Answer:
(a) the height in the open water column is: 18.15 (ft), (b) the gauge pressure acting on the bottom tank surface AB is: 8.73 (Psi) and (c) the absolute pressure of the air in the top of the tank with local atmospheric pressure at 14.7 Psi will be: 21.7 (Psi).
Explanation:
We need to apply the Pascal's Law, ([tex]P=Y*h[/tex]) where P is pressure, γ is the specific weight of the fluid and h is the height of the column of fluid, to find the pressure at differents points in the tank (see attached). First we need to assume a property of the water called specific weight as: 62.4 (lbf/ft^3) or 0.03612 (lbf/in^3). Then knowing that 1 feet as the same as 12 inches and appling Pascal's Law, we get: [tex]P_{c}=P_{gauge(air)}+Y_{H2O}*h_{c} =7+0.03612*24=7.87(Psi)[/tex] and the other hand, we have:[tex]P_{d} =Y*h=0.03612*h[/tex] and knowing that the pressure at point C and point D are the same, we can find the h as:[tex]P_{c}=P_{d}[/tex], so [tex]7.87 (Psi)=0.03612*h[/tex] getting h=217.75(in) or 18.15(ft) (a). Now similarly to find the gauge pressure acting on the bottom of the tank surface AB we can apply Pascal's Law as:[tex]P_{gaugeAB}=P_{gauge(air)}+Y_{H2O}*h_{AB} =7+0.03612*48=8.73(Psi)[/tex] (b). Finally we can find the absolute pressure of the air in the top of the tank, assuming a local atmospheric pressure as 14.7 (Psi) so:[tex]P_{abs(air)}=P_{gauge(air)} +P_{atm} =7+14.7=21.7(Psi)[/tex] (c).
