Chapter 15, Problem 67PQ

(a)

To determine

Volume flow rate between the two sections of the hose having a pressure difference $5.00\text{kPa}$.

Answer to Problem 67PQ

The volume flow rate between the two sections of the hose having a pressure difference $5.00\text{kPa}$ is $\underset{\_}{2.67\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}}$.

Explanation of Solution

Bernoulli’s theorem states that the energy stored in the olive oil flowing through both narrow and wide part of hose is constant.

Apply Bernoulli’s theorem in the given situation.

  $P_1+\frac{1}{2}\rho v_1{}^2+\rho g{y}_1=P_2+\frac{1}{2}\rho v_2{}^2+\rho g{y}_2$                                                                     (I)

Here, $P_1$ is the pressure of olive oil flowing through the wide part of hose, $\rho$ is the density of olive oil, $g$ is the acceleration due to gravity, $y_1$ is the height of oil in the wide part of hose, $v_1$ is the velocity of oil, $P_2$ is the pressure of oil flowing through the narrow part of hose, $v_2$ is the velocity of oil in narrow part of hose, and $y_2$ is the height of the oil in the narrow part of pipe.

Apply the continuity equation to find the relation between the velocities of oil in both parts of the hose.

  $A_1{v}_1=A_2{v}_2$                                                                                                          (II)

Here, $A_1$ is the area of wide part of hose and $A_2$ is the area of the narrow part of the hose.

Both wide and narrow parts of the hose are same height. Therefore, $y_1$ is equal to $y_2$.

Rewrite equation (I) to find $P_1-P_2$.

  $\begin{array}{c}{P}_1-P_2=\frac{1}{2}\rho v_2{}^2-\frac{1}{2}\rho v_1{}^2\\ \Delta P=\frac{1}{2}\rho v_2{}^2-\frac{1}{2}\rho v_1{}^2\end{array}$                                                                                            (III)

Here, $\Delta P$ is the pressure difference between the wide and narrow part of hose.

Write the equation to find the volume flow rate of the wide part of hose.

  $R_1=A_1{v}_1$                                                                                                            (IVa)

Here, $R_1$ is the volume flow rate of the wide part of hose.

Write the equation to find the volume flow rate of the narrow part of hose.

  $R_2=A_2{v}_2$                                                                                                           (IVb)

Here, $R_2$ is the volume flow rate of the wide part of hose.

Write the equation to find the area of the wide part of hose.

  $A_1=\pi r_1{}^2$                                                                                                           (V)

Here, $r_1$ is the radius of the wide part of hose.

Write the equation to find the area of the narrow part of hose.

  $A_2=\pi r_2{}^2$                                                                                                           (VI)

Here, $r_2$ is the radius of the narrow part of hose.

Write the equation to find the radius of wide part of hose.

  $r_1=\frac{d_1}{2}$                                                                                                                 (VII)

Here, $d_1$ is the diameter of the wide part of hose.

Write the equation to find the radius of narrow part of hose.

  $r_2=\frac{d_2}{2}$                                                                                                                  (VIII)

Here, $d_2$ is the diameter of the narrow part of hose.

Conclusion:

Substitute $3.00\text{cm}$ for $d_1$ in equation (VII) to get $r_1$.

  $\begin{array}{c}{r}_1=\frac{3.00\text{cm}}{2}\\ =1.50\text{cm}\end{array}$

Substitute $1.50\text{cm}$ for $r_1$ in equation (V) to get $A_1$.

  $A_1=\pi {\left(1.50\text{cm}\right)}^2$

Substitute $1.00\text{cm}$ for $d_2$ in equation (VIII) to get $r_2$.

  $\begin{array}{c}{r}_2=\frac{1.00\text{cm}}{2}\\ =0.500\text{cm}\end{array}$

Substitute $0.500\text{cm}$ for $r_2$ in (VI) to get $A_2$.

