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sleet_krkn [62]
3 years ago
15

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 st

ationary 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.
Physics
1 answer:
Nesterboy [21]3 years ago
8 0

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.

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