solar course 1 - lecture 3

3. Solar Radiation

• The Nature of Solar Radiation• Radiation on Earth’s Surface

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• Radiation on Earth’s Surface� Types of Solar Radiation� Measurement of Solar Radiation� Extraterrestrial/Terrestrial Solar Radiation

• Solar and Local Standard Time• Isogonic Chart• Sun Path Diagrams

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Sep 21

Fall Equinox

Polar Axis

24 hrs

Ecliptic Axis

= Solar declinationThe angle between the sun-earth line and the equatorial plane.

δEcliptic AxisPolar Axis

Motion of the Earth About the Sun

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Dec. 21

Winter Solstice June 21

Summer Solstice

March 21

Spring Equinox

95.90 x106 Miles89.83 x106 Miles

Ecliptic Plane

The tilt of the earth’s axis with respect to the Ecliptic Plane causes the changing seasons and the annual variation in the number of daylight and darkness hours.

The position of the earth in relation to the sun, the earth’s motion and the effect of land and water bodies establishes the prevailing weather patterns in different parts of the world.

Sun

Sun-Earth Lineδ

©2012 Stevens Institute of Technology 2

Approx. Dates Declination (Deg.)

June 22 +23.45⁰May 21, July 24 +20⁰

May 1, Aug. 12 +15⁰

Apr. 16, Aug. 28 +10⁰

Apr. 3, Sep. 10 +5⁰

Mar. 21, Sep. 23 0⁰

Mar. 8, Oct. 6 - 5⁰

Feb. 23, Oct. 20 - 10⁰

Feb. 9, Nov. 3 - 15⁰

Jan. 21, Nov. 22 - 20⁰

Dec. 22 - 23.45⁰

The table lists how far above and below the equator the sun is (declination) on the approx. dates.

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Dec. 22 - 23.45⁰

March 21 June 21 September 21 December 21

Equator

Tropic of Cancer

(Lat 23.5°N)

Arctic Circle

Tropic of Capricorn

(Lat 23.5°S)

Antarctic Circle

N

S

Earth Declination During the Year

Shows the Latitude at Which the Sun Would be Directly Overhead

The solar declination can be estimated by:

δS = 23.45⁰ sin[360(284+n)/365]⁰

Where, n = the day no. in the year

©2012 Stevens Institute of Technology

Source: Textbook Eq.2.23

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Radiant energy is received from the sun both directly as a beam component, diffused by scattering from the sky and reflection from the ground. Solar radiation is also called Insolation.

• Direct Radiation – travels from the sun to a point on earth with negligible change in direction

• Diffused Radiation – produced by scattering of sunlight by atmospheric components such as particulates, water vapor and aerosols. On a cloudy day, the radiation is 100% diffused

Solar Radiation

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the radiation is 100% diffused• Reflected Radiation – Diffused or direct radiation reflected from the foreground onto the

solar collector.

The solar energy is either converted directly into electricity or it must be converted to heat before it can be used for heating or cooling systems, including thermal conversion to electricity.

Units of solar radiation: Btu/hr-ft2

Watts/m2

Cal/cm2-minLangley/min (1.0 Langley = 1.0 gram-calorie/cm2)


A

BC

D

E

25%

* 23%

20%

Top of the

atmosphere

Reflected from

Diffused, on

the ground

Types of Solar Radiation

Sky Sky Diffused Clearness Radiation

Clear 1.0 12%Slightly Hazy 0.8 25%

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Solar Radiation Received at Earth’s Surface

* 27%

* 5%

Ground

Reflected from

the ground

Reflected from

the clouds

Absorbed in the

atmosphereDirect or

beam

radiation, on

the ground

* Approx. 55% of the total solar

radiation (429 Btu/hr-ft2)reaches earth or 100 – 200 Btu/hr-ft2

©2012 Stevens Institute of Technology

Slightly Hazy 0.8 25%Hazy 0.6 35%Overcast 0.4 55%

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Solar energy reaches the earth in the form of electromagnetic radiation extending from x-rays of 0.1µ to radio waves of 100 meters long.

99% of the energy is contained between 0.28µ and 4.96µ.

46% of the total energy is

contained in the visible

spectrum

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Wavelength characteristics of solar radiation are given for the edge of the atmosphere (dotted) and at the earth’s surface

The area under the dotted curve is the solar constant (429 Btu/hr-ft2) and the area under the solid curve is the solar energy at the earth’s surface of approx. 50% of the solar constant.

