Tap water flow and water diameter at end of it (Bernoulli eqzn )

AI Thread Summary
The discussion revolves around calculating the diameter of a water stream from a faucet using Bernoulli's equation and the continuity equation. The initial flow rate is determined to be 7.67 cm³/s, and the initial velocity (v1) is calculated as 0.10597 m/s based on the given diameter. To find the velocity at a point 13 cm below the faucet (v2), Bernoulli's equation is applied, resulting in v2 being approximately 1.596 m/s. The continuity equation is then used to relate the areas and velocities, ultimately leading to the determination of the stream's diameter at that point. The conversation emphasizes the importance of correctly applying these fluid dynamics principles to solve the problem.
~christina~
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Homework Statement


At a faucet, the diameter of the stream is 0.960cm. The stream fills a 125cm^3 container in 16.3s.
Find the diameter of the stream 13.0cm below the opening of the faucet.

http://img100.imageshack.us/img100/1258/15222430xr8.th.jpg

Homework Equations


Bernoulli's eqzn: P+ 1/2\rho v_1^2 +\rho gy = constant
or...P_1-P_2 = \rhog(y_2-y_1)= \rho gh

continuity eqzn: Av_1= Av_2

The Attempt at a Solution



um...I'm not sure how to go about doing this Q...

I did do:

Flow rate= Av_1= 125cm^3 / 16.3s = 7.67cm^3/s

after that I'm not sure about how to find the diameter.
I think I need the Area since Av= 7.67cm^3/s that I found and I guess I would go and find v for the first v1 at least.

I am confused as to:

1. is P1 the same as P2 (bottom of stream) ? If it is then would it cancel out?

2. would the v be the same? I think not...but if it isn't then how would I find v1 for when the water comes out of the faucet?

If someone could help me out I'd appreciate it alot.

Thanks :smile:
 
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well I worked on it for awhile...since yesterday and now I think.

Flow rate= Av1= 125cm^3/ 16.3s = 7.67cm^3/s

A_1=\pi r^2 =\pi d^2/4= \pie (0.960cm^2)/4= 0.007238cm^2

since A1V1= 0.007238cm^2
then V1= (7.67cm^3/s)/ (0.007238cm^2) = 0.10597m/s

I'm not sure how to find V2 though. (the v of the water at the bottom of the stream of water) It's open to the air so I would think gravity but I'm not sure..

Info I know:

v1= 0.10597m/s (from A1V1 and area)
A1= 7.67cm^3/s
y2-y1= 13cm => 0.13m

P_1 + 1/2 \rho v_1^2 + \rho gy = P_2 + 1/2 \rho v_2^2 + \rho gy
P1 and P2 cancel out I think so then I have:

1/2 \rho v_1^2 + \rho gy = 1/2 \rho v_2^2 + \rho gy


don't know:
v2= ?
A2=? => question asks for this


I wanted to get v2 through the A1V1= A2V2
but if I don't have V2 how can I find A2? UNLESS I have to find it using the Bernoulli's equation...I'm confused here


Can anyone help me out Please??
THANKS
 
~christina~ said:
I wanted to get v2 through the A1V1= A2V2
but if I don't have V2 how can I find A2? UNLESS I have to find it using the Bernoulli's equation...I'm confused here
You've got it. First find V2 using Bernoulli (which just tells you the increase in speed due to gravity), then you can use the continuity equation to find A2.
 
Doc Al said:
You've got it. First find V2 using Bernoulli (which just tells you the increase in speed due to gravity), then you can use the continuity equation to find A2.

Hm...I went and used this equation (V= Av\Delta t) (the thing is that I'm not sure it's the continuity equation...it includes time and the Volume.

If I used the continuity eqzn (A1v1= A2v2)
with what I found from the Bernoulli's eqzn which is:

v_2= \sqrt{2g(y2-y1)}
v_2= \sqrt{2(-9.8)(0.13m)}
v_2= 1.596m/s

now...I'm debating wheter to use V= Av_2 \Delta t or A_1v_1= A_2v_2

I used the other equation when working on the question and got:

V=Av_2 \Delta t
t= 16.3s
V= 125 cm^3 => 1.25m^3
A=?
v2= 1.596m/s

1.25m^3= \pie r^2(1.596m/s)(16.3s)
r= 0.12367

d= 0.2473m

I'm not sure how I'd do the question with A1v1=A2v2
since I do have
v2
A1
but do I know v1?

Thanks a lot :smile:
 
~christina~ said:
Hm...I went and used this equation (V= Av\Delta t) (the thing is that I'm not sure it's the continuity equation...it includes time and the Volume.
In the continuity equation, what does A1v1 represent? (Hint: They give you the flow rate for a reason!)


If I used the continuity eqzn (A1v1= A2v2)
with what I found from the Bernoulli's eqzn which is:

v_2= \sqrt{2g(y2-y1)}
v_2= \sqrt{2(-9.8)(0.13m)}
v_2= 1.596m/s
Looks like you assumed v1 = 0 when using Bernoulli. That can't be right. You need to figure out the initial speed from the given data.

now...I'm debating wheter to use V= Av_2 \Delta t or A_1v_1= A_2v_2

I used the other equation when working on the question and got:

V=Av_2 \Delta t
t= 16.3s
V= 125 cm^3 => 1.25m^3
A=?
v2= 1.596m/s

1.25m^3= \pie r^2(1.596m/s)(16.3s)
r= 0.12367

d= 0.2473m

I'm not sure how I'd do the question with A1v1=A2v2
since I do have
v2
A1
but do I know v1?
The analysis you did here is exactly what you have to do to find the initial speed. You are given the fow rate (Volume and time). Since you know the diameter (thus the area) at the top, you can use it to find the speed at the top (v1).

You're almost there. :wink:
 
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