Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

ME 312 – FLUID MECHANICS FOR ME
TOPIC 7: STEADY INCOMPRESSIBLE FLOW IN PRESSURE CONDUITS
DEFINITIONS:
PIPES – are closed conduits through which fluids or gases flow
...
Pipes are referred
to as conduits (usually circular) which flow full
...

LAMINAR FLOW – a type of steady flow in which the path of individual fluid particles does not cross or intersect
...

TURBULENT FLOW – a type of steady flow in which the path of individual fluid particles are irregular and continuously
cross each other
...

NOTE: Laminar flow in circular pipes can be maintained up to values of 𝑅𝑒 as high as 50,000
...
On the
other hand, it is practically impossible for turbulent flow in a straight pipe to persist at values of 𝑅𝑒 much below
2000, because any turbulence that is set up will be damped out by viscous friction
...
This is
represented by a Reynolds Number of 2,000
...

For pipes flowing full,
𝑹𝒆 =

𝒗𝑫𝝆 𝒗𝑫
=
𝝁
𝐯
v=

Eqn
...
2

Eqn
...
In circular pipes,
the velocity varies as the ordinates of a paraboloid of revolution with its
average velocity equal to one-half of its maximum velocity
...
4
𝒖 = 𝒗𝒄 −
𝒓
𝟒𝝁𝑳
Eqn
...
6

ℎ𝐿 = head lost in the pipe
𝐿 = pipe length
𝑟𝑜 = pipe radius
𝑣𝑐 = centerline or maximum velocity
𝜇 = absolute viscosity of the liquid
𝛾 = unit weight /specific weight of the fluid
𝑢 = velocity at distance r from the pipe center
𝑣 = average velocity

Turbulent Flow
The velocity distribution for turbulent flows varies with Reynolds
number, with zero velocity at the wall and increases more rapidly for a
short distance from the walls as compared to laminar flow
...
𝟕𝟓√ 𝐥𝐨𝐠
𝝆
𝒓𝒐 − 𝒓

Eqn
...
8

𝒖 = (𝟏 + 𝟏
...
9

The centerline or maximum velocity is given by

Where

𝑣𝑐 = 𝑣(1 + 1
...
10

𝝉𝒐
𝒗 = 𝒗𝒄 − 𝟑
...
11

𝜏𝑜 = maximum shearing stress in the pipe
𝑓 = friction factor
𝑣 = mean velocity

Shearing Stress in Pipes
Consider a mass of fluid of length 𝐿 and radius 𝑟 to move to the
right as shown in the figure
...
12

Multiplying the equation above by unit weight 𝛾
𝑝1 − 𝑝2
𝛾
𝜏𝑠
𝑥
2𝐿

𝑏𝑢𝑡

𝑝1 − 𝑝2
= ℎ𝐿 (ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠) 𝑡ℎ𝑒𝑛
𝛾

𝜸𝒉𝑳
Eqn
...

The maximum shearing stress, 𝜏𝑜 , is at the pipe wall (at 𝑥 = 𝑟)
...
14
Eqn
...
16

HEAD LOSSES IN PIPE FLOW
Major head loss – head loss caused by pipe friction along straight sections of pipe of uniform diameter and uniform
roughness
...
Darcy-Weishbach Formula (pipe-friction equation)
𝒉𝒇 =
Where

𝒇𝒍𝒗𝟐
𝟐𝒈𝑫

Eqn
...
18

Eqn
...
𝟎𝟖𝟐𝟔𝒇𝑳𝑸𝟐
𝑫𝟓

(𝑺𝑰 𝒖𝒏𝒊𝒕)

Where 𝑄 is the discharge
Value of f:
𝑓=

64 64𝜇
=
𝑅𝑒 𝑣𝐷𝜌

Eqn
...
21

128𝜇𝐿𝑄
𝜋𝜌𝑔𝐷 4

Eqn
...
21 with 𝐷 = 4𝑅

For turbulent flow:
1
...
For smooth pipes, 𝑅𝑒 between 3,000 and 100,000 (Blasius)
𝑓=

𝑓=

0
...
23

Eqn
...
25

3
...
80

√𝑓
4
...
3𝜀 (Karman)
1
Where

𝐷
= 2 log ( ) + 1
...
25

Eqn
...
6𝑣
𝜏
√ 𝜌𝑜
5
...
51
𝐷
= −2 log (
+
)
3
...
27

Eqn
...
Haaland formula – this is an alternate formula for Eqn
...
This varies less than 2% from Eqn
...

