Answer:
18.62 m/s
Explanation:
Given that:
A liquid with a density of 900 kg/m 3 is stored in a pressurized, closed storage tank.
Diameter of the tank = 10 m
The absolute pressure in the tank above the liquid is 200 kPa = 200, 000 Pa
At pressure of 200 kPa ; the final velocity = 0
Atmospheric pressure at 5cm = 101325 Pa
We are to calculate the initial velocity of a fluid jet when a 5cm diameter orifice is opened at point A?
By using Bernoulli's theorem between the shaded portion in the diagram;
we have:
[tex]Pa \ + \ \frac{1}{2} \ \delta \ v^2_1 \ + \ \delta gy_1 = P + \delta gy_2[/tex]
[tex]\frac{1}{2} \ \delta \ v^2_1 \ = P + \delta gy_2 - \ \delta gy_1 - Pa[/tex]
[tex]\frac{1}{2} \ \delta \ v^2_1 \ = \delta g(y_2 -y_1 )+ ( P - Pa )[/tex]
[tex]v_1 \ = \sqrt{ \frac {2 \ \delta g(y_2 -y_1 )+ 2 ( P - Pa )}{\delta}}[/tex]
where;
Pa = atmospheric pressure = 101325 Pa
[tex]\delta[/tex] = density of liquid = 900 kg/m³
[tex]v_1[/tex] = initial velocity = ???
g = 9.8 m/s²
[tex]y_1[/tex] = height of the hole from the buttom
[tex]y_2[/tex] = height of the liquid surface from the button
[tex]v_1 \ = \sqrt{ \frac {2*900*9.8(7 - 0.5 )+ 2 ( 200,000 - 101325 )}{900}}[/tex]
[tex]v_1 = 18.62 \ m/s[/tex]
Thus, the initial velocity of the fluid jet = 18.62 m/s
