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## Homework Statement

A water pipe tapers down from an initial radius of R

_{1}= 0.22 m

^{1}to a final radius of R

_{2}= 0.08 m

^{1}. The water flows at a velocity v

_{1}= 0.85 m

^{1}s

^{-1}in the larger section of pipe. The water pressure in the center of the larger section of the pipe is P

_{1}= 283640 Pa. Assume the density of water is 1000 kg

^{1}m

^{-3}.

## Homework Equations

(1) A ∝ r

^{2}

(2) p + ½·ρ·v

^{2}+ p·g·h = C

(3) A

_{1}·v

_{1}= A

_{2}·v

_{2}

## 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·v

^{2}= P

_{1}

ρ·v

^{2}= 2·P

_{1}

v

^{2}= 2·P

_{1}/ρ

v = √(2·P

_{1}/ρ)

Using Equation 3 in conjunction with the fact that A ∝ r

^{2}, I concluded that the final velocity must be (R

_{1})

^{2}·√(2·P

_{1}/ρ)/(R

_{2})

^{2}, however this isn't giving the correct numerical answer. Any thoughts?