Chapter 1.11, Problem 53P

The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine (a) the local atmospheric pressure and (b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0 85 at the same location.

To determine

The local atmospheric pressure in a liquid.

Answer to Problem 53P

The local atmospheric pressure in a liquid is $\underset{\_}{109.9535\text{kPa}}$.

Explanation of Solution

Show the free body diagram of the absolute pressure in water.

Determine the density of the liquid.

  $\rho =\text{SG}\times {\rho }_{{\text{H}}_{\text{2}}\text{O}}$                                     …… (I)

Here, the specific gravity is $\text{SG}$ and the density of water is ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$.

Write the expression of atmospheric pressure.

  $P_{\text{atm}}=P-\rho gh$                                          …… (II)

Here, the absolute pressure is $P$, the density is $\rho$, the height of the mercury column above the free surface is $h$, and acceleration of gravity is $g$.

Conclusion:

Substitute $0.85$ for $\text{SG}$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ in Equation (I).

  $\begin{array}{c}\rho =\left(0.85\right)\times \left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =850\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $185\text{kPa}$ for $P$, $850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $9\text{m}$ for $h$ in Equation (II).

  $\begin{array}{c}{P}_{\text{atm}}=\left(185\text{kPa}\right)-\left(850\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(9\text{m}\right)\\ =\left(185\text{kPa}\right)-\left(75046.5\text{N}/{\text{m}}^{\text{2}}\right)\times \left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =185\text{kPa}-75.0465\text{kPa}\\ =109.9535\text{kPa}\end{array}$

Thus, local atmospheric pressure in a liquid is $\underset{\_}{109.9535\text{kPa}}$.

To determine

The absolute pressure in a liquid.

Answer to Problem 53P

The absolute pressure in a liquid is $\underset{\_}{151.646\text{kPa}}$.

Explanation of Solution

Write the expression of absolute pressure.

  $P=P_{\text{atm}}+\rho gh$                                          …… (III)

Here, the absolute pressure is $P$, the density is $\rho$, the height of the mercury column above the free surface is $h$, and acceleration of gravity is $g$.

Conclusion:

Substitute $109.9535\text{kPa}$ for $P_{\text{atm}}$, $850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $5\text{m}$ for $h$ in Equation (III).

  $\begin{array}{c}P=\left(109.9535\text{kPa}\right)+\left(850\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(5\text{m}\right)\\ =\left(109.9535\text{kPa}\right)+\left(41692.5\text{N}/{\text{m}}^{\text{2}}\right)\times \left(\frac{1\text{kPa}}{1000\text{N}/{\text{m}}^{\text{2}}}\right)\\ =109.9535\text{kPa}+41.6825\text{kPa}\\ =\text{151}\text{.646}\text{kPa}\end{array}$

Thus, the absolute pressure in a liquid is $\underset{\_}{\text{151}\text{.646}\text{kPa}}$.