question on viscosity

Viscosity and Temperature
Simply put, at colder temperatures, a few degrees is a bigger change percentage wise than at higher temperatures. If you look at the equation for viscosity, you can see it as well.

viscosity = Ae^(k/T) where A and k are material specific constants and T is the absolute temperature.

can u explain the equation and relate it to something like honey for example?

Example
Sure, but I don't have the constants for honey, so I'll just make something up. You can still see how the equation changes. Please note that the equation I gave you is an approximation from empirical data. It is called the Andrade equation.

Let's make k=1000degR and A=1. At the freezing point of water (460degR or 0degF), the question works out to:

vis=1*e^(1000/460) = 8.8

Note that because the exponential term is greater than one, you get fast changes in value. If you plot e^x, you get a slow rise until you get to one, then a steeper rise. Now let's go to 40degF = 500degR.

vis=1*e^(1000/500) = 7.4

The viscosity drops by 1.4 Another 40degF makes the equation look like

vis=1*e^(1000/540) = 6.4

Only a drop of 1. Still, on a percentage basis, 1 is a big drop. The bigger the value of k, the bigger the effect with temperature since the exponent term is bigger. If the temperature is much higher than k, viscosity changes with temperature are very small. Fluids you think of as being viscous have high values of k and change rapidly at room temperature. Fluids like water have low values of k and don't change much

hmmm, how would u find a, e and k ? and what are they

Experimental determination
e is just the base of the natural log system. You can determine a and k experimentally using some fairly simple tests like measuring the time it takes to pour out a liquid through a certain sized funnel or measuring the time it takes for a ball to fall through the liquid. Do the test at three temperatures and perform a regression. You can find the exact techniques on the internet easily enough.

ive got the viscosities of honey from a viscometer at numerous temps, how would i go and calculate a and k from the results

Linear regression
Start with:

viscosity = A e^(k/T) Remember T in in absolute temperature units

Take the natural log ...

ln(viscosity) = ln A + k/T

Make two new variables with your data, y=ln(viscosity) and x=1/T. Now you have a set of x's and y's and the equation is ...

y = kx + ln A

Now do a simple linear regression to get a best fit solution for k and (ln A). If this isn't clear enough, post your data points and we'll work through it together.

im very sorry, im a grade 9 student, so im not too good with all of this stuff, so you need the viscosities? at 35 it was 5540 cP, at 40 it was 3570 cP, at 45 it was 2690 cP and at 50 it was 2400 cP (by the way those temps are centigrade)

Solution
Your data plus some calculated fields.
TdecC Vis TdegK 1/T ln(vis)
35 5540 308 0.003246753 8.61974978
40 3570 313 0.003194888 8.180320875
45 2690 318 0.003144654 7.897296473
50 2400 323 0.003095975 7.783224016

Put this into your favorite spreadsheet program or calculator and do a linear regression on the last two columns. (If you want a derivation of how to a least squares fit, start another thread.) You get viscosity = ln ( 5576 / T - 9.56). Note that when you plot it, it doesn't look completely straight. That is a function of your data which has some slop in it. Just about any method of measuring viscosity has a fair amount of error in it.

thanks, for all your help, i understand the stuff now