Question

A nozzle is use to connect two pipes. When the discharge is 0.1 m/s, the gage pressure at the flange A is 40 kPa, and at B is 5 kPa. Determine the vertical component of the anchoring force required to hold the nozzle in place. The nozzle has a weight of 200 N, and the volume of water in the nozzle is 0.012 m. Is the anchoring force upward or downward? Area = 0.01 m Nozzle - p = 40 kPa B A r ea = 0.02 m? 0.10 m/s

Answer 1

Vasingo B, - VCAGO / Aa = 0이 m2 , I A - 0-02 ma

Given data →

Density of water (P) = 1000 kg/m

Gravitational constant (g) = 9.81 m/s

Volume of water (V) = 0.012 m

Weight of water (W) = p*g* V

Weight of water (W) = 1000 * 9.81 * 0.012

$\textup{Weight of water}\left ( W_{w} \right )=117.72\textup{ N}$

Weight of nozzle (WN) = 200 N

Discharge flow rate (0) = 0.10 m/s

$\textup{mass flow rate}\left ( \dot{m} \right )=\rho *\dot{Q}$

$\textup{mass flow rate}\left ( \dot{m} \right )=1000*0.1=100\textup{ kg/s}$

Velocity at A (V) =

0.1 Velocity at A (V1) = = 5 m/s

$\textup{Velocity at B}\left ( V_{2} \right )=\frac{\dot{Q}}{A_{2}}$

$\textup{Velocity at A}\left ( V_{2} \right )=\frac{0.1}{0.01}=10\textup{ m/s}$

$\textup{Gauge pressure at A}\left ( P_{1} \right )=40*10^{3}\textup{ Pa}$

Gauge pressure at B(P2) = 5* 103 Pa

$\textup{Summation of all vertical forces}$

$\sum F_{y}=P_{1}A_{1}+\dot{m}*\left ( V_{1}-V_{2}\sin \left ( 30 \right ) \right )-P_{2}A_{2}\sin \left ( 30 \right )-W_{N}-W_{w}$

$\sum F_{y}=40*10^{3}*0.02+100*\left ( 5-10\sin \left ( 30 \right ) \right )-5*10^{3}*0.01-200-117.72$

F, = 432.28 N + Downward