Chapter 1.11, Problem 64P

A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the manometer arm with the (a) higher and (b) lower fluid level being attached to the tank.

To determine

The absolute pressure in the tank for the cases of manometer.

The absolute pressure in the tank for cases of the manometer arm with the higher.

Answer to Problem 64P

The absolute pressure in the tank for the cases of manometer is $\underset{\_}{1.26\text{psia}}$.

The absolute pressure in the tank for cases of the manometer arm with the higher is $\underset{\_}{11.44\text{psia}}$.

Explanation of Solution

Determine the density of fluid.

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

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

Write the expression for pressure difference in cases of the manometer arm.

$\Delta P={\rho }_{\text{fluid}}gh$ (II)

Here, the density of the fluid is ${\rho }_{\text{fluid}}$, the acceleration of gravity is $g$, and the height of the fluid in cases is $h$.

Write the expression for the absolute pressure in the tank for cases of the manometer arm with the higher.

$P_{\text{abs}}=P_{\text{atm}}-P_{\text{vac}}$ (III).

Here, the local atmospheric pressure is $P_{\text{atm}}$ and the vacuum pressure of the manometer is $P_{\text{vac}}$

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 $1.25$ for $S{G}_{\text{fluid}}$ and $62.4\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{w}}$ in Equation (I).

$\begin{array}{c}{\rho }_{\text{Hg}}=1.25\times 62.4\text{lbm}/{\text{ft}}^{\text{3}}\\ =78.0\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Substitute $78\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{\text{fluid}}$, $32.174\text{ft}/{\text{s}}^{\text{2}}$ for $g$, and $28/12\text{ft}$ for $h$ in Equation (II).

$\begin{array}{c}\Delta P=\left(78\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.174\text{ft}/{\text{s}}^{\text{2}}\right)\left(28/12\text{ft}\right)\\ =\left(78\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(32.174\text{ft}/{\text{s}}^{\text{2}}\right)\left(28/12\text{ft}\right)\times \left(\frac{1\text{lbf}}{32.174\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\left(\frac{1{\text{ft}}^{\text{2}}}{144{\text{in}}^{\text{2}}}\right)\\ =1.26\text{psia}\end{array}$

Substitute $12.7\text{psia}$ for $P_{\text{atm}}$ and $1.26\text{psia}$ for $P_{\text{vac}}$ in Equation (III).

$\begin{array}{c}{P}_{\text{abs}}=\left(12.7-1.26\right)\text{psia}\\ \text{=11}\text{.44}\text{psia}\end{array}$

Thus, the absolute pressure in the tank for cases of the manometer arm with the higher is $\underset{\_}{11.44\text{psia}}$.

To determine

The absolute pressure in the tank for cases of the manometer arm with the lower.

Answer to Problem 64P

The absolute pressure in the tank for cases of the manometer arm with the lower is $\underset{\_}{13.96\text{psia}}$.

Explanation of Solution

Write the expression for the absolute pressure in the tank for cases of the manometer arm with the lower.

$P_{\text{abs}}=P_{\text{gage}}+P_{\text{atm}}$ (IV).

Here, the local atmospheric pressure is $P_{\text{atm}}$ and the gage pressure of the manometer is $P_{\text{gage}}$

Conclusion:

Substitute $12.7\text{psia}$ for $P_{\text{atm}}$ and $1.26\text{psia}$ for $P_{\text{gage}}$ in Equation (III).

$\begin{array}{c}{P}_{\text{abs}}=\left(12.7+1.26\right)\text{psia}\\ \text{=13}\text{.96}\text{psia}\end{array}$

Thus, absolute pressure in the tank for cases of the manometer arm with the lower is $\underset{\_}{13.96\text{psia}}$.