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1. A constriction meter in a 2-inch diameter pipeline consists of a short section of 1.2-inch diameter pipe (as shown below).
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Answer #1

Ans a) Pressure head = P / \rhog

Velocity head = V2 /2g

Datum head = z

Since, all points are at equal elevation , datum head ,z = 0 we know EGL = Pressure head + Datum head & Velocity head HGL - Pressive head & Datum head Bessure at point 1, Pi= 1 upsi or 2

- V = 0.19 8.716 f418. 0.0218 for small chameter pipe, A= (1-2)² = 1.13 in?.. A=1-13in² of 0.00785 ft ². - v = 0.19 - 24.2 ft

I Loss coefficient k = 0.357 = 32.41+1-18= P2 & 9.09 +3-18. 38 = P2 = 21.32 ftch 88 ? V22 = 24. 2² 6 9.09 ft ag 2x32.2 S. Sex0.35+ 3.2 =3.55 Ex=0.35+ > 32.4+1-18 = pa + 4.18.41.18 = $z= 28.2347 12 = (8.716) 2 = 1.18 t. ag 2x32.2 o EGL level = 29.41 f

Ans (c) The EGL and HGL are curved due to sudden change in pressure and velocity head due to expansion and contraction of pipe in which fluid is flowing .

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