Chapter 3, Problem 55P

A multifluid container is connected to a U-tube. as shout in Fig. P3-55. For the given specific gravities and fluid column heights, determine the gage pressure at A. Also determine the height of a mercury column that would create the same pressure at A.

To determine

The gage pressure at $A$.

The height of mercury column that creates the same pressure.

Answer to Problem 55P

The gage pressure at point $\text{A}$ is $1.597\text{kPa}$ and the equivalent height of mercury column is $0.01197\text{m}$.

Explanation of Solution

Given information:

The following figure shows the arrangement of multifluid container connected to a U-tube manometer.

Figure-(1)

Write the expression for the density of oil using specific gravity.

  ${\rho }_{oil}=S{G}_{oil}{\rho }_w$....... (I)

Here, the density of oil is ${\rho }_{oil}$, specific gravity of oil is $S{G}_{oil},$ and density of water is ${\rho }_w$.

Write the expression for the density of glycerin using specific gravity.

  ${\rho }_{gly}=S{G}_{gly}{\rho }_w$....... (II)

Here, the density of glycerin is ${\rho }_{oil}$, specific gravity of glycerin is $S{G}_{oil},$ and density of water is ${\rho }_w$.

Write the expression for gage pressure at $A$ by equating the pressures in right limb and left limb.

  $\begin{array}{l}{P}_{right}=P_{left}\\ P_{gage}+{\rho }_{gly}g{h}_{gly}={\rho }_{oil}g{h}_{oil}+{\rho }_{w}g{h}_w+{\rho }_{gly}g{h}_{gly}\\ \frac{P_{gage}}{g}+{\rho }_{gly}{\left(h_{gly}\right)}_{right}={\rho }_{oil}{h}_{oil}+{\rho }_w{h}_w+{\rho }_{gly}{\left(h_{gly}\right)}_{left}\end{array}$....... (III)

Here, gage pressure is $P_{gage}$, density of glycerin is ${\rho }_{gly}$, height of glycerin column in right limb is ${\left(h_{gly}\right)}_{right}$, acceleration due to gravity is $g$, density of oil ${\rho }_{oil}$, height of oil column is $h_{oil}$, density of water is ${\rho }_w$, height of water column is $h_w,$ and height of glycerin column in left limb is ${\left(h_{gly}\right)}_{left}$.

Write the expression for height of mercury column that creates the same pressure at A

  $h_{Hg}=\frac{P_{gage}}{{\rho }_{Hg}g}$....... (IV)

Here, height of mercury column is $h_{Hg}$, gage pressure is $P_{gage}$, density of mercury is ${\rho }_g,$ and acceleration due to gravity is $g$.

Calculation:

Substitute $0.9$ for $S{G}_{oil}$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (I).

  $\begin{array}{c}{\rho }_{oil}=0.9\times 1000\text{kg}/{\text{m}}^{\text{3}}\\ =900\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.26$ for $S{G}_{gly}$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (II).

  $\begin{array}{c}{\rho }_{gly}=1.26\times 1000\text{kg}/{\text{m}}^{\text{3}}\\ =1260\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $900\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{oil}$, $1260\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{gly}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.9\text{m}$ for ${\left(h_{gly}\right)}_{right}$, $0.8\text{m}$ for $h_{oil}$, $0.35\text{m}$ for $h_w,$ and $0.18\text{m}$ for ${\left(h_{gly}\right)}_{left}$ in Equation (III).

  $\begin{array}{l}\left[\begin{array}{l}\frac{P_{gage}}{9.81\text{m}/{\text{s}}^{\text{2}}}\\ +\left(1260\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.9\text{m}\right)\end{array}\right]=\left[\begin{array}{l}\left(900\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.8\text{m}\right)\\ +\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.35\text{m}\right)\\ +\left(1260\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.18\text{m}\right)\end{array}\right]\\ \frac{P_{gage}}{9.81\text{m}/{\text{s}}^{\text{2}}}=\left[\begin{array}{l}\left(720\text{kg}/{\text{m}}^{\text{2}}\right)+\left(350\text{kg}/{\text{m}}^{\text{2}}\right)\\ +\left(226.8\text{kg}/{\text{m}}^{\text{2}}\right)-\left(1.134\text{kg}/{\text{m}}^{\text{2}}\right)\end{array}\right]\\ \frac{P_{gage}}{9.81\text{m}/{\text{s}}^{\text{2}}}=162.8\text{kg}/{\text{m}}^{\text{2}}\\ P_{gage}=162.8\text{kg}/{\text{m}}^{\text{2}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\end{array}$

  $\begin{array}{l}{P}_{gage}=1597.068\text{kg}/{\text{ms}}^{\text{2}}\\ P_{gage}=1597.068\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{kPa}}{1000\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ P_{gage}=1.597\text{kPa}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{Hg}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g,$ and $1.597\text{kPa}$ for $P_{gage}$ in Equation (IV).

  $\begin{array}{c}{h}_{Hg}=\frac{1.597\text{kPa}}{\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{1.597\text{kPa}\left(\frac{1000\text{kg}/{\text{ms}}^{\text{2}}}{1\text{kPa}}\right)}{133416\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\\ =\frac{1597\text{kg}/{\text{ms}}^{\text{2}}}{133416\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}\\ =0.01197\text{m}\end{array}$

Conclusion:

The gage pressure at point $\text{A}$ is $1.597\text{kPa}$ and the equivalent height of mercury column is $0.01197\text{m}$.