## The Calculation of Viscosity Blending Indices Using the Refutas Equation

## The Calculation of Viscosity Blending Indices Using the Refutas Equation

(OP)

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The following equation for calculating the Viscosity Blending Index (VBI) of heavy oils is often referred to as the Refutas method and I have seen it published in the technical literature in both of the forms below:

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

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

Seeing the equation published in the literature in both of the above forms leaves me confused. Does the equation require the use of the log to the base 10 or the natural log to the base e? Not a single one of the publications I read clearly states whether log base 10 or log base e is required.

Can anyone tell me which log to use and provide me with a reference that clearly and definitively verifies it.

I know that there are many other charts, tables, nomograms and equations for calculating Viscosity Blending Indices ... but I am only interested in the above equation.

Milton Beychok

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

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The following equation for calculating the Viscosity Blending Index (VBI) of heavy oils is often referred to as the Refutas method and I have seen it published in the technical literature in both of the forms below:

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

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

Seeing the equation published in the literature in both of the above forms leaves me confused. Does the equation require the use of the log to the base 10 or the natural log to the base e? Not a single one of the publications I read clearly states whether log base 10 or log base e is required.

Can anyone tell me which log to use and provide me with a reference that clearly and definitively verifies it.

I know that there are many other charts, tables, nomograms and equations for calculating Viscosity Blending Indices ... but I am only interested in the above equation.

Milton Beychok

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

.

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

www.ViscoAnalyser.com

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

Stachowiak and Batchelor, in their Engineering Tribology, Elsevier Tribology Series, 24, say the ASTM chart's ordinate is in log

_{10}log_{10}(cS+0.6).## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

_{e}(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: http:/

I then calculated the VBN values using log

_{e}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

_{e}, 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

_{e}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)

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## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

I may be wrong, but it seems both formulæ give equivalent results.

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

JMW

www.ViscoAnalyser.com

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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.

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

www.ViscoAnalyser.com

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

JMW

www.ViscoAnalyser.com

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

I made a quick check on your VBI values and found the VBI

_{10}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.

## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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)

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## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

_{10}.## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

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:

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)

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## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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 Hydrocarbonswhich is on a volume basis.## RE: The Calculation of Viscosity Blending Indices Using the Refutas Equation

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

^{e(VBI-10.975)/14.535}- 0.8 .... where e = 2.17183which makes it perfectly clear that using log

_{e}in the VBI equation and e in the above equation is the same as using log_{10}in the VBI equation and using 10 (rather than e) in the above equation.Milton Beychok

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

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