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For my Physics project, I decided to do an experiment with fluids. I'm using the Torricelli's law to verify that the value of acceleration due to gravity on earth is $9.81m/s^2$.

I am using the equation $v=\sqrt{2gh}$. Transposing, the equation becomes

$$g=\frac{v^2}{2h}$$

Now, I have a 1-liter plastic bottle. I have made a circular hole 3mm in radius, with center 2.5 cm from the base. I fill the bottle with water, note the height of the water column and then open the hole for 3 seconds. The water coming out is collected in a poly bag. After 3 seconds, I close the hole and again take the reading for height for water column.

Velocity is calculated by first taking the flow rate and then dividing by the area of the hole. I use the formula:

$$Q = \frac{mass \ \ of\ \ water\ \ collected}{time}$$

$$v = \frac{Q}{A}$$

(Q - Flow Rate; A - area of cross-section of hole) Area is calculated using: $$A = \pi r^2$$

Observations:

  1. Change in height of water column is about 1 - 1.1cm. So the value of $h$ is averaged.For a particular reading, the height of water column was 9.75cm above the hole initially. After 3 seconds, the final height of water is 8.65cm. So I took the value of $h$ as 9.2cm.

  2. The mass of the water collected in poly bag was measured using an electronic weighing machine, and it comes out to be 105 grams for the above reading.

I plug the values into the formula $g=\frac{v^2}{2h}$, but the values of $g$ are coming out way too off from the value of 9.81.

Now, after spending a lot of time figuring out the error, I have reached no good conclusion. All I can think is that such an experiment requires ideal conditions that are difficult to emulate. (probably impossible at my home)

So, I would like to ask if there any error in the method I am using, or are the errors actually due to the absence of ideal conditions? And is there any way to modify this experiment, so as to obtain more accurate values for $g$?

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  • $\begingroup$ Maybe the contraction factor (the famous "vena contracta") of the exit jet has to be taken into account. The area you should use is kA with k < 1, depending of the form of the exit. $\endgroup$ – Vincent Fraticelli Jan 14 at 18:51
  • $\begingroup$ Perhaps you could observe where the water stream first lands and treat it like a projectile motion problem? $\endgroup$ – Aaron Stevens Jan 14 at 19:03
  • $\begingroup$ What are the dimensions and shape of the 1-liter bottle? If it is a circular cylinder you can calculate the volume of water that exited by the change in fluid height, $V = (\pi D^2 \Delta h)/4$, where $D$ is the internal diameter of the bottle (in cm), $\Delta h$ is the change in fluid height in cm. You can compare this volume to the volume of liquid you collected in your poly bag, the volume being the fluid mass collected in grams divided by the fluid density (assume 1 gram/cm3). How does this compare? $\endgroup$ – Armadillo Jan 16 at 14:44
  • $\begingroup$ See the following thread for a detailed analysis of the unsteady state behavior: physicsforums.com/threads/velocity-of-efflux.868030/…. Posts #16 and #17. $\endgroup$ – Chet Miller Jan 21 at 15:14
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I suspect an error in method. But I'm not sure.

Bernoullis' Principle tells us that $P_0+\rho gh+\frac{1}{2}\rho v^2=const$.

$P_0$ is external pressure, $h$ is the height separating observation points in a fluid and $v$ is velocity of a fluid. At first the top of the fluid isn't moving. When the hole is opened, the top of the fluid starts descending at a certain rate and the water begins leaving the container at a certain, different rate.

Incompressibility of the fluid and tells us the conservation of mass tells us the area of the exit hole times the velocity of outflowing water is the velocity of the water at the top times the cross sectional area of the container.

$$P_0+\frac{1}{2}\rho v_{top}^2=P_0+\rho gh+\frac{1}{2}\rho v_{bot}^2$$

If $A$ is the average cross sectional area of the con tainer and $a$ is the area of the exit hole, we have $A\frac{dh}{dt}=av$.

The only speed we have at the top of the fluid is due to the descending height, so $v_{top}=\frac{dh}{dt.}$ It follows that $v=\frac{A}{a}\frac{dh}{dt}$.

