Chapter 2, Problem 2.12P

To determine

The height $h$ of the left-hand side of the tube above the tank surface.

Answer to Problem 2.12P

The height $h$ of the left-hand side of the tube above the tank surface is $8.0\text{cm}$.

Explanation of Solution

Write the expression of hydrostatic relation from oil surface to water surface.

$\left[p_{\text{atm}}+\left({\rho }_{\text{oil}}g{H}_1\right)-\left({\rho }_{\text{water}}g{H}_2\right)-p_{\text{atm}}\right]=0$ …… (I)

Here, the atmospheric pressure is $p_{\text{atm}}$, the density of oil is ${\rho }_{\text{oil}}$, the acceleration due to gravity is $g$, the density of water is ${\rho }_{\text{water}}$, the distance from top to oil-water interface on the left side of the tank is $H_1$ and the distance from oil-water interface to top of the right side of tank is $H_2$.

Write expression for the distance from top to oil-water interface on the left side of the tank.

$H_1=\left(h+h_1\right)$ …… (II)

Here, the distance from top to oil-water interface on the left side of the tank is $H_1$, the height of tube above tank surface on the left side is $h$ and the distance from surface of the tank to oil-water interface is $h_1$.

Write expression for the distance from top to oil-water interface on the right side of the tank.

$H_2=\left(h_1+h_2\right)$ …… (III)

Here, the distance from top to oil-water interface on the right side of the tank is $H_2$ and the height of tube above tank surface on the right side is $h_2$.

Substitute $\left(h+h_1\right)$ for $H_1$ and $\left(h_1+h_2\right)$ for $H_2$ in Equation (I).

$\left[p_{\text{atm}}+\left({\rho }_{\text{oil}}g\left(h+h_1\right)\right)-\left({\rho }_{\text{water}}g\left(h_1+h_2\right)\right)-p_{\text{atm}}\right]=0$ …… (IV)

Substitute $898\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{oil}}$, $12\text{cm}$ for $h_1$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{water}}$ and $6\text{cm}$ for $h_2$ in Equation (IV).

$\begin{array}{l}\left[\begin{array}{l}{p}_{\text{atm}}+\left(\left(898\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(h+\left(12\text{cm}\right)\right)\right)\\ -\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(\left(12\text{cm}\right)+\left(6\text{cm}\right)\right)\right)-p_{\text{atm}}\end{array}\right]=0\\ \left[\begin{array}{l}\left(898\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(h+\left(\left(12\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\right)\\ -\left(\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(\left(\left(12\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)+\left(\left(6\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)\right)\right)-p\end{array}\right]=0\\ \left(898\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(\left(h\right)+\left(0.12\text{m}\right)\right)=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(\left(0.12\text{m}\right)+\left(0.06\text{m}\right)\right)\\ \left(898\text{kg}/{\text{m}}^{\text{3}}\right)\left(h\right)+\left(107.76\text{kg}/{\text{m}}^2\right)=\left(179.64\text{kg}/{\text{m}}^2\right)\end{array}$

Further simplify the above Equation.

$\begin{array}{c}\left(898\text{kg}/{\text{m}}^{\text{3}}\right)\left(h\right)=\left(71.88\text{kg}/{\text{m}}^2\right)\\ \left(h\right)=\frac{\left(71.88\text{kg}/{\text{m}}^2\right)}{\left(898\text{kg}/{\text{m}}^{\text{3}}\right)}\\ =\left(0.080\text{m}\right)\left(\frac{100\text{cm}}{1\text{m}}\right)\\ \approx 8.0\text{cm}\end{array}$

Conclusion:

Thus, the height $h$ is $8.0\text{cm}$.