Chapter 2, Problem 53P

To determine

The absolute pressure in the pipeline.

Explanation of Solution

Given:

Local atmospheric pressure $\left(P_{\text{atm}}\right)$ is $14.2\text{psia}$.

Height of the mercury in the manometer $\left(h_{\text{Hg}}\right)$ is $6\text{in}\text{.}$.

Height of the water in the manometer $\left(h_{{\text{H}}_2\text{O}}\right)$ is $27\text{in}\text{.}$.

Height of the oil in the manometer $\left(h_{\text{oil}}\right)$ is $15\text{in}\text{.}$.

Density of the water $\left({\rho }_{{\text{H}}_{\text{2}}\text{O}}\right)$ is $62.4\text{lbm}/{\text{ft}}^{\text{3}}$.

Specific gravity of the mercury $\left({\text{SG}}_{\text{Hg}}\right)$ is $13.6$.

Specific gravity of the oil $\left({\text{SG}}_{\text{oil}}\right)$ is $0.69$.

Acceleration due to gravity $\left(g\right)$ is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Calculate the density of the mercury $\left({\rho }_{\text{Hg}}\right)$.

  $\begin{array}{c}{\rho }_{\text{Hg}}=\left({\rho }_{{\text{H}}_{\text{2}}\text{O}}\right)\left({\text{SG}}_{\text{Hg}}\right)\\ =\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(13.6\right)\\ =848.6\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the density of the oil $\left({\rho }_{\text{oil}}\right)$.

  $\begin{array}{c}{\rho }_{\text{oil}}=\left({\rho }_{{\text{H}}_{\text{2}}\text{O}}\right)\left({\text{SG}}_{\text{oil}}\right)\\ =\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(0.69\right)\\ =43.1\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the absolute pressure in the pipeline $\left(P_{\text{1}}\right)$.

  $P_1-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}-{\rho }_{{\text{H}}_{\text{2}}\text{O}}g{h}_{{\text{H}}_{\text{2}}\text{O}}=P_{\text{atm}}$

  $\begin{array}{c}{P}_1=P_{\text{atm}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{{\text{H}}_{\text{2}}\text{O}}g{h}_{{\text{H}}_{\text{2}}\text{O}}-{\rho }_{\text{oil}}g{h}_{\text{oil}}\\ =\left\{\begin{array}{l}\left(14.2\text{psia}\right)+\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(6\text{in}\text{.}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(27\text{in}\text{.}\right)\\ -\left(43.1\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(15\text{in}\text{.}\right)\end{array}\right\}\\ =\left\{\begin{array}{c}\left(14.2\text{psia}\right)+\left(848.6\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(\frac{6}{12}\text{ft}\right)\\ +\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(\frac{27}{12}\text{ft}\right)\\ -\left(43.1\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left(\frac{15}{12}\text{ft}\right)\end{array}\right\}\\ =\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(424.3\text{lbm}/{\text{ft}}^{\text{2}}\right)+\left(140.4\text{lbm}/{\text{ft}}^{\text{2}}\right)\\ -\left(53.875\text{lbm}/{\text{ft}}^{\text{2}}\right)\end{array}\right]\\ =\left\{\begin{array}{c}\left(14.2\text{psia}\right)+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(424.3\text{lbm}/{\text{ft}}^{\text{2}}\right)+\left(140.4\text{lbm}/{\text{ft}}^{\text{2}}\right)\\ -\left(53.875\text{lbm}/{\text{ft}}^{\text{2}}\right)\end{array}\right]\\ \left(\frac{1}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\end{array}\right\}\\ =17.7\text{psia}\end{array}$

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