Friction factor, f • Explicit form of Colebrook-White equation ε f = 0.0055[1 + 20000- + 106 1/3 ] D Re
Friction factor, f • Explicit form of Colebrook-White equation ε f = 0.0055[1 + 20000- + 106 1/3 ] D Re
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
A uniform pipeline, 5000 m long, 200 mm in
diameter and roughness size (ε) 0.03 mm, conveys water at
15 oC (ν = 1.13x10-6 m2/s) between two reservoirs, the
difference in water level is maintained at 50 m. In addition to
the entry loss of 0.5v2/2g and an exit loss k = 1.0. Determine the steady discharge between reservoirs using
a. Colebrook-White equation
b. Explicit function of f
![Friction factor, f
Explicit form of Colebrook-White equation
ƒ = 0.0055[1 + (20000 € + 100) ¹³1
1/3
f
D
]
Re](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ab1a6c1-f3e0-4699-9e77-e19b50268482%2F6e9f4cf5-7673-4552-a3bc-69748be88e5a%2Fnyw7s6_processed.png&w=3840&q=75)
Transcribed Image Text:Friction factor, f
Explicit form of Colebrook-White equation
ƒ = 0.0055[1 + (20000 € + 100) ¹³1
1/3
f
D
]
Re
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