Question

For the siphon, calculate (a) the volume flow rate of oil from the tank and (b) the pressures at points A-D.

Answer 1

(a)

Assume the top surface of the water to be section \(E\) and the jet where the water leaves the nozzle be section \(F\). Assume the section at \(F\) be the reference datum plane.

Write the Bemoulli's equation for sections \(E\) and \(F\). \(\frac{p_{E}}{\gamma}+\frac{v_{E}^{2}}{2 g}+z_{E}=\frac{p_{F}}{\gamma}+\frac{v_{F}^{2}}{2 g}+z_{F}\)

Here, \(p_{E}\) is the pressure at section \(E, p_{F}\) is the pressure at section \(F, v_{E}\) is velocity of flow

at \(E, v_{F}\) is the velocity of flow at \(F, \gamma\) is the specific weight of water and \(g\) is the acceleration due to gravity.

Since sections \(E\) and \(F\) are exposed to the atmosphere, the pressure head at this section is zero.

Rewrite equation (1), then the equation will be, \(0+0+z_{E}=0+\frac{v_{F}^{2}}{2 g}+0\)

$$ \begin{aligned} z_{E} &=\frac{v_{F}^{2}}{2 g} \\ v_{F}^{2} &=2 g z_{E} \\ &=2 \times 9.81 \times 10 \\ v_{F} &=14 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Calculate the volume rate flow of water by using the continuity equation. \(Q=A_{F} v_{F}\)

\(Q=\left(\frac{\pi D_{F}^{2}}{4}\right) v_{F}\)

$$ Q=\frac{\pi \times\left(25 \operatorname{mm} \times \frac{1 \mathrm{~m}}{1000 \mathrm{~mm}}\right)^{2}}{4} \times 14 \mathrm{~m} / \mathrm{s} $$

\(=6.87 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\)

Therefore, the volume rate of water through the system is \(6.87 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\)

\begin{tabular}{l}

4 \\

\hline

\end{tabular}

Calculate the area of cross-section of pipe at point \(A\) using the following equation:

$$ \begin{aligned} A &=\frac{\pi d_{A}^{2}}{4} \\ &=\frac{\pi(0.05)^{2}}{4} \\ &=1.963\left(10^{-3}\right) \mathrm{m}^{2} \end{aligned} $$

Calculate the velocity at point \(A\).

$$ \begin{aligned} Q &=A_{A} v_{A} \\ v_{A} &=\frac{Q}{A_{A}} \\ v_{A} &=\frac{6.87\left(10^{-3}\right)}{1.963\left(10^{-3}\right)} \\ &=3.499 \mathrm{~m} / \mathrm{s} \end{aligned} $$

Since the tube is of constant area, velocity of flow at \(A, B, C\) and \(D\) are same.

Write the Bemoulli's equation for sections \(E\) and \(A\). \(\frac{p_{E}}{\gamma}+\frac{v_{E}^{2}}{2 g}+z_{E}=\frac{p_{A}}{\gamma}+\frac{v_{A}^{2}}{2 g}+z_{A}\)

Since \(z_{E}=z_{A}\)

$$ \begin{array}{l} 0+0+z_{A}=\frac{p_{A}}{\gamma_{f}}+\frac{v_{A}^{2}}{2 g}+z_{A} \\ p_{A}=-\gamma_{f}\left(\frac{v_{A}^{2}}{2 g}\right) \\ p_{A}=-s g \times \gamma_{w}\left(\frac{v_{A}^{2}}{2 g}\right) \\ p_{A}=-0.86 \times 9810 \times \frac{3.499^{2}}{2 \times 9.81} \\ p_{A}=-5264.49 \mathrm{~N} / \mathrm{m}^{2} \\ p_{A}=-5.264 \mathrm{kPa} \end{array} $$

Note: \(\gamma_{w}\) is specific weight of water and sg is specific gravity of oil. \(\gamma_{w}=9810 \mathrm{~N} / \mathrm{m}^{3}\)

Write the Bemoulli's equation for sections \(E\) and \(B\).

\(\frac{p_{E}}{\gamma}+\frac{v_{E}^{2}}{2 g}+z_{E}=\frac{p_{B}}{\gamma_{f}}+\frac{v_{B}^{2}}{2 g}+z_{B}\)

\(0+0+0=\frac{p_{B}}{s g \times \gamma_{w}}+\frac{v_{B}^{2}}{2 g}+3\)

\(\frac{p_{B}}{s g \times \gamma_{w}}=-\left(\frac{3.499^{2}}{2 \times 9.81}+3\right)\)

\(\frac{p_{B}}{0.86 \times 9810}=-3.624\)

\(p_{B}=-30574.238 \mathrm{~Pa}\)

\(p_{B}=-30.574 \mathrm{kPa}\)

Therefore, the pressure at point \(B\) is \(-30.574 \mathrm{kPa}\).

At section \(C\), the kinetic head is the same as that of section \(A\) as both the sections are on the same elevation. And also the elevation heads for sections \(A\) and \(C\) are equal. Therefore, the pressure at both the sections is same. \(p_{c}=p_{4}=-5.264 \mathrm{kPa}\)

Therefore, the pressure at point \(C\) is \(-5.264 \mathrm{kPa}\).

Write the Bemoulli's equation for sections \(F\) and \(D\). \(\frac{p_{F}}{\gamma}+\frac{v_{F}^{2}}{2 g}+z_{F}=\frac{p_{D}}{\gamma_{f}}+\frac{v_{D}^{2}}{2 g}+z_{D}\)

\(0+\frac{v_{F}^{2}}{2 g}+0=\frac{p_{D}}{\gamma_{f}}+\frac{v_{D}^{2}}{2 g}+0\)

\(\frac{14^{2}}{2 \times 9.81}=\frac{p_{D}}{0.86 \times 9810}+\frac{3.499^{2}}{2 \times .981}\)

\(p_{D}=79015.5 \mathrm{~Pa}\)

\(p_{D}=79.015 \mathrm{kPa}\)

Therefore, the pressure at point \(D\) is \(79.015 \mathrm{kPa}\).