Problem Statement
A cylinder contains a fluid at a gauge pressure of \( 360 \, \text{KN/m}^2 \). Express this pressure in terms of a head of:
- Water.
- Mercury (specific gravity \( \text{sp gr} = 13.6 \)).
What would be the absolute pressure in the cylinder if the atmospheric pressure is \( 760 \, \text{mmHg} \)?
Solution
(a) Head in terms of water (\( \rho = 1000 \, \text{kg/m}^3 \)):
Given:
\( P = 360 \times 10^3 \, \text{N/m}^2 \)
Using the formula:
\( h = \frac{P}{\rho g} \)
Substitute values:
\( h = \frac{360 \times 10^3}{1000 \times 9.81} = 36.7 \, \text{m} \)
(b) Head in terms of mercury (\( \rho = 13.6 \times 1000 = 13600 \, \text{kg/m}^3 \)):
Using the formula:
\( h = \frac{P}{\rho g} \)
Substitute values:
\( h = \frac{360 \times 10^3}{13600 \times 9.81} = 2.7 \, \text{m} \)
Atmospheric Pressure (\( P_{\text{atm}} \)):
Atmospheric pressure in terms of mercury:
\( h = 760 \, \text{mmHg} = 0.76 \, \text{mHg} \)
Convert to pressure:
\( P_{\text{atm}} = \rho_{\text{mercury}} g h \)
\( P_{\text{atm}} = 13600 \times 9.81 \times 0.76 = 101396 \, \text{N/m}^2 = 101.3 \, \text{KN/m}^2 \)
Absolute Pressure (\( P_{\text{abs}} \)):
Using the formula:
\( P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} \)
Substitute values:
\( P_{\text{abs}} = 360 + 101.3 = 461.3 \, \text{KN/m}^2 \)
Explanation
- Head in terms of water: The height of a column of water is calculated by dividing the pressure by the product of water density and gravitational acceleration.
- Head in terms of mercury: Similar to water, but using the density of mercury derived from its specific gravity.
- Atmospheric pressure: Atmospheric pressure is converted from height in mercury to pressure using the density of mercury and gravitational acceleration.
- Absolute pressure: It is the sum of gauge pressure and atmospheric pressure, which represents the total pressure in the cylinder.
