Hi, I am doing physics coursework on finding viscosity of fluids by dropping a marble into fluids, finding terminal velocity, then using stoke's law to find viscosity. (using density of fluid, sphere, sphere diameter etc). I have completed all the practical, now just the write up However ... I could really do with someone's help one the formula I would use, has anybody done this before for their coursework? I know some people will say 'just google it,' but I have spent far too long doing so, and right now have dozens of tabs and each formula is a slight variation on the other. I honestly have no idea which is the best/right one (my physics teacher is ill till after the christmas holidays so I can't ask him) ... they all put the viscosity in a different units, some do not give the required unit of density for the calculation (kg/g per cm/m/mm), some have weird units I have never heard of, some use a terminal velocity, some use variable velocity, some involve Brenner’s Law due to the influence of the boundaries of the tube on the sphere, and some from a few university lecture notes go into further complexities!! Is there a simple formula that my measurements can easily 'slot' in to that someone can recommend, which will give me results which are not necessarily 100% spot on, but just a simple measure of viscosity that is near the value. Thanks!!!!!
Try this: http://www.engr.uky.edu/~egr101/ml/ML3.pdf Note, there are two different measures of viscosity, called kinematic viscosity and dynamic viscosity. See http://www.engineeringtoolbox.com/dynamic-absolute-kinematic-viscosity-d_412.html for how they are related and their units. The SI units for viscosity don't have any special names, but the CGS units are named as poise (kinematic) and stokes (dynamics). The commonly used "practical" units are centipoise and centistokes (= 1/100 or a poise and a stoke). FWIW the viscosity of water at room temperature is about 1 centipoise, and also 1 centistokes.
Hi I only know one formula for Stokes' Law V = (2/9) * ( [tex]\rho[/tex](Sphere) - [tex]\rho[/tex](Fluid) ) *(g/m) * R^{2} where m is the viscosity of the fluid This formula was derived using force equilibrium on the falling sphere Weight = Buoyancy Force + Drag Force and the units here are SI check this out : http://en.wikipedia.org/wiki/Stokes'_law I guess to ignore the boundary effect you should use a relatively big tank(or whatever you are keeping the fluid),I mean use something that its dimensions are bigger than the sphere in order to neglect the boundary effect and to use the previous formula Hope that helps :)
Hi everyone, thanks for your help!! Using this document (http://www.engr.uky.edu/~egr101/ml/ML3.pdf), on page 4 Fb+ Fd=mg where Fd= 6πμVd and where Fb= 4/3πr^3 × ρ ×g What are the units I need to use in the formula, as I have seen some densities with g/cm^3, kg/m^3, kg/cm^3 etc. Also, comparing formulas I have seen some radius measurements subbed into different formulas in m, cm, mm. Again, different variations on the theory use the volume in m^3, cm^3, and also one document I read had g as not as 9.8, but 9800 or 980 (I can't remember) in some other form. I know how to convert the units, it is just a case of which unit do I need to use in the above formulas to work out absolute (or dynamic) viscosity? This formula is meant give it to me in Pascal second, am I right? many thanks!
Why do you wanna make it hard for yourself?? As I told you,use the formula I gave you (which is also in that Wikipedia page) which give you the Dynamic Viscosity In the SI system the dynamic viscosity units are N s/m2, Pa s or kg/m.s and for the dimensions: Density : kg/m3 Gravity : m/s2 Velocity : m/s Viscosity : kg/m.s
and sorry for not paying attention to your post, I read the first one and scrolled to reply, not seeing it! apologies!
yeah I think so, the wiki formula is just a rearrangement of frictional force combined with the buoyant force balancing the gravitational force. To directly find the viscosity, I can just rearrange the wiki formula, pretty much just swapping the velocity and viscosity's 'position' in the equation. The only thing ... where does the 'difference in density' bit come from in the wiki formula? When this is rearranged ... Fb+ Fd=mg where Fd= 6πμVd and where Fb= 4/3πr^3 × ρ ×g ... have they calculated the 'm' of 'mg' using volume and density, and when that all gets rearranged we get ( density(Sphere) - density(Fluid) ) somehow? thanks
That's correct as Fb = (4/3)[tex]\Pi[/tex]*R^{3}*[tex]\rho[/tex](fluid)*g by rearranging the equation Fb + Fd = mg , you will get the difference in density.
