SOLAR GEOMETRY
By
Dr. S. SABOOR
D.M.E., B-Tech (Gold Medalist)., M-Tech (R&A/C)., PhD (Mech.)., M ASHRAE., M ASME
Senior Assistant Professor, SMEC, VIT, Vellore.
E-Mail: saboor.s@vit.ac.in
Website: http://shaiksaboor.wix.com/scientist-site
Department of Thermal and Energy Engineering
School of Mechanical Engineering
VIT University, Vellore, TN, India.
1
HOT
MODERATE
COLD
Angled sun (30O)
Overhead sun (90O) Angled sun (45O)
Spread over 1 unit Spread over 1.4 unit Spread over 2 unit
Lowest sun angle
Higher sun angles
North Pole:
• Northern hemisphere is close to sun
• Northern hemisphere has more sun angle
than southern
• Longer days in Northern hemisphere
• At 23.5 N (Tropical cancer) sun is directly
overhead at 90o and gives more heat
• North polar region is having full daylight 24
hrs
• North hemisphere experiences summer
South Pole:
• Southern hemisphere is far from sun
• Southern hemisphere has less solar angles
• Smaller days in southern hemisphere
• Southern polar region has full night of 24hrs
• Southern hemisphere experiences winter
South Pole:
• Southern hemisphere is close to sun
• Southern hemisphere has more sun angle
than northern
• Longer days in Southern hemisphere
• At 23.5 S (Tropical Capricorn) sun is directly
overhead at 90o and gives more heat
• South polar region is having full daylight 24
hrs
• southern hemisphere experiences summer
North Pole:
• Northern hemisphere is far from sun
• Northern hemisphere has less solar angles
• smaller days in northern hemisphere
• Northern polar region has full night of 24hrs
• Northern hemisphere experiences winter
On Equinox (Mar 21-22):
• All locations on earth experiences 12 hrs day and 12 hrs
night
• Sun is directly overhead at equator at 90o and gives more
heat.
10
Major countries in the Northern hemisphere entirely or partially:
• Russia
• Canada
• China
• United states
• India
• Kazakhstan
• Algeria
• Saudi Arabia
• Brazil
• Mexico
• Mongolia
Major countries in the Southern hemisphere entirely :
• Australia
• New Zealand
• Argentina
• Angola
• South Africa
• Bolivia
• Tanzania
• Namibia
• Mozambique
• Zambia
• Chile
70-80% Land
80-90% Population
30-40% Land
11
10-20% Population
N
E
W
S
Vellore/Coordinates
12.9165° N, 79.1325° E
N
E
W
S
indiaN STANDARD TIME
• Longitude
is
drawn
in
multiples
of
15°
(as 15 ° = 1 hour)=(360/24)
• There is a Standard longitude
for a country with which
standard/ watch times are
defined
• For
India,
the
standard
meridian = 82.5°E
• Hence, time difference between
Greenwich (0°) and India =
82.5/15
= 5.5 hours ahead of GMT
Coordinated Universal Time (abbreviated
UTC) is the primary time standard by which
the world regulates clocks and time.
Indian Standard Time is calculated on the basis
of 82.5° E Longitude, which is just west of the
town of Mirzapur, near Allahabad in the state of
Uttar Pradesh.
Solar time
• Time Based on the apparent angular motion
of the sun across the sky (v/s clock time)
• Solar noon is the time at which the sun
crosses the local meridian of the observer
– At solar noon, the sun will be at zenith of
the observer
• Movement of the sun is symmetrical about
solar noon, to an observer
• Solar Time does not coincide with the local
clock time.
Difference Between Watch time and
Solar Time OR local apparent time
(lat)
Solar time can be obtained from Watch time with two
corrections
•
First Correction arises due to difference between the
longitude of a location and meridian on which the watch time
is based.
•
It has a magnitude of 4 minutes for every degree
difference in longitude.
•
Second correction called “Equation of Time”.
•
EOT is the correction due to the fact that the earth orbit
and rate of rotation are subject to small variations. This
correction is based on experimental observations.
