The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the left limb is connected to a pipe in which a fluid of sp. gr. 0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 20 cm.

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U-Tube Manometer Pressure Calculation

Problem Statement

The right limb of a simple U-tube manometer containing mercury is open to the atmosphere while the left limb is connected to a pipe in which a fluid of sp. gr. 0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in the two limbs is 20 cm.

Given Data

  • Specific gravity of fluid in pipe, \(S_1 = 0.9\)
  • Specific gravity of mercury, \(S_2 = 13.6\)
  • Difference in mercury levels, \(h_2 = 20 \, \text{cm} = 0.2 \, \text{m}\)
  • Pipe center is 12 cm below the right limb mercury level.
  • Density of water, \(\rho_{\text{water}} = 1000 \, \text{kg/m}^3\)

Diagram

Diagram of the U-tube manometer setup

Solution

1. Calculate Densities and Heights

First, calculate the densities of the fluid and mercury from their specific gravities. All heights must be in meters.

$$ \rho_1 = S_1 \times \rho_{\text{water}} = 0.9 \times 1000 = 900 \, \text{kg/m}^3 $$ $$ \rho_2 = S_2 \times \rho_{\text{water}} = 13.6 \times 1000 = 13600 \, \text{kg/m}^3 $$

The height of the fluid column in the left limb above the datum line A-A (\(h_1\)) is the difference between the mercury levels minus the 12 cm offset.

$$ h_1 = (20 \, \text{cm} - 12 \, \text{cm}) = 8 \, \text{cm} = 0.08 \, \text{m} $$ $$ h_2 = 20 \, \text{cm} = 0.2 \, \text{m} $$

2. Apply the Manometric Equation

We can find the pressure in the pipe by balancing the pressures in the left and right limbs at the datum line A-A. The pressure at this level must be equal in both limbs.

$$ P_{\text{left limb at A-A}} = P_{\text{right limb at A-A}} $$ $$ p + \rho_1 g h_1 = \rho_2 g h_2 $$

Here, \(p\) is the gauge pressure of the fluid in the pipe.

3. Solve for Pipe Pressure (\(p\))

Rearrange the equation to solve for \(p\) and substitute the known values.

$$ p = \rho_2 g h_2 - \rho_1 g h_1 $$

Calculate the pressure from the right (mercury) and left (fluid) columns separately.

$$ P_{\text{right}} = 13600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.2 \, \text{m} = 26683.2 \, \text{N/m}^2 $$ $$ P_{\text{left}} = 900 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.08 \, \text{m} = 706.32 \, \text{N/m}^2 $$

Now, find the pipe pressure \(p\).

$$ p = 26683.2 - 706.32 $$ $$ p = 25976.88 \, \text{N/m}^2 $$
Final Result:

The pressure of the fluid in the pipe is approximately \( p \approx 25977 \, \text{N/m}^2 \) or \( 25.98 \, \text{kPa} \).

Explanation of Manometry

A U-tube manometer is a device used to measure the pressure of a fluid by balancing it against a column of another, heavier fluid (often mercury). The principle is based on the fact that the pressure at any two points at the same horizontal level within a continuous fluid at rest is equal.

By choosing a datum line (A-A) at the interface of the two fluids in the left limb, we can create a pressure equation. The total pressure pushing down on the left side of the datum must equal the total pressure pushing down on the right side. This allows us to solve for the unknown pressure in the pipe.

Physical Meaning

The result, \(25977 \, \text{N/m}^2\), is the gauge pressure of the fluid in the pipe, meaning it is the pressure above atmospheric pressure. This pressure is strong enough to support the weight of a small column of its own fluid (8 cm) and also push down the mercury in its limb, causing it to rise 20 cm higher in the opposite limb.

This demonstrates how a dense fluid like mercury can be used to measure much larger pressures without requiring an impractically tall column. A pressure that can support a 20 cm column of mercury is significant, equivalent to the pressure you would feel under approximately 2.65 meters of water.

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