Water flows steadily from the left pipe section (radius...

water flows steadily from the left pipe section (radius $r_{1}=2.00 R$ ), through the middle section (radius $R$ ), and into the right section (radius $r_{3}=3.00 R$ ). The speed of the water in the middle sec-tion is $0.500 \mathrm{~m} / \mathrm{s}$. What is the net work done on $0.400 \mathrm{~m}^{3}$ of the water as it moves from the left section to the right section?

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Key Concepts

Continuity Equation

The continuity equation is a fundamental principle in fluid dynamics stating that for an incompressible fluid, the mass flow rate remains constant along a streamline. This implies that the product of the cross-sectional area and the fluid velocity must be the same at all sections of a pipe, which explains how a change in the cross-sectional area (from a larger pipe to a smaller one, or vice versa) directly affects the speed of the fluid.

Work-Energy Principle

The work-energy principle in mechanics indicates that the net work done on an object or system is equal to the change in its kinetic energy. In the context of fluid flow, if a fluid element changes speed due to variations in the pipe’s cross-sectional area, this results in a corresponding change in kinetic energy, and thus the net work performed on the fluid can be determined by analyzing these kinetic energy differences.

Transcript

00:01 Right here we're given that the radius for this pipe the radius r1 is equal to 2r this radius is r radius and this radius is r3 is equal to 3r okay we're told that our um what else do i know here so we've got 0 .4 meters cubed of water and my velocity and the second part is 0 .5 meters per second.

01:05 0 .5 .00.

01:08 Okay, so we know that work is equal to negative of the delta p times v.

01:18 And let's do some figure out here.

01:20 So delta p from point one to three.

01:36 And let's call this point one, point two, and point three.

01:47 We're going to use bernoulli's equation, and that's p1 equals one half.

02:00 Whoops.

02:19 So we'll take p3 minus p1 equal this times one half times, and then we'll have vi minus v3.

02:36 And then we're going to find the relationships using this, a1, v1, equals a2, v2, equals a3, v3, b1 will equal a2 v2 over a1, that'll equal pi r squared, pi r squared times 0 .5 over pi pi times 2 r.

03:16 Okay, so here we'll get v1 because remember r is going to crancel out here, two there and two there.

03:27 So i'm going to get 0 .056 meters per second.

03:37 Now let's solve these into this equation.

03:44 Our delta p will equal 1 ,000 kilograms per meter cubed times one half times 0 .125.

03:59 I should have done my v3...

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