Solar Energy Engineering

Fundamentals of
Solar Energy Engineering
Naveed ur Rehman
www.naveedurrehman.com
Note: Reference information is available at Google Drive/Solar Energy Engineering

Famous quotes about Solar energy
"I’d put my money on the Sun and Solar Energy, what a source of
Power! I hope we don’t have to wait until oil and coal run out,
before we tackle that.“
Thomas Edison (Inventor of light bulb)
"I have no doubt that we will be successful in harnessing the
sun's energy. If sunbeams were weapons of war, we would have
had solar energy centuries ago."
George Porter (Nobel Prize winner in Chemistry, 1967)
2

"Because we are now running out of gas and oil, we must
prepare quickly for a third change, to strict conservation and to
the use of coal and permanent renewable energy sources, like
solar power."
Jimmy Carter (1977)
"We were delighted to have worked with Microsoft on its electric
solar system. Microsoft is effectively lowering operating costs,
reducing purchases of expensive peak electricity, and improving
the health and quality of life in California through its Silicon
Valley Campus solar power program.“
Dan Shuga (President of PowerLight)
3

“Solar energy, radiant light and heat from the sun, has
been harnessed by humans since ancient times using a
range of ever-evolving technologies. Solar energy
technologies include solar heating, solar photovoltaics,
solar thermal electricity and solar architecture, which
can make considerable contributions to solving some of
the most urgent problems the world now faces.”
"Solar Energy Perspectives: Executive Summary"
International Energy Agency (2011)
4

Contents
• Chapter #1: The Light
• Chapter #2: Solar Geometry
• Chapter #3: Solar Radiations
• Chapter #4: Flat-Plate Collectors
• Chapter #5: Concentrated Systems
• Chapter #6: Photovoltaic Systems
5

Recommended books
1. Solar Engineering of Thermal Processes
John A. Duffie & William A. Beckman
2. Solar Energy Engineering: Processes and
Systems
Soteris A. Kalogirou
3. Solar-thermal energy systems: analysis and
design
John R. Howell, Richard B. Bannerot &
Gary C. Vliet
4. Fundamentals of Renewable Energy
Processes
Aldo. V. Da. Rosa 6

Recommended websites
1. Power from the Sun
http://www.powerfromthesun.net/
2. PV Education
http://www.pveducation.org/
3. SolarWiki
http://solarwiki.ucdavis.edu/
7

Chapter #1
The Light
Naveed ur Rehman

Photon
• Today, quantum-mechanics
explains both the
observations of the wave
nature and the particle
nature of light.
• Light is a type of quantum-
mechanical particle, called
a Photon, which may also
be pictured as “wave-
packet”.
Chapter #1: The Light
9

Photon
• A photon is characterized by either a
wavelength (λ) or energy (E), such that:
or
h = Plank’s constant = 6.626 × 10 -34 joule·s
c = Speed of light = 2.998 × 108 m/s
Therefore, a short wavelength photon will posses
high energy content and vice versa.
10

Light Spectrum
11
Name Wave length (μm) Photon energy (eV)
Gamma rays > 10-6 > 106
X-rays 10-3 124 – 106
Ultraviolet (UV) 0.4 - 0.01 3.1 – 124
Visible (VIBGYOR) 0.75 – 0.4 1.65 – 3.1
Infrared (IR) 1000 – 0.75 0.0012 - 1.65

Light Spectrum
12

Photon Flux
13
• The photon flux is defined as the number of
photons per second per unit area:
• Note that the photon flux does not give
information about the characteristics of striking
photons i.e. energy or wavelength.

Power Density
14
• The power density is calculated by multiplying
the photon flux by the energy of a single
photon:
Where, q = Electronic charge = 1.6 x 10-19 joules

Photon Flux and Power Density
15
• The photon flux of high
energy (or short
wavelength) photons
needed to give a certain
power density will be
lower than the photon
flux of low energy (or long
wavelength) photons
required to give the same
power density.
Low
energy
photons
High
energy
photons

Spectral Irradiance
16
• The spectral irradiance (F) is given as a
function of wavelength λ, and gives the power
(energy per unit time) received by the surface
for a particular wavelength of light.
• It gives an idea of how much power is being
contributed from each wavelength.
• It is the most common way of characterizing a
light source.
or

Spectral Irradiance
17

Radiant Power Density
18
• The total power density emitted from a light
source can be calculated by integrating the
spectral irradiance over all wavelengths.
Where:
H = total power density emitted from the light
source = W/m2
F(λ) = spectral irradiance = W/m2μm
dλ = wavelength = μm

Blackbody Radiations
19
• A blackbody absorbs all radiation incident on
its surface and emits radiation based on its
temperature.
• Many commonly encountered light sources,
including the sun and incandescent light bulbs,
are closely modeled as "blackbody" emitters.
• The spectral irradiance from a blackbody is
given by Planck's radiation law, shown in the
following equation

20
• A blackbody absorbs all
radiation incident on its
surface and emits radiation
based on its temperature.
• Many commonly encountered
light sources, including the sun
and incandescent light bulbs,
are closely modeled as
"blackbody" emitters.

21
• The spectral irradiance from a blackbody is
given by Planck's radiation law:
Where:
λ = wavelength of light (m)
T = temperature of the blackbody (K)
F = spectral irradiance (W/m2m)
k = Boltzmann’s constant(1.380 × 10-23 joule/K)
c = speed of light (m/s), h = Plank’s constant (j.s)

22
• The total power density from a blackbody is
determined by integrating the spectral
irradiance over all wavelengths, which gives:
Where:
σ = Stefan-Boltzmann const. (5.67x10-8 W/m2K4)
T = temperature (K)

23
• The peak wavelength is the wavelength at
which the spectral irradiance is highest.
• It can be determined as:
• In other words, it is the wavelength where most
of the power is emitted.

