fluid force on vertical side of tank the weight density of water is 62.4

The gage pressure (kPa) at the bottom of the tank is most nearly A.4.7 B.20.2 C.25.6 D.46.1 Hint: Pressure due to each layer of liquid = h γ where h is the height of the liquid layer and γ is the specific weight of the liquid.

Density of brine is 1,030 kg/m3 and the density of oil is 880 kg/m3.

2.An open tank contains brine to a depth of 2 m and a 3-m layer of oil on top of the brine. If the specific gravity of seawater is 1.03 and the atmospheric pressure at this location is 14.8 psi, the absolute pressure (psi) at the bottom of the tank is most nearly A.5.4 B.20.2 C.26.8 D.27.2 Hint: Absolute pressure = gage pressure + atmospheric pressure where, gage pressure = γzh and γz = specific weight of seawater = SG x γw Solution: γz = 1.03 (62.4 lb/cu ft) = 64.27 lb/cu ft Gage pressure at the bottom of the tank = (12 ft) x (64.37 lb/cu ft) = 772.44 lb/sq ft = (772.44 lb/sq ft)/(144 sq in/sq ft)= 5.36 psi Absolute pressure = 14.8 psi + 5.36 psi= 20.16 psi Therefore, the key is (B). A tank is filled with seawater to a depth of 12 ft.

Fluid force on vertical side of tank the weight density of water is 62.4