Need desperate help on fluid dynamics

[gtabmx]

0.0251 = 3062.500000 x - 39.13118960 x + 0.125

0.0251 = 3062.500000 x - 39.13118960 x + 0.125

"->"
{x = 0.003525914211}, {x = 0.009251617087}

 

[Cyrus]

No, this is not correct.

Where did you get these numbers?

Intercept is ok, but the rest are wrong.

Where did this come from?

0.0251?
3062.500000 ??
39.13118960 ??

 

[gtabmx]

I used the equation and i plugged 25s into time and the corresponding height 0.0251m from my results and i simplified.

 

[Cyrus]

Ah, this is fine.

BUTTT, you should use data points as far as possible, i.e. use 35.

You will find that you get:

0.0035822
0.00551068

 

[gtabmx]

ok, so wat a couple of decmals so, wat are the units of lumbda?

and now wat? there must be a way to find the values before even doing the experiment?

 

[Cyrus]

You tell me.

Not a couple of decimals, a factor of 2.

 

[gtabmx]

It was a joke, i understand.

 

[gtabmx]

No units, makes me feel better i think

 

[gtabmx]

So now i know lambda has no units and u must do the experient to find it, that's not taht good since its not a general formula yet

 

[gtabmx]

Wait, lambda is not consistent, it has not units is in first term and units in second

 

[Cyrus]

Are you sure?

 

[gtabmx]

m = m/s^2 * ?^2 * s^2 is first term so m = m * ? no units
m = m^(1/2)/s * ? * m^(1/2) * s is second term so m = ? units

 

[Cyrus]

What is this?

Do a proper dimensional analysis, and take your time.

 

[gtabmx]

I DID IT 5 TIMES adn I was about to take a picture of it with my phone until i found out the qualiy sucks

 

[Cyrus]

Then do it a 6th time, but TAKE YOUR TIME!

Show me some work, try to make it cleaner.

 

[gtabmx]

same thing again, look, why can't u understand that i get the same thing?

 

[gtabmx]

WAIT 1/m 1/m no, oh now I'm even more confused

 

[Cyrus]

Here:

\frac{m}{s^2} s^2 * \lambda = m

Lambda is units of meter

\sqrt{ \frac{m}{s^2} {m}} * s *\lambda = m

lambda is in units of meter, and it better be.

come on, this is algebra 1. I warned you to TAKE YOUR TIME.

 

[gtabmx]

You know I have a feeling that's not right look at the final equations, each term has to be in meters after all the steps are carried out now the first temr is Y^2 times gravity times time squared which is m = Y^2 * m*s(-2) * s^2 so then s^(-2) outs s^2 and you have m = Y^2 * m no?

 

[gtabmx]

ohh
wait
wait wait

heres my deal, i made a stupid mistake probably due to the fact that I've been up till 2AM adn woke up at 6 for the past 3 days so I'm sorry

also, i get consistent, but NO UNITSyup sextuple checked no units as simplified its m = m * y^2 - m * y + m

u even proved it yourself

so your second tex simplifies to

m lambda = m

 

[Cyrus]

ahhh, I was sloppy. Sorry.

Lambda has the units of meters.

 

[gtabmx]

No unitsssssssss

 

[Cyrus]

Dammit, your right. What the hell am I telling you.

Sorry, it is unitless. I am trying to do a lab report and type and its not working out.

Ahahahahaa, stupid me.

At least one of us is paying attention.

 

[gtabmx]

Its ok, but look, where are we going with this, right now we, I, still have have no way of finding lambda unless I do the experiment, so I have no general law.

 

[gtabmx]

Alright, if its ok with you we will stop here and I will just use this to prove the data's relation to bernouilli's law and that it is in fact quadratic and that will be good enough, because let's face it, this is useless to either of us since its not in my course coverage and its not your lab.

I mean, i don't see how to get lamba but I know that there are many fluid properties whose units cancel out well, but frankly, we don't have time, energy, or enough infomation to do this right now, or at least I don't. I have to hand this in tomorow by 4 and i have classes so really, its between finishing almost complete or not finishing it at all. I thank you greatly for your help and hope you can get your stuff done to in spite of this fiasco

Thanks a lot,
Mike.

 

[Cyrus]

Maybe you can relate lambda to the cross section area somehow?

We know lambda is unitless, so that means it has to be a ratio of the cross sectional area to something else......maybe...

its worth a shot.

 

[gtabmx]

Well if you look at that formula i had posted as a mess before, I have tons of cross section areas in there, and if i have time maybe I'll take a crack at it. I'm sure cross sections would be in there since area of slot greatly affects flux rates and XSarea of bottle is what defines the amount of water knwoing what the height is. Would amke sense.

 

[gtabmx]

I'lll show you something...

 

[Cyrus]

We got it buddy.

 

[Cyrus]

Use this relationship and your gold:


\frac {dV}{dt}= A_c \sqrt{2gh}

Note: dV/dt is change in VOLUME per unit time. Plug in now your general equation for h and solve.

 

[gtabmx]

Here's something http://www.yousendit.com/transfer.php?action=download&ufid=852CA7CB14E5C3E2

solve for what, lambda of Volume, i see where you're going but I don't quite understand what you wnat to use that flux rate equation to get?