24

25

Solar Radiations Spectrum
26

Chapter #2
Solar Geometry
Naveed ur Rehman

The sun
• Effective blackbody
temperature of 5777 K
• Hot because of continuous
fusion reactions:
e.g. H + H → He + (Heat
Energy)
• Pores and sunspots on sun
surface
Chapter #2: Solar Geometry
28

The earth
• Very small as compared to sun
• Rotate about its own axis (day)
• Revolve around the sun in orbit (year)
29
Rotation Revolution

Earth-sun distance
• Mean earth-sun distance is 1au (149.5Mkm)
• It varies by ± 1.7%
• This variation is not responsible for earth’s
seasons
30
152 MKm 147 MKm

Earth’s geometry
Northern
Hemisphere
Southern
Hemisphere
31

Earth’s geometry
Locating position on earth:
Φ : Latitude
L : Longitude
Unit: Degrees
X
Prime-meridian
at Greenwich
(L=0°)
Equator
(ø=0°)
32

Earth’s geometry
Where is Karachi on earth?
Latitude (Φ) : 24.8508°N
Longitude (L) : 67.0181°E
Try: ”Latitude Karachi” at Google
Q1) In which hemisphere, Karachi is located?
Q2) To which direction from Greenwich, Karachi
is located?
X
N
S
EW
33

Exercise-1:
Earth’s geometry
Where the following cities are located?
1) Sydney (Australia)
2) Nairobi (Kenya)
3) Balingen (Germany)
4) Jeddah (Saudi Arabia)
5) Oregon (USA)
6) Greenwich (UK)
34

Magnetic compass directions
• The magnetic poles are not
at the geographic poles.
• Directions shown by a
magnetic compass are not
the “Geographic” directions.
• All solar engineering
calculations are based on
geographic directions!
35

Magnetic declination
• Magnetic declination is the
angle between geographic
north (Ng) and magnetic north
(Nm).
• Nm is M.D. away from Ng.
• Facts:
– M.D in Alberta (Canada) is
approx. 16°W
– For Karachi, M.D. is almost 0°!
To get M.D: http://magnetic-declination.com/
36

Magnetic declination
(e.g. 16°W or -16°
i.e. Nm is 16°W of Ng) 37
(e.g. 16°E or 16°
i.e. Nm is 16°E of Ng)

Date and day
• Date is represented by month and ‘i’
• Day is represented by ‘n’
38
Month nth day for ith date
January i
February 31 + i
March 59 + i
… …
December 334 + i
(See “Days in Year” in Reference Information)

Sun position from earth
• Sun rise in the east and set in the west
• “A” sees sun in south
• “B” sees sun in north
N
S
EW
A
B
39

Solar noon
40
Solar noon is the time when
sun is highest above the
horizon on that day

Solar altitude angle
E
αs
W
• Solar altitude angle (αs) is the angle between
horizontal and the line passing through sun
• It changes every hour and every day
S N
In northern hemisphere
41

Solar altitude angle at noon
E
αs,noon
W
Solar altitude angle is maximum at “Noon” for a
day, denoted by αs,noon
S N
42

• Zenith angle (θz) is the angle between vertical
and the line passing through sun
• θz = 90 – αs
Zenith angle
E
θz W
S N
43

Zenith angle at noon
• Zenith angle is minimum at “Noon” for a day,
denoted by θz,noon
• ϴz,noon = 90 – αs,noon
E
θz,noon W
S N
44

Air mass
• Another representation of solar altitude/zenith
angle.
• Air mass (A.M.) is the ratio of mass of
atmosphere through which beam passes, to the
mass it would pass through, if the sun were
directly overhead.
𝐴. 𝑀. = Τ1 cos 𝜃𝑧
If A.M.=1 => θz=0° (Sun is directly overhead)
If A.M.=2 => θz=60° (Sun is away, a lot of mass of air
is present between earth and sun)
45

Air mass
46
𝐴. 𝑀. = Τ1 cos 𝜃𝑧

Solar azimuth angle
E
γs
W
• In any hemisphere, solar azimuth angle (γs) is
the angular displacement of sun from south
• It is 0° due south, -ve in east, +ve in west
Morning
(γs = -ve)
Evening
(γs = +ve)
Noon
(γs = 0°)
S
47

Solar declination
December solstice
Northern
hemisphere is away
from sun
(Winter)
June solstice
Northern
hemisphere is
towards sun
(Summer)
March equinox
Equator faces sun directly
(Spring)
September equinox
Equator faces sun directly
(Autumn) 48
Important!

Solar declination (at solstice)
A
A
A sees sun in north.
B sees sun overhead.
C sees sun in south.
C
B
A sees sun in south.
B sees sun in more south.
C sees sun in much more south.
N N
SS
June solstice December solstice
B
C
(Noon)
49

Solar declination (at equinox)
A sees sun directly overhead
B sees sun in more south
C sees sun in much more south
Same situation happen during
September equinox.
March equinox
A
C
N
S
B
(Noon) 50

Solar declination
N
S
ø
ø ø
Latitude from frame of reference of horizontal
ground beneath feet
51

Solar declination
90 - φ
+23.45°
-23.45°
φ
W
NS
E
December
solstice
Equinox
June
solstice
Note: Altitude depends upon latitude but declination is independent.
Declination
angles
52

Exercise-2:
Solar declination and altitude angle
What is the altitude of sun at noon in Karachi
(Latitude=24.8508) on:
1) Equinox
2) June solstice
3) December solstice
53