  $A_2=\pi {\left(0.500\text{cm}\right)}^2$

Substitute $\pi {\left(1.50\text{cm}\right)}^2$ for $A_1$ and $\pi {\left(0.500\text{cm}\right)}^2$ for $A_2$ in equation (II) and solve for $v_2$.

  $\begin{array}{c}\pi {\left(1.50\text{cm}\right)}^2{v}_1=\pi {\left(0.500\text{cm}\right)}^2{v}_2\\ v_2=9v_1\end{array}$

Substitute $9v_1$ for $v_2$ and $875\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (III).

  $\begin{array}{c}\Delta P=\frac{1}{2}\rho {\left(9v_1\right)}^2-\frac{1}{2}\rho v_1{}^2\\ =\frac{1}{2}\left(875\text{kg}/{\text{m}}^{\text{3}}\right)\left(80v_1{}^2\right)\\ =35000v_1{}^2\end{array}$

Rearrange for $v_1$.

  $v_1=\sqrt{\frac{\Delta P}{35000}}$

Substitute $5\text{kPa}$ for $\Delta P$ in above equation to get $v_1$.

  $\begin{array}{c}{v}_1=\sqrt{\frac{5\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{35000}}\\ =0.378\text{m}/\text{s}\end{array}$

Substitute $0.378\text{m}/\text{s}$ for $v_1$ and $\pi {\left(1.50\text{cm}\right)}^2$ for $A_1$ in equation (IV a) and solve for $R_1$.

  $\begin{array}{c}{R}_1=\left(0.378\text{m}/\text{s}\right)\pi {\left(1.50\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =2.67\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Therefore, the volume flow rate between the two sections of the hose having a pressure difference $5.00\text{kPa}$ is $\underset{\_}{2.67\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}}$.

(b)

To determine

Volume flow rate between the two sections of the hose having a pressure difference $12.5\text{kPa}$.

Answer to Problem 67PQ

The volume flow rate between the two sections of the hose having a pressure difference $5.00\text{kPa}$ is $\underset{\_}{4.22\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}}$.

Explanation of Solution

Refer sub part (b).

Substitute $\pi {\left(1.50\text{cm}\right)}^2$ for $A_1$ and $\pi {\left(0.500\text{cm}\right)}^2$ for $A_2$ in equation (II) and solve for $v_1$.

  $\begin{array}{c}\pi {\left(1.50\text{cm}\right)}^2{v}_1=\pi {\left(0.500\text{cm}\right)}^2{v}_2\\ v_1=\frac{1}{9}{v}_2\end{array}$

Substitute $\frac{1}{9}{v}_2$ for $v_1$ and $875\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in equation (III).

  $\begin{array}{c}\Delta P=\frac{1}{2}\rho {\left(v_2\right)}^2-\frac{1}{2}\rho {\left(\frac{1}{9}{v}_2{}^2\right)}^2\\ =\left(875\text{kg}/{\text{m}}^{\text{3}}\right)\left(0.494v_2{}^2\right)\\ =432.25v_2{}^2\end{array}$

Rearrange for $v_1$.

  $v_2=\sqrt{\frac{\Delta P}{432.25}}$

Substitute $12.5\text{kPa}$ for $\Delta P$ in above equation to get $v_2$.

  $\begin{array}{c}{v}_2=\sqrt{\frac{12.5\text{kPa}\left(\frac{1\text{Pa}}{{10}^{-3}\text{kPa}}\right)}{432.25}}\\ =5.377\text{m}/\text{s}\end{array}$

Substitute $5.377\text{m}/\text{s}$ for $v_2$ and for $\pi {\left(0.500\text{cm}\right)}^2$ $A_2$ in equation (IV b) and solve for $R_2$.

  $\begin{array}{c}{R}_2=\left(5.377\text{m}/\text{s}\right)\pi {\left(0.500\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =4.22\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

Therefore, the volume flow rate between the two sections of the hose having a pressure difference $12.5\text{kPa}$ is $\underset{\_}{4.22\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}}$.