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Solar Energy Engineering

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An introductory level presentation for the undergraduate engineering students.

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Solar Energy Engineering

  1. 1. Fundamentals of Solar Energy Engineering Naveed ur Rehman www.naveedurrehman.com Note: Reference information is available at Google Drive/Solar Energy Engineering
  2. 2. 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
  3. 3. Famous quotes about Solar energy "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
  4. 4. Famous quotes about Solar energy “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
  5. 5. 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
  6. 6. 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
  7. 7. 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
  8. 8. Chapter #1 The Light Naveed ur Rehman www.naveedurrehman.com
  9. 9. 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
  10. 10. 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. Chapter #1: The Light 10
  11. 11. Light Spectrum Chapter #1: The Light 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
  12. 12. Light Spectrum Chapter #1: The Light 12
  13. 13. Photon Flux Chapter #1: The Light 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.
  14. 14. Power Density Chapter #1: The Light 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
  15. 15. Photon Flux and Power Density Chapter #1: The Light 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
  16. 16. Spectral Irradiance Chapter #1: The Light 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
  17. 17. Spectral Irradiance Chapter #1: The Light 17
  18. 18. Radiant Power Density Chapter #1: The Light 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
  19. 19. Blackbody Radiations Chapter #1: The Light 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. 20. Blackbody Radiations Chapter #1: The Light 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. 21. Blackbody Radiations Chapter #1: The Light 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. 22. Blackbody Radiations Chapter #1: The Light 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. 23. Blackbody Radiations Chapter #1: The Light 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. 24. Blackbody Radiations Chapter #1: The Light 24
  25. 25. Blackbody Radiations Chapter #1: The Light 25
  26. 26. Solar Radiations Spectrum Chapter #1: The Light 26
  27. 27. Chapter #2 Solar Geometry Naveed ur Rehman www.naveedurrehman.com
  28. 28. 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
  29. 29. The earth • Very small as compared to sun • Rotate about its own axis (day) • Revolve around the sun in orbit (year) Chapter #2: Solar Geometry 29 Rotation Revolution
  30. 30. 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 Chapter #2: Solar Geometry 30 152 MKm 147 MKm
  31. 31. Earth’s geometry Chapter #2: Solar Geometry Northern Hemisphere Southern Hemisphere 31
  32. 32. Earth’s geometry Locating position on earth: Chapter #2: Solar Geometry Φ : Latitude L : Longitude Unit: Degrees X Prime-meridian at Greenwich (L=0°) Equator (ø=0°) 32
  33. 33. 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? Chapter #2: Solar Geometry X N S EW 33
  34. 34. 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) Chapter #2: Solar Geometry 34
  35. 35. Magnetic compass directions Chapter #2: Solar Geometry • 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
  36. 36. Magnetic declination Chapter #2: Solar Geometry • 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
  37. 37. Magnetic declination Chapter #2: Solar Geometry (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)
  38. 38. Date and day Chapter #2: Solar Geometry • 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)
  39. 39. Sun position from earth Chapter #2: Solar Geometry • 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
  40. 40. Solar noon Chapter #2: Solar Geometry 40 Solar noon is the time when sun is highest above the horizon on that day
  41. 41. Solar altitude angle Chapter #2: Solar Geometry 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
  42. 42. Solar altitude angle at noon Chapter #2: Solar Geometry E αs,noon W Solar altitude angle is maximum at “Noon” for a day, denoted by αs,noon S N In northern hemisphere 42
  43. 43. • Zenith angle (θz) is the angle between vertical and the line passing through sun • θz = 90 – αs Zenith angle Chapter #2: Solar Geometry E θz W S N In northern hemisphere 43
  44. 44. Zenith angle at noon Chapter #2: Solar Geometry • Zenith angle is minimum at “Noon” for a day, denoted by θz,noon • ϴz,noon = 90 – αs,noon E θz,noon W S N In northern hemisphere 44
  45. 45. Air mass Chapter #2: Solar Geometry • 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
  46. 46. Air mass Chapter #2: Solar Geometry 46 𝐴. 𝑀. = Τ1 cos 𝜃𝑧
  47. 47. Solar azimuth angle Chapter #2: Solar Geometry 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
  48. 48. Solar declination Chapter #2: Solar Geometry 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!