The pressures cancel in the Bernoulli expression, then both sides can be divided by the density. Then double both sides to simplify the expressions.

$$(\frac{dh}{dt})^2=2gh+(\frac{A}{a}\frac{dh}{dt})^2$$

Rearranging: $$(\frac{dh}{dt})^2=\frac{2gh}{1-(\frac{A}{a})^2}$$

Take the square root. $$\frac{dh}{dt}=\frac{\sqrt{2gh}}{\sqrt{1-(\frac{A}{a})^2}}$$

and divide both sides by $2\sqrt{h}$.

$$\frac{d\sqrt{h}}{dt}=\sqrt{\frac{g}{2(1-(\frac{A}{a}^2))}}$$

So the rate of change in the square root of the height is aconstant. Coincidentally, this also happens if one assumes $v_{top}$ is always zero, but with a different constant, $\frac{a}{A}\sqrt{g/2}$.

Either way, solving the equation gives:

$$h=( \sqrt{h_0}-t\sqrt{\frac{g}{2(1-(\frac{A}{a})^2)}})^2$$

Where $h_0$ is the beginning height of the fluid above the hole.

So the relationship between change in height in time to $g$ is non-linear.

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  • $\begingroup$ So essentially you are saying the error could potentially come from an $h$ value that varies too much over time time of fluid collection? $\endgroup$ – Aaron Stevens Jan 14 at 19:30
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    $\begingroup$ I think there is an error in your resolution of the differential equation. We find $\frac{dh}{\sqrt{h}}=-Cdt$ which integrates into $2\left( \sqrt{{{h}_{0}}}-\sqrt{h} \right)=Ct$ Strictly speaking, the Bernoulli equation supposes permanent flow and therefore the rate of descent of the surface must be negligible compared to the speed of the jet. $\endgroup$ – Vincent Fraticelli Jan 14 at 19:34
  • $\begingroup$ Thanks for the tip about the solution. I think I've fixed it. I"m not sure how to address the Bernoulli deviation. I was thinking starting with $\nabla\cdot(\rho v)+\frac{\partial \rho}{\partial t}=0$ integrating both sides and applying Gauss law we get $\frac{\partial M}{\partial t}=-av$. Then the descent of the top is A(dh/dt). So A(dh/dt)=av. If $a/A<<1$ it might satify the condition. $\endgroup$ – R. Romero Jan 14 at 20:54
  • $\begingroup$ I understand that there was an error in the value of $h$ I was using. In the expression you derived for $h$, there is use of acceleration due to gravity $g$, which I cannot use in my calculation because the aim itself is to verify the value of $g$ as 9.81. So, the only method left to me is to use a container with large cross-sectional area, so that the difference in height of water is negligible. $\endgroup$ – Snehit Sah Jan 15 at 12:13
  • $\begingroup$ In experiments like this, the teacher typically wants you to graph the data, or some small mathematical change to it ,you collect over time looking for graphs that turn out to be straight lines. In this case, it looks like you'd get a straight line in your graph if you graph the square root of the height against time. Once you determine the slope, you can get at the value of g hiding in the data. $\endgroup$ – R. Romero Jan 15 at 15:29
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You can easily reduce the change in height of the water by using a bigger bottle, for instance a 5 or 10 liter plastic food container. Find out what you things your family buys in bulk from a supermarket, and you might get a suitable container for free.

A different source of error is in just making a simple hole in the container. If you think about the flow before the water exits the hole, it is being accelerated rapidly in all directions, not just horizontally. Because of its viscosity, this will dissipate energy and reduce the velocity out of the hole. In fact Torricelli noticed this effect, which makes the diameter of the water stream smaller than the hole, a short distance outside it. Google for "vena contracta." For a sharp edged hole, the effective area of the hole is about 0.63 times its actual area.

If applying that factor gets you closer to what you expect, then the problem is how to eliminate it without "cheating" and just quoting the correction factor from some website or textbook.

You could reduce the effect by making a "streamlined" funnel-shaped structure inside the bottle, so the water accelerates more gradually and smoothly towards the exit.

It might be hard to make that inside the bottle, but if you think about it you will get the same effect by making a bigger hole and attaching the funnel outside the bottle.

That will not completely fix the problem, because the viscosity of the water still means that the flow velocity around the edges of the hole will be smaller than in the center, but it should get you closer.

Don't forget that discovering that an experiment didn't work as expected, and improving it to make it work better, is much more like what happens doing real research than when things work first time!

This shows a "real world sized" streamlined hole to empty a water container: https://www.youtube.com/watch?v=EPJar2kh5zQ

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If you look in the literature, you will easily find contraction coefficients. One can even show by a nice balance of momentum that the reduction coefficient is 0.5 for a re-entrant mouthpiece (Borda mouthpiece). This case is treated in the Feynman. In the wikipedia ("vena contracta"), they give a coefficient of the order of 0.64 for a sharp orifice.

