Chapter 1.11, Problem 110RP

To determine

The altitude of the airplane from the ground level.

Answer to Problem 110RP

The altitude of the airplane from the ground level is $714\text{m}$.

Explanation of Solution

Write the formula to calculate pressure at the location of the airplane $\left(P_{\text{plane}}\right)$.

$P_{\text{plane}}={\left(\rho gh\right)}_{\text{plane}}$ (I)

Here, density of mercury is $\rho$, barometer reading at the airplane is $h$ and acceleration due to gravity is $g$.

Write the formula to calculate pressure at the location of the ground level $\left(P_{\text{ground}}\right)$.

$P_{\text{ground}}={\left(\rho gh\right)}_{\text{ground}}$ (II)

Here, barometer reading at the ground level is $h$.

Consider an air column between the airplane and ground.

Write the equation of force balance per unit base area.

$\begin{array}{l}\frac{W_{\text{air}}}{A}=P_{\text{ground}}-P_{\text{plane}}\\ {\left(\rho gh\right)}_{\text{air}}=P_{\text{ground}}-P_{\text{plane}}\end{array}$ (III)

Here, density of air is $\rho$, weight of air is $W_{\text{air}}$, cross-sectional area is $A$ and altitude of the airplane from the ground level is $h$.

Conclusion:

Convert the unit of pressure $P_{\text{plane}}$ from $\text{mmHg}$ to $\text{mHg}$.

$\begin{array}{c}{P}_{\text{plane}}=690\text{mmHg}\\ =690\text{mmHg}\times \left(\frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =0.690\text{mHg}\end{array}$

Convert the unit of pressure $P_{\text{ground}}$ from $\text{mmHg}$ to $\text{mHg}$.

$\begin{array}{c}{P}_{\text{ground}}=753\text{mmHg}\\ =753\text{mmHg}\times \left(\frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =0.753\text{mHg}\end{array}$

Substitute $13,600\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.690\text{m}$ for $h$ in Equation (I).

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

Substitute $13,600\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.753\text{m}$ for $h$ in Equation (II).

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

Substitute $1.20\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$,$92.06\text{kPa}$ for $P_{\text{plane}}$ and $100.46\text{kPa}$ for $P_{\text{ground}}$ in Equation (III).

$\begin{array}{l}\left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)=100.46\text{kPa}-92.06\text{kPa}\\ \left(1.20\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)=8.4\text{kPa}\\ h=714\text{m}\end{array}$

Thus, the altitude of the airplane from the ground level is $714\text{m}$.