Solar declination
• For any day in year, solar declination (δ) can be
calculated as:
𝛿 = 23.45 sin 360
284 + 𝑛
365
Where, n = numberth day of year
(See “Days in Year” in Reference Information)
• Maximum: 23.45 °, Minimum: -23.45°
• Solar declination angle represents “day”
• It is independent of time and location!
54

Solar declination
Days to
Remember δ
March, 21 0°
June, 21 +23.45°
September, 21 0°
December, 21 -23.45°
Can you prove this?
55
δ
n

Solar altitude and zenith at noon
• As solar declination (δ) is the function of day
(n) in year, therefore, solar altitude at noon
can be calculated as:
αs,noon = 90 – ø + δ
• Similarly zenith angle at noon can be
calculated as:
ϴz,noon = 90 – αs,noon= 90 – (90 – ø + δ)= ø - δ
56

Exercise-3:
Solar declination and altitude angle
What is the altitude of sun at noon in Karachi
(Latitude=24.8508) on:
1) January, 12
2) July, 23
3) November, 8
57

Solar time
• The time in your clock (local time) is not same
as “solar time”
• It is always “Noon” at 12:00pm solar time
Solar time “Noon” Local time (in your clock) 58

Solar time
The difference between solar time (ST) and local
time (LT) can be calculated as:
𝑆𝑇 − 𝐿𝑇 = 𝐸 −
4 × 𝑆𝐿 − 𝐿𝐿
60
Where,
ST: Solar time (in 24 hours format)
LT: Local time (in 24 hours format)
SL: Standard longitude (depends upon GMT)
LL: Local longitude (+ve for east, -ve for west)
E: Equation of time (in hours)
Try: http://www.powerfromthesun.net/soltimecalc.html59

Solar time
• Standard longitude (SL) can be calculated as:
SL = (𝐺𝑀𝑇 × 15)
• Where GMT is Greenwich Mean Time, roughly:
If LL > 0 (Eastward):
𝐺𝑀𝑇 = 𝑐𝑒𝑖𝑙 Τ𝐿𝐿 15
If LL < 0 (Westward):
𝐺𝑀𝑇 = −𝑓𝑙𝑜𝑜𝑟 Τ𝐿𝐿 15
• GMT for Karachi is 5, GMT for Tehran is 3.5.
• It is recommended to find GMT from standard
database e.g. http://wwp.greenwichmeantime.com/
60

Solar time
• The term Equation of time (E) is because of
earth’s tilt and orbit eccentricity.
• It can be calculated as:
𝐸 =
229.2
60
×
0.000075
+0.001868 cos 𝐵
−0.032077 sin 𝐵
−0.014615 cos 2𝐵
−0.04089 sin 2𝐵
61
Where,
𝐵 = Τ𝑛 − 1 360 365

Hour angle
• Hour angle (ω) is another representation of
solar time
• It can be calculated as:
𝜔 = (𝑆𝑇 − 12) × 15
(-ve before solar noon, +ve after solar noon)
11:00am
ω = -15°
12:00pm
ω = 0°
01:00pm
ω = +15°
62

Exercise-4:
Solar time from local time
What is the solar time and hour angle in
Karachi (Longitude=67.0181°E) on 8 November
2:35pm local time?
Hint:
Find in sequence LT, LL, GMT, SL, n, B, E, ST and finally ω
Remember! You can always find solar time from local
time if you are given with longitude and day
63

Exercise-5:
Local time from solar time
At what local time, sun will be at noon in
Karachi on 8 November?
Hint:
Solar time is given in terms of “noon”. Find in sequence
ST, LL, GMT, SL, n, B, E and finally ST
64

A plane at earth’s surface
• Tilt, pitch or slope angle: β (in degrees)
• Surface azimuth or orientation: γ (in degrees,
0° due south, -ve in east, +ve in west)
E
W
γ
S
β
(γ = -ve)
(γ = +ve)
(γ = 0) N
65

Summary of solar angles
66Can you write symbols of different solar angles shown in this diagram?

Interpretation of solar angles
Angle Interpretation
Latitude φ Site location
Declination δ Day (Sun position)
Hour angle ω Time (Sun position)
Solar altitude αs Sun direction (Sun position)
Zenith angle θz Sun direction (Sun position)
Solar azimuth γs Sun direction (Sun position)
Tilt angle β Plane direction
Surface azimuth γ Plane direction
1
2
3
4
67
Set#

Angle of incidence
Angle of incidence (θ) is the angle between
normal of plane and line which is meeting plane
and passing through the sun
E
W
S N
θ
68

Angle of incidence
• Angle of incidence (θ) depends upon:
– Site location (1): θ changes place to place
– Sun position (2/3): θ changes in every instant
of time and day
– Plane direction (4): θ changes if plane is
moved
• It is 0° for a plane directly facing sun and at
this angle, maximum solar radiations are
collected by plane.
69

Angle of incidence
If the sun position is known in terms of
declination (day) and hour angle, angle of
incidence (θ) can be calculated as:
cos 𝜃
= sin 𝛿 sin ∅ cos 𝛽 − sin 𝛿 cos ∅ sin 𝛽 cos 𝛾
+ cos 𝛿 cos ∅ cos 𝛽 cos 𝜔
+ cos 𝛿 sin ∅ sin 𝛽 cos 𝛾 cos 𝜔
+ cos 𝛿 sin 𝛽 sin 𝛾 sin 𝜔
70(Set 1+2+4)

Angle of incidence
If the sun position is known in terms of sun
direction (i.e. solar altitude/zenith and solar
azimuth angles), angle of incidence (θ) can be
calculated as:
cos 𝜃 = cos 𝜃𝑧 cos 𝛽 + sin 𝜃𝑧 sin 𝛽 cos 𝛾𝑠 − 𝛾
Remember, θz = 90 – αs
Note: Solar altitude/zenith angle and solar azimuth
angle depends upon location.
71(Set 1+3+4)

