×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Viscosity Temperature Relationship For Calibration Fluid

Viscosity Temperature Relationship For Calibration Fluid

Viscosity Temperature Relationship For Calibration Fluid

(OP)
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.

RE: Viscosity Temperature Relationship For Calibration Fluid

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

RE: Viscosity Temperature Relationship For Calibration Fluid

(OP)
As per ASTM D 341 - 93 Standard


RE: Viscosity Temperature Relationship For Calibration Fluid

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)' ?

RE: Viscosity Temperature Relationship For Calibration Fluid

Left out an equals sign. Correction in bold:

Quote (correction)

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

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

RE: Viscosity Temperature Relationship For Calibration Fluid

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

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

RE: Viscosity Temperature Relationship For Calibration Fluid

(OP)
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.

RE: Viscosity Temperature Relationship For Calibration Fluid

(OP)
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))

RE: Viscosity Temperature Relationship For Calibration Fluid

Quote:

T=[(100+459.67);(160+459.67)];
Doesn't look like Kelvin to me.

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

RE: Viscosity Temperature Relationship For Calibration Fluid

(OP)
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))

RE: Viscosity Temperature Relationship For Calibration Fluid

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

Quote:

A=[1 -log10(T(1)); 1 -log10(T(2))];
Does it recognize A as a 2x2 matrix... or is it creating a 2x1 matrix?

Quote:

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)' ?

RE: Viscosity Temperature Relationship For Calibration Fluid

(OP)
electricpete,

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

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources