## thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

## thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

(OP)

Here we use the following equation to find the percentage of the mixture Marine Fuel (let's call mf380) as well as the Marine Gas Oil (MGO we call mgo) that when making the mixture becomes a Marine Fuel 150 cSt 50 ° C (let's call mf150), for example.

It turns out that the formula used establishes that the products are the same temperature, but by the company's internal regulatory matters calculations are as follows when we make a mixture using the following products:

Marine Fuel MF380 = 380 cSt at 50 ° C and

MgO = Marine Gas Oil, usually 3 to 5 cSt at 40 ° C.

v, mf180 = 150 cSt at 50 ° C

v, M380 = 380 cSt at 50 ° C

v, MgO = 4 cSt at 40 ° C

a, mf380 = (1 + 1 / v, mf380) ^ (v, mf380) = (1 + 1/380) ^ (380) = 2.714713748

a, MgO = (1 + 1 / v, MgO) ^ (v, MgO) = (1 + 1/4) ^ (4) = 2.44140625

a, mf150 = (1 + 1 / v, mf150) ^ (v, mf150) = (1 + 1/150) ^ (150) = 2.709275911

b = ln the mf380, mf380 = ln = 2.714713748 0.996681408

b = ln the MgO, MgO = ln = 2.44140625 0.892574205

b = ln the mf150, mf150 = ln = 2.709275911 0.996681408

The formula employed, who do not know the name, is as follows:

A = 3.82227;

B = 79.20991;

w, w + mf380, MgO = 1

w, MgO = 1 - w, mf380

IMI = A * ln (ln (vi + ai - bi)) + B

IM m150 = 3.82227 * ln (ln (150 + a, mf150 - b mf150)) + 79.20991

IM m150 = 3.82227 * ln (ln (150 + 2.709275911-.996681408)) + 79.20991

IM, m150 = 85.37838814

IM M380 = 3.82227 * ln (ln (380 + 2.714713748 to 0.996681408)) + 79.20991

IM M380 = 3.82227 * ln (ln (380 + 2.714713748 to 0.996681408)) + 79.20991

IM, M380 = 86.02309508

IM MgO = 3.82227 * ln (ln (4 + 2.44140625 to .892574205)) + 79.20991

IM MgO = 3.82227 * ln (ln (4 + 2.44140625 to .892574205)) + 79.20991

IM MgO = 81.26854298

IM mf150 = w * IM mf380, mf380 + (1 - w mf380) * IM MgO

85.37838814 = w, mf380 + 86.02309508 * (1 - w, mf380) * 81.26854298

w = 0.8644021716 mf380

w, MgO = 0.355978284

Thus, the percentage of MF380 is 0.8644021716.

The MgO is 0.355978284.

The great problem, as seen, it is not always feasible to make conversion of MGO 4 cSt at 40 ° C to 50 ° C.

So, is there any formula that does not depend on the products are the same temperature? Since it is not feasible to further analysis to convert the MGO 4 cSt at 40 ° C to 50 ° C

It turns out that the formula used establishes that the products are the same temperature, but by the company's internal regulatory matters calculations are as follows when we make a mixture using the following products:

Marine Fuel MF380 = 380 cSt at 50 ° C and

MgO = Marine Gas Oil, usually 3 to 5 cSt at 40 ° C.

v, mf180 = 150 cSt at 50 ° C

v, M380 = 380 cSt at 50 ° C

v, MgO = 4 cSt at 40 ° C

a, mf380 = (1 + 1 / v, mf380) ^ (v, mf380) = (1 + 1/380) ^ (380) = 2.714713748

a, MgO = (1 + 1 / v, MgO) ^ (v, MgO) = (1 + 1/4) ^ (4) = 2.44140625

a, mf150 = (1 + 1 / v, mf150) ^ (v, mf150) = (1 + 1/150) ^ (150) = 2.709275911

b = ln the mf380, mf380 = ln = 2.714713748 0.996681408

b = ln the MgO, MgO = ln = 2.44140625 0.892574205

b = ln the mf150, mf150 = ln = 2.709275911 0.996681408

The formula employed, who do not know the name, is as follows:

A = 3.82227;

B = 79.20991;

w, w + mf380, MgO = 1

w, MgO = 1 - w, mf380

IMI = A * ln (ln (vi + ai - bi)) + B

IM m150 = 3.82227 * ln (ln (150 + a, mf150 - b mf150)) + 79.20991

IM m150 = 3.82227 * ln (ln (150 + 2.709275911-.996681408)) + 79.20991

IM, m150 = 85.37838814

IM M380 = 3.82227 * ln (ln (380 + 2.714713748 to 0.996681408)) + 79.20991

IM M380 = 3.82227 * ln (ln (380 + 2.714713748 to 0.996681408)) + 79.20991

IM, M380 = 86.02309508

IM MgO = 3.82227 * ln (ln (4 + 2.44140625 to .892574205)) + 79.20991

IM MgO = 3.82227 * ln (ln (4 + 2.44140625 to .892574205)) + 79.20991

IM MgO = 81.26854298

IM mf150 = w * IM mf380, mf380 + (1 - w mf380) * IM MgO

85.37838814 = w, mf380 + 86.02309508 * (1 - w, mf380) * 81.26854298

w = 0.8644021716 mf380

w, MgO = 0.355978284

Thus, the percentage of MF380 is 0.8644021716.

