Solar Radiation and Solar Geometry: Extraterrestrial and Ground Radiation, Instruments, Angles and Tilt Factor

This chapter, the first of two on solar thermal systems in the Engineering Sciences (XE) paper, answers the first part of the third topic of Section XE8.1 (Energy Resources and Conversion Technologies) of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Solar thermal systems: Solar radiation distribution and measurement, solar geometry, …”. It teaches extraterrestrial radiation, beam, diffuse and global radiation, the instruments that measure them, and solar geometry (declination, hour angle, zenith and incidence angles, day length, tilt factor). The numericals are extraterrestrial flux, declination, zenith angle, day length, solar time, tilt factor and radiation on a tilted plane.

1. The sun and extraterrestrial radiation

The sun radiates approximately like a blackbody at an effective temperature of about 5800 K, with most of its energy between 0.3 and 3 μm (ultraviolet, visible and near infrared), and the peak in the visible near 0.5 μm. The solar constant G_sc is the energy per unit time received on unit area of a surface perpendicular to the rays, outside the atmosphere, at the mean sun-earth distance. Its accepted value is about 1367 W/m². The earth's orbit is slightly elliptical, so the flux on a normal surface at the top of the atmosphere on day n of the year is

G_on = G_sc [1 + 0.033 cos(360 n/365)].

It is largest (about +3.3 per cent) in early January near n = 1, when the earth is closest to the sun, and smallest (about −3.3 per cent) in early July. Worked example for n = 185 with G_sc = 1367 W/m²: cos(360 × 185/365) = cos 182.5° = −0.999, so G_on = 1367 × (1 − 0.0330) = 1322 W/m². The effect of distance is small compared with the effect of the sun's altitude on what a given surface receives.

⚠️ Exam trap
The solar constant is on a surface normal to the rays. A horizontal surface at the top of the atmosphere receives G_on cos θz, with θz the zenith angle. Quoting 1367 W/m² as the flux on a horizontal surface is the usual error.

2. Beam, diffuse and global radiation at the ground

On its path through the atmosphere sunlight is absorbed (ozone in the ultraviolet, water vapour and carbon dioxide in the infrared) and scattered (by molecules and by dust and aerosol). Radiation that arrives without scattering, in a straight line from the sun, is beam (direct) radiation; scattered radiation arriving from the whole sky is diffuse radiation. On a horizontal surface global radiation G = G_b,n cos θz + G_d, where G_b,n is the beam normal to the rays and G_d is the diffuse part. The path length is measured by the air mass m = 1/cos θz for a plane atmosphere (m = 1 with the sun overhead, m = 2 at θz = 60°).

Daily totals use the clearness index K_T = H/H₀, the ratio of the daily global radiation on a horizontal surface to the extraterrestrial radiation on the same surface. Overcast skies have a low K_T and a high diffuse fraction. A monthly average of sunshine hours is related to radiation by a regression of the form H/H₀ = a + b (n_s/N), with n_s the bright sunshine hours and N the day length; the constants a and b are site-dependent and must be given.

ComponentMeaningDepends on
BeamUnscattered, directionalSun position, clear sky; falls to near zero under cloud
DiffuseScattered by atmosphere and clouds, from all of the skyCloud, aerosol, air mass; relatively larger on hazy days
GlobalBeam on the surface plus diffuse (plus ground-reflected on a tilted surface)Both of the above and the tilt
🎯 Why it matters
Concentrating collectors can use only beam radiation, because the optics need a definite direction. Flat plates use global radiation. That is why concentrators suit clear, dry regions and a flat plate also works in hazy ones.

3. Measurement of solar radiation

InstrumentMeasuresPrinciple and points
PyranometerGlobal (or diffuse) radiation on a horizontal plane, over the whole hemisphereBlackened thermopile under a glass dome (the temperature difference gives a voltage proportional to irradiance) or a photodiode type
Shaded pyranometerDiffuse radiationA shadow band or a small tracking disc blocks the sun's beam; a correction is applied for the sky that the band hides
PyrheliometerBeam radiation normal to the sun's raysA collimating tube with a narrow field of view on a sun tracker; only the sun and a small ring around it are seen
Sunshine recorderDuration of bright sunshine, in hoursA glass sphere focuses the sun to burn a trace on a card (Campbell-Stokes type); the length of the burn gives the hours

Global = beam on the horizontal + diffuse, so with a pyrheliometer and a shaded pyranometer the global value can be checked: G_h = G_b,n cos θz + G_d. A thermopile instrument has a calibrated sensitivity: for 10 μV per W/m², an output of 8.5 mV means 8.5 × 10⁻³/10 × 10⁻⁶ = 850 W/m². Keep the dome clean and level; calibration against a reference instrument is repeated periodically.