Cosmic λ X-Rays UV IR

Vis

ible Long

IRUHF VHFRadar SW Radio

10-9 10-7 10-5 10-2 10-1 101 103 105 107 109 1011 µm


Typical Commercially Available Solar Measuring Instruments

There are two different types of instruments for measuring solar radiation, the pyranometer and the pyrheliometer.

The pyranometer has a hemispherical view of the surroundings, which allows it to

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Thermal Detector (pyranometer)Thermo-couples generate a voltage proportional to the difference

between the hot and cold junctions. The hot junction is proportional to the incident

solar radiation.

Photovoltaic Detector (pyranometer)

Silicon solar cells generate a current proportional to the solar radiation

surroundings, which allows it to measure direct + diffused solar radiation. By using a shadow ring to block the direct sun’s rays, the pyranometer can also be used to only measure diffused sky radiation.

The pyrheliometer has a restricted view of the sun (about 5⁰), therefore it is used to measure direct or beam solar radiation by pointing it towards the sun.

Source: Textbook

©2012 Stevens Institute of Technology7

Solar Constant = 429.2 Btu/hr-ft2 (1.353 kW/m2) The radiation that strikes a 1 ft2

surface normal to the sun’s rays at the edge of the earth’s atmosphere (extraterrestrial) varies over the year.

Extraterrestrial Solar Radiation

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Effect of the time of year on the ratio of extraterrestrial radiation to the nominal solar constant (429 Btu/hr-ft2).

Approximately half the total extraterrestrialsolar radiation reaches earth or ≈ 100 – 200 Btu/hr-ft2

The monthly extraterrestrial solar radiation vs. latitude, shown in the Table of Slide #9, is the average radiation striking a horizontal surface at the edge of the earth’s atmosphere. The values are accurately calculated since there is no interference from meteorological or atmospheric conditions, thus, they are free of any diffused or reflected radiation. The radiation values are usually based on long term averages, over a period greater than 10 years.


Average Extraterrestrial Radiation On a Horizontal Surface vs. Lat N. - Based on a Solar Constant of 429 Btu/hr-ft2 (1.353kw/m2) -

CCSource: Textbook (Appendix 2) ©2012 Stevens Institute of Technology 9

Solar Radiation Data (Cont’d)

Hemispheric Solar Radiation

Measured solar radiation data is available from many sources throughout the world. Solar radiation data for the USA is primarily available from National Oceanic and Atmospheric Administration (NOAA) and the National Renewable Energy Laboratory (NREL). Slides #11- 24provide solar radiation data obtained from the references listed. As you will note, there are slight differences in the data reported for a given location. This is caused by the way the values were obtained; using remote sensing satellite data, on ground measurements made through local weather service operations and estimation of solar radiation data in the area under consideration when no measured data for that location exists.

CC©2012 Stevens Institute of Technology

Monthly Maps of Average Daily Solar Radiation

Slides #11 & #12 show a sampling of the monthly (January and February) average daily solar radiation on a horizontal surface for the US. These maps were published in 1964 by the US Department of Commerce Weather Bureau and are based on measurements taken at 116 stations over a period of 39 years. The maps show all usable solar radiation, direct and diffuse, measured on a horizontal surface on earth during a normal year. The data is given in Langleys, which can be converted to Btu/ft2 by multiplying the Langleys x 3.69.

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consideration when no measured data for that location exists.

To find solar radiation data anywhere around the world, find it in the NASA website: http://eosweb.larc.nasa.gov/sse/ Titled: Surface Meteorology and Solar EnergyA renewable energy resource web site sponsored by NASA's Earth Science Enterprise Program

Map of Average Daily Solar Radiation

(Beam + Diffused) on a Horizontal Surface

in January

(in Langleys)



(in Langleys)

1 Langley=3.69 Btu/day-ft2

For New Jersey:150 Langleys x 3.69 =

= 554 Btu/day-ft2

Source: Ref. 3, Pg. 378

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Map of Average Daily Solar Radiation

(Beam + Diffused) on a Horizontal Surface

in February

(in Langleys)



(in Langleys)

1 Langley=3.69 Btu/day-ft2

For New Jersey:225 Langleys x 3.69 =

= 830 Btu/day-ft2

Source: Ref. 3, Pg. 378

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Hourly Clear-Sky Solar Radiation on Ground (Btu/h-ft2)

(Beam + Sky Diffuse Radiation)

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Solar Position and Insolation Values for 40⁰⁰⁰⁰ N Lat.

