Question

A mercury barometer reads 745.0 mm on the roof of a building and 760.0 mm on the ground. Assuming a constant value of 1.29 kg/m3 for the density of air, determine the height of the building

Answer 1

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

$\Delta P \propto \rho \cdot \Delta h$

$\Delta P = k \cdot \rho \cdot \Delta h$

Where:

$\Delta P$ - Manometric pressure difference, measured in kilopascals.

$\rho$ - Fluid density, measured in kilograms per cubic meter.

$\Delta h$ - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

$\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}$

$\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}$

Where:

$\Delta h_{air}$ - Height difference of the air column, measured in meters.

$\Delta h_{Hg}$ - Height difference of the mercury column, measured in meters.

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\rho_{Hg}$ - Density of mercury, measured in kilograms per cubic meter.

If $\Delta h_{Hg} = 0.015\,m$, $\rho_{air} = 1.29\,\frac{kg}{m^{3}}$ and $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$, the height difference of the air column is:

$\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)$

$\Delta h_{air} = 158.140\,m$

The height of the building is 158.140 meters.

Answer 2

Answer:

158.13m

Explanation:

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ($P_{R}$)

Pressure measured at the ground level =  ($P_{G}$)

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

$P_{G}$ = $P_{R}$ + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) $P_{G}$ =  ρ$_{mercury}$ x h$_{(mercury)_{ground} }$ x g ------------(*)

But;

ρ$_{mercury}$ = density of mercury = 13600kg/m³  

h$_{(mercury)_{ground} }$ = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

$P_{G}$ =  13600 x 0.76 x 10

$P_{G}$ = 103360 Pa

(c) $P_{R}$ = ρ$_{mercury}$ x h$_{(mercury)_{roof} }$ x g       --------------(**)

But;

ρ$_{mercury}$ = density of mercury = 13600kg/m³  

h$_{(mercury)_{roof} }$ = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

$P_{R}$  =  13600 x 0.745 x 10

$P_{R}$  = 101320 Pa

(d) Now that we know the values of P, $P_{G}$ and $P_{R}$ , let's substitute them into equation (i) as follows;

$P_{G}$ = $P_{R}$ + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = $\frac{2040}{12.9}$

h = 158.13m

Therefore, the height of the building is 158.13m