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

[MoBaT]

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

 

[Simon Bridge]

Venturi effect?

 

[technician]

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)