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Steady Incompressible Flow in Pressure Conduits

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Laminar and Turbulent Flow in Pipes Laminar Paths of Particles don t obstruct each other Viscous forces are dominant Velocity of fluid particles only changes in ... – PowerPoint PPT presentation

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Title: Steady Incompressible Flow in Pressure Conduits

1
Steady Incompressible Flow in Pressure Conduits
2
Laminar and Turbulent Flow in Pipes
• Laminar
• Paths of Particles dont obstruct each other
• Viscous forces are dominant
• Velocity of fluid particles only changes in
magnitude
• Lateral component of velocity is zero
• Turbulent
• Paths do intersect each other
• Inertial forces are dominant
• Velocity of fluid particles change in magnitude
and direction
• Lateral components do exist.

3
Laminar and Turbulent Flow in Pipes
• If we measure the head loss in a given length of
uniform pipe at different velocities , we will
find that, as long as the velocity is low enough
to secure laminar flow, the head loss, due to
friction, is directly proportional to the
velocity, as shown in Fig. 8.1. But with
increasing velocity, at some point B, where
visual observation of dye injected in a
transparent tube would show that the flow changes
from laminar to turbulent, there will be an
abrupt increase in the rate at which the head
loss varies. If we plot the logarithms of these
two variables on linear scale or in other words,
if we plot the values directly on log-log paper,
we will find that, after passing a certain
transition region (BCA in Fig. 8.1), the lines
will have slopes ranging from about 1.75 to 2.

4
Laminar and Turbulent Flow in Pipes
• Thus we see that for laminar flow the drop in
energy due to friction varies as V, while for
turbulent flow the friction varies as Vn, where n
ranges from about 1.75 to 2. The lower value of
1.75 for turbulent flow occurs for pipes with
very smooth walls as the wall roughness
increases, the value of n increases up to its
maximum value of 2.
• If we gradually reduce the velocity from a high
value, the points will not return along line BC.
Instead, the points will lie along curve CA. We
call point B the higher critical point, and A the
lower critical point.
• However, velocity is not the only factor that
determines whether the flow is laminar or
turbulent. The criterion is Reynolds number.

5
Laminar and Turbulent Flow in Pipes
6
Reynolds Number
• Ratio of inertia forces to viscous forces is
called Reynolds number.
• Where we can use any consistent system of
units, because R is a dimensionless number.

7
Significance of Reynolds Number
• To investigate the development of Laminar and
Turbulent flow
• Investigate Critical Reynolds Number
• Develop a relationship between head loss (hL) and
velocity.

8
Critical Reynolds Number
• The upper critical Reynolds number, corresponding
to point B of Fig. 8.1, is really indeterminate
and depends on the care taken to prevent any
initial disturbance from effecting the flow. Its
value is normally about 4000, but experimenters
have maintained laminar flow in circular pipes up
to values of R as high as 50,000. However, in
such cases this type of flow is inherently
unstable, and the least disturbance will
transform it instantly into turbulent flow. On
the other hand, it is practically impossible for
turbulent flow in a straight pipe to persist at
values of R much below 2000, because any
turbulence that occurs is damped out by viscous
friction. This lower value is thus much more
definite than the higher one, and is the real
dividing point the two types of flow. So we
define this lower value as the true critical
Reynolds number.

9
Critical Reynolds Number
• It will be higher in a converging pipe and lower
in a diverging pipe than in a straight pipe.
Also, it will be less for flow in a curved pipe
than in a straight one, and even for a straight
uniform pipe it may be as low as 1000, where
there is excessive roughness. However, for normal
cases of flow in straight pipes of uniform
diameter and usual roughness, we can take the
critical value as
• Rcrit
2000

10
Problem
• Q In a refinery oil (? 1.8 x 10-5 m2/s) flows
through a 100-mm diameter pipe at 0.50 L/s. Is
the flow laminar or turbulent?
• Solution
• Q 0.50 Liter/s 0.0005 m3/s
• D 100 mm 0.1 m
• Q AV (pD2/4)V, V (4Q)/(pD2)
• V (4 x 0.0005)/(3.14 x 0.1 x0.1) 0.0637
m/s
• R (DV)/? (0.1 x 0.0637)/(1.8 x 10-5)
354
• Since R lt Rcrit2000, the flow is
laminar.

11
• For conduits having non-circular cross sections,
we need to use some value other than the diameter
for the linear dimension in the Reynolds number.
The characteristic dimension we use is the
• Rh A/P
• Where A is the cross sectional area of the
flowing fluid, and P is the wetted perimeter,
that portion of the perimeter of the cross
section where the fluid contacts the solid
boundary, and therefore where friction resistance
is exerted on the flowing fluid. For a circular
pipe flowing full,
• Full-pipe flow Rh (p r2)/(2pr)
r/2 D/4

12
• This discussion applies to either laminar or
turbulent flow and to any shape of cross section.
• Consider steady flow in a conduit of uniform
cross section A, not necessarily circular as
shown Fig. below. The pressures at sections 1 and
2 are p1 and p2, respectively. The distance
between the sections is L.

