Fluids: Water runs into a fountain

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Homework Help Overview

The problem involves determining the speed at which water exits a hole in a fountain, given a steady inflow rate. The subject area pertains to fluid dynamics and the application of the continuity equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the continuity equation, questioning the correct area calculations and the relationship between volume and speed. Some participants suggest visualizing the problem with a cylindrical model to aid understanding.

Discussion Status

There are multiple interpretations of the calculations involved, with some participants providing alternative methods to arrive at the speed of water exiting the hole. Guidance has been offered on calculating the area and relating it to the volume flow rate, but no consensus has been reached on the final answer.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the depth of discussion regarding methods and solutions. There are indications of potential calculation errors and assumptions about the geometry involved.

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Homework Statement


Water runs into a fountain, filling all the pipes, at a steady rate of 7.52×10−2 m^3/s

How fast will it shoot out of a hole 4.53cm in diameter?


Homework Equations


A1V1 = A2V2


The Attempt at a Solution



A1V1 = 7.52x10-2 m^3/s = A2V2
Divide by A2, which would be 4.53/2 (to get Radius) = 2.265 --> .02265m or 2.265x10-2

Now, for A2, am I suppose to square it to get m/s as my answer? I'm assuming so since V1=m/s so that must mean A1 = m^2.

If I do all of that, I get (7.52x10-2 m^3/s) / (2.265x10-2 m)^2 = 146 m/s which I think is wrong. Thanks for the help.
 
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Just try and picture it instead of hitting equations.
Imagine a sausage coming out a hole 4.53cm diameter (work out the area in m)
Now what length would that sausage have to be to have a volume of 7.52E-2 m^3
If that length comes out every second - that is your speed.
 
mgb_phys said:
Just try and picture it instead of hitting equations.
Imagine a sausage coming out a hole 4.53cm diameter (work out the area in m)
Now what length would that sausage have to be to have a volume of 7.52E-2 m^3
If that length comes out every second - that is your speed.
Ok, I can imagine the .0453m diameter hole, also the sausge coming out lol Now I'm trying to figure out the length.

So I would have Pi x r2 x h = Volume
H=L
L=7.52E-2 m^3 / pi*r^2 ?
 
Volume of a cylinder is just area * length.
Area = pi (0.0453/2)^2 = 0.00644m^2
The volume/sec = 7.52E-2 m^3
So length = volume/area = 7.5E-2/6.44E-3 = 11.6 m/s

(unless I got the sums wrong! )
 
mgb_phys said:
Volume of a cylinder is just area * length.
Area = pi (0.0453/2)^2 = 0.00644m^2
The volume/sec = 7.52E-2 m^3
So length = volume/area = 7.5E-2/6.44E-3 = 11.6 m/s

(unless I got the sums wrong! )
Ok
Volume of Cylinder = Area * Length
Area = pi x r^2
Divide .0453 by 2 to get r
Area = pi x (.0453/2)^2=.0016117077

V/A = L

7.52E-2 m^3 / 1.61E-3 = 46.7 m/s

EDIT: Thanks for your help. You seemed to sum something up wrong lol I got the right answer. Thanks again.!
 
Last edited:
oops you're right I forgot to divide the diameter by 2!
Always check your calculations - or at least always check mine!
 
mgb_phys said:
oops you're right I forgot to divide the diameter by 2!
Always check your calculations - or at least always check mine!
lol I also tried what I had put before[L=7.52E-2 m^3 / pi*r^2 ] and got the same # lol
 

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