Answer:
Volume flow rate: 2.71 cubic meters per second
Diameter at refinery: 0.714 m
Explanation:
The volume flow rate would be the product of pipe crossectional area and the flow speed:
[tex]\dot{V} = Av = \pi(d/2)^2v = \pi (0.563/2)^2 10.9 =2.71 m^3/s[/tex]
Assume this is steady flow, so the volume rate would be constant. So the volume rate at the refinery would be the same. Knowing its flow speed we can also calculate the cross-sectional area at this point:
[tex]A_2 = \dot{V}/v_2 = 2.71 / 6.77 = 0.4 m^2[/tex]
So the radius and diameter at the refinery is
[tex]\pi r_2^2 = A_2 = 0.4[/tex]
[tex]r_2^2 = A_2 / \pi = 0.128[/tex]
[tex]r_2 = \sqrt{0.128} = 0.357 m[/tex]
[tex]d_2 = r_2*2 = 0.357*2 = 0.714 m[/tex]
Answer:
a) 2.71 m³/s
b) 0.71 m
Explanation:
The flow rate, Q = πr²v
This is also referred to as the amount per second that flows through the pipe, so
Q = 3.142 * 0.2815² * 10.9
Q = 34.2478 * 0.07924
Q = 2.71 m³/s
Assuming that oil has not been added, subtracted or even compressed, then the flow rate Q, is the same everywhere. Thus,
πD²/4 * 6.77 = Q, knowing the value for our Q, we substitute
π/4 * D² * 6.77 = 2.71
D² * 6.77 = (2.71 * 4) / π
D² = 3.45 / 6.77
D² = 0.5096
D = √0.5096
D = 0.714 m
The diameter of the pipe is 0.71 m
