A 50 mm diameter syphon drawing oil of sp. gravity 0.82 from a reservoir is shown in Fig. 2.17. The head loss from point 1 to 2 is 1.5 m and from point 2 to 3 is 2.4 m. Determine flow rate of oil from syphon and also the pressure at point 2.

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Accepted Answer

Total head loss from 1 to 3 = [latex]1.5 + 2.4 =3.9 m[/latex]

Fix the datum line at point 3

[latex]S=\frac{ ho_{oil}}{ ho_{water}}[/latex]

[latex] ho_{oil} = S imes ho_{water} = 0.82 imes 10^{3} = 820 {kg}/{m^{3}}[/latex]

Applying Bernoull is equation between 1 and 3

[latex] \frac{P_{1}}{\gamma }+\frac{V^{2}_{1}}{2g}+z_{1}= \frac{P_{3}}{\gamma }+\frac{V^{2}_{3}}{2g}+z_{3}+h_{loss}[/latex]

The reservoir is considered large and therefore [latex]V_{1} = 0[/latex]
Also [latex]P_{1} = P_{3} = atmospheric pressure [/latex]

Substituting properly,

[latex]0 + 0 + 5 = 0 + \frac{V^{2}_{3}}{2g} + 0 + 3.9[/latex]

[latex] \frac{V^{2}_{3}}{2g} = 1.1 herefore V_{3} = \sqrt{2 imes 9.8 imes 1.1} = 4.65 {m}/{s}[/latex]

The velocity at exit is [latex] 4.65 {m}/{s}[/latex].

Diameter of the pipe = 50 mm is same; [latex] herefore V_{2} = V_{3} = 4.65 {m}/{s}[/latex]

Flow rate [latex] Q = a imes V =\frac{\pi}{4} imes \left(50 imes 10^{-3} ight)^{2} imes 4.65[/latex]

[latex] herefore Q = 9.12 imes 10^{-3} {m^{3}}/{s}[/latex]

Applying Bernoullis eq. between I and 2

[latex] \frac{P_{1}}{\gamma }+ \frac{V^{2}_{1}}{2g}+z_{1}= \frac{P_{2}}{\gamma }+ \frac{V^{2}_{2}}{2g}+z_{2}[/latex]

as [latex]P_{1} = 0[/latex] atmospheric and [latex]V_{1} = 0[/latex] large reservoir

[latex] 5 = \frac{P_{2}}{\gamma } + \frac{4.65^{2}}{2 imes 9.8 } + 7[/latex]

[latex] \frac{P_{2}}{\gamma } = 5 - 1.1 - 7 = - 3.1[/latex]

[latex] herefore P_{2} = 820 imes 9.8 imes \left(- 3.1 ight)[/latex]

[latex] P_{2} = 24.9 kPa[/latex] (negative)

Question 2.3: A 50 mm diameter syphon drawing oil of sp. gravity 0.82 from...

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