Question

Show all work and answer all parts of the problem please.

The 45 degree elbow nozzle discharges water into the atmosphere. The cross-sectional area of the elbow is 25 cm2 150 cm2 at the inlet and 25 cm2 at the exit. Velocity at point 1 and 2 are given as 2 m/s and 12 m/s 450 respectively. The gage pressure at 1 is 73 kPa 150 em2 30 cm The elevation difference between the centers of the exit and the inlet is 30 cm. The horizontal distance from point 1 and 2 is 50 cm. The mass of the elbow and the water in it is 50 kg. And the line of action of weight is 20 cm from point A. The density of water is 1000 kg/m3, and the flow rate of water is 30 kg/s 1)、 20 cm Water- 30.0 kg/s 50 cm Given A1 0.015 m2 A2-0.0025 m2 1000 kg/m'" m = 50 kg m = 30 kg/s What is the mom ent at A required to keep the system ½ = 12 m/s from rotating? (Ans: MRA 149 counter-clockwise) Pi.g 73 kPa r1 0.3 m r2 0.5 m r3 0.2 m 1) Steady, incompressible flow 2) Uniform flow 3) Pipe diameter is relatively smaller than moment arms Find MRA 4) P2 Patm

Answer 1

0b C-5 12 ml4 A 20cm 50x 4.8 410.5 N 150 cm 13 kPT be ealtue uc apply. z n, is net moment t A duc to the foiccs (moment dnc to wcight at point A is clockwist 50 -Ve

If )(Mv)e d+] D as the system '5 steady. The moment duc to the pse55w s is zcoo at inlcr as it passs tharugh point A whocas at cutler thc gage pxeas ure C.V C.s C.S point 'D is fsom whcse weight acts. wc nced to find Ac 20Cm AD=20cm. 0cm → C W A D 2 4-5 +VC (cou nta Clock) mxv2 x AD T should be プV