A tank contains water up to a height of 0.5 m above the base. An immiscible liquid of sp. gr. 0.8 is filled on top of the water up to a 1 m height.

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Pressure in a Tank with Layered Liquids

Problem Statement

A tank contains water up to a height of 0.5 m above the base. An immiscible liquid of sp. gr. 0.8 is filled on top of the water up to a 1 m height. Calculate: (i) the total pressure on one side of the tank, and (ii) the position of the centre of pressure for one side of the tank, which is 2 m wide.

Given Data

  • Depth of immiscible liquid, \( h_1 = 1.0 \, \text{m}\)
  • Depth of water, \( h_2 = 0.5 \, \text{m}\)
  • Width of tank, \( b = 2 \, \text{m}\)
  • Sp. gr. of liquid = 0.8, so its density \( \rho_1 = 800 \, \text{kg/m}^3 \)
  • Density of water, \( \rho_2 = 1000 \, \text{kg/m}^3 \)

Pressure Diagram

The pressure distribution on the side of the tank is composed of two parts due to the two different liquids.

Diagram showing the pressure distribution from two immiscible liquids

Solution

(i) Total Pressure (Force)

The total force is the sum of the forces from different sections of the pressure diagram. First, calculate the pressure at the interface (D) and at the base (B).

$$ P_D = \rho_1 g h_1 $$ $$ P_D = 800 \times 9.81 \times 1.0 $$ $$ P_D = 7848 \, \text{N/m}^2 $$ $$ P_B = P_D + \rho_2 g h_2 $$ $$ P_B = 7848 + (1000 \times 9.81 \times 0.5) $$ $$ P_B = 7848 + 4905 $$ $$ P_B = 12753 \, \text{N/m}^2 $$

Now, calculate the force from each section of the pressure diagram:

$$ F_1 \text{ (Force from top triangle ADE)} = (\frac{1}{2} \times P_D \times h_1) \times b $$ $$ F_1 = (\frac{1}{2} \times 7848 \times 1.0) \times 2 $$ $$ F_1 = 7848 \, \text{N} $$ $$ F_2 \text{ (Force from rectangle DBFE)} = (P_D \times h_2) \times b $$ $$ F_2 = (7848 \times 0.5) \times 2 $$ $$ F_2 = 7848 \, \text{N} $$ $$ F_3 \text{ (Force from bottom triangle EFC)} = (\frac{1}{2} \times (P_B - P_D) \times h_2) \times b $$ $$ F_3 = (\frac{1}{2} \times 4905 \times 0.5) \times 2 $$ $$ F_3 = 2452.5 \, \text{N} $$

The total pressure (force) is the sum of these forces:

$$ F_{total} = F_1 + F_2 + F_3 $$ $$ F_{total} = 7848 + 7848 + 2452.5 $$ $$ F_{total} = 18148.5 \, \text{N} $$

(ii) Centre of Pressure (\(h^*\))

The position of the centre of pressure is found by taking moments of the forces about the free surface (A).

$$ \text{Moment}_1 = F_1 \times (\frac{2}{3}h_1) $$ $$ \text{Moment}_1 = 7848 \times (\frac{2}{3} \times 1.0) \approx 5232 \, \text{N-m} $$ $$ \text{Moment}_2 = F_2 \times (h_1 + \frac{h_2}{2}) $$ $$ \text{Moment}_2 = 7848 \times (1.0 + 0.25) $$ $$ \text{Moment}_2 = 9810 \, \text{N-m} $$ $$ \text{Moment}_3 = F_3 \times (h_1 + \frac{2}{3}h_2) $$ $$ \text{Moment}_3 = 2452.5 \times (1.0 + \frac{2}{3} \times 0.5) $$ $$ \text{Moment}_3 \approx 3270 \, \text{N-m} $$

The total moment is the sum of individual moments:

$$ M_{total} = \text{Moment}_1 + \text{Moment}_2 + \text{Moment}_3 $$ $$ M_{total} = 5232 + 9810 + 3270 $$ $$ M_{total} = 18312 \, \text{N-m} $$

The centre of pressure \(h^*\) is the total moment divided by the total force:

$$ h^* = \frac{M_{total}}{F_{total}} $$ $$ h^* = \frac{18312}{18148.5} $$ $$ h^* \approx 1.009 \, \text{m} $$
Final Results:

Total Pressure (Force): \( F \approx 18148.5 \, \text{N} \)

Centre of Pressure: \( h^* \approx 1.009 \, \text{m} \) below the free surface

Explanation of Concepts

Immiscible Liquids: When liquids that do not mix are layered in a container, the pressure at any point is the sum of the pressures exerted by each liquid layer above it. The pressure increases linearly within each fluid layer, but the rate of increase (slope of the pressure diagram) changes at the interface between liquids according to their densities.

Pressure Diagram Method: For surfaces subjected to non-uniform pressure, like a tank wall with layered fluids, the total force can be found by calculating the "volume" of the pressure diagram (Area of diagram × width of the wall). The centre of pressure is the centroid of this pressure volume, which can be found by breaking the diagram into simple shapes and using the principle of moments.

Physical Meaning

The tank wall must withstand a total force of approximately 18,149 N, which is equivalent to the weight of about 1.85 metric tons. This force is distributed unevenly over the tank's side.

The resultant force acts at a single point, the centre of pressure, which is located 1.009 m below the top surface. This position is just below the interface of the two liquids (which is at 1 m). This is expected, as the denser water at the bottom contributes more significantly to the force moment. Knowing this exact point is vital for designing the tank's structural supports to prevent bulging, bending, or failure.

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