Question

The closed tank of fig. P.2.34 is filled with water and is 5 ft long. The pressure gage on the tank reads 7 psi. Determine: (a) the height, h, in the open water column, (b) the gage pressure acting on the bottom tank surface ab, and (c) the absolute pressure of the air in the top of the tank if the local atmospheric pressure is 14.7 ps

Answer 1

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, ($P=Y*h$) 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: $P_{c}=P_{gauge(air)}+Y_{H2O}*h_{c} =7+0.03612*24=7.87(Psi)$ and the other hand, we have:$P_{d} =Y*h=0.03612*h$ and knowing that the pressure at point C and point D are the same, we can find the h as:$P_{c}=P_{d}$, so $7.87 (Psi)=0.03612*h$ 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:$P_{gaugeAB}=P_{gauge(air)}+Y_{H2O}*h_{AB} =7+0.03612*48=8.73(Psi)$ (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:$P_{abs(air)}=P_{gauge(air)} +P_{atm} =7+14.7=21.7(Psi)$ (c).