* . Go Two hoses are connected to the same outlet using a...

$* .$ Go Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses A and B have the same length, but hose $\mathrm{B}$ has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille's law $\left[Q=\pi R^{4}\left(P_{2}-P_{1}\right) / 8 \eta L\right]$ applies to each. In this law, $P_{2}$ is the pressure upstream, $P_{1}$ is the pressure downstream, and $Q$ is the volume flow rate. The ratio of the radius of hose $\mathrm{B}$ to the radius of hose $\mathrm{A}$ is $R_{\mathrm{B}} / R_{\mathrm{A}}=1.50$. Find the ratio of the speed of the water in hose $\mathrm{B}$ to the speed in hose $\mathrm{A}$.

Key Concepts

Poiseuille's Law

This law describes the flow rate of a viscous fluid through a cylindrical pipe and shows that the flow rate is directly proportional to the fourth power of the radius of the pipe and the pressure difference, and inversely proportional to the viscosity of the fluid and the length of the pipe. It is key in analyzing laminar flow in systems where viscous effects are significant.

Relationship Between Flow Rate and Velocity

The average velocity of the fluid in a pipe is defined as the flow rate divided by the cross-sectional area. This concept is essential because even though the flow rate depends on a high power of the radius (R^4), the area depends on the square of the radius (R^2), leading to an average velocity that scales as the square of the radius for a given pressure drop.

Geometric Scaling in Circular Pipes

In circular pipes, geometric scaling tells us that if the radius changes, the change in cross-sectional area and flow resistance are not linear. While the flow rate increases with the fourth power of the radius due to the dynamics of Poiseuille flow, the increased area only grows with the square of the radius. This difference in scaling is crucial for understanding variations in velocity when the size of the conduit is altered.

Transcript

0:00 Hi there.

00:01 So for this problem we have the hoses a and b that have the same length, but host b has the larger radius.

00:12 Each is open to the atmosphere at the end where the water exits as is shown in this figure.

00:19 Water flows through both hoses as a bisquist fluid and post -cellulose law applies to each.

00:29 In this law, p2 is the pressure upstream, p1 is the pressure downstream, and q is the volume flow rate.

00:39 So the ratio of the radius of the host b to the radius of the host a is a given value and that it is that r b over r a the radius of b over the radius of a is equal to 1 .5 and for this problem we need to find the ratio of the speed of the water in host b to the speed in host a so for this problem we need to calculate that ratio that is the speed at the host b over the speed of host a.

01:32 So that's what we need to determine.

01:35 So the speed b of the water in the host is related to the volume flow rate by the following equation.

01:44 We know that the speed is equal to the flow rate over the cross -septional area.

01:54 Since the host is cylindrical, the radius with radius art, we will have that the area for this is going to be p times the radius square.

02:10 So for this we have that the speed is going to be equal to q over the area that we know is p times the radius square.

02:26 So we also know using poe -p -shieldy's law that the flow rate can be expresses as p times the radius to the four times the difference between the pressures over 8 times the viscosity times the length of these tubes.

02:57 So in postiolis law, the upstream pressure p2 is the same for both host.

03:04 Because it is the pressure at the outlet...

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