Question

Water flows in a shallow semicircular channel with inner and outer radii of $1\mathrm{m}$ and $2\mathrm{m}$ (see figure). At a point $P(r,\theta )$ in the channel, the flow is in the tangential direction (counterclockwise along circles), and it depends only on $r,$ the distance from the center of the semicircles. a. Express the region formed by the channel as a set in polar coordinates. b. Express the inflow and outflow regions of the channel as sets in polar coordinates. c. Suppose the tangential velocity of the water in $\mathrm{m}/\mathrm{s}$ is given by $v(r)=10r,$ for $1\le r\le 2.$ Is the velocity greater at $(1.5,\frac{\pi }{4})$ or $(1.2,\frac{3\pi }{4})?$ Explain. d. Suppose the tangential velocity of the water is given by $v(r)=\frac{20}{r},$ for $1\le r\le 2.$ Is the velocity greater at $(1.8,\frac{\pi }{6})$ or $(1.3,\frac{2\pi }{3})?$ Explain. e. The total amount of water that flows through the channel (across a cross section of the channel $\theta ={\theta }_0$ ) is proportional to ${\int }_1^{2}v(r)dr.$ Is the total flow through the channel greater for the flow in part (c) or (d)?

Short answer

Question: Determine which flow (either c or d) has a greater total amount of water passing through the channel, and compare the tangential velocity at points specified for both cases. Answer: The total flow through the channel is greater for the flow in part c, as 15 > 13.86. For part c with tangential velocity $v(r)=10r$, the velocity is greater at the point $(1.5,\frac{\pi }{4})$ with 15 m/s compared to 12 m/s at $(1.2,\frac{3\pi }{4})$. For part d with tangential velocity $v(r)=\frac{20}{r}$, the velocity is greater at the point $(1.3,\frac{2\pi }{3})$ with approximately 15.38 m/s compared to approximately 11.11 m/s at $(1.8,\frac{\pi }{6})$.

Step-by-step solution

Channel Region in Polar Coordinates

To represent the channel in polar coordinates, we consider the inner and outer radii mentioned in the problem: $1\le r\le 2$. So, the region formed by the channel can be expressed as the following set in polar coordinates: $$\{(r,\theta )\mid 1\le r\le 2\text{ and }0\le \theta \le \pi \}$$

Inflow and Outflow Regions in Polar Coordinates

The inflow region corresponds to the inner semicircular boundary, while the outflow region corresponds to the outer semicircular boundary. These regions can be expressed in polar coordinates as follows: - Inflow region: $\{(r,\theta )\mid r=1\text{ and }0\le \theta \le \pi \}$ - Outflow region: $\{(r,\theta )\mid r=2\text{ and }0\le \theta \le \pi \}$

Compare Tangential Velocities with $v(r)=10r$

We are given the tangential velocity of the water: $v(r)=10r$. To compare the velocity at the points $(1.5,\frac{\pi }{4})$ and $(1.2,\frac{3\pi }{4})$, we calculate the velocities at these points: - Velocity at $(1.5,\frac{\pi }{4})$ is $v(1.5)=10\cdot 1.5=15\text{m/s}$ - Velocity at $(1.2,\frac{3\pi }{4})$ is $v(1.2)=10\cdot 1.2=12\text{m/s}$ Therefore, the velocity is greater at the point $(1.5,\frac{\pi }{4})$.

Compare Tangential Velocities with $v(r)=\frac{20}{r}$

Now, we are given a different tangential velocity function: $v(r)=\frac{20}{r}$. To compare the velocity at the points $(1.8,\frac{\pi }{6})$ and $(1.3,\frac{2\pi }{3})$, we calculate the velocities at these points: - Velocity at $(1.8,\frac{\pi }{6})$ is $v(1.8)=\frac{20}{1.8}\approx 11.11\text{m/s}$ - Velocity at $(1.3,\frac{2\pi }{3})$ is $v(1.3)=\frac{20}{1.3}\approx 15.38\text{m/s}$ Therefore, the velocity is greater at the point $(1.3,\frac{2\pi }{3})$.

Compare Total Flow Through the Channel

We need to evaluate the following integral to find the total flow through the channel: ${\int }_1^{2}v(r)dr$. We will calculate this integral for both the velocity functions (parts c and d) and compare which flow is greater. For part c, $v(r)=10r$: $${\int }_1^{2}10rdr=10{\int }_1^{2}rdr=10{\left[\frac{1}{2}{r}^2\right]}_1^2=5(4-1)=15$$ For part d, $v(r)=\frac{20}{r}$: $${\int }_1^2\frac{20}{r}dr=20{\int }_1^2\frac{1}{r}dr=20{[\ln |r|]}_1^2=20(\ln 2-\ln 1)=20\ln 2\approx 13.86$$ The total flow through the channel is greater for the flow in part c, as $15>13.86$.

Key concepts

Tangential Velocity

Tangential velocity is a fundamental concept in understanding circular motion and involves the linear speed of an object moving along the circumference of a circular path. In the context of polar coordinates, this velocity depends solely on the radius from the center of the path. For the water flow problem, the tangential velocity is determined by functions of the radius $r$.

• In part (c) of the exercise, the velocity function is $v(r)=10r$, implying that velocity increases linearly with $r$.
• In part (d), the function $v(r)=\frac{20}{r}$ shows an inverse relationship, where velocity decreases as $r$ increases.

To determine which location has greater velocity, substitute the specific radii into each velocity function. For example, with $v(r)=10r$, point $(1.5,\frac{\pi }{4})$ has a tangential velocity of 15 m/s, exceeding that at $(1.2,\frac{3\pi }{4})$, which is 12 m/s. This illustrates how radial distance directly influences speed when the dependency on $r$ varies geometrically.

Integral Calculus

Integral calculus deals with the computation of integrals and is instrumental in determining quantities like area, volume, and total change. In fluid dynamics, integrals can calculate the total flow of fluid through a channel or any path.

To solve part (e) of the exercise, we use integral calculus to evaluate the integral ${\int }_1^{2}v(r)dr$, which represents the volumetric flow rate across a cross-sectional area of the channel. By integrating from the inner radius to the outer radius, we sum up the accumulated flow between these boundaries.

• For $v(r)=10r$, the integral evaluates to 15, which is achieved by performing $10{\int }_1^{2}rdr=5(4-1)$.
• For $v(r)=\frac{20}{r}$, the integral calculates to approximately 13.86, shown by $20(\ln 2-\ln 1)=20\ln 2$.

These integrations reveal not only the technique behind the calculation but indicate that the linear increase of velocity function leads to a larger total flow compared to an inverse relation.

Fluid Dynamics

Fluid dynamics is a branch of physics concerned with the movement of liquids and gases. It employs methods from mechanics and mathematics to model how fluids behave in different scenarios, such as in open channels or enclosed pipes.

In this exercise, fluid dynamics principles help describe the flow of water through a semicircular channel, where the flow direction is tangential. The core questions involve calculating flow rates and velocities as a response to varying radius.

• The parameters of the channel (inner and outer radii) shape both the flow path and boundary conditions, crucial to determining regions of inflow and outflow in polar coordinates.
• Understanding the tangential flow aids in typical applications of fluid dynamics, illustrating how tangential velocity influences overall flow which is pivotal in engineering and environmental sciences.

The problem encapsulates real-world challenges like optimizing water distribution or ensuring efficient drainage, displaying how fluid dynamics can be modeled using mathematical tools like polar coordinates and calculus.