Answer:
[tex]v_2[/tex] = 2.49 m/s
[tex]P_2[/tex] = 2.19 atm
Explanation:
By using continuity equation:
[tex]v_2[/tex] = ([tex]A_1[/tex][tex]v_1[/tex]) / [tex]A_2[/tex] = ([tex]d_1 / d_2[/tex])²[tex]v_1[/tex] = (5/2.8)² x 0.78 = 2.49 m/s
By using Bernoulli’s Equation:
[tex]P_1[/tex] + ρg[tex]h_1[/tex] + ½ρ ([tex]v_1[/tex] )²= [tex]P_2[/tex] + ρg[tex]h_2[/tex] + ½ρ([tex]v_2[/tex] )²
[[tex]v_1[/tex] ² -[tex]v_2[/tex] ²] /2 = ([tex]P_2[/tex]-[tex]P_1[/tex] )/ ρ + g([tex]h_2[/tex] - [tex]h_1[/tex])
(0.78²- 2.49²)/2 = ([tex]P_2[/tex]-[tex]P_1[/tex] )/ 1000 + (9.8 x 16)
-5.6= ([tex]P_2[/tex]-[tex]P_1[/tex] )/ 1000 + (156.8)
([tex]P_2[/tex]-[tex]P_1[/tex] ) = - 162400 Pa
[tex]P_2[/tex] = [tex]P_1[/tex] -162400, [tex]P_1[/tex] = 3.8atm = 385035 Pa
[tex]P_2[/tex] = 385035-162400 = 222635 Pa ( gauge pressure)
[tex]P_2[/tex] = 222635 Pa=> 2.19 atm
Answer:
The flow velocity is [tex]v_A = 2.4869 \ m/s[/tex]
The gauge pressure is [tex]P_B = 2.249 \ atm[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The gauge pressure [tex]P_A = 3.8 \ atm = 3.8 * 101325 = 385035 \ Pa[/tex]
The speed of flow is [tex]v_A = 0.78 m/s[/tex]
The diameter of the pipe is [tex]d = 5.0 cm = \frac{5}{100} = 0.05 \ m[/tex]
The diameter at the top floor is [tex]d_1 = \frac{2.8}{100} = 0.028 \ m[/tex]
The the height from the ground is [tex]h_B = 16 m[/tex]
So we are going to make some assumption
We would assume that the position on the street is A
and the position on the top floor is B
So from continuity equation the velocity of the flow in the street is
[tex]v_A = \frac{V_B * a_B}{a_A}[/tex]
Where
[tex]a_A[/tex] is the area of the pipe at the base
So
[tex]a_A = \frac{\pi d^2}{4}[/tex]
[tex]a_A = \frac{ 3.142 * (5)^2}{4}[/tex]
[tex]a_A =19.64 m^2[/tex]
and [tex]a_B[/tex] i the area of the pipe at the top floor
[tex]a_B = \frac{\pi d^2}{4}[/tex]
substituting values
[tex]a_B = \frac{ 3.142 * (2.8) }{4}[/tex]
[tex]a_B = 6.16 \ m^2[/tex]
So
[tex]v_A = \frac{0.78 * 19.64 }{6.16}[/tex]
[tex]v_A = \frac{0.78 * 19.64 }{6.16}[/tex]
[tex]v_A = 2.4869 \ m/s[/tex]
Applying Bernoulli's equation
[tex]P_A + \rho g h_A + \frac{1}{2} \rho v_A ^2 = P_2 + \rho g h_B + \frac{1}{2} \rho v_B^2[/tex]
Since the pipe started from the floor [tex]h_A = 0m[/tex]
Here the [tex]\rho[/tex] is the density of water with value [tex]\rho = 1000 \ kg/m^3[/tex]
Substituting values
[tex]385035 + (1000* 9.8 * 0 ) + \frac{1}{2} * 1000 * 0.78^2 = \\[/tex]
[tex]P_B + 1000 * 9.81 *16 + \frac{1}{2} * 1000 * 2.487^2[/tex]
[tex]P_B = 225446.6 \ Pa[/tex]
Converting back to atm
[tex]P_B = \frac{223446.6}{101325}[/tex]
[tex]P_B = 2.249 \ atm[/tex]
