# Simple Error Analysis for ratio of Flow Rates in a tube

I am trying to do simple error analysis but I am running into trouble. My experiment was to determine the ratio of flow rates of different pressure amounts on the tubes. My errors are:

1. $5\%$ error in measurement of $\mathrm{cm}^3$ from graduated cylinder, per trial
2. $1$ second error in time measurement per trial. (I used a stopwatch and rounded to nearest second)
3. Standard deviation of the overall trails.

(Is this a good time error? I rounded by values by the second so I am not entirely sure what my error would be, also accounting for things like human reaction time (but mostly just the rounding I guess).)

So let's say I have five trails, and I find that it takes ($24$,$25$,$25$,$24$, $24$) seconds for $1$ $\mathrm{cm}^3$ to drip from this very thin tube. (without errors stated above)

Now it gets a bit more complicated, I half the pressure of the water through the tube and measure the time again. I find that now it takes ($50$,$51$,$54$,$49$, $49$) seconds for $1\,\mathrm{cm}^3$ to drip. (again, without errors).

I take $1\,\mathrm{cm}^3$/(time) and have found the mean and dev of both of these sets, ($m: .0197885, d:.00078393$) and $(m:.041, d:.000912871$) respectively. (these are the means and deviations of the flow rate now). Now I divide the two means and get the ratio of the flow rates.

That's it, the thing is that I have no idea how the hell I would account for the errors through my calculations. Could someone show me for this particular set of data? I know I have to add the standard deviations of the data, but I don't know how to keep track of the time and grad cylinder errors.

If you were drawing a graph the error analysis would come out of the spread of the data points about the line of best fit. However, if you only have 2 data points (2 values of pressure) drawing a graph is not worthwhile.

The 5% error in measuring volume is already included in your measurement of the flow time, because an increase in the volume will cause an increase in the time for it to flow. Random timing error is also included, but not any systematic error such as a delay in starting of stopping the timer. It could take you perhaps $0.5s$ to react to an event and switch the timer on or off. However, if the same person starts and stops the timer these systematic delay errors will cancel out. Any residual error is random and should average out over the trials.

So all you need to do is find the mean times $t_1,t_2$ and standard deviations $\sigma_1, \sigma_2$ of trials for the same nominal amount of water to flow at the 2 pressure settings. The ratio $R$ of flow rates $Q$ is then
$R=\frac{Q_2}{Q_1}=\frac{t_1}{t_2}$.

The standard error in flow time $t$ is $\Delta t=\frac{\sigma}{N}$ where $N$ is number of repeated trials (runs). The standard error in $R$ is $\Delta R$ which is found from adding the squares of the fractional errors :
$|\frac{\Delta R}{R}|^2=|\frac{\Delta t_1}{t_1}|^2+|\frac{\Delta t_2}{t_2}|^2$.

Then you can report your result as $R \pm \Delta R$. Note that you should have at most 2 significant digits in your error $\Delta R$. In your case I would round $\Delta R$ to only 1 significant digit, and quote $R$ to the same number of decimal places.

I presume that you are changing the pressure by using different heights of water in a burette, say 25 and 50 cm, allowing the water level to fall only 1cm in each trial. If this is the case the ratio of pressures should be measured as the mean height of the water above the top of the narrow tube. Some error analysis will be required for this also, because the pressure might differ between each trial, and the ratio of pressures might not be 2:1 as you assume.

• Thank you for the help! You are totally right about the last part too, I didn't even think about that. Also I have multiple points for this relation, this is just a small subset of my data. – Shrodinger 2016 Oct 15 '16 at 18:14