Problem Statement
A pipe containing a liquid with a specific gravity of 0.8 is connected to a single-column micromanometer. The reservoir’s area is 60 times that of the tube. The manometer liquid is mercury. Find the pressure in the pipe.
- Specific gravity of oil (γoil) = 0.8
- Specific gravity of mercury (γmercury) = 13.6
- Area of reservoir = 60 × area of the tube
- Manometer liquid rise = 0.25 m
Solution
1. Initial Pressure Balance
When the manometer is not connected to the pipe, the mercury levels in the reservoir and tube are balanced. The pressure equation is:
2. Volume Conservation
Due to the applied pressure, the manometric liquid in the reservoir drops by \( \Delta y \), causing a rise of 0.25 m in the tube. Using volume conservation:
3. Pressure Balance with Applied Pressure
Equating pressure at the new levels:
Subtracting equation (a) from (b):
Substituting values:
Explanation
- Volume Conservation: The drop in mercury in the reservoir is equal to the rise in mercury in the tube, adjusted for their areas.
- Pressure Balance: The applied pressure causes a shift in the mercury levels, which is used to calculate the pressure in the pipe.
- Final Calculation: By combining the pressure and volume relationships, the exact pressure in the pipe is determined.
Physical Meaning
This numerical demonstrates the principles of fluid statics, specifically the interplay between pressure, volume conservation, and liquid levels in a micromanometer. These calculations are crucial for accurate pressure measurements in engineering systems.
