Chapter 2, Problem 101EP

To determine

The capillary drop of mercury in the tube.

Answer to Problem 101EP

The capillary drop in mercury is $0.874\text{in}$.

Explanation of Solution

Given information:

The diameter of glass tube is $0.018\text{in,}$ the contact angle with the glass is $140°,$ and the temperature of mercury is $68°\text{F}$.

Write the expression for the capillary drop.

  $h=\frac{4{\sigma }_s\cos \varphi }{\rho gD}$....... (I)

Here, capillary drop is $h$, the contact angle is $\varphi$, the density of air is $\rho$, acceleration due to gravity is $g,$ and diameter of the tube is $D$.

Refer to Table $2.4$ in the text book, to obtain the value of surface tension of mercury glass at $68°\text{F}\left(20°\text{C}\right)$ as ${\sigma }_s=0.440\text{}\text{N}/\text{m}$.

Write the formula for interpolation of two variables.

  $\rho =\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$....... (II)

Here, the temperature is denoted by variables $x,$ density is denoted by variables $y,$ and the density is $\rho$.

Calculation:

Refer to Table $\text{A-8E}$, "Properties of liquid metals" to obtain the values of $x_2$ as $68°\text{F}$, $x_1$ as $50°\text{F}$, $x_3$ as $100°\text{F}$.

Refer to Table $\text{A-8E}$, "Properties of liquid metals" to obtain the values of $y_1$ as $847.2\text{lbm}/{\text{ft}}^{\text{3}}$, $y_3$ as $842.9\text{lbm}/{\text{ft}}^{\text{3}}$.

Prepare the table for temperature and density of mercury.

Temperature, $°\text{F}$	Density, $\text{lbm}/{\text{ft}}^{\text{3}}$
$50\left(x_1\right)$	$847.2\left(y_1\right)$
$68\left(x_2\right)$	$?\left(\rho \right)$
$100\left(x_3\right)$	$842.9\left(y_3\right)$

Substitute $50°\text{F}$ for $x_1$, $68°\text{F}$ for $x_2$, $100°\text{F}$ for $x_3$, $847.2\text{lbm}/{\text{ft}}^{\text{3}}$ for $y_1$ and $842.9\text{lbm}/{\text{ft}}^{\text{3}}$ in Equation (II).

  $\begin{array}{c}\rho =\frac{\left(842.9\text{lbm}/{\text{ft}}^{\text{3}}-847.2\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(68°\text{F}-50°\text{F}\right)}{\left(100°\text{F}-50°\text{F}\right)}+847.2\text{lbm}/{\text{ft}}^{\text{3}}\\ =\left(-4.3\text{lbm}/{\text{ft}}^{\text{3}}\right)0.36+847.2\text{lbm}/{\text{ft}}^{\text{3}}\\ =-1.548\text{}\text{lbm}/{\text{ft}}^{\text{3}}+847.2\text{lbm}/{\text{ft}}^{\text{3}}\\ =845.65\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $0.440\text{N}/\text{m}$ for ${\sigma }_s$, $140°$ for $\varphi$, $845.65\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ and $0.18\text{in}$ in Equation (I).

  $\begin{array}{c}h=\frac{4\left(0.440\text{N}/\text{m}\right)\left(\cos 140°\right)}{\left(845.65\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.018\text{in}\right)}\\ =\frac{4\left(\left(0.440\text{N}/\text{m}\right)\times \left(\frac{0.2248\text{lbf}}{1\text{N}}\right)\times \left(\frac{1\text{m}}{3.28084\text{ft}}\right)\right)\left(\cos 140°\right)}{\left(845.65\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.018\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)}\left(\frac{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}{1\text{lbf}}\right)\\ =\frac{4\left(0.030148\text{lbf}/{\text{ft}}^{\text{3}}\right)\left(\cos 140°\right)}{\left(845.65\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(0.0015\text{ft}\right)}\left(\frac{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}{1\text{lbf}}\right)\\ =-0.07282\text{ft}\end{array}$

  $\begin{array}{c}h=-0.07282\text{ft}\left(\frac{12\text{ft}}{1\text{in}}\right)\\ =-0.874\text{in}\end{array}$

Here, the negative sign indicates the capillary drop instead of rise.

Conclusion:

The capillary drop in mercury is $0.874\text{in}$.