Question

1. - / 1 points

the temperature ( t ) (in degrees fahrenheit) of saturated steam increases as pressure increases. this relationship is approximated by the model
( t = 75.56 - 2.48x + 43.77sqrt{x}, quad 5 leq x leq 40 )
where ( x ) is the absolute pressure (in pounds per square inch).
the temperature of steam at sea level is ( 212^{circ} mathrm{f} ). find the absolute pressure (in lb per ( mathrm{in}^{2} )) at this temperature. use a graphing utility to verify your solution. (round your answer to two decimal places.)
per ( mathrm{in}^{2} )

Explanation:

Step1: Substitute \(T = 212\) into the equation

We have the equation \(T=75.56 - 2.48x+43.77\sqrt{x}\). Substituting \(T = 212\) gives \(212=75.56 - 2.48x+43.77\sqrt{x}\). Then, rearrange it to \(2.48x-43.77\sqrt{x}+212 - 75.56=0\), which simplifies to \(2.48x-43.77\sqrt{x}+136.44 = 0\). Let \(y=\sqrt{x}\), so the equation becomes \(2.48y^{2}-43.77y + 136.44=0\).

Step2: Use the quadratic formula

For a quadratic equation \(ay^{2}+by + c=0\) (\(a = 2.48\), \(b=-43.77\), \(c = 136.44\)), the quadratic formula is \(y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-43.77)^{2}-4\times2.48\times136.44\).

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Then \(y=\frac{43.77\pm\sqrt{564.3633}}{2\times2.48}=\frac{43.77\pm23.76}{4.96}\).
We have two solutions for \(y\):
\(y_1=\frac{43.77 + 23.76}{4.96}=\frac{67.53}{4.96}\approx13.62\) and \(y_2=\frac{43.77-23.76}{4.96}=\frac{20.01}{4.96}\approx4.04\).

Step3: Find \(x\) values

Since \(y = \sqrt{x}\), for \(y_1\approx13.62\), \(x_1=y_1^{2}\approx185.50\) (but \(x_1\) is out of the domain \(5\leq x\leq40\)). For \(y_2\approx4.04\), \(x_2=y_2^{2}\approx16.32\).

Answer:

\(16.32\)