Question

Fractional change in Volume of oil is 1 percent. when a pressure of 2x10^7
Nm^-2 is applied. Calculate the bulk modulus and its compressibility:​

Answer 1

Answer:

• Bulk Modulus = 2×10⁹Nm⁻²
• Compressibility = 5×10⁻¹⁰ m²N⁻¹

Explanation:

Topic

• Fluid Mechanics

According to the Question

• Fractional Change in volume = 1 %
• Pressure ,P = 2×10⁷ Nm⁻²

We have to calculate the bulk modulus & its compressibility .

Firstly we calculate here the fractional change .

$\boxed{\bf{Fractional ~Change = \frac{\delta\;V}{V}}}$

By putting the value we get

$\sf\implies\; 1\% = \frac{\delta\;V}{V}\\\\\sf\implies\; \frac{1}{100} = \frac{\delta\;V}{V}\\\\\sf\implies\; 0.01 = \frac{\delta\;V}{V} -------(i)$

Now, calculating the Bulk Modulus.

$\boxed{\bf{B = \bf\frac{P}{\frac{\delta\;V}{V} } }}-------(ii)$

Now , putting the value of equation (i) in equation (ii) we get

$\sf\implies\;B = \sf\frac{2\times\;10^{7}}{0.01} \\\\\sf\implies\;B = \sf\ 2\times10^{9}Nm^{-2}$

Now, calculating its Compressibility

Compressibility is calculated by

$\boxed{\bf{C = \frac{1}{B} }}$

putting the value we get

$\sf\implies\; C = \frac{1}{2\times10^{9}} \\\\\sf\implies\; C = \frac{0.5}{10^{9}} \\\\\sf\implies\; C = 5\times^{-10}m^{2}N^{-1}$

• Hence, Bulk modulus is 2×10⁹Nm⁻²

• Hence, its Compressibility is 5×10⁻¹⁰ m²N⁻¹

Answer 2

Information provided with us:

Fractional change in Volume

• ➡ 1 %

Applied Pressure

• ➡ 2 × 10⁷ N / m-²

What we have to calculate :

• ➡Calculate Bulk modulus (B) and its Compressibility factor (c)

We know that :

Bulk modulus formula is as given below –

Bulk modulus ➡ ( pressure applied / fractional change in volume)

$\rm \mapsto \: B= \dfrac{P}{ \dfrac{\triangle\:V}{V} }$

Where :

➡ B= Bulk Modulus

➡ P = Pressure

➡δ V = Change in volume

➡V = Original volume

It means ,

$\rm \mapsto \: \dfrac{\triangle\:V}{V}\times 100 = 1$

$\rm \mapsto \: \dfrac{\triangle\:V}{V} = \dfrac{1}{100}$

Now :

• ➡ As we know the formula for Bulk Modulus so by using that formula of bulk modulus and by putting the given values of δ V / V and Pressure (P) in it we get ,

Calculation :

$\rm \implies \: B= \dfrac{2 \times {10}^{7} }{ \dfrac{1}{100} }$

$\rm \implies \: B= 2 \times {10}^{7} \times 100$

$\rm \implies \: B = 2 \times {10}^{7} \times {10}^{2}$

$\rm \implies \: B = 2 \times {10}^{9} \: N / {m}^{2}$

Note :

• The units for bulk modulus is Pa or KPa and MPa as higher units.

Here :

• ➡ Compressibility factor = The ratio of change in volume to the change in pressure.

Or

Compressibility factor is also defined as:

$\rm \implies \: Compressibility= \dfrac{1}{Bulk \:Modulus }$

$\rm \implies \: c= \dfrac{1}{B}$

Where :

➡ c = Compressibility factor

➡ B = Bulk modulus

Now :

• ➡Place the obtained value of Bulk modulus in this formula and solve

$\rm \implies \: c= \dfrac{1}{2 \times {10}^{9} }$

$\rm \implies \: c= \dfrac{1}{2 \times {10}^{9} } \div \dfrac{10}{10}$

$\rm \implies \: c= \dfrac{1}{ \cancel{2} \times {10}^{9} } \div \dfrac{\cancel{10 {}}^{ \: 5} }{10}$

$\rm \implies \: c= \dfrac{5}{{10}^{10} }$

$\rm \implies \: c= 5 \times {10}^{ - 10} \: {m}^{2} / \: N$

Therefore :

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Bulk modulus (B)

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$\bf\implies \: B = 2 \times {10}^{9} \: N / {m}^{2}$

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Compressibility factor (c)

$\bf\implies \: c= 5 \times {10}^{ - 10} \: {m}^{2} / \: N$

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