of pipe and D is pipe diameter.
• However this equation is implicit and cannot be solved
directly. Using iteration (numerical method) is possible
but the method is tedious
22

23. 5.1 Major Losses (cont’)
• Hence, there are two option, (a), Moody chart, (b)
Modified Colebrook equation
f =0.025 ε/D=0.02
Re=100,000
• In Moody Chart, friction factor can be obtain by
knowing the roughness ratio, ε/D and Reynolds number
23

24. 5.1 Major Losses (cont’)
• Haaland modified Colebrook Equation (1983).
1  6.9  ε D 1.11 
= -1.8 log  +   (Modified for turbulent flow)
f  Re  3.7  
 
-2
  1.11

f =  -1.8 log  6.9 +  ε D 
   (simplified)
  Re  3.7  
  
• Easier to calculate
• The formula has about 2% error compare to original
equation
• Formula for major head loss for turbulent is the same
2
L Vave
hL ,major = f
D 2g
24

25. 5.1 Major Losses (cont’)
• In real pipe application (turbulent flow), friction is
unwanted because the rougher the surface, the higher
the friction
• Old piping system such as Cast Iron, GI, the
performance deteriorate through time as corrosion
reduce smoothness and size of internal pipe.
• New piping (HDPE), Anti-corrosion metal pipes,
smoothness maintain, thus performance is maintained
25

26. 5.1 Major Losses (example)
• Q5: Water at 40°C (ρ = 992.1 kg/m3 and µ = 0.653×10-3 kg/m.s) is flowing
steadily in a 5cm-diameter horizontal pipe made of stainless steel at a rate
of 300 l/min. Determine the pressure drop, the head loss, and the required
pumping power input for flow over a 50m-long section of the pipe.
Q 300 / (1000)/60
Q = AcVavg → Vavg = = = 2.55m/s
Ac π (0.05) / 4
2
ρVavg D 992.1(2.55)(0.05)
Re = = = 193710, Turbulent
µ 0.000653
Roughness ratio, ε D = 0.002mm / 50mm = 0.00004
from Moody Chart, ε D = 0.00004 & Re 1.94 ×105 , f ≈ 0.016
-2
  6.9  0.00004   
1.11
f =  −1.8log  +   = 0.016
 193710  3.7   
  
L ρVave
2
50 992.1(2.55) 2
∆PL = f = 0.016 × × = 51609 Pa
D 2 0.05 2
∆PL 51609
hL = = = 5.3 m
ρg 992.1(9.81)

W = Q∆PL = 300 / 1000 / 60 × 51609 = 258 Watt 26

27. 5.1 Major Losses (example cont’)
• Q6: Air at 40°C (ρ = 1.127 kg/m3, ν = 1.702×10-6 m2/s) is flowing steadily in
a 50cm-diameter horizontal pipe made of plastic at a rate of 30 l/min.
Determine the head loss, and the required pumping power input for flow
over a 150m-long section of the pipe.
Q 30 / (1000)/60
Q = AcVavg → Vavg = = = 2.55 ×10 −3 m/s
Ac π (0.5) / 4
2
Vavg D (2.55 × 10 −3 )(0.5)
Re = = = 748, Laminar
ν 0.000001702
64 64
f = = = 0.0856
Re 748
2
L Vave 150(2.55 ×10 −3 ) 2
hL = f = 0.0856 = 8.51×10 −6 m
D 2g 0.5(2)(9.81)
∆PL = ∆hL × ρg = 8.51×10 −6 (1.127)(9.81) = 9.41×10 −5 Pa

W = Q∆PL = 30 / 1000 / 60 × 9.41×10 −5 = 4.71×10 −8 Watt
27

28. 5.2 Minor Losses
• In piping system there are various fittings,
valves, bends, elbows, tees, inlets, exits,
enlargements, and contractions in addition to
the pipes.
• These component Interrupt the smoothness of
flow and cause additional losses
• Usually, these additional losses is called minor
losses and smaller than pipe losses .
• Rarely, minor losses will be greater than major
losses especially when the component installed
frequently along the pipe system (between
short distance)
28

