How to Calculate the Speed of Water Exiting a Shower Head with 24 Holes?

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Homework Help Overview

The discussion revolves around calculating the speed of water exiting a shower head with 24 holes, given the internal radius of the water line and the speed of water in the line. The problem involves concepts from fluid dynamics, specifically the continuity equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the continuity equation and the need to consider the total area of the 24 holes versus a single hole. There are attempts to calculate the speed using different interpretations of the area and flow rates.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the calculation of areas and the application of the continuity equation. Some participants have provided guidance on how to approach the calculations, while others express confusion about the correct application of the formulas.

Contextual Notes

Participants note the importance of squaring the radii when calculating areas and the implications of having multiple holes in the shower head. There is also mention of the conservation of mass in the context of fluid flow.

keemosabi
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Homework Statement


A water line with an internal radius of 6.1*10^-3 m is connected to a shower head that has 24 holes. The speed of the water in the line is 1.2 m/s.

(b) At what speed does the water leave one of the holes (effective radius = 4.6*10^-4 m) in the head?



Homework Equations


Av = Av


The Attempt at a Solution


6.1*10^-3(1.2)= 4.6*10^-4(v)
V = 175.851

Obviously this is wrong because I didn't use the 24 holes information, but where does that fit in?
 
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Isn't the area the total of the area of the 24 holes and not just the area of 1?
 
LowlyPion said:
Isn't the area the total of the area of the 24 holes and not just the area of 1?
I tried that and got 4220.42 m/s, which it says is also wrong.
 
Those are radiuses. And when you calculate the area you have to square them, since the area of the cross section of a cylinder is a circle, and the area of the circle is: [tex]R^2\pi[/tex].

And you have to do what LowlyPion said, that you consider all the 24 holes, which are also circles ;)
 
Thaakisfox said:
Those are radiuses. And when you calculate the area you have to square them, since the area of the cross section of a cylinder is a circle, and the area of the circle is: [tex]R^2\pi[/tex].

And you have to do what LowlyPion said, that you consider all the 24 holes, which are also circles ;)

I actually found the areas correctly when I did my work earlier. I just typed it in wrong.

What do you mean by consider the 24 holes?
 
You considered only one hole. Whereas there are 24 holes, so you have to divide your answer by 24, and you will get: [tex]v\approx 8.79 m/s[/tex]
 
Thaakisfox said:
You considered only one hole. Whereas there are 24 holes, so you have to divide your answer by 24, and you will get: [tex]v\approx 8.79 m/s[/tex]
How did you do that? I got 175/24 = 7.3
 
keemosabi said:
I tried that and got 4220.42 m/s, which it says is also wrong.

The equation of continuity that you are using is really the conservation of mass since the water is an incompressible fluid.

The area of the input pipe times flow velocity through it will equal the total area of all the output holes times the flow velocity. Discarding pi ...

Ri² * Vi = 24 * Ro² * Vo
 
I got 6.63. How did you get that?6.1*10^-6 (1.2) = 24 * (4.6*10^-8) (v)

I solved for v and got 6.63.
 
  • #10
you have to square 6.1 too i.e. [tex]6.1^2[/tex] and [tex]4.6^2[/tex]
 
  • #11
Thaakisfox said:
you have to square 6.1 too i.e. [tex]6.1^2[/tex] and [tex]4.6^2[/tex]
Thank you for the help. I get it now.
 

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