Question

QUESTION 5 A curved nozzle assembly that discharges to the atmosphere is shown. The nozzle mass is 3 kg and its internal volume is 0.002 m3. The fluid is water. Determine the reaction force in x-direction Rx in unit of N exerted by the nozzle on the coupling to the inlet pipe. P1 =231 kPa absolute. D1 = 7 cm, D2 = 3 cm, angle = 250, V1=4.5 m/s. Round to the nearest one decimal place. V PI D у Dz

Answer 1

The cross sectional area at inlet of the pipe is, 4 =D -0.07) =0.0038 m2 The cross sectional area at outlet of the pipe is, A - D TT = 4 TT (0.03) 4 = 7.1x10+ m² The outlet velocity of the pipe using continuity equation Q = Q2 V/ 4 =V24, A2 (4.5)(0.0038) 7.1x10+ = 24.1 m/s Apply momentum equation along x-direction Rş = { upV.da = ſul-pV.da)+ful-pV.da) Cs + = 4 (-PV, 4)+u2 (PV,A2) =(14)(-PV, 4)+(V, cos 25°)(0V242) An inlet of the pipe, there is no velocity component along x-direction, hence U = 0 R = 0+(V, cos 25°)(PV,4) = (V, cos 25°) (PV,A2) = (24.1cos 25°)(999(24.1)(7.1x104)) - = 411.96 N

Cs F Apply momentum equation along y-axis R-P4-W-Wit = vpV d4 - { vpV de R, -R4-mg-egy = { vpV de R, = B:4, - mg- pg+ { vpV.dt =R4-mg- pg+ ["( pv da) =R-4,-mg-pg + 1i(-pV1.4) + (-2V24) = R.4,-mg-pgū+(-1)(-PV, 4)+(-V sin 30°) (V, 4)---(1) =R4-mg-pgW+pV/24-(pvz.4, sin 30%) R = R4 – mg- ºgư+pV;24,- (pV} 4, sin 30°)) = (231x10*)(0.0038)–(3)(9.8)–(999)(9.8)(0.002)+(999(4.5)*(0.0038)–(999)( 24.1) (7.1x104 )sin 25°) = (231x10°)(0.0038)–(3)(9.8)–(999)(9.8)(0.002) +(999(4.5)* (0.0038)– (999)(24.1) (7.1x10 *)(0.422)) = 731.84 N