Problem Statement
What would the pressure in \( \text{kN/m}^2 \) be if the equivalent head is measured as \( 400 \, \text{mm} \) of:
- Mercury (\( \text{sp gr} = 13.6 \))
- Water
- Oil (\( \text{specific weight} = 7.9 \, \text{kN/m}^3 \))
- A liquid with \( \rho = 520 \, \text{kg/m}^3 \)
Solution
General Formula:
Pressure is given by:
\( P = \rho g h \)
Where:
- \( \rho \) = Density of the fluid (\( \text{kg/m}^3 \))
- \( g \) = Acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \))
- \( h \) = Height of the column (\( \text{m} \))
(a) In terms of mercury (\( \rho = 13600 \, \text{kg/m}^3 \)):
Using the formula:
\( P = \rho g h \)
Substitute values:
\( P = 13600 \times 9.81 \times 0.4 = 53366 \, \text{N/m}^2 = 53.366 \, \text{kN/m}^2 \)
(b) In terms of water (\( \rho = 1000 \, \text{kg/m}^3 \)):
Using the formula:
\( P = \rho g h \)
Substitute values:
\( P = 1000 \times 9.81 \times 0.4 = 3924 \, \text{N/m}^2 = 3.924 \, \text{kN/m}^2 \)
(c) In terms of oil (\( \gamma_{\text{oil}} = 7.9 \, \text{kN/m}^3 \)):
Pressure is directly calculated as:
\( P = \gamma_{\text{oil}} h \)
Substitute values:
\( P = 7.9 \times 0.4 = 3.16 \, \text{kN/m}^2 \)
(d) In terms of a liquid (\( \rho = 520 \, \text{kg/m}^3 \)):
Using the formula:
\( P = \rho g h \)
Substitute values:
\( P = 520 \times 9.81 \times 0.4 = 2040 \, \text{N/m}^2 = 2.04 \, \text{kN/m}^2 \)
Explanation
- Mercury: The density of mercury is calculated from its specific gravity (\( 13.6 \)) multiplied by the density of water. The resulting pressure is significantly higher due to the high density.
- Water: The pressure for water is lower, as it has a lower density compared to mercury.
- Oil: Pressure is calculated directly using its specific weight, which eliminates the need for gravity in the formula.
- Other Liquid: The pressure is derived using its density and standard formula. The low density results in the lowest pressure among all cases.
