Understanding Fluid Velocity in a Restricted Pipe

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SUMMARY

The discussion centers on calculating fluid velocity in a pipe with a restriction, specifically a 4.0 cm diameter pipe reducing to a 2.0 cm diameter. The initial velocity of the liquid is 1.0 m/s, leading to a calculated velocity of 4.0 m/s in the restriction based on the continuity equation for incompressible fluids. The key principle applied is the mass flow rate, expressed as ρVA, where ρ is fluid density, V is velocity, and A is cross-sectional area. The continuity equation V(in)A(in) = V(out)A(out) confirms the velocity increase in the restricted area.

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Homework Statement


Liquid flows through a 4.0 cm diameter pipe at 1.0 m/s. there is a 2.0 cm diameter restriction in the line. What is the liquids velocity in this restriction?

Homework Equations





The Attempt at a Solution


I know that solution is 4.0 m/s but in don't know how to get there. Thank you
 
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The concept one is concerned with here is continuity. For an incompressible fluid, the mass flow (rate) into a control volume must equal the mass flow (rate) out, otherwise mass would accumulate or decrease in the volume.

The mass flow rate is given by ρVA, where ρ is the fluid density, V is the mean (average) velocity and A is the cross-sectional area.

So we have ρVA(in) = ρVA(out), or ρVA(in) - ρVA(out) = 0, i.e. no net gain or loss of fluid (mass).

Now if ρ(in) = ρ(out), then the continuity equation becomes, V(in)A(in) = V(out)A(out).

Try apply this to given problem.
 


kmkindig said:
Liquid flows through a 4.0 cm diameter pipe at 1.0 m/s. there is a 2.0 cm diameter restriction in the line. What is the liquids velocity in this restriction?

I know the answer is 4.0 m/s but I am not for sure how to get there. A little help would be greatly appreciated. Thank you!

There is a principle that states Av=constant, where A is the area of the fluid's path at a given time and v is the velocity.
Hence the answer is not 4.0 m/s, unless I'm wrong.
 

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