Chapter 2, Problem 2.14P

To determine

The height of mercury column in water tube.

The height of oil column.

Explanation of Solution

Calculation:

Write the expression for density of the oil.

${\rho }_{\text{oil}}={\left(\text{SG}\right)}_{\text{oil}}\left({\rho }_{\text{water}}\right)$ …… (I)

Here, the density of oil is ${\rho }_{\text{oil}}$, the specific gravity of oil is ${\left(\text{SG}\right)}_{\text{oil}}$ and the density of water is $\left({\rho }_{\text{water}}\right)$.

Write the expression for pressure at bottom in mercury column.

$p_{\text{b,mercury}}=\left({\rho }_{\text{mercury}}\right)\left(g\right)\left(h_{\text{mercury}}\right)$ …… (II)

Here, the pressure at bottom in mercury column is $p_{\text{b,mercury}}$, the density of mercury is ${\rho }_{\text{mercury}}$, the acceleration due to gravity is $g$ and the height of mercury column is $h_{\text{mercury}}$.

Write the expression for pressure at the bottom of the water column.

$p_{\text{b,water}}=\left({\rho }_{\text{water}}\right)\left(g\right)\left(h_{\text{water}}\right)+\left({\rho }_{\text{mercury}}\right)\left(g\right)\left(h_1\right)$ …… (III)

Here, the pressure at bottom in water column is $p_{\text{b,water}}$, the height of water column is $h_{\text{water}}$ and height of mercury column in water tube is $h_1$.

Write the expression for pressure at bottom in oil column.

$p_{b,\text{oil}}=\left({\rho }_{\text{oil}}\right)\left(g\right)\left(h_2\right)+{\rho }_{\text{mercury}}\left(g\right)\left(h_3\right)$ …… (IV)

Here, the pressure at bottom in oil column is $p_{b,\text{oil}}$, the density of oil is ${\rho }_{\text{oil}}$, the height of oil column is $h_2$ and the height of mercury column in oil tube is $h_3$.

Substitute ${\left(\text{SG}\right)}_{\text{oil}}\left({\rho }_{\text{water}}\right)$ for ${\rho }_{\text{oil}}$ in Equation (IV).

$p_{b,\text{oil}}=\left({\left(\text{SG}\right)}_{\text{oil}}\left({\rho }_{\text{water}}\right)\right)\left(g\right)\left(h_2\right)+{\rho }_{\text{mercury}}\left(g\right)\left(h_3\right)$ …… (V)

The pressure in the liquid varies according to the depth of fluid. But, the hydrostatic pressure of the fluid is same at same depth.

The pressure at bottom in mercury column is equal to pressure at bottom in water column.

$p_{\text{b,mercury}}=p_{\text{b,water}}$ …… (VI)

The pressure at bottom in mercury column is equal to pressure at bottom in oil column.

$p_{\text{b,mercury}}=p_{b,\text{oil}}$ …… (VII)

Conclusion:

Substitute $13550\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$ and $8\text{cm}$ for $h_{\text{mercury}}$ in Equation (II).

$\begin{array}{c}{p}_{\text{b,mercury}}=\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(8\text{cm}\right)\\ =\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(8\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(0.08\text{m}\right)\end{array}$ …… (VIII)

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{water}}$, $27\text{cm}$ for $h_{\text{water}}$ and $13550\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$ in Equation (III).

$\begin{array}{c}{p}_{\text{b,water}}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(27\text{cm}\right)+\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(h_1\right)\\ =\left(g\right)\left[\begin{array}{l}\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(27\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ +\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_1\right)\end{array}\right]\\ =\left(g\right)\left[\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.27\text{m}\right)+\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_1\right)\right]\end{array}$ …. (IX)

Substitute $0.78$ for ${\left(\text{SG}\right)}_{\text{oil}}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{water}}$, $13550\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{mercury}}$ and $5\text{cm}$ for $h_3$ in Equation (IV).

$\begin{array}{c}{p}_{b,\text{oil}}=\left[\begin{array}{l}\left(\left(0.78\right)\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\right)\left(g\right)\left(h_2\right)\\ +\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\end{array}\right]\\ =\left(g\right)\left[\left(778.44\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_2\right)+\left(677.5\text{kg}/{\text{m}}^2\right)\right]\end{array}$ …… (X)

Substitute $\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(0.08\text{m}\right)$ for $p_{\text{b,mercury}}$ and $\left(g\right)\left[\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.27\text{m}\right)+\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_1\right)\right]$ for $p_{\text{b,water}}$ in Equation (VI).

$\begin{array}{l}\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(0.08\text{m}\right)=\left(g\right)\left[\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.27\text{m}\right)+\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_1\right)\right]\\ \left(1084\text{kg}/{\text{m}}^2\right)-\left(269.46\text{kg}/{\text{m}}^2\right)=\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_1\right)\\ \left(h_1\right)=\left(\frac{814.54\text{kg}/{\text{m}}^2}{\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)}\right)\\ =0.06011\text{m}\end{array}$

Further simplify.

$\begin{array}{c}\left(h_1\right)=\left(0.06011\text{m}\right)\left(\frac{100\text{cm}}{1\text{m}}\right)\\ \approx 6.0\text{cm}\end{array}$

Thus, the height $h_1$ of the mercury liquid column is $6.0\text{cm}$.

Substitute $\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(0.08\text{m}\right)$ for $p_{\text{b,mercury}}$ and $\left(g\right)\left[\left(778.44\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_2\right)+\left(677.5\text{kg}/{\text{m}}^2\right)\right]$ for $p_{b,\text{oil}}$ in Equation (VII).

$\begin{array}{l}\left(13550\text{kg}/{\text{m}}^{\text{3}}\right)\left(g\right)\left(0.08\text{m}\right)=\left(g\right)\left[\left(778.44\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_2\right)+\left(677.5\text{kg}/{\text{m}}^2\right)\right]\\ \left(1084\text{kg}/{\text{m}}^2\right)-\left(677.5\text{kg}/{\text{m}}^2\right)=\left(778.44\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_2\right)\\ \left(406.5\text{kg}/{\text{m}}^2\right)=\left(778.44\text{kg}/{\text{m}}^{\text{3}}\right)\left(h_2\right)\\ h_2=\left(\frac{406.5\text{kg}/{\text{m}}^2}{778.44\text{kg}/{\text{m}}^{\text{3}}}\right)\end{array}$

Further simplify.

$\begin{array}{l}{h}_2=\left(\frac{406.5\text{kg}/{\text{m}}^2}{778.44\text{kg}/{\text{m}}^{\text{3}}}\right)\\ =0.52219\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\\ \approx 52.219\text{cm}\end{array}$

Thus, the height $h_2$ of oil liquid column is $52.219\text{cm}$.