29. 5.2 Minor Losses (cont’)
29

30. 5.2 Minor Losses (cont’)
30

31. 5.2 Minor Losses (cont’)
• usually expressed in terms of
the loss coefficient, KL
• KL , is provided by manufacturer
and the value varies for different
components.
ρVavg
2
∆PL = K L
2
• Consider a component, valve,
the pressure losses is losses
due to valve minus losses by
imaginary pipe section without
valves
• Independent of Re, Reynolds number
31

32. 5.2 Minor Losses (cont’)
• Usually expressed in term of head losses, hL
2
Vavg
hL = K L = minor loss due to components
2g
• Also, In industrial application, the manufacturing data is
expressed in terms of equivalent length, Lequiv.
• for this method, simply add Lequiv to the total length of
pipe for the total head losses calculation.
• However, the former expression will be used thoroughly
2 2
Vavg Lequiv Vavg D
hL = K L = f → Lequiv = KL
2g D 2g f
32

33. 5.2 Minor Losses (cont’)
• Component with sharp edge such as
sharp edge exit has higher loss
coefficient compare to well rounded
• Sharp edge introduce recirculating flow
due to fluid unable to make sharp 90°
turn especially at high speed
• Same for sudden
expansion/contraction
33

34. 5.2 Minor Losses (cont’)
34

35. 5.2 Minor Losses (cont’)
• These loss coefficient depends on the manufacturer data
35

36. 5.2 Minor Losses (cont’)
36

37. 5.2 Minor Losses (cont’)
37

38. Ball valve
Globe valve
Swing check
Angle valve valve
38

39. 5.2 Minor Losses (cont’)
• Total head losses, hL, total
2
 L  Vavg
hL ,total = hL ,Major + hL ,Minor = f + ∑KL 
 D  2g
• Calculation of Minor losses is straight forward. If the
piping system consist same components such as bends,
2
 the loss coefficient with the number of the
simply multiply Li  Vavg,i
hL ,total = ∑ f i
 D + ∑K L i  2g 
same bends.  i 
(for pipe system with more than one pipe diameter)
Vavg is different for different pipe diameter
39

40. 6 Piping system
• Normally, as an engineer and consultant, piping system
for example water storage, sprinkler system, hose reel
system is the main concern.
• To solve piping system, extended Bernoulli equation is
required which the total losses is placed at the right hand
side (at point 2)
P (V1 ) 2 P2 (V2 ) 2
1
+ + z1 = + + z 2 + hL ,total
ρg 2g ρg 2g
• Piping system usually constructed to deliver fluid at
higher level or to create a pressurized system 40

41. 6 Piping system (cont’)
• Two principles in analyzing piping system which are
a) Conservation of mass throughout the system must
be satisfied
b) Pressure drop (and thus head loss) between two
junctions must be the same for all paths between the
two junctions
2
h
1
point 2, Just above water level,
point 1, Just above water level, P2=?, V2 = ?, Z2 = ?
P1=?, V1 = ?, Z1 = ?
41

42. 6 Piping system (example)
• Q7:A piping system delivering water at 25°C from tank 1
to tank 2. The system consist two 45º, a sharp entrance
and a sharp exit. The diameter of the stainless steel pipe
is 2cm and length of 55 m. Determine h so that the
flowrate is 83.3 L/min.
1
h=?m Tank 1
45°
2
Tank 2 45°
point 2, Just above water level, point 1, Just above water level,
P2=0, V2 = 0, Z2 = h P1=0, V1 = 0, Z1 = 0
42

