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Define fluid and its proerties
Typology: Summaries
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Training Package
in
Fluid Mechanics
Modular unit 5
Flow through Pipes
By
Risala A. Mohammed
Environmental & Pollution Engineering Department
2011
Bernoulli´s Equation
Consider a fluid flowing in a tube as shown in the figure
in a steady, incompressible, no viscous flow. We apply
the work-energy theorem to a sample of fluid initially
contained between point 1 and 2. During time Δt this sample moves along the tube to the region between
points 1 ´and 2 ´. So,
Wall forces = ΔK [Change of kinetic Energy]
Wall forces include gravitational and pressure forces. We neglecting internal frictional forces. Non-viscous flow.
No mechanical dissipation energy is considered.
Work of gravitational forces can be computed as the
variation of potential energy of sample. Considering the
continuity equation we can obtain
Change of kinetic energy of sample will be Fluid moving in a pipe that varies in both height and cross-sectional area. The net effect on the sample
during a time Δt is that a mass initially at height h 1 and
speed v 1 is transferred to a height h 2 with speed v 2
U ( m ) g ( h 1 h 2 ) g V ( h 1 h 2 )
( )( ) ( )
2 1
2 2 2
2 1 1
2 2 2
1 K m v v V v v
Bernoulli´s Equation
P gh v const
P gh v P gh v
2
2
1
2
2 2
1
2 2
2
2 1
1
1 1
Remarks about Bernoulli Equation
g
v
h
g
P
const
g
v h
g
P
2
2
2
2
2
Bernoulli´s Equation
The Bernoulli equation aids in solving problem in which the losses due to internal friction
(viscous flow) can be accounted experimentally by a determined coefficient. So, we can
write
1 2
2
2 2
1
2 2
2
2 1
1
(^1 1)
P gh v P gh v losses
A z hL g
p V z h
g
p V 2
2 2 2 1
2
1 1
2 2
T z hL g
p V z h
g
p V 2
2 2 2 1
2
1 1
2 2
Euler´s equation
When pump was used
hA= pump head
hL= head losses
When turbine was used
hT= turbine head
Friction Losses in a Pipe
For a constant-diameter horizontal pipe, the
extended Bernoulli equation yields
p p p gh
Head loss due to friction:
g
V
D
h f L 2
Our problem is now reduced to solving for Darcy friction factorf
Recall
Therefore
1 2
L
w
P 1 V P 2
Two general types of networks
Volume flow rate is constant. Head loss is the summation of parts
is the same
1 2 3
A B
12 , 300 N / m , 0. 62 Ns / m
Given: Glycerin@ 20
o C flows commercial steel pipe. Find: h
Solution:
m
γD
μLV h h
VD VD
z h γ
p z γ
p h
z γ
p z h γ
p
z γ
p
g
V z h α γ
p
g
V α
12 , 300 *( 0. 02 )
32 32 ( 0. 62 )( 1 )( 0. 6 )
5 (laminar)
1 * 10
6 * 0. 02 Re
( )
2 2
z ft
ft s A
K K K f
g
h K K K f
z h z
z γ
p
g
z h α γ
p
g
α
e b E
L e b E
L
L
2
1
2
2
1 2
2
2
2 2 1 2
1
2 1 1
Solution:
5 5
2 2
2
2 2
2
2
2 2 1 2
1
2 1 1
Re
x x
ft g
ft s A
K f g
h
z h γ
p
g
z h α γ
p
g
α
p b
p L
^
p
p
Sof = 0.