Source: Textbook(Table A2.6c)

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Climatic Design Conditions Tables(For United States and Many Other Parts of the World)

Ref. #6, Ch. #39

CC14©2012 Stevens Institute of Technology

Climatic Design Conditions Table(For Newark International Airport)

Ref. #6, Ch. #39

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Average Daily Solar Radiation for a Given Month on a Horizontal Surface, Ibh + Idh

(MJ/m2-day) , 1 J/m2-day = 88x10-6 Btu/ft2-day

CC©2012 Stevens Institute of TechnologySource: Table A2.3b, Textbook 16


Average Daily Solar Radiation for a Given Month on a Tilted Surface, Ib + Id(Btu/ft2-day), 1 J/m2-day = 88x10-6 Btu/ft2-day

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©2012 Stevens Institute of TechnologySource: Ref. #2, Appendix 6 17

Weather Data (Newark, NJ)

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HS = normal daily value of total hemispheric solar radiation i (Ibh + Idh ) incident on a horizontal surface (Btu/ft2 -day).

VS = normal daily value of total hemispheric solar radiation (Ibv + Idv ) incident on a vertical, south-facing surface (Btu/ft2 -day)

TA = (Tmin +Tmax)/2 where Tmin and Tmax are monthly (or annual) normal of daily minimum and maximum ambient temperatures (ºF)

Dxx = monthly (or annual) normal heating degree-days below the base temperature xx (ºF-days).

KT =average monthly (or annual) clearness ratio, i.e., the ratio of total hemispheric radiation incident on a horizontal surface to the

extraterrestrial radiation incident on a horizontal surface.

LD = LAT - DEC, latitude minus mid-month solar declination (degrees).

Source: Ref. #8,Appendix D

Weather Data (Boston, MA)

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HS = normal daily value of total hemispheric solar radiation (Ibh + Idh ) incident on a horizontal surface (Btu/ft2 -day).

VS = normal daily value of total hemispheric solar radiation (Ibv + Idv ) incident on a vertical, south-facing surface (Btu/ft2 -day)

TA = (Tmin +Tmax)/2 where Tmin and Tmax are monthly (or annual) normal of daily minimum and maximum ambient temperatures (ºF)

Dxx = monthly (or annual) normal heating degree-days below the base temperature xx (ºF-days).

KT =average monthly (or annual) clearness ratio, i.e., the ratio of total hemispheric radiation incident on a horizontal surface to the

extraterrestrial radiation incident on a horizontal surface.

LD = LAT - DEC, latitude minus mid-month solar declination (degrees).

Source: Ref. #8,Appendix D

Terrestrial Solar Radiation Data

The radiation data shown in the following table for Newark, NJ (Slide #24) were modeled using hourly values of direct beam and diffuse horizontal solar radiation from the National Solar Radiation Data Base (NSRDB), Ref. 4. The data base contains hourly values of measured or modeled solar radiation and meteorological data for 239 stations in the United States for a period of 30 years, from 1961 – 1990. The data presented shows monthly and yearly average terrestrial solar radiation values for flat-plate and concentrating collectors, and were computed as follows:

Total Solar Radiation on Tilted Flat Collector, Ic = (Direct Beam Radiation, Ib) + (Diffused Sky Radiation, I ) + (Radiation Reflected From the Surface in Front of Collector, I )

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Radiation, Id ) + (Radiation Reflected From the Surface in Front of Collector, Ir)

or Ic = Ib cos + Id + Ir where, = incident angle

The diffuse radiation was modeled by Perez (Ref. 5) as follows:

Id = Idh [0.5(1- F1)(1 + cos ᵦ) + F1 a/b + F2sinᵦ

where, Idh =diffused solar horizontal radiationF1 = circumsolar anisotropy (directional dependent) coefficient, function of sky conditionF2 = horizon/zenith anisotropy (directional dependent) coefficient, function of sky condition

ᵦ = tilt of the collector from the horizontal

a = 0 or the cosine of the incident angle, whichever is greaterb = 0.087 or the cosine of the solar zenith angle, whichever is greater

θ θ



The ground-reflected radiation used in modeling the total solar radiation received by a flat –plate collector is:

Ir = 0.5ρIh(1 - cos ᵦ)

Where, Ih = global horizontal radiation

ᵦ = tilt angle of the collector from the horizontal

ρ= surface reflectivity in front of collector (assumed to be 0.2, avg. for green vegetation and some

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ρ= surface reflectivity in front of collector (assumed to be 0.2, avg. for green vegetation and some

soil. If snow, the reflectivity ranges from 0.35 for old snow to 0.95 for dry new snow.)

