Study in compressible flow in pressure

conduits, laminar and turbulent flow

Course Co-ordinator:
Dr. Anupam Chowdhury
Professor
Department of Civil Engineering, RUET
Email: anupam@ce.ruet.ac.bd
 Flow of Viscous Fluid through Circular Pipe: Derivation of
shear stress equation-
• At low velocity the fluid moves in layers.
Topic: Flow in Pressure Conduits

• Each layer of fluid slides over the adjacent layer.

• For Newtonian fluids the shear stress, τ, is directly proportional
to the velocity gradient, 𝑑𝑢/𝑑𝑦, developed in the fluid.
• The proportionality constant, μ, is called the “dynamic” or
𝑑𝑢
“absolute”, viscosity. So shear stress, τ = µ
𝑑𝑦
ρ𝑉𝐷
• Expression for Reynolds number is given by: 𝑅𝑒 =
µ
• Here, ρ = Density of fluid flowing through the pipe ; V = Average
velocity of fluid; D = diameter of pipe; µ = Viscosity of fluid
Topic: Flow in Pressure Conduits

 Consider a horizontal pipe of radius R.

 The viscous fluid is flowing from left to right in the pipe as shown
in figure.
 Consider a fluid element of radius r sliding in a cylindrical fluid
element of radius (r+ dr)
 Let, length of fluid element = ∆x
 If p is the intensity of pressure on the face AB,
𝜕𝑝
then the intensity of pressure on the face CD will be p+ ∆𝑥
𝜕𝑥
Forces acting on the fluid Element
Topic: Flow in Pressure Conduits

Location of Force Type and Magnitude

Face AB Pressure force = 𝑝 ∗ 𝜋𝑟 2
Face CD 𝜕𝑝
Pressure force = (p+ ∆𝑥) ∗ 𝜋𝑟 2
𝜕𝑥
Surface of fluid element ABCD Shear force = τ ∗ 2𝜋𝑟∆𝑥
Topic: Flow in Pressure Conduits
Sum of all forces in the direction of flow must be zero
Topic: Flow in Pressure Conduits
Shear stress and Velocity distribution: Derivation of velocity eqn-

To obtain velocity distribution across a section,

𝑑𝑢
the value of shear stress τ = µ …………….(1)
𝑑𝑦
𝜕𝑝 𝑟
is substituted in equation τ = − ………(2)
𝜕𝑥 2
 𝑑𝑢
But in the relation, τ = µ ,y is measured from pipe wall.
𝑑𝑦
Hence, y = R −r and dy = − dr
𝑑𝑢 𝑑𝑢 𝑑𝑢
Topic: Flow in Pressure Conduits

So, shear stress, τ = µ = µ = −µ

𝑑𝑦 −𝑑𝑟 𝑑𝑟
Topic: Flow in Pressure Conduits
 Ratio of maximum velocity to average velocity
Condition:
When r = 0, velocity becomes maximum;
Topic: Flow in Pressure Conduits

So, maximum velocity, 𝑈𝑚𝑎𝑥

1 𝜕𝑝 2
=− [𝑅 − 02 ]
4µ 𝜕𝑥
1 𝜕𝑝
=− ∗ 𝑅2
4µ 𝜕𝑥
The average velocity 𝑢ത is obtained by,

𝐷𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑎𝑐𝑟𝑜𝑠𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛

𝑢ത =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 (𝜋𝑅2 )
Discharge Q across the section is obtained by considering the flow
Topic: Flow in Pressure Conduits
Discharge Q:
Topic: Flow in Pressure Conduits
Topic: Flow in Pressure Conduits Drop of Pressure for a given length L of the pipe (Hagen Poiseuille Formula)
 Practice Problem#18 (Rajput= Page 544)
A crude oil of viscosity 9 poise and specific gravity 0.90 is flowing through
a horizontal circular pipe of 60 mm diameter. If the pressure drop in 100 m
Topic: Flow in Pressure Conduits

length of the pipe is 1800 kN/𝑚2 , determine:

