# How to find exponential values

I have a power equation for flow between two circular disks on a common axis as stator and rotor, which follows,

$$P = \frac{n^a}{G}\tag{1}$$

Where:

• $P$ = power,
• $n$ = revolution (1200 rev/min),
• $G$ = gap between two plates (0-2.5 mm)

I have experimental data showing Power vs Gap curve at a constant revolution of 1200 rmp. How do I get the value of $a$ and how do I plot Eqn (1)?

The general procedure is to plot the data on log--log paper (old way), to plot $\log(\text{RHS})$ versus $\log(\text{LHS})$ on normal paper (alternate old way, supported on spreadsheets) or just hand the data to a robust minimization package and ask it for a power-law fit (new way).

The first two methods are equivalent and the work because

$$\begin{array} {} y &= bx^a \\ \log(y) &= \log(bx^a) \\ \log(y) &= \log(x^a) + \log(b) \\ \log(y) &= a \log(x) + \log(b) \end{array}$$

shows that the curve is transformed into a line---which is easy to fit with a ruler or a spreadsheet---where the power is the slope and the intercept is the logarithm of the coefficient.

By the way, the word "exponential" is generally reserved for relations like $y = \exp(x)$, which is managed in the old way by plotting on semi-log paper. You want "power-law".

Assuming $P=n^a/G~$: Then $a=\frac{\log (PG)}{\log n}$. Now take the mean or the median of the right hand side.

Assuming $P=n^{a/G}~$: Then $a=\frac{G\log (P)}{\log n}$. Now take the mean or the median of the right hand side.

Better estimates might be possible if you know more about a statistical model for the errors.

Assuming I understand your question $n^a$ is a constant in your experiment because you're only varying $G$. In that case just plot $P$ on the $y$ axis and $1/G$ on the $x$ axis, and the gradient of the line will be $n^a$.

But this is a poor way to calculate $a$. Really what you want is to keep $G$ fixed and measure $P$ while varying $n$. With those results you can use the method dmckee describes to calculate $a$, and you'll get a much more accurate result.