Let me give you the numbers.
On Friday, the total solar radiation measured at the University of Washington was .37 mega (million) joules (a unit of energy), breaking the previous record of .39. As I explained in my previous blog, this record was the result of the time of the year (December 20th is nearly at the winter solstice) and a deep, deep cloud cover from a super-moist atmospheric river.
I have termed a name for such a severe event: a dark storm.
It was an event reminiscent of something that would have been familiar to Pharaoh: the next to last plague when he refused to allow the Jews to leave Egypt.
The Pharaoh Learned About Terrible Darkness
Saturday was only a bit better, with a total of .76 megajoules, the third darkest day of 2019. Twice as much solar radiation as Friday....almost enough to think about finding your sunglasses.
Why the improvement? We still had the atmospheric river clouds over us, but thickness of the clouds and precipitation were less than the record-breaking conditions on Friday (see satellite image from Saturday around noon)
Solar radiation is already much higher today than either Friday or Saturday (see plot from the roof below through 11 AM)
Today? No black. A reassuring blue color for much of Washington.
Monday is more of a mixed bag. There will be clouds banking up on the western side of the Cascades, bringing darker conditions there. But some locations in western WA will be brighter and southern Oregon will be quite luminous.
Is there a trend in solar radiation reaching the surface? We really don't have decent long records, but WSU's AgWeather network has some sites going back 20 years. From what I can see, there is little evidence of significant long-term trend (see annual observations at Mt. Vernon,WA below--perhaps a slight increase at that station.) As part of my climate research simulations, solar radiation at the surface is projected....I will take a look at that soon.
Cliff,
ReplyDeleteHow are you calculating the total daily energy transfer? I can generate a graph that displays irrandiance in W/m^2 versus time as recorded by my PWS but I'm not sure how to integrate the area under the curve (there's no function, just a set of data points) in order to determine the total amount of energy transferred per unit area. I've noticed that it has been quite dark in NW Bellingham lately and would like to be able to compare my measurements with those from UW.
Graph the points in excel- use the “trendline” option to fit a curve to your datapoints, the equation will be displayed on your graph.
DeleteCan you use the trapezoidal rule?
DeleteIt looks like the darkest day of the recent "dark storm" in NW Bellingham was 12/19. I calculated 0.44MJ/m^2.
DeleteCorrection:
Delete12/18 was the darkest day in NW Bellingham.
P.S.
DeleteI used the trapezoidal rule to estimate this value which, as such, is likely an overestimation.
For comparison, the sunniest day of the month at my location was 12/15 on which I measured approximately 3MJ/m^2. There was nearly 1MJ/m^2 on the sunniest hour (1-2PM) of that day.
DeleteCorrection: The value is likely an underestimation due to the concave down curve.
DeleteIt was quite the Express and I got well over 5 inches of rain in north Bothell, but what happened to the Pineapples? The thermometer struggled to reach 50. I have seen it go over 60, even in January.
ReplyDeleteThat would seem to be significant info to have going forward for solar energy users as well as Ag folks. Perhaps you could find a guest blogger to explain how your info could be used by solar energy producers and users.
ReplyDeleteThere are some light meter and solar apps available for the iPhone, I suspect that some of them might do a passable job at measuring solar radiation. What kind of conversion would need to be done to conform to the mega joules per meter per day metric?
ReplyDelete1kWh - 3.6^6 joules
ReplyDeleteSo, 3.6 megajoules is one kWh
a joule is watts * seconds so 5 watts * 3600 seconds (in an hour) is 18,000 joules.
If you have a pyranometer app is going to be more accurate compared to a panel output app in kWr, as there is some efficiency coefficient kicking around for that. It could be integral calculus based as its not going to be the same coefficient after so many hours of service life and/or power outputs/ambient temp etc.
Cliff,
ReplyDeleteI'm interested in learning more about the surface solar radiation projections' effects.