For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the fig., find (a) the elevation of liquid surface in piezometer A, (b) the elevation of liquid surface in piezometer B, and (c) the total pressure at the bottom of the tank.

For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the fig., find (a) the elevation of liquid surface in piezometer A, (b) the elevation of liquid surface in piezometer B, and (c) the total pressure at the bottom of the tank.
For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the fig., find (a) the elevation of liquid surface in piezometer A, (b) the elevation of liquid surface in piezometer B, and (c) the total pressure at the bottom of the tank.

Problem Statement

For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the figure, find:

  1. (a) the elevation of liquid surface in piezometer A
  2. (b) the elevation of liquid surface in piezometer B
  3. (c) the total pressure at the bottom of the tank

Solution

Given:

  • Specific weight of liquid A (\( \gamma_A \)) = \( 0.72 \times 9810 = 7063.2 \, \text{N/m}^3 \)
  • Specific weight of liquid B (\( \gamma_B \)) = \( 2.36 \times 9810 = 23151.6 \, \text{N/m}^3 \)

(a) Elevation of Liquid Surface in Piezometer A:

The elevation of the liquid surface in piezometer A is equal to the elevation of liquid A in the tank:

Elevation = \( 2.1 \, \text{m} \)

(b) Elevation of Liquid Surface in Piezometer B:

Pressure due to liquid A at the interface:

\( P_A = \gamma_A \cdot h_{\text{liquid A}} \)

Substitute the values:

\( P_A = 7063.2 \times (2.1 – 0.3) \)

\( P_A = 12713.8 \, \text{N/m}^2 \)

Equivalent head for liquid B due to \( P_A \):

\( h_A = \frac{P_A}{\gamma_B} \)

Substitute the values:

\( h_A = \frac{12713.8}{23151.6} \)

\( h_A = 0.55 \, \text{m} \)

Elevation of the liquid surface in piezometer B:

Elevation = Elevation of liquid B + \( h_A \)

Elevation = \( 0.3 + 0.55 \)

Elevation = \( 0.85 \, \text{m} \)

(c) Total Pressure at the Bottom of the Tank:

The total pressure at the bottom is the sum of the pressures due to liquids A and B:

\( P_{\text{total}} = \gamma_A \cdot h_{\text{liquid A}} + \gamma_B \cdot h_{\text{liquid B}} \)

Substitute the values:

\( P_{\text{total}} = 7063.2 \cdot (2.1 – 0.3) + 23151.6 \cdot 0.3 \)

\( P_{\text{total}} = 19659 \, \text{N/m}^2 \)

Explanation

This problem involves calculating the elevation of liquids in piezometers and the total pressure at the bottom of a tank containing immiscible liquids:

  1. The elevation in piezometer A reflects the height of liquid A in the tank.
  2. The elevation in piezometer B accounts for the pressure at the interface and converts it to an equivalent height of liquid B.
  3. The total pressure at the bottom is the sum of the contributions from the two liquids, based on their respective heights and specific weights.

Physical Meaning

  1. Piezometer Elevations: The elevations indicate the hydrostatic pressure at specific points, converted into equivalent liquid heights.
  2. Specific Weights: The specific weights of the liquids determine the pressure contributions for given heights.
  3. Total Pressure: The pressure at the bottom reflects the combined effects of the two immiscible liquids, providing a measure of the force per unit area.

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