32µഥ
𝑢𝐿
1. Rate of flow of oil (Q = Area* Velocity; Pressure drop, ∆p = )
𝐷2
2. Center-line velocity ( 𝑈𝑚𝑎𝑥 = 2ത
𝑢)
𝜕𝑝 𝑅
3. Total frictional drag over 100 m length ( F = τ0 ∗ 𝜋𝐷𝐿; τ0 = − )
𝜕𝑥 2
𝑑𝑢
4. Velocity gradient at the pipe wall (τ0 = µ )
𝑑𝑦
1 𝜕𝑝
5. Velocity and shear stress at 8 mm from the wall (u = − [𝑅2 − 𝑟2] )
4µ 𝜕𝑥
τ0 τ
*For similar triangle relationship, =
𝑅 𝑟
Topic: Flow in Pressure Conduits
Topic: Flow in Pressure Conduits
Topic: Flow in Pressure Conduits
 Practice Problem#19
A laminar flow is taking place in a pipe of diameter 200 mm. The
maximum velocity is 1.5 m/s. Find the mean velocity and radius at
Topic: Flow in Pressure Conduits

which this occurs. Also calculate the velocity at 4 cm from the wall
of the pipe.
Solution:
Given information,
D = 200 mm = 0.2 m ; R = 0.1 m = 10 cm
𝑈𝑚𝑎𝑥 = 1.5 𝑚/𝑠
y = 4 cm.
𝑈𝑚𝑎𝑥
We know, ഥ
=2;
𝑢
So, mean velocity = 0.75 m/s (answer)
 Solution
Velocity at any distance r from the center is given by,
𝑟 2
u = 𝑈𝑚𝑎𝑥 [1 − ( ) ]
Topic: Flow in Pressure Conduits

𝑅
𝑟 2
or, 0.75 = 1.5 × [1−( ) ]
0.1
or, r = 70.7 mm (answer)
Velocity at y = 4 cm from the pipe wall;
y=R−r
or, 0.04 = 0.1 − r
or, r = 0.06 m
Now, u = 𝑈𝑚𝑎𝑥 [1
𝑟 2
−( ) ]
Velocity at y = 4 cm
𝑅
6 2 or r = 6 cm from the
u = 1.5 ∗ [1 −( ) ]
10 pipe wall
u = 0.96 m/s (answer)
 Loss of Head due to friction in Viscous flow
Hagen Poiseuille formula:
32µഥ
𝑢𝐿
Topic: Flow in Pressure Conduits

Loss of pressure head, ℎ𝑓 =

ρ𝑔𝐷2

Darcy Weisbach formula:

4𝑓𝐿 𝑉 2
Loss of head due to friction, ℎ𝑓 =
𝐷 2𝑔
Velocity in pipe is always average velocity, that is, V = 𝑢ത
Equating, we get,
32µഥ
𝑢𝐿 ഥ2
4𝑓𝐿 𝑢
=
ρ𝑔𝐷2 𝐷 2𝑔
 Loss of Head due to friction in Viscous flow
Friction factor, f
Topic: Flow in Pressure Conduits

32µഥ
uL∗D∗2g
=
ρgD2 ∗4L∗ഥ
u2

16µ
= ഥρ𝐷
𝑢

16
= 𝑢ഥ ρ 𝐷
µ
16
=
𝑅𝑒
ഥρ𝐷
𝑢
[By definition, Reynolds Number, 𝑅𝑒 = ]
µ
 Practice Problem#20
Water is flowing through a 200 mm diameter pipe with co-
Topic: Flow in Pressure Conduits

efficient of friction f = 0.04 ; The shear stress at a point 40 mm

from the pipe axis is 0.00981 N/𝑐𝑚2 . Calculate shear stress at
the pipe wall.
Solution:

First, find whether the flow is viscous or not.

