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UPDATE - I update this question today(28.12.2022), sorry for being not specific earlier.

I have a bunch of solar panels, each one is connected to an inverter. For each solar panel I have two sensors, a wind and a west sensor. From every inverter, I collect the power output.

The readings are collected for every 5 minutes, and I have a dataset over 7 days.

I know that the power can be estimated using the equation: $$\rho_A=irr_s*A_e*(1+\eta_T*(T_m-25°C))$$ where,

$\rho_A$ the expected poweroutput
$irr_s$ the solare irradiation on the PV-Panel
$A_e$ the effective are of the PV-Panel (a Product of the effective effeciency of the Panel and the actual area),
$\eta_T$ the Temperaturecoefficient of the Panel and
$T_m$ the Modul temperature

Here is what I did so far: I only used the west sensor as I thought the wind sensor has little effect on the power output of a solar panel(common sense?). Then, I grouped the data by hour and then day, for example, for on the 22.12.2022 i got 24 readings, and so on.

I rearranged the equation, to get something like Ax + Bxy + C = 0, see here:

$$A_e * irr_s + A_e * \eta_T * (irr_s * T_m + irr_s * 25) - rho_A = 0$$
if I say, A = $irr_s$ and B = $irr_s * T_m + irr_s * 25$ and C = $-rho_A$,

Now I calculated the two unknowns, $A_e$ and $\eta_T$. I used matrixes to solve different pairs of equations using:

Let us assume the two pairs look like this:
$A_1x + B_1xy + C_1 = 0$ and
$A_2x + B_2xy + C_2 = 0$

M1 = $$\begin{matrix} A_1 & B_1 \\ A_2 & B_2 \\ \end{matrix} $$

and

M2 = $$\begin{matrix} C_1 \\ C_2 \\ \end{matrix} $$

ROOTS = $$\begin{matrix} R_1 \\ R_2 \\ \end{matrix} $$

Then, $ROOTS = M1^{-1}.M2$

I plotted them on a graph, see below: Click here

Do you think my approach is good? How else could one approach this problem? Thank you for your inputs in advance!

rish
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    Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. – José Carlos Santos Dec 27 '22 at 13:09
  • thanks done!... – rish Dec 27 '22 at 13:26
  • Can you try a multiple linear regression (may be with a cross term) ? – Claude Leibovici Dec 27 '22 at 13:58
  • The solution space will be a curve of $A_e$ against $\eta _T$ values (or vice versa). If you fix one of the variables you can find the unique value of the other with straightforward algebra. – Paul Dec 27 '22 at 14:03
  • I got a linear relationship between $A_e$ abd $\eta_T$, on what basis can I fix the value of one of them? – rish Dec 27 '22 at 14:06
  • As Claude's comment suggests, maybe you are not trying to find solutions of this equation at all? Maybe you are trying to find the best $A_e$ and $\eta _T$ values that fit your other data? – Paul Dec 27 '22 at 14:06
  • @ClaudeLeibovici, the model is already present. What benefit do I have by performing multiple linear regression? – rish Dec 27 '22 at 14:10
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    If you fix a value of $\eta _T$ then $A_e =\frac{\rho_A}{irr_s(1+\eta_T*(T_m-25°C))}$. There is not a linear relationship between the two variables so I don't know what you have graphed. Maybe you have asked for one thing (roots) but mean something else, I don't know. – Paul Dec 27 '22 at 14:11
  • Well, your new edit explains why you get a dependency between $A_e$ and $\eta_T$. Your $R_1$ and $R_2$ are NOT $A_e$ and $\eta_T$. They are $A_e$ and $A_e\eta_T$. To find $\eta_T$, you have to take the ratio $\eta_T = \frac{R_2}{R_1}$. – Paul Sinclair Dec 29 '22 at 04:20

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First. let me simplify the notation by replacing $\rho_A$ and $irr_s$ with their ratio $r = \dfrac{\rho_A}{irr_s}$ and $T_m$ with $T = T_m - 25\ ^\circ C$. And because carrying around a useless subscript is just a nuisance, call the other two variables just $A$ and $\eta$. Then your equations have the form

$$r_i = A(1 + \eta T_i)$$

Where $i = 0,1,2, ...$ represents the various measurements you've taken. For two measurements , let $K = \dfrac{r_i - r_j}{T_i - T_j}$ Then $$A = r_i - KT_i = r_j - KT_j\\\eta = \frac KA$$

I assume this is what you did. If so, then the most likely explanation of your graph is that the points on the line represent pairs of measurements where $T_i$ and $T_j$ were significantly different, while the outliers are pairs where $T_i$ and $T_j$ are close in value. The scatter you see is simply cancellation error. But that is a guess. There are other possible causes for the outliers.

In any case, the line is a strong indication that the outliers are the result of error, and $A$ and $\eta$ are not constants, but vary with each other, apparently with the relationship $$A = 30(1 -\eta)$$ Apparently the efficiency of the panels depends on the Temperature Coefficient, which itself is not constant, but depends on the other state variables. That dependence is not visible in this graph. You would need to plot $\eta$ against $T$ or $r$ to see that.

Paul Sinclair
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  • Thanks for you input, I udated the question recently. Perhaps, now you can answer it better. – rish Dec 28 '22 at 19:43