Our PV system's power peaks around 9 am, goes down during noon and then goes back up again around 3 pm.

[gadfly56]

How do you figure that? The sun is the highest in the sky at noon on the summer solstice, which would be the midday when the angle of the insolation would be the closest to normal incidence to the modules.

That all depends on what was chosen as the default angle. The swing from high to low over the course of the year is 46 degrees, centered over your latitude. You are only normal to the modules when you are at that default angle. Because the sun traces an arc across the sky, there are usually two times during the day when the sun is normal to the module, assuming accurate east-west tracking. From the point of view of the module, the sun rises in front of it, reaches a peak, and then sets in front of it, following a vertical line. Lets not get into the picky details of how there is only one infinitesimally thin line on the module for which this view is perfectly true. So, assuming that the panel is not set for either extreme (solstice), the sun will be normal to the panel twice a day, once going up and once going down. If you set the panel normal to the latitude, then on the equinoxes you will get only one max on those days. Likewise, at any other angle, there will be two days per year with only one maximum, but I believe the equinoxes are the furthest apart those two days can be.

 

[wwhitney]

That all depends on what was chosen as the default angle.

The diagram says "1.5 degrees maximum N-S tilt from horizontal" and "90 degrees tracker range of motion." I assume that means that in the middle of its range, the tracker points the panel straight up, and that at all other positions the panel normal vector has an azimuth of east or west, with an elevation angle between 45 degrees and 90 degrees.

Cheers, Wayne

 

[ggunn]

That all depends on what was chosen as the default angle. The swing from high to low over the course of the year is 46 degrees, centered over your latitude. You are only normal to the modules when you are at that default angle. Because the sun traces an arc across the sky, there are usually two times during the day when the sun is normal to the module, assuming accurate east-west tracking. From the point of view of the module, the sun rises in front of it, reaches a peak, and then sets in front of it, following a vertical line. Lets not get into the picky details of how there is only one infinitesimally thin line on the module for which this view is perfectly true. So, assuming that the panel is not set for either extreme (solstice), the sun will be normal to the panel twice a day, once going up and once going down. If you set the panel normal to the latitude, then on the equinoxes you will get only one max on those days. Likewise, at any other angle, there will be two days per year with only one maximum, but I believe the equinoxes are the furthest apart those two days can be.

But his modules are never in that "ideal" orientation; when they are not flat (parallel to the ground) they are tilted due east or west.

Can we at least agree that the saddle shape of his production curve is because at noon his modules are pointed straight up?

 

[gadfly56]

But his modules are never in that "ideal" orientation; when they are not flat (parallel to the ground) they are tilted due east or west.

Can we at least agree that the saddle shape of his production curve is because at noon his modules are pointed straight up?

Maybe you can help me understand the OP's tracker, because it seems bonkers to me. the panels are flat or they can tilt eastward. Westward rotation seems to be blocked by the equipment. South is out of the paper and north is into the paper. It doesn't look like you can tilt the panels towards the south. I doubt I could design something worse on purpose, so I must be misunderstanding something. Is this actually a two-axis tracker?

 

[ggunn]

Maybe you can help me understand the OP's tracker, because it seems bonkers to me. the panels are flat or they can tilt eastward. Westward rotation seems to be blocked by the equipment. South is out of the paper and north is into the paper. It doesn't look like you can tilt the panels towards the south. I doubt I could design something worse on purpose, so I must be misunderstanding something. Is this actually a two-axis tracker?

The diagram shows the tracker in the center of its range; it can tilt to the east or west at a maximum 45 degrees. The plane of rotation is congruent with the "paper", i.e., the modules can tilt only due east or due west and they point straight up in the middle of the day. This is typical of many single axis trackers.

 

[ggunn]

The diagram shows the tracker in the center of its range; it can tilt to the east or west at a maximum 45 degrees. The plane of rotation is congruent with the "paper", i.e., the modules can tilt only due east or due west and they point straight up in the middle of the day. This is typical of many single axis trackers.

The arrays are very long lines of modules running N to S that are all turned by a single motor. Whether their increased production over that of static arrays is worth the cost of the hardware and upkeep in the final analysis is a question left to the developer, but you can play around with them in PVWatts to get an idea of where they might be practical. The farther north in the northern hemisphere you build them the more advantage they have in the mornings and evenings but the less they have in the middle of the day.

 

[solarken]

How long has this system been in operation? Does it appear on Google satellite images yet? Address?

 

[jaggedben]

How do you figure that? The sun is the highest in the sky at noon on the summer solstice, which would be the midday when the angle of the insolation would be the closest to normal incidence to the modules.

At mid to high latitudes I believe you get the following.

Summer Solstice: sun rises and sets north of due east and west, and reaches a normal to the array (peak output) mid morning and afternoon when it crosses the due east-west line. Since in the middle of the day the sun is somewhat south of overhead, you'll still see the dip in the OPs graph.

Around the equinoxes, the sun will start closer to due east and west at sunrise/set, so peak output will be closer to those times than at the solstice, and the dip will be greater in the middle of the day than at the summer solstice, because the sun will be farther from normal.

