An open tank contains water up to a depth of 2 m and above it an oil of sp. gr. 0.9 for a depth of 1 m. Find the pressure intensity (i) at the interface of the two liquids, and (ii) at the bottom of the tank.

Pressure in Layered Fluids

Problem Statement

Given Data

Depth of water, $Z_{\text{water}} = 2 \, \text{m}$
Depth of oil, $Z_{\text{oil}} = 1 \, \text{m}$
Specific gravity of oil, $S_{\text{oil}} = 0.9$
Density of water, $\rho_{\text{water}} = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Diagram

Solution

1. Calculate the Density of Oil

First, find the density of the oil using its specific gravity.

$$ \rho_{\text{oil}} = S_{\text{oil}} \times \rho_{\text{water}} $$ $$ \rho_{\text{oil}} = 0.9 \times 1000 \, \text{kg/m}^3 $$ $$ \rho_{\text{oil}} = 900 \, \text{kg/m}^3 $$

(i) Pressure at the Interface

The pressure at the interface between the oil and water is due only to the column of oil above it.

$$ p_{\text{interface}} = \rho_{\text{oil}} \times g \times Z_{\text{oil}} $$ $$ p_{\text{interface}} = 900 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 1 \, \text{m} $$ $$ p_{\text{interface}} = 8829 \, \text{N/m}^2 $$

(ii) Pressure at the Bottom of the Tank

The pressure at the bottom is the sum of the pressure from the oil layer and the pressure from the water layer.

$$ p_{\text{bottom}} = p_{\text{interface}} + p_{\text{water}} $$ $$ p_{\text{bottom}} = (\rho_{\text{oil}} g Z_{\text{oil}}) + (\rho_{\text{water}} g Z_{\text{water}}) $$

We already calculated the pressure from the oil ($8829 \, \text{N/m}^2$). Now we calculate the pressure from the water.

$$ p_{\text{water}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 2 \, \text{m} $$ $$ p_{\text{water}} = 19620 \, \text{N/m}^2 $$

Finally, add the two pressures together.

$$ p_{\text{bottom}} = 8829 \, \text{N/m}^2 + 19620 \, \text{N/m}^2 $$ $$ p_{\text{bottom}} = 28449 \, \text{N/m}^2 $$

Final Results:

(i) Pressure at the interface: $ 8829 \, \text{N/m}^2 $

(ii) Pressure at the bottom: $ 28449 \, \text{N/m}^2 $

Explanation of Pressure in Layered Fluids

When a tank contains multiple layers of immiscible (non-mixing) fluids, the total pressure at any point is the sum of the pressures exerted by each layer of fluid above that point. The pressure at the very top surface is atmospheric pressure (which we treat as zero gauge pressure in this problem).

The pressure at the interface between two fluids is determined solely by the weight of the fluid in the top layer. The pressure at the bottom of the tank is the pressure from the top layer plus the pressure from the second layer, and so on.

Physical Meaning

This problem illustrates how pressure increases with depth in a fluid system. The pressure at the bottom of the tank ($28449 \, \text{N/m}^2$) is significantly higher than the pressure at the oil-water interface ($8829 \, \text{N/m}^2$) because it must support the weight of both the oil and the water column above it.

This principle is fundamental in many fields, including geology (for understanding pressures within the Earth’s crust), oceanography (for calculating pressure at different sea depths), and engineering (for designing dams, submarines, and storage tanks that can withstand the total pressure of the fluids they contain).

An open tank contains water up to a depth of 2 m and above it an oil of sp. gr. 0.9 for a depth of 1 m. Find the pressure intensity (i) at the interface of the two liquids, and (ii) at the bottom of the tank.

Problem Statement

Given Data

Diagram

Solution

1. Calculate the Density of Oil

(i) Pressure at the Interface

(ii) Pressure at the Bottom of the Tank

Explanation of Pressure in Layered Fluids

Physical Meaning

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