Upload
nguyentu
View
225
Download
2
Embed Size (px)
Citation preview
2
Theoretical Discussion
Fluid flow in pipes is of considerable importance in process.
•Animals and Plants circulation systems.
•In our homes.
•City water.
•Irrigation system.
•Sewer water system
Fluid could be a single phase: liquid or gases,
Mixtures of gases, liquids and solids
NonNewtonian fluids such as polymer melts, mayonnaise
Newtonian fluids like in your experiment (water)
3
1 2
1
1
2
2
2
3
5
4
4
Basic Components of a Pipe System
1. Individual straight pipes
2. Pipe connectors
3. Flow rate control devices (valves)
4. Inlet and outlet
5. Pump or turbines that add energy
or remove energy from the fluid
Items 2,3 and 4 are often called pipe components.
4
Bernoulli’s Equation - Revisit
For INVISCID, INCOMRESSIBLE, STEADY and
IRROTATIONAL flows, we have
i.e.
0)2
(2
V
gzP
streamlineaalongconstV
gzP ,2
2
5
Real Pipe Flows
For real pipe flows, we need to consider
The pressure drop in a pipe system due to the viscous
nature of real fluid, , including
(1) pressure drop in straight pipes
i.e. major loss,
(2) pressure drop in pipe components
i.e. minor loss,
Energy input and output by pumps or turbines,
LP
workP
majorLP ,
orLP min,
6
Real Pipe Flows – Energy Equation
• The energy equation for a real pipe system:
where
• These equations have significant implications to engineering design of a pipe system
0)2
(2
Lwork PPV
gzP
orLmajorLL PPP min,,
8
Flow Through Circular Conduits
Consider the steady flow of a fluid of constant density in
fully developed flow through a horizontal pipe and
visualize a disk of fluid of radius r and length dL moving
as a free body. Since the fluid posses a viscosity, a shear
force opposing the flow will exist at the edge of the disk
9
Boundary layer buildup in a pipe
Pipe
Entrance
v v v
Because of the share force near the pipe wall, a boundary layer
forms on the inside surface and occupies a large portion of the flow
area as the distance downstream from the pipe entrance increase. At
some value of this distance the boundary layer fills the flow area.
The velocity profile becomes independent of the axis in the direction
of flow, and the flow is said to be fully developed.
10
Laminar vs. Turbulent Flow
• Laminar flow: Fluid flows in smooth layers
and the shear stress is the result of
microscopic action of the molecules.
• Turbulent flow is characterized by large
scale, observable fluctuations in the fluid and
flow properties are the result of these
fluctuations.
12
Indication of Laminar or Turbulent Flow
The term flowrate should be replaced by Reynolds number, ,where V is the average velocity in the pipe.
It is not only the fluid velocity that determines the character of the flow – its density, viscosity, and the pipe size are of equal importance.
For general engineering purpose, the flow in a round pipe
Laminar
Transitional if 2100< Re < 4000
Turbulent
/VDRe
2100Re
4000>Re
13
Reynolds Number The Reynolds number can be used as a criterion to distinguish
between laminar and turbulent flow:
For very high Reynolds numbers, viscous forces are
negligible: inviscid flow
For very low Reynolds numbers (Re<<1) viscous forces
are dominant: creeping flow
(5.1)
DV Re
14
Fully Developed Flow
Flow in the entrance region of a pipe is complex.
Once the velocity profile no longer changes, we have reached fully
developed flow. Mathematically dV/dx = 0
Typical entrance length, 20 D < Le < 30 D
20
Forces acting on a fluid
The forces acting on a fluid are divided into two groups:
• Body forces act without physical contact. They act on
every mass element of the body and are proportional to
its total mass. Examples are gravity and electromagnetic
forces
• Surface forces require physical contact (i.e. surface
contact) with surroundings for transmission. Pressure
and stresses are surface forces.
21
Stresses In fluid mechanics it is convenient to define a force per unit area called
a stress (same units as pressure).
• Normal stress acts perpendicular to the surface (F=normal force).
Tensile causes elongation Compressive causes shrinkage
(Pressure is the most important
example of a compressive stress)
F F F F A A
A
Fnormalnormal
22
Stresses • Shear stress acts tangentially to the surface
(F=tangential or shear force).
F
F A
A
Fshearshear
Recall from chapter 2:
A fluid is defined as a substance that deforms continuously
when acted on by a shearing stress of any magnitude.
