In a circular tube the diameter changes abruptly from D1 = 2 m to D2 = 3 m. The flow velocity in the part with smaller diameter is vi = 3 m/s. Determine if for water in the both parts of the tube there is laminar flow or tubulent flow. The kinematic viscosity of water is v= 1.24. 10^-6

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AskewTronics AskewTronics · Answer 1 · 2019-09-21T04:07:03+02:00

Answer:

The flow is turbulent at both the parts of the tube.

Explanation:

Given:

Water is flowing in circular tube.

Inlet diameter is [tex]d_{1}= 2[/tex]m.

Outlet diameter is [tex]d_{2}= 3[/tex]m.

Inlet velocity is [tex]V_{1}= 3[/tex] m/s.

Kinematic viscosity is [tex]\nu =1.24\times 10^{-6}[/tex] m²/s.

Concept:

Apply continuity equation to find the velocity at outlet.

Apply Reynolds number equation for flow condition.

Step1

Apply continuity equation for outlet velocity as follows:

[tex]A_{1}V_{1}=A_{2}V_{2}[/tex]

[tex]\frac{\pi}{4}d^{2}_{1}V_{1}=\frac{\pi}{4}d^{2}_{2}V_{2}[/tex]

Substitute the values in the above equation as follows:

[tex]\frac{\pi}{4}2^{2}\times 3=\frac{\pi}{4}3^{2}V_{2}[/tex]

[tex]2^{2}\times 3=3^{2}V_{2}[/tex]

[tex]V_{2}=\frac{4}{3}[/tex] m/s.

Step2

Apply Reynolds number formula for the flow condition at inlet as follows:

[tex]Re=\frac{v_{1}d_{1}}{\nu }[/tex]

[tex]Re=\frac{2\times 3}{1.24\times 10^{-6}}[/tex]

Re=4838709.677

Apply Reynolds number formula for the flow condition at outlet as follows:

[tex]Re=\frac{v_{2}d_{2}}{\nu }[/tex]

[tex]Re=\frac{\frac{4}{3}\times 3}{1.24\times 10^{-6}}[/tex]

Re=3225806.452

Thus, the Reynolds number is greater than 2000. Hence the flow is turbulent.

Hence, the flow is turbulent at both the parts of the tube.