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I have the following two equations:

\begin{align} \frac{\lambda}{u_L} = 3.2693\left(\frac{\nu}{\epsilon}\right)^{1/2 }\end{align}

\begin{align} u_L = \left(\epsilon \lambda\right)^{1/3}\end{align}

I have tried to solve for $\lambda$ as follows:

\begin{align} \frac{\lambda}{\left(\epsilon \lambda\right)^{1/3}} = 3.2693\left(\frac{\nu}{\epsilon}\right)^{1/2}\end{align}

combining variables on the LHS: \begin{align} \frac{{\lambda}^{2/3}}{\epsilon^{1/3}} = 3.2693\left(\frac{\nu}{\epsilon}\right)^{1/2}\end{align}

Multiplying both sides by ${\epsilon}^{1/3}$

\begin{align} {\lambda}^{2/3} = 3.2693\left(\frac{\nu^3}{\epsilon^3}\right)^{1/6}\left({\epsilon^2}\right)^{1/6}\end{align}

This simplifies to: \begin{align} {\lambda}^{2/3} = 3.2693\left(\frac{\nu^3}{\epsilon}\right)^{1/6}\end{align}

Raising both sides to the 3/2 power yields: \begin{align} {\lambda} = 3.2693^{3/2}\left(\frac{\nu^3}{\epsilon}\right)^{3/12}\end{align}

Finally, \begin{align} {\lambda} = 5.911\left(\frac{\nu^3}{\epsilon}\right)^{1/4}\end{align}

However, when I use this equation to calculate the value of $\lambda$ from the data, I'm not getting values equal to an alternative equation for calculating $\lambda$, which alternative equation I am certain is correct.

Does anyone see an error; I can't seem to find one.

rdemyan
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  • I can't see any errors in your work. Does the alternative equation use the same variables? I think it's a bit difficult to answer your question beyond a "that looks OK" sort of thing, and it's a bit ridiculous posting that in the answer box. – Suzu Hirose Jun 13 '22 at 23:40
  • Well.... what's the alternative equation for calculating $\lambda$, what did you get with that equation, and why do you think the results are different? – fleablood Jun 13 '22 at 23:41
  • Take a simple case, for example $\nu=\epsilon=1$ and see which formula is right. – dxiv Jun 13 '22 at 23:45
  • I misspoke. The alternative equation is actually the first equation in my post. Let me explain. The first equation is a curve fit to data. The data provided is $\epsilon$, $\nu$ and $\lambda$. $u_L$ is then calculated from equation 2. Then $\frac{\lambda}{u_L}$ is plotted versus $\left(\frac{\nu}{\epsilon}\right)^{1/2 }$ yielding the curve fit equation (equation 1). The R-squared coefficient for equation 1 at 0.984 is quite high and the fit is visually very good. – rdemyan Jun 14 '22 at 14:19
  • So, I thought I could replace uL in equation 1 with equation 2. The result is the derivation shown in my original post. However, as a check I plotted λ versus $\left(\frac{\nu^3}{\epsilon}\right)^{1/4}$ and the fit was visually relatively poor. The coefficient from this curve fit is 6.2063 instead of 5.911 and the R-squared coefficient is only 0.7515. So, I thought maybe I had an error in the derivation of the last eqn. in my post. I'm not sure I understand what the issue is, but I guess it has something to do with the fact that the original curve fit used a calculated value for $u_L$?? – rdemyan Jun 14 '22 at 14:37

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