Homework Help: This long uncertainty viscosity problem where I need to get the error

1. Sep 13, 2014

psstudent

Any help with this basic college problem.

1. The problem statement, all variables and given/known data

The following readings were obtained in an experiment to find the coefficient of viscosity of water n by measuring the rate of flow of water through a horizontal capillary tube.

Diameter of capillary tube = 1.40 mm
Length of capillary tube, l = 21.0 cm

Pressure difference between end of tube, p =18.5 cm water rate of water flow, v/t = .74 cm^3/s

If the diameter of the tube is measured to +/- .05mm, the length to +/- 1mm, the pressure difference to +/- 2 mm, water rate of flow to +/- 0.01cm^3/s and the equation used is

V/T= pi p r^3/8nl

Where r= radius of the tube, calculate the percentage for the value of the coefficient of viscosity of water.

2. Relevant equations

V/T= pi p r^3/8nl

3. The attempt at a solution

So here is where i come in. Now i tried to tackle this question in parts. First i tried to go about this in a away to not have to use delta margin of error formulas because frankly i dont understand them. The problem has to do with viscosity but the topic we are on in physics is really uncertainty so i dont have much of a background solving viscosity related problems.

The first thing i did was try to convert every thing into meters sooo

Pressure difference: 18.5cm = .185m
Length of capillarity tube = 21 cm = .21m
diameter of capillarity tube = 1.4mm = .0014m
Radius = .7 mm = .0007m

Rate of water flow = .74cm^3/s = .00000074 m^3/s

Margins of error
Diameter = .05 mm = .00005m
Length= 1mm = .001m
pressure difference = 2mm = .002m
Rate of change =.01cm^3/s = .00000001 cm^3/s

Then me and my friend both tried to rearrange the formula and we got n =pi p r^3/8 * rate of water * length. This maybe where we went wrong. Anyway my methodology was to use the maximum possible values on one calculation using the plus/minus and the minimum possible values and then averaging them off. So i ended up with .00016076 as the maximum value and .000159755 as the minimum value, i then averaged them and found the magnitude between each value to find the plus or minus , which gave me .000160275 =/- .00000052. I then divided the .000160275 by .00000052 and multiplied by 100 to give me a .32% fractional error. Now i think my answer is wrong because a .32 fractional error seems too small and i think when i transposed the formula I made an error. My friend also tried the problem he used (as i said above) the same formula and ended up with a final answer of .000160275 =/- 2.2 * 10^-5, which he said gave him an error percent of 13% which to me sounds more plausible, but i suspect both of our answers are wrong as we both ended up with our answers in seconds, ( i have researched but can't find a coherent answer for what the unit should be).

Any help would be greatly appreciated I tried to make this as compact as possible but ive exhausted so many avenues of rechecking this problem so i have a lot to write lol. Thank you for your time.

Last edited: Sep 13, 2014
2. Sep 13, 2014

composyte

My recommendation would be to use the delta margin of error formulas because they are the fastest and only way to receive the correct answer the book is looking for.. That being said I have done you the favor of putting the correct formula you should use for this particular problem..

δq=$\sqrt{((\frac{∂q}{∂x}δx)^{2}+.....+(\frac{∂q}{∂z}δz)^{2})}$

where x,......,z are measured with uncertainties δx,......,δz and these values are used to compute q(x,......,z) and then the uncertainty for the value q (or n in the case of your problem) is δq and is found in the previously shown formula...

All that must be done is take the partial derivatives of every variable with respect to n that is, ∂n/∂p,∂n/∂t,..... use these values along with the uncertainties of the given values and you will have your answer...

let me know if this helps. Also let me know if you require assistance understanding the equation (ie. where it came from, how its used, etc)

3. Sep 13, 2014

Staff: Mentor

You may wish to look up the Hagen–Poiseuille equation, also known as Poiseuille's Law. That's what you're dealing with here. When you do, take a very close look at the exponent on the radius (or diameter) variable

I suspect that your pressure difference, given in cm of water, should be converted to Pascals...

