To determine Draw the energy grade line and hydraulic grade line. Explanation Given: Diameter of the pipe ( D ) is 2 in. Pressure at A ( p A ) is 4.5 psi. Discharge ( Q ) is 120 gpm . Calculation: Determine the discharge. Q = ( 120 gal min ) ( 1 min 60 s ) ( 1 ft 3 7.48 gal ) = 0.2674 ft 3 / s Determine the velocity. V = Q A = 0.2674 ft 3 / s π ( 1 in . ( 1 ft 12 in . ) ) 2 = 12.26 ft / s Determine the Reynolds number. Re = V D v w = ( 12.26 ft / s ) ( 2 in . ( 1 ft 12 in . ) ) 14.1 × 10 − 6 ft 2 / s = 1.45 × 10 5 Determine the relative roughness. ε D D = 0.00015 ft ( 2 in . ( 1 ft 12 in . ) ) = 0.0009 Determine the major head loss. ( h L ) major = f L D V 2 2 g = 0.0212 ( 34 ft ( 2 in . ( 1 ft 12 in . ) ) ) [ ( 12.26 ft / s ) 2 2 ( 32.2 ft / s 2 ) ] = 10.087 ft Determine the minor head loss. ( h L ) minor = ∑ K L V 2 2 g = ( 2 ( 0.9 ) + 0.19 ) [ ( 12.26 ft / s ) 2 2 ( 32.2 ft / s 2 ) ] = 4.641 ft Apply energy equation between A and E . p A γ w + V A 2 2 g + z A + h pump = p E γ + V E 2 2 g + z E + h turb + h L { ( 4.50 psi ) ( 12 in . 1 ft ) 2 ( 1.94 slug / ft 3 ) ( 32.2 ft / s 2 ) + V 2 2 g + 0 + h pump = 0 + V 2 2 g + 20 ft + 0 + ( 10.087 ft + 4.641 ft ) } h pump = 24.355 ft Determine the power supplied by the pump. W ˙ O = γ w Q h pump = ( 1.94 slug / ft 3 ) ( 32.2 ft / s 2 ) ( 0.2674 ft 3 / s ) ( 24.355 ft ) = 406.79 ft ⋅ lb / s ( 1 hp 550 ft ⋅ lb / s ) = 0.74 hp Determine the velocity head. V 2 2 g = ( 12

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A fluid is a substance that is continuously deformed by the application of shear stress. The rate of deformation of a fluid depends on its viscosity. The fluid tries to flow when it interacts with some other system.

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Chapter 10, Problem 64P

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