Chapter 3, Problem 3.132P

To determine

The exit velocity of siphon tube with friction head losses.

The volume flow rate of the siphon tube with friction head losses.

The variation between the exit velocity and volume flow rate with and without friction head losses.

Answer to Problem 3.132P

The exit velocity of siphon tube with friction head losses $1.614\text{m}/\text{s}$.

The volume flow rate of the siphon tube with friction losses head $126.69{\text{cm}}^{\text{3}}/\text{s}$.

The velocity and volume flow rate through the siphon tube is more if there is no head loss in the tube.

Explanation of Solution

Given information:

The diameter of the tube is $0.001\text{m}$, the length of the siphon tube is $2\text{m}$ The height of the water level is $60\text{cm}$, the height of the siphon tube above the water surface from the ground level is 90 cm, the height of the siphon tube with the ground at the exit is $-0.25\text{m}$.

The Figure-(1) show different height at different sections.

Figure-(1)

Here, the section (1) is for the top of the water level in the left region, the section (2) is at the exit of the siphon tube

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=h_f\end{array}$ …… (I)

Here, the pressure at section (1) is $p_1$, the velocity at section (1) is $V_1$, the height of the section (1) from the datum is $z_1$, the height of the section (2) from the datum is $z_2$, the density of water is $\rho$, the head loss due to friction is $h_f$ and the acceleration due to gravity is $g$.

Since the fluid at section (1) is stationary, so the velocity at the sections (1) is zero.

$V_1=0$

Since the pressure at the section (1) and section (2) is atmospheric, so the difference between the pressures is zero.

$p_1-p_2=0$

Write the expression for friction head loss in the siphon tube.

$h_f=5.4\left(\frac{V_2^2}{2g}\right)$

Here, the friction head loss in the siphon tube is $h_f$.

Substitute $0$ for $p_1-p_3$ and $0$ for $V_1$, $5.4\left(\frac{V_2^2}{2g}\right)$ for $h_f$ in Equation (I).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{\left(0\right)}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_3\right)=5.4\left(\frac{V_2^2}{2g}\right)\\ -\frac{V_2^2}{2g}+\left(z_1-z_3\right)=5.4\left(\frac{V_2^2}{2g}\right)\\ \left(z_1-z_3\right)=5.4\left(\frac{V_2^2}{2g}\right)+\frac{V_2^2}{2g}\\ \left(z_1-z_3\right)=\left(5.4+1\right)\left(\frac{V_2^2}{2g}\right)\end{array}$

$V_2^2=\frac{2g\left(z_1-z_3\right)}{6.4}$ ... (II)

Here, the exit velocity at section (2) with friction head losses is $V_2$.

Substitute $0$ for $p_1-p_3$ and $0$ for $V_1$, ${V^{\prime }}_2$ for $V_2$, $0$ for $h_f$ in Equation (I).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{\left(0\right)}{2g}-\frac{{V^{\prime }}_{}^{}}{}\right)\end{array}$

2 2 2g+z1−z2=0 V ′222g=z1−z2V′22=2gz1−z2 ... (III)

Here, the exit velocity at section (2) without friction head losses is ${V^{\prime }}_2^{}$

Write the expression for the flow rate at section (2) with friction losses.

$Q=A{V}_2$ …… (IV)

Here, the flow rate is $Q$, the siphon cross-sectional section (2) is $A$ and the velocity with friction head losses at section (2) is $V_2$.

Write the expression for the flow rate at section (2) without friction head losses.

$Q^{\prime }=A{V^{\prime }}_2$ …… (V)

Here, the flow rate is $Q^{\prime }$, the siphon cross-sectional section (2) is $A$ and the velocity without friction head losses at section (2) is ${V^{\prime }}_2$.

Calculation:

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.6\text{m}$ for $z_1$, $-0.25\text{m}$ for $z_2$ in Equation (II).

$\begin{array}{l}{V}_2^2=\frac{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.6\text{m}-\left(-0.25\text{m}\right)\right)}{6.4}\\ V_2^2=2.605{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\\ V_2=\sqrt{2.605{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ V_2=1.614\text{m}/\text{s}\end{array}$

Substitute $4.08\text{m}/\text{s}$ for $V_2$ and $\left(\frac{\pi }{4}\times d^2\right)$ for $A$ in Equation (IV)

$Q=\left(\frac{\pi }{4}\times d^2\right)V_2$

Here diameter of the siphon tube is $d=0.001\text{m}$.

$\begin{array}{c}Q=\left(\frac{\pi }{4}\times {\left(0.001\text{m}\right)}^2\right)\left(1.614\text{m}/\text{s}\right)\\ =\left(\frac{\pi }{4}\times \left(1\times {10}^{-6}{\text{m}}^2\right)\right)\left(1.614\text{m}/\text{s}\right)\\ =\left(7.85\times {10}^{-7}{\text{m}}^2\right)\left(1.614\text{m}/\text{s}\right)\\ =\left(126.69\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}\right)\left(\frac{1{\text{cm}}^{\text{3}}/\text{s}}{{10}^{-6}{\text{m}}^{\text{3}}/\text{s}}\right)\end{array}$

$Q=126.69{\text{cm}}^{\text{3}}/\text{s}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0.6\text{m}$ for $z_1$, $-0.25\text{m}$ for $z_2$ in Equation (III).

$\begin{array}{l}{V^{\prime }}_2^2=2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(0.6\text{m}-\left(-0.25\text{m}\right)\right)\\ {V^{\prime }}_2^2=16.67{\text{m}}^2/{\text{s}}^{\text{2}}\\ {V^{\prime }}_2=\sqrt{16.67{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ {V^{\prime }}_2=4.08\text{m}/\text{s}\end{array}$

Substitute $4.08\text{m}/\text{s}$ for ${V^{\prime }}_2$ and $\left(\frac{\pi }{4}\times d^2\right)$ for $A$ in Equation (V)

$Q^{\prime }=\left(\frac{\pi }{4}\times d^2\right){V^{\prime }}_2$

Here diameter of the siphon tube is $d=0.001\text{m}$.

$\begin{array}{c}{Q}^{\prime }=\left(\frac{\pi }{4}\times {\left(0.001\text{m}\right)}^2\right)4.08\text{m}/\text{s}\\ =\left(\frac{\pi }{4}\times \left(1\times {10}^{-6}{\text{m}}^2\right)\right)4.08\text{m}/\text{s}\\ =\left(7.85\times {10}^{-7}{m}^2\right)4.08\text{m}/\text{s}\\ =320.28\times {10}^{-6}{\text{m}}^{\text{3}}/\text{s}\times \left(\frac{1{\text{cm}}^{\text{3}}/\text{s}}{{10}^{-6}{\text{m}}^{\text{3}}/\text{s}}\right)\end{array}$

$Q^{\prime }=320.28{\text{cm}}^{\text{3}}/\text{s}$

Conclusion:

The exit velocity of siphon tube with friction head losses $1.614\text{m}/\text{s}$.

The volume flow rate of the siphon tube with friction losses head $126.69{\text{cm}}^{\text{3}}/\text{s}$.

The velocity and volume flow rate through the siphon tube is more if there is no head loss in the tube.