TURBULENT FLOW IN CIRCULAR PIPES 1

Learning Outcome

1) Derive the formula for

friction factor. 3) Calculate head and energy
loss due to friction in pipes.
Flow Profile
Major Loss
Minor Loss
2) Use friction factor chart

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 Osborne Reynolds Experiment

A flow can be Laminar, Turbulent or Transitional in

nature. This becomes a very important classification
of flows and is brought out vividly by the experiment
conducted by Osborne Reynolds (1842 - 1912).

Into a flow through a glass tube he injected a dye to

observe the nature of flow. When the speeds were
small the flow seemed to follow a straight line path
(with a slight blurring due to dye diffusion).

As the flow speed was increased the dye fluctuates and

one observes intermittent bursts.

As the flow speed is further increased the dye is blurred

and seems to fill the entire pipe. These are what we call
Laminar, Transitional and Turbulent Flows.
 Turbulent Flow 3

Turbulent Reynolds Number Re >4000

Most flow encountered in engineering practice are turbulent – how turbulences
affects wall shear stress

Turbulent flow characterized by random and rapid fluctuations of swirling

regions of fluid (eddies) throughout the flow.

The swirling eddies transport mass, momentum and energy to other regions
of flow more rapidly from molecular diffusion – higher values of friction,
heat and mass transfer

The chaotic fluctuations of fluid particles in turbulent flow play a dominant role
in pressure drop thus energy losses.
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 Fully Developed Pipe Flow
Recall, for simple shear flows u=u(y),
 =  du/dy where y is measured from wall

Thus, in fully developed pipe flow,

 = - du/dr where r is measured from centre

w = shear stress at the

wall, acting on the fluid

w,turbulent > w,laminar

Turbulent Flow Velocity Profile

Turbulent flow along a wall consist of 4 regions

– characterized by the distance from the wall
1.Viscous/laminar/linear/wall sublayer – very
linear, flow is streamlined
2.Buffer layer – flow is still dominated by
viscous effect
Turbulent layer 3.Overlap/transition layer – turbulent effects
more significant but not dominant
4.Outer/turbulent layer – turbulent effects
Overlap layer
dominate over viscous effect

Buffer layer
Viscous sublayer
Reynolds Number (Re)
𝜌𝑑𝑣
𝑅𝑒 =
𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝜇
𝑑 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑢𝑖𝑡 (𝑝𝑖𝑝𝑒, 𝑡𝑢𝑏𝑒 etc..)
𝑣 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝜇 = 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 ~
-

For non-rounded pipes, apply the

hydraulic diameter, Dh

Dh = 4(A)/P
A = cross-section area
P = wetted perimeter
 Energy Losses Due to - sudden expansion
- sudden contraction
Due to friction
- bends
- pipe fittings
- obstruction
Major Losses
Minor Losses

Energy Losses Head Losses

ℎ𝐿, 𝑡𝑜𝑡𝑎𝑙 = ℎ𝐿, 𝑚𝑎𝑗𝑜𝑟 + ℎ𝐿, 𝑚𝑖𝑛𝑜𝑟

 turbulent

me
Laminar

FLOw
" -

Re Re 2300? Re 14000 Rex400

=2300

low/slow moderate/medium
relacing high
A pipe and the fluid flowing have the following properties:

Water density, & = 1000kg/m3

Pipe diameter, d = 0.5m
Dynamic viscosity, M = 0.55x10^3 Ns/m2

What is the maximum velocity, u when the flow is laminar and maximum velocity for turbulent

Re: POU0
3,4000 + tur 4
Re

? Me + lam #e: 2300

2300

'
IN =

kgm/s

Laminar Turbulent

Re
pVD
=

Re
pVD
=

M
M

10003 100.3x103
1300 = sm

ig( 1.0.0is
4000 = 1000 r

V =
m/s
v = m/s
 Major Losses
Let’s consider: Uniform horizontal pipe and steady flow

F1
P1 P2
Flow direction
F1

1 l 2

Let: 𝑃1 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 1 𝑃2 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 2

𝑢1 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 1 𝑢2 = 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑎𝑡 2

𝑙 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 1 & 2

𝑑 = 𝑝𝑖𝑝𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟

ℎ𝑓 = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛

𝐹1 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
 Major Losses

Lets consider: Uniform horizontal pipe and steady flow

F1
P1 P2
Flow direction
F1

1 l 2
friction
Applying Bernoulli’s Equation between 1 and 2:
H
𝑃1 𝑢1 2
𝑃2 𝑢2 2 𝑧1 = 𝑧2
up t
+ + 𝑧1 = + + 𝑧2 + ℎ𝑓
P 𝜌𝑔 2𝑔 𝜌𝑔 2𝑔 𝑢1 = 𝑢2
purp
𝑃1 𝑃2 𝑃1 𝑃2
= + ℎ𝑓 ℎ𝑓 = − → 𝐷 ℎ𝑓 = head loss due to friction
𝜌𝑔 𝜌𝑔 𝜌𝑔 𝜌𝑔
Pressure reduces in the direction of flow due to the frictional resistance
 Major Losses

Frictional resistance , F1
2
= frictional resistance per unit wetted area per unit (velocity) x wetted area x (velocity)2

𝐹1 = 𝑓 ∗ × 𝜋𝑑𝑙 × 𝑢 2

Force Balance:
Pressure Force at cross-section 1 = 𝑃1 𝐴
Pressure Force at cross-section 2 = 𝑃2 𝐴
Frictional Force = F1

where A = cross-sectional area of the pipe.