Since concentrating solar collectors have a small field of view, they do not collect diffuse or reflected radiation. Thus, the solar radiation received by concentrating collectors is:

Ic = Ib cos θ



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The above flat-plate collector types are considered in the Solar Radiation Data



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The above concentrating collector types are considered in the Solar Radiation Data


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Source: Solar Radiation Data Manual for Flat-Plate and Concentrating Collectors, National Renewable Energy Laboratory, 1994 (Ref. 4)

Solar Radiation Data Shown (Ic = Ib + Id + Ir)

Note: 1kwh/m2/day = 317.2 Btu/ft2-day


Normal to Tilted

Collector

Normal to Horizontal

Plane

Sun

Altitude Angle, α = Between the solar beam and

the horizontal plane.

Solar Azimuth Angle, αS = Between the True South line

and the horizontal solar beam projection

= Collector tilt from the horizontal

Incident Angle, i = Between the solar beam and normal to the tilted plane.

Incident Angle, i

ᵝ

Solar radiation

striking the

Ibh

Ibc

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True East

Horizontal Plane

ᵝ

Altitude Angle, α

Definition of solar and surface angles for a non-south-facing tilted surface

Zenith Angle, z = Between the normal to the horizontal plane and the solar beam

z = 90º - αSource: Textbook, Eq. 2.24Solar beam projected

onto the horizontal plane

striking the

horizontal plane


Normal to Horizontal

Plane

Sun

Incident Angle, i

Solar radiation

striking the

horizontal plane

bc

cos sin

Ii cos

sin

==

=

=

i

III

I

I

I

bcbhb

b

b

bh

α

α

Ibh

Ibc

Normal to Tilted

Collector

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Horizontal Plane

ᵝ

Altitude Angle, α

Solar beam projected

onto the horizontal plane

Radiation Striking a Surface at Given Incident Angles

bR

Factor sin

cosor

sin

cos

cos sin

=

=

=

Tilti

α

iII

i

bhbc

α

α


( )

SSS

S

S

PS

hLL

iFor

i

coscoscossinsin

cossinsincoscosII and

cossinsincoscoscos ,0

)0 south, faces panel (if AngleAzimuth Panel

cossinsincoscoscos

bhbc

P

PP

+

+=

+==

==

+−=

δδ

βαβαα

βαβααα

αα

βαβααα

Solar Radiation (Cont’d)

Source: Eq. 2.44, Textbook

Where, hs = solar hour angle


SSS

SS

SS

S

SSS

hδβ)(Lδβ)(Li

sametheyeldsthat

hLoftermsinwrittenbecanisurfacefacingsouthaFor

LLi

hwith

hLL

When

coscoscossinsincos

:isequation The n.computatio less with results

and , , cos ,

)sin(sin)cos(coscos

:in results noon)(solar 0 when above), (see 2.44 Eq.

coscoscossinsinsin

:Textbook from 2.28 Eq. combining

−+−=

−+−=

=

+=

δβ

δβδβ

δδα

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Source: Eq. 2.28, Textbook ˅

˅

Solar Radiation (Cont’d)

Solar altitude, α and solar azimuth, αs angles can be calculated from the law of cosines:

sin α = sin L sin δs + cos L cos δs cos hs Source: Textbook, Eq. 22.8

The altitude angle, α depends on solar declination, δs the latitude angle, L and the solar hour angle, hs

The solar hour angle, hs is defined below.

The solar azimuth angle, αs can also be calculated in relation to three angles; the solar declination, δs the solar hour angle, hs and the altitude angle, α

sin α = cos δ sin h / cos α Source: Textbook, Eq. 22.9

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sin αs = cos δs sin hs / cos α Source: Textbook, Eq. 22.9

At solar noon, hs = 0 and αs = 0, the above equation can be further simplified as:

α = 90º – (L - δs)

The solar hour angle, is based on the nominal time of 24 hrs the sun moves 360º around the earth or 360º/24 = 15º/hr. or (24hrs x 60)/360º = 4 minutes/degree.

