Answer:
0.59 [tex]\frac{ft^3}{s}[/tex]
Explanation:
First, to hold the plate against the water stream, forces must be the same. For that reason, we have to define the force exerted by the water jet in terms of what we need to determine (volume flow rate). Therefore, we need the following definitions:
[tex]F_R: force \ required \ to \ hold \ the \ plate\\F_J: force \ exerted \ by \ the \ jet\\w: volume\ flow \ rate\\v: velocity \\A: area\\F_J=\frac{P_J}{A}\\A= \frac{w}{v}\\P_J =\frac{1}{2} \rho v^2[/tex]
Then, if we do the force balance and we replace each term by its definition, we can get:
[tex]F_J=F_R\\P_J A=F_R\\\frac{1}{2} \rho v^2 \frac{w}{v}=F_R\\\frac{1}{2} \rho v w=F_R\\w=\frac{2F_R}{\rho v}\\w=\frac{2 \times 500 lbf}{62.4\frac{lbm}{ft^3} 27\frac{ft}{s}}\\w=0.59 \frac{ft^3}{s}[/tex]
To hold the plate against the water stream, the water flow must be equal to 0.59 [tex]\frac{ft^3}{s}[/tex]
The volume flow rate of the water is 9.548 cubic feet per second.
In this question we must apply the principle of linear momentum conservation to a water jet-wall system.
Since the system is at equilibrium, the horizontal force applied on the wall ([tex]F[/tex]), in pounds-force, must have the same magnitude of the force from the water jet:
[tex]F = \frac{\rho\cdot \dot V \cdot v}{g_{c}}[/tex] (1)
Where:
If we know that [tex]F = 500\,lbf[/tex], [tex]g_{c} = 32.174\,\frac{\frac{lbm\cdot ft}{s^{2}} }{lbf}[/tex], [tex]\rho = 62.4\,\frac{lbm}{ft^{3}}[/tex] and [tex]v = 27\,\frac{ft}{s}[/tex], then the volume flow rate of the water is:
[tex]\dot V = \frac{F\cdot g_{c}}{\rho\cdot v}[/tex]
[tex]\dot V = \frac{(500\,lbf)\cdot \left(32.174\,\frac{\frac{lbm\cdot ft}{s^{2}} }{lbf} \right)}{\left(62.4\,\frac{lbm}{ft^{3}} \right)\cdot \left(27\,\frac{m}{s} \right)}[/tex]
[tex]\dot V = 9.548\,\frac{ft^{3}}{s}[/tex]
The volume flow rate of the water is 9.548 cubic feet per second. [tex]\blacksquare[/tex]
To learn more on linear momentum conservation, we kindly invite to check this verified question: https://brainly.com/question/2141713