4. Solar geometry: declination, hour angle, zenith angle

The earth's axis is tilted 23.45° to its orbital plane, so the angle between the sun's rays and the equatorial plane, the declination δ, changes between +23.45° (about 21 June) and −23.45° (about 22 December), and is zero at the equinoxes. A good approximation (Cooper's equation) is δ = 23.45° sin[360 (284 + n)/365]. Worked example: n = 100 gives 360 × 384/365 = 378.7°, equal to 18.7°, so δ = 23.45 × sin 18.7° = 7.5°.

The hour angle ω is the angular displacement of the sun east or west of the local meridian because of the earth's rotation, 15° per hour, zero at solar noon, negative before noon and positive after: ω = 15° × (solar time − 12 h). At 10:00 solar time ω = −30°; at 15:30, ω = +52.5°. Solar time differs from clock time: solar time = standard time + 4 (L_loc − L_std) + E, with the minutes from the longitude difference in degrees east (4 minutes per degree) and E the equation of time in minutes (within about ±16 minutes). For L_loc = 77.2° E, L_std = 82.5° E and E = +3 min, 12:00 standard time is 12:00 − 21.2 + 3 = 11:41.8 solar time.

For a horizontal surface at latitude φ (north positive) the zenith angle θz, between the sun and the vertical, satisfies cos θz = sin φ sin δ + cos φ cos δ cos ω. Worked example: φ = 25°, δ = 15°, ω = −30° (10:00 solar time): cos θz = sin 25° sin 15° + cos 25° cos 15° cos 30° = 0.1094 + 0.7581 = 0.8675, so θz = 29.8°. The solar altitude is 90° − θz. At solar noon (ω = 0) this reduces to θz = |φ − δ|.

5. Sunrise, sunset and day length

At sunrise and sunset θz = 90°, so cos ω_s = −tan φ tan δ, where ω_s is the sunset hour angle (positive; sunrise is at −ω_s). The day length is N = 2 ω_s/15 hours with ω_s in degrees (or N = (24/π) ω_s with ω_s in radians). Worked example: φ = 40°, δ = 20°: tan 40° = 0.8391, tan 20° = 0.3640, cos ω_s = −0.3054, ω_s = 107.8°, so N = 2 × 107.8/15 = 14.4 hours.

  • δ = 0 (equinox): ω_s = 90° at every latitude, so day and night are each 12 hours.
  • Summer in the hemisphere (φ and δ of the same sign): tan φ tan δ > 0, ω_s > 90°, the day is longer than 12 hours.
  • Winter (opposite signs): the day is shorter than 12 hours.
  • On the equator (φ = 0) the day is 12 hours at all times of the year.

When |tan φ tan δ| exceeds 1 the sun does not set (midnight sun) or does not rise (polar night); there is no real ω_s. Atmospheric refraction and the finite diameter of the sun add a few minutes to real days and are neglected in the formula.

6. Incidence angle on a tilted surface and the tilt factor

A collector is tilted at slope β from the horizontal and faces a surface azimuth γ (zero facing the equator). The incidence angle θ between the beam and the surface normal is given by a long general expression; for the usual case of a surface facing the equator (γ = 0) it reduces to cos θ = sin(φ − β) sin δ + cos(φ − β) cos δ cos ω. This is the horizontal-surface formula with φ replaced by φ − β: tilting a surface toward the equator by β is like moving it to latitude φ − β. At solar noon θ = |φ − β − δ|.

The tilt factor for beam radiation, R_b = G_b,tilt/G_b,horizontal = cos θ/cos θz. Worked example at solar noon: φ = 30°, β = 30°, δ = −10°. θ = |30 − 30 + 10| = 10° and θz = |30 + 10| = 40°, so R_b = cos 10°/cos 40° = 0.9848/0.7660 = 1.29. In winter an equator-facing tilted surface catches much more beam than a horizontal one.

