Figure below shows a pipe containing a liquid of sp gr 0.8 connected to a single column micromanometer. The area of reservoir is 60 times that of the tube.

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.

Given data:

Specific gravity of oil (γ_oil) = 0.8
Specific gravity of mercury (γ_mercury) = 13.6
Area of reservoir = 60 × area of the tube
Manometer liquid rise = 0.25 m

Solution

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:

\( \gamma_{\text{oil}} \cdot y = \gamma_{\text{mercury}} \cdot h_1 \quad \text{(a)} \)

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:

\( \text{Area of reservoir} \cdot \Delta y = \text{Area of tube} \cdot 0.25 \) \( 60 \cdot \text{Area of tube} \cdot \Delta y = \text{Area of tube} \cdot 0.25 \) \( \Delta y = 0.0041 \; \text{m} \)

Pressure Balance with Applied Pressure

Equating pressure at the new levels:

\( P_A + \gamma_{\text{oil}} \cdot (y + \Delta y) = \gamma_{\text{mercury}} \cdot (0.25 + h_1 + \Delta y) \quad \text{(b)} \)

Subtracting equation (a) from (b):

\( P_A = \gamma_{\text{mercury}} \cdot (0.25 + \Delta y) - \gamma_{\text{oil}} \cdot \Delta y \)

Substituting values:

\( P_A = 13.6 \cdot 9810 \cdot (0.25 + 0.0041) - 0.8 \cdot 9810 \cdot 0.0041 \) \( P_A = 33868 \; \text{N/m}^2 \)

Pressure in the pipe:\( P_A = 33868 \; \text{N/m}^2 \)

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.

Figure below shows a pipe containing a liquid of sp gr 0.8 connected to a single column micromanometer. The area of reservoir is 60 times that of the tube. The manometer liquid is mercury. Find the pressure in the pipe.