Answer:
a
The velocity at which the oil leaves the pump is [tex]v_1 = 5.15 m/s[/tex]
b
The radius of the horizontal pipe [tex]r_2 = 0.01244 m[/tex]
Explanation:
From the question we are told that
The rate of flow of unfiltered olive oil is [tex]v = 3.0 m/s[/tex]
The pressure on the pump is [tex]P_1 = 88 kPa = 88 *10^{3} \ Pa[/tex]
The radius of the pipe is [tex]r = 9.5 = \frac{9.5}{1000} = 0.0095 \ m[/tex]
Generally we can define this motion with Bernoulli's Equation as
[tex]P_1 + \frac{1}{2} \rho v_1 ^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2 + \rho g h_2[/tex]
[tex]P_2[/tex] is the pressure inside the pipe which is the atmospheric pressure which has a value of [tex]P_2 = 1.01 *10^{5} \ Pa[/tex]
[tex]\rho[/tex] is the density of olive oil which has a value of [tex]\rho = 980 kg/m^3[/tex]
[tex]g[/tex] is the acceleration due to gravity
h the height of the pipe since from the question we are not told that it is place on anything hence the height is the diameter of the pipe
[tex]v_1 \ and \ v_2[/tex] are the speed at which it leaves the pump and the speed at which it flows in the pipe respectively
Making [tex]v_2[/tex] the subject of the equation
[tex]v_1 = \sqrt{[\frac{2}{\rho} (P_2 -P_1 - \frac{1}{2} v^2_2)]}[/tex]
Substituting values
[tex]v_1 = \sqrt{[\frac{2}{980} (1.01 *10^5 -88 *10^{3} - \frac{1}{2}(3.0)^2)]}[/tex]
[tex]v_1 = 5.15 m/s[/tex]
Using continuity equation to define the motion of this fluid we have
[tex]A_1 v_1 = A_2 v_2[/tex]
Where [tex]A_1[/tex] is the area of the pump which is circular so mathematically it is
[tex]A_1 = \pi r^2_1[/tex]
[tex]A_2[/tex] is the area of the horizontal pipe which is circular so mathematically it is
[tex]A_2 = \pi r^2_2[/tex]
So the continuity equation becomes
[tex]\pi r^2 _1 v_1 = \pi r^2 _2 v_2[/tex]
Making [tex]r_2[/tex] subject of formula
[tex]r_2 = \sqrt{\frac{r_1^2 v_1}{v_2} }[/tex]
Substituting values
[tex]r_2 = \sqrt{\frac{ 0.0095^2 5.15 }{3.0} }[/tex]
[tex]r_2 = 0.01244 m[/tex]