𝑃1 𝐴 − 𝑃2 𝐴 − 𝐹1 = 0
𝑃1 𝐴 − 𝑃2 𝐴 = 𝐹1
𝑃1 − 𝑃2 𝐴 = 𝐹1
 Major Losses
∗ 2
∗ 2 𝐹1 𝑓 𝜋𝑑𝑙𝑢
𝑃1 − 𝑃2 𝐴 = 𝐹1 𝐹1 = 𝑓 𝜋𝑑𝑙𝑢 𝑃1 − 𝑃2 = =
𝐴 𝐴
∗ 2
𝑓 𝜋𝑑𝑙𝑢 →B
𝜋 2
𝑑
4
∗ 2
𝑓 4𝑙𝑢
f

=
𝑑
𝑃1 𝑃2
ℎ𝑓 = − → 𝐷 From D : 𝑃1 −𝑃2 = 𝜌𝑔ℎ𝑓
𝜌𝑔 𝜌𝑔
2
𝑓 4𝑙𝑢 ∗ 2
∗
4𝑙𝑢
D=B 𝜌𝑔ℎ𝑓 = ℎ𝑓 = 𝑓
𝑑 𝜌𝑔𝑑
1
 4+(t)
nf
Major Losses
=

2
∗
4𝑙𝑢
ℎ𝑓 = 𝑓
𝜌𝑔𝑑
𝑓 ∗
𝑓 where f = friction factor (Darcy)
=
𝜌 8

𝑓 𝑓 𝑅
∗ = 2=ϕ where R = Shear Stress
𝑓 =𝜌 8 𝜌𝑢
8

2
𝑙𝑢 2 𝑙𝑢
ℎ𝑓 = 𝑓 ℎ𝑓 = 8ϕ
𝑑 2𝑔 𝑑 2𝑔
 Darcy Weisbach Equation
2
𝐿𝑢
ℎ𝑓 = 𝑓 ∆𝑃 = 𝜌𝑔∆ℎ = 𝜌𝑔ℎ𝑓
𝑑 2𝑔 2
𝐿𝑢
= 𝜌𝑔 𝑓
𝑑 2𝑔
𝐿 𝜌𝑢 2
∆P = 𝑓
𝑑 2
Where:
f = friction factor (Darcy friction factor, Moody Chart)
L = pipe length, m
d = pipe diameter, m
u = average flow velocity, m/s
3
ρ = density of the fluid, 𝑘𝑔Τ𝑚
g = acceleration of gravity, m/s2
 The Moody Chart

The friction factor in fully developed turbulent pipe flow depends on the
Reynolds number and the relative roughness

Value of ε is determined experimentally by using artificially roughened

surfaces (by gluing sand grains in the inner of pipes) – by Prandtl’s
student.

Friction factor was calculated from the measurements of flow rate and
pressure drop.
Colebrook equation:
The Moody
Chart
 Equivalent roughness values for new commercial pipes*

- . .
The Moody Chart
Observations from Moody
chart:

➢For laminar flow, f ↓

with ↑ Re; independent
of surface roughness.

➢f – minimum for smooth

pipe & ↑ with roughness.

➢Transition region –
shaded area.

➢>> Re – nearly
horizontal independent
of Re
 Given EORe -One t

*******
⑰ t,
Given
②
Re

ED t**
③civent,
CHART A: Pipe friction chart 𝛷Re2 versus Re for various values of e/d where, 𝛷Re2 =
 Example
A pipe of 0.5 m diameter and 1200 m long is used to deliver an oil of density 950
& kg/m3and
viscosity 0.01 Ns/m2 at 0.4 m3/s. If the roughness of the pipe surface is 0.5 mm, determine
~

the pressure drop due to frictional losses across this pipeline.

DP: Pe-Pr
The same pipeline is used to deliver a second oil of density 980 kg/m and viscosity
3

0.02 Ns/m2 to the refinery. If the same pressure drop is to be maintained for optimum
operation, determine the required volumetric flow rate of the second oil in the pipe.
a Avv
= o5 2.On M/s
I
=
=

turbulent
Y flow
Re =

ND 4)(0.5)
=
= 9.67x10*
0.01
Dp =

<S
f1 =0.01 0.023X
=

1200x(95 2.04%
OX

moodypart 40.023 =
189117. 152 Pa
 109117P Q1
b) DP
HdZ
=

Re =
PUD T
-

qM80(v)(0.5) 9.67x10* =

0.02
490V =
199C
y =
3.94

&= AU = 0.190X 3.92

=
0.77m"/s
 pipe of diameter and at
through long galvanized
water at 15c flow a 200m steel 250 mm

m/s. Given Kinematic 1.14x10.5m%s and

average surface roughness
0.225
viscosity of water is

for galvanized steel =

0.15mm. Determine the loss of head due to friction.
hint:
I find U from a= AU
&= AU
②
③
Find Re
Find Relative roughness
ht=tg
0.225-16
a
I use equation
-

0.225

Re =

t v 0.011
=

=1000 (0.011) (0.25) 5

* = 2.41x10
5