16
Friction factor, f =
𝑅𝑒
16 16
Reynolds number , 𝑅𝑒 = = = 400 (𝐿𝑎𝑚𝑖𝑛𝑎𝑟 𝑓𝑙𝑜𝑤)
𝑓 0.04
 Solution
Let, shear stress at pipe wall = τ0 τ0
Topic: Flow in Pressure Conduits

From similar triangle relationship,

0.00981 τ0
=
40 100

100 mm
τ = 0.00981

τ0 = 0.0245 N/𝑐𝑚2

40 mm
 Turbulent Flow
• In a pipe, turbulent flow occurs when Re > 4000.
• In a turbulent flow, fluid motion is irregular and chaotic and
Topic: Turbulent Flow

there is complete mixing of fluid due to collision of fluid

masses with one another.
• The fluid masses are interchanged between adjacent layers.
• As the fluid masses in adjacent layers have different
velocities, interchange of fluid masses between the adjacent
layers is accompanied by a transfer of momentum.
• The shear in turbulent flow is mainly due to momentum transfer.
Topic: Turbulent Flow
Transfer of Momentum in Turbulent flow
 Velocity distribution curve for Laminar and Turbulent flow
• The velocity distribution in turbulent flow is more uniform than in
laminar flow.
• In turbulent flow, the velocity gradients near the boundary wall
Topic: Turbulent Flow

shall be quite large resulting in more shear.

• In turbulent flow, the flatness of velocity distribution curve in the
core region away from the wall is because of mixing of fluid
layers and exchange of momentum between them.
• The velocity distribution which is paraboloid in laminar flow tends
to follow logarithmic variation in turbulent flow.
• Random orientation of fluid particles in a turbulent flow gives rise
to additional stresses, called Reynolds stresses.
Topic: Turbulent Flow
Velocity Distribution Curve
 Laminar-Turbulent
A laminar flow changes to turbulent flow when-
1. Velocity is increased.
2. Diameter of pipe is increased.
Topic: Turbulent Flow

3. Viscosity of fluid is decreased

Frictional resistance for turbulent flow is:
1. Proportional to 𝑉 𝑛 ; where n varies from 1.5 to 2.0.
2. Proportional to the density of fluid.
3. Proportional to the area of surface in contact.
4. Independent of pressure.
5. Dependent on the nature of surface in contact.
 Loss of Head due to Friction in Pipe flow (Darcy Equation)
Notation Parameter
p1 Intensity of Pressure at section 1−1
𝑉1 Velocity of flow at section 1−1
Topic: Turbulent Flow

𝑝2, 𝑉2 Intensity of pressure and Velocity of

flow at section 2−2 respectively.
L Length of the pipe between section
1−1 and 2−2
d Diameter of the pipe
𝑓′ Frictional resistance per unit wetted
area per unit velocity
ℎ𝑓 Loss of head due to friction
 Loss of Head due to Friction in Pipe flow (Darcy Equation)
Forces acting on the fluid between section 1−1 and 2−2;

1. 1. Pressure force at section 1 −1 =p1∗A

Topic: Turbulent Flow

2. 2. Pressure force at section 2 −2 = p2∗A

3. 3. Frictional resistance F1
Resolving all the forces in horizontal direction:
p1A − p2A − F1 = 0
A (p1 − p2) −F1 = 0
(p1− p2) =F1/A………………………. (1)
 ApplyingBernoulli'sequationatsection1−1and2−2;

Let, ℎ𝑓 = loss of head due to friction.

𝑝1 𝑉12 𝑝2 𝑉22
Topic: Turbulent Flow

+ + 𝑧1 = + + 𝑧2 + ℎ𝑓
ρ𝑔 2𝑔 ρ𝑔 2𝑔
z1 = z2 ; 𝑉1 = 𝑉2 ;
𝑝1 𝑝2
ℎ𝑓 = −
ρ𝑔 ρ𝑔
𝑝1 − 𝑝2 = ρ𝑔ℎ𝑓 ………………….. (2)
 Frictional Resistance, 𝐹1
Frictional resistance
= frictional resistance per unit wetted area per unit velocity∗ wetted area
∗ 𝑉2
Topic: Turbulent Flow

= 𝑓 ′ ∗ 𝜋𝑑𝐿 ∗ 𝑉 2
= 𝑓 ′ ∗ P∗ 𝐿 ∗ 𝑉 2 (Perimeter , P = 𝜋𝑑 )
From equation 1 and 2, we get,
F1
ρ𝑔ℎ𝑓 =
A
𝑓 ′ 𝑃𝐿𝑉 2
ρ𝑔ℎ𝑓 =
A
𝑓 ′ 𝑃𝐿𝑉 2
ℎ𝑓 =
Aρ𝑔
Darcy Weisbach Equation
Hydraulic mean depth,
𝑃 𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 4
R= = =
𝐴 𝐴𝑟𝑒𝑎 𝑑

