Thermodynamics The magnitude of the range R of exhaust...

Thermodynamics The magnitude of the range $R$ of exhaust temperatures (in degrees Fahrenheit) in an experimental diesel engine is approximated by the model $$R=\frac{5000(4-3 x)}{(11-7 x)(7-4 x)}, \quad 0< x \leq 1$$ where $x$ is the relative load (in foot-pounds). $$\begin{array}{l}{\text { (a) Write the partial fraction decomposition of the }} \\ {\text { equation. }} \\ {\text { (b) The decomposition in part (a) is the difference of }} \\ {\text { two fractions. The absolute values of the terms give }} \\ {\text { the expected maximum and minimum temperatures }} \\ {\text { of the exhaust gases for different loads. }}\end{array}$$ $$\begin{array}{l}{\text { (c) Use a graphing utility to graph each equation from }} \\ {\text { part (b) in the same viewing window. }} \\ {\text { (d) Determine the expected maximum and minimum }} \\ {\text { temperatures for a relative load of } 0.5 .}\end{array}$$

Thermodynamics The magnitude of the range $R$
of exhaust temperatures (in degrees Fahrenheit) in an
experimental diesel engine is approximated by the
model
$$R=\frac{5000(4-3 x)}{(11-7 x)(7-4 x)}, \quad 0< x \leq 1$$
where $x$ is the relative load (in foot-pounds).
$$\begin{array}{l}{\text { (a) Write the partial fraction decomposition of the }} \\ {\text { equation. }} \\ {\text { (b) The decomposition in part (a) is the difference of }} \\ {\text { two fractions. The absolute values of the terms give }} \\ {\text { the expected maximum and minimum temperatures }} \\ {\text { of the exhaust gases for different loads. }}\end{array}$$
$$\begin{array}{l}{\text { (c) Use a graphing utility to graph each equation from }} \\ {\text { part (b) in the same viewing window. }} \\ {\text { (d) Determine the expected maximum and minimum }} \\ {\text { temperatures for a relative load of } 0.5 .}\end{array}$$

Key Concepts

Mathematical Modeling in Thermodynamics

Mathematical modeling applies theoretical equations to represent and predict the behavior of physical systems. In thermodynamics, such models help bridge the gap between abstract mathematical expressions and real-world phenomena, enabling engineers to simulate, analyze, and optimize variables like temperature ranges under different operational conditions.

Graphical Analysis of Functions

Graphing functions is a powerful tool for visualizing the behavior of equations across different variable inputs. By using graphing utilities, one can easily identify characteristics like maximum and minimum values, asymptotic behavior, and overall trends, which aids in the interpretation and validation of mathematical models.

Extreme Value Analysis

Extreme value analysis involves determining the maximum and minimum values of a function over a given domain. This concept is critical in applications where identifying peak performance or limiting parameters is essential, and it often integrates both algebraic methods and graphical interpretations.

Partial Fraction Decomposition

This is an algebraic technique that breaks a complex rational function into a sum or difference of simpler fractions. It is especially useful in calculus for integration, solving differential equations, and analyzing functions by separating them into components whose behaviors are easier to study and interpret.

Rational Function Analysis

Rational functions, which are ratios of polynomials, exhibit behaviors such as asymptotes, discontinuities, and varied intercepts. Analyzing these functions involves understanding these features to predict trends, determine domains, and examine limits, all of which are fundamental when modeling real-world phenomena.

Transcript

00:01 For today, let's take a look at the partial fraction decomposition of the following function.

00:06 We have 5 ,000, multiplying the expression 4 minus 3x, divided by 11 minus 7x, multiplying 7 minus 4x, for which x lies between 0 and 1.

00:25 For questions such as these, we usually begin by factorizing the denominator, but since our problem has got its denominator already presented in factor form, we will simply assign a partial fraction with a constant numerator to each factor of the denominator and write it as follows.

00:45 The first constant a divided by the first factor 11 minus 7x plus the second constant b divided by the second factor 7 minus 4x.

00:59 Now the next step would be to multiply both sides of the equation by the least common denominator and that breaks us to this equation 5 ,000 multiplying 4 minus 3x equal to a times 7 minus 4x plus b multiplying 11 minus 7x now we would have to substitute values of x that can help us determine the constant a and b values of x that are usually convenient are those that make the factors equal to zero so for the first factor here 7 minus 4 x equal to 0 x is giving us 7 over 4 for 11 minus 7 x equal to 0 x is giving us 11 over 7 now we just have to substitute these values of x and that will help us find a constant and b but i don't want us to get into trouble with all this fraction so let's try another method with this method we're simply going to expand the expressions and group terms without x and terms with x so we have 20 ,000 minus 15 ,000 x equal to 7a minus 4a x plus 11b minus 7bx.

02:35 So we simply group numbers without x and then terms with x.

02:45 So i have here 7a 7a plus 11b minus 11b minus let's factor...

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