Problem

Small rectangular tank full of Fresh water upto a height of 0.5m above base. Oil of relative density 0.8 is floating in settled condition above water, the height of Oil is 1.0m (Example: a small sludge tank in Engine Room) Calculate total pressure on one bulkhead of tank and find center of pressure from top of the tank.

Answer

To understand the pressure diagram and its use here : Pressure at Point 1 = $h \times \rho \times g = 0$ Pressure at point 2 = $h_1 \times \rho \times g = h_1 \times 0.8 T/m^2$

Therefore triangle ABK represents a pressure diagram If we make one triangle for each vertical at $P_1 and P_2$. Volume of wedge formed by triangle ABK across BB that is 2m gives us the thrust.

1. Thrust due to Oil only $= AB \times \frac {1}{2} \times 1 \times 0.8 \times 2 = 0.8 tons$

2. Thrust on BBCC (due to water only)$= 0.5 \times 1.0 \times 0.5 \times \frac{1}{2}\times 2= 0.25 tons$

**3.**thrust on BKMC

• (oil) = Pressure at BK is same as at CM . = (h_1 \times 0.8 = 0.8 Tm^-2)
• (water) = $0.8 tm^-2 \times 2 \times 0.5 = 0.8 tons$

Therefore Final result $= 1+2+3 = 0.8 + 0.25 +0.8 = 1.85 tons$