Question

Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in the tank if h1 = 20 cm, h2 = 30 cm, h3 = 46 cm. Given r for water, oil and Hg to be 1000 kg/m3, 800 kg/m3, and 13, 600 kg/m3, respectively.

Answer 1

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, $P_{atm}$ = 101 Kpa

$P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}$

$P_{gauge} = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3$

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

$P_{gauge} = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge} = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge} = 110,298.26 pa$

Therefore, the gauge pressure of air is 110 kpa