![]() The height of the water column is 0.3 m, and the height of the manometer fluid column is h. The density of the manometer fluid is given as 2.5 times the density of water, so its density is 2.5 * 1000 kg/m^3 = 2500 kg/m^3. The pressure at point B is due to the water column above it and the manometer fluid column below it. Again, we can neglect the density of air. Next, let's calculate the pressure at point B, which is the air-manometer fluid interface. Plugging these values into the formula, we get: P_A = 1000 kg/m^3 * 9.8 m/s^2 * 0.3 m = 2940 Pa.ģ. The density of water is approximately 1000 kg/m^3, and the height of the water column is 0.3 m. ![]() Therefore, the pressure at point A is only due to the water column above it. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus we can determine the pressure at the air-water interface. The gage pressure of air in the tank is to be determined. The density of air is very low compared to water, so we can neglect it. Solution The pressure in a pressurized water tank is measured by a multi-fluid manometer. ![]() We can use the hydrostatic pressure formula: P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.Ģ. ![]() First, let's calculate the pressure at point A, which is the air-water interface. ![]()
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