  49. 49. Solar declination (at solstice) Chapter #2: Solar Geometry 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
  50. 50. Solar declination (at equinox) Chapter #2: Solar Geometry 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
  51. 51. Solar declination Chapter #2: Solar Geometry N S ø ø ø Latitude from frame of reference of horizontal ground beneath feet 51
  52. 52. Solar declination Chapter #2: Solar Geometry 90 - φ +23.45° -23.45° φ W NS E December solstice Equinox June solstice Note: Altitude depends upon latitude but declination is independent. In northern hemisphere Declination angles 52
  53. 53. 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 Chapter #2: Solar Geometry 53
  54. 54. 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! Chapter #2: Solar Geometry 54
  55. 55. Solar declination Chapter #2: Solar Geometry 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
  56. 56. 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 Chapter #2: Solar Geometry 56
  57. 57. 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 Chapter #2: Solar Geometry 57
  58. 58. Solar time • The time in your clock (local time) is not same as “solar time” • It is always “Noon” at 12:00pm solar time Chapter #2: Solar Geometry Solar time “Noon” Local time (in your clock) 58
  59. 59. 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) Chapter #2: Solar Geometry Try: http://www.powerfromthesun.net/soltimecalc.html59
  60. 60. 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/ Chapter #2: Solar Geometry 60
  61. 61. Solar time • The term Equation of time (E) is because of earth’s tilt and orbit eccentricity. • It can be calculated as: Chapter #2: Solar Geometry 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
  62. 62. 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) Chapter #2: Solar Geometry 11:00am ω = -15° 12:00pm ω = 0° 01:00pm ω = +15° 62
  63. 63. 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 Chapter #2: Solar Geometry 63
  64. 64. 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 Chapter #2: Solar Geometry 64
  65. 65. 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) Chapter #2: Solar Geometry E W γ S β (γ = -ve) (γ = +ve) (γ = 0) N 65
  66. 66. Summary of solar angles Chapter #2: Solar Geometry 66Can you write symbols of different solar angles shown in this diagram?
  67. 67. Interpretation of solar angles Chapter #2: Solar Geometry 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#
  68. 68. 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 Chapter #2: Solar Geometry E W S N θ 68
  69. 69. 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. Chapter #2: Solar Geometry 69
  70. 70. 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 Chapter #2: Solar Geometry 70(Set 1+2+4)
  71. 71. 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. Chapter #2: Solar Geometry 71(Set 1+3+4)
  72. 72. 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. Chapter #2: Solar Geometry 72
  73. 73. 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: Chapter #2: Solar Geometry 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.
  74. 74. 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 Chapter #2: Solar Geometry 74
  75. 75. 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 ∅ Chapter #2: Solar Geometry 75
  76. 76. Sun path diagram or sun charts Chapter #2: Solar Geometry Note: These diagrams are different for different latitudes. αs γs 76 -150° -120° -90° - 60° -30° 0° 30° 60° 90° 120° 150°
  77. 77. 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. Chapter #2: Solar Geometry 77
  78. 78. Shadow analysis (objects at distance) Chapter #2: Solar Geometry • 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
  79. 79. Inclinometer Chapter #2: Solar Geometry A simple tool for finding azimuths and altitudes of objects http://rimstar.org/renewnrg/solar_site_survey_shading_location.htm 79
  80. 80. Shadow analysis using sun charts Chapter #2: Solar Geometry αs γs 80 -150° -120° -90° - 60° -30° 0° 30° 60° 90° 120° 150°
  81. 81. 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. Chapter #2: Solar Geometry cos 𝜔𝑠 = − tan ∅ tan 𝛿 𝑁 = 2 15 𝜔𝑠 81
  82. 82. 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. Chapter #2: Solar Geometry 82
  83. 83. 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. Chapter #2: Solar Geometry 83
  84. 84. Profile angle • It is denoted by αp and can be calculated as follow: Chapter #2: Solar Geometry 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).