Watch time
• Based on the longitude.
• Depends on standard meridian of that country
• Time difference between the local time (based on the
local meridian) and standard time (based on standard
meridian) will be 4 (Lst - Llo), in minutes,
where,
Lst is the standard longitude and
Llo is the local longitude
Solar time = Watch Time - 4 (Lst – Llo)
(in minutes)
Equation of Time
• As the earth moves around the sun, solar time changes
slightly with respect to local standard time
• It is the correction due to the fact that the earth orbit and
rate of rotation are subject to small variations.
• This time difference is called the Equation of Time (EoT)
• The equation of time can also be calculated from the
following correlation.
EOT = 229.18*[0.000075 + (0.001868*cosB) –
(0.032077*sinB) –(0.014615*cos2B) – (0.04089*sin2B)]
• From some correlations, the Equation of time can be
predicted by,
EoT = 9.87 sin 2B - 7.53 Cos B - 1.5 Sin B
Solar Time
Days Vs EOT
20
Days Vs Constants
400
EOT1
B1
EOT2
B2
15
Angle (B) in minutes
time in minutes
300
10
5
0
-5
200
100
0
-10
-15
0
100
200
Days
300
400
-100
0
100
200
Days
300
B1=(360*(n-81)/364);
B2=(360*(n-1)/365)
EOT1=9.87 sin 2B 1- 7.53 Cos B1 - 1.5 Sin B1
EOT2=229.18*[0.000075 + (0.001868*cosB2) – (0.032077*sinB2) –
(0.014615*cos2B2) – (0.04089*sin2B2)]
400
Equation of Time
Equation of Time
B = 360*(n-81)/364 (in degrees)
n = Day of the year, (1< n< 365)
Note:
For India (east of Greenwich),
Solar time = Watch Time - 4 (Lst - Llo) + EoT (in minutes)
For America (west of Greenwich),
Solar time = Watch Time + 4 (Lst - Llo) + EoT (in minutes)
Note:
In general, the ‘-’ sign is applicable to Eastern hemisphere,
and ‘+’ is applicable to Western hemisphere.
Calculate solar time on Feb. 2, 10.30 a.m.
at Vellore (12.9833°N, 79.1833°E)
Solar time = Watch Time - 4 (Lst - Llo) + EoT
EOT = 229.18*[0.000075 + (0.001868*cosB) – (0.032077*sinB) –
(0.014615*cos2B) – (0.04089*sin2B)]
Where, B=(360*(n-1)/365
n
EoT
= 33 (on Feb. 2)
= -13.5 min. (on February)
• Solar time = 10.30 – 4*(82.5-79.1833) + (-13.5)
= 10 hours 03 minutes
Earth centric co-ordinate system
1. Consider a globe
2. Cut out a quarter
3. Connect centre of the earth to centre of the sun
4. Include an equatorial plane
5. Include a meridional plane which goes through line going
to centre of the sun (Insolation line)
6. Solar Angles
i.) Declination angle: the angle between the insolation line and the equatorial plane is
called declination angle
ii.) Latitude angle: the angle between the line joining any place on earth surface and centre of
the earth to the equatorial plane is called latitude angle.
What is Latitude angle of Madison of 43◦ N :
+ 43 deg
iii.) Hour angle: the angle between –y co-ordinate axis and meridional plane is called hour angle
• This angle is directly related to the time of the day. Insolation line moves from east to west due
to earth’s rotation. In every four minutes, one meridional plane will be travelled (1 Deg Longitude) .