Exercise-6:
Angle of incidence
Calculate angle of incidence on a plane located
in Karachi (Latitude=24.8508°N,
Longitude=67.018°E) at 10:30am (solar time)
on February 13, if the plane is tilted 45° from
horizontal and pointed 15° west of south.
Hint:
Convert given data into solar angles and then check
which equation for calculating θ suits best.
For geeks!
Solve the same problem if 10:30am is local time.
72

Special cases for angle of incidence
• If the plane is laid horizontal (β=0°)
–Equation is independent of γ (rotate!)
–θ becomes θz because normal to the plane
becomes vertical, hence:
cos 𝜃𝑧 = cos ∅ cos 𝛿 cos 𝜔 + sin ∅ sin 𝛿
Remember, θz = 90 – αs
73
Note: Solar altitude/zenith angle depends
upon location, day and hour.

Exercise-7:
Special cases for angle of incidence
Reduce equation for calculating angle of
incidence for the following special cases:
1. Plane is facing south
2. Plane is vertical
3. Vertical plane is facing south
4. A plane facing south and is tilted at angle
equals to latitude
74

Solar altitude and azimuth angle
Solar altitude angle (αs) can be calculated as:
sin𝛼 𝑠 = cos ∅ cos 𝛿 cos 𝜔 + sin ∅ sin 𝛿
Solar azimuth angle (γs) can be calculated as:
𝛾𝑠 = sign 𝜔 cos−1
cos 𝜃𝑧 sin ∅ − sin 𝛿
sin 𝜃𝑧 cos ∅
75

Sun path diagram or sun charts
Note: These diagrams are different for different latitudes.
αs
γs 76
-150° -120° -90° - 60° -30° 0° 30° 60° 90° 120° 150°

Exercise-8:
Sun path diagram or sun charts
Draw a sun path diagram for Karachi with lines
of June, 21 and December, 21.
Hint:
You need to calculate αs and γs for all hour angles of the
days mentioned in question.
77

Shadow analysis (objects at distance)
• Shadow analysis for objects at distance (e.g. trees,
buildings, poles etc.) is done to find:
– Those moments (hours and days) in year when
plane will not see sun.
– Loss in total energy collection due to above.
• Mainly, following things are required:
– Sun charts for site location
– Inclinometer
– Compass and information of M.D.
78

Inclinometer
A simple tool for finding azimuths and altitudes of objects
http://rimstar.org/renewnrg/solar_site_survey_shading_location.htm 79

Shadow analysis using sun charts
αs
γs 80
-150° -120° -90° - 60° -30° 0° 30° 60° 90° 120° 150°

Sunset hour angle and daylight hours
• Sunset occurs when θ z = 90° (or αs = 0°). Sunset hour
angle (ωs) can be calculated as:
• Number of daylight hours (N) can be calculated as:
For half-day (sunrise to noon or noon to sunrise),
number of daylight hours will be half of above.
cos 𝜔𝑠 = − tan ∅ tan 𝛿
𝑁 =
2
15
𝜔𝑠
81

Exercise-9:
Sunset hour angle and daylight hours
What is the sunset time (solar) on August, 14 in
Karachi (Latitude=24.8508°N) and Balingen
(Latitude=48.2753°N)? Also calculate number
of daylight hours for each city.
For geeks!
Also convert solar time (sunset hour angle) to local
time. You will need longitudes of these places.
82

Profile angle
It is the angle through which a plane that is initially
horizontal must be rotated about an axis in the plane of
the given surface in order to include the sun.
83

Profile angle
• It is denoted by αp and can be calculated as follow:
84
tan 𝛼 𝑝 =
tan 𝛼 𝑠
cos 𝛾𝑠 − 𝛾
• It is used in calculating shade of one collector (row) on
to the next collector (row).
• In this way, profile angle can also be used in calculating
the minimum distance between collector (rows).

Profile angle
• Collector-B will be in shade of collector-A, only when:
85
𝛼 𝑝 < ҧ𝛽

Exercise-10:
Profile angle and shading
According to figure, for a 25° profile angle, will
the collector-B be in the shade of collector-A?
86

Angles for tracking surfaces
• Some solar collectors "track" the sun
by moving in prescribed ways to
minimize the angle of incidence of
beam radiation on their surfaces and
thus maximize the incident beam
radiation.
• Tracking the sun is much more
essential in concentrating systems
e.g. parabolic troughs and dishes.
(See “Tracking surfaces” in Reference
Information)
87

Chapter #3
Solar Radiations
Naveed ur Rehman

Types of solar radiations
1. Types by components:
Total = Beam + Diffuse
or Direct or Sky
Chapter #3: Solar Radiations
89

Types of solar radiations
2. Types by terrestre:
Extraterrestrial Terrestrial
• Solar radiations
received on earth
without the presence
of atmosphere OR solar
radiations received
outside earth
atmosphere.
• We always calculate
these radiations.
• Solar radiations
received on earth in
the presence of
atmosphere.
• We can measure or
estimate these
radiations. Ready
databases are also
available e.g. TMY. 90

Measurement of solar radiations
1. Magnitude of solar radiations:
Irradiance Irradiation/Insolation
• Rate of
energy
(power)
received
per unit
area
• Symbol: G
• Unit: W/m2
Energy received per unit area in a
given time
Hourly: I
Unit: J/m2
Monthly avg. daily: H
Unit: J/m2
Daily: H
Unit: J/m2
91

Measurement of solar radiations
2. Tilt (β) and orientation (γ) of measuring
instrument:
– Horizontal (β=0°, irrespective of γ)
– Normal to sun (β=θz, γ= γs)
– Tilt (any β, γ is usually 0°)
– Latitude (β=ø, γ is usually 0°)
92

