1. ## Question 12

Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
which is 1 cm wide.

How wide is the stream 30 cm below the tap (assuming that the stream
does not break up due to surface tension induced instability).

RonL

2. Hey Cap'n.

No one bit on this problem, so I will give my farthings worth.

I think I have seen a problem like this before, a while back.

I think the width would be proportional to how far it is from the tap.

If I remember right the formula is $\displaystyle R(y)=ky^{\frac{-1}{4}}$

Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume $\displaystyle {\pi}(R(y))^{2}h$

Because of gravity and all that, the rate this disk is passing y is:

$\displaystyle F={\pi}(R(y))^{2}h\sqrt{19.6y}$

Where F is the rate the water is leaving the tap.

Now, solve for R(y).

Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

This does look like an interesting little problem, though.

3. Originally Posted by galactus
Hey Cap'n.

No one bit on this problem, so I will give my farthings worth.

I think I have seen a problem like this before, a while back.

I think the width would be proportional to how far it is from the tap.

If I remember right the formula is $\displaystyle R(y)=ky^{\frac{-1}{4}}$

Where R is the radius at some point y down the stream and y is the distance down the stream form the tap.

We could probably stack an infinite number of itty-bitty disks of thickness h up the stream with volume $\displaystyle {\pi}(R(y))^{2}h$

Because of gravity and all that, the rate this disk is passing y is:

$\displaystyle F={\pi}(R(y))^{2}h\sqrt{19.6y}$

Where F is the rate the water is leaving the tap.

Now, solve for R(y).

Now, maybe we could integrate. Maybe construct a DE. That's all I have for now.

This does look like an interesting little problem, though.
The problem of the instability of the stream is more interesting, It is a
problem related to the (in)stability of broad planetary ring systems to break
up into ringlets.

RonL

4. ## Question 12 Solution

Originally Posted by CaptainBlack
Water leaves a tap (faucet) in a steady stream at a speed of 20 cm/s
which is 1 cm wide.

How wide is the stream 30 cm below the tap (assuming that the stream
does not break up due to surface tension induced instability).

RonL
Question 12: Solution

The mass flow rate is a constant so at any point Av is that same
as at any other point, where A is the crossectional area of the stream
and v is the speed of the stream.

A mass element undergoes an accelaration of g m/s^2 (sign convention
is positive downwards), and so at time t its speed is:

v(t)=g.t+v(0)

and it has fallen a distance:

s= g.t^2/2+v(0)t.

So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

0.3=9.81 t^2/2+0.2t,

the roots of this are t~=-0.2685, and t~=0.22776. The first
of these roots is unphysical so we are left with t~=0.22776 seconds.

Its speed when it has fallen 0.3 m is:

v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

So now we have from the constancy of the mass flow rate past any point:

pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

which gives:

D~=0.002872 m, or D~=0.29 cm.

RonL

5. Originally Posted by CaptainBlack
Question 12: Solution

The mass flow rate is a constant so at any point Av is that same
as at any other point, where A is the crossectional area of the stream
and v is the speed of the stream.

A mass element undergoes an accelaration of g m/s^2 (sign convention
is positive downwards), and so at time t its speed is:

v(t)=g.t+v(0)

and it has fallen a distance:

s= g.t^2/2+v(0)t.

So the time to fall 0.3 m, is a root of (taking g=9.81 m/s^2):

0.3=9.81 t^2/2+0.2t,

the roots of this are t~=-0.2685, and t~=0.22776. The first
of these roots is unphysical so we are left with t~=0.22776 seconds.

Its speed when it has fallen 0.3 m is:

v(0.22776)=9.81 0.22776 + 0.2 ~= 2.434 m/s

So now we have from the constancy of the mass flow rate past any point:

pi*(0.01/2)^2*0.2=pi*(D/2)^2*2.434

which gives:

D~=0.002872 m, or D~=0.29 cm.

RonL
Might I point out that this is an application of Bernoulli's equation (an extremely simplified version of fluid dynamics: the flow for a non-viscous incompressible irrotational fluid) along with the constant mass flow-rate condition. I typically assign something like it in my Freshman Physics course.

-Dan