A simple siphon system is shown in the following drawing. The diameter of the tu

Question

A simple siphon system is shown in the following drawing. The diameter of the tube is 3.5 cm. the tank is open to the atmosphere and can be considered to be at sea level. Assume the fluid is water and is an ideal incompressible fluid. Further assume that there is no energy loss in the tube. The tank is large enough that the overall velocity at point 1 of the fluid is negligible, v_1 = 0. The pipe is open to the atmosphere at point 3 as well and can be assumed to be at the same atmospheric pressure as point 1. Find the speed of the fluid as it exits the tube at point 3. Find the volumetric flow rate out of the tube at point 3.

Explanation / Answer

Given that

The diameter of tube=3.5 cm

Radius of the tube =1.75 cm

This problem belongs to principle of Bernoulli's theorem

So the principle

P+dgh+1/2dv^2= constant

Where p= pressure

d=density of fluid

H height

Vis velocity of flow of liquid

Now we find velocity at point 3

(A) and

Pressure p1=p2=p

Velocity v1=0

Height h1=1.4m

Height h2=3.4 m

Now substitute the values in Bernoulli's principle

P1+dgh1+1/2dv1^2 = p2+dgh2+1/2dv2^2

[P1-P2]+dg(h1-h2)=1/2d(v2^2--v1^2)

[P-P]+dg(1.4-3.4) =1/2d(V2^2 --0)

0+dg(-2)=1/2dV2^2

V2=[2g2]

=4×9.8 =6.3 m/s

The velocity at point 3 is 6.3 m/s

(b) and

Basing on equation of continuity

We find the volume of liquid flowing per second through a pipe

The volumetric flow rate through a pipe=AV2

=r^2×v2

=3.14×[1.75]^2×10^-4×6.3

=60.8×10^-4 m^3/s

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