Chapter 2, Problem 50P

To determine

The percentage error caused by Boussinesq approximation.

Answer to Problem 50P

The percentage error caused by Boussinesq approximation is $6\%$.

Explanation of Solution

Given information:

The pressure of the air is $95\text{kPa}$, the initial temperature is $20°\text{C}$, the final temperature is $60°\text{C}$, and the midway temperature is $40°\text{C}$.

Write the expression for the Boussineq approximation for $20\text{°C}$.

  ${\rho }_{20}={\rho }_0\left[1-\beta \left(T-T_0\right)\right]$   ..... (I)

Here, the density of fluid at $20°\text{C}$ is ${\rho }_{20}$, the density of air is ${\rho }_0$, the temperature is $T$, the midway temperature is $T_0$ and the coefficient of volume expansion is $\beta$.

Write the expression for the Boussineq approximation for $60\text{°C}$.

  ${\rho }_{60}={\rho }_0\left[1-\beta \left(T-T_0\right)\right]$   ..... (II)

Here, the density of fluid at $60°\text{C}$ is ${\rho }_{60}$.

Write the expression for the pressure at $20°\text{C}$ using ideal gas law.

  $P={\left({\rho }_{20}\right)}_{a}RT$   ..... (III)

Here, the pressure is $P$ the actual density at $20°\text{C}$ is ${\left({\rho }_{20}\right)}_a$.and the universal gas constant is $R$.

Write the expression for the pressure at $60°\text{C}$ using ideal gas law.

  $P={\left({\rho }_{60}\right)}_{a}RT$   ..... (IV)

Here, the actual density at $60°\text{C}$ is ${\left({\rho }_{60}\right)}_a$.

Write the expression the percentage error at $20°\text{C}$.

  $\%\text{error}=\left(\frac{{\rho }_{20}-{\left({\rho }_{20}\right)}_a}{{\left({\rho }_{20}\right)}_a}\right)\times 100$   ..... (V)

Write the expression the percentage error at $60°\text{C}$.

  $\%\text{error}=\left(\frac{{\rho }_{60}-{\left({\rho }_{60}\right)}_a}{{\left({\rho }_{60}\right)}_a}\right)\times 100$   ..... (VI)

Calculation:

Convert the value of midway temperature in Kelvin.

  $\begin{array}{c}{T}_0=40°\text{C}\\ =\left(40°\text{C}+273.15\right)\text{K}\\ \text{=313}\text{.15}\text{K}\end{array}$

Refer Boussinesq approximation data at $313.5\text{K}$ to obtain the values of ${\rho }_o$ as $1.127\text{kg}/{\text{m}}^{\text{3}}$ and $\beta$ as $3.2\times {10}^{-3}/\text{K}$.

Substitute $1.127\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_0$, $3.2\times {10}^{-3}/\text{K}$ for $\beta$, $20°\text{C}$ for $T$ and $313.5\text{K}$ for $T_0$ in Equation (I).

  $\begin{array}{c}{\rho }_{20}=\left(1.127\text{kg}/{\text{m}}^{\text{3}}\right)\left[1-\left(3.2\times {10}^{-3}/\text{K}\right)\left(20°\text{C}-313.15\text{K}\right)\right]\\ =\left(1.127\text{kg}/{\text{m}}^{\text{3}}\right)\left[1-\left(3.2\times {10}^{-3}/\text{K}\right)\left(\left(20°\text{C}+273.15\right)\text{K}-313.15\text{K}\right)\right]\\ =1.199\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.127\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_0$, $3.2\times {10}^{-3}/\text{K}$ for $\beta$, $60°\text{C}$ for $T$ and $313.5\text{K}$ for $T_0$ in Equation (II).

  $\begin{array}{c}{\rho }_{60}=\left(1.127\text{kg}/{\text{m}}^{\text{3}}\right)\left[1-\left(3.2\times {10}^{-3}/\text{K}\right)\left(60°\text{C}-313.15\text{K}\right)\right]\\ =\left(1.127\text{kg}/{\text{m}}^{\text{3}}\right)\left[1-\left(3.2\times {10}^{-3}/\text{K}\right)\left(\left(60°\text{C}+273.15\right)\text{K}-313.15\text{K}\right)\right]\\ =1.054\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $95\text{kPa}$ for $P$, $0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $20°\text{C}$ for $T$ in Equation (III).

  $\begin{array}{l}95\text{kPa}={\left({\rho }_{20}\right)}_a\times \left(0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}\right)\times \left(20°\text{C}\right)\\ {\left({\rho }_{20}\right)}_a=\frac{\left(95\text{kPa}\times \frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\right)}{\left(\left(0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}\right)\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}\right)\times \left(20°\text{C}+273.15\right)\text{K}}\\ {\left({\rho }_{20}\right)}_a=1.129\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $95\text{kPa}$ for $P$, $0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}$ for $R$ and $60°\text{C}$ for $T$ in Equation (IV).

  $\begin{array}{l}95\text{kPa}={\left({\rho }_{60}\right)}_a\times \left(0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}\right)\times \left(60°\text{C}\right)\\ {\left({\rho }_{60}\right)}_a=\frac{\left(95\text{kPa}\times \frac{1000\text{N}/{\text{m}}^2}{1\text{kPa}}\right)}{\left(\left(0.287\times {10}^3\text{J}/\text{kg}\cdot \text{K}\right)\times \frac{1\text{N}\cdot \text{m}}{1\text{J}}\right)\times \left(60°\text{C}+273.15\right)\text{K}}\\ {\left({\rho }_{60}\right)}_a=0.993\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.199\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{20}$ and $1.129\text{kg}/{\text{m}}^{\text{3}}$ for ${\left({\rho }_{20}\right)}_a$ in Equation (V).

  $\begin{array}{c}\%\text{error}=\left(\frac{1.199\text{kg}/{\text{m}}^{\text{3}}-1.129\text{kg}/{\text{m}}^{\text{3}}}{1.129\text{kg}/{\text{m}}^{\text{3}}}\right)\times 100\\ =0.06\times 100\\ =6\%\end{array}$

Substitute $1.054\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{60}$ and $0.993\text{kg}/{\text{m}}^{\text{3}}$ for ${\left({\rho }_{60}\right)}_a$ in Equation (VI).

  $\begin{array}{c}\%\text{error}=\left(\frac{1.054\text{kg}/{\text{m}}^{\text{3}}-0.993\text{kg}/{\text{m}}^{\text{3}}}{0.993\text{kg}/{\text{m}}^{\text{3}}}\right)\times 100\\ =0.06\times 100\\ =6\%\end{array}$

Conclusion:

The percentage error caused by Boussinesq approximation is $6\%$.