A differential manometer is connected at the two points A and B of two pipes as shown in the figure. The pipe A contains a liquid of sp. gr. = 1.5 while pipe B contains a liquid of sp. gr. = 0.9. The pressures at A and B are 1 kgf/cm² and 1.80 kgf/cm² respectively. Find the difference in mercury level in the differential manometer.

Differential Manometer Calculation

Problem Statement

Given Data

Sp. gr. of liquid at A, $S_A = 1.5 \implies \rho_A = 1500 \, \text{kg/m}^3$
Sp. gr. of liquid at B, $S_B = 0.9 \implies \rho_B = 900 \, \text{kg/m}^3$
Pressure at A, $P_A = 1 \, \text{kgf/cm}^2$
Pressure at B, $P_B = 1.8 \, \text{kgf/cm}^2$
Manometer fluid: Mercury ($\rho_{Hg} = 13600 \, \text{kg/m}^3$)

Diagram

Solution

1. Convert Pressures to SI Units (N/m²)

We need to convert the pressures from kgf/cm² to N/m².

$$ P_A = 1 \, \frac{\text{kgf}}{\text{cm}^2} = 1 \times 10^4 \, \frac{\text{kgf}}{\text{m}^2} $$ $$ P_A = 10^4 \times 9.81 \, \frac{\text{N}}{\text{m}^2} = 98100 \, \text{N/m}^2 $$

$$ P_B = 1.8 \, \frac{\text{kgf}}{\text{cm}^2} = 1.8 \times 10^4 \, \frac{\text{kgf}}{\text{m}^2} $$ $$ P_B = 1.8 \times 10^4 \times 9.81 \, \frac{\text{N}}{\text{m}^2} = 176580 \, \text{N/m}^2 $$

2. Set up the Manometric Equation

We establish a datum line at the lower mercury level, X-X. The total pressure at this level must be the same in both the left and right limbs.

$$ P_{\text{left limb at X-X}} = P_{\text{right limb at X-X}} $$

Pressure in the left limb at X-X is the sum of pressure at A, plus the pressure from the 5m column of liquid A, plus the pressure from the mercury column of height 'h'.

$$ P_{\text{left}} = P_A + \rho_A g (3+2) + \rho_{Hg} g h $$

Pressure in the right limb at X-X is the sum of pressure at B, plus the pressure from the column of liquid B of height (h+2)m.

$$ P_{\text{right}} = P_B + \rho_B g (h+2) $$

3. Solve for the Mercury Level Difference (h)

Now, we set the left and right pressure equations equal and solve for 'h'.

$$ P_A + \rho_A g (5) + \rho_{Hg} g h = P_B + \rho_B g (h+2) $$

Substitute the known values.

$$ 98100 + (1500 \times 9.81 \times 5) + (13600 \times 9.81 \times h) = 176580 + (900 \times 9.81 \times (h+2)) $$

To simplify, we can divide the entire equation by 9.81.

$$ \frac{98100}{9.81} + (1500 \times 5) + 13600h = \frac{176580}{9.81} + 900(h+2) $$ $$ 10000 + 7500 + 13600h = 18000 + 900h + 1800 $$ $$ 17500 + 13600h = 19800 + 900h $$

Now, group the 'h' terms and the constant terms.

$$ 13600h - 900h = 19800 - 17500 $$ $$ 12700h = 2300 $$ $$ h = \frac{2300}{12700} \approx 0.1811 \, \text{m} $$

Final Result:

The difference in mercury level is $ h \approx 18.11 \, \text{cm} $.

Explanation of Differential Manometry

A differential manometer is used to measure the pressure difference between two points, which may be in the same pipe or in two different pipes. Unlike a simple manometer that measures pressure relative to the atmosphere, a differential manometer measures one pressure relative to another.

The principle remains the same: the pressures at a common datum line within a continuous fluid are equal. The manometric equation is formed by starting at one point (e.g., pipe A), adding or subtracting the hydrostatic pressures of each fluid column as you move through the manometer to the second point (pipe B). In this case, we balanced the total pressure from each pipe down to the datum line.

Physical Meaning

The result, $h = 18.11 \, \text{cm}$, represents the height difference in the mercury columns needed to balance the complex system of pressures. The pressure in pipe B ($1.8 \, \text{kgf/cm}^2$) is significantly higher than in pipe A ($1 \, \text{kgf/cm}^2$).

However, the liquid in pipe A is much denser (sp. gr. 1.5 vs 0.9) and is at a higher elevation relative to the manometer fluid. The calculation shows how all these factors—pipe pressures, fluid densities, and column heights—combine to produce a final, measurable difference in the manometer. The mercury column must be higher on the left side to help the lower-pressure system (Pipe A's side) balance the higher-pressure system (Pipe B's side).