Water Pressure in One Pipe Flowing to Another

[TMO]

Homework Statement

A water pipe tapers down from an initial radius of R₁ = 0.22 m¹ to a final radius of R₂ = 0.08 m¹. The water flows at a velocity v₁ = 0.85 m¹s^-1 in the larger section of pipe. The water pressure in the center of the larger section of the pipe is P₁ = 283640 Pa. Assume the density of water is 1000 kg¹m^-3.

Homework Equations

(1) A ∝ r²
(2) p + ½·ρ·v² + p·g·h = C
(3) A₁·v₁ = A₂·v₂

The Attempt at a Solution

I assumed that since the water remains at roughly the same height, and that the only thing changing is the velocity, and that pressure in Bernoulli's equation is conserved, that one should try to compute the velocity from this.

½·ρ·v² = P₁
ρ·v² = 2·P₁
v² = 2·P₁/ρ
v = √(2·P₁/ρ)

Using Equation 3 in conjunction with the fact that A ∝ r², I concluded that the final velocity must be (R₁)²·√(2·P₁/ρ)/(R₂)², however this isn't giving the correct numerical answer. Any thoughts?

 

[LawrenceC]

You are using Bernoulli's equation incorrectly. Write the equation as

P1 + .5*rho*(V1)^2 = P2 + .5*rho*(V2)^2

and evaluate for P2.