ps4 solutions - MIT Department of Mechanical Engineering...

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MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Problem 4.4 This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin Q x A ( x ) A 1 A 2 nozzle atmosphere, p a A nozzle with exit area A 2 is mounted at the end of a pipe of area A 1 , as shown. The nozzle converges gradually, and we assum that the flow in it is (i) approximately uniform over any particular station x , (ii) incompressible, and (iii) inviscid. Gravitational effects are, furthermore, taken as negligible. The volume flow rate in the nozzle is given as Q and the ambient pressure is p a . (a) Derive an expression for the gage pressure at a station where the area is A ( x ). (b) Show, by integrating the x -component of the pressure force on the nozzle’s interior walls, that the net x -component of force on the nozzle due to the flow is independent of the specific nozzle contour and is given by F = ρQ 2 ( A 1 - A 2 ) 2 2 A 1 A 2 2 (c) The expression in (b) predicts that F is in the positive x -direction regardless of whether the nozzle is converging ( A 2 < A 1 ) or diverging ( A 2 > A 1 ). Explain. Solution: Given: Q , A 1 , A 2 are constants. (a) By mass conservation, (mass in) = ρv 1 A 1 = ρv ( x ) A ( x ) = (mass out) Since there is no change in the mass inside the CV: v 1 A 1 = Q = v ( x ) A ( x ) v ( x ) = Q A ( x ) Apply Bernoulli’s equation along a stream line from station 1 to 2: Q x A ( x ) A 1 A 2 1 2 2.25 Advanced Fluid Mechanics 1 Copyright c ± 2006, MIT
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Inviscid Flows A.H. Shapiro and A.A. Sonin 4.4 Note that all the assumption required for Bernoulli have been satisfied: (a) inviscid (b) along a streamline (c) steady (d) constant density (e) no work/energy input or loss p ( x ) + 1 2 ρv ( x ) 2 | {z } station 1 = p a + 1 2 ρv 2 2 | {z } station 2 Therefore, p g ( x ) = p ( x ) - p a = 1 2 ρ ( v 2 2 - v 2 ) p g ( x ) = 1 2 ρQ 2 ± 1 A 2 2 - 1 A ( x ) 2 ! (4.4a) (b) Integrate the pressure along the nozzle to obtain the x -component of pressure force Q x A ( x ) A 1 A 2 z 1 projected vertical area of the inner wall = A 1 - A F x = Z 2 1 dF x = Z 2 1 p g ( x ) d (projected vertical area) = Z A 2 A 1 p g d ( A 1 - A ) = - Z A 2 A 1 p g dA We can reverse the integration limits to get rid of the minus sign in front and substitute Eq. (4.4a): F x = Z A 1 A 2 p g dA = ρQ 2 Z A 1 A 2 ² 1 A 2 2 - 1 A 2 ³ dA F x = ρQ 2 2 ( A 1 - A 2 ) 2 A 1 A 2 2 (4.4b) (c) For ( A 2 < A 1 ), p g ( x ) is positive. Since the pressure is greater on the inside than the outside, the net pressure force acts on the inner wall. A 1 A 2 Q For ( A 2 > A 1 ), p g ( x ) is negative. Thus, the net pres- sure force acts on the outer wall, still pointing to the right. A 1 A 2 Q ± Problem Solution by Sungyon Lee, Fall 2005 2.25 Advanced Fluid Mechanics 2 Copyright c ± 2006, MIT
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Inviscid Flows A.H. Shapiro and A.A. Sonin 4.7 Problem 4.7 This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin An engine carburetor consists of a duct of diameter D with a small metering jet of diameter d through which fuel enters from the float bowl. The fuel in the bowl is at the same level as the jet. The engine draws air into the duct from the ambient atmosphere, which is at pressure p a and density ρ a . The fuel stream breaks
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ps4 solutions - MIT Department of Mechanical Engineering...

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