A U-tube differential manometer connects two pressure pipes A and B. Pipe A contains carbon tetrachloride having a specific gravity 1.594 under a pressure of 11.772 N/cm². Pipe B contains oil of sp. gr. 0.8 under a pressure of 11.772 N/cm². The pipe A lies 2.5 m above pipe B. Find the difference of pressure measured by mercury as fluid filling U-tube.

U-Tube Manometer with Two Fluids

Problem Statement

Given Data & Constants

Pressure in Pipe A, $P_A = 11.772 \, \text{N/cm}^2$
Pressure in Pipe B, $P_B = 11.772 \, \text{N/cm}^2$
Specific gravity of CCl₄ (Pipe A), $S_A = 1.594$
Specific gravity of Oil (Pipe B), $S_B = 0.8$
Elevation difference, $z_A - z_B = 2.5 \, \text{m}$
Specific gravity of mercury, $S_m = 13.6$
Density of water, $\rho_w = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Convert Units and Calculate Densities

First, we convert the pressure to standard units (N/m²) and calculate the densities of the fluids.

$$ P_A = P_B = 11.772 \, \frac{\text{N}}{\text{cm}^2} \times \frac{10000 \, \text{cm}^2}{1 \, \text{m}^2} = 117720 \, \text{N/m}^2 $$ $$ \rho_A (\text{CCl}_4) = S_A \times \rho_w = 1.594 \times 1000 = 1594 \, \text{kg/m}^3 $$ $$ \rho_B (\text{Oil}) = S_B \times \rho_w = 0.8 \times 1000 = 800 \, \text{kg/m}^3 $$ $$ \rho_m (\text{Mercury}) = S_m \times \rho_w = 13.6 \times 1000 = 13600 \, \text{kg/m}^3 $$

2. Set up the Manometer Pressure Balance Equation

To solve this, we balance the piezometric pressure between the two pipes with the pressure exerted by the manometer fluid. The difference in piezometric pressure between A and B must be balanced by the head of the manometric fluid. Let $h$ be the difference in mercury levels.

Difference in Piezometric Pressure = Manometer Head

$$ (P_A + \rho_A g z_A) - (P_B + \rho_B g z_B) = h g (\rho_m - \rho_B) $$

Here, the term $(\rho_m - \rho_B)$ is used because the oil from pipe B is the fluid being displaced by the mercury column in the right limb of the manometer.

3. Solve for the Mercury Level Difference (h)

Let's set the elevation of pipe B as the datum, so $z_B = 0$ and $z_A = 2.5 \, \text{m}$. Since $P_A = P_B$, the pressure terms cancel out.

$$ (\rho_A g z_A) - (\rho_B g z_B) = h g (\rho_m - \rho_B) $$ $$ \text{Dividing by g gives:} $$ $$ \rho_A z_A - \rho_B z_B = h (\rho_m - \rho_B) $$ $$ (1594)(2.5) - (800)(0) = h (13600 - 800) $$ $$ 3985 = h (12800) $$ $$ h = \frac{3985}{12800} $$ $$ h \approx 0.3113 \, \text{m} $$

Final Result:

The difference in the mercury level is $ h \approx 31.13 \, \text{cm} $.

Explanation of the Pressure Balance

Even though the gauge pressures in Pipe A and Pipe B are identical, they are at different elevations. The total energy at each point, represented by the piezometric head ($P + \rho g z$), is different.

Pipe A is 2.5 meters higher and contains a much denser fluid (CCl₄). This gives it a significantly higher piezometric pressure than Pipe B. This difference in total pressure must be balanced by the heavy column of mercury in the U-tube. The resulting 31.13 cm difference in mercury level is what's required to hold back the higher energy from the A-side.