Bernoulli's Equation and Fluid Mechanics?

[wmrunner24]

Water flows through a 0.30m radius pipe at the rate of 0.20m^2/s. The pressure in the pipe is atmospheric. The pipe slants downhill and feeds into a second pipe with a radius of 0.15m, positioned 0.60m lower. What is the gauge pressure in the second pipe?

So, what I've figured from the problem and what I've learned so far about flow continuity and Bernoulli's principle gave me this.

FR=A₁v₁
FR=A₂v₂

Flow rate = Area of a Cross-section * Velocity

So, I can find the velocity for Bernoulli's equation.

P₁ + \rhogh₁ + 1/2\rhov₁² = P₂ + \rhogh₂ + 1/2\rhov₂²

And since there's no change in height at the first pipe cross-section, \rhogh₁= 0, right?

P₁ + 1/2\rhov₁² = P₂ + \rhogh₂ + 1/2\rhov₂²

So, then I solve for P₂, but since it wants gauge pressure, I have to subtract atmospheric pressure, which means that I can remove P₁ from the equation, because the problem statement says it equals atmospheric pressure, right?

1/2\rhov₁² - 1/2\rhov₂² - \rhogh₂ = P_g

All I'm really looking for here is a logic check for this much of it. Am I right?
All help is greatly appreciated.

 

[Brian_C]

That looks correct. There is actually no reason to set h1=0 because the problem gave you the height difference between the two pipes.

 

[wmrunner24]

Oh? I was under the impression that whenever you wrote something like \rhogh₁, it was assumed to be the change in height, like in gravitational potential energy. I said h₁ was zero because as you moved left to right, at the first cross-section, there was no change yet so it was 0. But, if I understand what you're saying, I factor to get h₁ - h₂ together and replace it with \Deltah, right? Which actually seems to make more sense...