Solving Water Pressure Problems: Guidance for Parts B and C

AI Thread Summary
The discussion centers on solving a physics problem involving water pressure in a pipe system. For part (a), the pressure at the exit opening is identified as atmospheric pressure, approximately 1.01E5 Pa. In parts (b) and (c), participants express uncertainty about applying Bernoulli's principle to determine the pressure at the connection point and the exit speed of the water. The relationship between the areas and velocities of the pipe and exit is established using the equation A1v1 = A2v2, leading to the conclusion that v2 can be expressed in terms of v1. Overall, the conversation emphasizes the application of Bernoulli's equation and the continuity equation to solve for unknown pressures and velocities in the system.
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Could anyone offer me some guidance on parts b and c of the following question?

The pressure in a uniform pipe of flowing water is enough to hold up 4.00 m of water in an open, vertical tube as shown in Figure P.60. Assume that water is an ideal fluid. The pipe leads 3.00 m upstairs and empties into the open air through an opening with an area one-fourth that of the uniform pipe.

(a) At the exit opening of the pipe, what is the pressure?
(b) What is the pressure at the point where the vertical tube is connected to the pipe?
(c) What is the exit speed of the water from the pipe?


this is the picture of the problem:
http://img346.imageshack.us/img346/5326/p11607ht.gif"

for part (a), i figured b/c the water is open to the air, its pressure is 1 atm (approx. 1.01E5 Pa).

for parts (b) and (c), i think i need to somehow use the equation:
P1 + (1/2)*density*v1^2 + density*g*y1 = P2 + (1/2)*density*v2^2 + density*g*y2

but I'm not sure how to get the velocity or pressure at the point under the 4m opening.

any ideas/hints for me?

thanks!
 
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i figured this one out, looks like i was on the right track afterall!
 
I am doing this exact same problem and am also unsure of how to use Bernoulli's principle for b) and c).
Help please.
 
but I'm not sure how to get the velocity or pressure at the point under the 4m opening.
Pressure P1 = Pa + rho*g*h where Pa atmospheric pressure, rho is the density of water and h = 4m. P2 = Pa
 
I've gotten the pressure at the opening= P(atmosphere)
I've gotten the pressure under the 4 m opening= 140565=P(atmosphere)+1000(9.81)(4 m)

What is the exit speed of the water from the pipe?? In my case,
The pipe leads 3.00 m upstairs and empties into the open air through an opening with an area one-third that of the uniform pipe."

so we know that A1v1=A2v2 so A1v1=1/3A1v2 and thus 3v1=v2 where v2 is the exit speed and v1=speed in the pipe

how do I set up

Bernoulli's principle to get v1 and thus v2?

P1 + (1/2)*density*v1^2 + density*g*y1 = P2 + (1/2)*density*v2^2 + density*g*y2
 
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