𝑓′ 𝑓
Putting = ; 𝑤ℎ𝑒𝑟𝑒 𝑓 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛 𝑎𝑠 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑤𝑒 𝑔𝑒𝑡,
ρ 2

Loss of head due to friction in pipes flow,

4𝑓𝐿 𝑉 2
ℎ𝑓 =
𝑑 2𝑔
Expression for Co-efficient of Friction in terms of shear stress

Forces acting on the fluid between section 1−1 and 2−2;

1. Pressure force at section 1−1 = 𝑝1 ∗ 𝐴
2. Pressure force at section 2−2 = 𝑝2 ∗ 𝐴
3. Frictional resistance 𝐹1
Expression for Co-efficient of Friction in terms of shear stress

Forces acting on fluid between sections 1-1- & 2-2 is given by,
𝑝1 𝐴 − 𝑝2 𝐴 − 𝐹1 = 0
1
(𝑝1 −𝑝2 ) ∗ 𝜋𝑑 2 = 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 𝑑𝑢𝑒 𝑡𝑜 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠τ0
4
1
(𝑝1 −𝑝2 ) ∗ 𝜋𝑑 2 = shear stress ∗ surface area
4
1
(𝑝1 −𝑝2 ) ∗ 𝜋𝑑 2 = τ0 ∗ 𝜋dL
4
4τ0 ∗ 𝜋dL 4τ0 ∗ L
(𝑝1 −𝑝2 ) = 2 = …………………………. (1)
𝜋𝑑 𝑑
 Expression for Co-efficient of Friction in terms of shear stress
From Darcy-Weisbach equation,
4𝑓𝐿 𝑉 2
ℎ𝑓 =
Topic: Turbulent Flow

𝑑 2𝑔
𝑝1 − 𝑝2 4𝑓𝐿 𝑉 2
=
ρ𝑔 𝑑 2𝑔
4𝑓𝐿 𝑉 2
𝑝1 − 𝑝2 = ∗ ρ𝑔 ………………….(2)
𝑑 2𝑔

Equating, we get,
2τ0
Friction factor , f =
ρ𝑉 2
 Velocity distribution in turbulent flow:
Prandtl’s Universal Distribution Equation,
𝑢𝑚𝑎𝑥 − 𝑢 𝑅
Topic: Turbulent Flow

= 2.5 ln[ ]
𝑢𝑓 𝑦
Here,
𝑢𝑚𝑎𝑥 = Center line velocity
u = local velocity at distance y
τ0
𝑢𝑓 = Shear friction velocity =
ρ
 Practice Problem#21 (Bansal= 445)
Determine the wall shearing stress in a pipe of diameter 100
mm which carries water. The velocities at the pipe center and 30
mm from pipe center are 2 m/s and 1.5 m/s respectively. The
Topic: Turbulent Flow

flow in pipe is given as turbulent.

Solution:
𝑢𝑚𝑎𝑥 −𝑢 𝑅
Formula: = 2.5 ln[ ]
𝑢𝑓 𝑦
R = 50 mm ; r = 30 mm.
y = R−r
y = 20 mm
Velocity at pipe center, 𝑢𝑚𝑎𝑥 = 2 𝑚/𝑠
Velocity at distance y = 20 mm from pipe wall, u = 1.5 m/s.
 Solution
Applying the formula:
Topic: Turbulent Flow

Shear friction velocity,

𝑢𝑓 = 0.2185 𝑚/𝑠
τ0
If τ0 𝑖𝑠 𝑡ℎ𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑡 𝑝𝑖𝑝𝑒 𝑤𝑎𝑙𝑙, 𝑢𝑓 =
ρ

τ0 = 𝑢𝑓2 ∗ ρ = 0.2185 ∗ 0.2185 ∗ 1000 = 47.676 𝑁/𝑚2