Question

Water flows steadily through a curved duct that turns the flow through angle = 135 degrees, 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 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 1 = 1.01 at the duct inlet and 2 = 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.

V Bi P1.gage A V2, B2, P2.gage 42 Figure 3

Answer 1

V = 6 m/s P. 78.47 kea. Fy gauge B=1.01. A,=0.025 m o 135 = 65.23kla 12 = ? Prgange Br 1.03 A = 0.05m² Draws Control volume (FBD): (Brivi) PA, (gauge) Ry Ju5o AY us (Byma) 45 P₂A2 (gange) Bo momentum Correction factor 0 ON POCO X2

now, in x-direction - man flow rate in = PANI = Pe A₂ V 2 ń = 103x 0.025 x 6 150 kg/s. in = PA₂V2. V₂ = 150 3 m/s. 103x0.05 V2 = 3 m/s from momentum flux equation (oma V2 - B, m, V.) x-direction EFx -Rx+P,A, + P2Ą Cos (45) Convention. tve -ve m[Cf. V. Grcusi) - ( -Rx+ (78.47X10 PX 0.025) +(65.23x102x0.05). 150[ (103x3 cos (43) - (1.01x6) - Rx + 1961.75 + 3261.5 = -581.25 Rx 5804.50 N Rx - 5.804 KN ON POCO X2

the component of force in x-direction Гу 5.804 KN now, in y-direction: Efy m (BVA - B.V.) y-direction -0.2 Since there is no flow at inlet in y-direction Viyo (Bividy (P.A.) -0 Efy y-direction, (m P, va) y.direction 1) - -ve -Ry + (P2A, sin (45) I tve MB V2 Sin (45) - 103 -Ry + (65.23 x 0.05xsin (457) (150) * J.03 x 3 Sin (45) - Ry 327.74 - 230 6.228 - 1978.488 - Ry Ry = 1978. 488 N Ry = 1.978 KN .. The component of force Fy = 1.978 kN
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