Re= pvD AP = μ 2 fpv L D S=ε/D fm = 4f Vertical Lift (V.L.) Concept Example: A liquid flows in a pipe, calculate friction pressure loss when the fluid's flow rate = 0.02 m3/s; density = 850 kg/m3; viscosity = 0.1 poise; pipe inner ID = 10 cm; pipe roughness = 0.1 mm; pipe length = 1000 m.
Re= pvD AP = μ 2 fpv L D S=ε/D fm = 4f Vertical Lift (V.L.) Concept Example: A liquid flows in a pipe, calculate friction pressure loss when the fluid's flow rate = 0.02 m3/s; density = 850 kg/m3; viscosity = 0.1 poise; pipe inner ID = 10 cm; pipe roughness = 0.1 mm; pipe length = 1000 m.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![Re=
pvD
AP
=
μ
2 fpv L
D
S=ε/D
fm = 4f](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3041687-8ecc-4d3e-a323-3d11a576ec95%2F41e2bf55-5f7c-4a67-9244-79dc2cdddf8b%2Fnrxyly_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Re=
pvD
AP
=
μ
2 fpv L
D
S=ε/D
fm = 4f
![Vertical Lift (V.L.) Concept
Example: A liquid flows in a pipe, calculate friction pressure loss when the fluid's
flow rate = 0.02 m3/s; density = 850 kg/m3; viscosity = 0.1 poise; pipe inner ID = 10
cm; pipe roughness = 0.1 mm; pipe length = 1000 m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3041687-8ecc-4d3e-a323-3d11a576ec95%2F41e2bf55-5f7c-4a67-9244-79dc2cdddf8b%2Fdyfdaf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Vertical Lift (V.L.) Concept
Example: A liquid flows in a pipe, calculate friction pressure loss when the fluid's
flow rate = 0.02 m3/s; density = 850 kg/m3; viscosity = 0.1 poise; pipe inner ID = 10
cm; pipe roughness = 0.1 mm; pipe length = 1000 m.
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