Problem Statement
The pressure intensity at a point in a fluid is given as 3.924 N/cm². Find the corresponding height of the fluid when the fluid is: (a) water, and (b) oil of specific gravity 0.9.
Given Data
- Pressure Intensity, \(p = 3.924 \, \text{N/cm}^2\)
- Specific Gravity of Oil, \(S.G._{oil} = 0.9\)
Solution
1. Convert Pressure to SI Units (N/m²)
To work with standard units, we convert the pressure.
(a) Corresponding Height of Water
The relationship between pressure and height (head) is \(p = \rho g h\). For water, \(\rho_w = 1000 \, \text{kg/m}^3\).
(b) Corresponding Height of Oil
First, we find the density of the oil.
Now, we calculate the height of the oil column.
(a) The corresponding height of water is \( h = 4.0 \, \text{m} \).
(b) The corresponding height of oil is \( h \approx 4.44 \, \text{m} \).
Explanation of Pressure Head
The height of a fluid column that exerts a certain pressure at its base is known as the pressure head. It’s a convenient way to express pressure in terms of the height of a specific fluid. The fundamental relationship, derived from the hydrostatic equation, is:
$$ p = \rho g h \quad \text{or} \quad h = \frac{p}{\rho g} $$This shows that for a given pressure \(p\), the height \(h\) is inversely proportional to the density \(\rho\) of the fluid.
Physical Meaning
The results demonstrate that to create the same pressure of 3.924 N/cm², you need a taller column of oil (4.44 m) than you do of water (4.0 m). This is because oil is less dense than water.
Since each cubic metre of oil weighs less than a cubic metre of water, you need to stack the oil higher to achieve the same total weight (and thus pressure) over a given area at the base. This concept is fundamental in manometry and hydraulics, where different fluids are used to measure or transmit pressure.