For total radiation on the tilted surface the isotropic-sky model is used: G_T = G_b R_b + G_d (1 + cos β)/2 + G ρ (1 − cos β)/2, where G_b, G_d and G are the horizontal beam, diffuse and global values, and ρ is the ground reflectance. The factors (1 + cos β)/2 and (1 − cos β)/2 are the fractions of sky and ground seen by the surface. Worked example: G_b = 600, G_d = 100, G = 700 W/m² (horizontal), R_b = 1.5, β = 30°, ρ = 0.2: G_T = 900 + 100 × 0.933 + 700 × 0.2 × 0.067 = 900 + 93.3 + 9.4 = 1003 W/m².

⚠️ Exam trap
R_b multiplies only the horizontal beam; do not multiply the global value by it. Also, keep ω negative before solar noon, and check that cos θ is positive: a negative cos θ means the sun is behind the surface and the beam on it is zero, not negative.

Key takeaways

  • G_sc = 1367 W/m² is the flux normal to the rays outside the atmosphere; G_on = G_sc[1 + 0.033 cos(360n/365)] varies by ±3.3 per cent, and a horizontal surface receives G_on cos θz.
  • Global = beam × cos θz + diffuse; air mass is 1/cos θz; K_T = H/H₀. Concentrators use only beam radiation; flat plates use beam, diffuse and reflected.
  • Pyranometer measures global (diffuse with a shade), pyrheliometer measures beam normal on a tracker, a sunshine recorder gives bright-sunshine hours.
  • δ = 23.45° sin[360(284 + n)/365]; ω = 15° (solar time − 12), negative before noon; cos θz = sin φ sin δ + cos φ cos δ cos ω; solar time = standard time + 4(L_loc − L_std) + E.
  • Sunset hour angle cos ω_s = −tan φ tan δ; day length N = 2ω_s/15 hours; 12 hours at equinox and at the equator.
  • For an equator-facing surface, cos θ = sin(φ − β) sin δ + cos(φ − β) cos δ cos ω and R_b = cos θ/cos θz; total tilted radiation uses the isotropic model G_b R_b + G_d(1 + cos β)/2 + Gρ(1 − cos β)/2.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The solar constant is defined as the energy received per unit time per unit area on:

    1. a horizontal surface at sea level on a perfectly clear day at the time of solar noon, after atmospheric absorption
    2. a surface perpendicular to the sun's rays outside the atmosphere, at the mean sun-earth distance
    3. a horizontal surface at the top of the atmosphere at the equator on the day of the equinox, averaged over the day
    4. a surface perpendicular to the sun's rays at the ground after the sunlight has been attenuated by the atmosphere
    Show answer

    Answer: B — a surface perpendicular to the sun's rays outside the atmosphere, at the mean sun-earth distance

    The definition is a surface normal to the rays, outside the atmosphere, at the mean distance. A horizontal surface or a ground-level surface gives a different number because of the cosine of the zenith angle and attenuation.

  2. Using G_on = G_sc[1 + 0.033 cos(360 n/365)], on which of the following does the extraterrestrial flux on a normal surface reach its maximum, and why?

    1. Early January, because the earth is then closest to the sun in its elliptical orbit
    2. Early July, because the northern hemisphere is then tilted toward the sun and receives its longest days
    3. 21 June, because the declination of the sun is then at its greatest value of 23.45 degrees
    4. At the two equinoxes, because day and night are then of equal length everywhere
    Show answer

    Answer: A — Early January, because the earth is then closest to the sun in its elliptical orbit

    The cosine term equals +1 at n = 1 (early January) when the earth is nearest the sun, giving +3.3 per cent. The seasons come from axial tilt, which changes the angle of incidence and day length, not the flux on a normal surface at the top of the atmosphere.

  3. For a plane parallel atmosphere, the air mass when the zenith angle of the sun is 60° is:

    1. 0.5
    2. 1.0
    3. 1.73
    4. 2.0
    Show answer

    Answer: D — 2.0

    m = 1/cos θz = 1/cos 60° = 1/0.5 = 2.0. The value 0.5 is cos 60° itself, taken instead of its reciprocal.

  4. Which instrument, mounted on a sun-tracker with a narrow collimating tube, measures the beam radiation normal to the sun's rays?

