You are Here: Home

Announcements Posted on
Uni student? You could win a Samsung Galaxy S7 or £500 of travel vouchers by taking this quick survey 05-05-2016
Talking about ISA/EMPA specifics is against our guidelines - read more here 28-04-2016
1. Hi,
i have to plot a graph from which i can derive viscosity. the best way that i can think of is to use this equation

where:
Vt = [(2g)(d1-d2)/9µ] r^2

Vt = termical velocity of fall
g = acceleration of gravity
r = "equivalent" radius of particle
dl = density of sphere
d2 = density of medium
µ = viscosity of medium

if i were to draw a graph of terminal velocity versus radius squared...
k= [(2g)(d1-d2)/9µ]
that bit that bit would be the gradient of the graph since it is a constant.
but since in my experiment i have to use different radius's of the sphere, does that still mean the the density of the different spheres will still make k a constant? how does that work? also if i plot that, will it be a straight line graph since k/ the gradient is a constant?

2. (Original post by rogla)
Hi,
i have to plot a graph from which i can derive viscosity. the best way that i can think of is to use this equation

where:
Vt = [(2g)(d1-d2)/9µ] r^2

Vt = termical velocity of fall
g = acceleration of gravity
r = "equivalent" radius of particle
dl = density of sphere
d2 = density of medium
µ = viscosity of medium

if i were to draw a graph of terminal velocity versus radius squared...
k= [(2g)(d1-d2)/9µ]
that bit that bit would be the gradient of the graph since it is a constant.
but since in my experiment i have to use different radius's of the sphere, does that still mean the the density of the different spheres will still make k a constant? how does that work? also if i plot that, will it be a straight line graph since k/ the gradient is a constant?

If you dont understand what im saying please ask, and I or someone smarter than me will explain.

k is not dependent on r. so it will stay constant. If you didnt change anything graph would just be a straight line with a gradient of zero. in short k will still be constant.

finally. yes, it should be constant. and the gradient will give you an answer for k. but i hasten to add that this is only the case if the variables are linear which they are.

please ask if that dosnt make sence. also the best thing to do in this case it to try it to see if it works.
3. This is exam coursework. You must acknowledge all such helpyou receive.
4. ok so what needs to go on the axis for this graph?
Have a bit more confidence!
6. Hopefully this will help;
Since if you plot Vt=k r^2, the gradient of the graph will give you k, then you rearrange the formula given in the brief(the long one) to make viscosity(heta) the subject, you say you know all the values so substitute all of them and calculate to see if the viscosity matches the viscosity you know, thats stokes law proven, the only thing nagging me is that I cant imagine that the graph will give me a straight line since when y= Constant(variable^2) its normally always a parabola.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: May 18, 2009
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR