Chapter 2, Problem 2.13P

To determine

The height $h$ of the third liquid in the right leg of the tank.

Answer to Problem 2.13P

The height $h$ of the third liquid in the right leg of the tank is $1.52\text{m}$.

Explanation of Solution

Write the expression for specific weight of the liquid.

${\gamma }_{\text{liquid}}={\left(\text{SG}\right)}_{\text{liquid}}\left({\gamma }_{\text{water}}\right)$ …… (I)

Here, the specific weight of liquid is ${\gamma }_{\text{liquid}}$, the specific gravity of liquid is ${\left(\text{SG}\right)}_{\text{liquid}}$ and specific weight of water is $\left({\gamma }_{\text{water}}\right)$.

Write the expression for total column height.

$h_t=\left(h_{\text{water}}\right)+\left(h_{{}_{\text{liquid}}}\right)$ …… (II)

Here, the total column height is $h_t$, the height of water column is $h_{\text{water}}$ and the height of liquid column is $h_{{}_{\text{liquid}}}$.

Write the hydrostatic formula.

$\left[\begin{array}{l}\left({\gamma }_{\text{water}}\right)\left(h_{\text{water}}\right)+\left({\gamma }_{\text{liquid}}\right)\left(h_{{}_{\text{liquid}}}\right)-\left({\gamma }_{\text{liquid}}\right)\left(h\right)\\ -\left({\gamma }_{\text{gasoline}}\right)\left(\left(h_t\right)-h\right)\end{array}\right]=0$ …… (III)

Here, the specific weight of gasoline is ${\gamma }_{\text{gasoline}}$ and the height from liquid to liquid-gasoline interface is $h$.

Substitute ${\left(\text{SG}\right)}_{\text{liquid}}\left({\gamma }_{\text{water}}\right)$ for ${\gamma }_{\text{liquid}}$ and $\left(h_{\text{water}}\right)+\left(h_{{}_{\text{liquid}}}\right)$ for $h_t$ in Equation (III).

$\left[\begin{array}{l}\left({\gamma }_{\text{water}}\right)\left(h_{\text{water}}\right)+\left({\left(\text{SG}\right)}_{\text{liquid}}\left({\gamma }_{\text{water}}\right)\right)\left(h_{{}_{\text{liquid}}}\right)-\left({\left(\text{SG}\right)}_{\text{liquid}}\left({\gamma }_{\text{water}}\right)\right)\left(h\right)\\ -\left({\gamma }_{\text{gasoline}}\right)\left(\left(\left(h_{\text{water}}\right)+\left(h_{{}_{\text{liquid}}}\right)\right)-h\right)\end{array}\right]=0$ …… (IV)

Substitute $9790\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{water}}$, $1.5\text{m}$ for $h_{\text{water}}$, $1.60$ for ${\left(\text{SG}\right)}_{\text{liquid}}$, $1\text{m}$ for $h_{{}_{\text{liquid}}}$ and $6670\text{N}/{\text{m}}^{\text{3}}$ for ${\gamma }_{\text{gasoline}}$ in Equation (IV).

$\begin{array}{l}\left[\begin{array}{l}\left(9790\text{N}/{\text{m}}^{\text{3}}\right)\left(1.5\text{m}\right)+\left(\left(1.60\right)\left(9790\text{N}/{\text{m}}^{\text{3}}\right)\right)\left(1\text{m}\right)\\ -\left(\left(1.60\right)\left(9790\text{N}/{\text{m}}^{\text{3}}\right)\right)\left(h\right)-\left(6670\text{N}/{\text{m}}^{\text{3}}\right)\left(\left(\left(1.5\text{m}\right)+\left(1\text{m}\right)\right)-h\right)\end{array}\right]=0\\ \left[\left(30349\text{N}/{\text{m}}^2\right)-\left(16675\text{N}/{\text{m}}^2\right)-\left(8994\text{N}/{\text{m}}^{\text{3}}\right)\left(h\right)\right]=0\\ \left(13674\text{N}/{\text{m}}^2\right)=\left(8994\text{N}/{\text{m}}^{\text{3}}\right)\left(h\right)\\ h=\left(\frac{13674\text{N}/{\text{m}}^2}{8994\text{N}/{\text{m}}^{\text{3}}}\right)\end{array}$

Further simplify the above equation.

$\begin{array}{c}h=\left(\frac{13674\text{N}/{\text{m}}^2}{8994\text{N}/{\text{m}}^{\text{3}}}\right)\\ =1.52034\text{m}\\ \approx 1.52\text{m}\end{array}$

Conclusion:

Thus, the height of the third liquid in the right leg of the tank is $1.52\text{m}$.