What is hour angle at 9:30 AM :
What is hour angle at 6:30 PM:
-37.5 deg
97.5 deg
Sunset hour angle at sunset (at zenith angle= 90o) as in Fig. can be computed by
the following relation:
Incident angle,
To Measure the day length (Smax):
As Day length (Smax) = (2/15)* (ωs)
In hours
To measure the sun rise time (SRT):
As Sun rise time (SRT)
= [12 – (0.5*Smax)]
In hours
To Measure the Sun Rise time in Watch (IST):
Sunrise Watch Time
= Solar Time + [ 4 (Long Std – Longlocal) + EOT (in minutes)]
To measure the sun Set time (SST):
As Sun Set time
= (0.5* Smax)
To Measure the Sun Set Time in Watch (IST):
Sunrise Watch Time
= Solar Time + [ 4 (Long Std – Longlocal) + EOT (in minutes)]
Numericals
1. Find the Sun rise time and Sun Set time, day
length @ VIT University on 28th July 2019.
Take the latitude and longitude @ VIT is,
Latitude (Ø) = 12.98330 N
Longitude = 79.18330 E
Numericals
Answer: Pb.No:1.
To measure the day number: (As on 28th July 2019):
n = (31+28+31+30+31+30+28) = 209
Day number (n) = 209
To Measure the Value of B:
As B = (360/364)*(n-81)
= (360/364)*(209-81)
B = 126.593
To Measure the Value of Equation of Time (EOT) as on 28 th July 2019):
EOT = 9.87 Sin 2B – 7.53CosB – 1.5 SinB
= 9.87 Sin(2*126.293) – 7.53 Cos126.593 – 1.5 Sin 126.593
EOT = -6.12 minutes.
Numericals
To measure the declination angle ( ):
As Declination angle ( ) = 23.45 Sin [360*(284+n)/365)]
= 23.45 Sin [360*(284+209)/365)]
Declination angle ( )
= 18.91195 (Degrees)
To measure the Solar hour angle: (ωs):
As hour angle (ωS)
hour angle (ωS)
= Cos-1[ -tan( ) tan(Ø)]
= Cos-1[ - tan(18.91195) tan(12.9833)]
= 94.52946 (Degrees)
To Measure the day length (Smax):
As Day length (Smax) = (2/15)* (ωs)
= (2/15)*(94.52946)
Day length (Smax)
= 12.604 (hours)
Numericals
To measure the sun rise time (SRT):
As Sun rise time (SRT)
= [12 – (0.5*Smax)]
= [12- (0.5*12.604)]
Sun rise time (SRT) = 5.698 Hours (Solar) = 5 hours 42 minutes
To Measure the Sun Rise time (IST):
Watch Time
Sun Rise Time (Watch)
= Solar Time +[ 4 (Long Std – Longlocal) - EOT (in minutes)]
= 5 hours 42 minutes+ [4 (82.5-79.1833) + (-6.12)]
= 5.49 AM
(As per internet SRT = 5.52 AM)
To measure the sun Set time (SST):
As Sun Set time
= (0.5* Smax)
= (0.5* 12.604)
Sun Set time
= 6.308 hours (Solar) = 6 hours 18 Minutes.
To Measure the Sun Set Time in (IST):
Watch Time
= Solar Time +[ 4 (Long Std – Longlocal) - EOT (in minutes)]
= 6 hours 19 minutes+ [4 (82.5-79.133) + (-6.12)]
Sun Set time (Watch)
= 6. 27 PM
(As per internet SST = 6.29 PM)
Numericals
2. Find the Sun rise time and Sun Set time, day
length @ VIT University on 3rd August 2012.
Take the latitude and longitude @ VIT is,
Latitude (Ø) = 12059’00 N = 12.9833 N
Longitude = 79008’00 E = 79.1833 E
Numericals
Answer: Pb.No:1.
To measure the day number: (As on 3rd August 2012):
n = (31+29+31+30+31+30+31+3) = 216
Day number (n) = 216
To Measure the Value of B:
As B = (360/365)*(n-81)
= (360/365)*(216-81)
B = 133.1506 = 133.1510
To Measure the Value of Equation of Time (EOT) as on 28 th July 2011):
EOT = 9.87 Sin 2B – 7.53CosB – 1.5 SinB
= 9.87 Sin(2*133.151) – 7.53 Cos133.151 – 1.5 Sin 133.151
EOT = -5.79 minutes.
