Minimizing the energy required to break a cylinder of fluid into droplets

[happyparticle]

Homework Statement

We consider a cylinder of incompressible fluid with length L and cross-sectional area A $\ll$ L which breaks into N spherical droplets of radius r. Determine the value of N for which this energy is minimized.

Relevant Equations

$$E_g = T N 4 \pi (\frac{3 A L }{4 \pi N})^{2/3}$$
$$E_c = T 2\sqrt{\pi A} L$$

I'm trying to determine the value of N for which this energy is minimized for the droplets.
I found that the energy for N droplets is $$E_g = T N 4 \pi (\frac{3 A L }{4 \pi N})^{2/3}$$.
At first I thought of deriving the energy by the number N and set it to zero. However, it does not work since I get $$N=0$$.

Then, I thought about the volume which must remains the same, because the fluid is incompressible and by the conservation of the mass.
I get $$N = \frac{3AL}{4 \pi r^3}$$, but I feel that is not what it is expected.

 

[Steve4Physics]

Homework Statement: We consider a cylinder of incompressible fluid with length L and cross-sectional area A $\ll$ L

You can't compare the values of an area and a length as they have different dimensions.

which breaks into N spherical droplets of radius r.

Since the initial and final total volumes are equal:
$AL = N \frac 43 \pi r^3$ giving $N = \frac {3AL}{4 \pi r^3}$ as you calculated.

Determine the value of N for which this energy is minimized.

It does not make sense to ask this.

The value of N is completely determined by the values of A, L and r. So there is only one possible value for N (unless I've misunderestood what you are asking).

 

[happyparticle]

The value of N is completely determined by the values of A, L and r. So there is only one possible value for N (unless I've misunderestood what you are asking).

At the end there will be N droplets. Thus, it must be possible to find the value of N.

I should have written $\sqrt{A} \ll L$

 

[vela]

At the end there will be N droplets. Thus, it must be possible to find the value of N.

What Steve is saying is that if you're given $A$ and $L$, you know the total volume of the fluid, and if you're given $r$, you know the volume of the droplets, so $N = V_{\rm total}/V_{\rm droplet}$ is completely determined, regardless of the energy.

Did you post the problem statement as given word for word?

 

[happyparticle]

What Steve is saying is that if you're given $A$ and $L$, you know the total volume of the fluid, and if you're given $r$, you know the volume of the droplets, so $N = V_{\rm total}/V_{\rm droplet}$ is completely determined, regardless of the energy.

Did you post the problem statement as given word for word?

First I had to show that (Energy of N droplets) < (Energy of cylinder), then I'm asked to find the value of N for which the energy of the droplets will be minimal. I might be wrong, but I guess I have to find the value of N for which the energy of droplets will be the lowest. Maybe I'm wrong with the volume conservation.

 

[Steve4Physics]

What is the original (complete, word-for-word) question?

 

[happyparticle]

We consider a cylinder of incompressible fluid of length L and section a ( √ a ≪ L), which beads into N spherical droplets of radius R.

Determine the value of N for which this energy will be minimal.

 

[Steve4Physics]

We consider a cylinder of incompressible fluid of length L and section a ( √ a ≪ L), which beads into N spherical droplets of radius R.

Determine the value of N for which this energy will be minimal.

The question is faulty - for the reason already explained in posts #2 and #4.

Minor edit.

 

[happyparticle]

Alright, I trust you. So If I understand correctly N is independent of the energy. Thus, for a Volume (V), there will always be the same number of droplets?

What puzzles me is that we can clearly see that the energy for droplets will be less than the energy of the cylinder and I was pretty sure that there was a maximum value for N which the energy of droplets will be minimized.

$E_g = T N 4 \pi (\frac{3 A L }{4 \pi N})^{2/3}$

 

[Steve4Physics]

Alright, I trust you.

You shouldn't! You need to see it for yourself. If there is still (genuine?) doubt, maybe expressing it as a similar simpler problem will help.

Consider a solid block measuring 1 cm x 1 cm x 20 cm.

You want to divide this into N cubes each measuring 1 cm x 1 cm x 1 cm.

You end up with 20 cubes each measuring 1 cm x 1 cm x 1 cm. So N= 20.

(Unless you can describe a way where you don’t end up with N=20.)

So If I understand correctly N is independent of the energy. Thus, for a Volume (V), there will always be the same number of droplets?

Yes - for a given choice of droplet radius.

What puzzles me is that we can clearly see that the energy for droplets will be less than the energy of the cylinder

No. The surface energy of the droplets will be greater than the surface energy of the cylinder. Because the total surface area of the droplets is greater than the surface area of the cylinder.

I won't be replying further so I wish you seasonal greetings.