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Siphoning A Clogged Sink (Bernoulli's Principle)

  • Thread starter GDGirl
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Homework Statement


You need to siphon water from a clogged sink. The sink has an area of 0.496 m2 and is filled to a height of 4.0 cm. Your siphon tube rises 50 cm above the bottom of the sink and then descends h = 93 cm to a pail as shown above. The siphon tube has a diameter of 1.57 cm.
a. Assuming that the water enters the siphon with almost zero velocity, calculate its velocity when it enters the pail.
Help: Use Bernoulli's Principle.

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/18/02/HW19_5.jpg [Broken]


Homework Equations


P1+1/2[tex]\rho[/tex]v12+[tex]\rho[/tex]gy1= P2+1/2[tex]\rho[/tex]v22+[tex]\rho[/tex]gy2
equations derived from similar examples in class:
v1=sqrt(2gh) when v2 (or whatever sub number theother velocity is) is negligible.
or v1=sqrt((2gh)/(1-(A12/A22)))


The Attempt at a Solution


I have tried everything I can think of. I tried finding the cross-sectional area of the tube and plugging that area in as A1 in the final equation I listed. I square that, divide it by the square of the area of the sink, subtract that from one and divide my answer for 2gh (2*9.8*(.5+.93)=28.028) by that and take the square root and I get 5.298 m/s which is incorrect.
When I simply plug my numbers into sqrt(2gh) I get 5.294, which is also incorrect.
Then, I've tried plugging numbers into the entire equation for Bernoulli's Principle. However, I was kind of BSing the pressures since I'm not sure what the pressure would be at the start of the tube and what it would be at the end of the tube. So I'm disregarding that attempt.
So generally, I'm completely stuck on this, can someone give me a hand?
 
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Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
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Can you not just use conservation of energy?
 
  • #3
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Rather, if I understood it better. I don't know how that applies to what I'm working on here (I tend to have trouble making connections between things like that)
It's not like our work gets checked. I'm just going with the equation that the help suggests.
 
Last edited:
  • #4
LowlyPion
Homework Helper
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(2*9.8*(.5+.93)=28.028)
You added the heights?

Isn't the difference in height just .43 m?
 
  • #5
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Yes, I did.
Since I need the velocity at the bottom of the tube... I wasn't sure if I should just use one or the other of the heights given or take the height at the entirety of the tube... I'll try the difference of heights though. :/
 
  • #6
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Oh, that worked out perfectly... I didn't think about taking the difference of the height ^^'
 

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