𝜀 1
...
9
= −1
...
7
𝑒
√𝑓
1

Eqn
...
This chart is identical to Eqn
...
Manning Formula
The Manning formula is one of the best-known open channel formulas and is commonly used in pipes
...
30
(𝑆𝐼 𝑢𝑛𝑖𝑡𝑠)
𝑣 = 𝑅3𝑆2
𝑛
1
...
31

ℎ𝑓
𝐿

to Eqn
...
35 𝑛2 𝐿𝑣 2
4
𝐷3

(𝑆𝐼 𝑢𝑛𝑖𝑡𝑠)

Eqn
...
29𝑛2 𝐿𝑄 2
16
𝐷3

Eqn
...
Hazen Williams Formula
The Hazen Williams formula is widely used in waterworks industry
...
) and velocities less than 3 m/s
...

English Units:

Eqn
...
318𝐶1 𝑅0
...
54
For circular pipes flowing full, this formula becomes
𝑄 = 0
...
23 𝑆 0
...
35

SI units
Eqn
...
849𝐶1 𝑅0
...
54
For circular pipes flowing full, this formula becomes
𝑄 = 0
...
63 𝑆 0
...
67𝐿𝑄1
...
85 𝐷4
...
37

MINOR HEAD LOSS
Minor losses are caused by the changes in direction or velocity of flow
...
These losses can usually be neglected if the length of the
pipeline is greater than 1500 times the pipe’s diameter
...

A
...
38
ℎ𝐿 =
, 𝑚𝑒𝑡𝑒𝑟𝑠
2𝑔
Where
𝑣1 = velocity before enlargement, m/s
𝑣2 = velocity before enlargement, m/s
Another equation for head loss caused by sudden enlargement was determined experimentally by Archer, and
given as:
(𝑣1 − 𝑣2 )1
...
39
ℎ𝐿 =
, 𝑚𝑒𝑡𝑒𝑟𝑠
2𝑔
A special application of Eqn
...
39 is the discharge from a pipe into a reservoir
...

B
...
40

C
...
41

𝐾𝑐 = the coefficient of sudden contraction
𝑣 = velocity in smaller pipe

Where

D2/D1
KC

𝑣2
2𝑔

0
...
50

0
...
45

0
...
42

0
...
4
0
...
6
0
...
39
0
...
33
0
...
22
Loss coefficients for Sudden Contraction

0
...
15

0
...
06

0
...
00

D
...
42
2𝑔
The approximate values of 𝐾 are given in the table below
...
For these reasons, manufacturer’s data are the best source for loss coefficients
...
For very smooth pipes, it is better to use the 𝐾 values when determining the loss through fittings
...


𝐻𝐿 = 𝐻 = ℎ𝐿𝑚𝑖𝑛𝑜𝑟 + ℎ𝑓

Eqn
...
44
Eqn
...
Neglecting minor losses, the head lost becomes
𝐻𝐿 = ℎ𝑓1 + ℎ𝑓2 + ℎ𝑓3
If, however, it is desired to include minor losses, a solution may be made first by neglecting them and correcting the
results
...
The sum of the flow in pipes 2 and 3 equals the flow in
pipes 1 and 4
...


The necessary equations for the system are:
𝑄1 = 𝑄4
𝑄1 = 𝑄2 + 𝑄3
ℎ𝑓2 = ℎ𝑓3
𝐻𝐿𝐴𝐵 = ℎ𝑓1 + ℎ𝑓2 + ℎ𝑓4

→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 3
→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 4

NOTE: The number of equations needed to solve the problem must be equal to the number of pipes
...
869 and dynamic viscosity of 0
...
The pipe is 50 m long and 150 mm in diameter
...