Representation of solar radiations
• Symbols:
–Irradiance: G
–Irradiations:
I (hourly), H (daily), H (monthly average daily)
• Subscripts:
–Ex.terr.: o Terrestrial: -
–Beam: b Diffuse: d Total -
–Normal: n Tilt: T Horizontal -
93

Exercise-1
Representation of solar radiations
What are these symbols representing?
1. Go
2. Gn
3. Gon
4. GT
5. GoT
6. G
7. Gb
1. Io
2. In
3. Ion
4. IT
5. IoT
6. I
7. Ib
1. Ho
2. Hn
3. Hon
4. HT
5. HoT
6. H
7. Hb
1. Ho
2. Hn
3. Hon
4. HT
5. HoT
6. H
7. Hb
A B C D
94

Extraterrestrial solar radiations
Solar
constant
(Gsc)
Irradiance
at normal
(Gon)
Irradiance at
horizontal
(Go)
Mathematical integration…
Hourly
irradiations on
horizontal
(Io)
Daily
irradiations on
horizontal
(Ho)
Monthly avg.
daily irrad. on
horizontal
(Ho)
95

Solar constant (Gsc)
Extraterrestrial solar radiations received at
normal, when earth is at an average distance
(1 au) away from sun.
𝐺𝑠𝑐 = 1367 Τ𝑊 𝑚2
Adopted by World Radiation Center (WRC)
Gsc96

Spectral dist. of ext. terr. sol. rad.
97

Spectral dist. of ext. terr. sol. rad.
98(See “Fraction of Solar Irradiance” in Reference Information)
To calculate solar irradiance in
particular wavelength range:
1. Take the difference of f0-λ
given against the
wavelengths.
2. Multiply the difference by
solar constant (=1367W/m2)
3. This can also be done by
taking difference of Gsc, λ

Ex.terr. irradiance at normal
normal. It deviates from GSC as the earth
move near or away from the sun.
𝐺 𝑜𝑛 = 𝐺𝑠𝑐 1 + 0.033 cos
360𝑛
365
Gon99

Example-2
Ex.terr. irradiance at normal
What is the extraterrestrial solar irradiance
at normal on the following days of year:
1. April, 1
2. October, 1
3. June, 10
4. December, 10
100

Ex.terr. irradiance on horizontal
Go101
horizontal. It is derived from Gon and
therefore, it deviates from GSC as the earth
move near or away from the sun.
𝐺 𝑜 = 𝐺 𝑜𝑛 × cos ∅ cos 𝛿 cos 𝜔 + sin ∅ sin 𝛿

Example-3
Ex.terr. irradiance on horizontal
What is the extraterrestrial solar irradiance
on horizontal at 10:00am on February 19, in
1. Karachi (Latitude=24.8508°N)
2. Balingen (Latitude=48.2753°N)
3. Nairobi (Latitude=1.2833°S)
102

Ex.terr. hourly irradiation on
horizontal
𝐼 𝑜
=
12 × 3600
𝜋
𝐺𝑠𝑐 × 1 + 0.033 cos
360𝑛
365
× ቈcos ∅ cos 𝛿 sin 𝜔2 − sin 𝜔1
Io103

Example-4
Ex.terr. hourly irradiation on horizontal
What is the extraterrestrial hourly solar
irradiations on horizontal between 10:00am
and 11:00am on February 19, in Karachi
(Latitude=24.8508°N)
104

Ex.terr. daily irradiation on
horizontal
𝐻 𝑜
=
24 × 3600
𝜋
𝐺𝑠𝑐 × 1 + 0.033 cos
360𝑛
365
× cos ∅ cos 𝛿 sin 𝜔𝑠 +
𝜋𝜔𝑠
180
sin ∅ sin 𝛿
Ho105

Example-5
Ex.terr. daily irradiation on horizontal
What is the extraterrestrial daily solar
irradiations on horizontal on February 19, in
Karachi (Latitude=24.8508°N)
106

Ex.terr. monthly average daily
irradiation on horizontal
ഥ𝐻 𝑜
=
24 × 3600
𝜋
𝐺𝑠𝑐 × 1 + 0.033 cos
360𝑛
365
× cos ∅ cos 𝛿 sin 𝜔𝑠 +
𝜋𝜔𝑠
180
sin ∅ sin 𝛿
Where day and time dependent parameters
are calculated on average day of a particular
month i.e. 𝑛 = ത𝑛
Ho107

Example-6
Ex.terr. mon. avg. daily irrad. on horiz.
What is the extraterrestrial monthly average
daily solar irradiations on horizontal in the
month of February, in Karachi
(Latitude=24.8508°N)
108

Terrestrial radiations
Can be…
• measured by instruments
• obtained from databases e.g. TMY, NASA
SSE etc.
• estimated by different correlations
109

Terrestrial radiations measurement
• Total irradiance can
be measured using
Pyranometer
• Diffuse irradiance can
be measured using
Pyranometer with
shading ring
110

Terrestrial radiations measurement
• Beam irradiance can
be measured using
Pyrheliometer
• Beam irradiance can
also be measured
by taking difference
in readings of
pyranometer with
and without
shadow band:
beam = total - diffuse
111

Terrestrial radiations databases
1. NASA SSE:
Monthly average daily total irradiation on
horizontal surface ( ഥ𝐻) can be obtained from
NASA Surface meteorology and Solar Energy
(SSE) Database, accessible from:
http://eosweb.larc.nasa.gov/sse/RETScreen/
(See “NASA SSE” in Reference Information)
112

