Problem Statement
In the left-hand tank, the air pressure is \(-225 \, \text{mm of Hg}\). Determine the elevation of the gauge liquid in the right-hand column at point A.
Solution
Given:
- Air pressure in the left-hand tank: \(-225 \, \text{mm Hg} = -0.225 \, \text{m Hg}\)
- Specific weight of water (\( \gamma \)): \( 9810 \, \text{N/m}^3 \)
- Specific weight of oil (\( \gamma_{\text{oil}} \)): \( 0.8 \times 9810 = 7848 \, \text{N/m}^3 \)
- Specific weight of mercury (\( \gamma_{\text{m}} \)): \( 13600 \times 9.81 = 133416 \, \text{N/m}^3 \)
- Specific weight of liquid (\( \gamma_{\text{liquid}} \)): \( 1.6 \times 9810 = 15696 \, \text{N/m}^3 \)
Air Pressure:
The air pressure in the left-hand tank is given by:
\( P_{\text{air}} = -0.225 \times \gamma_{\text{m}} \)
Substitute the values:
\( P_{\text{air}} = -0.225 \times 133416 \)
Final Value:
\( P_{\text{air}} = -30018.6 \, \text{N/m}^2 \)
Pressure Equation:
From the pressure balance equation at point A:
\( 233.5 = 101.3 + \gamma_{\text{oil}} h_{\text{oil}} + \gamma_{\text{water}} h_{\text{water}} + \gamma_{\text{mercury}} h_{\text{mercury}} \)
Substitute the values and simplify:
\( 233.5 = 101.3 + (8.829 \times 1.5) + (9.81 \times 0.2) + (133.416 \times 0.4) \)
Simplify further to find the result.
Elevation at A:
The elevation at point A is:
\( \text{Elevation at A} = 106 – h \)
Substitute the value of \( h \):
\( \text{Elevation at A} = 106 – 5.96 \)
Final Value:
\( \text{Elevation at A} = 100.04 \, \text{m} \)
Explanation
This problem determines the elevation of a liquid in a gauge using hydrostatic principles:
- The air pressure in the left-hand tank contributes negatively to the overall pressure due to its value being below atmospheric pressure.
- The balance of pressures includes contributions from oil, liquid, and water columns, requiring the calculation of an equivalent liquid height.
- The resulting elevation at point A accounts for the reduction caused by the calculated liquid height.
Physical Meaning
- Gauge Liquid Elevation: This represents the height of the liquid in the column relative to a reference point, adjusted for pressure contributions.
- Specific Weight: Specific weight impacts the pressure exerted by each fluid column in the system.
- Negative Pressure: The negative air pressure in the left-hand tank highlights the effect of sub-atmospheric conditions on the overall system.