  85. 85. Profile angle • Collector-B will be in shade of collector-A, only when: Chapter #2: Solar Geometry 85 𝛼 𝑝 < ҧ𝛽
  86. 86. 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? Chapter #2: Solar Geometry 86
  87. 87. 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) Chapter #2: Solar Geometry 87
  88. 88. Chapter #3 Solar Radiations Naveed ur Rehman www.naveedurrehman.com
  89. 89. Types of solar radiations 1. Types by components: Total = Beam + Diffuse or Direct or Sky Chapter #3: Solar Radiations 89
  90. 90. Types of solar radiations 2. Types by terrestre: Extraterrestrial Terrestrial Chapter #3: Solar Radiations • 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
  91. 91. Measurement of solar radiations 1. Magnitude of solar radiations: Irradiance Irradiation/Insolation Chapter #3: Solar Radiations • 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
  92. 92. 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°) Chapter #3: Solar Radiations 92
  93. 93. 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 - Chapter #3: Solar Radiations 93
  94. 94. Exercise-1 Representation of solar radiations What are these symbols representing? Chapter #3: Solar Radiations 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
  95. 95. Extraterrestrial solar radiations Chapter #3: 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
  96. 96. 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) Chapter #3: Solar Radiations Gsc96
  97. 97. Spectral dist. of ext. terr. sol. rad. Chapter #3: Solar Radiations 97
  98. 98. Spectral dist. of ext. terr. sol. rad. Chapter #3: Solar Radiations 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
  99. 99. Ex.terr. irradiance at normal Extraterrestrial solar radiations received at normal. It deviates from GSC as the earth move near or away from the sun. 𝐺 𝑜𝑛 = 𝐺𝑠𝑐 1 + 0.033 cos 360𝑛 365 Chapter #3: Solar Radiations Gon99
  100. 100. 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 Chapter #3: Solar Radiations 100
  101. 101. Ex.terr. irradiance on horizontal Chapter #3: Solar Radiations Go101 Extraterrestrial solar radiations received at 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
  102. 102. 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) Chapter #3: Solar Radiations 102
  103. 103. 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 Chapter #3: Solar Radiations Io103
  104. 104. 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) Chapter #3: Solar Radiations 104
  105. 105. 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 Chapter #3: Solar Radiations Ho105
  106. 106. 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) Chapter #3: Solar Radiations 106
  107. 107. 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 Chapter #3: Solar Radiations Ho107
  108. 108. 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) Chapter #3: Solar Radiations 108
  109. 109. Terrestrial radiations Can be… • measured by instruments • obtained from databases e.g. TMY, NASA SSE etc. • estimated by different correlations Chapter #3: Solar Radiations 109
  110. 110. Terrestrial radiations measurement • Total irradiance can be measured using Pyranometer Chapter #3: Solar Radiations • Diffuse irradiance can be measured using Pyranometer with shading ring 110
  111. 111. Terrestrial radiations measurement • Beam irradiance can be measured using Pyrheliometer Chapter #3: Solar Radiations • Beam irradiance can also be measured by taking difference in readings of pyranometer with and without shadow band: beam = total - diffuse 111
  112. 112. Terrestrial radiations databases Chapter #3: Solar Radiations 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
  113. 113. Terrestrial radiations databases Chapter #3: Solar Radiations 2. TMY files: Information about hourly solar radiations can be obtained from Typical Meteorological Year files. (See “TMY” section in Reference Information) 113
  114. 114. Terrestrial irradiation estimation Chapter #3: Solar Radiations • Angstrom-type regression equations are generally used: ഥ𝐻 ഥ𝐻 𝑜 = 𝑎 + 𝑏 ത𝑛 ഥ𝑁 (See “Terrestrial Radiations Estimations” section in Reference Information) 114
  115. 115. Terrestrial irradiation estimation Chapter #3: Solar Radiations 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
  116. 116. Example-7 Terrestrial irradiation estimation What is the terrestrial monthly average daily irradiations on horizontal in the month of February, in Karachi (Latitude=24.8508°N) Chapter #3: Solar Radiations 116
  117. 117. Clearness index Chapter #3: Solar Radiations • 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
  118. 118. Clearness index Chapter #3: Solar Radiations 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
  119. 119. Frequency distribution of various radiation levels Chapter #3: Solar Radiations Example: For a month having average clearness index 0.4, 50% and less days will have clearness index of ≈0.4 119
  120. 120. 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? Chapter #3: Solar Radiations 120
