Viscosity as a function of temperature

1. May 13, 2014

MexChemE

1. The problem statement, all variables and given/known data
Using the provided data for viscosity coefficients of C2H5OH (ethanol) as a function of temperature, find the constants A and B for this substance using the following equation.
Viscosity coefficients of ethanol:
T (K) - $\eta$ (cP)
273.15 - 1.773
293.15 - 1.2
313.15 - 0.834
333.15 - 0.592

2. Relevant equations
$$\log \eta = \frac{A}{T} + B$$

3. The attempt at a solution
So, I start by plugging two sets of data into the equation in order to solve them algebraically:
$$\log (1.773) = \frac{A}{273.15} + B$$
$$\log (1.2) = \frac{A}{293.15} + B$$
I clear B from the first equation and plug it into the second equation:
$$B=0.248 - \frac{A}{273.15}$$
$$0.079 = \frac{A}{293.15} + 0.248 - \frac{A}{273.15}$$
Solve for A:
$$-0.169= \frac{273.15A-293.15A}{80073.922}$$
$$-20A=-13532.492$$
$$A=676.624$$
Since I solved the second equation for A, I plug it into the first equation to solve for B:
$$0.248=2.477+B$$
$$B=-2.229$$
So, the viscosity coefficient as a function of temperature of ethanol is:
$$\log \eta_{{C_2} {H_5} {OH}} = \frac{676.624}{T}-2.229$$
I tested the above formula and it yields very close results to those of the chart provided above for T=273.15 and T=293.15, but as I increase T the error gets bigger. I don't know if this is normal because I didn't use all the decimals, or I made a mistake in my math. Thanks in advance, PF! And sorry for the long post.

2. May 13, 2014

SteamKing

Staff Emeritus
Well, you derived A and B using only the first two data points. Are you really surprised that these values of A and B may not be accurate for the other data points?

The fallacy in your method is that the relationship between temperature and viscosity is not linear. For a linear relationship, the slope of the line is the same no matter where your are on the line.

Trying to determine A and B, you should make a plot of your data. Can you draw a straight line or any other smooth curve which passes thru all the points, or must you try to adjust the curve so that it comes close to each point? If the latter, this is how you decide whether you need to do a regression analysis, so that you can produce the curve which has the 'best' fit (defined as the one which minimizes the error between the curve and the data points).

http://en.wikipedia.org/wiki/Regression_analysis

3. May 13, 2014

256bits

The equation is of the form
y = mx +b

from which, if the data points are plotted, and a straight line is obtained, it is trivial to determine m and b.

4. May 13, 2014

Staff: Mentor

Plot a graph of ln(η) vs 1/T using all your data. Draw your best straight line. Determine the equation for the line to find A and B. The slope will be A.

Chet

5. May 13, 2014

MexChemE

My professor told us that the graph of logarithm of viscosity versus reciprocal temperature is indeed linear.

We're not supposed to use software yet, but I did give it a try, although it didn't help much. The only software I have access to right now is Excel, and it has no way to implement a reciprocal scale for T, not that I know, so the linear regression done by Excel was of no use for me.

I started over, now using T=273.15 and T=333.15. My results were a bit more closer this time. I was told this degree of uncertainty was okay for pencil and paper calculations, is that true? Thank you everybody for your responses!

6. May 13, 2014

SteamKing

Staff Emeritus
Professors can tell you a lot of things, some of which may even be true. However, you go to school to learn how to think for yourself, and to acquire the tools and knowledge which allow you to verify whatever claims come your way.

You can analyze this data using Excel; you just haven't done it correctly. I've checked the log of the viscosity for the given data set (using both common and natural logs). It's close to a straight line, but not quite. Maybe close enough for government work, but that's about it. If you plot the data with the proper scale, you can remove all doubt.