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Unformatted text preview: Chapter 13 Momentum Analysis of Flow Systems Review Problems 1359 Water is flowing into and discharging from a pipe Usection with a secondary discharge section normal to return flow. Net x and y forces at the two flanges that connect the pipes are to be determined. Assumptions 1 The flow is steady, onedimensional, and incompressible. 2 The weight of the Uturn and the water in it is negligible. Properties We take the density of water to be 1000 kg/m 3 . Analysis The flow velocities of the 3 streams are m/s 51 . ] 4 / m) 05 . ( )[ kg/m (1000 kg/s 30 ) 4 / ( 2 3 2 1 1 1 1 1 = = = = D m A m V m/s 80 . 2 ] 4 / m) 10 . ( )[ kg/m (1000 kg/s 22 ) 4 / ( 2 3 2 2 2 2 2 2 = = = = D m A m V m/s 3 . 11 ] 4 / m) 03 . ( )[ kg/m (1000 kg/s 8 ) 4 / ( 2 3 2 3 3 3 3 3 = = = = D m A m V We take the entire Usection as the control volume. We designate the horizontal coordinate by x with the direction of incoming flow as being the positive direction and the vertical coordinate by y . The momentum equation for steady onedimensional flow is i i e e m m F V V  = . We let the x and y components of the anchoring force of the cone be F Rx and F Ry , and assume them to be in the positive directions. Then the momentum equations along the x and y axes become ) ( 3 3 3 3 1 1 2 2 2 2 1 1 1 1 2 2 2 2 1 1 V V V V V V m F m F m m A P A P F m m A P A P F Ry Ry Rx Rx = = + =  = + + Substituting the given values, N 666 = =   = N k F Rx 666 . m/s kg 1000 kN 1 m/s) 1 kg/s)(0.5 30 ( m/s kg 1000 kN 1 m/s) kg/s)(2.8 22 ( 4 m) (0.10 ] kN/m ) 100 150 [( 4 m) (0.05 ] kN/m ) 100 200 [( 2 2 2 2 2 2 N 90.4 = = 2 m/s kg 1 N 1 m/s) 3 kg/s)(11. 8 ( Ry F The negative value for F Rx indicates the assumed direction is wrong, and should be reversed. Therefore, a force of 666 N acts on the flanges in the opposite direction. A vertical force of 90.4 N acts on the flange in the vertical direction. Discussion To assess the significance of gravity forces, we estimate the weight of the weight of water in the Uturn and compare it to the vertical force. Assuming the length of the Uturn to be 0.5 m and the average diameter to be 7.5 cm, the mass of the water becomes kg 2 . 2 m) (0.5 4 m) (0.075 ) kg/m 1000 ( 4 2 3 2 = = = = = L D AL V m whose weight is 2.2 9.81 = 22 N, which is much less than 90.4 N, but still significant. Therefore, disregarding the gravitational effects is a reasonable assumption if great accuracy is not required. 1360 A fireman was hit by a nozzle held by a tripod with a rated holding force. The accident is to be investigated by calculating the water velocity, the flow rate, and the nozzle velocity....
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 Spring '08
 Chung

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