At the winter solstice, the sun will rise and set south of due east and west, so I believe the angle to normal changes less throughout the day than in the other scenarios, since it never reaches normal at all. Thus less dip or no dip in winter. But admittedly in this case I'm not sure if the angle is closer to normal at noon or at sunrise/set. Wayne would have to do the math for me.

I should also say that for these scenarios in my head I'm imagining the tracker has closer to 180 degrees of motion, which in reality it doesn't. Harder to intuit how that would affect production.

 

[wwhitney]

Wayne would have to do the math for me.

So, if the sun is at an elevation angle E above the horizon, and at an azimuth A, where 0 degrees is towards the equator (not sure if that's a standard choice or not), then the coordinates of the unit vector S pointing to the sun can be taken to be
S=(sin A cos E, cos A cos E, sin E).

The above formulation implies positive y is towards the equator, and positive x coincides with positive A. The tracker unit normal T is more naturally parameterized by the elevation angle E' relative to the direction (1,0,0), in which case
T = (cos E', 0, sin E').

In that formulation E' varies from 0 to 180 degrees, although the tracker under discussion can only only do E' from 45 degrees to 135 degrees. Then the geometric factor F for solar irradiance on the panel is given by the dot product
F = T dot S = sin A cos E cos E' + sin E sin E'

The choice of E' for which F will be maximized will occur when
dF/dE' = 0, or
sin A cos E sin E' = sin E cos E', or
tan E' = tan E / sin A.

Here we see that when A = 0, i.e. the sun is towards the equator, tan E' should be infinite, i.e. E' = 90 degrees, and having the tracker straight up maximizes the geometric factor (as must be true by symmetry). Then for A > 0 we have E' < 90 degrees, and for A < 0 we get E' > 90 degrees.

Note that unless the sun is due east or west (sin A = +/- 1), the tracker elevation angle (in the usual sense, only equal to E' for positive A) will not match the sun elevation angle. Instead the tracker elevation angle will be higher (tan E' larger in magnitude) as |sin A| < 1.

Now we should substitute the solution for E' into the formula for F to get an expression for F in terms of A and E, but the trig is a bit too messy for me at this hour. Once you have that, you can figure out the formula for A and E over the course of a day for a given day to get the shape of the graph in the OP.

Cheers, Wayne

 

[ggunn]

So, if the sun is at an elevation angle E above the horizon, and at an azimuth A, where 0 degrees is towards the equator (not sure if that's a standard choice or not), then the coordinates of the unit vector S pointing to the sun can be taken to be
S=(sin A cos E, cos A cos E, sin E).

The above formulation implies positive y is towards the equator, and positive x coincides with positive A. The tracker unit normal T is more naturally parameterized by the elevation angle E' relative to the direction (1,0,0), in which case
T = (cos E', 0, sin E').

In that formulation E' varies from 0 to 180 degrees, although the tracker under discussion can only only do E' from 45 degrees to 135 degrees. Then the geometric factor F for solar irradiance on the panel is given by the dot product
F = T dot S = sin A cos E cos E' + sin E sin E'

The choice of E' for which F will be maximized will occur when
dF/dE' = 0, or
sin A cos E sin E' = sin E cos E', or
tan E' = tan E / sin A.

Here we see that when A = 0, i.e. the sun is towards the equator, tan E' should be infinite, i.e. E' = 90 degrees, and having the tracker straight up maximizes the geometric factor (as must be true by symmetry). Then for A > 0 we have E' < 90 degrees, and for A < 0 we get E' > 90 degrees.

Note that unless the sun is due east or west (sin A = +/- 1), the tracker elevation angle (in the usual sense, only equal to E' for positive A) will not match the sun elevation angle. Instead the tracker elevation angle will be higher (tan E' larger in magnitude) as |sin A| < 1.

Now we should substitute the solution for E' into the formula for F to get an expression for F in terms of A and E, but the trig is a bit too messy for me at this hour. Once you have that, you can figure out the formula for A and E over the course of a day for a given day to get the shape of the graph in the OP.

Cheers, Wayne

OK, I won't be slogging through all that math, but it is apparent to me that my initial comment that the the dip in production in the middle of the day is due to the modules pointing straight up at noon was correct.

 

[wwhitney]

Now we should substitute the solution for E' into the formula for F to get an expression for F in terms of A and E, but the trig is a bit too messy for me at this hour.

A bit of algebra gives that the geometric factor F (aka the cosine of the angle between the sun and the tracker normal)

F = sqrt(sin² A cos² E + sin² E)

Which is pleasingly fairly simple, and gives the expected result of 1 when A = +/- 90 degrees, and of sin E when A = 0 degrees.

So now you can combine that with equations for the sun's elevation angle E and azimuth A (modified if my choice of A=0 is nonstandard) to see the expected shape of the curve in the OP.

Certainly it is true that outside the tropics in the summer the sun will be above the horizon when A = +/- 90 degrees, and F = 1; but that the sun will never reach E = 90 degrees, so at noon F = sin E < 1.

Cheers, Wayne