The units associated with this viscosity coefficient should be, I think, Pa*s, or Pascal-seconds, and usually scaled to milli-Pascal-seconds, or mPas for wimpy fluids like water

4. Sep 13, 2014

Staff: Mentor

What is the % uncertainty on r, l, p, and v/l?

Solve the equation V/T= pi p r^3/8nl for the viscosity in terms of the other parameters.

If x is the % uncertainty on r, what is the percent uncertainty on r^3?

If you know the percent uncertainties in all the parameters in your equation except viscosity, do you know how to determine the % uncertainty on the viscosity?

Chet

5. Sep 13, 2014

psstudent

@composyte

"Also let me know if you require assistance understanding the equation (ie. where it came from, how its used, etc)"

I believe i understand how the equation is used we are using a similar one in my physics class i would like to know where it came from if possible.

6. Sep 13, 2014

psstudent

Hi gneil thanks for your response. I looked up the equation you said pay attention to the exponents , I see that it is 4 instead of 2? Is that what you wanted me to discover, so is the equation given in the question essentially wrong? I realize, what you are saying about the pascal*seconds, so I need to find pascals, but how do I do that from the info given? Use the standard density of water and gravity to give me pascals? Thanks for your reply

7. Sep 13, 2014

Staff: Mentor

That's a 4 instead of a 3 for the radius exponent. Must be a typo in your lab materials. Yes, use the density of water and g to determine the pressure at the bottom of the water column of the given height.

FYI, there's a lab from the University of Toronto described "faraday.physics.utoronto.ca/IYearLab/viscosit.pdf" [Broken] (a PDF file) which outlines what your question is all about. Note the formulas!

Last edited by a moderator: May 6, 2017
8. Sep 13, 2014

Staff: Mentor

The approach Chestermiller suggested is by far the easiest method to get the relative uncertainty of the viscosity - you do not even have to calculate this viscosity value itself!

9. Sep 13, 2014

psstudent

Mfb, wouldn't Chestermillers approach require me to also find the pressure? So i find the percentage error of each given quantity in the question and then what? Thanks Mfb for the reply.

10. Sep 13, 2014

Staff: Mentor

You just need the relative uncertainties. As pressure comes from a known height with a known uncertainty (both given as length), you don't even have to calculate an actual pressure.
Quadratic sum. One of the most basic formulas you should have.

11. Sep 14, 2014

composyte

Yea so basically all you need to do is plug in the median values of each variable to find what exactly n is equal to (or in my equation, q) once you do this, calculate the partial derivatives of each variable with respect to n and then use the formula. the δ sign is simply for uncertainty, not percent uncertainty...So the final value you will obtain will be n$\pmδn$, hope that helps!

As for the formula's derivation I found a link
http://people.westminstercollege.edu/faculty/kphilippi/An_Introduction_To_Error_Analysis_The_Study_Of_Uncertainties_In_Physical_Measurements_Taylor_John.pdf [Broken]

its a pdf file of the book 'Introduction to Error Analysis' the derivation is on page 73-75...its a fairly straightforward and intuitive derivation...And honestly if this is for a course in experimental uncertainty I would use this pdf for a quick reference occasionally.

Last edited by a moderator: May 6, 2017
12. Sep 14, 2014

psstudent

Ok thanks, now I understand you formula, but im still caught up on the δq as the numerator, what would I put as the value for δq? would I just put "1" or what because i dont have a value for n and the only way i would be able to get a value is to find pressure and then solve the formula which as one person said is wrong already.

Last edited by a moderator: May 6, 2017
13. Sep 14, 2014

psstudent

So I used the standard density. That would give me 1000 kg/m^3 X 10 m/s^2 X .185 m. which equals 1850 pa. Isnt that rather large though? Thanks for your reply these answers really help, seeing as the formula in the question seems to have been a typo in the text book.

