Physics, 11.04.2020 03:05 Greekfreak

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 from: xdimplesjoon

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.

v_1A_1 = v_2A_2

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 = $\pi \frac{d_1^2}{4}$

A₂ = $\pi \frac{d_2^2}{4}$ π

Let's substitute

$v_1 \pi \frac{d_1^2}{4} = v_2 \pi \frac{d_2^2}{4} \\d_2 = \sqrt{\frac{v_1}{v_2} } \ d_1$

Let's calculate

d₂ = $\sqrt{\frac{13.5}{5.83} } \ 0.573$

d₂ = 0.872 m

Flow is constant throughout the pipeline

Q = v₁ A₁

Q = $v_1 \pi \frac{d_1^2}{4}$

Let's calculate

Q = $13.5 \ \pi \ \frac{0.573^2}{4}$

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

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Answer from: iliketurtures

Answer with Explanation:

We are given that

Diameter of pipe, d_1=0.573 m

v_1=13.5 m/s

v_2=5.83 m/s

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

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

By equation of continuity

A_1v_1=A_2v_2

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

d_2=0.87 m

$d_2=0.87\time 100=87 cm$

1 m=100 cm

Answer from: Quest

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is

answer

1,500

kg • m/s.

Answer from: Quest

What even is this. where is the question

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