Terrestrial radiations databases
2. TMY files:
Information about hourly solar radiations can be
obtained from Typical Meteorological Year files.
(See “TMY” section in Reference Information)
113

Terrestrial irradiation estimation
• Angstrom-type regression equations are
generally used:
ഥ𝐻
ഥ𝐻 𝑜
= 𝑎 + 𝑏
ത𝑛
ഥ𝑁
(See “Terrestrial Radiations Estimations” section in Reference
Information)
114

For Karachi:
ഥ𝐻
ഥ𝐻 𝑜
= 0.324 + 0.405
ത𝑛
ഥ𝑁
Where,
ത𝑛 is the representation of cloud
cover and ഥ𝑁 is the day length of
average day of month.
115

Example-7
What is the terrestrial monthly average
daily irradiations on horizontal in the month
of February, in Karachi (Latitude=24.8508°N)
116

Clearness index
• A ratio which mathematically represents sky
clearness.
=1 (clear day)
<1 (not clear day)
• Used for finding:
–frequency distribution of various radiation
levels
–diffuse components from total irradiations
117

Clearness index
1. Hourly clearness index:
𝑘 𝑇 =
𝐼
𝐼 𝑜
2. Daily clearness index:
𝐾 𝑇 =
𝐻
𝐻 𝑜
3. Monthly average daily clearness index:
ഥ𝐾 𝑇 =
ഥ𝐻
ഥ𝐻 𝑜
118

Frequency distribution of various
radiation levels
Example:
For a month
having average
clearness index
0.4, 50% and
less days will
have clearness
index of ≈0.4
119

Example-8
Freq. dist. of various rad. levels
What is the fraction of time, the days in
month of February in Karachi will have
clearness index less than or equals to 0.4?
120

Diffuse component of hourly
irradiation (on horizontal)
Orgill and Holland correlation:
𝐼 𝑑
𝐼
= ቐ
1 − 0.249𝑘 𝑇, 𝑘 𝑇 ≤ 0.35
1.557 − 1.84𝑘 𝑇, 0.35 < 𝑘 𝑇 < 0.75
0.177, 𝑘 𝑇 ≥ 0.75
Erbs et al. (1982)
121

Example-9
Diff. comp. of hourly irrad. (on horiz.)
How much is the hourly beam irradiations
when total hourly extraterrestrial and total
hourly terrestrial irradiations are 12 MJ/m2
and 6 MJ/m2?
122

Diffuse component of daily
Collares-Pereira and Rabl correlation:
𝐻 𝑑
𝐻
=
0.99, 𝐾 𝑇 ≤ 0.17
1.188 − 2.272𝐾 𝑇
+9.473𝐾 𝑇
2
−21.865𝐾 𝑇
3
+14.648𝐾 𝑇
4
, 0.17 < 𝐾 𝑇 < 0.75
−0.54𝐾 𝑇 + 0.632, 0.75 < 𝐾 𝑇 < 0.8
0.2, 𝐾 𝑇 ≥ 0.8
123

Diffuse component of monthly average
daily irradiation (on horizontal)
Collares-Pereira and Rabl correlation:
ഥ𝐻 𝑑
ഥ𝐻
= 0.775 + 0.00606 𝜔𝑠 − 90
− ሾ0.505
124

Hourly total irradiation from daily
For any mid-point (ω) of an hour,
𝐼 = 𝑟𝑡 𝐻
According to Collares-Pereira and Rabl:
𝑟𝑡 =
𝜋
24
𝑎 + 𝑏 cos 𝜔
cos 𝜔 − cos 𝜔𝑠
sin 𝜔𝑠 −
𝜋𝜔𝑠
180
cos 𝜔𝑠
Where,
𝑎 = 0.409 + 0.5016 sin 𝜔𝑠 − 60
𝑏 = 0.6609 − 0.4767 sin 𝜔𝑠 − 60
125

Hourly diffuse irradiations from daily
diffuse irradiation (on horizontal)
For any mid-point (ω) of an hour,
𝐼 𝑑 = 𝑟𝑑 𝐻 𝑑
From Liu and Jordan:
𝑟𝑑 =
𝜋
24
cos 𝜔 − cos 𝜔𝑠
sin 𝜔𝑠 −
𝜋𝜔𝑠
180
cos 𝜔𝑠
126

Air mass and radiations
• Terrestrial radiations depends upon the path
length travelled through atmosphere. Hence,
these radiations can be characterized by air
mass (AM).
• Extraterrestrial solar radiations are
symbolized as AM0.
• For different air masses, spectral distribution
of solar radiations is different.
127

128

• The standard spectrum at the Earth's surface
generally used are:
– AM1.5G, (G = global)
– AM1.5D (D = direct radiation only)
• AM1.5D = 28% of AM0
18% (absorption) + 10% (scattering).
• AM1.5G = 110% AM1.5D = 970 W/m2.
129

130

Radiations on a tilted plane
To calculate radiations on a tilted plane,
following information are required:
• tilt angle
• total, beam and diffused components of
radiations on horizontal (at least two of
these)
• diffuse sky assumptions (isotropic or
anisotropic)
• calculation model
131

Diffuse sky assumptions
132

Diffuse sky assumptions
Diffuse radiations consist of three parts:
1. Isotropic (represented by: iso)
2. Circumsolar brightening (represented by : cs)
3. Horizon brightening (represented by : hz)
There are two types of diffuse sky assumptions:
1. Isotropic sky (iso)
2. Anisotropic sky (iso + cs, iso + cs + hz)
133