Also mfb said you use the quadratic sum, I know he said it was basic but I have never heard of it ( i might have but it could be called another name, i dont live in america) so if anyone could explain what it is and how I use it that would be great. Unless he just means the quadratic formula , because anything quadratic must pertain to squares I believe.

Last edited by a moderator: May 6, 2017
14. Sep 14, 2014

composyte

okay so now you have the pressure and all other values for the variables of the equation. All that must be done is plug in these values to find n. Once this is done calculate the partial derivatives and use the formula I posted to determine the value of δn and your good to go. And I assume, although I could be wrong, the quadratic sum mfb is talking about is the same formula I provided you with. (notice its a quadratic formula with uncertainties).

15. Sep 14, 2014

Staff: Mentor

About 2% of the atmospheric pressure. It is right.

I don't live in America either. It's just the square root of the sum of squares.
$\sqrt{a^2+b^2+\dots}$
If a and b are relative uncertainties...

16. Sep 14, 2014

Staff: Mentor

I believe it's also known as adding (or summing) in quadrature. One does the same thing to find the magnitude of a vector from its individual components.

17. Sep 14, 2014

psstudent

Just want to preface this by saying thanks to everyone who has helped so far. The hw is due tomorrow and even though I feel like im close, I dont think ill get the correct answer. I tried two last rolls of the dice and this is what I came up with:(please tell me if any of the values I got for my answers are the correct answer to the question)

The first attempt is pretty much my original but with the pressure calculated in the formula instead of just the p which was in just plain old meters. I used the maximum and minimum values of each quantity (used the plus minus) to try to find the value of n with its plus and minus in the end I got n as = .001233098 +/- .000234416. I then divided the margin of error by the average and multiplied by 100 to get the percentage error, .000234416/.001233098 * 100 to get roughly 19%.

My second attempt to get the answer was through the use of mfb's method which was using the quadratic sum:

So i found the percentage error of each of the given values (except n, because it wasnt given the question ) which gave me √1.08%2^2 + .47%^2 + .71%^2 + 1.35%^2 which gave me an answer of .01927 %

Note: I think it was another poster who asked if I knew what to do with the r^4 particularly the exponent In the above try ^ I just squared the radius, but I also did the summation above with the radius to the 4th power as well and got : .0179 %

Also @composite, I like your formula but I dont understand what you want me to put as the numerator for the equation,"q", I know q in your equation is my "n" but if I use your equation it would have no value for n to put as the numerator.

Anyways I would really appreciate all you guys have done, i dont think ill make it in time for the due date tomorrow morning unless it's just a simple thing I need to tweak , as always any help would be very appreciated.

18. Sep 14, 2014

Staff: Mentor

That would mean your uncertainties are all correlated - they are not.

You made a calculation error here, the final result cannot be smaller than the largest contribution (1.35%?) - did you take the square root of the sum?. You also forgot that the radius goes with the 4th power.

An even smaller value? No, that is wrong. If r has an uncertainty of 1% (made-up number), what is the uncertainty of r^4? Here you can use the approach of calculating lower/higher values if you like, as in r*r*r*r, all r-factors vary in the same way (because they are all the same...).

19. Sep 14, 2014

psstudent

thanks, Mfb, yes you're right there was a error I believe it was 1.927 % not .019%.

20. Oct 3, 2014

psstudent

Just want to say thanks to everybody , and a little update: I finally did get the correct answer. Thank you!

21. Jun 5, 2017

Pearl_King

Can
Would You please post up the correct Final calculation and Result ?

Last edited: Jun 5, 2017
22. Jun 6, 2017

Staff: Mentor

Hi Pearl_King,

Welcome to Physics Forums!

That's not how things work here. You need to show your own attempt at solution, show us what you've tried, before others can help you. We don't directly provide solutions for homework problems, but can offer guidance on your own work.

23. Jun 6, 2017

Staff: Mentor

This thread is 3 years old and psstudent's last post was in 2014.