    1. Pyranometer, which responds to the radiation arriving from the whole sky dome
    2. Campbell-Stokes sunshine recorder, which burns a trace on a card while the sun shines
    3. Pyrheliometer, which views only the sun and a small ring of sky around it
    4. Albedometer, which compares the readings of upward and downward facing sensors
    Show answer

    Answer: C — Pyrheliometer, which views only the sun and a small ring of sky around it

    A pyrheliometer has a narrow field of view aimed at the sun and so measures beam normal radiation. A pyranometer sees the whole hemisphere and therefore measures global radiation.

  5. A pyranometer fitted with a shadow band or tracking shading disc is used to measure which quantity?

    1. Diffuse radiation on a horizontal surface, from the sky only
    2. Beam radiation normal to the sun's rays, received directly from the solar disc
    3. Duration of bright sunshine in hours during the day
    4. Net long-wave radiation lost from the ground to the sky
    Show answer

    Answer: A — Diffuse radiation on a horizontal surface, from the sky only

    Blocking the sun's disc leaves only scattered sky radiation, which is the diffuse component. Beam radiation is measured by the pyrheliometer, and sunshine hours by the sunshine recorder.

  6. A thermopile pyranometer has a calibrated sensitivity of 10 μV per W/m². Its output reads 8.5 mV. The irradiance is:

    1. 85 W/m²
    2. 1176 W/m²
    3. 850 W/m²
    4. 8.5 W/m²
    Show answer

    Answer: C — 850 W/m²

    Irradiance = output/sensitivity = 8.5 × 10⁻³ V/(10 × 10⁻⁶ V per W/m²) = 850 W/m². Multiplying the two (instead of dividing) or slipping a factor of ten gives the other values.

  7. What is the hour angle of the sun at 15:30 solar time?

    1. +52.5°
    2. −52.5°
    3. +37.5°
    4. +232.5°
    Show answer

    Answer: A — +52.5°

    ω = 15° × (15.5 − 12) = +52.5°; the hour angle is positive after solar noon. The negative sign is for the morning, and 37.5° would be for 14:30.

  8. A place at longitude 77.2° E uses a standard meridian of 82.5° E. At 12:00 standard time the equation of time is +3 minutes. Using solar time = standard time + 4(L_loc − L_std) + E, with longitudes in degrees east, the solar time is closest to:

    1. 12:24
    2. 11:36
    3. 12:18
    4. 11:42
    Show answer

    Answer: D — 11:42

    4 × (77.2 − 82.5) = −21.2 minutes (the place is west of the standard meridian, so its sun is behind the clock), and E = +3, so solar time = 12:00 − 21.2 + 3 = 11:41.8, which is 11:42. Reversing the longitude sign gives 12:24, and reversing the sign of E gives 11:36.

  9. At the equinox (declination zero), the length of the day, neglecting refraction and the size of the sun's disc, at a latitude of 45° N is:

    1. Greater than 12 hours, since the latitude of 45° is high and the sun stays up longer
    2. Exactly 12 hours, since cos ω_s = −tan φ tan 0 = 0 gives ω_s = 90°
    3. Less than 12 hours, since the sun is low in the sky at that latitude
    4. Equal to 12 hours only on the equator itself, and different at every other latitude
    Show answer

    Answer: B — Exactly 12 hours, since cos ω_s = −tan φ tan 0 = 0 gives ω_s = 90°

    With δ = 0, cos ω_s = 0, ω_s = 90° and N = 2 × 90/15 = 12 hours at every latitude. High latitude changes day length only when δ is non-zero.

  10. On a horizontal surface the beam, diffuse and global radiation are 600, 100 and 700 W/m². For a collector tilted at 30° with R_b = 1.5 and a ground reflectance 0.2, the isotropic-sky model gives a total radiation closest to:

    1. 900 W/m²
    2. 1003 W/m²
    3. 1050 W/m²
    4. 1200 W/m²
    Show answer

    Answer: B — 1003 W/m²

    G_T = 600 × 1.5 + 100 × (1 + cos 30°)/2 + 700 × 0.2 × (1 − cos 30°)/2 = 900 + 93.3 + 9.4 = 1003 W/m². The value 900 is the beam term alone, and multiplying the whole 700 by R_b gives 1050 plus extras, which is wrong.

