Question

Oil with a specific gravity of 0.80 forms a layer 0.90m deep in an open tank that is otherwise filled with water. The total depth of water and oil is 3m. What is the gage pressure at the bottom of the tank?

Answer 1

Given,

Specific gravity of oil,  $S_o$ = 0.80

Therefore, density of oil, $\rho_o$ =

=  

S, * density of water

= 0.80 * 1000

= 800 kg/

3

Depth of oil,  $h_1$ = 0.9 m

Total depth of water and oil, H = 3 m

Therefore, depth of water in the tank, $h_2$ =3 - 0.9 = 2.1 m

We are asked to determine the gauge pressure at the bottom of tank.

Gauge pressure refers to that pressure which is in excess of the atmosphere pressure. This is found by the use of Hydrostatic law.

According to this law, the equation for gauge pressure at any point in a fluid at rest is given by,

$P = h\rho g$

where, 'h' is the depth of point in fluid under consideration,

$\rho$ is the density of fluid, and

'g' is the acceleration due to gravity.

Based on the data given in question, we should first draw the cross section of the open tank.

This is obtained as follows:

9:00 80 OIL h=0.۹ m 3 * - ATER لط h= = 2.1 m لا

To calculate pressure at bottom of tank, we must consider effect of both oil and water.

Hence, total pressure at bottom of tank will be,

$P = h_1\rho_og \ +\ h_2\rho_wg$

where, $\rho_w$ is the density of water.

We have to then substitute the given values in above equation. Then, it shall be obtained as:

Hence, gauge pressure at bottom of the tank is  27.66 kN/m^2.