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Flow in Siphon

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A siphon consisting of a 3 cm diameter tube is used to drain water from a tank. The outlet end of the tube is 2 m below the water surface in the tank. Neglecting friction, calculate the discharge. If the peak point of the siphon is 1.4 m above the water surface in the tank, estimate the pressure of fluid at the point of siphon.

**Data:**

Dia of tube = 3 cm = 0.03 m

Formula:

Bernoulli's equation for frictionless flow is

\(\displaystyle \frac{p}{\rho g} + \frac{v^2}{2g} + z = \text{ constant} \)

Discharge (Q) = cross sectional area x velocity

Calculations:

Applying Bernoulli's equation for the points 1 and 3, ( i.e. comparing the energy levels for the fluid in the tank surface and at the discharge point of tube)

p_{1 }= 0 N/m^{2}(g)

p_{3 }= 0 N/m^{2}(g)

z_{1} = 0 m

z_{3} = -2 m

Since the rate of fall of liquid level in the tank is almost negligible,

v_{1} = 0 m/sec.

Therefore,

0 + 0 + 0 = 0 + (v_{3}^{2} / 2g) - 2

v_{3} = (2 x 2g)^{0.5} = 6.265 m/sec

Q = (π/4)D^{2} v = (π/4) x 0.03^{2} x 6.265 = 0.00443
m^{3}/sec = **15.94
m**^{3}/hr

Applying Bernoulli's equation for the points 1 and 2, ( i.e. comparing the energy levels for the fluid at the tank surface to the peak point of siphon)

p_{1 }= 0 N/m^{2}(g)

z_{2} = 1.4 m

v_{2 = }v_{3 }= 6.265 m/sec (since the cross sectional area of sections 2 and 3 are the same)

0 + 0 + 0 = p_{2} / (ρg) + 6.265^{2} / (2g) + 1.4

p_{2} / (ρg) = -3.4 m

p_{2 }= -3.4 x 1000 x 9.812 N/m^{2}(g) = -33360.8 N/m^{2}(g)

Absolute pressure at point 2 = 101325 - 33360.8 = **67964.2 N/m**^{2}(a)

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Last Modified on: 01-May-2024

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