Problem Statement
Determine the pressure difference between two points A and B in the figure.
Solution
Given Data:
- Water: \(\gamma = 9.81 \, \text{kN/m}^3\)
- Mercury: \(\gamma_{\text{m}} = 13.6 \times 9.81 = 133.416 \, \text{kN/m}^3\)
- Benzene: \(\gamma_{\text{B}} = 0.88 \times 9.81 = 8.6328 \, \text{kN/m}^3\)
- Kerosene: \(\gamma_{\text{K}} = 0.82 \times 9.81 = 8.0442 \, \text{kN/m}^3\)
- Air: \(\gamma_{\text{air}} = 0.01177 \, \text{kN/m}^3\)
Solution Steps:
Starting from point A, we can write the pressure balance equation:
\begin{align*}
P_A &+ \gamma_{\text{B}} \times 0.21 – \gamma_{\text{m}} \times 0.09 \\
&- \gamma_{\text{K}} \times (0.41 – 0.09) \\
&+ \gamma \times (0.41 – 0.15) – \gamma_{\text{air}} \times 0.1 = P_B
\end{align*}
Substituting the values:
\begin{align*}
P_A &+ 8.6328 \times 0.21 – 133.416 \times 0.09 \\
&- 8.0442 \times 0.32 + 9.81 \times 0.26 \\
&- 0.01177 \times 0.1 = P_B
\end{align*}
Final result:
\(P_A – P_B = 10.22 \, \text{kPa}\)
Explanation
This problem demonstrates the calculation of pressure difference in a multi-fluid system:
- The pressure balance equation accounts for all fluid contributions (benzene, mercury, kerosene, water, and air).
- Each fluid’s contribution is calculated using its specific weight and height difference.
- The final pressure difference shows the net effect of all fluid columns.
Physical Meaning
- Pressure Difference: The calculated value of 10.22 kPa represents the net pressure difference caused by the various fluid columns.
- Fluid Properties: The significant contribution of mercury (due to its high specific weight) compared to lighter fluids like air demonstrates how fluid density affects pressure.
- Hydrostatic Equilibrium: The system demonstrates how multiple fluids maintain balance through pressure differences.
