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Fluid Mechanics 08 Classifying Flow We can classify the flow by calculate Reynolds No When The Reynolds No can be calculated from one equation of the following Developing and Fully Developed Flow The distance required for flow to develop is called the entrance length (Le). For flow entering a circular pipe from a reservoir the Le can be calculated from the following equation:- Summary In summary, flow is classified into four categories:laminar developing, laminar fully developed turbulent developing turbulent fully developed. The key to classification is to calculate the Reynolds number Example Consider fluid flowing in a round tube of length 1 m and diameter 5 mm. Classify the flow as laminar or turbulent and calculate the entrance length for (a) air (50°C) with a speed of 12 m/s and (b) Water (15°C) with a mass flow rate of 8 g/s . Solution 1. Air (50°C), ν = 1.79 × 10-5 m2/s. 2. Water (15°C), μ = 1.14 × 10-3 N*s/m2. ∗ 12 ∗ 0.005 = = = 3350 −5 1.79 ∗ 10 The flow is turbulent for air = 50 ∗ = 50 ∗ 0.005 = 0.25 4 ∗ ʹ 4 ∗ 0.008 = = Π ∗ ∗ µ 3.14 ∗ 0.005 ∗ 1.14 ∗ 10−3 = 1787 The flow is laminar for water = 0.05 ∗ Re ∗ D = 0.05 ∗ 1787 ∗ 0.005 = 0.447m Specifying Pipe Sizes Pipe Head Loss Component head loss is associated with flow through devices such as valves, bends, and tees. Pipe head loss is associated with fully developed flow, and it is caused by shear stresses that act on the flowing fluid. Note that pipe head loss is sometimes called major head loss, and component head loss is sometimes called minor head loss. Darcy-Weisbach Equation To use Darcy-Weisbach Equation the flow should be fully developed and steady. The DarcyWeisbach equation is used for either laminar flow or turbulent flow and for either round pipes or nonround conduits such as a rectangular duct. Friction Factor For Laminar flow :For turbulent flow 64 = 0.25 = 5.74 2 (log10 ( + 0.9 )) 3.7 ∗ Or from Moody Chart Moody Chart Roughness Examples Oil (S = 0.85) with a kinematic viscosity of 6 × 10-4 m2/s flows in a 15 cm pipe at a rate of 0.020 m3/s. What is the head loss per 100 m length of pipe? Solution 0.02 ∗ 4 = = = 1.13/ 2 3.14 ∗ 0.15 ∗ 1.13 ∗ 0.15 = = = 283 −4 6 ∗ 10 64 64 = = = 0.226 283 2 ℎ = ∗ ∗ 2∗ 100 1.132 = 0.226 ∗ ∗ = 9.83 0.02 2 ∗ 9.81 Example Water (T = 20°C) flows at a rate of 0.05 m3/s in a 20 cm asphalted cast-iron pipe. What is the head loss per kilometer of pipe? Solution 0.05 ∗ 4 1.59 = = = 2 3.14 ∗ 0.2 ∗ 1.59 ∗ 0.2 = = = 318471 −6 1 ∗ 10 The flow is turbulent From table e=0.12mm=0.00012m 0.00012 = = 0.0006 0.2 From Moody Chart = 0.019 = 0.25 = = 5.74 2 (log10 ( + 0.9 )) 3.7 ∗ 0.25 0.0006 5.74 (log10 +( ))^2 0.9 3.7 318471 0.25 2 = 0.0188 == 0.019 (log10 0.000226) 2 ℎ = ∗ ( ) ∗ ( ) 2∗ =12.2m =0.019 ∗ 1000 0.2 ∗ 1.592 ( ) 2∗9.81