Chapter 1.11, Problem 69P

The pressure in a natural gas pipeline is measured by the manometer shown in Fig. P1–67E with one of the arms open to the atmosphere where the local atmospheric pressure is 14.2 psia. Determine the absolute pressure in the pipeline.

To determine

The absolute pressure in the pipeline.

Answer to Problem 69P

The absolute pressure in the pipeline is $\underset{\_}{18.1\text{psia}}$.

Explanation of Solution

Determine the density of mercury.

${\rho }_{\text{Hg}}=S{G}_{\text{Hg}}\times {\rho }_{\text{w}}$ (I)

Here, the specific gravity of the mercury is $S{G}_{\text{Hg}}$ and the density of the water is ${\rho }_{\text{w}}$.

Write the expression of pressure in a double U-tube manometer with one arms open to the atmosphere.

$\begin{array}{l}{P}_1-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{water}}g{h}_{\text{water}}=P_{\text{atm}}\\ P_1=P_{\text{atm}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{water}}g{h}_{\text{water}}\end{array}$ (III)

Here, the absolute pressure in the pipeline is $P_{\text{1}}$, the pressure in the atmosphere is $P_{\text{atm}}$, the acceleration of gravity is $g$, the density of the mercury is ${\rho }_{\text{Hg}}$, the height of the mercury is $h_{\text{Hg}}$, and the density of water is ${\rho }_{\text{water}}$, and the height of the water is $h_{\text{water}}$.

Conclusion:

From the Table A-3E (a) “Properties of common liquids, solids, and foods” to obtain the value for density of water as $62.4\text{lbm}/{\text{ft}}^{\text{3}}$.

Substitute $13.6$ for $S{G}_{\text{Hg}}$ and $62.4\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{w}}$ in Equation (I).

$\begin{array}{c}{\rho }_{\text{Hg}}=13.6\times 62.4\text{lbm}/{\text{ft}}^{\text{3}}\\ =848.6\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $14.2\text{psia}$ for $P_{\text{atm}}$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $848.6\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{Hg}}$, $6/12\text{ft}$ for $h_{\text{Hg}}$, $62.4\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{water}}$, and $27/12\text{ft}$ for $h_{\text{water}}$ in Equation (III).

$\begin{array}{c}{P}_1=\left[\begin{array}{l}\left(14.2\text{psia}\right)+\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(6/12\text{ft}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(27/12\text{ft}\right)\end{array}\right]\\ =\left[\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(6/12\text{ft}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(27/12\text{ft}\right)\end{array}\right]\right]\\ =\left[\begin{array}{l}\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\left(424.3\text{lbm}/{\text{ft}}^{\text{2}}\right)+\left(140.4\text{lbm}/{\text{ft}}^{\text{2}}\right)\right]\\ \times \left(\frac{1}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\end{array}\right]\\ =18.1\text{psia}\end{array}$

Thus, the absolute pressure in the pipeline is $\underset{\_}{18.1\text{psia}}$.