Energy and Environment II
HW #2
Due Friday, January 18, 2002
Answers in red
- Solar noon and solar time:
Clocks set for solar time for a particular site will read exactly 12:00 noon when the sun is true south (or true north). This is solar noon. The sun is "as high" in the sky as it will get that day for the site. The solar incidence angle (qz) between the perpendicular to the horizontal and the sun beam is minimum for the day. For this period of the year (ie, about January 15th), in terms of Pacific Standard Time, when is it solar noon in Seattle? Clearly show how you arrived at your answer.
There is more than one way to do this problem:
- According to the Sunday (1/13/02) Seattle Times, the sun rose at 7:53 and set at 16:43. Taking the mid point as solar noon, one obtains solar noon at 12:18 PST on 1/13/02.
- A similar result is obtained by looking at the UW home page for the sunrise and sunset information.
- From the rredc.nrel.gov web site, one obtains for Seattle for 1/13/90: sunrise = 8 – 24/1414 = 7.9830 = 7:59; sunset = 16 + 825/1414 = 16.5835 = 16:35. The mid point is 12:17 PST.
Where are you? Imagine it is January 1, 2003. The weather is clear, so you can see your shadow. You are standing on flat, horizontal ground. The elevation of the land is fairly high, about 1 mile. There is a very famous mountain about 200 km south southeast from your location. The length of your shadow is 40% of your height – your shadow is as short as possible for the day. It is 1:23 AM (ie, 23 minutes after 1AM) solar time in Seattle. Where are you? Clearly show how you arrived at your answer. Note the longitude of Seattle is West 122.30 degrees. You may need to use an atlas. There are some on the web, eg, http://atlas.dhs.org/RealWorld/Applicatio/WorldAtlas
The time at your location is solar noon, since your shadow is as short as possible for the day. The time difference between your location and Seattle is 12-1:23 = 10:37 = 10.62 hours. Since each hour is equal to 15 degrees of rotation of the earth, your longitude is 10.62 x 15 = 159.30 degrees east of Seattle. That is, your longitude is 122.30-159.30 = -37.0 degrees = East 37.0 degrees.
Your latitude is found from the length of your shadow. The tangent of the solar incidence angle is 0.40. That is, tan
qz = 0.4, and thus qz = 21.8 degrees. At solar noon, qz = f – d or qz = d – f. By the equation, d = 23.5sin[360°
(284 + n)/365], with n = 1 for January 1, one obtains d = -23.061 degrees. Thus, the latitude angle is f = 21.8 + (-23.061) = -1.26 degrees = South 1.26 degrees. The other possibility is f = -23.061 – (21.8) = -44.86 = South 44.86 degrees.
East 37.0 degrees, South 1.26 degrees places you at Nairobi, Kenya. The other possibility places you in the Indian Ocean – it doesn’t meet the other conditions given.
Solar energy for Honolulu for May 25th. Hourly solar energy flux data for May 25, 1990 for Honolulu are given in the first part of the MS EXCEL file I’m sending you (342.02.HW2.data.xls).
- Define
the solar radiation data listed in each of the columns F, G, H, I, and J. For example, one of the columns is GLOBAL (for the ground).
Column F = extraterrestrial solar energy flux, component perpendicular to ground.
Column G = extraterrestrial solar energy flux parallel to sun’s rays = "solar constant for the day".
Column H = global (total) solar energy flux incident on a horizontal surface on the ground.
Column I = beam solar energy flux at ground level.
Column J = diffuse solar energy flux on a horizontal surface on the ground.
To the right of the columns a plot is shown. This is a plot of the extraterrestrial solar energy flux component perpendicular to the ground (ie, received by a 1 meter square surface parallel to the ground at the edge of the earth’s atmosphere). Note the peak value of this solar energy flux (w/m2). What does this value tell you about the position of the sun in the sky at solar noon on May 25th in Honolulu? Would one see their shadow (assuming cloud is not blocking the sun)? Provide answer and explanation.
The peak value of the extraterrestrial solar flux perpendicular to the ground is 1331 watts per sq meter, identical to the solar constant for the day. Thus, the noontime sun is directly overhead. There are no shadows. This occurs because Honolulu lies south of the Tropic of Cancer.
Suppose you made a new plot, dividing the solar energy flux data in my plot by 1331 w/m2, the "solar constant" for the day. Your new plot would be the cosine of a very important angle. What is this angle called? Write the angle’s equation in terms of sines and cosines of other angles.
This is the solar incidence angle,
qz. The equation is:
cosqz = sinfsind + cosfcosdcosw
Farther to the right, see the plot of the global, beam, and diffuse solar energy flux at ground level versus the solar time. Using bullet format, succinctly point out four major and interesting features of the data plotted.
