Help Me Solve a Difficult Physics Problem!

[Erik_at_DTU]

Hey,

One of my homework tasks is, embarrasing, too hard for me to tackle without some help. So I help somebody can help me with the following problem:

TASK:
"You have a filled glass of water. A straw is stuck partly down into the glass (a depth d from the surface), the other end of the straw is on the outside (a distance h below the surface). The surface of the water is at a height y1. The glass' cross section area is A1, and the straw's cross section area is A2. We assume that A1 is much larger than A2, and that there goes a "flow line" from the water surface until the end (the one with the distance h from the surface) of the straw." Illustration:

http://www.student.dtu.dk/~s052861/fysikprob.jpg

a)
Decide the velocity of the water, that flows out ouf the straw, when the glass is filled (the surface of the water is at y1.

b)
Show, that the height y of the water, can be expressed by this equation: dy/dt = - A2/A1 * sqrt(2 * g * (y - y2)).

c)
How long will it go until the water stops flowing out of the straw?


So, my problem is that I really don't now where to start. I've tried to think out a way to apply Bernoullis equation, or to calculate the pressure at the entrance of the straw and using that in some way... But I really need a hint about what I should focus on!

 

[Astronuc]

This is a siphon problem.

One has to determine the 'head' between y1-d and y2.

The atmospheric pressure is the same at y1 and y2 - same atmosphere.

The pressure at the inlet of pipe (at y1-d) is atmospheric pressure + the pressure due to height of water d.

One needs to determine the differential pressure between inlet and outlet of the glass tube.

Also there is a 'head' due to the height of water h-d in the downward leg.

 

[Erik_at_DTU]

Okey, so using the that, I get that the speed at the end of the pipe is:

v = sqrt(2dg-2gh+(v0)^2)

But I've gotten this far before, my problem is to determine the speed v0 when the water is entering the straw... My guess is that it can be determined by some relationship between y1, A1 and A2... But how?

 

[Astronuc]

The mean speed across some flow area can be found by

[itex]\dot{m} = \rho * V * A[ /itex] where [itex]\dot{m}[ /tex] is the mass flow rate, $\rho[ /tex] is the density, V is the mean flow speed, and A is the cross-sectional area of the flow channel.<br /> <br /> <i>Well, Nuts! LaTeX is still not working.</i>$[/itex][/itex]

 

[Erik_at_DTU]

I've tried that as well! But then I run into the problem of determining the speed of the water before it enters the straw (in other words: the speed of the flow in the glass)...

 

[Astronuc]

The glass' cross section area is A1, and the straw's cross section area is A2. We assume that A1 is much smaller thanA2,

Actually, A1 >> A2, otherwise one has to use (A1-A2) with respect to the mass flowrate in the glass beaker (upper reservoir).

From continuity, we know that mass flow into the glass straw = mass flow out of the glass straw.

Now the mass flow into the glass straw = rho * -dy/dt * A1 = rho * V * A2. The densities (rho) cancel, so

-dy/dt * A1 = V * A2

or dy/dt = - A2/A1 * V. Now one must find V.

This will help for part a and b.

http://en.wikipedia.org/wiki/Siphon#Bernoulli.27s_equation

To find the time, solve the differential equation, obtain y(t) and solve for the time to change by height, d.

 

[Erik_at_DTU]

Ok

Thanks, think I can solve it now (will try later today). About A1>>A2, I had just misspelled, but I have corrected it now!

 

[Erik_at_DTU]

Just one short question, I've ended up with a equation for the time t (question c), could you disconfirm or confirm it? My answer is:

$t=1/2*d*A1/A2*sqrt(2)/sqrt(g*(h-d))$

 

[Astronuc]

$$\frac{dy}{dt}\,=\,-\frac{A_2}{A_1}\,\sqrt{2g(y-y_2)}$$

Now at time t=0, the water is at elevation $y_1$, and at t_f the water is at elevation $y_1\,-\,d$

Rewriting the differental equation in integral form with limits, one obtains

$$-\frac{A_1}{\sqrt{2g}A_2}\,\int_{y_1}^{y_1-d}\frac{dy}{\sqrt{y-y_2}} = \int_{0}^{t_f} dt$$

Solution would be, remembering h = y₁ - y₂,

$$t_f\,=\,\frac{A_1}{\sqrt{2g}A_2}\,\left[\sqrt{h}\,-\,\sqrt{h-d}\right]$$