Numericals
To measure the day number: (As on 28th July 2011):
n = (31+29+31+30+31+30+31+3) = 216
Day number (n) = 216
To measure the declination angle ( ):
As Declination angle ( ) = 23.45 Sin [360*(284+n)/365)]
= 23.45 Sin [360*(284+216)/365)]
Declination angle ( )
= 17.1081 (Degrees)
To measure the Solar hour angle: (ωs):
As hour angle (ωS)
hour angle (ωS)
= Cos-1[ -tan( ) tan(Ø)]
= Cos-1[ - tan(17.1081) tan(12.9833)]
= 94.0694 (Degrees)
To Measure the day length (Smax):
As Day length (Smax) = (2/15)* (ωs)
= (2/15)*(94.0694)
Day length (Smax) = 12.5426 (hours)
Numericals
To measure the sun rise time (SRT):
As Sun rise time (SRT)
= [12 – (0.5*Smax)]
= [12- (0.5*12.5426)]
Sun rise time (SRT) = 5.7287 Hours (Solar) = 5 hours 44 minutes
To Measure the Sun Rise time (IST):
Watch Time
Sun Rise Time
= Solar Time + [ 4 (Long Std – Longlocal) - EOT (in minutes)]
= 5 hours 44 minutes+ [4 (82.5-79.1833) - (-5.79)]
= 6.03 AM
(As per internet SRT = 5.59 AM)
To measure the sun Set time (SST):
As Sun Set time
= (0.5* Smax)
= (0.5* 12.5426)
Sun Set time
= 6.2713 hours (Solar) = 6 hours 16 Minutes.
To Measure the Sun Set Time in (IST):
Watch Time
= Solar Time + [ 4 (Long Std – Longlocal) - EOT (in minutes)]
= 6 hours 16 minutes+ [4 (82.5-79.1833) - (-5.79)]
Sun Set time
= 6. 35 PM
(As per internet SST = 6.39 PM)
Local centric co-ordinate system
1. Consider a point of interest on meridional plane and
connect that point with centre of the earth.
2. Place a plane (Horizon plane) tangential to meridional plane at the
locality considered.
3. The axis which is normal to harizon plane and passing through
centre of the earth is called zenith axis (z).
4. The axis tangential to the meridion going down south is called south
axis (S)
5. The axis E is towards east and parallel to east of earth coordinate
system (E)
Zenith
East
South
i.) Zenith angle: the angle that insolation line makes with zenith axis (vertical axis) is called zenith angle
Zenith
Zenith angle can be computed by the following relation:
Calculate Zenith angle for the latitude 43 deg N at a 9:30 am on February 13th
ii.) solar Azimuth angle: The angle between the point on the insolation line projected onto the horizon
plane and the south axis is called azimuth angle.
Solar Azimuth angle can be computed by the following relation:
γs is negative when the hour angle is negative and positive when the hour angle is
positive. The sign function in the above Equation equal to +1 if ω is positive and is
equal to −1 if ω is negative
Calculate solar azimuth angle for the latitude 43 deg N at a 9:30 am on February 13th
αs
Declination angle
Hour angle
Latitude angle
Zenith angle
Solar Azimuth angle
Solar altitude angle
Surface azimuth angle
iii.) Solar altitude angle (αs): The angle between the insolation line and the horizontal axis is
called Solar altitude angle. Opposite to zenith angle.
iii.) Angle of incidence: The angle between the insolation line on surface and normal to that
surface
Angle of incidence can be computed by the following relation for any angles:
Angle of incidence can be computed by the following relation for vertical surfaces:
Angle of incidence becomes zenith angle for horizontal surfaces β=0 and it can be calculated by
β= Slope of the harizon plane or surface
iv.) Surface azimuth angle: The deviation of the projection .
Air Mass (m)
• Used as a measure of the distance
travelled by beam radiation through
the atmosphere before it reaches a
location on the earth’s surface.
• ratio of the optical thickness of the
atmosphere through which beam
radiation passes to the optical
thickness if the sun were at the zenith.
• Where,
• AM0 is Air mass Zero is corresponds
to extraterrestrial radiation.
