Chapter 2 · 6 hours
Solar energy
Practice questions
Practice questions and answers
6 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.
- Practice · 5 marks
Explain the movement of the earth around the sun and the cause of seasons. Define solar constant, air mass, beam, diffuse and global radiation.
Answer
Movement of the earth
The earth makes two motions: rotation about its own axis once in 24 hours (causes day and night) and revolution around the sun in an elliptical orbit once in 365.25 days. The axis is tilted by 23.45° from the normal to the orbital plane and keeps a fixed direction in space.
The sun-earth distance varies from 1.471×10¹¹ m (about 3 January) to 1.521×10¹¹ m (about 4 July), a mean of 1.496×10¹¹ m (1 AU).
Cause of seasons
Because of the fixed tilt, the angle between the sun's rays and the surface at a place changes through the year. The declination (angle of the sun north or south of the equator) varies from +23.45° on 21 June (summer solstice) to −23.45° on 22 December, and is zero at the equinoxes (21 March, 23 September). When the northern hemisphere leans towards the sun it has longer days and more direct rays, i.e. summer. Distance from the sun is not the cause.
Declination (Cooper): , = day of year.
Definitions
- Solar constant : energy from the sun per unit time on a unit area perpendicular to the rays, at the mean earth-sun distance, outside the atmosphere. Value = 1367 W/m² (WRC value).
- Air mass : ratio of the path length of rays through the atmosphere to the path length when the sun is overhead. (for ). Air mass zero (AM0) is outside the atmosphere, AM1.5 is the standard for solar cells.
- Beam (direct) radiation: reaches the surface without scattering.
- Diffuse radiation: scattered by air molecules, dust and clouds and arriving from all directions.
- Global (total) radiation: sum of beam and diffuse on a surface, .
- Practice · 5 marks
Describe the working of a pyranometer, a pyrheliometer and a sunshine recorder. State what each one measures.
Answer
Solar radiation instruments convert incident radiation into a measurable signal, usually heat or voltage.
Pyranometer
Measures global radiation (beam + diffuse) on a horizontal surface; with a shade ring or disc it measures diffuse radiation only.
The Eppley type has a black and white sensing surface (thermopile) under two glass domes. Black parts absorb radiation and get hotter than the white parts, so the thermopile produces a small voltage (about 10 µV per W/m²) proportional to the radiation. The domes protect from wind and rain, and a desiccant keeps them dry. A photodiode type is cheaper but spectrally selective.
glass domes
____
/ \
|B|W|B|W| <- thermopile hot (black)
|______| and cold (white) junctions
heat sink --> voltage out
Pyrheliometer
Measures beam (direct) radiation at normal incidence. A long collimating tube with a narrow field of view (about 5–6°) is kept pointed at the sun by a tracker. The radiation falls on a thermopile or a blackened receiver. In the Angstrom compensation type, one of two strips is shaded and heated electrically until the temperatures equal; the electrical power then equals the absorbed radiation. It is also the reference instrument for calibration.
Sunshine recorder
Measures the duration of bright sunshine (hours), not intensity. In the Campbell-Stokes recorder a solid glass sphere focuses the sun's rays onto a card strip; the burn trace length gives the sunshine hours. The data are used with Angstrom regression to estimate radiation where no pyranometer is available.
| Instrument | Measures | Principle |
|---|---|---|
| Pyranometer | Global/diffuse, W/m² | Thermopile |
| Pyrheliometer | Beam, W/m² | Collimated thermopile |
| Sunshine recorder | Hours of sunshine | Burn trace |
- Practice · 8 marks
Calculate the declination, hour angle, solar altitude angle, zenith angle, solar azimuth angle, sunrise hour angle and day length at a place of latitude 27.7° N at 10:00 hours solar time on 15 May.
Answer
Data
N; 15 May is day number . Solar time = 10:00.
Declination
Hour angle
15° per hour from solar noon, negative in the morning:
Zenith and altitude
Solar azimuth
measured from south, towards the east (morning). That is, azimuth is 78.5° east of south.
Sunrise hour angle and day length
Sunrise is at solar time.
