Answer:
The value is [tex]F_t = 76024 \ N[/tex]
Explanation:
From the question we are told that
The flow rate is [tex]v_1 = 5 \ m/s[/tex]
The entrance diameter is [tex]d = 0.7 \ m[/tex]
The exit diameter is evaluated as [tex]d_e = 0.6 * 0.7 = 0.42 \ m[/tex]
The entrance gauge pressure is [tex]P_g = 350 \ kPa = 350*10^{3} \ Pa[/tex]
The droped gauge pressure is [tex]P_d = 50 \ kPa = 50*10^{3} \ Pa[/tex]
The presure at the exist is evaluated as [tex]P_e =(350 - 50 ) \ kPa = 300*10^{3} \ Pa[/tex]
Generally the entrance cross-sectional area is mathematically represented as
[tex]A = \pi \frac{d^2}{4}[/tex]
[tex]A = 3.142 \frac{0.7^2}{4}[/tex]
[tex]A =0.385 \ m^2[/tex]
Generally the exit cross-sectional area is mathematically represented as
[tex]A_e = \pi \frac{d_e^2}{4}[/tex]
[tex]A_e = 3.142 \frac{0.42^2}{4}[/tex]
[tex]A_e =0.139\ m^2[/tex]
Generally from the continuity equation
[tex]v_1 * A = v_2 * A_e[/tex]
=> [tex]v_2 = \frac{v_1 * A}{A_e}[/tex]
=> [tex]v_2 = \frac{5 * 0.385}{0.139}[/tex]
=> [tex]v_2 = 13.8 m/s [/tex]
Generally the net force in horizontal axis is equivalent to the net momentum change in the horizontal direction
So
[tex]P_g * A + F_t - P_e * A_e = P_d [ A_e * v_2^2 - A* v_1^2][/tex]
Here [tex]F_t[/tex] is the horizontal thrust on the fitting
So
[tex]350*10^{3} * 0.385 + F_t -0.385 * 0.139 = 50*10^{3} [ 0.385* 13.8^2 - 0.385* 5^2][/tex]