Given:

𝑠𝑔 = 0
...
0814 𝑃𝑎 − 𝑠
𝐷 = 150 𝑚𝑚 = 0
...

Use the Reynolds Number formula to identify the type of flow
...
15 𝑚) (1000 𝑚3 ) (0
...
0814 𝑃𝑎 − 𝑠
𝑅𝑒 = 1601 < 2000 (𝑙𝑎𝑚𝑖𝑛𝑎𝑟)
We will use the Darcy-Weishbach Formula to solve for the head lost due to friction
𝑓𝐿𝑣 2
ℎ𝑓 =
2𝑔𝐷
We need to calculate first the value of 𝑓
...
04
𝑅𝑒 1601
Substituting the values in the Darcy-Weishbach Formula
𝑚 2
0
...
𝟔𝟖 𝒎
𝑚
2 (9
...
15 𝑚)
𝑠
(b) The shearing stress can be calculated using
𝑁
𝛾ℎ𝐿 𝐷 𝑠𝑔(𝛾𝑤𝑎𝑡𝑒𝑟 )ℎ𝑓 𝐷 0
...
68 𝑚)(0
...
𝟑𝟓 𝑷𝒂
4𝐿
4𝐿
4 (50 𝑚)
𝒎
2) Determine the (a) shear stress at the walls of a 300-mm-diameter pipe when water flowing causes a head loss of
5 m in a 90-m pipe length, (b) the shear velocity, and (c) the shear stress at 50 mm from the centerline of the
pipe
...
3 𝑚

ℎ𝐿 = 5 𝑚

𝐿 = 90 𝑚

Solution:
(a) For the Shear stress at the walls use Eqn
...
3 𝑚)
𝜏𝑂 =
=
= 𝟒𝟎
...
16 to calculate the shear stress

𝜏𝑜
40
...
𝟐𝟎𝟐 𝒎/𝒔
𝜌
1000 𝑘𝑔/𝑚3
(c) Use Eqn
...
05 𝑚 from the pipe center
𝑁
9810 3 (5 𝑚)
𝛾ℎ𝐿
𝑚
(0
...
𝟔𝟑 𝑷𝒂
𝜏𝑠 =
𝑥=
2𝐿
2(90 𝑚)
3) What commercial size of new cast iron pipe shall be used to carry 4,490 gpm with a lost of head of 10
...
019
Given:

1 𝑚𝑖𝑛
1 𝑚3
)
(
) = 0
...
28 𝑓𝑡) = 1609
...
56 𝑓𝑡 (
) = 3
...
28 𝑓𝑡
𝑔𝑎𝑙

3
...
019

Solution:
Use Darcy-Weishbach Formula (Eqn
...
0826𝑓𝐿𝑄 2
𝐷5

2

𝑚3
0
...
019)(1609
...
284 𝑠 )
3
...
𝟓𝟕𝟔 𝒎 = 𝟓𝟕𝟔 𝒎𝒎
4) Water flows from a tank through 160 feet of 4 inches diameter
pipe and then discharges into air as shown in the figure below
...
Assume 𝑛 = 0
...
Determine the following: (a) the velocity of
flow in the pipe in fps; (b) the total head lost in the pipe in feet;
(c) the pressure at the top of the tank in psi
...
013

Solution:
(a) Velocity of flow in the pipe
𝑓𝑡 3
12 𝑠
𝑄
𝑄 = 𝐴𝑣
→ 𝑣= =
2 = 𝟏𝟑𝟕
...
Since the formula is for SI units, convert the given
values to SI
1

𝐿 = 160 𝑓𝑡 = 48
...
1016

𝑄 = 12

𝑓𝑡 3
𝑠

= 0
...
29𝑛 𝐿𝑄
16
𝐷3

2

=

𝑚3
10
...
013)2 (48
...
3401 𝑠 )

𝐻𝐿 = 1942
...
1016) 3

3
...
𝟖𝟔 𝒇𝒕
1𝑚

(c) For the pressure in the tank, make an energy equation between A and C
𝐸𝐴 − 𝐻𝐿 = 𝐸𝐵

