Chapter 1.11, Problem 70P

Repeat Prob. 1–67E by replacing air with oil with a specific gravity of 0.69.

To determine

The absolute pressure in the pipeline.

Answer to Problem 70P

The absolute pressure in the pipeline is $\underset{\_}{17.7\text{psia}}$.

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 ${\rho }_{\text{w}}$.

Determine the density of oil.

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

Here, the specific gravity of the oil is $S{G}_{\text{oil}}$.

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

$\begin{array}{l}{P}_1-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}+{\rho }_{\text{water}}g{h}_{\text{water}}=P_{\text{atm}}\\ P_1=P_{\text{atm}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{water}}g{h}_{\text{water}}-{\rho }_{\text{oil}}g{h}_{\text{oil}}\end{array}$ (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 $h_{\text{Hg}}$, the height of the water is $h_{\text{water}}$, the density of the oil is ${\rho }_{\text{oil}}$, and the height of the oil is $h_{\text{oil}}$.

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 $0.69$ for $S{G}_{\text{oil}}$ and $62.4\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{w}}$ in Equation (II).

$\begin{array}{c}{\rho }_{\text{Hg}}=0.69\times 62.4\text{lbm}/{\text{ft}}^{\text{3}}\\ =43.1\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

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 ${\rho }_{\text{water}}$, $27/12\text{ft}$ for $h_{\text{water}}$, $43.1\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{oil}}$, and $15/12\text{ft}$ for $h_{\text{oil}}$ in Equation (III).

$\begin{array}{c}{P}_1=\left[\begin{array}{l}\left(14.2\text{psia}\right)+\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(6/12\text{ft}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(27/12\text{ft}\right)\\ -\left(43.1\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(15/12\text{ft}\right)\end{array}\right]\\ =\left[\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(6/12\text{ft}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(27/12\text{ft}\right)\\ -\left(43.1\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(15/12\text{ft}\right)\end{array}\right]\right]\\ =\left[\begin{array}{l}\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(424.3\text{lbm}/{\text{ft}}^{\text{2}}\right)+\left(140.4\text{lbm}/{\text{ft}}^{\text{2}}\right)\\ -\left(53.875\text{lbm}/{\text{ft}}^{\text{2}}\right)\end{array}\right]\\ \times \left(\frac{1}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\end{array}\right]\\ =17.7\text{psia}\end{array}$

Thus, the absolute pressure in the pipeline is $\underset{\_}{17.7\text{psia}}$.