pt

2. 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 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.





 VD
VL
LV
V
L
F
F
R
V
,
2
2
1











 where
DV
DV
R ,

7. Significance of Reynolds Number
 To investigate the development of Laminar and Turbulent
flow
 Investigate Critical Reynold’s 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 (ν = 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 = (πD2/4)V, V = (4Q)/(πD2)
V = (4 x 0.0005)/(3.14 x 0.1 x0.1) = 0.0637 m/s
R = (DV)/ν = (0.1 x 0.0637)/(1.8 x 10-5) = 354
Since R < Rcrit=2000, the flow is laminar.

11. Hydraulic Radius
 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 hydraulic radius, defined as
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 = (π r2)/(2πr) = r/2 = D/4

12. Friction Head Loss in Conduits
 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. Friction Head Loss in Conduits
 For equilibrium in steady flow, the summation of forces acting
on any fluid element must be equal to zero (i.e., ΣF=ma=0).
Thus, in the direction of flow,
p1A – p2A – γLAsinα – τ’0(PL) = 0 …………… (1)
where we define τ’0, the average shear stress (average shear
force per unit area) at the conduit wall.
Nothing that sinα = (z2 – z1)/L and dividing each term in eq.
(1) by γA gives,
p1A/(γA) – p2A/(γA) – γLA(z2 - z1)/(γAL) = τ’0(PL)/(γA)
p1/γ – p2/γ – z2 + z1 = τ’0(PL)/(γA) ……………. (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/γ) – (z2 + p2/γ) ………………….. (3)

14. Friction Head Loss in Conduits
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 = τ’0L/(Rhγ) ……………… (4)
This equation is applicable to any shape of uniform cross
section, regardless of whether the flow is laminar or turbulent.

15. Friction Head Loss in Conduits
For a smooth-walled conduit, where we can neglect wall
roughness, we might assume that the average fluid shear stress
τ’0 at the wall is some function of ρ, μ, V and some
characteristic linear dimension, which we will here take as the
hydraulic radius Rh. Thus
τ’0 = f(ρ, μ, V, Rh)
Using the pi theorem of dimensional analysis to better
determine the form of this relationship, we choose ρ, Rh and V
as primary variables, so that
∏1 = μ ρa1 Rh
b1 Vc1
∏2 = τ’0 ρa2 Rh
b2 Vc2
With the dimensions of the variables being ML-1T-1 for μ,
ML-1T-2 for τ’0, ML-3 for ρ, L for Rh, and LT-1 for V,

16. Friction Head Loss in Conduits
the dimensions for ∏1 are
∏1 = μ ρa1 Rh
b1 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 ρ-1 Rh
-1 V-1
∏1 = μ /(ρ Rh V) = R-1
where (ρ Rh V)/μ is a Reynolds number with Rh as the
characteristic length.

17. Friction Head Loss in Conduits
the dimensions for ∏2 are
∏2 = τ’0 ρa2 Rh
b2 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 = τ’0 ρ-1 V-2
∏2 = τ’0 /(ρV2)
We can write ∏2 = Ф( ∏1
-1), which results in
τ’0 = ρ V2 Ф (R)

18. Friction Head Loss in Conduits
Setting the dimensionless term Ф (R) = ½ Cf , this yields
τ’0 = Cf ρ V2/2
Where Cf = average friction-drag coefficient for total surface
(dimensionless)
Inserting this value of τ’0 and γ = ρg, in eq. (4), which is
hf = τ’0L/(Rhγ) , 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 for pipe-friction head loss,
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 = τ’0L/(Rhγ) = 2τ0L/(r0γ)
where the local shear stress at the wall, τ0, is equal to the
average shear stress τ’0 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 τ0 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 = τ0L/(Rhγ), we have
τ0= hf (Rhγ)/L, substituting eq.(7) and Rh = D/4 into this, we
obtain
τ0= f (L/D)(V2/2g)(D/4)(γ/L)
τ0= (f /4) γ (V2/2g) or τ0= (f /4) ρ (V2/2) where γ=ρg
With this equation, we can compute τ0 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.
R
DV
where
R
DV
f 




,
64
64

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:
γ = 0.32 lb/ft3
V = 94 ft/s
f = 0.0171
g = 32.2 ft/s2
τ0= ?
τ0= (f /4) γ (V2/2g)
τ0= (0.0171/4)(0.32)(94x94)/(2x32.2)
τ0= 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:
γ = 38 N/m3
V = 35 m/s
f = 0.0154
g = 9.81 m/s2
τ0= ?
τ0= (f /4) γ (V2/2g)
τ0= (0.0154/4)(38)(35x35)/(2x9.81)
τ0= 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 head loss per unit length.
 Solution:
ν = 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 = (πD2/4)V, V = (4Q)/(πD2)
V = (4 x 0.00064)/(3.14 x 0.1 x 0.1) = 0.0815 m/s
R = (DV)/ν = (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 hydraulic radius.
For a circular pipe flowing full, that
Rh = A/P = (π D2/4)/(πD) = 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 = (DVρ)/μ = (4RhVρ)/μ = (4RhV)/ν
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 return to
established flow. Usually such fittings are so far apart that fully
developed flow is common; but when they are close enough it
is possible that established flow never occurs.