Chapter 3, Problem 57P

Consider the system shown in Fig. P3-51. If a change of 0.9 kPa in the pressure of air causes the brine-mercury interface in the ratio to right column to drop by 5 mm in the brine level in the right column while the pressure in the bine pipe remains constant, determine.

To determine

The ratio of $A_2/A_1$.

Answer to Problem 57P

The ratio of $\frac{A_2}{A_1}$ is $0.429$.

Explanation of Solution

Given information:

The change in pressure is $0.9\text{kpa}$ and the drop in the right column in the brine level in the right column is $5\text{mm}$ and the pressure in the brine pipe remains constant.

The following figure shows the arrangement of the liquids in the differential tube.

Figure-(1)

Write the expression for equating the pressure of the fluids in both the limbs initially.

  $\begin{array}{l}{P}_{A1}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_1=P_B+{\left({\rho }_{br}g{h}_{br}\right)}_1\\ P_{A1}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_1-{\left({\rho }_{br}g{h}_{br}\right)}_1=P_B\end{array}$....... (I)

Here, the initial pressure of air is $P_{A1}$, the density of water is ${\rho }_w$, the acceleration due to gravity is $g$, height of water level is $h_w$, density of mercury is ${\rho }_{Hg}$, height of mercury level is $h_{Hg}$, density of brine is ${\rho }_{br}$, the pressure of brine in the right limb is $P_B,$ and height of brine level is $h_{br}$.

Pressure in the left side of the limb is equal to the pressure in the right limb.

Write the expression for equating the pressure of the fluid in both the limbs after the pressure drop of air.

  $\begin{array}{l}{P}_{A2}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_2=P_B+{\left({\rho }_{br}g{h}_{br}\right)}_2\\ P_{A2}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_2-{\left({\rho }_{br}g{h}_{br}\right)}_1=P_B\end{array}$....... (II)

Here, the final pressure of the air is $P_{A2}$.

Substitute $P_{A1}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_1-{\left({\rho }_{br}g{h}_{br}\right)}_1$ for $P_B$ in Equation (II).

  $\begin{array}{l}\left(\begin{array}{c}{P}_{A2}+\\ \left({\rho }_{w}g{h}_w\right)\\ +{\left({\rho }_{Hg}g{h}_{Hg}\right)}_2\\ -{\left({\rho }_{br}g{h}_{br}\right)}_1\end{array}\right)=P_{A1}+\left({\rho }_{w}g{h}_w\right)+{\left({\rho }_{Hg}g{h}_{Hg}\right)}_1-{\left({\rho }_{br}g{h}_{br}\right)}_1\\ P_{A1}-P_{A2}=\left({\rho }_{Hg}g\Delta h_{Hg}\right)-\left({\rho }_{br}g\Delta h_{br}\right)\\ P_{A1}-P_{A2}=\left(S{G}_{Hg}{\rho }_{w}g\Delta h_{Hg}\right)-\left(S{G}_{Hg}{\rho }_{w}g\Delta h_{br}\right)\\ \frac{P_{A1}-P_{A2}}{{\rho }_{w}g}=\left(S{G}_{Hg}\Delta h_{Hg}\right)-\left(S{G}_{Hg}\Delta h_{br}\right)\end{array}$....... (III)

Here, the change in differential mercury height is $\Delta h_{Hg}$ and the change in brine column height is $\Delta h_{br}$.

Write the equation for the volume of brine as it remains constant.

  $\begin{array}{l}{A}_1{\left(\Delta h_{br}\right)}_{left}=A_2{\left(\Delta h_{br}\right)}_{right}\\ {\left(\Delta h_{br}\right)}_{left}=\frac{A_2}{A_1}{\left(\Delta h_{br}\right)}_{right}\end{array}$....... (IV)

Write the expression change of mercury level in the arrangement.

  $\Delta h_{Hg}={\left(\Delta h_{br}\right)}_{left}+{\left(\Delta h_{br}\right)}_{right}$....... (V)

Substitute $\frac{A_2}{A_1}{\left(\Delta h_{br}\right)}_{right}$ for ${\left(\Delta h_{br}\right)}_{left}$ in Equation (V).

  $\begin{array}{c}\Delta h_{Hg}=\left(\Delta h_{br}\right)+\left(\Delta h_{br}\left(\frac{A_2}{A_1}\right)\right)\\ =\Delta h_{br}\left(1+\frac{A_2}{A_1}\right)\end{array}$

Substitute $\Delta h_{br}\left(1+\frac{A_2}{A_1}\right)$ for $\Delta h_{Hg}$ in Equation (III).

  $\frac{P_{A1}-P_{A2}}{{\rho }_{w}g}=\left(S{G}_{Hg}\Delta h_{br}\left(1+\frac{A_2}{A_1}\right)\right)-\left(S{G}_{Hg}\Delta h_{br}\right)$....... (VI)

Here, the pressure difference is $P_{A1}-P_{A2}$, the specific gravity of mercury is $S{G}_{Hg}$ and specific gravity of brine liquid is $S{G}_{br}$.

Calculation:

Substitute $0.9\text{kPa}$ for $P_{A1}-P_{A2}$, $13.56$ for $S{G}_{Hg}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $5\text{mm}$ for $\Delta h_{br}$ and $1.1$ for $S{G}_{br}$ in Equation (VI).

  $\begin{array}{l}\frac{\left(0.9\text{kPa}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}=13.6\times 5\text{mm}\left(1+\frac{A_2}{A_1}\right)-1.1\times 5\text{mm}\\ \frac{\left(0.9\text{kPa}\times \frac{1000\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{ms}}^{\text{2}}$}\right.}{1\text{kPa}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}=\left[\begin{array}{l}13.6\times 5\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(1+\frac{A_2}{A_1}\right)-\\ 1.1\times 5\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\end{array}\right]\\ \frac{\left(900\text{kg}/{\text{ms}}^{\text{2}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}=13.6\times 0.005\text{m}\left(1+\frac{A_2}{A_1}\right)-1.1\times 0.005\text{m}\\ \text{0}\text{.0917}\text{m=}\left(\text{0}\text{.068}\text{m}\right)\left(1+\frac{A_2}{A_1}\right)-0.0055\text{m}\end{array}$

  $\begin{array}{l}0.0917\text{m}-0.068\text{m}+0.0055\text{m}=\left(0.068\text{m}\right)\frac{A_2}{A_1}\\ 0.0292\text{m}=\left(0.068\text{m}\right)\frac{A_2}{A_1}\\ \frac{A_2}{A_1}=\frac{0.0292}{0.068}\\ \frac{A_2}{A_1}=0.429\end{array}$

Conclusion:

The ratio of $\frac{A_2}{A_1}$ is $0.429$.