General calculation model
𝑋 𝑇 = 𝑋 𝑏 𝑅 𝑏 + 𝑋 𝑑,𝑖𝑠𝑜 𝐹𝑐−𝑠 + 𝑋 𝑑,𝑐𝑠 𝑅 𝑏 + 𝑋 𝑑,ℎ𝑧 𝐹𝑐−ℎ𝑧 + 𝑋𝜌 𝑔 𝐹𝑐−𝑔
Where,
• X, Xb, Xd: total, beam and diffuse radiations
(irradiance or irradiation) on horizontal
• iso, cs and hz: isotropic, circumsolar and horizon
brightening parts of diffuse radiations
• Rb: beam radiations on tilt to horizontal ratio
• Fc-s, Fc-hz and Fc-g: shape factors from collector to sky,
horizon and ground respectively
• ρg: albedo
134

Calculation models
1. Liu and Jordan (LJ) model (iso, 𝛾=0°, 𝐼)
2. Liu and Jordan (LJ) model (iso, 𝛾=0°, ഥ𝐻)
3. Hay and Davies (HD) model (iso+cs, 𝛾=0°, 𝐼)
4. Hay, Davies, Klucher and Reindl (HDKR) model
(iso+cs+hz, 𝛾=0°, 𝐼)
5. Perez model (iso+cs+hz, 𝛾=0°,𝐼)
6. Klein and Theilacker (K-T) model (iso+cs, 𝛾=0°, ഥ𝐻)
7. Klein and Theilacker (K-T) model (iso+cs, ഥ𝐻)
(See “Sky models” in Reference Information) 135

Optimum tilt angle
136

Chapter #4
Flat-Plate Collectors
Naveed ur Rehman

Introduction
1. Flat-plate collectors are special type of
heat-exchangers
2. Energy is transferred to fluid from a
distant source of radiant energy
3. Incident solar radiations is not more
than 1100 W/m2 and is also variable
4. Designed for applications requiring
energy delivery up to 100°C above
ambient temperature.
138
Chapter #4:Flat-Plate Collectors

Introduction
1. Use both beam and diffuse solar
radiation
2. Do not require sun tracking and thus
require low maintenance
3. Major applications: solar water heating,
building heating, air conditioning and
industrial process heat.
139

140

141

142

143
In (cold)
Out (hot)
To tap

144
Installation of flat-plate collectors at Mechanical Engineering Department,
NED University of Engg. & Tech., Pakistan

Heat transfer: Fundamental
Heat transfer, in general:
145
𝑞 = 𝑄/𝐴 = Τ𝑇1 − 𝑇2 𝑅 = Τ∆𝑇 𝑅 = 𝑈∆𝑇[W/m2]
Where,
𝑇1 > 𝑇2: Heat is transferred from higher to lower
temperature
∆𝑇 is the temperature difference [K]
R is the thermal resistance [m2K/W]
A is the heat transfer area [m2]
U is overall H.T. coeff. U=1/R [W/m2K]

Heat transfer: Circuits
Resistances in series:
146
R1 R2T1
T2
q=q1=q2
Resistances in parallel:
T1 T2
q=q1+q2
R2
R1
𝑅 =
1
ൗ1
𝑅1
+ ൗ1
𝑅2
𝑈 = ൗ1
𝑅1
+ ൗ1
𝑅2
𝑅 = 𝑅1 + 𝑅2
𝑈 =
1
𝑅1 + 𝑅2
q2
q1
q2
q1

Example-1
Heat transfer: Circuits
Determine the heat transfer per unit area(q)
and overall heat transfer coefficient (U) for the
following circuit:
147
R1 R2T1 T2
R4
R3
q

Heat transfer: Radiation
Radiation heat transfer between
two infinite parallel plates:
148
𝑹 𝒓 = Τ𝟏 𝒉 𝒓
and,
𝒉 𝒓 =
𝝈 𝑻 𝟏
𝟐
+ 𝑻 𝟐
𝟐
𝑻 𝟏 + 𝑻 𝟐
𝟏
𝜺 𝟏
+
𝟏
𝜺 𝟐
− 𝟏
Where,
𝜎 = 5.67 × 10−8
W/m2K4
𝜖 is the emissivity of a plate

Heat transfer: Radiation
Radiation heat transfer between a
small object surrounded by a large
enclosure:
149
𝑹 𝒓 = Τ𝟏 𝒉 𝒓
and,
𝒉 𝒓 =
𝝈 𝑻 𝟏
𝟐
+ 𝑻 𝟐
𝟐
𝑻 𝟏 + 𝑻 𝟐
Τ𝟏 𝜺
= 𝜺𝝈 𝑻 𝟏
𝟐
+ 𝑻 𝟐
𝟐
𝑻 𝟏 + 𝑻 𝟐 [W/m2K]

Heat transfer: Sky Temperature
1. Sky temperature is denoted by Ts
2. Generally, Ts = Ta may be assumed
because sky temperature does not make
much difference in evaluating collector
performance.
3. For a bit more accuracy:
In hot climates: Ts = Ta+ 5°C
In cold climates: Ts = Ta+ 10°C
150

Heat transfer: Convection
Convection heat transfer between parallel plates:
151
𝑹 𝒄 = Τ𝟏 𝒉 𝒄 and 𝒉 𝒄 = Τ𝑵 𝒖 𝒌 𝑳
Where,
𝑵 𝒖 = 𝟏 +
𝟏. 𝟒𝟒 𝟏 −
𝟏𝟕𝟎𝟖 𝐬𝐢𝐧 𝟏. 𝟖𝜷 𝟏.𝟔
𝑹 𝒂 𝒄𝒐𝒔 𝜷
𝟏 −
𝟏𝟕𝟎𝟖
𝑹 𝒂 𝒄𝒐𝒔 𝜷
+
+
𝑹 𝒂 𝐜𝐨𝐬 𝜷
𝟓𝟖𝟑𝟎
Τ𝟏 𝟑
− 𝟏
+
Note: Above is valid for tilt angles between 0° to 75°.
‘+’ indicates that only positive values are to be
considered. Negative values should be discarded.

