07-Solar Resource Part 2 ECEGR 452 Renewable Energy Systems
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07-Solar Resource Part 2
ECEGR 452
Renewable Energy Systems
7/21/2019 Solar Resource Part2
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Overview
• Angle of Incidence Components
• Effect of Declination
• Effect of Latitude
• Effect of Tilt
•
Effect of Hour Angle• Hours of Day Light
Dr. Louie 2
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Introduction
• Last lecture we determined that the angle ofincidence affects the irradiance received by asurface
• We now investigate the variables that affect theangle of incidence
Dr. Louie 3
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Introduction
• Angle of incidence depends on many factors,including:
Tilt of the surface (already discussed)
Latitude (f)
Declination angle (d)
Surface azimuth angle (g)
Hour angle (w)
Dr. Louie 4
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Introduction
Dr. Louie 5
N
ES
W
Zenith
qz
b
g
We will assume that g = 0For horizontal surfaces:
q = qz
Normal totilted surface
q
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Effect of Declination Angle
•
Earth is tilted on an axis, which causes seasons• Axis is tilted at 23.5o
• Declination ( ): angular position of the sun atsolar noon wrt the plane of the equator (degrees)
Dr. Louie 6
JuneDecember
23.5o
δ
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Effect of Declination Angle
Dr. Louie 7
δ
δ
summer
δ
δ
winter
negative
declination
positive
declination
For Northern Hemisphere
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Effect of Declination Angle
•
Declination angle is zero during the equinoxes
Dr. Louie 8
March September
viewed from the sun viewed from the sun
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Effect of Declination Angle
• Declination is computed as:
• where
d0 = 23.5o
Dr. Louie 9
0
360 284
365sin d
d d
(where does the 284 come from?)
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Effect of Declination Angle
• Summer solstice:
• Winter solstice:• Spring equinox:
• Autumn equinox:
Dr. Louie 10
0 23 5. od d
0 23 5. o
d d 0d
0d
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Effect of Declination Angle
• Northern Hemisphere: the axial tilt increases thedaylight hours in March-September
• Southern Hemisphere: the axial tilt increases thedaylight hours in the September-March
• More daylight hours means more daily insolation
Dr. Louie 11
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Effect of Declination Angle
• Daylight on April 9th, 2012 at 13:57:25
Dr. Louie 12
Source: time.gov
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Effect of Declination Angle
• Declination affects zenith angle
• Assume solar noon (sun directly overhead)
• Assume the surface is at the equator (latitude=0o)
spring and autumn equinox:
summer solstice: winter solstice:
Dr. Louie 13
G0n
March
June
G0n
z q
0q o
z
23 5
. z q
23 5. z
q
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Effect of Latitude
• Let
f: latitude of the surface (degrees)
• Assume North is positive, South is negative
• -90 f 90
Dr. Louie 14
equatorf
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Effect of Latitude
• Assume:
declination = 0o (i.e. Spring/Autumn Equinox)
Sun directly overhead (solar noon)
Horizontal surface is at latitude f (degrees)
• It follows that q qz
= f and G0
= G0n
cos(f)
Dr. Louie 15
Gon
equator
q
f
Earth tilted outof the paper
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Effect of Latitude
• Combining the effects of declination and latitude
Assume solar noon (sun directly overhead) Assume the surface is horizontal (q qz )
• Using trigonometry:
q qz f d
cos(q) = cos(qz) = sin(d)sin(f ) + cos(d)cos(f )
Dr. Louie 16
Gon
equator
q
f
d
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Example
• What is the irradiance for a horizontal surface at
the top of the atmosphere (extraterrestrial)above Seattle, Washington (latitude 47.60) onJanuary 23 at solar noon? Account for intra-yearirradiance variation.
Dr. Louie 17
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Example
• What is the irradiance for a horizontal surface at
the top of the atmosphere (extraterrestrial)above Seattle, Washington (latitude 47.60) onJanuary 23 at solar noon?
Dr. Louie 18
0
23
1 0 034 2 1408 6365
360 284 360 284 2323 5 19 75
365 365
( ) . cos .
( ) ( )sin . sin .
d d
on sc
d
d G d G
d
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Example
• What is the irradiance for a horizontal surface at
the top of the atmosphere (extraterrestrial)above Seattle, Washington (latitude 47.60) onJanuary 23 at solar noon?