43. 6 Piping system (example)
Q 83.3 / (60000)
Q = AcVavg → Vavg = = = 4.42m/s
Ac π (0.01) 2
ρVavg D 997(4.42)(0.02)
Re = = = 98916, Turbulent
µ 0.891×10 -3
Stainless steel, Roughness , ε = 0.002mm = 2 × 10 −6 m
Roughness ratio , ε / D = 2 × 10 − 6 ÷ 0.02 = 0.0001
-2
  6.9  0.0001 
1.11

f =  −1.8log  +   = 0.018
 98916  3.7  
  
2
 L  Vavg  55  (4.42) 2
hL ,major = f  =  0.018  = 49.29 m H 2O
 D  2g  0.02  2(9.81)
43

44. 6 Piping system (example)
2
Vavg (4.42) 2
hL ,minor = ( ∑K L ) = ( 0.5 + 2 × 0.4 +1) = 2.29 m H 2O
2g 2(9.81)
hL ,total = hL ,Major + hL ,Minor = 49.29 + 2.29 = 51.58m
0 0
2 0 02 Reference point
P (V1 ) P2 (V2 )
1
+ + z1 = + + z 2 + hL ,total
ρg 2g ρg 2g
0 + 0 + h = 0 + 0 + 0 + hL ,total
h = 51.58 m
44

45. 6 Piping system (example)
• Q8:A piping system delivering water at 25°C from pressurised tank 1 to tank
2. Initial plastic pipe diameter is 2 cm has a sharp entrance(inlet), fully open
globe valve and two 90º smooth flanged bend, and a sudden expansion at
the 1/3 of the pipe length. After expansion, Galvanized Iron (GI) pipe
diameter is installed with 8 cm diameter and along the GI pipe, there are two
90º miter bend, a fully open angle valve and a sharp exit. The total pipe
length is 75 m. Determine gauge pressure at tank 1 required to deliver
water at 226 liter/hour.
2
GI pipe
Tank 2
h=25m
1
Tank 1
Plastic point 2, Just above water level,
pipe P2=0, V2 = 0, Z2 = h
point 1, Just above water level, 45
P1≠0, V1 = 0, Z1 = 0

46. 6 Piping system (example)
Q 226 / (3600 × 1000)
Q = Ac1Vavg1 → Vavg1 = = = 0.2m/s
Ac1 π (0.01) 2
Q 226 / (3600000)
Q = Ac 2Vavg2 → Vavg2 = = = 0.0125m/s
Ac 2 π (0.04) 2
ρVavg1D 997(0.2)(0.02)
Re1 = = = 4475, Turbulent
µ 0.891× 10 -3
ρVavg2 D 997(0.0125)(0.08)
Re 2 = = = 1119, Laminar
µ 0.891×10-3
Plastic pipe, Roughness , ε = 0, smooth pipes
-2
  1.11

f1 =  − 1.8log  6.9 +  0 
   = 0.039 (turbulent friction factor)
  4475  3.7  
  
64 64
f2 = = = 0.057 (laminar friction factor)
Re 1119 46

47. 6 Piping system (example)
2 2
 D12   0.02 2 
K L ,sudden expansion = 1 − 2  = 1 −
   0.08 2  = 0.88

 D2   
2
 L1  Vavg1
hL ,total plastic =  f1
 D + K L ,inlet + K L ,g.valve + 2 K L ,smooth flange bend + K L ,sudden expansion  2 g

 1 
2
 75 / 3  0.2
hL ,total plastic =  0.039 + 0.5 + 10 + 2(0.3) + 0.88  = 0.124m
 0.02  2(9.81)
2
 L2  Vavg2
hL ,total GI  f2
= + 2 K L ,miter bend + K L ,angle valve + K L ,sharp exit 
 2g
 D2 
 75 − 75 / 3  0.0125
2
hL ,total GI =  0.057 + 2(1.1) + 5 + 1 = 0.00035m
 0.08  2(9.81)
hL ,total = hL ,total plastic + hL ,total GI = 0.124 + 0.00035 = 0.12435m
47

48. 6 Piping system (example)
Reference point
0
2 0 02
P (V1 ) P2 (V2 )
1
+ + z1 = + + z 2 + hL ,total
ρg 2g ρg 2g
P1
+ 0 + 0 = 0 + 0 + 25 + 0.12435
ρg
P = 25.12435( 997 )( 9.81) = 245730Pa
1
Most of the pressure at tank 1 is to overcome
the elevation difference as losses is very small
48

49. 49