The effect on the measurement of $g$ is far from negligible: $\frac{dg}{g}=2\frac{dv}{v}=-2\frac{dA}{A}$ So, roughly, if $\frac{dA}{A}=0.3$ then $\frac{dg}{g}=-0.6$ that is to say a value 60% too low !

Finally, as indicated in the previous answer, the hardest part is to develop the experiment. When I tried, the residual kinetic moment of the water and the asymmetries of the hole made the jet leave laterally.

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I think you are actually pretty close. The image below is a screen shot of some MS Excel calculations. The first 4 columns on the left show your experimentally measured values along with some calculated values.

enter image description here

To the bottom right are the defined nomenclature used for the measured and calculated values. E.g., the value rho (index 8) is your assumed fluid density. The value Q (index 9) is the volumetric flow rate $=\frac{\dot m}{\rho}$. The value C is the coefficient of discharge of the orifice that @alephzero had mentioned. Using Torricellis law as you have assumes $C=1$.

The slightly more general equation for an orifice discharging freely into atmosphere while flow is at steady-state (i.e. there is no change in the flow rate with time) is $$\tag{1} Q=CA\sqrt{2gh}$$

where $h$ is the constant fluid height above the hole (orifice).

Rearranging for $g$ we have $$\tag{2} g=(Q/CA)^2/(2h)$$

Taking $CA=A_{eff}$, then we have $v=Q/A_{eff}$, and then we have your equation $$\tag{3} g=v^2/(2h)$$

As you can see, using the values you give results in an acceleration of gravity (g_calc, index 16) of ~833 cm/s^2, a 15% difference (this result is shown at the top right of the figure above).

There will be an uncertainty associated with each value we use as inputs for calculation. Near the top-right in the figure is a table of values that were independently changed to minimize the difference between the calculated g and the true g. For example, keeping all the inputs you see shown on the left unchanged other than for hbar (index 4). To minimize the difference an hbar value of 7.82 cm is needed. Changing hbar back to 9.2 cm, the mass collected was then modified, resulting in a needed value of 113.9 grams to minimize the difference. This process was carried through for the values you see in the table.

As @R. Romero has commented, we can plot the data and look for a straight line. Experimentally what you can do is perform a number of trails (measurements) with different starting water heights (h1, index 1). Eqn(1) can be rearranged into the form of a line $y = mx + b$ as shown here:

$$\tag{4} (Q/CA)^2 = g(2h)$$

Here, $y = (Q/CA)^2$, $x = 2h$, your acceleration of gravity $g$ is your slope ($m$), and your y-intercept $b$ is assumed to be zero (as assumed by your Torricellis model).

Take many more measurements following the exact same procedure as you did, only this time change your starting height (h1) to different values. Then plot $y = (Q/CA)^2$ as a function of $x = 2h$ (here we assume $h$ = hbar (index 4)). Perform a least-squares linear regression (do not assume the line intercepts the origin (0,0)). The slope of this line will be a better estimate of your experimentally determined gravitational acceleration.

Apply the proper error bars to your data points (i.e., the uncertainties to these values). Your regressed line should be encompassed within these error bars. E.g., the figure below is an example, where a fictitious second data point of $y = (Q/CA)^2=0$ and $x = 2h=0$ is assumed. The linear best fit shows the slope (our gravity acceleration) to be 832.79 cm/s2.

enter image description here

To get a better idea of what the discharge coefficient $C$ of the hole you made in your container you could perform a steady-state experiment. One way to do this would be to continually add water into your container at the same rate it exits the hole at the bottom of your container. An example of a setup to do this is shown in the figure below. Allow flow to achieve steady-state and then make a measurement. Solve for $C$ by rearranging Eqn(1), assume g = 981 cm/s2.

Like previously described, you can make multiple measurements by changing the height of the 'overflow hole' shown in the figure, and plot $Q$ as a function of $A\sqrt{2gh}$. The slope of your linear least-squares regression will be your best estimate of $C$. Use this value for $C$ in your analysis of $(Q/CA)^2$ as a function of $x = 2h$ (i.e. when trying to experimentally determine g).

enter image description here

Please let us know if you conduct more trials (measurements). It would be fun for us to look at the data.

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  • $\begingroup$ Thanks! I did the experiment again, with an orifice radius of 1mm to minimize the change in height of water surface and a weighing machine of lower least count - 1gm (earlier, I was using a weighing scale of least count 5gm). After multiple readings, I could reduce the average error percentage to 1.01%. $\endgroup$ – Snehit Sah Jan 26 at 10:09

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