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Water (specific gravity, G, is 1.0 and the kinematic viscosity, ν, is 1.14 *10-6 m2 /s) flows through a piping system with a flow rate of 0.9 m3/s. The pipe is 2800 m long and 600 mm inside diameter. The pipe has three gate valves, two globe valves, one swing check valve, three standard tees through-branch, and four 90o standard elbows. Calculate the following: c- The Darcy friction factor if the pipe is fully rough and has an absolute internal roughness, e, of 0.01 mm. d- The frictional head loss in units of (m) using the Darcy equation.

Water (specific gravity, G, is 1.0 and the kinematic viscosity, ν, is 1.14 *10-6 m2 /s) flows through a piping system with a flow rate of 0.9 m3/s. The pipe is 2800 m long and 600 mm inside diameter. The pipe has three gate valves, two globe valves, one swing check valve, three standard tees through-branch, and four 90o standard elbows. Calculate the following: c- The Darcy friction factor if the pipe is fully rough and has an absolute internal roughness, e, of 0.01 mm. d- The frictional head loss in units of (m) using the Darcy equation.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Water (specific gravity, G, is 1.0 and the kinematic viscosity, ν, is 1.14 *10-6 m2 /s) flows through a piping system with a flow rate of 0.9 m3/s. The pipe is 2800 m long and 600 mm inside diameter. The pipe has three gate valves, two globe valves, one swing check valve, three standard tees
through-branch, and four 90o standard elbows. Calculate the following:

c- The Darcy friction factor if the pipe is fully rough and has an absolute internal roughness, e, of 0.01 mm.
d- The frictional head loss in units of (m) using the Darcy equation.

Use the following equations if you need to, 64 f = Re 2.51 -2 log Re f 1 1 2.51 e = -2 log 3.7D 1 -2 log 3.7D, Ref e
Transcribed Image Text:Use the following equations if you need to, 64 f = Re 2.51 -2 log Re f 1 1 2.51 e = -2 log 3.7D 1 -2 log 3.7D, Ref e
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