Find out the differential reading 'h' of an inverted U-tube manometer containing oil of specific gravity 0.7 as the manometric fluid when connected across pipes A and B as shown in the figure, conveying liquids of specific gravities 1.2 and 1.0 and immiscible with manometric fluid. Pipes A and B are located at the same level and assume the pressures at A and B to be equal.

Inverted Manometer with Equal Pipe Pressures

Problem Statement

Given Data

Sp. gr. of liquid in pipe A, $S_A = 1.2 \implies \rho_A = 1200 \, \text{kg/m}^3$
Sp. gr. of liquid in pipe B, $S_B = 1.0 \implies \rho_B = 1000 \, \text{kg/m}^3$
Sp. gr. of manometer oil, $S_{oil} = 0.7 \implies \rho_{oil} = 700 \, \text{kg/m}^3$
Pressures at pipes are equal: $P_A = P_B$

Diagram

Solution

1. Set up the Manometric Equation

We establish a datum line at the lower oil-liquid interface, X-X, in the left limb. The pressure at this level in the left limb must equal the pressure at the same level in the right limb.

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

Pressure in the left limb at X-X is the pressure at A minus the pressure from the 30 cm column of liquid A.

$$ P_{\text{left}} = P_A - \rho_A g (0.3) $$

Pressure in the right limb at X-X is the pressure at B, minus the pressure from the 30 cm column of liquid B, minus the pressure from the oil column of height 'h'.

$$ P_{\text{right}} = P_B - \rho_B g (0.3) - \rho_{oil} g h $$

2. Solve for the Differential Reading (h)

Now, we set the left and right pressure equations equal.

$$ P_A - \rho_A g (0.3) = P_B - \rho_B g (0.3) - \rho_{oil} g h $$

The problem states that the pressures in the pipes are equal ($P_A = P_B$). Therefore, these terms cancel out from both sides.

$$ - \rho_A g (0.3) = - \rho_B g (0.3) - \rho_{oil} g h $$

We can divide the entire equation by $g$ and multiply by -1 to simplify.

$$ \rho_A (0.3) = \rho_B (0.3) + \rho_{oil} h $$

Rearrange to solve for 'h' and substitute the density values.

$$ \rho_{oil} h = \rho_A (0.3) - \rho_B (0.3) $$ $$ 700 h = 1200(0.3) - 1000(0.3) $$ $$ 700 h = 360 - 300 $$ $$ 700 h = 60 $$ $$ h = \frac{60}{700} \approx 0.0857 \, \text{m} $$

Final Result:

The differential reading is $ h \approx 8.57 \, \text{cm} $.

Explanation of Inverted Manometry

An inverted U-tube manometer uses a light fluid (like oil) to measure pressure differences between two points, typically in a liquid system. The key principle is balancing the pressures at a common horizontal datum line. For an inverted manometer, pressure is subtracted as you move upwards through a fluid column ($P_{bottom} - \rho g h = P_{top}$).

In this unique case, the pressures in the pipes ($P_A$ and $P_B$) are equal. This means the differential reading 'h' is not caused by a pressure difference in the pipes, but rather by the difference in the weights of the two liquid columns in the manometer limbs.

Physical Meaning

The result, $h = 8.57 \, \text{cm}$, shows that even when the pressures in the connected pipes are identical, a differential reading can still exist if the fluids in the pipes have different densities.

The liquid in pipe A (sp. gr. 1.2) is denser than the liquid in pipe B (sp. gr. 1.0). Therefore, the 30 cm column of liquid A exerts more downward pressure than the 30 cm column of liquid B. To balance this, the lighter oil in the manometer is pushed down on the right side and up on the left side, creating the reading 'h'. The pressure exerted by this 8.57 cm oil column perfectly compensates for the pressure difference created by the different densities of the pipe liquids.