Determine the bulk modulus of elasticity of a liquid if the pressure of the liquid is increased from 70 N/cm² to 130 N/cm². The volume of the liquid decreases by 0.15 percent.

Bulk Modulus of Elasticity Calculation

Problem Statement

Determine the bulk modulus of elasticity of a liquid if the pressure of the liquid is increased from 70 N/cm² to 130 N/cm². The volume of the liquid decreases by 0.15 percent.

Given Data

  • Initial Pressure, \(p_1 = 70 \, \text{N/cm}^2\)
  • Final Pressure, \(p_2 = 130 \, \text{N/cm}^2\)
  • Percentage Decrease in Volume = 0.15%

Solution

1. Calculate the Change in Pressure (\(dp\))

The change in pressure is the difference between the final and initial pressures.

$$ dp = p_2 - p_1 = 130 \, \text{N/cm}^2 - 70 \, \text{N/cm}^2 = 60 \, \text{N/cm}^2 $$

2. Determine the Volumetric Strain (\(dV/V\))

The volume decreases by 0.15%, so the volumetric strain is negative.

$$ \frac{dV}{V} = -\frac{0.15}{100} = -0.0015 $$

3. Calculate the Bulk Modulus (\(K\))

The bulk modulus of elasticity is defined as the ratio of the change in pressure to the negative of the volumetric strain.

$$ K = \frac{dp}{-\frac{dV}{V}} $$ $$ K = \frac{60 \, \text{N/cm}^2}{-(-0.0015)} = \frac{60}{0.0015} \, \text{N/cm}^2 $$ $$ K = 40,000 \, \text{N/cm}^2 = 4 \times 10^4 \, \text{N/cm}^2 $$
Final Result:

The bulk modulus of elasticity of the liquid is \( K = 4 \times 10^4 \, \text{N/cm}^2 \).

Explanation of Bulk Modulus

The Bulk Modulus of Elasticity (\(K\)) is a measure of a substance's resistance to uniform compression. It describes how much pressure is needed to cause a given fractional decrease in volume.

The defining formula is: $$ K = -V \frac{dp}{dV} $$ Rearranged for calculation, it becomes \( K = \frac{dp}{-dV/V} \). The negative sign is crucial because an increase in pressure (\(dp > 0\)) results in a decrease in volume (\(dV < 0\)). The negative sign ensures that the bulk modulus \(K\) is always a positive value.

Physical Meaning

A high bulk modulus indicates that a substance is difficult to compress. Liquids, like the one in this problem, generally have very high bulk moduli and are often considered to be "incompressible" for many practical engineering applications. For instance, water has a bulk modulus of approximately \(2.2 \times 10^9\) N/m² (or \(2.2 \times 10^5\) N/cm²).

This property is the fundamental principle behind hydraulics. Because liquid is nearly incompressible, pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel (Pascal's Principle). This allows for the multiplication of force in systems like hydraulic lifts, brakes, and jacks.

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