Viscosity Temperature Relationship For Calibration Fluid 1

[rocketscientist1]

The problem goes down to 0.30cST. Using the equations below to find out coeficients A and B, Z always turns out to be 1, and then the viscosity is always the same.

What could be causing this?

log10 (log10 Z) = A - B log10 T(K)
where Z is given by

Z = (v + 0.7 + C − D + E) 2E7 to 0.30 cSt

The additional constants are:
C = exp (−1.14883 − 2.65868v),
D = exp (−0.0038138 − 12.5645v),
E = exp (5.46491 − 37.6289v),

The two known viscosities and temperatures used to find out coeficients A and B, I read from the kinematic viscosity versus temperature graph, where the kinematic viscosity axis is logarithmic.

 

[cloa]

Are these constants experimentally derived or something from a book or paper- which one?

 

[rocketscientist1]

As per ASTM D 341 - 93 Standard

 

[electricpete]

Given v1, v2 at temperatures T1, T2
Compute
Z1 = (v1 + 0.7 + C((v1)) − D((v1)) + E((v1))) 2E7
Z2 = (v1 + 0.7 + C((v2)) − D((v2)) + E((v2))) 2E7

Now we have two equations from which to solve A and B:
Equation 1: log10 (log10 Z1) = A - B log10 T1)
Equation 2: log10 (log10 Z2) = A - B log10 (T2)

Are you saying Z1=Z2=1?
Left hand side = Log(Log(1)) = Log(0) -infinity
Hmmm.
Can you post the values of (v1, T1), (v2,T2)?

Seems like bad math or bad data or perhaps working in a region of the relationship that is very sensitive to small errors.

Just thinking out loud, 0.3 cSt is the limit of the equation you cited. How about using the next relationship which is supposed to be good down to 0.24cST:
Z = (v + 0.7 + C − D + E − F + G) *2E7 7cSt


=====================================
(2B)+(2B)' ?

 

[electricpete]

Left out an equals sign. Correction in bold:

Left hand side = Log(Log(1)) = Log(0)= -infinity

=====================================
(2B)+(2B)' ?

 

[electricpete]

F = exp (13.0458 − 74.6851v)
G = exp (37.4619 − 192.643v)

=====================================
(2B)+(2B)' ?

 

[rocketscientist1]

electricpete,

The two known viscosities and temperatures I used are ( 1 cSt, 100 deg F) and ( 0.7 cSt, 160 deg F). I converted the temperature from deg F to Kelvin and found out constants A and B. However, when I use A and B to calculate Z and obtain viscosity for different temperatures, Z always turns out to be 1 and so the viscosity always turns out to be the same for any temperature.

 

[rocketscientist1]

Here is the code I am using:

v=[1;0.7];
T=[(100+459.67);(160+459.67)];
for i=1:2
C=exp(-1.14883-2.65868*v(i));
D=exp(-0.0038138-12.5645*v(i));
E=exp(5.46491-37.6289*v(i));
Z(i)=v(i)+0.7+C-D+E;
end
A=[1 -log10(T(1)); 1 -log10(T(2))];
b=[log10(log10(Z(1))); log10(log10(Z(2)))];
x=b\A;
temp=input('Insert Temperature in deg F ')
z=10^(10^(x(1)-(x(2)*log10(temp+459.67))))
visc=(z-0.7)-exp(-0.7487-(3.295*(z-0.7))+(0.6119*(z-0.7)^2)-(0.3193*(z-0.7)^3))

 

[electricpete]

T=[(100+459.67);(160+459.67)];

Doesn't look like Kelvin to me.


=====================================
(2B)+(2B)' ?

 

[rocketscientist1]

How about now? I still get z=1 and visc=0.1156 for any input temperature ...
v=[1;0.7];
T=[((100+459.67)*5/9);((160+459.67)*5/9)];
for i=1:2
C=exp(-1.14883-2.65868*v(i));
D=exp(-0.0038138-12.5645*v(i));
E=exp(5.46491-37.6289*v(i));
Z(i)=v(i)+0.7+C-D+E;
end
A=[1 -log10(T(1)); 1 -log10(T(2))];
b=[log10(log10(Z(1))); log10(log10(Z(2)))];
x=b\A;
temp=input('Insert Temperature in deg F ')
z=10^(10^(x(1)-x(2)*log10((temp+459.67)*5/9)))
visc=(z-0.7)-exp(-0.7487-(3.295*(z-0.7))+(0.6119*(z-0.7)^2)-(0.3193*(z-0.7)^3))

 

[electricpete]

I don't have access to Matlab at the moment. Just some questions:

A=[1 -log10(T(1)); 1 -log10(T(2))];

Does it recognize A as a 2x2 matrix... or is it creating a 2x1 matrix?

x=b\A;

should it be: x=A\b ?

What are the numerical value of the results (coefficients A and B) stored in x?
(if they don't satisfy each of the original equations, obviously we made a mistake along the way)

=====================================
(2B)+(2B)' ?

 

[rocketscientist1]

electricpete,

Thank you! It was a math error(x=A\b not b\A). Thank you!