Question

The water in a tank is pressurized by air, and the pressure is measured by a multifluid manometer as shown in Figure 1. Determine the gage pressure of air in the tank if h1=26cm, h2=45cm, and h3-58cm. The specific gravity of mercury and oil are given to be 13.6 and 0.85 respectively.

Answer 1

Liver: h= 26 cm = 0.26 m, h2 = 45 cm= 0,45m h3 = 58 cm = 0.58 m S.G : mercury CHg) = 13.6, oil 0.85 now Sorgevēloty i pow x1 3.6 = 13600 legim 3 3 ove lovo X 0.853 850 kg Swaler = 1ooo kg/m3 Let Pressure at boint ) is P, Cabsolule ) Pressure at point (a) is P2 = Patm atmosphere pressure Starting with the pressure at point l) at the an. water interface and moving along the tube, by adding or subtracting the pgh terms untill we reach point 2, we kan while. : P t witer ghi + foie ghz-P mercury gha = Parm of Pi - Patm = 9 ( Prestomaths - Pusterta - Love 12) Substituting the required values and noting at - G+ : Sage (Pigage = 9.81 13600 X 0.58 - 1000 X 0.26 - 850 X 0.45 9.81 Jage = = 11078,355 Nim? 71.08 i la - fins