Hi, I have completed the experiment, yet now trying to work out a decent viscosity is proving a nightmare, and using my calculator and also http://goo.gl/dt71T, my results for viscosity are always way off. Is there any chance anyone could look over these figures and help me work out which part of my experiment has gone so horribly wrong?! Ball, diameter 2cm, mass 30g, density roughly 7000kg/m^3 Time to fall (in seconds) 0.8m (80cm) in a tube of the following liquids (time is an average of 10) Water 1.293 Veg Oil 2.617 Olive Oil 2.792 Motor Oil 4.99 Glycerol 20.77 Syrup 794.85 Corresponding Fluid density 998 915 921 1250 888 1495 The tube diameter is 2.5cm, so is this just too thin compared to the ball for a reliable measure to be calculated? Thanks! :D
I agree the results look like nonsense. I think you have two big problems with the experiment. Stokes Law gives the force on a sphere in an infinite volume of fluid, at low velocities. In your case 1. The ball is blocking up 64% of the area of the tube, so the flow pattern will be very different from the assumptions. if you just about the relative areas (64% ball and 36% water) , the water must be flowing upwards past the ball 64/36 = 1.8 times faster than the ball is moving downwards. Stokes assumes that upwards water velocity is zero. In my PDF link there is an approximate correction formula for this effect mu_c = mu[1 - 2.104 d/D + 2.09 (d/D)^3 - 0.95 (d/D)^5] But that is only accurate when d/D is fairly small. In your case d/D = 2/2.5 = 0.8 and the some of the "correction terms" are bigger than the measured value. mu_c = mu[1 - 1.68 + 1.07 - .31] = 0.08 mu The answer of 0.08 is probably nonsense, except that it shows assumptions behind the formulas do not apply. 2. The other problem is your velocity is much too high, at least for water. In fluid dyamics there is an important non-dimensional number called Reynolds number. Its physical meaning is the the ratio of the inertia forces (mass x acceleration) to the viscous forces on an object. For a sphere it is Re = rho V d / mu Using the correct value of mu for water (about 0.01 in CGS units) and your measured values, Re = 1 x 60 x 2 / 0.01 = 12000 Re > 3000 means the fluid flow will be fully turbulent. Unfortunately Stokes Law only applies for completely laminar flow, when Re < 1. So using a bigger tube won't fix the problem, because you need to reduce the speed by a lot. The way to do that is use a much smaller ball, preferably with a lower density as well. Try a 1mm diameter ball bearing. A small plastic sphere might work even better.
Now you are thinking like a mechanical engineer and not a fluid dynamicist. For pipe flow, the flow is laminar until [itex]$Re_{D} \approx 2300$[/itex] and is fully turbulent typically by [itex]$Re_{D} \approx 4000$[/itex]. That only holds for pipe flow. For external flows, that number is orders of magnitude higher (e.g. transition doesn't occur on an airfoil until well into the millions). Actually, this flow will be laminar. Stoke's Law applies just fine. The bigger issue is that the OP is not measuring terminal velocity, he/she is measuring an average velocity from top to bottom. You have to start your time after the ball has reached its terminal velocity or you are measuring the wrong thing. That should happen fairly quickly though. Edit: Sorry if my Reynolds numbers are showing up funny. The LaTeX interpreter is acting goofy.
Actually I've spent (or mis-spent) quite a lot of my life worrying about external flows round aircraft nacelles and internal flows in engine oil systems. I can successfully identify which is which at least 9 times out of 10. You are right that Stokes Law applies to external laminar flow, but you are wrong about the OP's actual experiment, which was done on a pipe with a stonking great flow restrictor and turbulence generator inside it (namely, the ball bearing). Airfoils are not usually spherical either, so the transition Re for a typical airfoil doesn't apply. Look up a graph of the drag coefficient of a sphere against Re, if you don't believe me.
His experiment is in a "pipe" but the fluid is not moving with respect to that pipe (except in the annulus between the pipe and the sphere). On top of that, the Reynolds number of importance in a pipe flow is that which is based on the pipe diameter. The correlations would be entirely different for Reynolds number based on the ball diameter. The flow will be separated (for most of these cases), but not turbulent. I realize airfoils are not spherical, and I don't claim that transition Reynolds number would be the same for a sphere. My claim is that an external flow is a fundamentally different phenomenon in many respects, including with respect to transition to turbulence. I have a very hard time believing that, in this situation, the necessary instabilities would develop and grow fast enough to be turbulent before the sphere has passed. In order for the flow to transition, there needs to be such an instability. It won't be an inviscid instability because there is (as far as I can see) no inflection point in the boundary layer. It won't be a Görtler instability, and in fact, the convex surface would actually stabilize the system. The only remaining 2-D, viscous instability is the Tollmein-Schlicting instability, and those grow very slowly.
I suggest you stop theorizing, get a piece of pipe and a ball bearing, put something in the fluid to visualize the flow pattern, and do the experiment.
I think I find AlephZero correct @carnocs3m5 : I Suggested you before to use something relatively bigger in dimensions than the sphere 2.5 cm & 2 cm are pretty much close don't you think ?? may be you can use a tub(for water of course) and for the other fluids you can use a smaller sphere. Plus are you sure you are measuring the terminal velocity correctly ? Do the experiment again and show us the results
I will be honest and say a lot of he previous discussion (argument :D) was fairly confusing to me, although I get he general gist of it! So to confirm, the fact that the ball is blocking up 64% of the area of the tube is quite important, and I also understand the terminal velocity factor. Although the fall of the ball in the (see through) pipe was 0.8m, I left space between the ball entering he fluid and my start line for a (nearly) terminal velocity to be met. I suppose this was not enough space, and my terminal velocity is the thinner fluids was wrong. The golden syrup one was a bit different, in the sense that terminal velocity was definitely met as it had so bloody long to reach it. Hopefully I will have time to do another experiment of some sorts this week, not sure how it will go though! thanks :D
Well if you had yor start line offset then you definitely did a pretty good job taking account of the initial acceleration. It may be a good idea to offset your finish point since the presence of the bottom of the tube will affect the motion as well but that isn't going to be as important. The bottom line is that blockage is significant. It will likely offset your results by a certain factor, and while the relationship between all the fluids' viscosities will stay the same (syrup will still be the most viscous, water the least), the values you get will be off. Turbulence won't be an issue, especially in the more viscous fluids, so that blockage will be the primary concern. I would have to say that your initial results don't look terrible. The syrup seems a bit inflated but that very well may be a blockage issue since I would expect blockage to be more important for the more viscous fluids.