• AM1 is Air mass One is corresponds
to case of sun at Zenith.
• AM2 is Air Mass Two is corresponds
to case of Zenith angle of 600.
m = (Cos qz)-1
Air mass
AM  1
cos q z
Where,
X
θz is Zenith angle
Distance traveled by sun rays
to reach earth’s surface, or
Air Mass
AM0 (extra terrestrial)
Y
Atmosphere
Earth
Solar radiation flux reaching
the surface (W /m2)
1376
AM1 (sun at overhead position)
1105
AM1.5 (sun at about 48o from
overhead position)
1000
AM 2 (sun at about 60o from
overhead position)
894
1. Declination angle
𝐝𝐢𝐚 = 𝟐𝟑. 𝟒𝟓𝐬𝐢𝐧
𝟑𝟔𝟎(𝟐𝟖𝟒 + 𝐧𝐝 )
𝟑𝟔𝟓
2. Solar altitude angle
𝐬𝐢𝐧𝛃 = 𝐜𝐨𝐬𝐥𝐜𝐨𝐬𝐝𝐢𝐚 𝐜𝐨𝐬𝐡 + 𝐬𝐢𝐧𝐥𝐬𝐢𝐧𝐝𝐢𝐚
β= 90-zenith angle
3. Solar azimuthal angle
𝐬𝐢𝐧𝛃𝐬𝐢𝐧𝐥 − 𝐬𝐢𝐧𝐝𝐢𝐚
𝐜𝐨𝐬γ𝐬 =
𝐜𝐨𝐬𝛃𝐜𝐨𝐬𝐥
Interms of zenith angle
4. Surface solar azimuth angle
𝜸 = γ𝐬 − 𝜳
Surface orientations and azimuths, measured from south
Orientation
Surface azimuth 𝛹
N
1800
NE
-1350
E
-900
SE
-450
S
00
SW
450
W
900
NW
1350
5. Angle of Incidence
𝐜𝐨𝐬𝛉 = 𝐜𝐨𝐬𝛃𝐜𝐨𝐬𝛄𝑺𝒊𝒏𝐤 − 𝐬𝐢𝐧𝛃𝑪𝒐𝒔𝐤
k= window tilt angle from horizontal. For horizontal window k= 0 deg; for vertical window k= 90
deg.
6. At the earth’s surface on a clear day solar irradiance at clear atmosphere is given by
𝐀𝟏
𝐈𝐃𝐍 =
𝐞𝐱𝐩 𝐁𝟏 𝐬𝐢𝐧𝛃
Table 1 indicates the hourly solar radiation values on 21 st of every month in different climatic regions in India. These values
were used to find the direct, diffuse and ground reflected radiation values on any surface.
Table 1. Constants A1, B1 and C1 obtained for predicting hourly solar radiation in India .
Day
Jan. 21st
Feb.21st
Mar.21st
Apr.21st
May.21st
Jun. 21st
Jul. 21st
Aug.21st
Sep. 21st
Oct. 21st
Nov.21st
Dec.21st
A1 (W/m2)
Solar radiation in absence of
atmosphere
610.00
652.20
667.86
613.35
558.39
340.71
232.87
240.80
426.21
584.73
616.60
622.52
B1
Atmospheric extinction
coefficient
0.000
0.010
0.036
0.121
0.200
0.428
0.171
0.148
0.074
0.020
0.008
0.000
C1
Sky radiation
coefficient
0.242
0.249
0.299
0.395
0.495
1.058
1.611
1.624
0.688
0.366
0.253
0.243
Numericals
Pb.No:2. Find the day length, Sun rise time and Sunset
time at VIT, Vellore on 18th January, 2012 and also
determine the following angles at 4.20 pm.
Assume Tilt angle of 12.60 (towards south)
a. Hour angles (ω)
•
b. Incident angle (θ)
•
c. Zenith Angle (θz)
•
d. Altitude angle (α)
•
e. Solar Azimuth angle (γs)
•
f. Surface Azimuth angle (γ)
•
g. Air Mass (AM)
Numericals
Answer: Pb.No:2.
