Chapter 2, Problem 2.19P

To determine

The free-surface height in the right leg.

The free-surface height in the left leg.

Explanation of Solution

Write the expression for the volume of the right end leg of U-tube.

$V_r=\frac{\pi }{4}{d}^{2}h$ …… (I)

Here, the volume is $V_r$, the diameter of the leg is $d$ and the height of the water column is $h$.

The extra water added to the U-tube will displace the mercury, so the height of the column will be changed.

Write the expression for the final pressure in the left leg of U-tube.

$p_L=p_a+{\gamma }_m\left(h_l+l_b-L\right)$ …… (II)

Here, the pressure in the left leg is $p_L$, the atmospheric pressure is $p_a$, the specific weight of mercury is ${\gamma }_m$, the length of the displaced mercury column is $L$, the height of the mercury column in the left leg is $h_l$ and the length of the mercury column at the base is $l_b$.

Write the expression for the final pressure in the right leg of U-tube.

$p_r=p_a+{\gamma }_{w}h+{\gamma }_{m}L$ …… (III)

Here, the pressure in the right leg is $p_r$, the specific weight of water is ${\gamma }_w$.

Since all the liquid column of the U-tube is in equilibrium, the pressure in the right and left leg are same

Equate Equation (II) and Equation (III).

$\begin{array}{c}{p}_a+{\gamma }_m\left(h_l+l_b-L\right)=p_a+{\gamma }_{w}h+{\gamma }_{m}L\\ {\gamma }_m\left(h_l+l_b-L\right)={\gamma }_{w}h+{\gamma }_{m}L\end{array}$ …… (IV)

Write the expression for the final height of the right leg.

$z_r=L+h$ …… (V)

Write the expression for the final height of the left leg.

$z_l=h_l+l_b-L$ …… (VI)

Conclusion:

Refer to the Table 2.1 “Specific Weight of Some Common Fluids” to obtain the specific weight of the water as $9790\text{N}/{\text{m}}^3$.

Refer to the Table 2.1 “Specific Weight of Some Common Fluids” to obtain the specific weight of the mercury as $133100\text{N}/{\text{m}}^3$.

Substitute $20{\text{cm}}^3$ for $V_r$ and $1\text{cm}$ for $h$ in Equation (I).

$\begin{array}{c}\left(20{\text{cm}}^3\right)=\frac{\pi }{4}{\left(1\text{cm}\right)}^{2}h\\ h=\frac{4}{\pi }\frac{20{\text{cm}}^3}{1{\text{cm}}^2}\\ =25.46\text{cm}\end{array}$

Substitute $133100\text{N}/{\text{m}}^3$ for ${\gamma }_m$, $9790\text{N}/{\text{m}}^3$ for ${\gamma }_w$, $10\text{cm}$ for $h_l$, $10\text{cm}$ for $l_b$ and $25.46\text{cm}$ for $h$ in Equation (II) and solve for $L$ in Equation (IV).

$\begin{array}{l}\left(133100\text{N}/{\text{m}}^3\right)\left[\left(10\text{cm}\right)+\left(10\text{cm}\right)-L\right]=\left[\begin{array}{l}\left(9790\text{N}/{\text{m}}^3\right)\left(25.46\text{cm}\right)\\ +\left(133100\text{N}/{\text{m}}^3\right)L\end{array}\right]\\ \left(133100\text{N}/{\text{m}}^3\right)\left[\begin{array}{l}\left(10\text{cm}\right)\left(\frac{1\text{m}}{\text{100}\text{cm}}\right)\\ +\left(10\text{cm}\right)\left(\frac{1\text{m}}{\text{100}\text{cm}}\right)-L\end{array}\right]=\left[\begin{array}{l}\left\{\begin{array}{l}\left(9790\text{N}/{\text{m}}^3\right)\\ \times \left(25.46\text{cm}\right)\left(\frac{1\text{m}}{\text{100}\text{cm}}\right)\end{array}\right\}\\ +\left(133100\text{N}/{\text{m}}^3\right)L\end{array}\right]\\ L=0.0906\text{m}\end{array}$

Substitute $0.0906\text{m}$ for $L$ and $25.46\text{cm}$ for $h$ in Equation (V).

$\begin{array}{c}{z}_r=\left(0.0906\text{m}\right)+\left(25.46\text{cm}\right)\\ =\left(0.0906\text{m}\right)\left(\frac{100\text{cm}}{\text{1}\text{m}}\right)+\left(25.46\text{cm}\right)\\ =34.52\text{cm}\end{array}$

Thus, the free-surface height in the right leg is $34.52\text{cm}$.

Substitute $0.0906\text{m}$ for $L$, $10\text{cm}$ for $l_b$ and $10\text{cm}$ for $h_l$ in Equation (VI).

$\begin{array}{c}{z}_l=\left(10\text{cm}\right)+\left(10\text{cm}\right)-\left(0.0906\text{m}\right)\\ =20\text{cm}-\left(0.0906\text{m}\right)\left(\frac{100\text{cm}}{\text{1}\text{m}}\right)\\ =\text{10}\text{.94}\text{cm}\end{array}$

Thus, the free-surface height in the left leg is $10.94\text{cm}$.