Answer to Question #130095 in Mechanics | Relativity for Ariyo Emmanuel

Question

The water in a tank is pressurized by air,
and the pressure is measured by a multifluid manometer as shown in the Fig. The
tank is located on a mountain at an
altitude of 1400 m where the atmospheric
pressure is 85.6 kPa. Determine the air
pressure in the tank if ℎ1 = 0.1 m, ℎ2 =
0.2 m, and ℎ3 = 0.35 m. Take the
densities of water, oil, and mercury to be
1000 kg/m3
, 850 kg/m3
, and 13,600
kg/m3
, respectively?

Answer 1

Solution

given data:-

atmospheric pressure( P_atm)=85.6kPa

P_atm =85600Pa

density of water$(\rho_w)=1000kg/m^3$

density of oil$(\rho_o)=850kg/m^3$

density of mercury$(\rho_m)=13600kg/m^3$

height of water column(h₁)= 0.1m

height of oil column from water level(h₂)=0.2m

height of mercury column (h₃)=0.35m

figure can be drawn like this using data of question

pressure at point B from left side of figure can be written

$P_B =P_{air}+\rho_ogh_1+\rho_wgh_2$ ........(eq.1)

pressure at point B from right side of figure can be written

$P_B =P_{atm}+\rho_mgh_3$ ................(eq.2)

from equation 1 and 2

$P_{air}+\rho_ogh_1+\rho_wgh_2=P_{atm}+\rho_mgh_3$

$P_{air}=P_{atm}+\rho_mgh_3-\rho_ogh_1-\rho_wgh_2$ ...(eq.3)

$\rho_mgh_3=13600\times9.8\times0.35=46648Pa$

$\rho_ogh_1=850\times9.8\times0.1=833Pa$

$\rho_wgh_2=1000\times9.8\times0.2=1960Pa$

by putting the value in eq.3

$P_{air}=85600+46648-833-1960$

=129455Pa

$\colorbox{aqua}{$P_{air}=129.45KPa$}$

therefore pressure of air in the tank is 129.45KPa.