1. 20. Pipe Flow 2
CH EN 374: Fluid Mechanics

2. Pipes, Pipes, Pipes Pipe flow is important in industrial processes
What we’ve learned so far:
Comparison with drag and porous flow
Friction factor in round, smooth pipes
Low Re
High Re
Over next few lectures we will learn about:
Non-circular ducts
Rough pipe material
Bends, breaks, fittings, etc.
Pipes in series and parallel
Pumps and turbines

3. Non-Circular Pipes

4. Noncircular Pipes 𝑅𝑒= 𝜌𝑣𝐷 𝜇 Δ𝑃=2𝜌𝑓 𝐿 𝐷 𝑣 2
Δ𝑃=2𝜌𝑓 𝐿 𝐷 𝑣 2
In non-circular pipes, what do we use for D?

5. 𝐷 𝐻 = 4 𝐴 𝑐 𝑝 Hydraulic Diameter 𝐴 𝐶 : cross-sectional area
𝐷 𝐻 = 4 𝐴 𝑐 𝑝
𝐴 𝐶 : cross-sectional area
𝑝: wetted-perimeter

6. Hydraulic Diameter
𝐴 𝐶 = 𝜋 4 𝐷 2 D
𝑝=𝜋𝐷
𝜋 4 𝐷 2 𝜋𝐷 =
𝐷 𝐻 = 4 𝐴 𝑐 𝑝 = 𝐷

7. Hydraulic Diameter
Hydraulic diameters for these and other shapes can be found on Table 2.1 in your book.

8. Hydraulic Diameter and Friction Factor
For turbulent flow, the equation for friction factor as a function of Re does not depend on 𝐷 𝐻
The turbulent velocity profile is largely independent of pipe shape
For laminar flow, 𝐷 𝐻 does affect friction factor.
Laminar flow dominated by large-scale flow features like pipe shape

9. Laminar Friction Factor
𝑓= 𝑐 𝑅𝑒
Geometry c
circle 16
slit 24
rectangle
𝛼 −6𝛼 𝑛=0 ∞ 𝑡𝑎𝑛ℎ λ 𝑛 𝛼 λ 𝑛 −1
λ 𝑛 = 𝑛 𝜋
𝛼= 𝑎 𝑏 ≤1
annulus
16 1−κ 2 𝑙𝑛κ 1− κ κ 2 𝑙𝑛κ
κ= 𝐷 1 𝐷 2 D1 D2 a b

10. Friction Factor for Laminar Flow in a Rectangular Ductb
𝜶 ( 𝒂 𝒃 )
c (where 𝒇=𝒄/𝑹𝒆) 1
14.23
1/2
15.55
1/4
17.09
1/5
18.23
1/6
19.70
1/8
20.58
𝒃 𝒈𝒐𝒆𝒔 𝒕𝒐 ∞ 24

11. Rough Pipes

12. Wall Roughness Wall roughness has characteristic length k.
For laminar flow, we can use the same 𝑓 as for a smooth pipe because 𝑘≪𝐷
But turbulent flow has small-scale features (eddies), that dominate

13. Dimensional Analysis 𝑓=𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛(𝑅𝑒,𝑘/𝐷)
For turbulent flow in a rough-walled pipe, 𝜏 𝑤 depends on 𝜌,𝜇, 𝑢, 𝐷, 𝑎𝑛𝑑 𝑘. (Remember, 𝑘 is a length.)
Find the Π’s governing this flow, then check with your neighbor. Did you get the same ones?
Π 1 = 2 𝜏 𝑤 𝜌 𝑢 2 =𝑓 Π 2 = 𝜌𝑢𝐷 𝜇 =𝑅𝑒 Π 3 = 𝑘 𝐷
𝑓=𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛(𝑅𝑒,𝑘/𝐷)

14. Colebrook-White: 1 𝑓 =−4𝑙𝑜𝑔 1.26 𝑅𝑒 𝑓 + 1 3.7 𝑘 𝐷

15. Effective Roughness (Typical)
Material
k (mm)
Glass and drawn metals*
0.0015
Commercial steel or wrought iron
0.046
Asphalt-Coated Cast Iron
0.12
Galvanized Iron
0.15
Cast Iron
0.26
Concrete
*includes copper, brass, lead…
Plastic is assumed smooth!

16. Example: Determining Flow Through a Pipe
A 12-cm-diameter, 800m-long cast iron pipe runs down a 8m hill. It is open to the atmosphere at both ends. At what rate will water (𝜌=1000 𝑘𝑔 𝑚 3 ,ν =1𝑥 10 −6 𝑚 2 𝑠 ) flow down the pipe?

17. Pipes and Pumping Power

18. Pumping Power 𝑊 𝑝𝑢𝑚𝑝 = 𝑉 Δ𝑃
The power needed by a pump is related to pressure drop.
In your own words, why does this make sense?
How would you go about deriving this relationship?
Power needed to move fluid through pipe is:
𝑊 𝑝𝑢𝑚𝑝 = 𝑉 Δ𝑃