𝑚3
𝑠

𝑣𝐴2 𝑃𝐴
𝑣𝐶2 𝑃𝐶
+ + 𝑧𝐴 − 𝐻𝐿 =
+ + 𝑧𝐶
2𝑔 𝛾
2𝑔 𝛾
𝑓𝑡 2
(137
...
86 𝑓𝑡 =
+ 0 + 100 𝑓𝑡
𝑓𝑡
𝛾
2 (32
...
48 𝑓𝑡
→ 𝑃𝐴 = 6754
...
4 3 ) = 421,479
...
43

𝑙𝑏𝑓
𝑓𝑡 2
(
) = 𝟐𝟗𝟐𝟔
...
9) and dynamic viscosity of 𝜇 = 0
...
If the head loss is 6 m, determine: (a) the mean velocity of flow; (b) the type of flow; (c) the friction
factor 𝑓; (d) velocity at the centerline of the pipe; (e) the shear stress at the wall of the pipe and (f) the velocity
50 mm from the center of the pipe
...
90

𝐿 = 50 𝑚
Solution:
(a) Mean velocity

𝐿

1 𝑚3

𝜇 = 0
...
06

𝐷 = 120 𝑚𝑚 = 0
...
06 𝑠
𝑄
𝑄
𝑣= =
=
= 𝟓
...
12 𝑚)2
𝐴 𝜋𝐷 2
4
4
(b) Type of flow
𝑘𝑔
𝑚
𝑣𝐷𝜌 𝑣𝐷𝜌𝑤𝑎𝑡𝑒𝑟 𝑠𝑔 5
...
12 𝑚) (1000 𝑚3 ) (0
...
04 𝑃𝑎 − 𝑠
𝑅𝑒 = 14,337 > 2000 (𝒕𝒖𝒓𝒃𝒖𝒍𝒆𝒏𝒕 𝒇𝒍𝒐𝒘)
(c) Friction factor
0
...
0826𝐿𝑄 2

6 𝑚(0
...
0826 (50 𝑚) (0
...
𝟎𝟏𝟎𝟎𝟒

(d) Centerline velocity
1

𝑣𝑐 = 𝑣 (1 + 1
...
31

1
𝑚
[1 + 1
...
01004)2 ]
𝑠

𝒗𝒄 = 𝟔
...
90)(0
...
31 )
3
𝑠
𝑚
= 𝟑𝟏
...
75√ log
𝜌
𝑟𝑜 − 𝑟

𝑢 = 6
...
85 𝑃𝑎
60 𝑚𝑚
− 5
...
90)
𝑚
𝒖 = 𝟓
...
The flow of water through the pipes is 150 L/s with a total
frictional loss of 15 m
...
02 for both pipes
...
3 𝑚

𝐻𝐿 = 15 𝑚

𝐿

𝑚3

𝑄 = 150 𝑠 (1000 𝐿) = 0
...
15

𝑚3
𝑠

𝐻𝐿 = ℎ𝑓1 + ℎ𝑓2
𝐻𝐿 =

0
...
0826𝑓2 𝐿2 𝑄22
𝐷25

2

2

𝑚3
𝑚3
0
...
02)(300 𝑚) (0
...
0826(0
...
15 𝑠 )
15 𝑚 =
+
0
...
𝟐𝟓𝟓 𝒎 = 𝟐𝟓𝟓 𝒎𝒎

7) Two pipes 1 and 2 having the same length and diameter are in parallel
...
75

𝑚3
𝑠

Solution:
For pipes in parallel, the head losses are equal
ℎ𝑓1 = ℎ𝑓2
0
...
0826𝑓2 𝐿2 𝑄22
+
𝐷15
𝐷25
Since 𝐿1 = 𝐿2 , 𝐷1 = 𝐷2 𝑎𝑛𝑑 𝑓2 = 2𝑓1 the equation above becomes
𝑓1 𝑄12 = 2𝑓1 𝑄22

𝑄12 = 2𝑄22

→

2

𝑄12
𝑄22 = √
2

𝑚3
(0
...
𝟓𝟑
= 𝟓𝟑𝟎
2
𝒔
𝒔



---