Chapter 14, Problem 40SP

Fuel oil of density $820{\text{ kg/m}}^{\text{3}}$ flows through a venturi meter having a throat diameter of 4.0 cm and an entrance diameter of 8.0 cm. The pressure drop between entrance and throat is 16 cm of mercury. Find the flow. The density of mercury is 13 600 ${\text{kg/m}}^{\text{3}}$.

To determine

The rate of flow if a fuel oil flows through a venturi meter and the pressure drop between the entrance and the throat is $16\text{ cm}$ of mercury.

Answer to Problem 40SP

Solution:

$9.3\times {10}^{-3}{\text{ m}}^3\text{/s}$

Explanation of Solution

Given data:

The density of fuel oil is $820{\text{ kg/m}}^3$.

The diameter of the throat of ventureis $4.0\text{ cm}$.

The entrance diameter of the venture is $\text{8}\text{.0 cm}$.

The density of mercury is ${\text{13600 kg/m}}^3$.

The dropin pressure between entrance and throat of the venture is $16\text{ cm of mercury}$.

Formula used:

Write the expression for the equation of continuity:

$J=A_1{v}_1=A_2{v}_2$

Here, $v_1$ and $v_2$ are the average speed of the fluids at the entrance and the exit section of the venture, respectively, $A_1$ is the cross-sectional area of the entrance point of the venture, $A_2$ is the cross-sectional area of the throatat the exit point of the venture, and $J$ is the rate of the discharge.

Write the expression for Bernoulli’s equation:

$P_1+\frac{1}{2}{\rho }_1{v}_1{}^2+h_1{\rho }_{2}g=P_2+\frac{1}{2}{\rho }_2{v}_2{}^2+h_2{\rho }_{2}g$

Here, $P_1$ is the absolute pressure at the entrance of the venture, ${\rho }_1$ is the density of the fluid at the entrance of the venture, $v_1$ is the average velocity of the oil at the entrance of the venture, $h_1$ is the height of the venture’s entrance from the datum surface, $P_2$ is the absolute pressure at the throat of the venture, ${\rho }_2$ is the density of the oil at the throat of the venture, $v_2$ is the average velocity of the oil at the throat of the venture, and $h_2$ is the height of the throat from the datum surface.

Write the expression for area:

$A=\frac{\pi }{4}{d}^2$

Here, $d$ is the diameter.

Write the expression forabsolute pressure:

$P=\rho gh$

Here, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, and $h$ is the height of the column.

Explanation:

Recall the expression for entrance cross-sectional area of the venture:

$A_1=\frac{\pi }{4}{d}_1{}^2$

Here, $A_1$ is the area of theentrance cross-sectional area of the venture and $d_1$ is the entrance diameter of the venture.

Substitute $\text{8 cm}$ for $d_1$

$A_1=\frac{\pi }{4}{\left(\text{8 cm}\right)}^2$

Recall the expression for throat cross-sectional area of the venture:

$A_2=\frac{\pi }{4}{d}_2{}^2$

Here, $A_2$ is the area of the throat cross-sectional area of the venture and $d_2$ is the diameter of the throat.

Substitute $\text{4 cm}$ for $d_2$

$A_2=\frac{\pi }{4}{\left(4\text{ cm}\right)}^2$

Recall the expression for the equation of continuity:

$A_1{v}_1=A_2{v}_2$

Substitute $\frac{\pi }{4}{\left(8\text{ cm}\right)}^2$ for $A_1$ and $\frac{\pi }{4}{\left(4\text{ cm}\right)}^2$ for $A_2$

$\begin{array}{c}\frac{\pi }{4}{\left(8\text{ cm}\right)}^2{v}_1=\frac{\pi }{4}{\left(4\text{cm}\right)}^2{v}_2\\ 64v_1=16v_2\\ v_2=4v_1\end{array}$

Calculate the value of the difference in pressure:

Recall the expression for drop in pressure:

$P_1-P_2=\rho gh$

Here, $P_1-P_2$ is the dropin pressure, $\rho$ is the density of the mercury, and $h$ is the height of the column.

Substitute $13,600{\text{ kg/m}}^3$ for $\rho$, $16\text{ cm}$ for $h$, and $9.81{\text{ m/s}}^2$ for $g$

$\begin{array}{c}{P}_1-P_2=\left(13,600{\text{ kg/m}}^3\right)\left(9.81{\text{ m/s}}^2\right)\left(16\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)\right)\\ =21,346.56\text{ Pa}\end{array}$

Recall the expression for Bernoulli’s equation:

$P_1+\frac{1}{2}{\rho }_1{v}_1{}^2+h_1\rho g=P_2+\frac{1}{2}\rho v_2{}^2+h_2{\rho }_{2}g$

Here, fluid is same at the entrance and the exit point of the venture. Therefore, ${\rho }_1={\rho }_2=\rho$.

Substitute $\rho$ for ${\rho }_1$ and $\rho$ for ${\rho }_2$

$P_1+\frac{1}{2}\rho v_1{}^2+h_1\rho g=P_2+\frac{1}{2}\rho v_2{}^2+h_2\rho g$

Consider the height of the throat and the entrance point is same. Therefore, $h_1=h_2$.

Substitute $h_1$ for $h_2$

$\begin{array}{c}{P}_1+\frac{1}{2}\rho v_1{}^2+h_1\rho g=P_2+\frac{1}{2}\rho v_2{}^2+h_1\rho g\\ P_1-P_2=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)\end{array}$

Substitute $820{\text{ kg/m}}^3$ for $\rho$, $21346.56\text{ Pa}$ for $P_1-P_2$, and $4v_1$ for $v_2$

$\begin{array}{c}21,346.56\text{ Pa}=\frac{1}{2}\left(820{\text{ kg/m}}^3\right)\left({\left(4v_1\right)}^2-v_1{}^2\right)\\ 52.06=\left(16v_1{}^2-v_1{}^2\right)\\ 52.06=15v_1{}^2\\ v_1=1.86\text{ m/s}\end{array}$

Recall the expression for the discharge of the oil at the entrance point of the venture:

$J=A_1{v}_1$

Substitute $\frac{\pi }{4}{\left(8\text{ cm}\right)}^2$ for $A_1$ and $\text{1}\text{.86 m/s}$ for $v_1$

$\begin{array}{c}J=\frac{\pi }{4}{\left(8\text{ cm}\right)}^2{\left(\frac{1\text{ m}}{100\text{ cm}}\right)}^2\left(1.86\text{ m/s}\right)\\ =9.3\times {10}^{-3}{\text{ m}}^3\text{/s}\end{array}$

Conclusion:

Therefore, theflow rate of the oil is $9.3\times {10}^{-3}{\text{ m}}^3\text{/s}$.