Problem Statement
The tank in the figure contains oil of specific gravity \( 0.75 \). Determine the reading of gauge A in \( \text{N/m}^2 \).
Solution
Given:
- Specific gravity of oil = \( 0.75 \)
- Specific weight of oil (\( \gamma_{\text{oil}} \)) = \( 0.75 \times 9810 = 7357.5 \, \text{N/m}^3 \)
- Specific weight of mercury (\( \gamma_{\text{m}} \)) = \( 13600 \times 9.81 = 133416 \, \text{N/m}^3 \)
- Take atmospheric pressure as 0 for gauge pressure.
Pressure at Gauge A (\( P_A \)):
Using the hydrostatic pressure equation and starting from point X, neglecting air:
\( P_A = 0 – \gamma_{\text{m}} h_{\text{m}} + \gamma_{\text{oil}} h_{\text{oil}} \)
Substitute the values:
\( P_A = – (133416 \times 0.2) + (7357.5 \times 3) \)
Final Value:
\( P_A = -4610.7 \, \text{N/m}^2 \)
Explanation
This problem involves determining the gauge pressure at a specific point within a tank containing oil and mercury:
- The pressure at gauge A is influenced by the height of the mercury and oil columns, taking into account their respective specific weights.
- Since atmospheric pressure is taken as 0 (gauge pressure), the equation simplifies to include only the contributions from mercury and oil.
- The negative value indicates a pressure below the atmospheric reference point, relative to the system’s configuration.
Physical Meaning
- Specific Gravity: Specific gravity relates the density of a fluid to that of water, directly affecting the pressure it exerts for a given height.
- Hydrostatic Pressure: The pressure at a point within a fluid column is determined by the height and specific weight of the fluid above it.
- Gauge Pressure: Gauge pressure measures pressure relative to atmospheric pressure, simplifying the analysis by setting atmospheric pressure to 0.
