Chapter 2, Problem 84RQ

To determine

The altitude of the plane from the ground.

Explanation of Solution

Given:

Density of the air $\left({\rho }_{\text{air}}\right)$ is $1.20\text{kg}/{\text{m}}^{\text{3}}$.

Density of the mercury $\left({\rho }_{\text{Hg}}\right)$ is $13,600\text{kg}/{\text{m}}^{\text{3}}$.

Barometric reading at the ground $\left(h_{\text{ground}}\right)$ is $753\text{mmHg}$.

Barometric reading at the plane $\left(h_{\text{plane}}\right)$ is $690\text{mmHg}$.

Acceleration due to gravity $\left(g\right)$ is $9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the atmospheric pressure at the location of the plane $\left(P_{\text{plane}}\right)$.

  $\begin{array}{c}{P}_{\text{plane}}={\left(\rho gh\right)}_{\text{plane}}\\ ={\rho }_{\text{Hg}}g{h}_{\text{plane}}\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(690\text{mmHg}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(690\text{mmHg}\times \frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.690\text{mHg}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.690\text{mHg}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{\text{1000}\text{N}/{\text{m}}^{\text{2}}}\right)\\ =92.06\text{kPa}\end{array}$

Calculate the atmospheric pressure at the location of the ground $\left(P_{\text{ground}}\right)$.

  $\begin{array}{c}{P}_{\text{ground}}={\left(\rho gh\right)}_{\text{ground}}\\ ={\rho }_{\text{Hg}}g{h}_{\text{ground}}\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(753\text{mmHg}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(753\text{mmHg}\times \frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.753\text{mHg}\right)\\ =\left(13,600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.753\text{mHg}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{\text{1000}\text{N}/{\text{m}}^{\text{2}}}\right)\\ =100.46\text{kPa}\end{array}$

Calculate the altitude of the plane from the ground $\left(h\right)$.

  $\begin{array}{l}\frac{W_{\text{air}}}{A}=P_{\text{ground}}-P_{\text{plane}}\\ {\left(\rho gh\right)}_{\text{air}}=P_{\text{ground}}-P_{\text{plane}}\\ {\rho }_{\text{air}}gh=P_{\text{ground}}-P_{\text{plane}}\\ \left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)h=100.46\text{kPa}-92.06\text{kPa}\\ \left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)h\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{\text{1000}\text{N}/{\text{m}}^{\text{2}}}\right)=8.4\text{kPa}\end{array}$

  $h=\text{714}\text{m}$

Thus, the altitude of the plane from the ground is $\underset{\_}{\text{714}\text{m}}$.