Laminar Flow

1. LAMINAR FLOW

2. Contents
• Flow of viscous fluid in circular pipes- Hagen
Poiseuile Law
• Flow of viscous fluid between two parallel
plates

3. Hagen Poiseuille Law
• Fig shows a horizontal circular pipe of radius 𝑅,
having laminar flow of fluid through it.
• Consider a small concentric cylinder of radius 𝑟
and length 𝑑𝑥 as a free body.

4. 1) Shear stress distribution
2) Velocity profile distribution
3) Relation between average and maximum velocity
4) Pressure drop difference for a length of pipe

5. 1) Shear stress distribution:
If 𝜏 is the shear stress, the shear force F is given by:
𝐹 = τ 𝑥 2π𝑟 𝑥 𝑑𝑥
Let 𝑝 be the intensity of pressure at left end and
the intensity of pressure at the right end be
𝑝 +
𝜕𝑝
𝜕𝑥
. 𝑑𝑥
Thus the forces acting on the fluid element are:
1) The shear force, τ 𝑥 2π𝑟 𝑥 𝑑𝑥 on the surface of
fluid element
2) The pressure force, 𝑝 𝑥 π𝑟2
on the left end.
3) The pressure force, 𝑝 +
𝜕𝑝
𝜕𝑥
. 𝑑𝑥 π𝑟2
on the right
end

6. For steady flow, the net force on the cylinder must
be zero.
∴ 𝑝 𝑥 π𝑟2
− 𝑝 +
𝜕𝑝
𝜕𝑥
. 𝑑𝑥 π𝑟2
− τ 𝑥 2π𝑟 𝑥 𝑑𝑥 = 0
𝑝 𝑥 π𝑟2 − 𝑝 𝑥 π𝑟2 −
𝜕𝑝
𝜕𝑥
. 𝑑𝑥. π𝑟2 − τ 𝑥 2π𝑟 𝑥 𝑑𝑥 = 0
−
𝜕𝑝
𝜕𝑥
. 𝑑𝑥. π𝑟2
− τ 𝑥 2π𝑟 𝑥 𝑑𝑥 = 0
−
𝜕𝑝
𝜕𝑥
. 𝑑𝑥. π𝑟2
= τ 𝑥 2π𝑟 𝑥 𝑑𝑥
τ = −
𝜕𝑝
𝜕𝑥
𝑟
2
………………(1)
Eqn (1) represents shear stress distribution

7. τ = −
𝜕𝑝
𝜕𝑥
𝑟
2

8. 2) Velocity distribution:
From Newton’s law of viscosity,
τ = μ
𝑑𝑢
𝑑𝑦
… … … … . . 𝑎
• In this eqn, the distance 𝑦 is measured from the
boundary.
• The radial distance 𝑟 is related to distance 𝑦 by the
relation:
𝑦 = 𝑅 − 𝑟 𝑜𝑟 𝑑𝑦 = −𝑑𝑟
• The eqn (a) becomes τ = − μ
𝑑𝑢
𝑑𝑟
……………(2)
• Comparing two values of 𝜏 from eqn (1) & (2) we
have:
− μ
𝑑𝑢
𝑑𝑟
= −
𝜕𝑝
𝜕𝑥
𝑟
2

9. − μ
𝑑𝑢
𝑑𝑟
= −
𝜕𝑝
𝜕𝑥
𝑟
2
𝑑𝑢 =
1
2μ
𝜕𝑝
𝜕𝑥
𝑟. 𝑑𝑟
Integrating the above eqn w.r.t ‘r’, we get:
𝑢 =
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑟2
+ 𝐶 … … … … 3
Where 𝐶 is the constant of integration and its value
is obtained from the boundary condition:
At, 𝑟 = 𝑅, 𝑢 = 0
∴ 0 =
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑅2
+ 𝐶
𝐶 = −
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑅2
… (b)

10. Substituting this value of C in eqn (3) we get:
𝑢 =
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑟2
+ 𝐶….(3)
𝐶 = −
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑅2
……(b)
𝑢 =
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑟2
−
1
4μ
.
𝜕𝑝
𝜕𝑥
.𝑅2
𝑢 = −
1
4μ
.
𝜕𝑝
𝜕𝑥
𝑅2
− 𝑟2
………………(4)
The eqn (4) represents velocity distribution

11. 3) Relation b/n average velocity 𝑢 and maximum
velocity 𝑢𝑚𝑎𝑥 :
The maximum velocity occurs at the centre and is
given by 𝑢𝑚𝑎𝑥 = −
1
4μ
.
𝜕𝑝
𝜕𝑥
𝑅2
…………(5) (𝑟 = 0)
From eqn (4) & (5) we have,
𝑢 = 𝑢𝑚𝑎𝑥 1 −
𝑟
𝑅
2
… … (6)

12. • The discharge through an elementary ring of
thickness 𝑑𝑟 at radial distances 𝑟 is given by :
𝑑𝑄 = 𝑢 𝑥 2π𝑟 𝑥 𝑑𝑟
𝑑𝑄 = 𝑢𝑚𝑎𝑥 1 −
𝑟
𝑅
2
2𝜋𝑟 𝑥 𝑑𝑟
Total discharge, 𝑄 = 𝑑𝑄
𝑄 =
0
𝑅
𝑢𝑚𝑎𝑥 1 −
𝑟
𝑅
2
2𝜋𝑟 𝑥 𝑑𝑟
𝑄 = 2𝜋𝑢𝑚𝑎𝑥
0
𝑅
𝑟 −
𝑟3
𝑅2
𝑑𝑟
𝑄 = 2𝜋𝑢𝑚𝑎𝑥
𝑟2
2
−
𝑟4
4𝑅2
𝑅
0

13. 𝑄 = 2𝜋𝑢𝑚𝑎𝑥
𝑅2
2
−
𝑅4
4𝑅2
𝑄 =
π
2
𝑢𝑚𝑎𝑥 𝑅2
Average velocity of flow,
𝑢 =
𝑄
𝐴
=
π
2
𝑢𝑚𝑎𝑥 𝑅2
π𝑅2 =
𝑢𝑚𝑎𝑥
2
…………………. (7)
Eqn (7) shows that the avg velocity is one-half the
max velocity.
𝑢 =
𝑢𝑚𝑎𝑥
2
Relation b/n average and maximum velocity

14. 4) Pressure drop in a pipe for a length
Substituting the value of 𝑢𝑚𝑎𝑥 𝑓𝑟𝑜𝑚 𝑒𝑞𝑛 5 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑢𝑚𝑎𝑥 = −
1
4μ
.
𝜕𝑝
𝜕𝑥
𝑅2
….(5)
Eqn (7) 𝑢 =
𝑢𝑚𝑎𝑥
2
substituting (5) in (7) we get
𝑢 = −
1
8μ
.
𝜕𝑝
𝜕𝑥
𝑅2
or −𝜕𝑝 =
8μ𝑢
𝑅2 . 𝜕𝑥
The pressure diff b/n two sections 1 & 2 at distance
𝑥1𝑎𝑛𝑑 𝑥2 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦:

15. −
𝑝1
𝑝2
𝜕𝑝 =
8μ𝑢
𝑅2
𝑥1
𝑥2
𝜕𝑥
𝑝1 − 𝑝2 =
8μ𝑢
𝑅2 𝑥1 − 𝑥2
𝑝1 − 𝑝2 =
8μ𝑢𝐿
𝑅2
𝑝1 − 𝑝2 =
32μ𝑢𝐿
𝐷2 …………..(8)
Eqn (8) is known as Hagen-Poiseuille Equation