A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 cm ( 0.573 m) and the flow speed of the petroleum is 13.5 m/s. At the refinery, the petroleum flows at 5.83 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

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Answer with Explanation:

We are given that

Diameter of pipe,[tex]d_1=0.573 m[/tex]

[tex]v_1=13.5 m/s[/tex]

[tex]v_2=5.83 m/s[/tex]

Volume flow rate of the petroleum along the pipe=[tex]Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})[/tex]

[tex]Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s[/tex]

By equation of continuity

[tex]A_1v_1=A_2v_2[/tex]

[tex]\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2[/tex]

[tex]d^2_2=\frac{v_1}{v_2}d^2_1[/tex]

[tex]d_2=\sqrt{\frac{v_1}{v_2}}d_1[/tex]

[tex]d_2=0.573\sqrt{\frac{13.5}{5.83}}[/tex]

[tex]d_2=0.87 m[/tex]

[tex]d_2=0.87\time 100=87 cm[/tex]

1 m=100 cm

The continuity equation allows to find the results the questions about the pipes and the flow are:

  • The flow is: Q = 3.48 m³ / s
  • The diameter of the pipe in the refinery is: d₂ = 0.872 m

Given parameters

  • Pipe diameter d = 57.3 cm = 0.573 m
  • The velocity of the oil is v₁ = 13.5 m / s
  • The speed at the refinery v₂ = 5.83 m / s

To find

  • The flow.
  • The diameter of the refinery pipe.

Fluid mechanics studies the movement of fluids, liquids and gases in different systems, the continuity equation is the conservation of mass for fluids.

        [tex]v_1A_1 = v_2A_2[/tex]

Where the subscripts 1 and 2 represent two points of interest, v is the velocity and A is the area.

Let's use the subscript 1 for the pipeline and the subscript 2 for the refinery, let's find the areas.

         A₁ = π r₁² = [tex]\pi \frac{d_1^2}{4}[/tex]  

         A₂ = [tex]\pi \frac{d_2^2}{4}[/tex]π  

Let's substitute

       [tex]v_1 \pi \frac{d_1^2}{4} = v_2 \pi \frac{d_2^2}{4} \\d_2 = \sqrt{\frac{v_1}{v_2} } \ d_1[/tex]

Let's calculate

       d₂ = [tex]\sqrt{\frac{13.5}{5.83} } \ 0.573[/tex]  

       d₂ = 0.872 m

Flow is constant throughout the pipeline

       Q = v₁ A₁

       Q = [tex]v_1 \pi \frac{d_1^2}{4}[/tex]

Let's calculate

        Q = [tex]13.5 \ \pi \ \frac{0.573^2}{4}[/tex]  

         Q = 3.48 m³ / s

In conclusion using the continuity equation we can find the results the questions about the pipes and the flow are:

  • The flow is: Q = 3.48 m³ / s
  • The diameter of the pipe in the refinery is: d₂ = 0.872 m

Learn more here: brainly.com/question/14747327