Chapter 1.11, Problem 59P

The basic barometer can be used to measure the height of a building. If the barometric readings at the top and at the bottom of a building are 675 and 695 mmHg, respectively, determine the height of the building. Take the densities of air and mercury to be 1.18 kg/m³ and 13.600 kg/m³, respectively.

To determine

The height of the building.

Answer to Problem 59P

The height of the building is $\underset{\_}{231\text{m}}$.

Explanation of Solution

Show the free body diagram of the building.

Write the expression of pressure at the top.

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

Here, the density is $\rho$, the vertical distance of the location in water from the free surface is $h$, and acceleration of gravity is $g$.

Write the expression of pressure at the top.

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

Write the expression of an air column between the top and bottom of the mountain and writing a force balance per unit base area.

$\begin{array}{l}{W}_{\text{air}}/A=P_{\text{bottom}}-P_{\text{top}}\\ {\left(\rho gh\right)}_{air}=P_{\text{bottom}}-P_{\text{top}}\end{array}$ (III)

Conclusion:

Write the unit conversion from mm of Hg to m of Hg.

For $P_{\text{top}}$,

$\begin{array}{c}{P}_{\text{top}}=675\text{mm}\text{of}\text{Hg}\\ =675\text{mm}\text{of}\text{Hg}\times \left(\frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =0.675\text{m}\text{of}\text{Hg}\end{array}$

For $P_{\text{bottom}}$,

$\begin{array}{c}{P}_{\text{bottom}}=695\text{mm}\text{of}\text{Hg}\\ =695\text{mm}\text{of}\text{Hg}\times \left(\frac{1\text{m}\text{of}\text{Hg}}{1000\text{mm}\text{of}\text{Hg}}\right)\\ =0.695\text{m}\text{of}\text{Hg}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0.675\text{m}$ for $h$ in Equation (I).

$\begin{array}{c}{P}_{\text{top}}=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.675\text{m}\right)\\ =\left(90055.8\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =90.055\text{kPa}\\ \simeq \text{90}\text{.06}\text{kPa}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $0.695\text{m}$ for $h$ in Equation (I).

$\begin{array}{c}{P}_{\text{top}}=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.695\text{m}\right)\\ =\left(92724.12\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =92.72\text{kPa}\end{array}$

Substitute $1.18\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $92.72\text{kPa}$ for $P_{\text{bottom}}$, and $90.06\text{kPa}$ for $P_{\text{top}}$ in Equation (I).

$\begin{array}{l}\left(1.18\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)=\left(92.72\text{kPa}\right)-\left(90.06\text{kPa}\right)\\ \left(11.5758\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)\times \left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)=\left(2.66\text{kPa}\right)\\ h=231\text{m}\end{array}$

Thus, the height of the building is $\underset{\_}{231\text{m}}$.