152
𝑅 𝑎 =
𝑔𝛽′∆𝑇𝐿3
𝜗𝛼
also 𝑃𝑟 = Τ𝜗 𝛼
Where,
Fluid properties are evaluated at mean temperature
Ra Rayleigh number
Pr Prandtl number
L plate spacing
k thermal conductivity
g gravitational constant
β‘ volumetric coefficient of expansion
for ideal gas, β‘ = 1/T [K-1]
𝜗, α kinematic viscosity and thermal diffusivity

Example-2
Find the convection heat transfer coefficient
between two parallel plates separated by 25 mm
with a 45° tilt. The lower plate is at 70°C and the
upper plate is 50°C.
Hint:
Calculate k, T, β’, 𝜗, α at mean temperature using air
thermal properties table. Then calculate Ra, Nu and
finally h.
153

Heat transfer: Conduction
Conduction heat transfer through a
material:
154
𝑹 𝒌 = Τ𝑳 𝒌
Where,
L material thickness [m]
k thermal conductivity [W/mK]

General energy balance equation
155
In steady-state:
Useful Energy = Incoming Energy – Energy Loss [W]
𝑸 𝒖 = 𝑨 𝒄 𝑺 − 𝑼 𝑳 𝑻 𝒑𝒎 − 𝑻 𝒂
Ac = Collector area [m2]
Tpm = Absorber plate temp. [K]
Ta = Ambient temp. [K]
UL = Overall heat loss coeff. [W/m2K]
Qu = Useful Energy [W]
SAc = Incoming (Solar) Energy [W]
AcUL(Tpm-Ta) = Energy Loss [W]
Incoming Energy
Useful Energy
Energy
Loss

Thermal network diagram
156
Ta
Ta
Ambient (a)
Cover (c)
Structure (b)
QuS
Rr(c-a) Rc(c-a)
Rr(p-c) Rc(p-c)
Rr(b-a) Rc(b-a)
Rk(p-b)
a
c
p
b
a
Plate (p)
Top losses
Bottom losses
qloss (top)
qloss (bottom)

Thermal network diagram
157
Ta
Ta
QuS
Rr(c-a) Rc(c-a)
Rr(p-c) Rc(p-c)
Rr(b-a) Rc(b-a)
Rk(p-b)
a
c
p
b
a
R(c-a) =1/(1/Rr(c-a) + 1/Rc(c-a))
R(p-c) =1/(1/Rr(p-c) + 1/Rc(p-c))
Ta
Tc
Tp
R(b-a) =1/(1/Rr(b-a) + 1/Rc(b-a))
Rk(p-b)
Tb
Ta
QuS

Cover temperature
158
R(c-a)
R(p-c)
Ta
Tc
Tp
R(b-a)
Rk(p-b)
Tb
Ta
QuS
1. Ambient and plate temperatures are
generally known.
2. Utop can be calculated as:
Utop= 1/(R(c-a) +R(p-c))
3. From energy balance:
qp-c = qp-a
(Tp-Tc)/R(p-c) = Utop(Tp-Ta)
=>Tc = Tp- Utop (Tp-Ta)x R(p-c)

Thermal resistances
159
Rr(c-a) =1/hr(c-a) = 1/εcσ(Ta
2+Tc
2) (Ta+Tc)
Rc(c-a) = 1/hc(c-a) = 1/hw
Rr(p-c) = 1/hr(p-c) = 1/[σ(Tc
2+Tp
2) (Tc+Tp)/(1/εc+ 1/εp-1)]
Rc(p-c) =1/hc(p-c) = 1/hc
Rk(p-b) = L/k
Rr(b-a) =1/hr(b-a) = 1/εbσ(Ta
2+Tb
2) (Ta+Tb)
Rc(b-a) =1/hc(b-a) = 1/hw

Numerical problem
160
Calculate the top loss coefficient for an absorber with a
single glass cover having the following specifications:
Plate-to-cover spacing: 25mm
Plate emittance: 0.95
Ambient air and sky temperature: 10°C
Wind heat transfer coefficient: 10 W/m2°C
Mean plate temperature: 100°C
Collector tilt: 45°
Glass emittance: 0.88

Solution methodology
161
Utop= 1/(R(c-a) + R(p-c))
R(c-a) =1/(1/Rr(c-a) + 1/Rc(c-a)) R(p-c) =1/(1/Rr(p-c) + 1/Rc(p-c))
Rr(c-a) =1/hr(c-a)
= 1/εcσ(Ta
2+Tc
2) (Ta+Tc)
Rc(c-a) = 1/hc(c-a) = 1/hw
Rr(p-c) = 1/hr(p-c)
= 1/[σ(Tc
2+Tp
2) (Tc+Tp)/(1/εc+ 1/εp-1)]
Rc(p-c) =1/hc(p-c) = 1/hc
Cover to ambient Plate to cover
Radiation Radiation
Convection Convection
Assume Tc between ambient and absorber
plate temperature
Tc = Tp- Utop (Tp-Ta)x R(p-c)
!Assumption validation

Problem solution
162
As per methodology, first assume Tc:
Tc = 35°C
And find following:
Rc(p-c) = 0.2681 m2°C/W
Rc(c-a) = 0.1000 m2°C/W
Rr(p-c) = 0.3136 m2°C/W
Rr(c-a) = 0.1938 m2°C/W
Utop = 6.49 W/m2°C
Validate assumption:
Tc = 48.5°C ≠ 35°C
Therefore, restart with:
Tc = 48.5°C
After few iterations, when assumption is validated:
Utop = 6.62 W/m2°C

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