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p01106 - μ is μ = ρν = p 998 kg m 3 Q p 1 00 10 − 6...

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1.106. CHAPTER 1, PROBLEM 106 127 1.106 Chapter 1, Problem 106 Problem: Water is flowing very slowly through a drain pipe of diameter D = 5 cm and length L = 120 m. The pressure difference between inlet and outlet is ( p 1 p 2 ) = 0.04 kPa, and the temperature is 20 o C. Compute the maximum velocity and the Reynolds number of the water flowing in the pipe. Solution: From the laminar pipe-flow solution, the maximum velocity, u m , and Reynolds number, Re , are given by u m = ( p 1 p 2 ) 4 μL R 2 and Re = ρ u m R μ where R is pipe radius, μ is fluid viscosity and ρ is fluid density. For this problem, we are given ( p 1 p 2 ) = 0.04 kPa, R = 2.5 cm and L = 120 m. Also, since the fluid is water at 20 o C, density is ρ = 998 kg/m 3 and kinematic viscosity is ν = 1.00 · 10 6 m 2 /sec. Therefore, the molecular viscosity,
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Unformatted text preview: μ , is μ = ρν = p 998 kg / m 3 Q p 1 . 00 · 10 − 6 m 2 / sec Q = 9 . 98 · 10 − 4 kg / (m · sec) To simplify the calculations, it is worthwhile to express the pressure difference in terms of the basic units, viz., 1 Pa = 1 N/m 2 = 1 (kg · m/sec 2 )/m 2 = 1 kg/(m · sec 2 ). Thus, the pressure difference and the radius are p 1 − p 2 = 40 kg / p m · sec 2 Q and R = 0 . 025 m Therefore, the maximum velocity is u m = [40 kg / (m · sec 2 )] (0 . 025 m) 2 4 [9 . 98 · 10 − 4 kg / (m · sec)] (120 m) = 0 . 052 m / sec and the Reynolds number is Re = p 998 kg / m 3 Q (0 . 052 m / sec) (0 . 025 m) 9 . 98 · 10 − 4 kg / (m · sec) = 1300...
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