Chapter 14, Problem 24P

(a)

To determine

The volume flow rate as a function of $\Delta P$.

Answer to Problem 24P

The volume flow rate as a function of $\Delta P$ is $I_{\text{V}}=3.933\times {10}^{-6}\sqrt{\Delta P}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

Density of liquid is $850\text{kg}/{\text{m}}^{\text{3}}$, radius of the first horizontal tube is $1.0\text{cm}$ and the radius of second horizontal tube is $0.50\text{cm}$.

The cross section area of first pipe is,

$A_1=\pi r_1^2$

$A_1$ is the cross section area of first pipe.

$r_1$ is the radius of the first pipe.

The cross section area of the second pipe is,

$A_2=\pi r_2^2$

$A_2$ is the cross section area of second pipe.

$r_2$ is the radius of the second pipe.

From the equation of continuity,

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

$v_1$ is the speed of liquid in first pipe.

$v_2$ is the speed of liquid in second pipe.

Substitute $\pi r_1^2$ for $A_1$ and $\pi r_2^2$ for $A_2$ in the above equation.

$\begin{array}{c}\pi r_1^2{v}_1=\pi r_2^2{v}_2\\ v_2=v_1{\left(\frac{r_1}{r_2}\right)}^2\end{array}$

Substitute $1.0\text{cm}$ for $r_1$ and $0.50\text{cm}$ for $r_2$ in the above equation.

$\begin{array}{c}{v}_2=v_1{\left(\frac{1.0\text{cm}}{0.50\text{cm}}\right)}^2\\ =4v_1\end{array}$ (I)

Apply Bernoulli’s equation for two pipes,

$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Since both the pipes are at same elevation then,

$h_1=h_2=h$

Therefore,

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

Substitute $4v_1$ for $v_2$ in the above equation.

$\begin{array}{c}\Delta P=\frac{1}{2}\rho \left({\left(4v_1\right)}^2-v_1^2\right)\\ =\frac{15}{2}\rho v_1^2\\ v_1={\left(\frac{2\Delta P}{15\rho }\right)}^{\frac{1}{2}}\end{array}$ (II)

Formula to calculate the volume flow rate is,

$I_{\text{V}}=A_1{v}_1$

$I_{\text{V}}$ is the volume flow rate.

Substitute $\pi r_1^2$ for $A_1$ and ${\left(\frac{2\Delta P}{15\rho }\right)}^{\frac{1}{2}}$ for $v_1$ in the above equation.

$I_{\text{V}}=\pi r_1^2{\left(\frac{2\Delta P}{15\rho }\right)}^{\frac{1}{2}}$

Substitute $1.0\text{cm}$ for $r_1$ and $850\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ in the above equation.

$\begin{array}{c}{I}_{\text{V}}=3.14\times {\left(1.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}^2{\left(\frac{2\Delta P}{15\times 850\text{kg}/{\text{m}}^{\text{3}}}\right)}^{\frac{1}{2}}\\ =3.933\times {10}^{-6}\sqrt{\Delta P}{\text{m}}^{\text{3}}/\text{s}\end{array}$ (III)

Conclusion:

Therefore, the volume flow rate as a function of $\Delta P$ is $I_{\text{V}}=3.933\times {10}^{-6}\sqrt{\Delta P}{\text{m}}^{\text{3}}/\text{s}$.

(b)

To determine

The volume flow rate for $\Delta P=6.0\text{kPa}$.

Answer to Problem 24P

The volume flow rate for $\Delta P=6.0\text{kPa}$ is $3.046\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

Density of liquid is $850\text{kg}/{\text{m}}^{\text{3}}$, radius of the first horizontal tube is $1.0\text{cm}$ and the radius of second horizontal tube is $0.50\text{cm}$.

Substitute $6.0\text{kPa}$ for $\Delta P$ in equation (III).

$\begin{array}{c}{I}_{\text{V}}=3.933\times {10}^{-6}\sqrt{\left(6.0\text{kPa}\times \frac{{10}^3\text{Pa}}{1\text{kPa}}\right)}{\text{m}}^{\text{3}}/\text{s}\\ =3.046\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

Therefore, the volume flow rate for $\Delta P=6.0\text{kPa}$ is $3.046\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}$.

(c)

To determine

The volume flow rate for $\Delta P=12.0\text{kPa}$.

Answer to Problem 24P

The volume flow rate for $\Delta P=12.0\text{kPa}$ is $4.31\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

Density of liquid is $850\text{kg}/{\text{m}}^{\text{3}}$, radius of the first horizontal tube is $1.0\text{cm}$ and the radius of second horizontal tube is $0.50\text{cm}$.

Substitute $12.0\text{kPa}$ for $\Delta P$ in equation (III).

$\begin{array}{c}{I}_{\text{V}}=3.933\times {10}^{-6}\sqrt{\left(12.0\text{kPa}\times \frac{{10}^3\text{Pa}}{1\text{kPa}}\right)}{\text{m}}^{\text{3}}/\text{s}\\ =4.31\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

Therefore, the volume flow rate for $\Delta P=12.0\text{kPa}$ is $4.31\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}$.