A single column manometer is connected to a pipe containing a liquid of sp. gr. 0.9 as shown in the figure. Find the pressure in the pipe if the area of the reservoir is 100 times the area of the tube for the manometer reading shown. The specific gravity of mercury is 13.6.

Single Column Manometer Calculation

Problem Statement

Given Data

Sp. gr. of pipe liquid, $S_1 = 0.9$
Sp. gr. of mercury, $S_2 = 13.6$
Area Ratio, $A/a = 100$
Height of liquid from pipe center, $h_1 = 20 \, \text{cm} = 0.2 \, \text{m}$
Rise of mercury in right limb, $h_2 = 40 \, \text{cm} = 0.4 \, \text{m}$

Solution

1. Define Densities

Density of pipe liquid ($\rho_1$):

$$ \rho_1 = S_1 \times \rho_{\text{water}} $$ $$ \rho_1 = 0.9 \times 1000 \, \text{kg/m}^3 = 900 \, \text{kg/m}^3 $$

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. Calculate Fall in Reservoir Level ($\Delta h$)

When the mercury rises by $h_2$ in the narrow tube, the level in the wide reservoir must fall by a small amount $\Delta h$. By conservation of volume:

$$ A \times \Delta h = a \times h_2 $$ $$ \Delta h = \frac{a}{A} h_2 $$ $$ \Delta h = \frac{1}{100} \times 0.4 \, \text{m} $$ $$ \Delta h = 0.004 \, \text{m} $$

3. Set up the Manometer Equation

Let's set the datum line at the final, lower mercury level in the reservoir. The pressure on both sides of the system at this datum line must be equal.

$$ p_{\text{left at datum}} = p_{\text{right at datum}} $$ $$ p_{\text{pipe}} + \rho_1 g (h_1 + \Delta h) = \rho_2 g (h_2 + \Delta h) $$

Rearranging to solve for the pressure in the pipe, $p_{\text{pipe}}$:

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

4. Calculate the Pipe Pressure

Now we substitute the known values into the equation.

$$ p_{\text{pipe}} = [13600 \times 9.81 \times (0.4 + 0.004)] - [900 \times 9.81 \times (0.2 + 0.004)] $$ $$ p_{\text{pipe}} = [133416 \times 0.404] - [8829 \times 0.204] $$ $$ p_{\text{pipe}} = 53898 - 1801.1 $$ $$ p_{\text{pipe}} = 52096.9 \, \text{N/m}^2 $$

Converting to N/cm²:

$$ p_{\text{pipe}} = 52096.9 \, \text{N/m}^2 \times \frac{1 \, \text{cm}^2}{100^2 \, \text{m}^2} $$ $$ p_{\text{pipe}} \approx 5.21 \, \text{N/cm}^2 $$

Final Result:

The pressure in the pipe is approximately $ 52,097 \, \text{N/m}^2 $ or $ 5.21 \, \text{N/cm}^2 $.

Explanation of the Single Column Manometer

A single column manometer is a modified U-tube manometer with a large reservoir on one side. Its main advantage is that the change in the liquid level in the reservoir ($\Delta h$) is very small and can often be neglected. However, for high precision, as in this problem, it must be calculated.

The pressure in the pipe pushes the liquid down in the reservoir, causing the manometric fluid (mercury) to rise significantly in the narrow tube. By balancing the hydrostatic pressures at a common reference point (the final level in the reservoir), we can accurately determine the unknown pipe pressure.

Physical Meaning

The final result of 5.21 N/cm² is the gauge pressure in the pipe. This means the pressure inside the pipe is 52,097 N/m² higher than the local atmospheric pressure.

The calculation demonstrates the sensitivity of this type of manometer. The large area ratio (100:1) means that even a tiny, almost unnoticeable drop in the reservoir level ($\Delta h = 4$ mm) corresponds to a large, easily readable rise in the narrow tube ($h_2 = 400$ mm). This design allows for more precise readings of small to moderate pressure differences compared to a standard U-tube manometer with equal-sized limbs.

A single column manometer is connected to a pipe containing a liquid of sp. gr. 0.9 as shown in the figure. Find the pressure in the pipe if the area of the reservoir is 100 times the area of the tube for the manometer reading shown. The specific gravity of mercury is 13.6.

Problem Statement

Given Data