  11. Which statements about radiation measurement are correct?

    1. A pyranometer measures radiation received over the whole hemisphere on its horizontal sensor
    2. A pyrheliometer must be kept pointed at the sun, so it is mounted on a tracker
    3. A sunshine recorder gives the intensity of radiation in W/m²
    4. Beam radiation on a horizontal plane is the pyrheliometer reading without any angle correction
    Show answer

    Answer: A — A pyranometer measures radiation received over the whole hemisphere on its horizontal sensor; B — A pyrheliometer must be kept pointed at the sun, so it is mounted on a tracker

    The pyranometer integrates over the hemisphere and the pyrheliometer must track the sun. A sunshine recorder gives hours, not W/m², and the beam on a horizontal plane is the normal reading multiplied by cos θz.

  12. Which statements about solar angles are correct?

    1. The declination varies between +23.45° and −23.45° over a year
    2. The hour angle changes by 15° each hour and is negative in the morning
    3. At solar noon the zenith angle on a horizontal surface equals |φ − δ|
    4. The solar altitude is equal to the zenith angle
    Show answer

    Answer: A — The declination varies between +23.45° and −23.45° over a year; B — The hour angle changes by 15° each hour and is negative in the morning; C — At solar noon the zenith angle on a horizontal surface equals |φ − δ|

    All three are standard results. The altitude is the complement of the zenith angle (90° − θz), not equal to it, so the statement equating altitude with zenith angle is false.

  13. For a surface facing the equator at latitude φ, tilted by β from the horizontal, which statements are correct?

    1. The incidence angle is given by the horizontal-surface formula with φ replaced by φ − β
    2. At solar noon, θ = |φ − β − δ|
    3. The beam tilt factor R_b is cos θz/cos θ
    4. R_b multiplies the global radiation on the horizontal to give the tilted total
    Show answer

    Answer: A — The incidence angle is given by the horizontal-surface formula with φ replaced by φ − β; B — At solar noon, θ = |φ − β − δ|

    The first two follow from tilting toward the equator being equivalent to a shift in latitude. R_b is cos θ/cos θz (the ratio is inverted in the statement that gives cos θz/cos θ), and it multiplies only the beam, so the statement about global radiation is false.

  14. Take G_sc = 1367 W/m² and G_on = G_sc[1 + 0.033 cos(360 n/365)]. Find the extraterrestrial flux on a surface normal to the sun's rays on day n = 185, in W/m², to the nearest integer.

    Numerical answer — type the value.

    Show answer

    Answer: 1322

    360 × 185/365 = 182.466°; cos 182.466° = −0.99907. G_on = 1367 × (1 − 0.033 × 0.99907) = 1367 × 0.96703 = 1321.9, so 1322 W/m². Python check: 1321.93.

  15. Using δ = 23.45° sin[360 (284 + n)/365], find the declination of the sun on day n = 100 of the year, in degrees, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 7.5

    The angle is 360 × 384/365 = 378.74°, which is 18.74° after removing 360°. sin 18.74° = 0.3213, so δ = 23.45 × 0.3213 = 7.53°, or 7.5°. Python check: 7.5338.

  16. A horizontal surface is at latitude 25° N on a day when the declination is 15°. Find the zenith angle of the sun at 10:00 solar time, in degrees, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 29.8

    ω = 15 × (10 − 12) = −30°. cos θz = sin 25° sin 15° + cos 25° cos 15° cos(−30°) = 0.42262 × 0.25882 + 0.90631 × 0.96593 × 0.86603 = 0.10938 + 0.75814 = 0.86752. θz = arccos 0.86752 = 29.83°, or 29.8°.

  17. Find the length of the day, in hours, at latitude 40° N when the declination is 20°, to one decimal place. Neglect atmospheric refraction and the size of the sun's disc.

    Numerical answer — type the value.

    Show answer

    Answer: 14.4

    cos ω_s = −tan 40° tan 20° = −0.83910 × 0.36397 = −0.30542, so ω_s = 107.78°. N = 2 × 107.78/15 = 14.37 hours, or 14.4 hours.

  18. At solar noon at latitude 30° N, the declination is −10°. A surface faces due south, the equator-facing direction, with a tilt of 30° from the horizontal. Find the beam tilt factor R_b = cos θ/cos θz, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 1.29

    At noon θ = |φ − β − δ| = |30 − 30 + 10| = 10° and θz = |φ − δ| = 40°. R_b = cos 10°/cos 40° = 0.98481/0.76604 = 1.2856, or 1.29.