Peak global energy flux exceeds 1000 w/m2 (at about 11 hr)
Maximum diffuse solar energy flux is about 200 w/m2.
The day is not totally clear – cloudiness is indicated for 14 hr solar time.
The shoulders on the beam energy flux curve are interesting. They appear in both morning and afternoon at about 5 hours relative to solar noon. At this time the solar angle relative to the horizon is about 20 degrees.
The transmissivity of the atmosphere above Honolulu on this day may be determined by dividing the ground-level beam radiation by the extraterrestrial beam radiation. Did at least 70% of the solar constant radiation reach the ground as beam radiation? Show your calculation.
No
, the maximum transmissivity of the beam radiation is about 68%. This occurred at about 11 hr. The calculation: 908 x 100/1331 = 68.2%. (Of course, if you consider the global solar energy flux, the percentage received by the ground is greater than 70%. The maximum is 1022 x 100/1331 = 76.8%.)
Make a plot of the diffuse solar energy flux divided by the global solar energy flux. Restrict your ratio to the red/yellow sector on the spread sheet. Comment on your plot: generally, how does the diffuse/global ratio vary over the day (when the sky is clear or mostly clear-as the case for 5/25/90 in Honolulu).
The plot shows a characteristic shape – essentially 100% of the solar energy flux received by the horizontal ground near sunrise and sunset is diffuse radiation scattered from the sky, while in midday on a sunny day, the ratio reaches its minimum -- with the sun overhead in Honolulu, the midday percentage drops to about 15%.
- Comparing daily average solar energy.
Now look at the second part of 342.02.HW2.data. That is, look at the average daily data for Honolulu, Seattle, and Phoenix. These data are shown for each month and for the year. (The data are 30-year averages.) To the right of the Honolulu data, I’ve made a plot of the global data (watt-hours/m2/day) for the three cities and for eastern Montana (which has about the same latitude as Seattle, but is pretty sunny).
- Look at the yearly value for AETRN. It is the same for the three cities (16405 watt-hours/m2/day). Explain why it is independent of location. (Note that a very famous value appears if you divide by 12.0)
Division by 12 gives 1367 w/m2, the solar constant. All parts of the planet receive this, and all parts of the planet receive on average for the year 12 hours of daylight, 12 hours of night.
Now look at the yearly values for AVETR. These vary with location. Divide by 24. Explain the variation with respect to location and relative to the familiar value of 342 w/m2.
The average sun angle is different for different parts of the planet. At the equator the average value of
qz is much greater than at the poles. Thus, the yearly average value of the component of the extraterrestrial solar energy flux perpendicular to the ground varies inversely with latitude. For the three cities examined the vales are:
Honolulu (21.33 degees): 395 w/m2 > 342 w/m2 average for planet.
Phoenix (33.43 degrees): 359 w/m2 > 342 w/m2 average for planet.
Seattle (47.45 degrees): 300 w/m2 < 342 w/m2 average for planet.
Note: the average of 342 w/m2 occurs at about North 38 degrees latitude.
Now look at the monthly AVGLO data. Divide by 24. Do any of the locations have a 6-months hourly average exceeding 300 w/m2 This was mentioned in class as indicative of a very sunny location?
Yes
: Phoenix has a 6-months average of about 7300-7400 watt-hours/m2/day. (7200 corresponds to 300 w/m2 average.)
Look at my plot of the three cities plus eastern Montana. In bullet format, briefly state four distinct differences or similarities between the three cities.
The daily solar energy for Honolulu shows less variation with the time of year than the daily solar energies for Phoenix and Seattle. The closer one gets to the equator, the lower the variation. At the poles, the variation between summer and winter is very great.
In mid summer, Seattle receives almost as much daily solar energy as Honolulu – our sunshine is less intense than that of Honolulu, but our days are longer.
Phoenix receives significantly more daily solar energy than Honolulu, except in the winter.
Seattle receives very little daily solar energy in the winter. There is a very large variation in the daily solar energy from summer to winter. Thus, organizations and individuals interested in installing solar energy systems in Western Washington (and Oregon and British Columbia) should a situation that mainly requires the solar energy in the March-September period.
Seattle cannot do much about its location (47.5 degrees latitude). What if there were fewer clouds and rain – about much how increase in the yearly solar energy might be possible? Express your answer as a percentage increase, and show how you obtain it.
The data for eastern Montana indicate how much solar energy one might expect at 47.5 degrees latitude if the sky were relatively clear a good part of the year. It looks like about 20% more solar energy could be obtained over the year compared to (gloomy) Seattle.