The activation can be determined from either the gradient of tf vs 1/T, or, the gradient of emin vs. 1/T (strain rate vs 1/T). See attached first couple of paragraphs.

Dear Veronica Gray

Thank you for your answer and the detailed reference attachment.

I have come across certain calculations in which strain rate vs 1/T have been plotted and Boltzmann constant k is used in the place of Gas constant R. By computing the values i found contradictions in the activation energy values. I have attached a sample plot in which i am experiencing this issue (Please see to the Plot to find the apparent activation energy at constant stress) the result activation energy values does not match with the expected standard values.

Dear Modassir Akhtar

Thank you for your links. I have been referring these links some weeks back.

for more detailed discussion, of you see my reply to Dr. Veronica Gray, i have attached a plot of ln strain rate vs 1/T. It is familiar that from this plot, the gradient will give the value of -Q/R or -Q/k where R and k is Gas constant and Boltzmann constant respectively. Do u have any idea of calculating the activation energies from the above plot and it is with respect to R or k?

The gradient of the plots shown is -Q/R with R=8.314 J/mol. Take the gradient OF Ln(emin)) vs 1/T and multiply it by R (you are now in J/mol), but then divide it by 1000 to get to kJ/mol. You should expect a spread of values as creep inherently has a lot of scatter. For you its about 130-180 kJ/mol (I think from a quick look) with activation energy increasing with stress (see the paper I attached before about my thoughts on this). A different approach is to use the apparent activation energy which is the same process but rather than evaluating at constant stress you use a constant normalised stress (e.g. test stress/UTS). This has produced less variation for us and more realistic activation energy numbers

Dear Veronica Gray

Again Thanking your for your detailed answer.

From your answer, the value u got finally which is about 130-180 KJ/mol is the expected range of activation energy. But from the above graph i am unable to get the same value what u have got. I am going wrong some where wrong in the calculation. Also in the above sample plot, the author have used Boltzmann constant (k) in place of R. Therefore, gradient gives -Q/k and when calculating with k, also the expected final value is not got.

From your above answer i am unable to get what is 'Ln(emin)),

A sample of calculation how i have proceeded from the above plot for line at constant stress 5.9

the y axis scale is already in log scale

Gradient=change in y/change in x

= (1E-13)-(1e-07)/(1.5*1e-03)-(1.05*1e-03),

= -0.00022

Now by multiplying by R

= -0.00022*8.314

-Q = - 1.82 E-03

I am missing some calculation in finding the activation energy.

Please help me to fix it.

Thank you!

I lifted the points using a plot digitizer so they are roughly right.

Look at the Excel file.

Dear Veronica Gray

Thanks a lot much for your explanation.!

I am now clearly able to calculate and understand the activation energy calculation.

But finally i am confused in the usage of R or k. The case which i was given was to use the Boltzmann constant. But the required value is only obtained by using R. A little of confusion still exists in this constant area.

Thanks a lot once again.!