An oil of sp. gr. 0.9 is contained in a vessel. At a point the height of oil is 40 m. Find the corresponding height of water at the point.

Equivalent Pressure Head Calculation

Problem Statement

An oil of sp. gr. 0.9 is contained in a vessel. At a point the height of oil is 40 m. Find the corresponding height of water at the point.

Given Data

  • Specific gravity of oil, \(S_{\text{oil}} = 0.9\)
  • Height of oil column, \(Z_{\text{oil}} = 40 \, \text{m}\)
  • Density of water, \(\rho_{\text{water}} = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

The core principle is that the pressure exerted by the 40 m column of oil is equal to the pressure exerted by the corresponding column of water.

1. Calculate the Density of Oil

The density of the oil (\(\rho_{\text{oil}}\)) is its specific gravity multiplied by the density of water.

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

2. Calculate the Pressure Exerted by the Oil

Using the hydrostatic pressure formula \(p = \rho g Z\).

$$ p = \rho_{\text{oil}} \times g \times Z_{\text{oil}} $$ $$ p = 900 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 40 \, \text{m} $$ $$ p = 353160 \, \text{N/m}^2 $$

3. Find the Corresponding Height of Water

Rearrange the pressure formula to solve for the height of water (\(Z_{\text{water}}\)) that would produce this same pressure.

$$ Z_{\text{water}} = \frac{p}{\rho_{\text{water}} \times g} $$ $$ Z_{\text{water}} = \frac{353160 \, \text{N/m}^2}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} $$ $$ Z_{\text{water}} = \frac{353160}{9810} \, \text{m} $$ $$ Z_{\text{water}} = 36 \, \text{m} $$
Final Result:

The corresponding height of water is \( 36 \, \text{m} \).

Explanation of Pressure Head

Pressure Head is the height of a liquid column that corresponds to a particular pressure. The core idea is that the pressure at the base of any two liquid columns is identical if the product of their density, gravity, and height (\(\rho g Z\)) is the same. This allows us to express a pressure value in terms of an equivalent height of a standard fluid, usually water.

The relationship is: \(p = \rho_1 g Z_1 = \rho_2 g Z_2\). By cancelling gravity (\(g\)), we get \(\rho_1 Z_1 = \rho_2 Z_2\). This shows that the height of a liquid column needed to produce a certain pressure is inversely proportional to its density.

Physical Meaning

The result shows that a 40-meter column of oil exerts the same pressure at its base as a 36-meter column of water. This makes sense because the oil is less dense than water (90% as dense). To create the same amount of pressure (weight per unit area), you need a taller column of the lighter liquid (oil) compared to the shorter column of the heavier liquid (water).

This concept is crucial in manometry and fluid mechanics for comparing pressures in different systems using a common reference, like “meters of water”.

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