From Internet,
Katpadi is located at latitude of 12.98330 N and
longitude of 79.18330 E.
Step No:1: To measure the day number: (As on 18th
January 2012):
n = 18, Therefore,
Day number (n) = 18
Numericals
Step No:2: To Measure the Value of B:
As B = (360/365)*(n-81)
= (360/365)*(18-81)
B = - 62.302
Step No:3: To Measure the Value of Equation of Time
(EOT) as on 18th January 2012):
EOT = 9.87 Sin 2B – 7.53CosB – 1.5 SinB
= 9.87 Sin(2*(-62.302) – 7.53 Cos(-62.302) – 1.5 Sin (-62.302)
EOT = - 10.296 minutes.
Numericals
Step No:4: To measure the declination angle ( ):
As Declination angle (δ) = 23.45 Sin [360*(284+n)/365)]
= 23.45 Sin [360*(284+18)/365)]
Declination angle ( ) = - 20.731 (Degrees)
Step No:5: To measure the Solar hour angle: (ωs):
As Solar hour angle (ωS) = Cos-1[ - tan( ) tan(Ø-β)]
= Cos-1[ - tan(-20.731) tan(12.98-12.916)]
Solar hour angle (ωS) =
(Degrees)
Numericals
Step No:6: To Measure the day length (Smax):
As Day length (Smax) = (2/15)* (ωs)
= (2/15)*(84.995)
Day length (Smax) = 11.333 (hours)
Numericals
Step No:7: To measure the sun rise time (SRT):
As Sun rise time (SRT) = [12 – (0.5*Smax)]
= [12 -(0.5*11.333)]
Sun rise time (SRT) = 6.333 Hours (Solar)
= 6 hours 20 minutes (Solar)
Step No:8: To measure the sun rise time in (IST):
Watch Time = Solar Time + 4 (Long Std – Longlocal) - EOT
(in minutes)
= 6 hours 20 minutes+ [4*(82.5 -79.13) –(-10.30)]
Sun Rise Time = 6.43AM
(As per internet SRT = 6.40 AM)
Numericals
Step No:9: To measure the sun Set time (SST):
As Sun rise time (SRT) = [ (0.5*Smax)]
= [(0.5*11.333)]
Sun rise time (SRT) = 5.667 Hours (Solar)
= 5 hours 40 minutes (Solar)
Step No:10: To measure the sun Set time in (IST):
Watch Time = Solar Time + 4 (Long Std – Longlocal) - EOT
(in minutes)
= 5 hours 40 minutes+ [4*(82.5 -79.13) –(-10.30)]
Sun Rise Time = 6.03 PM
(As per internet SST = 6.06 PM)
Numericals
Step No:11: To Measure the Local Apparent Time (Solar
time):
LAT (Solar Time) = Watch Time - 4*(LongStd – LongLoc) + EOT
(In minutes)
= 16. 20 hours - 4* (82.5 – 79.13) + (-10.30)
Local Apparent time (Solar Time) = 15.57 hours
Step No:12: To Measure the hour angle (ω):
Hour angle (ω)
= 15*(LAT - 12)
= 15*(15.57 – 12)
Hour angle (ω) = 53.550 (Degrees)
Numericals
Step No:13: To measure the surface Azimuth angle (γ):
