Chapter 4, Problem 22TP

Solution

To Calculate: The air density at $10000$ feet above sea level and at the temperature of $60°F$ .

(A) $D\approx 1.394\times {10}^{-3}$ pounds per cubic foot.

(B) $h\approx 507.25$ feet.

(C) $P\propto D$ .

(D) $-10\text{\%}$ .

Given: The air density model is given as $D=0.0761\cdot e^{-0.0004h}$ and also given as the pressure model $P=14.7\cdot e^{-0.0004h}$ .

Where;

  $D=$ air density in pounds per cubic.

  $P=$ air pressure in pounds per square inch.

  $h=$ height above sea level in feet.

Concept used:

(1) If a function $y=f\left(x\right)$ passes through a point $\left(x_0,y_0\right)$ , then the point $\left(x_0,y_0\right)$ satisfies the equation of the function, that is, $y_0=f\left(x_0\right)$ .

(2) An exponential model or function is as follows:

  $y=a\cdot {(b)}^x$ ; where $a\ne 0$ .

Calculation:

(A)Firstly take the air density model $D=0.0761\cdot e^{-0.0004h}$ and give height $(h)=10000$ feet, so substituted this value in the air density model.

  $\begin{array}{c}D=0.761\cdot e^{-0.0004\left(10000\right)}\\ =0.761\cdot e^{-4}\\ \approx 0.761\cdot \left(0.1832\right)\\ \approx 0.001394\end{array}$

Thus,

  $D\approx 1.394\times {10}^{-3}$ pounds per cubic foot.

(B)Now simplify the air model $P=14.7\cdot e^{-0.0004h}$ . Here $P$ is the air pressure in pounds per square inch and $h$ the height above sea level in feet.

Then, rewrite the above air pressure model.

  $\frac{P}{14.7}=e^{-0.0004\cdot h}$

Take on both sides a logarithm with a natural base $e$ .

Therefore,

  $\begin{array}{c}\ln \left(\frac{P}{14.7}\right)=\ln \left(e^{-0.0004\cdot h}\right)\\ \ln \left(\frac{P}{14.7}\right)=-0.0004\cdot h\ln e\left[\ln a^b=b\ln a\right]\\ \ln \left(\frac{P}{14.7}\right)=-0.0004\cdot h\left[\ln e=1\right]\end{array}$

  $h=\frac{\ln \left(\frac{P}{14.7}\right)}{-0.0004}$   ....... (1)

And also given the air pressure $P=12$ pounds per square inch.

  $\begin{array}{c}h=\frac{\ln \left(\frac{12}{14.7}\right)}{-0.0004}\\ h=\frac{\ln \left(0.8163\right)}{-0.0004}\\ h\approx \frac{-0.2029}{-0.0004}\left[\ln (0.1863)\approx -0.2029\right]\end{array}$

Thus,

  $h\approx 507.25$ feet.

(C) Similarly, again take the air density model $D=0.0761\cdot e^{-0.0004h}$ . Where $D$ be the air density and $h$ the height above sea level in feet.

Then, rewrite the above air density model.

  $\frac{D}{0.0761}=e^{-0.0004\cdot h}$

Take on both sides a logarithm with a natural base $e$ .

Therefore,

  $\begin{array}{c}\ln \left(\frac{D}{0.0761}\right)=\ln \left(e^{-0.0004\cdot h}\right)\\ \ln \left(\frac{D}{0.0761}\right)=-0.0004\cdot h\ln e\left[\ln a^b=b\ln a\right]\\ \ln \left(\frac{D}{0.0761}\right)=-0.0004\cdot h\left[\ln e=1\right]\end{array}$

Thus,

  $h=\frac{\ln \left(\frac{D}{0.0761}\right)}{-0.0004}$

  ....... (2)

So equation (1) and equation (2), equate $P$ and, $D$ to each other and simplify :

  $\begin{array}{c}\frac{\ln \left(\frac{P}{14.7}\right)}{-0.0004}=\frac{\ln \left(\frac{D}{0.0761}\right)}{-0.0004}\\ \ln \left(\frac{P}{14.7}\right)=\ln \left(\frac{D}{0.0761}\right)\\ \frac{P}{14.7}=\frac{D}{0.0761}\\ P=\frac{14.7}{0.0761}D\end{array}$

Therefore,

  $P=193.167\cdot D$

Hence,

  $P\propto D$ .

Air pressure is directly proportional to air density. If the air density is increased then the air pressure will also increase, and vice versa.

(D) To check and explain the general rule which says that for every $80$ meter increase in height there is a $1\%$ decrease in air pressure.

The air pressure model is given as follows:

  $P=14.7\cdot e^{-0.0004h}$

Solve for $P$ , then we put $h=0$ ; in the air pressure model.

  $\begin{array}{c}{P}_0=14.7\cdot e^{-0.0004\cdot \left(0\right)}\\ P_0=14.7\cdot e^0\\ P_0=14.7\left[a^0=1\right]\end{array}$

Also, know that;

  $\begin{array}{c}\because 1\text{ meter}=3.28084\text{ feet}\\ \therefore \text{80 meter=80}\cdot 3.28084\text{ feet}\end{array}$

  $\text{80 meter=262}\text{.467 feet}$

So,

  $\begin{array}{c}{P}_{80}=14.7\cdot e^{-0.0004\cdot \left(262.467\right)}\\ P_{80}=14.7\cdot e^{-0.1049}\\ P_{80}\approx 14.7\cdot 0.9004\left[e^{-0.1049}\approx 0.9004\right]\\ P_{80}\approx 13.23\end{array}$

The difference between $0$ to $80$ meters is $1\%$ .

  $\begin{array}{c}\text{percentage of air pressure difference =}\frac{P_{final}-P_{initial}}{P_{initial}}\times 100\%\\ =\frac{13.23-14.7}{14.7}\times 100\text{ \%}\\ \text{=}\frac{-1.47}{14.7}\times 100\text{ \%}\\ \text{=}-\text{0}\text{.1}\times 100\text{ \%}\\ =-10\text{\%}\end{array}$

We obtained a $10\text{\%}$ decrease in air pressure when the altitude is increased from $0$ to $80$ meters this disagrees with the general rule which states that there is only a $1\%$ decrease in air pressure for every $80$ meter increase in height.

Conclusion: Hence, the air density $D\approx 1.394\times {10}^{-3}$ pounds per cubic foot at height $(h)=10000$ feet above sea level.

And air pressure model is true for the height $h\approx 507.25$ feet.

If the air density is increased then the air pressure will also increase or air pressure is directly proportional to air density.

Disagree with the general rule which states that there is only a $1\%$ decrease in air pressure for every $80$ meter increase in height.