The Calculation of Viscosity Blending Indices Using the Refutas Equation 1

Did you refer to the ASTM D341 standard?
It shows the graphical method for determining blend ratios but it also cites calculation methods....the standard might let you determine which answer agrees with the ASTM D341 since you get about a 2:1 error in VBI depending on if you use ln or Log.
That then brings up another question which i haven't had satisfactorily answered which is if the temperature viscosity relationship shown in ASTM D341 is Log or Ln?

JMW

Stachowiak and Batchelor, in their Engineering Tribology, Elsevier Tribology Series, 24, say the ASTM chart's ordinate is in log[sub]10[/sub]log[sub]10[/sub](cS+0.6).

I found the answer to my question. The Refutas equation for calculating Viscosity Blending Numbers (VBN) in my original posting definitely uses the natural logarithm Log[sub]e[/sub] (often but not always denoted as ln).

I found a number of tabulated data on the viscosity (in centistokes) at a given temperature and the correspomding VBN at the same temperature for a good many different refinery streams in this patent: which was assigned to Exxon Research and Engineering.

I then calculated the VBN values using log[sub]e[/sub] in the Refutas equation and obtained exactly the same values as tabulated in the patent.

It is interesting that the patent presents the Refutas equation (their equation 13) as VBI = 14.534 × log[log(v + 0.8)] + 10.975 with no explanation of whether base 10 or base e applies. Since my calculations show that the tabulated values of VBN were obtained using log[sub]e[/sub], I can only conclude that some people use log to mean the natural logarithm. That is what led to my confusion in the first place.

JMW, thanks for your suggestion that I try calculating VBN's (for published values of viscosity and VBN) using log[sub]e[/sub] to find if that is the correct basis. It certainly helped me clear up my confusion.


Milton Beychok
(Visit me at www.air-dispersion.com)
.

I may be wrong, but it seems both formul[æ] give equivalent results.

That's what someone suggested to me about the ASTM D341 equation but the VBI is about a 2:1 difference between the two versions (if I didn't make a mistake in Excel).

As an outcome of another thread (on blending) someone sent me a copy of their spreadsheet based on the Louis and refutas method.

JMW

To jmw, either one of us is mistaken. I've done several exercises and found equal blend viscosity values using both equations. Kindly check and let us know of your findings.

Hm.
Using excel for the VBI calcs (VBI = 14.534 × ln[ln(v + 0.8)] + 10.975

VBI = 14.534 × log[log(v + 0.8)] + 10.975) I get:
Visc VBI (ln) VBI (log 10) v+0.8 ln lnln lnln*14.534 l10 loglog loglogx 14.534
1 3.25 2.32 1.80 0.59 -0.53 -7.72 0.26 -0.59 -8.62
10 23.57 11.14 10.80 2.38 0.87 12.60 1.03 0.01 0.21
100 33.20 15.32 100.80 4.61 1.53 22.22 2.00 0.30 4.39
1000 39.07 17.87 1000.80 6.91 1.93 28.09 3.00 0.48 6.94
10000 43.25 19.69 10000.80 9.21 2.22 32.27 4.00 0.60 8.75
But if you refer to the ASTM D341 equation of log.log(v+0.7) = A-B.logT where v is kinematic viscosity at T degrees kelvin, I haven't done the calculations in both. n the spreadsheet i often link to I use log base 10 and it seems to correspond with the real world very well.

JMW

PS sorry about the columns not lining up properly, I tried adding spaces but can't quite manage it. That'll teach me to paste straight from excel.

JMW

I made a quick check on your VBI values and found the VBI[sub]10[/sub] lower by 0.04 in all cases.

Now when making blends, say 1:1, I find the same blend viscosities using both VBI's.

Please check my assumptions and correct me if I'm wrong.

25362:

Using VBI = 14.534 × ln[ln(v + 0.8)] + 10.975

v = 1 cSt
VBI = 14.534 × ln[ln(1.8) + 10.975 = 14.534 × (-0.53) + 10.975 = 3.25

v = 1000 cSt
VBI = 14.534 × ln[ln(1000.8) + 10.975 = 14.534 × (1.93) + 10.975 = 39.07

Using VBI = 14.534 × log[log(v + 0.8)] + 10.975

v = 1 cSt
VBI = 14.534 × log[log(1.8) + 10.975 = 14.534 × (-0.59) + 10.975 = 2.36

v = 1000 cSt
VBI = 14.534 × log[log(1000.8) + 10.975 = 14.534 × (0.48) + 10.975 = 17.91

Thus, for v = 1 cSt, the difference is 3.25 versus 2.36 and for v = 1000 cSt, the difference is 39.07 versus 17.91. These are almost identical with jmw's spreadsheet.

All done usung my HP32SII scientific calculator.

Milton Beychok
(Visit me at www.air-dispersion.com)
.

To mbeychok, of course the VBI's should differ by simple mathematical logic, not so the kinematic viscosities of the blend. Try a 50:50 blend of 1 cS and 1000 cS using either equation, you'll find the same answer with the ln as with the log[sub]10[/sub].

25362:

My original posting was only concerened with the determination of VBI values using the posted equation. In your 10:43 AM response, you said:

I made a quick check on your VBI values and found the VBI[sub]10[/sub] lower by 0.04 in all cases.

Click to expand...

As both jmw and I have shown, the VBI values differ by a great deal more than that.

Now, in your 19:13 PM response, you said:

...of course the VBI's should differ by simple mathematical logic ...

Click to expand...

and you went on to change the subject and are discussing the viscosity of blends (rather than the determination of VBI values) which somewhat confuses the subject that I raised in my original post.

How about giving us the complete, step-by-step details of your calculations and please spell out whether you are talking of a 50:50 blend by volume or by weight?

Milton Beychok
(Visit me at www.air-dispersion.com)
.

To mbeychok. Your first question, as I understood it, referred to which of both formulas was the correct one. Until proven wrong, my answer would be: both provide the same result for viscosities of blends, which is, in fact, their purpose.

As a quick example let’s assume a 1:1 blend of oils having 1 cst and 1000 cSt using the blending indices posted by you:


Parameter ln basis log10 basis

VBI, 1 cSt 3.25 2.36
VBI, 1000 cSt 39.07 17.91
1:1 blend VBI 21.16 10.135
Deduct 10.975 10.185 -0.84
Divide by 14.534 0.7008 -0.0578
First power 2.015 0.8753
Second power 7.50 7.50
Deduct 0.8 cSt 6.7 cSt 6.7 cSt


The result using the formulas may differ from the kin visc. of an actual blend, but it appears that both give the same results.

As for whether on a weight or volume basis, I prefer to use Maxwell’s procedure in his Data Book on Hydrocarbons which is on a volume basis.

25362:

I apologise, you are correct. It took me a while to get it through my thick head, but I finally came up with the invert of the VBI equation in my original post, which is

v = e[sup]e[sup](VBI-10.975)/14.535[/sup][/sup] - 0.8 .... where e = 2.17183

which makes it perfectly clear that using log[sub]e[/sub] in the VBI equation and e in the above equation is the same as using log[sub]10[/sub] in the VBI equation and using 10 (rather than e) in the above equation.



Milton Beychok
(Visit me at www.air-dispersion.com)
.