13
• For equilibrium in steady flow, the summation of
forces acting on any fluid element must be equal
to zero (i.e., SFma0). Thus, in the direction
of flow,
• p1A p2A ?LAsina t0(PL) 0 (1)
• where we define t0, the average shear stress
(average shear force per unit area) at the
conduit wall.
• Nothing that sina (z2 z1)/L and dividing
each term in eq. (1) by ?A gives,
• p1A/(?A) p2A/(?A) ?LA(z2 - z1)/(?AL)
t0(PL)/(?A)
• p1/? p2/? z2 z1 t0(PL)/(?A) .
(2)
• From the left hand sketch of the Fig., we can
see that the head loss due to friction at the
wetted perimeter is
• hf (z1 p1/?) (z2 p2/?) .. (3)

14
• The eq.(3) equation indicates that hf depends
only on the values of z and p on the centerline,
and so it is the same regardless of the size of
the cross-sectional area A. Substituting hf from
eq.(3) and replacing A/P by Rh in eq.(2), we get,
• hf t0L/(Rh?)
(4)
• This equation is applicable to any shape of
uniform cross section, regardless of whether the
flow is laminar or turbulent.

15
• For a smooth-walled conduit, where we can
neglect wall roughness, we might assume that the
average fluid shear stress t0 at the wall is
some function of ?, µ, V and some characteristic
linear dimension, which we will here take as the
• t0 f(?, µ, V, Rh)
• Using the pi theorem of dimensional analysis
to better determine the form of this
relationship, we choose ?, Rh and V as primary
variables, so that
• ?1 µ ?a1 Rhb1 Vc1
• ?2 t0 ?a2 Rhb2 Vc2
• With the dimensions of the variables being
ML-1T-1 for µ, ML-1T-2 for t0, ML-3 for ?,
L for Rh, and LT-1 for V,

16
• the dimensions for ?1 are
• ?1 µ ?a1 Rhb1 Vc1
• M0L0T0 (ML-1T-1) (ML-3)a1 (L)b1 (LT-1)c1
• For M 0 1 a1
• For L 0 -1 3a1 b1 c1
• For T 0 -1 c1
• The solution of these simultaneous equations
is
• a1 b1 c1 -1, from which
• ?1 µ ?-1 Rh-1 V-1
• ?1 µ /(? Rh V) R-1
• where (? Rh V)/µ is a Reynolds number with Rh
as the characteristic length.

17
• the dimensions for ?2 are
• ?2 t0 ?a2 Rhb2 Vc2
• M0L0T0 (ML-1T-2) (ML-3)a2 (L)b2 (LT-1)c2
• For M 0 1 a2
• For L 0 -1 3a2 b2 c2
• For T 0 -2 c2
• The solution of these simultaneous equations
is
• a2 -1, c2 -2, b2 0, from which
• ?2 t0 ?-1 V-2
• ?2 t0 /(?V2)
• We can write ?2 ?( ?1-1), which results in
• t0 ? V2 ? (R)
• .

18
• Setting the dimensionless term ? (R) ½ Cf ,
this yields
• t0 Cf ? V2/2
• Where Cf average friction-drag coefficient
for total surface (dimensionless)
• Inserting this value of t0 and ? ?g, in
eq. (4), which is
• hf t0L/(Rh?) , we get
• hf Cf (L/Rh)(V2/2g) (5)
• which can apply to any shape of smooth-walled
cross section. From this equation , we may easily
obtain an expression for the slope of the energy
line,
• S hf / L Cf /Rh (V2/2g) (6)
• which we also know as the energy gradient.

19
Friction in Circular Conduits
• Head loss due to friction, hf Cf
(L/Rh)(V2/2g)
• Energy gradient, S hf / L Cf /Rh (V2/2g)
• For a circular pipe flowing full, Rh D/4,
and
• f 4Cf , where f is friction factor (also
some times called the Darcy friction factor) is
dimensionless and some function of Reynolds
number.
• Substituting values of Rh and Cf into above
equations, we obtain (for both smooth-walled and
rough-walled conduits) the well known equation
• Circular pipe flowing full (laminar or
turbulent flow)
• hf f (L/D) (V2/2g) . (7)
• and hf /L S f /D (V2/2g) .. (8)

20
Friction in Circular Conduits
• For a circular pipe flowing full, by
substituting Rh r0/2, where r0 is the radius of
the pipe in the eq. (4), we get
• hf t0L/(Rh?) 2t0L/(r0?)
• where the local shear stress at the wall, t0,
is equal to the average shear stress t0 because
of symmetry.

21
Friction in Circular Conduits
• The shear stress is zero at the center of the
pipe and increases linearly with the radius to a
maximum value t0 at the wall as shown in Fig.
8.3. This is true regardless of whether the flow
is laminar or turbulent.
• From eq.(4) hf t0L/(Rh?), we have
• t0 hf (Rh?)/L, substituting eq.(7) and Rh
D/4 into this, we obtain
• t0 f (L/D)(V2/2g)(D/4)(?/L)
• t0 (f /4) ? (V2/2g) or t0 (f /4) ?
(V2/2) where ??g
• With this equation, we can compute t0 for
flow in a circular pipe for any experimentally
determined value of f.