Dr. Louie 19
20
47 6 19 75 67 4
1408 6 0 385 542
. . .
cos . .
z
n z W G G
m
q
q
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Effect of Declination Angle
• At large values of (f – d , the angle of incidence
is large (cosine effect is significant)
• How can we compensate for this?
Dr. Louie 20
q (f – d
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Surface Orientation
• Tilt the surface
• Want the surface to be normal to the irradiance
b = (f-d) (Northern Hemisphere)
Want angle of incidence to be zero
Dr. Louie 21
b
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Surface Orientation
• Tilt should equal latitude during equinox
• As increases, less tilt needed
At solar noon: cos(q) = cos(f-d-b)
• In the southern hemisphere:
cos(q) = cos(-f+d-b)
Dr. Louie 22
March
latitude
δ
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Surface Orientation
Dr. Louie 23
b
f d
Surface is normal to Gwhen b = f - d
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Surface Orientation
• General rule of thumb: tilt a PV panel at the
latitude
Normal to irradiance on equinoxes
Too much tilt in summer
Too little tilt in winter
Dr. Louie 24
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Where in the world are these PV panels?
Dr. Louie 25
Singapore
Snohomish
Ellensburg
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Surface Orientation
• cos(q) = cos(f-d-b)
Note: cos(w+z) = cos(w)cos(z) – sin(w)sin(z)
Note: sin(w+z) = sin(w)cos(z)+cos(w)sin(z)
• cos(f-d-b)= cos(q + x) [set x =-d-b]
• cos(f + x)= cos(f)cos(x) – sin(f)sin(x)
Dr. Louie 26
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Surface Orientation
= cos(f)[cos(d)cos(b) – sin(d)sin(b)] – sin(f)sin(x)
[back substituting for the remaining x =-d-b]
=cos(f)[cos(d)cos(b) – sin(d)sin(b)] – sin(f)sin(-d -b)
[Using sin(w+z) = sin(w)cos(z)+cos(w)sin(z)]=cos(f)[cos(d)cos(b) – sin(d)sin(b)]
– sin(f)[sin(-b)cos(-d)+cos(-b)sin(-d )]
Dr. Louie 28
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Surface Orientation
=cos(f)[cos(d)cos(b) – sin(d)sin(b)]
– sin(f)[sin(-b)cos(-d)+cos(-b)sin(-d )]
[multiplying out]
=cos(f)cos(d)cos(b) – cos(f)sin(d)sin(b)
– sin(f)sin(-b )cos(-d) - sin(f)cos(-b)sin(-d )
[using cos(-u) =cos(u) and sin(-u) = -sin(u)]
cos(q)=cos(f)cos(d)cos(b) – cos(f)sin(d)sin(b)
+ sin(f)sin(b)cos(d ) + sin(f)cos(b)sin(d )
Dr. Louie 29
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Surface Orientation
• Extraterrestrial irradiance accounting for the tilt,
latitude and declination of a surface at solarnoon:
G0T = G0ncos(q) = G0ncos(f-d-b)
= G0n[cos(f)cos(d)cos(b)
– cos(f)sin(d)sin(b)
+ sin(f)sin(b)cos(d)
+ sin(f)cos(b)sin(d )]
Dr. Louie 30
Important result
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Surface Orientation
• Compute the extraterrestrial irradiance on a
vertical surface above 30o N on April 15 at solarnoon.
Hint: April 15 is the 105th day of the year
Dr. Louie 31
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Surface Orientation
• Compute the extraterrestrial irradiance on a
vertical surface above 30o N on April 15 at solarnoon.
Hint: April 15 is the 105th day of the year
Dr. Louie 32
30
90
f
b
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Surface Orientation
• Compute the extraterrestrial irradiance on a
vertical surface above 30o N on April 15 at solarnoon.
Hint: April 15 is the 105th day of the year
Dr. Louie 33
2
0
1051 0 033 2 1356 4
365
360 284 360 284 10523 5 9 4
365 365
. cos . W/m
sin . sin .
d d
on sc G d G
d
30
90
f
b
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Surface Orientation
G0T = G0n[cos(f)cos(d)cos(b)
– cos(f)sin(d)sin(b)
+ sin(f)sin(b)cos(d)
+ sin(f)cos(b)sin(d )] = 476 W/m2
or G0T = G0ncos(f-d-b) = 476 W/m2
Dr. Louie 34
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Effect of Hour Angle
•
We want to relate this angle to time• How many degrees does the Earth rotate each
hour?
