Finding the Width of a Nuclear Power Plant Cooling Tower Using Hyperbola

Introduction:

Cooling towers in nuclear power plants are essential for dissipating excess heat. These towers have a unique shape represented by a hyperbola. In this problem, we are given specific dimensions of a cooling tower and tasked with finding its width at a particular height. Let's dive into the solution step by step.

Step 1: Establishing the Hyperbola Equation

We assume the center of the hyperbola as origin (0, 0). The smallest diameter of the cooling tower is 194 feet, making \(2a = 194\). Solving for \(a\), we get \(a = 97\). The equation of the hyperbola in this context is:

\[ \frac{x^2}{97^2} - \frac{y^2}{b^2} = 1 \quad \text{(i)} \]

Step 2: Utilizing the Larger Diameter and Radius

The larger diameter is 308 feet, so the radius is \(x = 154\) which 401 feet below the origin. Using the point \((x, y) = (154, -401)\) in equation (i), we have:

\[ \frac{154^2}{97^2} - \frac{(-401)^2}{b^2} = 1 \]

Solving for \(b^2\), we find \(b^2 = 10619.12\), which implies \(b = \sqrt{10619.12}\).

Step 3: Calculating the Width at 248 Feet Height

At a height of 248 feet above the ground, \(y = -153\) and \(b^2 = 10619.12)\. Substituting these values into equation (i), we get:

\[ \frac{x^2}{97^2} - \frac{(-153)^2}{10619.12} = 1 \implies x^2=(1 + \frac {153^2}{10619.12})(97^2)= 30150.3873 \]

\[\implies x = 173.6386 \implies 2x = 2 \times 173.6386 = 347.3 ft \]

Therefore the width of the cooling tower at height 248 feet above the ground is 347.3 feet

Conclusion:

By following these steps and applying the properties of the hyperbola, we can accurately determine the width of the nuclear power plant cooling tower at any given height, essential for structural considerations and efficient heat dissipation.

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