Problem 55 - Algebra and Trigonometry with Analytic Geometry

Question

When a hot gas exits a cylindrical smokestack, its velocity varies throughout a circular cross section of the smokestack, with the gas near the center of the cross section having a greater velocity than the gas near the perimeter. This phenomenon can be described by the formula $$ V=V_{\max }\left[1-\left(\frac{r}{r_{0}}\right)^{2}\right] $$ where \(V_{\max }\) is the maximum velocity of the gas, \(r_{0}\) is the radius of the smokestack, and \(V\) is the velocity of the gas at a distance \(r\) from the center of the circular cross section. Solve this formula for \(r\).

Step-by-Step Solution

Answer

The formula solved for \(r\) is \(r = r_0 \sqrt{1 - \frac{V}{V_{\max}}}\).

Step 1: Isolate the square term

First, let's rewrite the equation to isolate the squared term: \[ V = V_{\max} \left[1 - \left(\frac{r}{r_{0}}\right)^{2}\right] \] \[ \Rightarrow \frac{V}{V_{\max}} = 1 - \left(\frac{r}{r_{0}}\right)^{2} \] This step involves transposing terms to get everything related to \(r\) on one side.

Step 2: Move terms involving velocity

Next, rearrange the equation to move the velocity term to the other side:\[ \left(\frac{r}{r_{0}}\right)^{2} = 1 - \frac{V}{V_{\max}} \] This helps get \(r\) closer to isolation.

Step 3: Solve for \(r/r_0\)

Take the square root on both sides to solve for \(\frac{r}{r_{0}}\):\[ \frac{r}{r_{0}} = \sqrt{1 - \frac{V}{V_{\max}}} \] This eliminates the square from the \(r/r_0\) term.

Step 4: Isolate \(r\)

Finally, multiply through by \(r_{0}\) to solve for \(r\):\[ r = r_{0} \cdot \sqrt{1 - \frac{V}{V_{\max}}} \] This is the final expression for \(r\) in terms of \(V\), \(V_{\max}\), and \(r_0\).

Key Concepts

Velocity Distribution

The concept of velocity distribution is apparent in many fluid dynamics scenarios, such as hot gases exiting a cylindrical smokestack. In essence, velocity distribution describes how the speed of a fluid varies at different points across a cross-sectional area. For cylindrical flows, it's observed that the fluid near the center tends to move faster than fluid near the edges or walls. This is due to frictional forces at the boundaries slowing down the fluid's velocity. Understanding velocity distribution is crucial since it affects the design and operation of devices like chimneys, pipes, and turbines. Engineers use specific formulas to model this distribution. In the given formula \( V = V_{\text{max}} \left[1-\left(\frac{r}{r_{0}}\right)^{2}\right] \), the velocity \( V \) decreases as the distance \( r \) from the center increases, highlighting how velocity distribution changes radially outward. This behavior impacts emissions, heat dissipation, and even noise levels in industrial settings.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height value. Useful in many engineering and physics problems, this system is ideal for analyzing phenomena occurring in cylindrical shapes, like smokestacks. Instead of using the traditional Cartesian coordinates \((x, y, z)\), cylindrical coordinates utilize \((r, \theta, z)\):

• \( r \) is the radial distance from the origin to the point in the plane.
• \( \theta \) is the polar angle, measured counterclockwise from the positive x-axis.
• \( z \) is the height above the plane; similar to the Cartesian z-coordinate.

In the exercise, only \( r \) is used to calculate velocity distribution, indicating a focus on radial data rather than angular and vertical components. This approach simplifies the problem by concentrating on how changes occur radially, which is essential for understanding flow dynamics in circular paths.

Algebraic Manipulation

Algebraic manipulation is a mathematical process used to rearrange and simplify equations. It is crucial when solving for particular variables, as seen in the exercise where the goal is to isolate \( r \). Starting with the equation:\[V = V_{\max} \left[1-\left(\frac{r}{r_{0}}\right)^{2}\right]\]The solution involves several algebraic steps: - First, we reorganize to isolate terms involving \( r \).- Then, we transform the equation incrementally until \( r \) is isolated. This includes steps such as taking square roots and rearranging terms for clarity.These manipulations require a clear understanding of algebraic rules:

• Isolating a variable usually starts by eliminating addition or subtraction surrounding it, followed by removing multiplying or dividing factors.
• Taking a square root is an effective way to solve for a variable squared, though care must be taken regarding positive and negative roots.

These fundamental skills are vital for delving deeper into calculus and other advanced mathematical operations.