Answer: , , , altitude , azimuth east of south, , day length h.
- Practice · 6 marks
Describe the construction and working of a liquid flat plate collector. Explain how a thermosyphon solar water heater works, with a sketch.
Answer
A flat plate collector (FPC) absorbs both beam and diffuse radiation on a flat surface and transfers the heat to a fluid, without concentrating. It is used for temperatures up to about 80–100 °C.
Construction
- Absorber plate: copper or aluminium sheet with tubes (risers) attached to it, coated with a black or selective coating (high absorptance, low emittance).
- Transparent cover: one or two glass sheets; they pass short-wave solar radiation but block long-wave re-radiation from the plate (greenhouse effect) and reduce convection loss.
- Insulation: glass wool or polyurethane (50–75 mm) at back and sides.
- Casing (box): aluminium or galvanised steel, weatherproof.
- Headers: connect the riser tubes to inlet and outlet pipes.
sun rays \ \ \
===== glass cover ======
air gap
~~~ absorber plate + tubes ~~~
///// insulation //////////////
|________ casing ____________|
Working
Radiation passes the glass and heats the plate. Water in the tubes takes this heat and leaves hotter. The useful heat is given by the Hottel-Whillier-Bliss equation , and the efficiency is .
Thermosyphon water heater
______________
| storage tank |<--- hot water out
|______________|
^ |
hot | | cold
| ____ v
| |FPC |
|_|____|<---- cold feed
The storage tank is placed above the collector. Water heated in the collector becomes less dense and rises to the top of the tank, while the colder, denser water from the bottom of the tank flows down to the collector inlet. This natural circulation (thermosyphon) needs no pump and runs only while the sun shines. The tank must be at least 0.3–0.5 m above the collector top, and pipes must be short and sloped without air traps.
- Practice · 6 marks
A flat plate collector of area 2 m² has and W/m²K. The solar radiation incident on the collector plane is 800 W/m², the ambient temperature is 20 °C and water enters at 30 °C with a flow rate of 0.03 kg/s ( J/kgK). Calculate (a) the useful heat gain, (b) the collector efficiency, (c) the outlet water temperature, and (d) the maximum value of at which the collector still gives useful heat at this radiation.
Answer
Hottel-Whillier-Bliss equation
(a) Useful heat
(b) Efficiency
(c) Outlet temperature
(d) Stagnation condition
Useful heat becomes zero when :
Answer: W; ; °C; useful heat is zero when K (the collector is then at stagnation).
- Practice · 4+3 marks
(a) Why are concentrating collectors used? Classify solar concentrators and compare them with flat plate collectors. (b) A parabolic trough has an aperture width of 3.0 m and a tubular receiver of outer diameter 5 cm. Find the geometric concentration ratio, taking the receiver as a cylinder, and comment on it.
Answer
(a) Why and types
A concentrator uses mirrors or lenses to focus beam radiation from a large aperture onto a small absorber (receiver). The small receiver area gives low heat loss, so high temperatures (100–1500 °C) and good efficiency at high temperature are obtained, and less expensive absorber material is needed per unit of heat. They are used for process heat, industrial steam and solar thermal power plants.
Classification
- By focus: line focus (parabolic trough, linear Fresnel) and point focus (parabolic dish, central receiver/heliostat field).
- By optics: reflecting (mirrors) or refracting (Fresnel lens).
- By tracking: single axis (trough) or two axis (dish, tower).
| Point | Flat plate | Concentrator |
|---|---|---|
| Radiation used | Beam + diffuse | Beam only |
| Tracking | Not needed | Needed |
| Concentration ratio | 1 | 10 to above 1000 |
| Temperature | up to 100 °C | 100 to 1500 °C |
| Cost and maintenance | Low | High |
| Working in cloud | Works | Poor |
(b) Concentration ratio
If the aperture area is taken as the net opening, . So the receiver is irradiated with about 19 times the intensity falling on the aperture. Such a value is typical of a trough working at around 200–300 °C; the ideal limit for a line-focus 2-D concentrator is (about 213 for a half acceptance angle of 0.27°).
Answer: .
Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.
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