Question

Water flows steadily through a curved duct that turns the flow through angle @= 135º, as shown in Fig. 3. The cross-sectional area of the duct changes from A1 = 0.025 m² at the inlet to A2 = 0.05 m’ at the outlet. The average velocity at the duct inlet is V1 = 6 m/s. The momentum flux correction factor may be taken as B1 = 1.01 at the duct inlet and B2 = 1.03 at the its outlet. The Inlet gage pressure is P1,gage = 78.47 kPa and outlet gage pressure is P2,gage = 65.23 kPa. Determine the horizontal and vertical components of the force exerted by the water on the walls of the duct. Neglect the weight of the duct and the water in it. Draw a sketch to illustrate the problem solution and show all your work (write formulae, substitutions with units, detailed calculations). (8 points) V B P1.gage A1 V2, B2, P2,gage 42 Figure 3

Answer 1

sol P. gauge A B, XPAN Fy control volume B₂ X PA₂ ur -P2 gauge Az Fx - Horizontal Force exerted by water on the wall of Duct Fy a vertical Force exerted by water on the wall of Duct Newton's Second Law Net Force in x-direction = rate of change of moment in x-direction Fx = -B2 x 8A₂ 2 Cosas l gauge Az-(-P, gauge A, + F. 8 Aur) - 8 A2 V 2 costs - Pegange A2+ P, gauge A, ISPAN Net Force in y-direction rate of change of moment in y-direction Fy = - Pegauge Asinas -P₂PAzua sings. Fy= -22 gauge A₂ sinas - P28 AzVisints - ③

Given Data - 135 = 45° A = 0.025m A = 0.05m v=6m/sec apply conservation of mass 8. Av = 8₂ M₂ V2 6X0,025 = 3 m/sec 0.05 8 = 1000kg / m3 P gange = 78.47 kpa. = 1.01 B2= 1.03 P2 gange = 65.23 kPa 3 . 1.03x103 0.05 x 3 cos 45o _ 45.23 x 0.05 X 10' 78.47 x 0.025 x10 - 1.01*10*x0.025x6 Fa = - 462.507 N ove sign indicates to direction is opposite then what we assume meome -ve , x-direction 1.03 xto x 0.05 x 3'xsins Fy = - 65.23 0.05 x10 - Fy = -3589.25N -ve sign indicates Fy should downward direction