Chapter 2, Problem 85RQ

To determine

The absolute pressure at the center of the pipe.

Explanation of Solution

Given:

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

Height of the water in the manometer $\left(h_{\text{w}}\right)$ is $20\text{in}\text{.}$.

Height of the mercury in the manometer $\left(h_{\text{Hg}}\right)$ is $25\text{in}\text{.}$.

Height of the oil on left arm of the manometer $\left(h_{\text{oil}}\right)$ is $60\text{in}\text{.}$.

Height of the oil on left arm of the manometer $\left(h_{\text{oil}}\right)$ is $30\text{in}\text{.}$.

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

Specific gravity of the oil $\left({\text{SG}}_{\text{oil}}\right)$ is $0.80$.

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

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

Calculation:

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

  $\begin{array}{c}{\rho }_{\text{oil}}=\left({\rho }_{\text{w}}\right)\left({\text{SG}}_{\text{oil}}\right)\\ =\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\text{0}\text{.80}\right)\\ =49.92\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

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

  $\begin{array}{c}{\rho }_{\text{Hg}}=\left({\rho }_{\text{w}}\right)\left({\text{SG}}_{\text{Hg}}\right)\\ =\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\text{13}\text{.6}\right)\\ =848.64\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Calculate the absolute pressure at the center of the pipe $\left(P\right)$.

  $P-{\rho }_{\text{w}}g{h}_{\text{w}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}-{\rho }_{\text{Hg}}g{h}_{\text{Hg}}-{\rho }_{\text{oil}}g{h}_{\text{oil}}=P_{\text{atm}}$

  $\begin{array}{c}P=P_{\text{atm}}+{\rho }_{\text{w}}g{h}_{\text{w}}-{\rho }_{\text{oil}}g{h}_{\text{oil}}+{\rho }_{\text{Hg}}g{h}_{\text{Hg}}+{\rho }_{\text{oil}}g{h}_{\text{oil}}\\ =P_{\text{atm}}+g\left({\rho }_{\text{w}}{h}_{\text{w}}-{\rho }_{\text{oil}}{h}_{\text{oil}}+{\rho }_{\text{Hg}}{h}_{\text{Hg}}+{\rho }_{\text{oil}}{h}_{\text{oil}}\right)\\ =14.2psia+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(20\text{in}\text{.}\right)-\left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(60\text{in}\text{.}\right)+\\ \left(848.64\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(25\text{in}\text{.}\right)+\left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(30\text{in}\text{.}\right)\end{array}\right]\\ =14.2psia+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{20}{12}\text{ft}\right)-\left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{60}{12}\text{ft}\right)+\\ \left(848.64\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{25}{12}\text{ft}\right)+\left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{30}{12}\text{ft}\right)\end{array}\right]\\ =14.2psia+\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\left[\begin{array}{l}\left(62.4\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{20}{12}\text{ft}\right)-\\ \left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{60}{12}\text{ft}\right)+\\ \left(848.64\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{25}{12}\text{ft}\right)+\\ \left(49.92\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(\frac{30}{12}\text{ft}\right)\end{array}\right]\left(\frac{1\text{lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}{\text{ft}}^{\text{2}}}{144\text{in}{\text{.}}^{\text{2}}}\right)\\ =26.4\text{psia}\end{array}$

Thus, the absolute pressure at the center of the pipe is $\underset{\_}{26.4\text{psia}}$.