Determining the coefficient of viscosity of water (Graph Problem)

[WhiteWolf98]

Homework Statement

So, the question is based around an experiment to determine the coefficient of viscosity for water (by capillary flow).
Part of the experiment involved producing a graph of Q against h. This graph is supposed to be linear, and is supposed to pass through the origin.
As can be the seen in the equation, the gradient of the graph is supposed to be used in the calculation. The problem is my graph doesn't go through the origin, so I was wondering what to do about it. Thanks

Homework Equations

η=ρgπr^4/8l*1/slope

The Attempt at a Solution

 

[haruspex]

my graph doesn't go through the origin

Can you post the data in cut-and-pastable form? Failing that, can you post the graph?

 

[WhiteWolf98]

The Results

The Graph

 

[Chestermiller]

I plotted the data of h vs Q, and did a straight line fit to the data. It seems to me the straight line fit passes very close to the origin.

 

[WhiteWolf98]

Oh yeah, well, it cuts the y-axis at 8×10^-7. It is very, very close, but I was wondering if that had any effect on the value I obtain for the coefficient of viscosity because the value I calculated was very far off from the actual value

 

[Chestermiller]

Oh yeah, well, it cuts the y-axis at 8×10^-7. It is very, very close, but I was wondering if that had any effect on the value I obtain for the coefficient of viscosity because the value I calculated was very far off from the actual value

You didn't really expect it to pass exactly through the origin, did you.

Let's see your calculation.

 

[WhiteWolf98]

Uhm, maybe I did, oops. Ehehe

Calculation:
ρ=10^3
g=9.81
r=0.65×10^-3
L=15.5×10^-2

Hence, overall calculation is:

η=[(10^3×9.81×π×(0.65×10^-3)^4)/8(15.5×10^-2)]×1/1×10^-7=0.0444 Pa⋅s (to 3 s.f.)

 

[Chestermiller]

From your data, I get a viscosity of 0.0018 Pa.s

 

[WhiteWolf98]

Could you show me the calculation please?

 

[Chestermiller]

It looks like your curve fit is way off. I get a curve fit of: $y=-2.38\times 10^{-8}+2.233\times 10^{-6}x$

 

[haruspex]

it cuts the y-axis at 8×10^-7

It cuts the y-axis of the graph you posted at that value, but that axis is not at x=0.

 

[Charles Link]

It looks like your curve fit is way off. I get a curve fit of: $y=-2.38\times 10^{-8}+2.233\times 10^{-6}x$

Edit: I agree.

 

[WhiteWolf98]

Alrighty! So first of all, apologies for the late reply. I took your points into consideration, and decided just to draw the graph by hand. Using that graph, I managed to obtain a value for the coefficient of viscosity as $2.06 × 10^{-3} Pa⋅s$. The actual value is $8.90 × 10^{−4} Pa·s$ according to this source. That's still over a 100% percentage error though; is that really the best I can do? Thank you all for your help!

 

[Chestermiller]

Alrighty! So first of all, apologies for the late reply. I took your points into consideration, and decided just to draw the graph by hand. Using that graph, I managed to obtain a value for the coefficient of viscosity as $2.06 × 10^{-3} Pa⋅s$. The actual value is $8.90 × 10^{−4} Pa·s$ according to this source. That's still over a 100% percentage error though; is that really the best I can do? Thank you all for your help!

It's hard to say without seeing in person the actual experimental procedure being performed, and the apparatus. There are other things you might look at like the hydrodynamic entrance effect (probably not important). This could be ascertained by varying the length of the tube. Also, there seems to be a lot of scatter in the data, and this might be improved by better technique.

 

[WhiteWolf98]

It's hard to say without seeing in person the actual experimental procedure being performed, and the apparatus. There are other things you might look at like the hydrodynamic entrance effect (probably not important). This could be ascertained by varying the length of the tube. Also, there seems to be a lot of scatter in the data, and this might be improved by better technique.

Understood, guess that's finally sorted; and somehow, my line of best fit for my hand-drawn graph went exactly through the origin. Thank you Chestermiller, and everyone else for your input in solving this. Have a good day

 

[Ray Vickson]

Understood, guess that's finally sorted; and somehow, my line of best fit for my hand-drawn graph went exactly through the origin. Thank you Chestermiller, and everyone else for your input in solving this. Have a good day

I would be suspicious of a best-fit-line that goes exactly through the origin. The chance of that happening is essentially zero, because there are always experimental errors (random or otherwlse) in any real experiment, and that will almost always give you a line that does not pass exactly through (0,0). However, if the errors are truly random (and "statistically independent) there are statistical tests you can perform to check if the true (unknown and unobservable) intercept could really be zero. That is different saying that any particular best-fit-line goes through (0,0).

 

[haruspex]

That's still over a 100% percentage error though

Take a look at your relevant equation. Can you see a parameter to which the result will be highly sensitive? How accurately is that parameter known in your experiment?