A tank contains water up to a height of 1 m above the base. An immiscible liquid of sp. gr. 0.8 is filled on the top of water up to 1.5 m height. Calculate: (i) total pressure on one side of the tank, (ii) the position of centre of pressure for one side of the tank, which is 3 m wide.

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Immiscible Liquids Pressure Problem

Problem Statement

A tank contains water up to a height of 1 m above the base. An immiscible liquid of sp. gr. 0.8 is filled on the top of water up to 1.5 m height. Calculate: (i) total pressure on one side of the tank, (ii) the position of centre of pressure for one side of the tank, which is 3 m wide.

Given Data

  • Top liquid (oil): S.G. = 0.8, \( \rho_o = 800 \, \text{kg/m}^3 \), height \(h_o = 1.5 \, \text{m}\)
  • Bottom liquid (water): \( \rho_w = 1000 \, \text{kg/m}^3 \), height \(h_w = 1.0 \, \text{m}\)
  • Width of tank side, \( b = 3 \, \text{m} \)

Solution

The total force is the sum of the force on the area in contact with oil (\(F_1\)) and the force on the area in contact with water (\(F_2\)).

1. Force on Upper Area (\(F_1\), wetted by oil)

$$ A_1 = b \times h_o = 3 \times 1.5 = 4.5 \, \text{m}^2 $$ $$ \bar{h}_1 = \frac{h_o}{2} = \frac{1.5}{2} = 0.75 \, \text{m} $$ $$ F_1 = \rho_o g A_1 \bar{h}_1 $$ $$ F_1 = 800 \times 9.81 \times 4.5 \times 0.75 $$ $$ F_1 = 26487 \, \text{N} $$

2. Force on Lower Area (\(F_2\), wetted by water)

The pressure on this area is from the 1.5 m of oil above (surcharge) plus the pressure from the water itself. The force \(F_2\) is the area of the trapezoidal pressure diagram on this section, multiplied by the tank width.

$$ \text{Pressure at oil-water interface, } p_{int} = \rho_o g h_o $$ $$ p_{int} = 800 \times 9.81 \times 1.5 = 11772 \, \text{N/m}^2 $$
$$ \text{Pressure at tank base, } p_{base} = p_{int} + \rho_w g h_w $$ $$ p_{base} = 11772 + (1000 \times 9.81 \times 1.0) $$ $$ p_{base} = 11772 + 9810 = 21582 \, \text{N/m}^2 $$
$$ A_2 = b \times h_w = 3 \times 1.0 = 3.0 \, \text{m}^2 $$ $$ F_2 = \left( \frac{p_{int} + p_{base}}{2} \right) \times A_2 $$ $$ F_2 = \left( \frac{11772 + 21582}{2} \right) \times 3 $$ $$ F_2 = 16677 \times 3 = 50031 \, \text{N} $$

(i) Total Pressure (Total Force, \(F_R\))

$$ F_R = F_1 + F_2 $$ $$ F_R = 26487 + 50031 $$ $$ F_R = 76518 \, \text{N} $$

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

We find the total centre of pressure by taking moments of the component forces about the free surface of the oil.

Position of \(F_1\), (\(h_1^*\)):

$$ h_1^* = \frac{2}{3} h_o = \frac{2}{3} \times 1.5 = 1.0 \, \text{m} \text{ from surface} $$

To find the position of \(F_2\), we can split it into a rectangular pressure block (from oil) and a triangular block (from water).

$$ \text{Force from surcharge, } F_{2a} = p_{int} \times A_2 $$ $$ F_{2a} = 11772 \times 3 $$ $$ F_{2a} = 35316 \, \text{N} $$
$$ \text{Force from water, } F_{2b} = \frac{1}{2} (\rho_w g h_w) A_2 $$ $$ F_{2b} = 0.5 \times (1000 \times 9.81 \times 1.0) \times 3 $$ $$ F_{2b} = 0.5 \times 9810 \times 3 $$ $$ F_{2b} = 14715 \, \text{N} $$

Positions of these component forces from the oil surface:

$$ h_{2a}^* = h_o + \frac{h_w}{2} = 1.5 + \frac{1.0}{2} = 2.0 \, \text{m} $$ $$ h_{2b}^* = h_o + \frac{2}{3}h_w = 1.5 + \frac{2}{3}(1.0) \approx 2.167 \, \text{m} $$

Now, we take the sum of moments and divide by the total force.

$$ h^* = \frac{(F_1 \times h_1^*) + (F_{2a} \times h_{2a}^*) + (F_{2b} \times h_{2b}^*)}{F_R} $$ $$ h^* = \frac{(26487 \times 1.0) + (35316 \times 2.0) + (14715 \times 2.167)}{76518} $$ $$ h^* = \frac{26487 + 70632 + 31885.5}{76518} $$ $$ h^* = \frac{129004.5}{76518} \approx 1.686 \, \text{m} $$
Final Results:

(i) Total Pressure (Force): \( F_R \approx 76.52 \, \text{kN} \).

(ii) Position of Centre of Pressure: \( h^* \approx 1.686 \, \text{m} \) below the free surface of the oil.

Explanation of Concepts

Layered (Immiscible) Fluids: When dealing with fluids that don't mix, the pressure at any point is the sum of the pressures exerted by all the fluid layers above it. The pressure increases linearly within each fluid layer, but the rate of increase (the slope of the pressure diagram) changes at the interface between fluids due to their different densities.

Pressure Surcharge: The top layer of oil exerts pressure on the water below. This pressure, calculated at the oil-water interface, acts as a uniform "surcharge" over the entire surface area wetted by the water. The total pressure on the water-wetted area is this surcharge plus the additional hydrostatic pressure from the water itself.

Moment Method for Centre of Pressure: The simplest way to find the overall centre of pressure is to break the total force into simpler components (\(F_1\), \(F_{2a}\), \(F_{2b}\)). We find where each of these component forces acts. Then, by summing the moments of these forces about a reference point (like the top free surface) and dividing by the total force, we can find the single point (\(h^*\)) where the total resultant force acts.

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