Therefore, hs = 15º x hours from local solar noon Source: Textbook, Eq. 22.5

Sunrise and sunset times can be calculated by finding the hour angle, hs when α = 0

Therefore, in the above Eq. sin α = sin L sin δs + cos L cos δs cos hs, solving for hs with α = 0 results in Sunset/Sunrise Hour Angle, hss or hsr = ± cos-1(-tan L x tan δs) = Hour-angle from solar noon

Sunrise time / Sunset time = 12:00 noon ± (hss or hsr)(4 minutes/degree)

Solar Radiation on Tilted Surfaces

Ic = Ib,c + Id,c + Ir,c Assume Ir,c = 0

Where, Ic = total solar radiation striking collectorIb,c = solar beam radiation striking collectorId,c = sky diffuse radiation striking collectorIr,c = ground reflected solar radiation on collector

Ib,c = Ib(cos i) (Eq. 2.43, Textbook) where, Ib = solar beam radiation normal to sun’s raysI = CI cos2(β/2) (Eq. 2.45, Textbook) where, C = empirical sky diffuse factor (Slide #30)


b,c b b

Id,c = CIbcos2(β/2) (Eq. 2.45, Textbook) where, C = empirical sky diffuse factor (Slide #30)

Then, Ic = Ibcos i + CIbcos2(β/2) = Ib[cos i + Ccos2(β/2)]

Ih = Ibsinα + CIb = Ib(sinα + C) (Eq. 2.41, Textbook) where Ih = Ib,h + Id,h

or Ib = Ih/(sinα + C)

Therefore,

Where, β = collector tilt angleC = empirical sky diffuse factor (Slide #30)

sin α = sinL sinδs + cosL cosδs coshs

cos i = cosα cos(αs – αp) sinβ + sinα cosβ

+

+=

2coscos

sin

2 β

αCi

C

II h

c

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Solar Radiation on Tilted Surfaces (cont’d)

Sky diffuse factor (C) for the 21st day of each month for average atmospheric conditions at sea level for the United States


Source: Extracted from Table 2.4, Textbook

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Solar and Local Standard TimeMany of the solar energy calculations require the use of solar time, which differs from local standard time. Local time is the same in the entire time zone whereas solar time relates to the position of the sun with respect to the observer and that depends on the longitude where solar time is calculated. The relationship between solar time and local standard time can be expressed by the following equation:

Solar Time (ST) = Local Std. Time (LST) + Equation of Time (Eqt)/60 + (Long sm – Long local)/15

where, Solar Time (ST) and Local Std. Time (LST) are in hours.Equation of Time (Eqt) = The difference between solar time and local standard time,

as determined by a sundial (minutes).

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as determined by a sundial (minutes).Long sm = Longitude of the standard time meridian for the observer’s time zone

The longitude of the std. time meridians for North America are

Atlantic: 60º, Eastern: 75º, Central: 90º, Mountain: 105º, Pacific: 120º, Yukon:135º, Alaska-Hawaii:150º

Long local = Longitude of the observer’s position. Find from: http://www.ngdc.noaa.gov/geomagmodels/struts/declZipE;jsessionid=7BF5368E4D8BF89C7E4107053AB574D6

An approximate formula for the Equation of Time (Eqt), in minutes , is:

Eqt = -14.2 sin[π(n + 7)/111] for day no. in the year, “n”, between n = 1 and 106Eqt = 4.0 sin[π(n - 106)/59] for day no. in the year, “n”, between n = 107 and 166Eqt = - 6.5 sin[π(n - 166)/80] for day no. in the year,“n”, between n = 167 and 246Eqt = 16.4 sin[π(n - 247)/113] for day no. in the year, “n”, between n = 247 and 365

Note: The sin angle is in radians

A more precise Equation of Time (Eqt) can be found in the solar position program “solpos”, which can be down-loaded from the NREL website.


