Get to know about Pipes working and Pressure loss and head loss also Laminar Flow in noncircular pipers and Moodies Chart Inclines pipes study

1. Fluid Mechanics II
By
Dr. Jawad Sarwar
Assistant Professor
Department of Mechanical Engineering,
University of Engineering & Technology Lahore, Pakistan

2. Laminar flow in pipes
1
Consider a ring shaped differential volume element of
radius , r, thickness dr, and length dx oriented
coaxially with the pipe. A force balance on the volume
element in the flow direction gives:
Which indicates that in a fully developed flow in a
horizontal pipe, the viscous and pressure forces
balance each other.
τ = 𝜇 Τdu dy
𝑉𝑎𝑣𝑔 = −
𝑅2
8𝜇
𝑑𝑃
𝑑𝑥
𝑢 𝑟 = 2𝑉𝑎𝑣𝑔 1 −
𝑟2
𝑅2
2𝜋𝑟𝑑𝑟𝑃 𝑥 − 2𝜋𝑟𝑑𝑟𝑃 𝑥+𝑑𝑥 + 2𝜋𝑟𝑑𝑥𝜏 𝑟 − 2𝜋𝑟𝑑𝑥𝜏 𝑟+𝑑𝑟 = 0
𝑑𝑃
𝑑𝑥
= −
2𝜏 𝑤
𝑅
𝑢 𝑚𝑎𝑥 = 2𝑉𝑎𝑣𝑔

3. Pressure drop and Head Loss
2
𝑑𝑃
𝑑𝑥
=
𝑃2 − 𝑃1
𝐿
=
∆𝑃
𝐿
𝑉𝑎𝑣𝑔 = −
𝑅2
8𝜇
𝑑𝑃
𝑑𝑥
∆𝑃 = 𝑃1 − 𝑃2 =
8𝜇𝐿𝑉𝑎𝑣𝑔
𝑅2
=
32𝜇𝐿𝑉𝑎𝑣𝑔
𝐷2
Laminar flow
∆𝑃𝐿= 𝑓
𝐿
𝐷
𝜌𝑉𝑎𝑣𝑔
2
2
Darcy friction factor
Or
Darcy-Weisbach friction factor
Dynamic pressure
𝑓 =
64𝜇
𝜌𝑉𝑎𝑣𝑔 𝐷
=
64
Re
For fully developed laminar flow in a
circular pipe

4. 3
Pressure drop and Head Loss
ℎ 𝐿 =
∆𝑃
𝜌𝑔
= 𝑓
𝐿
𝐷
𝑉𝑎𝑣𝑔
2
2𝑔
Head Loss:
Note: Head loss represents the additional height that the fluid needs to be
raised by a pump in order to overcome the frictional losses in the pipe.
∆𝑃 =
32𝜇𝐿𝑉𝑎𝑣𝑔
𝐷2
For horizontal pipe
𝑉𝑎𝑣𝑔 =
∆𝑃𝐷2
32𝜇𝐿
ሶ𝑊𝑝𝑢𝑚𝑝,𝐿 = ሶ𝑉∆𝑃𝐿= ሶ𝑉𝜌𝑔ℎ 𝐿 = ሶ𝑚𝑔ℎ 𝐿Pumping power:
ሶ𝑉 =
∆𝑃𝐷2
32𝜇𝐿
𝜋𝐷2
4
=
∆𝑃𝜋𝐷4
128𝜇𝐿
ሶ𝑉 = 𝑉𝑎𝑣𝑔 𝐴
Poiseuille’s Law
Volume Flow rate, pressure drop and thus the required pumping power
is proportional to:
1. Length of the pipe
2. Viscosity of the fluid

5. Inclined Pipes
4
𝑉𝑎𝑣𝑔 =
∆𝑃 − 𝜌𝑔𝐿 sin 𝜃 𝐷2
32𝜇𝐿
ሶ𝑉 =
∆𝑃 − 𝜌𝑔𝐿 sin 𝜃 𝜋𝐷4
128𝜇𝐿 Uphill flow
𝜽 > 𝟎, 𝐬𝐢𝐧 𝜽 > 𝟎
Downhill flow
𝜽 < 𝟎, 𝐬𝐢𝐧 𝜽 < 𝟎
Additional Readings:
Examples 8-1 and 8-2
Included for class tasks

6. Laminar Flow in noncircular pipes
5

7. 6
Turbulent flow in pipes
Turbulent flow is characterized by random and rapid fluctuations of swirling
regions of fluid, called eddies, throughout the flow.
Turbulent flow along a wall consist of four regions:
1) Viscous sublayer
2) Buffer layer
3) Overlap or transition layer (inertial sublayer)
4) Outer or turbulent layer
Viscous dominated effects
Inertial dominated effects

8. 7
The Moody Chart
1
𝑓
= −2.0 𝑙𝑜𝑔
Τ𝜀 𝐷
3.7
+
2.51
𝑅𝑒 𝑓
Colebrook equation (Implicit equation)
Friction factor for turbulent flow
Consider a following parameters for a turbulent flow
𝜀
𝐷
= 0.02
Re = 107
1. Find friction factor using Colebrook equation:
Surface roughness
Diameter
1
𝑓
≅ −1.8 log
6.9
𝑅𝑒
+
Τ𝜀 𝐷
3.7
1.11
Explicit equation by E. Haaland
2. Find friction factor using explicit equation:
3. Find friction factor using Moody’s chart:
Additional Readings:
Example 8-3 included for class tasks

9. 8

10. 9

11. 10

12. 11

13. 12

14. 13

15. 14

16. 15

17. 16