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democritus

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A water pipeline with the water under a pressure of 5 atm is being pumped into a building through a pipe of radius .2m with a velocity of 5 m/s. The pipe then proceeds to climb the 20m tall building and narrow to a radius of .02m. What is the pressure and velocity of the water at the top of the building?

Now it makes sense to me to take the ratio of the areas, which ends up being 100:1 and applying it to the equation of continuity, giving me a velocity of 500 m/s at the top. When I plus this into Bernoulli's equation,

506500 + .5(1000)(5)^2 + 0(9.8)(1000) =

P2 + (.5)(1000)(500)^2 + (20)(9.8)(1000)

And I end up with the wonderful pressure of around -1.2 x 10^8 Pascals, which obviously can't be the real pressure for the water. I've been racking my brain trying to figure out why exactly this situation is impossible, assuming that water is an ideal nonviscous incompressible fluid.

So I can't figure out why in the real world, the pressure should turn out to be some negative number, and it seems like this would happen for any large differences in velocity in Bernoulli's equation. I've been looking around for an explanation of this, as our AP Physics class spent half an hour trying to reason out logically why this would happen, and our Physics teacher couldn't explain it either.

Any help would be greatly appreciated.

-J Stoncius