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Unformatted text preview: ME 309 Fall 2008 Section 2 Homework # 34 Due Monday 24 November 2008 Problem 341 Consider a conical diffuser with inlet and outlet diameters of 20 and 40 mm respectively and a length of 0.1 m . The diffuser is fed by a pipe containing fully developed pipe flow. The average velocity in the pipe is 5 m/s . The working fluid is water at a density of 997 gg/m 3 and a viscosity of 8.9 e4 kg/ms . The pressure at the inlet section is 50,000 Pa . Gravity effects are negligible over this short length. Calculate: The ideal pressure recovery, C p_ideal ; The actual pressure recovery C p_actual ;(from Fig. 8.16); The ideal and actual static pressure at Station 2; The actual and ideal stagnation pressure at Station 2; and The minor loss coefficient for the actual pressure recovery (neglecting major head loss); Compare this pressure recovery coefficient with the value for a sudden expansion (Fig. 8.15) Solution : Known : Water, D 1 = 0.02 m , D 2 = 0.04 m , , s m u / 5 , L = 0.1 m, 997 = , 4 9 . 8 = e Find : Ideal and actual pressure rise in diffuser. Sketch: Above Assumptions : Steady flow, incompressible Analysis : Mass conservation 3 2 1 / 3133 . 4 5 * 02 . * * 997 m kg A u m = = = & Conservation of mass: 2 2 2 1 1 1 A u A u = velocity ratio: 2 2 2 1 1 2 D D u u = (Can start with mechanical energy equationnote it gives same results as starting from 1 2 0.1 m conservation of energy with constant temperature assumption) Conservation of energy: ( ) 1 2 h h m W Q = & & & No heat addition, no work done: 1 2 h h = Expanding stagnation enthalpy: 2 2 2 1 1 2 2 2 u h u h + = + and expanding enthalpy: ( ) 2 2 2 1 2 2 1 2 1 2 = + + u u p p T T c v For isothermal flow, 1 2 T T = , and using continuity: 1 2 2 1 2 2 2 1 1 2 =  + u u u p p 1 2 4 2 4 1 2 1 1 2 =  + D D u p p p C D D u p p = = 4 2 4 1 2 1 1 2 1 2 / and, 9375 . 40 20 1 1 4 4 2 4 1 =  = = p C D D Pa u p ideal 11684 2 9375 . 2 1 = = p 2_ideal = 50,000 + 11684 = 61,684 Pa Ideal stagnation pressure Station 2: 2 2 _ 2 _ 2 2 1 u p p ideal ideal + = = 62,462 Pa Ideal stagnation pressure Station 1: 2 1 1 1 2 1 u p p + = = 62,462 Pa Note the stagnation pressure is conserved as it should be for an ideal process Acual pressure rise: Find pChart C 25 . 2 30 45 2 2 1 2 2 1 2 = = = D D A A and 33 . 3 03 . 1 . 1 = = D L From Fig. 8.16: Nondimensional length: N/R 1 = 0.1*2/0.020 = 10. Angle: = Tan1 ( R/L ) = Tan1 ((0.04 0.02)/(2.*0.1) = Tan1 (0.1) = 2 = 5.71 degrees; Area ratio: = 4. And from the figure: 70 . = pChart C Actual pressure rise: Pa u p ideal 8274 2 70 ....
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 Spring '08
 MERKLE

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