3
$\begingroup$

For physics class we've done this experiment where we would measure the speed of the water escaping from a water bottle through a hole in a side of it. We would calculate the speed through kinematics by measuring the distance from the bottle the stream would reach.

The data obtained are

h (cm) --- v (cm/s)
9.8    --- 135.7
5      --- 82.4
3.9    --- 70.8
2.9    --- 56.2
1.4    --- 33.2

But when analyzing the data the speeds of the streams don't follow Torricelli's law, $v=\sqrt{2gh}$, as they should. Instead they roughly follow another power-law formula, $v=26\ h^{0.8}$, where v is the escape velocity and h is the height of the fluid surface from the hole.

I've been staring at the data for days now and I don't know if we've measured wrongly the experimental data or if there is something wrong with the things I've been doing.

$\endgroup$
  • $\begingroup$ Would you mind posting your data and showing what you have tried? It will help me and others answer your question $\endgroup$ – InertialObserver Jan 5 '19 at 21:48
  • 1
    $\begingroup$ Thanks for adding part of your data. Since your velocities are inferred from the measuring the distance traveled by the streams before they hit the ground, shouldn't your raw data contain only distances? I would expect a height above the ground and a horizontal distance for each data point. $\endgroup$ – rob Jan 5 '19 at 21:57
  • 1
    $\begingroup$ Unfortunately, in a lot of these school experiments there are lots of other factors that get in the way. For example, there might have been sizable air resistance or viscosity effects. When I was in high school, I once measured the concentration of a solution to be $937\%$, so this is actually pretty good. $\endgroup$ – knzhou Jan 5 '19 at 21:58
  • $\begingroup$ Also, did you make all these measurements simultaneously (e.g. from a photograph)? If not, you should estimate how much the water level dropped while you measured. $\endgroup$ – rob Jan 5 '19 at 22:00
  • $\begingroup$ We decided the heights first (based on some marks on the plastic bottle) and then lowered the water level to each mark, blocked the hole and try to measure the distance of the stream when we unblocked it (probably a cause of more errors as the stream didn't move perfectly right after putting your finger away) $\endgroup$ – Diego Jan 5 '19 at 22:04
1
$\begingroup$

It might have been better to start with the water at a height above that which you wished to take the measurement and then waited until the water level reached the required height and then taking the measurement?
Doing this would reduce the "cause of more errors as the stream didn't move perfectly right after putting your finger away".

Perhaps a better way of estimating the speed at which the water emerges from the hole is to measure the flow rate (volume emerging per second) which is equal to the speed of the water $\times$ the cross-sectional area of the hole?
To try and get reasonable accuracy the cross-sectional area of the reservoir should be much larger than the cross-sectional area of the hole so that the level of water in the reservoir does not fall by very much whilst the flow rate is being measured and the speed of the water at the top of the reservoir is much less than the speed of the water as it emerges from the hole.

The experiment is one which is done quite often and here is one example of the experimental results on the Internet.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ unluckily I don't have access anymore to the experiment and I can only work with the data we measured at that time, so I can't work with the cross-sectional area of the hole $\endgroup$ – Diego Jan 6 '19 at 18:02
  • $\begingroup$ @Diego there does seem to be a lot of errors that you would have to justify then in your calculations, to find a really close number to the expected model. $\endgroup$ – Karthik Jan 6 '19 at 21:19
0
$\begingroup$

We would calculate the speed through kinematics by measuring the distance from the bottle the stream would reach.

How exactly did you calculate velocity in this measurement?

Fluid particles trace a parabolic path in this situation, and this path follows the equation

$$ y = \frac{x^2}{2R} = g\frac{x^2}{2V^2}$$

where the coordinate system is shown by (taken from my answer here):

enter image description here

The velocity is therefore given by

$$ V = \sqrt{\frac{g}{2y}} x$$

where $y$ is the height that the stream has fallen (i.e., the height of the hole), and $x$ is a horizontal distance (i.e., the distance on the table that the stream lands). From Torricelli's law, this suggests that $\frac{x}{\sqrt{y}} = 2\sqrt{h}$. Maybe you could measure $\frac{x}{\sqrt{y}}$ to see where your disagreement is?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If the experimenter kept the distance through which the water fell after leaving the hole $y$ constant then $V\propto x$ which is possibly what was done? $\endgroup$ – Farcher Jan 5 '19 at 23:23
  • $\begingroup$ @Farcher Oh yeah, this could be a way to get $V$ by measuring $x$. I'm not sure if OP did that. $\endgroup$ – Drew Jan 6 '19 at 2:18
  • $\begingroup$ That is exactly how i calculated the velocity $\endgroup$ – Diego Jan 6 '19 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.