Surface Azimuth angle (γ)= 00 (Degrees)
(Due to receiver facing south)
Step No:14: To measure the Zenith angle (θz):
Zenith angle = Cos-1[ (Cos . CosØ. Cosω)+ (Sin . SinØ)]
= Cos-1[ ((Cos(-20.73) Cos(12.98) Cos(53.55)) +
(Sin(-20.73) Sin(12.98))]
Zenith angle = θZ = 57.140 (Degrees)
Numericals
Step No:15: To measure the Altitude Angle ( ):
Altitude angle ( ) = 90 – θz = (90 – 57.14)
Altitude angle ( ) = 32.860 (Degrees)
Step No:16: To measure the Solar Azimuth angle (γs):
Solar Azimuth angle (γs) = Cos -1[(Cosθz SinØ - Sinδ) /
(SinθzCosØ)]
Solar Azimuth angle (γs) =
Cos -1[(Cos(57.14)Sin(12.98)–Sin(-20.73)/(Sin(57.14) Cos(12.98)
Solar Azimuth angle (γs) = 56.340 (Degrees)
Step No:16: To measure the incident angle (θ):
(θ) = Cos-1[(Sinδ SinØ Cosβ) – (Sinδ CosØ Sinβ Cosγ) +
+ (Cosδ CosØ Cosβ Cosω) + (Cosδ SinØ Sinβ Cosγ Cosω) +
(Cosδ Sinβ Sinγ Sinω)]
Where, Surface Azimuth angle (γ) = 0 (Due to facing south)
then, the
(θ) = Cos-1[ (Sinδ SinØ Cosβ) –(Sinδ CosØ Sinβ) +
+ (Cosδ CosØ Cosβ Cosω) + (Cosδ SinØ Sinβ Cosω)]
= Cos-1[((Sin(-20.73)*Sin(12.98)*Cos(12.60)) –
(Sin(-20.73)*Cos(12.98)*Sin(12.60)) +
(Cos(-20.73)*Cos(12.98)*Cos(12.60)*Cos(53.55)) +
(Cos (-20.73)*Sin(12.98)*Sin(12.60)*Cos(53.55))]
= Cos-1[(-0.077589) – (-0.07524) + (0.52842) + (0.02723)]
Incident angle (θ) = 56.410 (Degrees)
Step No:17: To measure the Air Mass (AM):
Air mass = (Cos θz)-1
= (Cos (57.14)) -1
Air mass = 1.84
Numericals
Pb.No:2. Find the day length, Sun rise time and Sunset
time at VIT, Vellore on 03rd August, 2012 and also
determine the following angles at 10.15 AM.
a. Hour angles (ω)
b. Incident angle (θ)
c. Zenith Angle (θz)
d. Altitude angle (α)
e. Solar Azimuth angle (γs)
f. Surface Azimuth angle (γ)
g. Air Mass (AM)
Numericals
Answer: Pb.No:2.
From Internet,
Katpadi is located at latitude of 12.98330 N and
longitude of 79.18330 E.
Step No:1: To measure the day number: (As on 03rd
August 2012):
n = (31+29+31+30+31+30+31+3), Therefore,
Day number (n) = 216
Numericals
Step No:2: To Measure the Value of B:
As B = (360/365)*(216-81)
= (360/365)*(216-81)
B = 133.1510
Step No:3: To Measure the Value of Equation of Time
(EOT) as on 3rd August 2012):
EOT = 9.87 Sin 2B – 7.53CosB – 1.5 SinB
= 9.87 Sin(2*(133.151) – 7.53 Cos(133.151) – 1.5 Sin (133.151)
EOT = - 5.79 minutes.