 

[Cyrus]

Try it. I was going to but I did not know your bottle diameter.

 

[gtabmx]

Oh no problem, its just going to take me an hour more, I mean its 1:14AM here but who cares :P

 

[Cyrus]

My typical bed time is 4 am, because I get distracted by questions like these.

 

[gtabmx]

Did you get the picture and my question about what I am solving for?

I wake at 6AM everyday... that's in 5hrs, wow

 

[Cyrus]

Yeah, but I don't have time right now to go through that big mess. Sorry. I really gota get work done on this lab report.

 

[gtabmx]

OK honestly, I'm out for tonight, and as close as we are, its too late and I'll do it tomorow, goodnight and thanks so much

And sorry I took up your time, I'm sure, trust me, I'm sure your stuff was more important than mine, so thanks, but I just want to tell you, as much as you may think I don't know wat I'm doing and I am clumsy with Algebra 1 lol, I actually hold the top scores in many of my classes that I actually LEARNT THE INFO IN and I am extremely interested in science, but i get frustrated too easily when i don't follow. hehe and i too aways, maybe too often, jump to help others when I know I have too too much on my own hands to do so. Anyway, I'm off to bed and thanks a lot for the help.

Mike

 

[Cyrus]

OK honestly, I'm out for tonight, and as close as we are, its too late and I'll do it tomorow, goodnight and thanks so much

And sorry I took up your time, I'm sure, trust me, I'm sure your stuff was more important than mine, so thanks, but I just want to tell you, as much as you may think I don't know wat I'm doing and I am clumsy with Algebra 1 lol, I actually hold the top scores in many of my classes that I actually LEARNT THE INFO IN and I am extremely interested in science, but i get frustrated too easily when i don't follow. hehe and i too aways, maybe too often, jump to help others when I know I have too too much on my own hands to do so. Anyway, I'm off to bed and thanks a lot for the help.

Mike

I don't think your bad at school, I am just busting your ass to make sure you pay attention.

If you have a constant, it can't have two values of units! No matter what.


Post the solution when your teacher gives it to you guys.

 

[siddharth]

This is the same question as posted here?
If so, this result is fairly easy to show, isn't it?

If the density is assumed to be constant

v_1A_1=v_2A_2

and bernoulli's equation gives

\frac{\rho \left(v_2^2-v_1^2 \right)}{2} = \rho g h

where v_1=-\frac{dh}{dt}

 

[gtabmx]

yes, but that formula oversimplifies the problem since it ignores the fact that there is no input so the flux rate ot the hole drops as volume drops, and volume drop rate is not consatnt and relies of the past flux rates and how much water was previously lost.

Anyway, cyrusabdollahi, don't worry, I think we have 90% of the answer but if you think my professor is going to give us the solution then I can't stress enough how muh of a jokester he is to never do so.

 

[siddharth]

yes, but that formula oversimplifies the problem since it ignores the fact that there is no input so the flux rate ot the hole drops as volume drops, and volume drop rate is not consatnt and relies of the past flux rates and how much water was previously lost.

Er.. what? I don't understand. Could you elaborate?

 

[gtabmx]

SOLVED! I will explain when I get home. Thanks everyone!

 

[vanesch]

If you use Bernoulli's law in this case, we simply have:

v^2/2 + h g = constant
(the density of the fluid drops out).
If we are at the surface, v = 0, and h = d, so constant = g d
If we are at the hole, h = 0 and hence v = sqrt(2 d g)

So the velocity of the fluid that squirts out of a hole at depth d from the surface of the fluid, equals v = sqrt(2 d g)

This is independent of the area of the hole or even the density of the fluid.

However, the fluid flux will be equal to this velocity times the area of the hole A:
phi = v.A = A sqrt(2 d g).

Now, in a time dt, a fluid flux phi will correspond to a volume dV = phi.dt of water that has escaped. This will then lower the surface of the water with an amount - d d (silly notation, meaning that depth d has diminished by d d). If the section of your bottle equals S, we obtain that when a volume dV has been taken away, we have that d d . S = - dV

So: d d. S = - dV = - phi.dt = - A.sqrt(2 d g).dt

which gives you a differential equation for the function d(t):

(hum, I did a bit too much here, you should work this out yourself...)

I almost gave the solution in my quoted post (actually I did, and then I quickly erased it for it is against the policy of PF). Given all the work and the solution, I think it is fair now to go to the final solution.

Now that we got this far, once we have the differential equation:

d d. S = - A. sqrt(2 d g).dt

with:
d the fluid level (in meters, say) above the hole, S the cross section of the bottle (horizontal area in m^2), A the cross section of the hole (also m^2), g the gravitational acceleration (9.81 m/s^2), we can write this as:

d d / sqrt(d) = - A/S sqrt(2 g) dt

and integrate, which gives us:

sqrt(d) + cte = - A/S sqrt(g/2) t

at t = 0, we have d = d0 (the initial fluid level), so cte = - sqrt(d0)

sqrt(d) = sqrt(d0) - A/S sqrt(g/2) t

Or:

d(t) = (sqrt(d0) - A/S sqrt(g/2) t)^2

which gives you the evolution of the fluid level above the hole d(t) as a function of t...

It has the form of a parabola, and I hope I didn't make any silly mistake.