  121. 121. Diffuse component of hourly irradiation (on horizontal) Chapter #3: Solar Radiations 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
  122. 122. 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? Chapter #3: Solar Radiations 122
  123. 123. Diffuse component of daily irradiation (on horizontal) Chapter #3: Solar Radiations 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
  124. 124. Diffuse component of monthly average daily irradiation (on horizontal) Chapter #3: Solar Radiations Collares-Pereira and Rabl correlation: ഥ𝐻 𝑑 ഥ𝐻 = 0.775 + 0.00606 𝜔𝑠 − 90 − ሾ0.505 124
  125. 125. Hourly total irradiation from daily irradiation (on horizontal) Chapter #3: Solar Radiations 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
  126. 126. Hourly diffuse irradiations from daily diffuse irradiation (on horizontal) Chapter #3: Solar Radiations 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
  127. 127. Air mass and radiations Chapter #3: Solar 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. 128. Air mass and radiations Chapter #3: Solar Radiations 128
  129. 129. Air mass and radiations Chapter #3: Solar Radiations • 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. 130. Air mass and radiations Chapter #3: Solar Radiations 130
  131. 131. Radiations on a tilted plane Chapter #3: Solar Radiations 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
  132. 132. Diffuse sky assumptions Chapter #3: Solar Radiations 132
  133. 133. Diffuse sky assumptions Chapter #3: Solar Radiations 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
  134. 134. General calculation model Chapter #3: Solar Radiations 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
  135. 135. Calculation models Chapter #3: Solar Radiations 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
  136. 136. Optimum tilt angle Chapter #3: Solar Radiations 136
  137. 137. Chapter #4 Flat-Plate Collectors Naveed ur Rehman www.naveedurrehman.com
  138. 138. 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
  139. 139. 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 Chapter #4:Flat-Plate Collectors
  140. 140. 140 Chapter #4:Flat-Plate Collectors
  141. 141. 141 Chapter #4:Flat-Plate Collectors
  142. 142. 142 Chapter #4:Flat-Plate Collectors
  143. 143. 143 Chapter #4:Flat-Plate Collectors In (cold) Out (hot) To tap
  144. 144. 144 Chapter #4:Flat-Plate Collectors Installation of flat-plate collectors at Mechanical Engineering Department, NED University of Engg. & Tech., Pakistan
  145. 145. 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] Chapter #4:Flat-Plate Collectors
  146. 146. 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 Chapter #4:Flat-Plate Collectors q2 q1 q2 q1
  147. 147. 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 Chapter #4:Flat-Plate Collectors
  148. 148. Heat transfer: Radiation Radiation heat transfer between two infinite parallel plates: Chapter #4:Flat-Plate Collectors 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
  149. 149. Heat transfer: Radiation Radiation heat transfer between a small object surrounded by a large enclosure: Chapter #4:Flat-Plate Collectors 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]
  150. 150. 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 Chapter #4:Flat-Plate Collectors 150
  151. 151. Heat transfer: Convection Convection heat transfer between parallel plates: Chapter #4:Flat-Plate Collectors 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. 152. Heat transfer: Convection Chapter #4:Flat-Plate Collectors 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
  153. 153. Example-2 Heat transfer: Convection 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 Chapter #4:Flat-Plate Collectors
  154. 154. Heat transfer: Conduction Conduction heat transfer through a material: Chapter #4:Flat-Plate Collectors 154 𝑹 𝒌 = Τ𝑳 𝒌 Where, L material thickness [m] k thermal conductivity [W/mK]
  155. 155. 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] Chapter #4:Flat-Plate Collectors Incoming Energy Useful Energy Energy Loss
  156. 156. Thermal network diagram 156 Chapter #4:Flat-Plate Collectors 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)
  157. 157. Thermal network diagram 157 Chapter #4:Flat-Plate Collectors 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
  158. 158. Cover temperature 158 Chapter #4:Flat-Plate Collectors 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)
  159. 159. Thermal resistances 159 Chapter #4:Flat-Plate Collectors 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
  160. 160. Numerical problem 160 Chapter #4:Flat-Plate Collectors 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
  161. 161. Solution methodology 161 Chapter #4:Flat-Plate Collectors 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
  162. 162. Problem solution 162 Chapter #4:Flat-Plate Collectors 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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