Numericals
Step No:4: To measure the declination angle ( ):
As Declination angle (δ) = 23.45 Sin [360*(284+n)/365)]
= 23.45 Sin [360*(284+216)/365)]
Declination angle ( ) = 17.1081 (Degrees)
Step No:5: To measure the Solar hour angle: (ωs):
As Solar hour angle (ωS) = Cos-1[ -tan( ) tan(Ø)]
= Cos-1[ - tan(17.1081) tan(12.98)]
Solar hour angle (ωS) = 94.0684 (Degrees)
Numericals
Step No:6: To Measure the day length (Smax):
As Day length (Smax) = (2/15)* (ωs)
= (2/15)*(94.0683)
Day length (Smax) = 12.5424 (hours)
Numericals
Step No:7: To measure the sun rise time (SRT):
As Sun rise time (SRT) = [12 – (0.5*Smax)]
= [12 -(0.5*12.5424)]
Sun rise time (SRT) = 5.7288 Hours (Solar)
= 5 hours 43 minutes (Solar)
Step No:8: To measure the sun rise time in (IST):
Watch Time = Solar Time + 4 (Long Std – Longlocal) - EOT
(in minutes)
= 5 hours 43 minutes+ [4*(82.5 -79.1833) –(-5.79)]
Sun Rise Time = 6.02AM
(As per internet SRT = 6.02 AM)
Numericals
Step No:9: To measure the sun Set time (SST):
As Sun rise time (SRT) = [ (0.5*Smax)]
= [(0.5*12.5424)]
Sun rise time (SRT) = 6.2712 Hours (Solar)
= 6 hours 16 minutes (Solar)
Step No:10: To measure the sun Set time in (IST):
Watch Time = Solar Time + 4 (Long Std – Longlocal) - EOT
(in minutes)
= 6 hours 16 minutes+ [4*(82.5 -79.13) –(-5.79)]
Sun Rise Time = 6.35 PM
(As per internet SST = 6.35 PM)
Numericals
Step No:11: To Measure the Local Apparent Time (Solar
time):
LAT (Solar Time) = Watch Time - 4*(LongStd – LongLoc) + EOT
(In minutes)
= 10. 15 hours - 4* (82.5 – 79.13) + (-5.76)
Local Apparent time (Solar Time) = 09.56 hours
Step No:12: To Measure the hour angle (ω):
Hour angle (ω)
= 15*(LAT - 12)
= 15*(09.56 – 12)
Hour angle (ω) = -36.60 (Degrees)
Numericals
Step No:13: To measure the surface Azimuth angle (γ):
Surface Azimuth angle (γ)= 00 (Degrees)
(Due to receiver facing south)
Step No:14: To measure the Zenith angle (θz):
Zenith angle = Cos-1[ (Cos . CosØ. Cosω)+ (Sin . SinØ)]
+
= Cos-1[ ((Cos(17.1081) Cos(12.9833) Cos(-36.6))+
(Sin(17.1081) Sin(12.9833))]
Zenith angle = θZ = 35.5340 (Degrees)
Numericals
Step No:15: To measure the Altitude Angle ( ):
Altitude angle ( ) = 90 – θz = (90 – 35.534)
Altitude angle ( ) = 54.46590 (Degrees)
Step No:16: To measure the Solar Azimuth angle (γs):
Solar Azimuth angle (γs) = Cos -1[(Cosθz SinØ - Sinδ) /
(SinθzCosØ)]
Solar Azimuth angle (γs) =
Cos -1[(Cos35.534*Sin12.9833 – Sin17.1081)/(Sin35.534*
Cos12.9833)]
Solar Azimuth angle (γs) = 56.340 (Degrees)
Step No:16: To measure the incident angle (θ):
(θ) = Cos-1[(Sinδ SinØ Cosβ) – (Sinδ CosØ Sinβ Cosγ) +
+ (Cosδ CosØ Cosβ Cosω) + (Cosδ SinØ Sinβ Cosγ Cosω) +
(Cosδ Sinβ Sinγ Sinω)]
Where, Surface Azimuth angle (γ) = 0 (Due to facing south)
then, the
(θ) = Cos-1[ (Sinδ SinØ Cosβ) –(Sinδ CosØ Sinβ) +
+ (Cosδ CosØ Cosβ Cosω) + (Cosδ SinØ Sinβ Cosω)]
= Cos-1[((Sin(-20.73)*Sin(12.98)*Cos(12.60)) –
(Sin(-20.73)*Cos(12.98)*Sin(12.60)) +
(Cos(-20.73)*Cos(12.98)*Cos(12.60)*Cos(53.55)) +
(Cos (-20.73)*Sin(12.98)*Sin(12.60)*Cos(53.55))]
= Cos-1[(-0.077589) – (-0.07524) + (0.52842) + (0.02723)]
Incident angle (θ) = 56.410 (Degrees)
Step No:17: To measure the Air Mass (AM):
Air mass = (Cos θz)-1
= (Cos (35.534)) -1
Air mass = 1.22