Question

Water flows steadily through a curved duct that turns the flow through angle = 1359, as shown in Fig. 3. The cross-sectional area of the duct changes from A1 = 0.025 m’ at the inlet to Az = 0.05 m² at the outlet. The average velocity at the duct inlet is V1 = 6 m/s. The momentum flux correction factor may be taken as B - 1.01 at the duct inlet and B = 1.03 at the its outlet. The Inlet gage pressure is P = 78.47 kPa and outlet gage pressure is P2sage = 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 В. P1.page AL V2. B2. P: A Figure 3

please help...add sketch

Answer 1

P2 gauge Az sinus 80, Ⓡ = 45 Ai = 0.025 m² = -ve Means fy should downward. Net force in y-direction Rate of change of Momentum in y direction Az Sinus B2 S AZU? sinus-o fy fy = -P2 gauge Bz JA ZV? sin 45 Given o = 135 Az = 0.05 m2 6 mis. Apply conservation of mass, SA, VI =fz AzV2 Vz 6x0.023 3 mls. B. = 1.01 Pi = 78.47 Kla I = 1ooo kylm? ß2 1.03 P2 65.23 kpa. gauge 1.03 x103 x 0.05% 32C03450-65.23*0.05**10'+ 78.47 x 0.0258163 1.01 X 103 Xo.025 x 62 462.5076 N sign fx direction is opposite then what we Means -ve, x-dirn. fy P2 gauge Az - B2 J Az Vz² sinus Fy = -65.23 x 0.052 1.03 x103 x 0.05 x 137 x sinus =-3589,25 N gauge fx 2 Fx 11 - ve Mean assume Ify

Pigauge A, < B, x SA, u,2 1 Ex ) D=135 control volume. $-45 B ₂ X 9 AzV₂² Fx Horizontal i force exerted by water on the wall of Duct. Pzgauge Az fy Vertical force exerted by water on the wall of Duct. how Newton's second Net force in x-direction = Rate of change of Momentum. in ae dirn fer fx=-B278 Az V 2 Casus = Pz gauge Az A, + BISA,V, 2) т (-Pi gauge J: Fx -B2 f Az U₂² Cajus - P2 gauge Azt Pigauge A, B, 8A, V,? .