Solar and Local Standard Time (Cont’d)

Example: Find the solar time for Pittsburgh, PA on July 21 at 11:00 am LST (Local Std. Time)

Longitude = 80º (for Pittsburgh, PA)Standard Time Meridian = 75º (Eastern)Local Standard Time = 11:00 am

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Solar time (ST) = Local Std. Time (LST) + Equation of Time (Eqt)/60 + (Long sm - Long local)/15

Eqt = - 6.5 sin[π(n - 166)/80] (From Slide #31 for “n”, between n = 167 and 246, July 21 = 202th day)= - 6.5 sin[π(202 – 166)/80] Note: Angle in radians= - 6.42 minutes

Solar Time (ST) = 11.0 - 6.42/60 + (75 – 80)/15= 11.0 - 0.107 – 0.333= 10.56 hours or 10:34 am

If a given location is precisely over the standard meridian than Solar Time (ST) ≈ Local Std. Time (LST)


True N

True STrue S

Magnetic N

W DeclinationE Declination

The Isogonic Chart shows magnetic compass

True NNORTH AMERICA

Magnetic S

(-)(+)

Magnetic S

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The Isogonic Chart shows magnetic compass deviation from true north. Solar insolation table values are always shown for true south, therefore the south indicated by the magnetic compass must always be adjusted according to the Isogonic Chart, otherwise some loss of insolation may result; especially if the collectors are facing east of true south. For northern New Jersey, as an example, the Isogonic Chart shows a magnetic deviation of about -13º west of true north. True south must then be 13º west of indicated compass south.In this case, solar panels oriented to an uncorrected south compass reading would truly be facing about 13º east of true south.Some tolerance, up to 15º is acceptable. One should favor west of true south rather than east of true south.


Magnetic Declination (Cont’d)

The differences between magnetic north and true north (magnetic declination) are caused by deposits of iron ore or magnetite deep in the Earth's crust. Long term movements of these deposits result in slow changes to the magnetic field strength and direction, at the same point on Earth. The magnetic declination in a given area will change slowly over time, possibly as much as 2-2.5 degrees every hundred years or so, depending upon how far from the magnetic poles it is.

True north is the direction in which the north pole is located along the Earth's rotational axis.On the other hand, magnetic north is the direction toward which the compass needle points.


The magnetic declination for any location in the US can be found from:http://www.ngdc.noaa.gov/geomagmodels/struts/declZipE;jsessionid=7BF5368E4D8BF89C7E4107053AB574D6

The magnetic declination for any location on the globe can be found from:www.geo-orbit.org/sizepgs/magmapsp.html

Sample Problem:

If a compass points magnetic north in Phoenix, AZ, how many degrees east or west of the true north-south line is it pointing?

From the Isogonic Chart , the magnetic declination is 12º East. That is, the compass points the magnetic north to the East of the true north-south line.

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The Sun-Path Diagram

In order to determine the solar heat gain on a particular “surface” and the effects of the sun on shading the surface, it is important to know the sun’s position in the sky at any time of day during any month of the year. The Sun-Path Diagram provides such information as to the location of the sun at any solar time of day, during any month of the year, at any location (latitude). Two different versions of Sun-Path Diagram/Charts are presented here, which give similar results. For ease of presentation, both types of charts assume that the earth is stationary and the sun revolves around the earth.

The first chart is the projection of the sun’s movement across the sky on a horizontal plane and

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The first chart is the projection of the sun’s movement across the sky on a horizontal plane and is called a Sun-Path Diagram. The concentric circles on the diagram are solar altitude angles and the azimuth angles are shown around the outer circle, measured in 10⁰ increments from

due south. The declination angles are shown on the horizontal lines and the solar hours on the vertical lines. The Sun-Path Diagram can be used to determine the times of sunrise and sunset, length of day and more importantly, for shading calculations. The Sun-Path Diagram will be used in the lecture that follows to determine if a particular obstruction at the site prevents the sun’s direct rays from reaching the “solar collector”.

The Sun-Path Diagram that follows was reproduced from the Textbook and shows the path of the sun for a latitude 40⁰N location, such as Newark, NJ. Since the sun’s path varies according

to the location on earth, different Sun-Path Diagrams are required for different latitudes. The textbook contains Sun-Path Diagrams for latitudes 25⁰N to 50⁰ N.


Azimuth Angle, αs

Altitude Angle, α

Horizon

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Declination Angle, δs

with respect to dates

shown below

Sun-Path Diagram (40⁰⁰⁰⁰ N Lat)(Horizontal Projection)

Source: Textbook

(Appendix 2)


The Sun-Path Chart

The second chart, less complicated, is based on the vertical projection of the sun traveling across the sky as viewed from earth, similar to its movement across a skydome, and is called a Sun-Path Chart.

