0

I have an equation of the following form

$$\frac{2L}{v} = A\left(\frac{\nu}{\epsilon}\right)^{0.5}Re^y$$

I would like to plot this in order to obtain the constant "A" and the exponent "y". $L, v, \nu, \epsilon, Re$ are all variables. I tried plotting $\frac{2L}{v}$ versus $\left(\frac{\nu}{\epsilon}\right)^{0.5}Re$, but this gave a bad fit (R2 = 0.85). I sort of already know what the exponent should be and based on that number, the fit is really good R2> 0.97. Although I have good values for A and y now, I'm looking to improve on them which is why I thought I would try to plot the parameters I've already mentioned. I'm not sure what I'm doing wrong. Even a response that says I am doing this the correct way would be appreciated.

rdemyan
  • 97

1 Answers1

0

Try plotting $\ln \left( \frac{2 L}{v} \right)$ against $\ln {\rm Re}$. This is because

$$\ln \left( \frac{2 L}{v} \right) = \ln \left( A \left( \frac{\nu}{\varepsilon} \right)^{0.5} {\rm Re}^y \right) = \ln A + 0.5 \ln \left( \frac{\nu}{\varepsilon} \right) + y \ln {\rm Re}$$

If the model is appropriate, you should see a near-linear relationship, where the slope of the graph is $y$ and the vertical intercept is $\ln A + 0.5 \ln \left( \frac{\nu}{\varepsilon} \right)$.

K. Jiang
  • 7,210
  • I plotted $\ln \left( \frac{2 L}{v} \right) - 0.5 \ln \left( \frac{\nu}{\epsilon} \right)$ against ln Re and obtained coefficients for A any that seemed to be good. However, when I used these coefficients, the R2 value of a linear fit through the data is 0.9458 which is lower than the R2 fit (0.947) for the coefficients I mentioned earlier that I thought were pretty good but maybe not the best. I realize now, however, that I am not asking the right question. I need to repost. – rdemyan Feb 20 '23 at 21:31