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...
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)))]...
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...
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...