Dr. Louie 35
36015
24
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Dr. Louie 36
N
ES
W
Zenith
qz
b
g
We will assume that g = 0For horizontal surfaces:
q = qz
Normal totilted surface
q
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Effect of Hour Angle
•
We define the hour angle,ω
, as:
h local civil time (hours)
λ longitude (degrees)
λ zone longitude of the meridian defining the localtime (degrees)
• w: angle that the Earth has rotated since solarnoon
Dr. Louie 37
15 12w zone
h
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Effect of Hour Angle
• UTC (Coordinated Universal Time) is defined at 0o
longitude
• Seattle is 8 hours behind UTC during standardtime
zone is then 8 x 15o = 120o W
• During Day Light Savings Time (roughly March – Nov) we are 7 hours behind UTC
zone is then 7 x 15o = 105o W
• For a more accurate calculation use the Equation
of Time
• We will assume that solar time = civil time
( zone = 0)
Dr. Louie 38
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Effect of Hour Angle
• Hour Angle is:
negative in the morning (before solar noon)
positive in the evening (after solar noon)
Dr. Louie 39
w
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Effect of Hour Angle
• If f = d = 0 and b = 0, then
cos(q) =cos(w)
Dr. Louie 40
w
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Angle of Incidence
• Derivation of the angle of incidence is more
difficult, so the result is provided cos(q) =sin(d)sin(f)cos(b)
-sin(d)cos(f)sin(b)
+cos(d)cos(f)cos(b)cos(w)
+cos(d)sin(f)sin(b)cos(w)
Dr. Louie 41
Important result
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Simplifications
• If b = 0 (no tilt), then qz = q and
cos(q) =sin(d)sin(f)+cos(d)cos(f)cos(w)
• For surfaces tilted at their latitude
• cos(q) =cos(d)cos(w)
For surfaces at solar noon
• cos(q) = cos(f-d-b)
Dr. Louie 42
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Angle of Incidence
• Note: cos(q) must be greater than or equal to 0,
otherwise the sun is shining on the rear of thesurface (set the value to 0)
• Note: angle of incidence equations do notaccount for the Earth blocking the sun’s
irradiance Try: w =180, b = 90, f =0 and d=1 (sunny at
midnight!)
• Only use the angle of incidence for daylight hours
Dr. Louie 43
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Astronomy Trivia
•
How many hours of daylight are there in Seattleduring the spring equinox?
A. 6
B. 10
C. 12
D. 14
E. 16
F. 18
Dr. Louie 44
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Astronomy Trivia
•
How many hours of daylight are there in Seattleduring the spring equinox?
A. 6
B. 10
C. 12
D. 14
E. 16
F. 18
Dr. Louie 45
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Hours of Day Light
• Daylight hours vary depending on latitude and
declination• For a horizontal surface the sun sets (G = 0)
when q = 90o
• Find w such that:
cos(q) =sin(d)sin(f)+cos(d)cos(f)cos(w) = 0
• Solving yields:
cos(ws) = -tan(d)tan(f)
ws: sunset angle
Dr. Louie 46
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Hours of Day Light
• Since every 150 is one hour:
Hours of daylight is:
Dr. Louie 47
12
15cos tan tan N d f
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Effect of Hour Angle
• Visualization
Dr. Louie 48
During Equinox,Sunrise at -90o Sunset 90o
w w
Looking downon the North pole
In Summer:Sunrise <-90o
Sunset >90o
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Side Note
How did Eratosthenes estimate the circumference in
the third century BCE?
Dr. Louie 49
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Dr. Louie 50
Welcome to
Syene
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Dr. Louie 51
June 21
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Dr. Louie 53
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Dr. Louie 54
Syene f = 23o
d = -23.5o
June 21
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Dr. Louie 55
Welcome to
Alexandria
Syene
500 miles South
North
7.2o From another angle
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Dr. Louie 56
Syene
angles exaggerated
x
x
Alexandria
x
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• Therefore, Syene and Alexandria are 7.2o of
latitude apart Syene: 24o N, 33o E
Alexandria: 31o N, 30o E
• Distance between Syene and Alexandria: 500
miles• (7.2/360)C = 500 miles
=> C = 25,000
Actual circumference: ~24,900 miles