22
Friction in Circular Conduits
• For laminar flow under pressure in a circular
pipe,
• We may use the pipe-friction equation (7) with
this value of f as given by the above equation.

23
Problem
• Q Stream with a specific weight of 0.32 lb/ft3
is flowing with a velocity of 94 ft/s through a
circular pipe with f 0.0171. What is the shear
stress at the pipe wall?
• Solution
• ? 0.32 lb/ft3
• V 94 ft/s
• f 0.0171
• g 32.2 ft/s2
• t0 ?
• t0 (f /4) ? (V2/2g)
• t0 (0.0171/4)(0.32)(94x94)/(2x32.2)
• t0 0.187 lb/ft2

24
Problem
• Q Stream with a specific weight of 38 N/m3 is
flowing with a velocity of 35 m/s through a
circular pipe with f 0.0154. What is the shear
stress at the pipe wall?
• Solution
• ? 38 N/m3
• V 35 m/s
• f 0.0154
• g 9.81 m/s2
• t0 ?
• t0 (f /4) ? (V2/2g)
• t0 (0.0154/4)(38)(35x35)/(2x9.81)
• t0 9.13 N/m2

25
Problem
• Q Oil of viscosity 0.00038 m2/s flows in a 100mm
diameter pipe at a rate of 0.64 L/s. Find the
• Solution
• ? 0.00038 m2/s
• D 100 mm 0.1 m
• Q 0.64 L/s 0.00064 m3/s
• g 9.81 m/s2
• hf /L ?
• Q AV (pD2/4)V, V (4Q)/(pD2)
• V (4 x 0.00064)/(3.14 x 0.1 x 0.1)
0.0815 m/s
• R (DV)/? (0.1 x 0.0815)/(0.00038)
21.45

26
Problem
• f 64 / R 64/21.45 2.983
• hf /L S f /D (V2/2g)
• hf /L (2.983/0.1)(0.0815x0.0815)/(2x9.81)
• hf /L 0.010 m/m

27
Friction in Non-Circular Conduits
• Most closed conduits we use in engineering
practice are of circular cross section however
we do occasionally use rectangular ducts and
cross sections of other geometry. We can modify
many of the equations for application to non
circular sections by using the concept of
• For a circular pipe flowing full, that
• Rh A/P (p D2/4)/(pD) D/4
• D 4 Rh
• This provides us with an equivalent diameter,
which we can substitute into eq. (7) to yield
• hf f (L/4Rh)(V2/2g)

28
Friction in Non-Circular Conduits
• and when substitute into equation of Reynolds
number, we get
• R (DV?)/µ (4RhV?)/µ (4RhV)/?
• This approach gives reasonably accurate
results for turbulent flow, but the results are
poor for laminar flow, because in such flow
viscous action causes frictional phenomena
throughout the body of the fluid, while in
turbulent flow the frictional effect occurs
largely in the region close to the wall i.e., it
depends on the wetted perimeter.

29
Entrance Conditions in Laminar Flow
• In the case of a pipe leading from a
reservoir, if the entrance is rounded so as to
avoid any initial disturbance of the entering
stream, all particles will start to flow with the
same velocity, except for a very thin film in
contact with the wall. Particles in contact with
the wall have zero velocity and with the slight
exception, the velocity is uniform across the
diameter.

30
Entrance Conditions in Laminar Flow
• As the fluid progresses along the pipe,
friction origination from the wall slows down the
streamlines in the vicinity of the wall, but
since Q is constant for successive sections, the
velocity in the center must accelerate, until the
final velocity profile is a parabola as shown in
Fig. 8.3. Theoretically, this requires an
infinite distance, but both theory and
observation have established that the maximum
velocity in the center of the pipe wall reach 99
of its ultimate value in a distance
• Le 0.058 RD
• We call this distance the entrance length. For
a critical value of R 2000, the entrance length
Le equals 116 pipe diameters. In other cases of
laminar flow with Reynolds number less than 2000,
the distance Le will be correspondingly less in
accordance with the above equation.

31
Entrance Conditions in Laminar Flow
• Within the entrance length the flow is
unestablished that is the velocity profile is
changing. In this region, we can visualize the
flow as consisting of a central inviscid core in
which there are no frictional effects, i.e., the
flow is uniform, and an outer, annular zone
extending from the core to the pipe wall. This
outer zone increases in thickness as it moves
along the wall, and is known as the boundary
layer. Viscosity in the boundary layer acts to
transmit the effect of boundary shear inwardly
into the flow. At section AB the boundary layer
has grown until it occupies the entire cross
section of the pipe. At this point, for laminar
flow, the velocity profile is a perfect parabola.
Beyond section AB, for the same straight pipe the
velocity profile does not change, and the flow is
known as (laminar) established flow or (laminar)
fully developed flow.

32
Entrance Conditions in Laminar Flow
• The flow will continue as fully developed so
long as no change occurs to the straight pipe
surface. When a change occurs, such as at a bend
or other pipe fitting, the velocity profile will
deform and will require some more flow length to