Chapter 2, Problem 42P

To determine

The height of the building.

Explanation of Solution

Given:

Barometer reading at the bottom of the building $\left(h_{\text{bottom}}\right)$ is $695\text{mmHg}$.

Barometer reading at the top of the building $\left(h_{\text{top}}\right)$ is $675\text{mmHg}$.

Density of air $\left(\rho \right)$ is $1.18\text{kg}/{\text{m}}^{\text{3}}$.

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

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

Calculation:

Calculate the pressure at the top of the building $\left(P_{\text{top}}\right)$.

  $\begin{array}{c}{P}_{\text{top}}={\rho }_{\text{Hg}}g{h}_{\text{top}}\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(675\text{mmHg}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(675\text{mmHg}\right)\left(\frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.675\text{m}\right)\\ =\left(90055.8\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =\text{90}\text{.06}\text{kPa}\end{array}$

Calculate the pressure at the bottom of the building $\left(P_{\text{bottom}}\right)$.

  $\begin{array}{c}{P}_{\text{bottom}}={\rho }_{\text{Hg}}g{h}_{\text{bottom}}\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(695\text{mmHg}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(695\text{mmHg}\right)\left(\frac{1\text{mHg}}{1000\text{mmHg}}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.695\text{m}\right)\\ =\left(92724.12\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =92.72\text{kPa}\end{array}$

Calculate the height of the building $\left(h\right)$ using the force balance per unit base area relation.

  $\begin{array}{l}\frac{W_{\text{air}}}{A}=P_{\text{bottom}}-P_{\text{top}}\\ {\left(\rho gh\right)}_{air}=P_{\text{bottom}}-P_{\text{top}}\\ \left(1.18\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)=92.72\text{kPa}-90.06\text{kPa}\\ \left(11.5758\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(h\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)=2.66\text{kPa}\\ h=231\text{m}\end{array}$

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