Chapter 2, Problem 64P

To determine

The area ratio of $A_2/A_1$.

Explanation of Solution

Given:

Specific gravity of the brine $\left({\text{SG}}_{\text{br}}\right)$ is $1.1$.

Specific gravity of the mercury $\left({\text{SG}}_{\text{Hg}}\right)$ is $13.56$.

Change in differential brine column height $\left(\Delta h_{\text{br}}\right)$ is $\text{5}\text{mm}$.

Density of the water $\left({\rho }_{\text{w}}\right)$ is $1000\text{kg}/{\text{m}}^{\text{3}}$.

Acceleration due to gravity is $\left(g\right)$$9.81\text{m}/{\text{s}}^{\text{2}}$.

Calculation:

Write the expression for pressure balance in U-tube manometer before the change in pressure of air.

  $P_{A1}+{\rho }_{\text{w}}g{h}_{\text{w}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg,1}}+{\rho }_{\text{br}}g{h}_{\text{br,1}}=P_B$        (I)

Here, pressure at area 1 is $P_{A1}$, column height of the water is $h_{\text{w}}$, density of mercury is ${\rho }_{\text{Hg}}$, column height of the mercury at area 1 is $h_{\text{Hg,1}}$, density of brine is ${\rho }_{\text{br}}$, column height of the brine at area 1 is $h_{\text{br,1}}$ and pressure at brine pipe is $P_B$.

Write the expression for pressure balance in U-tube manometer after the change in pressure of air.

  $P_{A2}+{\rho }_{\text{w}}g{h}_{\text{w}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg,2}}+{\rho }_{\text{br}}g{h}_{\text{br,2}}=P_B$        (II)

Here, pressure at area 2 is $P_{A2}$, column height of the mercury at area 2 is $h_{\text{Hg,2}}$ and column height of the brine at area 2 is $h_{\text{br,2}}$.

Simplify Equations (I) and (II).

  $\begin{array}{l}{P}_{A1}-P_{A2}+{\rho }_{\text{Hg}}g\Delta h_{\text{Hg}}-{\rho }_{\text{br}}g\Delta h_{\text{br}}=0\\ \frac{P_{A1}-P_{A2}}{{\rho }_{\text{w}}g}={\text{SG}}_{\text{Hg}}\Delta h_{\text{Hg}}-{\text{SG}}_{\text{br}}\Delta h_{\text{br}}\end{array}$

Here, change in differential mercury column height is $\Delta h_{\text{Hg}}$.

Write the expression for the change in differential mercury column height.

  $\begin{array}{c}\Delta h_{\text{Hg}}=\Delta h_{\text{Hg,right}}+\Delta h_{\text{Hg,rleft}}\\ =\Delta h_{\text{br}}+\Delta h_{\text{br}}{A}_2/A_1\\ =\Delta h_{\text{br}}\left(1+A_2/A_1\right)\end{array}$

Here, area 1 is $A_1$ and area 2 is $A_2$.

Calculate area ratio of $A_2/A_1$.

  $\begin{array}{c}\frac{P_{A1}-P_{A2}}{{\rho }_{\text{w}}g}={\text{SG}}_{\text{Hg}}\Delta h_{\text{Hg}}-{\text{SG}}_{\text{br}}\Delta h_{\text{br}}\\ ={\text{SG}}_{\text{Hg}}\left[\Delta h_{\text{br}}\left(1+A_2/A_1\right)\right]-{\text{SG}}_{\text{br}}\Delta h_{\text{br}}\end{array}$

  $\begin{array}{c}{A}_2/A_1=\frac{\frac{P_{A1}-P_{A2}}{{\rho }_{\text{w}}g}+{\text{SG}}_{\text{br}}\Delta h_{\text{br}}}{{\text{SG}}_{\text{Hg}}\Delta h_{\text{br}}}\\ A_2/A_1=\frac{\left[\frac{0.7\text{kPa}-0\text{kPa}}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right]+\left(1.1\times 5\text{mm}\right)}{\left(13.56\right)\left(5\text{mm}\right)}-1\end{array}$

  $\begin{array}{c}{A}_2/A_1=\frac{\left[\frac{\left(0.7\text{kPa}\right)\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{}\text{kPa}}\right)\left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{1\text{N}/{\text{m}}^{\text{2}}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right]+\left(1.1\times 0.005\text{m}\right)}{\left(13.56\right)\left(0.005\text{m}\right)}-1\\ =\frac{\left(0.0703156\text{m}\right)+\left(0.0055\text{m}\right)}{0.0678\text{m}}-1\\ =1.13356-1\\ =0.13356\\ \simeq 0.134\end{array}$

Thus, the area ratio of $A_2/A_1$ is $\underset{\_}{0.134}$.