Water discharges from a horizontal cylindrical pipe at the...

Water discharges from a horizontal cylindrical pipe at the rate of 465 $\mathrm{cm}^{3} / \mathrm{s}$ . At a point in the pipe where the radius is $2.05 \mathrm{cm},$ the absolute pressure is $1.60 \times 10^{5} \mathrm{Pa} .$ What is the pipe's radius at a constriction if the pressure there is reduced to $1.20 \times 10^{5} \mathrm{Pa}$ ?

Key Concepts

Continuity Equation

The continuity equation in fluid dynamics states that for an incompressible, steady flow, the mass (or volume) flow rate remains constant along a streamline. This means that if the cross-sectional area of the flow changes, the fluid's velocity must adjust inversely in order to maintain a constant volumetric flow rate. This concept is critical when analyzing flows through pipes that have varying diameters, as seen in many practical applications, including pipe constrictions.

Bernoulli's Principle

Bernoulli's principle relates the pressure, velocity, and elevation in a moving fluid by asserting that the total mechanical energy per unit volume is conserved along a streamline in an ideal fluid. In horizontal flows, where elevation changes are negligible, any decrease in pressure will be associated with an increase in the fluid velocity, and vice versa. This relationship is essential for understanding how pressure differences in a pipe lead to variations in flow speed, particularly when there are changes in the pipe's cross-sectional area.

Transcript

00:01 Okay, so in this problem we have a cylindrical pipe and we have a continuous flow of water through this pipe.

00:08 And we want to calculate what is the second radius of the pipe knowing only the pressure.

00:14 So the first thing we can use to calculate this is the equation of continuity.

00:21 Because it's the most simple equation for flowing of fluids.

00:28 So we can say that the v, oops, let's put a v in here, we can say that the v1 times the initial area is going to be equal the second speed times the second area, cross -sectional area.

00:49 And we know because the problem says that this rate is of four, this should be a four, 465 times 10 to the minus 6 meters cubic per second.

01:11 So that is the rate of the flowing of the water through this pipe.

01:16 And we want to calculate what is the second radius in here.

01:21 So the problem is we do not have neither the speed or the radius.

01:27 Therefore we cannot use this equation directly.

01:30 So we need to think about other equations because the only information that we have is the pressure.

01:38 So let's think about the bernouille's equation.

01:42 So the bernouise equation is going to be p1 plus the half whole v1 square, which is the connecting energy of the water.

01:57 It's going to be equal the second pressure plus the connecting energy.

02:02 Of the water in this point, which is whole v2 square.

02:09 So what's we gonna do here? if we rearrange this bernoulli's equation, we can say that, let's put in here, we can say that the final speed, the v2, is going to be equal two times, actually two divided by whole, that multiplies the pressure one, minus pressure pressure 2 plus the v1 is square and all this and this square root so that's the first equation we can use the equation to calculate the v2 but who is v1 well let's look to this first equation of continuity here we can use this side with this side...

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