23
Shear Stress Profile
Force balance on cylindrical fluid element:
rx
PP
xrrPrP
2)(
02
21
2
2
2
1
x
r
o r
(5.2)
1 2
xrrPrP 2 2
2
2
1
24
Shear Stress profile
x
PP 21
2
r
Re-arrange eq 6.2, shows shear stress varies linearly with r, see eq below:
At the wall (r = ro):
x
PPW
21o
2
r
o
W
oW r
ror
r
r
Shear stress is a function of
the radial coordinate
or D
xPP W
4
)( 21
(5.3)
(5.4b)
(5.5)
(5.4a)
25
Case one: Laminar flow
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. Where the key
issue is the relation between flow rate and pressure drop.
If the flow of fluid is:
a. Newtonian
b. Isothermal
c. Incompressible (dose not depend on the pressure)
d. Steady flow (independent on time).
e. Laminar flow (the velocity has only one single component)
26
Shear Flow
NO-SLIP CONDITION: The fluid “sticks” to the solid boundaries.
The velocity of the fluid touching each plate is the same as that of the
plate (Vo for the top plate, and V= 0 for the bottom plate).
The velocity profile is a straight line: The velocity varies uniformly from
0 to Vo
Vo
A
yo
Fluid
x
y
F
= F/A
o
o
o
o
y
V
dy
dV and y
y
V)y(V
27
Shear Flow The force, F is proportional to the velocity Vo, the area in contact with
the fluid (A) and inversely proportional to the gap between plates (yo):
y
V AF
o
o
Recall, shear stress, = F / A
y
V
o
o
In the limit of small deformations the ratio Vo/yo can be replaced by the
velocity gradient dV/dy (or dU/dy):
dy
dV
Rate of shearing strain or shear rate:
dy
dV
28
Newton’s law of Viscosity Newton’s law of viscosity
dy
dV
m Viscosity [N/m2 . s =Pa . s]
n = /ρ : Kinematic viscosity [m2/s]
Newtonian fluids: Fluids which obey Newton’s law: Shearing stress is
linearly related to the rate of shearing strain.
(5.6)
The viscosity of a fluid measures its resistance to flow under an
applied shear stress.
29
Laminar Flow: Velocity profile Let’s consider again the flow of a fluid inside a pipe.
In cylindrical coordinates (6.6) can be written:
dr
dV By combining with eq (5.3)
and integrating:
2
o
2
o21
r
r1
x4
r )P-P()r(V
Velocity profile is parabolic
x
)P-P(
4
rr)r(V 21
22
o
[5.8 (a)]
[5.8 (b)]
x
PP 21
2
r
30
Laminar Flow: Velocity profile Minimum velocity, V=0 at the pipe wall (r=ro)
Maximum velocity Vmax at pipe centerline (located at r =0) and
x4
r )P-P(V
2
o21max
The velocity profile can be written:
2
o
maxr
r1 V)r(V
[5.8 (c)]
(5.9)
0dr
du
32
Hagen-Poiseuille Law
dA )r(VQThe volumetric flowrate through the pipe is:
x128
)PP( D
x8
)PP(r Q 21
4
o21
4
o
Or using average velocity:
x8
r )P-P(
r
Q
A
QV
2
o21
2
o
ave
(5.10)
(5.11)
And because of (5.9) :
2
VV max
ave (5.12)
dA=2πrdr
33
Example: Laminar Pipe Flow
An oil with a viscosity of μ= 0.40 (N/m2)·s and density ρ= 900 kg/m3
flows in a pipe of diameter D= 0.20m . What pressure drop, p1-p2, is
needed to produce a flowrate of Q=2.0×10-5 m3/s if the pipe is
horizontal with length of 10 m (ie let x1=0 and then x2=10 m) ?
Solution: First, we have to examine the flow type
210087.2/VDRe
s/m0637.0A
QV
The flow is laminar flow, then we can use Hagen-Poiseuille Law
kPaD
Qppp 4.20
128421
x
PPQ
128
)(D 21
4
o
34
Example:
pressure of oil in a pipe which discharges into the atmosphere is
measured at a certain location. Assume laminar flow conditions.
The pressure at the distance 15 m found 88 kPa.
(a) The flow rates are to be determined? Proof the flow is laminar.
Properties: The density and dynamic viscosity of oil are given to be = 876
kg/m3 and = 0.24 kg/ms.
(b) If the pipe is designed to produce uphill flow with an inclination of 8,
find flow rates
(c) If the pipe is designed to produce down hill flow with an inclination of 8,
find flow rates
35
2422
21
m 10767.14/m) 015.0(4/
kPa 4788135
DA
PPP
c
/sm 101.62 35
kPa 1
N/m 1000
N 1
m/skg 1
m) s)(15kg/m 24.0(128
m) (0.015kPa) 47(
128
2244
L
DPQ
0.7skg/m 24.0
m) m/s)(0.015 127.0)(kg/m 876(Re
m/s 127.0m 101.767
/sm.1024.2
3
24-
35
VD
AV
c
Q َ
Solution: (a)
Re < 2100 thus it is a laminar flow