Question 12

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December 25th 2006, 11:31 PM
CaptainBlack

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
December 29th 2006, 02:03 PM
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 $R(y)=ky^{\frac{-1}{4}}$

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 ${\pi}(R(y))^{2}h$

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

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

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.
December 29th 2006, 02:09 PM
CaptainBlack

Quote:

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 $R(y)=ky^{\frac{-1}{4}}$

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 ${\pi}(R(y))^{2}h$

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

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

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
December 31st 2006, 09:54 PM
CaptainBlack

Question 12 Solution

Quote:

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
January 1st 2007, 10:38 AM
topsquark

Quote:

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