Question

A 100 ft3 /s water jet (density 62.4 lbm/ft3 ) is moving in the positive x-direction with velocity 20 ft/s. The stream hits a stationary splitter, such that half of the flow is diverted upward at 45° and the other half is directed downward at 45°, and both streams have a final speed of 18 ft/s. Disregarding gravitational and viscous effects, determine the force required to hold the splitter in place against the water force.

Answer 1

Answer:

Explanation:

Given

Initial Flow of water$=100 ft^3/s$

density $\rho =62.4 lbm/ft^3$

Initial velocity$(v_i)=20 ft/s$

Final velocity $(v_f)=18 ft/s$

Initial mass flow rate$=\rho Av=\rho Q$

$m_i=62.4\times 100=62.4\times 10^2 lbm/s$

Let area of cross section of Two streams after striking be A

Conserving Flow

$Q_i=Q_1+Q_2$

$100=A\times 18+A\times 18$

$A=\frac{50}{18} ft^2$

or Flow rate after splitter splits in to half

$Q_i=2Q_f$

conserving momentum in x direction

Final momentum in x direction $=m\times v$

$P_f=2\times \rho \frac{Q}{2}\times 18\times \cos 45$

$P_i=\rho Q\times 20$

Change in momentum$=P_i-P_f$

$=20\rho \times Q-\frac{18\rho Q}{\sqrt{2}}$

$=7.28\times 62.4\times 100$

$=45,427.2 lbm-ft/s^2$

F=1409.24 lbf

change of momentum in Y direction is  zero as the two flows oppose thier motion after striking with splitter.