Chapter 1.11, Problem 81P

Consider the system shown in Fig. 1–77. If a change of 0.7 kPa in the pressure of air causes the brine–mercury interface in the right column to drop by 5 mm in the brine level in the right column while the pressure in the brine pipe remains constant, determine the ratio of A₂/A₁.

To determine

The ratio of $A_2/A_1$ for U-tube manometer.

Answer to Problem 81P

The ratio of $A_2/A_1$ for U-tube manometer is $\underset{\_}{0.134}$.

Explanation of Solution

Write the expression of pressure in U-tube manometer with before the pressure change 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, the pressure in area 1 is $P_{A1}$, the density of water is ${\rho }_{\text{w}}$, the acceleration of gravity is $g$, the height of the water is $h_{\text{w}}$, the density of mercury is ${\rho }_{\text{Hg}}$, the height of the mercury at area 1 is $h_{\text{Hg,1}}$, the density of brine is ${\rho }_{\text{br}}$, the height of the brine at area 1 is $h_{\text{br,1}}$, and the pressure at brine pipe is $P_B$.

Write the expression of pressure in U-tube manometer with after the pressure change 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, the pressure in area 2 is $P_{A2}$, the height of the mercury at area 2 is $h_{\text{Hg,2}}$, and the height of the brine at area 2 is $h_{\text{br,2}}$.

Simplify the Equation (I) and Equation (II).

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

Here, the specific gravity of mercury is $S{G}_{\text{Hg}}$, the specific gravity of brine is $S{G}_{\text{br}}$, the change in specific enthalpy for brine is $\Delta h_{\text{br}}$, and the change in specific enthalpy for mercury is $\Delta h_{\text{Hg}}$.

Determine the change in specific enthalpy of mercury.

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

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

Substitute Equation (IV) into Equation (III).

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

Conclusion:

Substitute $0.7\text{kPa}$ for $P_{A1}$, $5\text{mm}$ for $\Delta h_{\text{br}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $13.56$ for $S{G}_{\text{Hg}}$, and $1.1$ for $S{G}_{\text{br}}$ in Equation (V).

$\begin{array}{l}\frac{0.7\text{kPa}-\left(0\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}=\left(13.56\right)\left(5\text{mm}\right)\left(1+A_2/A_1\right)-\left(1.1\times 5\text{mm}\right)\\ \frac{0.7\text{kPa}\times \left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{}\text{kPa}}\right)\left(\frac{1\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}}{1\text{N}/{\text{m}}^{\text{2}}}\right)}{\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}=\left[\begin{array}{l}\left(13.56\right)\left(5\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\\ \times \left(1+A_2/A_1\right)\\ -\left(1.1\times 5\text{mm}\times \left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)\end{array}\right]\\ 0.071356\text{m}=\left[\begin{array}{l}\left(0.0678\text{m}\right)\\ \times \left(1+A_2/A_1\right)\\ -\left(0.0055\text{m}\right)\end{array}\right]\\ A_2/A_1=0.13356\end{array}$

$A_2/A_1=0.134$

Thus, the ratio of $A_2/A_1$ for U-tube manometer is $\underset{\_}{0.134}$.