A simple U-tube manometer containing mercury is connected to a pipe in which a fluid of specific gravity 0.8 and having vacuum pressure is flowing. The other end of the manometer is open to the atmosphere. Find the vacuum pressure in the pipe, if the difference of mercury level in the two limbs is 40 cm and the height of fluid in the left limb from the centre of the pipe is 15 cm below.

U-Tube Manometer Vacuum Pressure

Problem Statement

Given Data

Sp. gr. of fluid, $S_1 = 0.8$
Sp. gr. of mercury, $S_2 = 13.6$
Difference in mercury level, $h_2 = 40 \, \text{cm} = 0.4 \, \text{m}$
Height of fluid from pipe center, $h_1 = 15 \, \text{cm} = 0.15 \, \text{m}$

Solution

1. Define Densities

Density of the fluid ($\rho_1$):

$$ \rho_1 = S_1 \times \rho_{\text{water}} $$ $$ \rho_1 = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 $$

Density of mercury ($\rho_2$):

$$ \rho_2 = S_2 \times \rho_{\text{water}} $$ $$ \rho_2 = 13.6 \times 1000 \, \text{kg/m}^3 = 13600 \, \text{kg/m}^3 $$

2. Set up the Manometer Equation

For a vacuum, the atmospheric pressure pushes the mercury up on the pipe side. Let's set the datum line at the lower mercury level (the side open to the atmosphere). The pressure on both sides of the tube at this datum line must be equal.

$$ p_{\text{left at datum}} = p_{\text{right at datum}} $$ $$ p_{\text{pipe}} + \rho_1 g h_1 + \rho_2 g h_2 = p_{\text{atm}} $$

We need to find the vacuum pressure, which is the gauge pressure ($p_{\text{gauge}} = p_{\text{pipe}} - p_{\text{atm}}$).

$$ p_{\text{gauge}} = - (\rho_1 g h_1 + \rho_2 g h_2) $$

3. Calculate the Vacuum Pressure

Now we substitute the known values into the equation.

$$ p_{\text{gauge}} = - \left[ (800 \times 9.81 \times 0.15) + (13600 \times 9.81 \times 0.4) \right] $$ $$ p_{\text{gauge}} = - [ (1177.2) + (53366.4) ] $$ $$ p_{\text{gauge}} = - 54543.6 \, \text{N/m}^2 $$

Converting to N/cm²:

$$ p_{\text{gauge}} = -54543.6 \, \text{N/m}^2 \times \frac{1 \, \text{cm}^2}{100^2 \, \text{m}^2} $$ $$ p_{\text{gauge}} \approx -5.454 \, \text{N/cm}^2 $$

Final Result:

The vacuum pressure in the pipe is $ -54,543.6 \, \text{N/m}^2 $ or $ -5.454 \, \text{N/cm}^2 $.

Explanation of the Manometer Principle

A U-tube manometer measures pressure differences. In this case, it measures the difference between the pipe pressure and the atmospheric pressure.

Because the pipe has a vacuum pressure (pressure lower than atmospheric), the higher atmospheric pressure on the open end pushes the mercury down on that side and up on the pipe side, as shown in the diagram. To find the pipe pressure, we can balance the pressures at a common reference point (datum). By choosing the datum at the lowest mercury level (the atmospheric side), we can state that the atmospheric pressure on the right must be equal to the sum of all pressures on the left: the pipe pressure, the pressure from the fluid column ($h_1$), and the pressure from the mercury column difference ($h_2$).

Physical Meaning

The calculated pressure of -54,543.6 N/m² is a gauge pressure. The negative sign is critical—it confirms that the pressure in the pipe is below the surrounding atmospheric pressure, which is the definition of a vacuum.

This value represents the magnitude of the suction. It means the absolute pressure inside the pipe is 54,543.6 N/m² less than the local atmospheric pressure. This type of measurement is essential for monitoring and controlling industrial processes that operate under vacuum, such as vacuum distillation, freeze-drying, or in the suction lines of pumps.