Problem Statement
A manometer connected to a pipe indicates a negative gauge pressure of \( -50 \, \text{mmHg} \). What is the absolute pressure in the pipe in \( \text{N/m}^2 \) if the atmospheric pressure is \( 1 \, \text{bar} \)?
Solution
Given:
- Atmospheric pressure (\( P_{\text{atm}} \)) = \( 1 \, \text{bar} = 1 \times 10^5 \, \text{N/m}^2 \)
- Head (\( h \)) = \( -50 \, \text{mmHg} = -0.05 \, \text{mHg} \)
Density of Mercury (\( \rho \)):
\( \rho = \text{sp gr} \times \text{density of water} \)
\( \rho = 13.6 \times 1000 = 13600 \, \text{kg/m}^3 \)
Gauge Pressure (\( P_{\text{gauge}} \)):
Using the formula:
\( P_{\text{gauge}} = \rho g h \)
Substitute values:
\( P_{\text{gauge}} = -13600 \times 9.81 \times 0.05 = -6671 \, \text{N/m}^2 \)
Absolute Pressure (\( P_{\text{abs}} \)):
Using the formula:
\( P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} \)
Substitute values:
\( P_{\text{abs}} = -6671 + 1 \times 10^5 = 93329 \, \text{N/m}^2 = 93.3 \, \text{kN/m}^2 \)
Explanation
- Gauge Pressure: The gauge pressure is calculated using the negative height of mercury (\( h \)), resulting in a negative value indicating a vacuum.
- Atmospheric Pressure: The atmospheric pressure is converted to \( \text{N/m}^2 \) for compatibility with the gauge pressure calculation.
- Absolute Pressure: The absolute pressure is the sum of the atmospheric pressure and the gauge pressure, providing the total pressure in the pipe.
- This calculation ensures that negative gauge pressures are properly incorporated to find the actual (absolute) pressure.
