#### Transcript Document

```Fluid Mechanics 08
Classifying Flow
We can classify the flow by calculate Reynolds No
When
The Reynolds No can be calculated from one
equation of the following
Developing and Fully Developed Flow
The distance required for flow to develop is
called the entrance length (Le).
For flow entering a circular pipe from a
reservoir the Le can be calculated from the
following equation:-
Summary
In summary, flow is classified into four
categories:laminar developing,
laminar fully developed
turbulent developing
turbulent fully developed.
The key to classification is to calculate the
Reynolds number
Example
Consider fluid flowing in a round tube of length
1 m and diameter 5 mm. Classify the flow as
laminar or turbulent and calculate the entrance
length for (a) air (50°C) with a speed of 12 m/s
and
(b) Water (15°C) with a mass flow rate of 8 g/s .
Solution
1. Air (50°C), ν = 1.79 × 10-5 m2/s.
2. Water (15°C), μ = 1.14 × 10-3 N*s/m2.
𝑉∗𝐷
12 ∗ 0.005
𝑅𝑒 =
=
= 3350
−5
𝑣
1.79 ∗ 10
The flow is turbulent for air
𝐿𝑒 = 50 ∗ 𝐷 = 50 ∗ 0.005 = 0.25𝑚
4 ∗ 𝑚ʹ
4 ∗ 0.008
𝑅𝑒 =
=
Π ∗ 𝐷 ∗ µ 3.14 ∗ 0.005 ∗ 1.14 ∗ 10−3
= 1787
The flow is laminar for water
𝐿𝑒 = 0.05 ∗ Re ∗ D = 0.05 ∗ 1787 ∗ 0.005
= 0.447m
Specifying Pipe Sizes
Component head loss is associated with flow
through devices such as valves, bends, and tees.
Pipe head loss is associated with fully developed
flow, and it is caused by shear stresses that act on
the flowing fluid.
Note that pipe head loss is sometimes called
Darcy-Weisbach Equation
To use Darcy-Weisbach Equation the flow should
be fully developed and steady. The DarcyWeisbach equation is used for either laminar
flow or turbulent flow and for either round pipes
or nonround conduits such as a rectangular duct.
Friction Factor
For Laminar flow :For turbulent flow
64
𝑓=
𝑅𝑒
0.25
𝑓=
𝑒
5.74 2
(log10 (
+ 0.9 ))
3.7 ∗ 𝐷 𝑅𝑒
Or from Moody Chart
Moody Chart
Roughness
Examples
Oil (S = 0.85) with a kinematic viscosity of 6 ×
10-4 m2/s flows in a 15 cm pipe at a rate of
0.020 m3/s. What is the head loss per 100 m
length of pipe?
Solution
𝑄
0.02 ∗ 4
𝑣= =
= 1.13𝑚/𝑠
2
𝐴 3.14 ∗ 0.15
𝑉 ∗ 𝐷 1.13 ∗ 0.15
𝑅𝑒 =
=
= 283
−4
𝑣
6 ∗ 10
64
64
𝑓=
=
= 0.226
𝑅𝑒 283
𝐿
𝑣2
ℎ𝐿 = 𝑓 ∗
∗
𝐷
2∗𝑔
100
1.132
= 0.226 ∗
∗
= 9.83𝑚
0.02
2 ∗ 9.81
Example
Water (T = 20°C) flows at a rate of 0.05 m3/s in
a 20 cm asphalted cast-iron pipe. What is the
head loss per kilometer of pipe?
Solution
𝑄
0.05 ∗ 4
1.59𝑚
𝑣= =
=
2
𝐴 3.14 ∗ 0.2
𝑠
𝑉 ∗ 𝐷 1.59 ∗ 0.2
𝑅𝑒 =
=
= 318471
−6
𝑣
1 ∗ 10
The flow is turbulent
From table e=0.12mm=0.00012m
𝑒 0.00012
=
= 0.0006
𝐷
0.2
From Moody Chart
𝑓 = 0.019
=
0.25
𝑓=
=
𝑒
5.74 2
(log10 (
+ 0.9 ))
3.7 ∗ 𝐷 𝑅𝑒
0.25
0.0006
5.74
(log10
+(
))^2
0.9
3.7
318471
0.25
2 = 0.0188 == 0.019
(log10 0.000226)
𝐿
𝑣2
ℎ𝐿 = 𝑓 ∗ ( ) ∗ ( )
𝐷
2∗𝑔
=12.2m
=0.019 ∗
1000
0.2
∗
1.592
(
)
2∗9.81
```