What Is the Speed and Pressure Difference in a Constricted Tube?

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SUMMARY

The discussion focuses on fluid dynamics in a constricted tube, specifically applying Bernoulli's principle to determine the speed and pressure difference of water flowing through a tube that narrows from a radius of 4.00 cm to 2.50 cm. The initial speed of water in the larger tube is 3.50 m/s, resulting in a calculated speed of 8.96 m/s in the smaller tube using the equation A1V1 = A2V2. The pressure difference ΔP is derived from the formula ΔP = 0.5ρ(v2^2 - v1^2), illustrating the relationship between kinetic energy and pressure in fluid flow.

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Homework Statement


A constricted horizontal tube of radius r1 = 4.00cm tapers to a tube of radius r2 = 2.50cm. If water flows at a speed 3.50m/s in the larger tube, (A) find its speed in the smaller tube. (b) Find the water pressure difference ΔP=ΔP1-ΔP2 in kPa and in atm.


Homework Equations



I have no clue.

The Attempt at a Solution



Succeeded in doing part A.

A1V1 = A2V2

V2 = (A1V1)/A2 = (pi(4)^2(3.50)) / (pi(2.50)^2) = 8.96 m/s
 
Last edited:
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Venturi effect?
 
You need to use Bernoulli's principle . The water speeds up and therefore gains KE.
This gain in KE comes from a decrease in PE (pressure)

ΔP = 0.5ρ(v2^2 - v1^2) ρ = density of water
(this is bernoulli's principle in its simplest form with the tube horizontal)
 
Last edited:

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