p0734

# p0734 - 046 mm 75 mm = 0 0006 Using the Colebrook formula...

This preview shows page 1. Sign up to view the full content.

7.34. CHAPTER 7, PROBLEM 34 991 7.34 Chapter 7, Problem 34 Problem: The head loss for a steel pipe of length L =2kmis h L = 96.68 m. Water with kinematic viscosity ν =10 6 m 2 /sec flows through the pipe, whose diameter, D , is 75 mm. What is the average flow velocity, u ? HINT: Compute Re D f and use the Colebrook formula. Solution: First, note that by definition h L = f u 2 2 g L D and Re D = uD ν Combining these equations yields Re D ± f = uD ν ² 2 gDh L u 2 L = D 3 / 2 ν ² 2 gh L L For the given values Re D ± f = (0 . 075 m) 1 . 5 10 6 m 2 / sec > ³ ³ : 2 p 9 . 807 m / sec 2 Q (96 . 68 m) 2000 m =2 . 0 · 10 4 Now, for steel, k s = 0.046 mm. So, k s D = 0 .
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 046 mm 75 mm = 0 . 0006 Using the Colebrook formula, there follows 1 √ f = − 2 log 10 X k s /D 3 . 7 + 2 . 51 Re D √ f ~ = − 2 log 10 w . 0006 3 . 7 + 2 . 51 2 · 10 4 W = 7 . 0822 Thus, the friction factor is f = 0 . 00199 Now, the head loss is h L = f u 2 2 g L D = ⇒ u = ² 2 gh L D fL Therefore, the average flow velocity is u = > ³ ³ : 2 p 9 . 807 m / sec 2 Q (96 . 68 m)(0 . 075 m) (0 . 0199)(2000 m) = 1 . 89 m sec...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online