FTFS Chap13 P045

# FTFS Chap13 P045 - Chapter 13 Momentum Analysis of Flow...

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Chapter 13 Momentum Analysis of Flow Systems Angular Momentum Equation 13-45C The angular momentum equation is obtained by replacing B in the Reynolds transport theorem by the total angular momentum sys H , and b by the angular momentum per unit mass V r . 13-46C The angular momentum equation in this case is expressed as V m r I where is the angular acceleration of the control volume, and r is the position vector from the axis of rotation to any point on the line of action of F . 13-47C The angular momentum equation in this case is expressed as V m r I where is the angular acceleration of the control volume, and r is the position vector from the axis of rotation to any point on the line of action of F . 13 - 28

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Chapter 13 Momentum Analysis of Flow Systems 13-48 Water is pumped through a piping section. The moment acting on the elbow for the cases of downward and upward discharge is to be determined. Assumptions 1 The flow is steady and uniform. 2 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 3 Effects of water falling down during upward discharge is disregarded. Properties We take the density of water to be 1000 kg/m 3 . Analysis We take the entire pipe as the control volume, and designate the inlet by 1 and the outlet by 2. We also take the x and y coordinates as shown. The control volume and the reference frame are fixed. The conservation of mass equation for this one-inlet one-outlet steady flow system is m m m 2 1 , and V V V 2 1 since A c = constant. The mass flow rate and the weight of the horizontal section of the pipe are kg/s 24 . 45 ) m/s 4 ]( 4 / m) 12 . 0 ( )[ kg/m (1000 2 3 V c A m N/m 3 . 294 m/s kg 1 N 1 ) m/s 81 . 9 )( m 2 )( kg/m (15 2 2 mg W ( a ) Downward discharge : To determine the moment acting on the pipe at point A , we need to take the moment of all forces and momentum flows about that point. This is a steady and uniform flow problem, and all forces and momentum flows are in the same plane. Therefore, the angular momentum equation in this case can be expressed as in out V V m r m r M where r is the moment arm, all moments in the counterclockwise direction are positive, and all in the clockwise direction are negative. The free body diagram of the pipe section is given in the figure. Noting that the moments of all forces and momentum flows passing through point A are zero, the only force that will yield a moment about point A is the weight W of the horizontal pipe section, and the only momentum flow that will yield a moment is the exit stream (both are negative since both moments are in the clockwise direction). Then the angular momentum equation about point A becomes 2 2 1 V m r W r M A Solving for M A and substituting, m N 70.0 2 2 2 1 m/s kg 1 N 1 m/s) kg/s)(4 m)(45.54 (2 - N) m)(294.3 1 ( V m r W r M A
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## This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.

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FTFS Chap13 P045 - Chapter 13 Momentum Analysis of Flow...

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