Answer:
The force is [tex]F= 46.25kN[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
At Equilibrium the summation of the of force on the vertical axis is zero
i.e [tex]\sum F_y =0[/tex]
=> [tex]F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)[/tex]
[tex]v_2[/tex] is the is the speed of water at the nozzle which can be mathematically evaluated as
[tex]v_2 = \frac{R}{A_n}[/tex]
substituting [tex]0.5m^3/s[/tex] for R and [tex]\frac{\pi}{4}(12*\frac{1m}{100} )^2[/tex] for [tex]A_n[/tex]
[tex]v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }[/tex]
[tex]= 44.23 m/s[/tex]
[tex]v_1[/tex] is the is the speed of water at the pipe which can be mathematically evaluated as
[tex]v_1 = \frac{R}{A_p}[/tex]
substituting [tex]0.5m^3/s[/tex] for R and [tex]\frac{\pi}{4}(30*\frac{1m}{100} )^2[/tex] for [tex]A_p[/tex]
[tex]v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }[/tex]
[tex]= 7.07 m/s[/tex]
[tex]\rho[/tex] is he density of water with value [tex]\rho =1000 kg /m^3[/tex]
Substituting values into the equation above
[tex]F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)[/tex]
[tex]= 21.99kN[/tex]
At Equilibrium the summation of the of force on the horizontal axis is zero
i.e [tex]\sum F_x =0[/tex]
=> [tex]F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)[/tex]
Since The speed at both A and B nozzle are the same then [tex]v_2[/tex] remains the same
Substituting values
[tex]F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)[/tex]
=> [tex]F_x = 40.69kN[/tex]
Hence the force acting on the flange bolts required to hold the nozzle in place is
[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]= \sqrt{40.69 ^2 + 21.99^2}[/tex]
[tex]F= 46.25kN[/tex]
