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# 741 different types of minor losses minor losses are

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7.4.1 Different Types of Minor Losses Minor losses are losses due to the inclusion of a pipe fittings in a pipeline. Some examples are entrances or exit of a pipe expansions or contractions of a pipe bends, elbow and tees valves of open or partially closed gradual expansions or contractions Minor losses is given by h L = K* v g 2 2 (7.17) where K is a constant

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-17 Component K a. Elbows Regular 90 ° , flanged 0.3 Regular 90 ° , threaded 1.5 Long radius 90 ° , flanged 0.2 Long radius 90 ° , threaded 0.7 Long radius 45 ° , flanged 0.2 Regular 45 ° , threaded 0.4 b. 180 ° return bends 180 ° return bends, flanged 0.2 180 ° return bends, threaded 1.5 c. Tees Line flow, flanged 0.2 Line flow, threaded 0.9 Branch flow, flanged 1.0 Branch flow, threaded 2.0 d. Valves Globe, fully open 10 Gate, fully open 0.15 Ball valve, fully open 0.05 e. Others Entrance loss 0.5 Exit loss 1.0 7.4.2 Modified Bernoulli’s Equation The original Bernoulli’s equation should be extended to include the friction loss and minor losses. i.e. total energy at 1 = total energy at 2 + energy loss on the way 2 2 2 2 1 2 1 1 z + g 2 v + p = z + g 2 v + p γ γ + Σ f L d v g i i i 2 2 2 + Σ K v g i 2 2 2 (7.18)
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-18 Worked examples: 1. Find the discharge through the pipe in the figure below. The minor loss coefficient for entrance is 0.5. The pipe diameter is 15 mm and the pipe roughness produces a friction factor of 0.025. 1 2 150m 15m Answer Applying Bernoulli’s equation between pt.1 and 2 2 2 2 2 1 2 1 1 z + g 2 v + p = z + g 2 v + p γ γ + fL d v g 2 2 2 + K v g 2 2 2 15 = (1 + K + fL d )* v g 2 2 2 = (1+0.5+0.025*150/0.015) * v g 2 2 2 = 251.5 v g 2 2 2 or v 2 = 1.082 m/s Hence Q = A 2 *v 2 = 1.082* π *0.015 2 /4 m 3 /s = 0.1912 L/s

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Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-19 2. Find the discharge through the pipe in the figure below for H = 20 m. The minor loss coefficients for entrance, elbows and globe valve are 0.5, 0.8 (each) and 10 respectively. The kinematic viscosity of water is 1.02*10 -6 m 2 /s. 40m 30m 40m 20m elbows globe valve dia. = 150mm ε = 0.0003 1 2 Answer Using 2 2 2 2 1 2 1 1 z + g 2 v + p = z + g 2 v + p γ γ + Σ f L d v g i i i 2 2 2 + Σ K v g i 2 2 2 Σ f L d v g i i i 2 2 2 = f d v g L i 2 2 2 Σ = f v 015 2 981 30 20 40 2 2 . * . ( ) + + = 30.58 f v 2 2 Σ K v g i 2 2 2 = v g K i 2 2 2 Σ = v 2 2 2 9 81 05 2 08 10 * . ( . * . ) + + = 0.617 v 2 2 As P 1 -P 2 = 0, v 1 = 0, z 1 -z 2 = 20 m, i.e. 20 = v g 2 2 2 + Σ f L d v g i i i 2 2 2 + Σ K v g i 2 2 2 = (0.051+30.58f + 0.617)v 2 2 = (0.668+30.58f) v 2 2
Fluid Mechanics Chapter 7 – Steady Flow in Pipes P.7-20 or v 2 = 547 1 4578 . . + f Since Re = 1.47*10 5 v 2 ; ε /D = 0.002 f v 2 (m/s) Re f cal. 0.030 3.551 5.2*10 5 0.0230 0.0230 3.818 5.6*10 5 0.0235 0.0235 3.797 5.6*10 5 0.0235 (ok) v 2 = 3.797 m/s Since Q = A 2 *v 2 = π *0.15 2 /4 * 3.797 m 3 /s = 0.067 m 3 /s

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