The MgO is 0.355978284.

The great problem, as seen, it is not always feasible to make conversion of MGO 4 cSt at 40 ° C to 50 ° C.

So, is there any formula that does not depend on the products are the same temperature? Since it is not feasible to further analysis to convert the MGO 4 cSt at 40 ° C to 50 ° C

## RE: thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek

## RE: thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

(MGO we call mgo) that when making the mixture becomes a Marine Fuel 150 cSt 50 ° C (let's call mf150), for example.

It turns out that the formula used establishes that the products are the same temperature, but by the company's internal regulatory

matters calculations are as follows when we make a mixture using the following products:

Marine Fuel MF380 = 380 cSt at 50 ° C and

Mgo = Marine Gas Oil, usually 3 to 5 cSt at 40 ° C.

v,mf180 = 150 cSt at 50 °C

v,mf380 = 380 cSt at 50 °C

v,mgo = 4 cSt at 40 °C

a,mf380 = (1 + 1 / v,mf380) ^ (v,mf380) = (1 + 1/380) ^ (380) = 2.714713748

a,mgo = (1 + 1 / v,mgo) ^ (v,mgo) = (1 + 1/4) ^ (4) = 2.44140625

a,mf150 = (1 + 1 / v,mf150) ^ (v,mf150) = (1 + 1/150) ^ (150) = 2.709275911

b,mf380 = ln (a,mf380) = ln(2.714713748) = 0.996681408

b,mgo = ln (a,mgo) = ln(2.44140625) = 0.892574205

b,mf150 = ln (a,mf150) = ln(2.709275911) = 0.996681408

The formula employed, who do not know the name, is as follows:

A = 3.82227;

B = 79.20991;

w,mf380 + w,mgo = 1

w,mgo = 1 - w,mf380

IMi = A * ln (ln (vi + ai - bi)) + B

IM,mf150 = 3.82227 * ln (ln (150 + a,mf150 - b,mf150)) + 79.20991

IM,mf150 = 3.82227 * ln (ln (150 + 2.709275911-0.996681408)) + 79.20991

IM,mf150 = 85.37838814

IM,mf380 = 3.82227 * ln (ln (380 + a,mf380 - b,mf380)) + 79.20991

IM,mf380 = 3.82227 * ln (ln (380 + 2.714713748 - 0.996681408)) + 79.20991

IM,mf380 = 86.02309508

IM,mgo = 3.82227 * ln (ln (4 + 2.44140625 - 0.892574205)) + 79.20991

IM,mgo = 3.82227 * ln (ln (4 + 2.44140625 - 0.892574205)) + 79.20991

IM,mgo = 81.26854298

IM,mf150 = w,mf380 * IM,mf380 + (1 - w,mf380) * IM,mgo

85.37838814 = w,mf380 * 86.02309508 * (1 - w,mf380) * 81.26854298

w,mf380 = 0.8644021716

w,mgo = 0.355978284

Thus, the percentage of MF380 is 0.8644021716.

The MGO is 0.355978284.

The great problem, as seen, it is not always feasible to make conversion of MGO 4 cSt at 40 °C to 50 °C.

So, is there any formula that does not depend on the products are the same temperature? Since it is not feasible to further analysis

to convert the MGO 4 cSt at 40 °C to 50 °C

## RE: thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

For example, the program BCALC.EXE (BunkerCalc Shell for Windows V1.0), Module "Blending Ratios" calculates the percentage of mixture using different temperatures.

But to do so must have some algorithm that converts the temperature of products (Residual Fuel and Fuel Diluent) to the target temperature of the mixture. Certainly must have the constants A and B determined already in formula:

log10 (log10 (v + 0.7)) = A - B * log10 (T + 273.15)

v = viscosity in cSt

t = temperature in ° C

log10 = logarithm to the base 10

But as I said, that is not feasible in every mix I have to calculate a viscosity at two different temperatures ever. It would be practical, even with some loss of accuracy, using values of A and B tabulated for both Residual Fuel well as for Fuel Diluent.

I look forward, if possible, help!

Since now, thank you all!

## RE: thread483-198846 - The Calculation of Viscosity Blending Indices Using the Refutas Equation

"People will work for you with blood and sweat and tears if they work for what they believe in......" - Simon Sinek