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

Question #130095
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?
1
2020-08-26T11:24:33-0400

Solution

given data:-

atmospheric pressure( Patm)=85.6kPa

Patm =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.

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