The heavy curved lines represent the sun’s path across the sky on the 21st of a given month. The horizontal lines show the solar altitude angles and the vertical lines are the solar azimuth angles. It should be noted that the solar azimuth angles, shown are measured east or west from true south, which is determined by the observer’s location on the Isogonic

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or west from true south, which is determined by the observer’s location on the Isogonic Chart. That is, if the observer is located on the East side of the 0⁰ variation line, shown on

the Isogonic Chart, then the magnetic north will be pointing west of true north and magnetic south will be pointing east of true south by the number of degrees indicated.

Since the chart shows solar time, convert the standard time or daylight savings time, at the observer’s location, to solar time, using the equation shown in the previous slides.

The Sun-Path Chart that follows was reproduced from Ref. 3 and shows the path of the sun for the 21st day of a given month for a latitude 40⁰N location, such as Newark, NJ. Since the

sun’s path varies according to the location on earth, different Sun-Path Charts are required for different latitudes. Reference 3 contains more Sun-Path Charts, all for the United States (excluding Alaska) and southern Canada, for latitudes 28⁰ to 56⁰ N.


Vertical Projection Sun-Path Chart (40⁰⁰⁰⁰ N Lat)- Locate the Position of the Sun at Any Time of Day, During Any Month -

, α

Sun’s Path

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Typical Solar

Time

Azimuth

Angle,αs

True South True WestTrue EastSource: Ref. 3

Horizon


Sample Problem #1:

Calculate the declination (δs), the zenith angle (z), and the azimuth angle (αs) of the sun for New York City (Latitude 40.77ºN) on October 1 at 2:00 pm solar time.

δs = 23.45⁰ sin[360(284+n)/365]⁰ Where, n = the day no. in the year, Oct. 1 = 274th day

δs = 23.45⁰ sin[360(284+274)/365]⁰ (see Slide #3)

δs = - 4.22º

z = 90º - α (see Slide #25)

CC©2012 Stevens Institute of Technology39

z = 90º - α (see Slide #25)

Since, sin α = sin L sin δs + cos L cos δs cos hs (see Slide #28)and Solar Hour Angle, hs= 15º x Hours from local solar noon = 15 x 2.0 = 30º (see Slide #28)

Then, sin α = sin 40.77º sin (- 4.22º) + cos 40.77º cos (- 4.22º) cos 30º and Altitude Angle, α = 37.3º

Thus, Zenith Angle, z = 90º - 37.3º = 52.7º

sin αs = cos δs sin hs / cos α (see Slide #28)

sin αs = cos(-4.22º) sin 30º / cos 37.3º

Azimuth Angle, αs = 38.8º

Sample Problem #2:

Predict the hourly beam and diffuse radiation on a horizontal surface for Newark, NJ(latitude 40⁰N) on September 21 at 1:00 pm local std. time, on a clear day.

Convert the LST to solar time.Longitude = 74.2ºStandard Time Meridian = 75º (Eastern)Local Standard Time = 1:00 pm

Solar time (ST) = Local Std. Time (LST) + Equation of Time (E )/60 + (Long - Long )/15


Solar time (ST) = Local Std. Time (LST) + Equation of Time (Eqt)/60 + (Long sm - Long local)/15

Eqt = 16.4 sin[π(n - 247)/113] (From Slide #31 for “n”, between n = 247 and 365, Sep. 21 = 264th day)= 16.4 sin[π(264 - 247)/113] Note: Angle in radians= 7.46 minutes

Solar Time (ST) = 1.0 + 7.46/60 + (75 - 74.2)/15= 1.0 + 0.124 + 0.053= 1.177 pm or 1:11 pm

From Table on Slide #13 the Hourly Clear-Sky Solar Radiation (Beam + Sky Diffuse Radiation) for latitude 40⁰N and a solar time of 1:11 pm is 239 Btu/hr-ft2

40

Homework Problems – Lecture 3

3.1 What are the altitude and azimuth angles of the sun on the 21st day of May at Newark, NJ at 11:00 am EST? Do not use the Sun-Path Diagram to solve the problem.What is the magnetic declination of the azimuth angle at this location?

3.2 At what time does the sun set in Calcutta, India (Lat. 22º34’N) on May 1 and December 1?

CC

3.3 If a compass points magnetic north in Cleveland, Ohio, how many degrees east or west of true north-south line is it pointing?

3.4 Predict the hourly beam and diffuse radiation on a horizontal surface for Denver, CO (Lat. 40⁰N) on September 9 at 9:30 am (LST) on a clear day.

Study the Slides and Read Ch. #2 in “Principles of Solar Engineering”


solar course 1 - lecture 3

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