Chapter 3, Problem 46P

Repeat Prob. 3-48 by replacing the air with oil whose specific gravity is 0.72.

To determine

The pressure difference between two pipelines.

Answer to Problem 46P

The pressure difference is $10.336\text{kPa}$.The air column can be ignored since the density of air very low.

Explanation of Solution

Given information:

The density of seawater is $1035\text{kg}/{\text{m}}^{\text{3}}$.

The following figure represents freshwater and seawater flowing in parallel horizontal pipelines which are connected to each other by a double U-tube manometer.

  

Figure-(1)

Write the equation for the pressure difference of sea water and fresh water.

  $\Delta P=P_{\text{Hg}}+P_{\text{oil}}-P_{fw}-P_{sw}$   ...... (I)

Here, the pressure difference is $\Delta P$, the pressure on mercury is $P_{Hg}$, the pressure on oil is $P_{\text{oil}}$, pressure on freshwater is $P_{fw}$ and pressure on seawater is $P_{sw}$.

Write the expression for pressure on oil.

  $\begin{array}{c}{P}_{\text{oil}}={\rho }_{\text{oil}}\times g\times h_{\text{oil}}\\ P_{\text{oil}}=\left(S{G}_{\text{oil}}\times {\rho }_w\right)\times g\times h_{\text{oil}}\end{array}$   ...... (II)

Here, density of oil is ${\rho }_{\text{oil}}$, acceleration due to gravity is $g,$ height of oil column is $h_{\text{oil}}$, specific gravity of oil is $S{G}_{\text{oil}},$ and density of water is ${\rho }_w$.

Write the expression for density of mercury.

  ${\rho }_{\text{Hg}}=S{G}_{\text{Hg}}\times {\rho }_w$   ...... (III)

Here, specific gravity of mercury is $S{G}_{\text{Hg}}$ and density of water is ${\rho }_w$.

Write the expression for pressure on mercury.

  $P_{\text{Hg}}={\rho }_{\text{Hg}}\times g\times h_{\text{Hg}}$   ...... (IV)

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

Write the expression for pressure on seawater.

  $P_{sw}={\rho }_{sw}\times g\times h_{sw}$...... (V)

Here, pressure on seawater is $P_{sw}$, density of seawater is ${\rho }_{sw}$, acceleration due to gravity is $g$, height of seawater column is $h_{sw}$.

Write the expression for pressure on seawater.

  $P_{fw}={\rho }_{fw}\times g\times h_{fw}$...... (VI)

Here, pressure on freshwater is $P_{fw}$, density of freshwater is ${\rho }_{fw}$, acceleration due to gravity is $g$, height of freshwater column is $h_{fw}$.

Calculation:

Substitute $0.72$ for ${\rho }_{\text{oil}}$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $70\text{cm}$ for $h_{\text{oil}}$ in Equation (II).

  $\begin{array}{c}{P}_{\text{oil}}=0.72\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(70\text{cm}\right)\\ =\left(0.72\times 1000\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(70\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =\left(720\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\times \left(0.7\text{m}\right)\\ =4944.2\frac{\text{kg}}{{\text{ms}}^{\text{2}}}\end{array}$

  $\begin{array}{c}{p}_{\text{oil}}=4944.2\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =4944.2\text{Pa}\left(\frac{1\text{Pa}}{1000\text{kPa}}\right)\\ =4.944\text{kPa}\end{array}$

Substitute $13.6$ for $S{G}_{\text{Hg}}$ and $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$ in Equation (III).

  $\begin{array}{c}{\rho }_{\text{Hg}}=\left(13.6\right)\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\\ =13600\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $13600\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{Hg}}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $10\text{cm}$ for $h_{\text{Hg}}$ in Equation (IV).

  $\begin{array}{c}{P}_{\text{Hg}}=\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(10\text{cm}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(10\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =\left(13600\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.1\text{m}\right)\\ =13341.6\text{kg}/{\text{ms}}^{\text{2}}\end{array}$

  $\begin{array}{c}{P}_{\text{Hg}}=13341.6\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =13341.6\text{Pa}\left(\frac{1\text{Pa}}{1000\text{kPa}}\right)\\ =13.34\text{kPa}\end{array}$

Substitute $1035\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{sw}$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30\text{cm}$ for $h_{sw}$ in Equation (V).

  $\begin{array}{c}{P}_{sw}=\left(1035\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{cm}\right)\\ =\left(1035\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(30\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =\left(1035\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.3\text{m}\right)\\ =3045.0055\text{kg}/{\text{ms}}^{\text{2}}\end{array}$

  $\begin{array}{c}{P}_{sw}=3045.0055\text{kg}/{\text{ms}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =3045.0055\text{Pa}\left(\frac{1\text{Pa}}{1000\text{kPa}}\right)\\ =3.045\text{kPa}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{fw}$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $50\text{cm}$ for $h_{fw}$ in Equation (VI).

  $\begin{array}{c}{P}_{fw}=\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(50\text{cm}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(50\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.5\text{m}\right)\\ =4905\text{kg}/{\text{ms}}^{\text{2}}\end{array}$

  $\begin{array}{c}{P}_{fw}=4905\text{kPa}\left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =4905\text{kPa}\left(\frac{1\text{Pa}}{1000\text{kPa}}\right)\\ =4.9\text{kPa}\end{array}$

Substitute $13.34\text{kPa}$ for $P_{\text{Hg}}$, $4.944\text{}\text{kPa}$ for $P_{\text{oil}}$, $4.905\text{kPa}$ for $P_{fw}$ and $3.045\text{kPa}$ for $P_{sw}$ in Equation (I)/

  $\begin{array}{c}\Delta P=13.34\text{kPa+4}\text{.944}\text{kPa+}4.905\text{kPa+}3.045\text{kPa}\\ \text{=10}\text{.336}\text{kPa}\end{array}$

Conclusion:

The pressure difference is $10.336\text{kPa}$.The air column can be ignored since the density of air very low.