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22-Mec-B3 Energy Conversion and Power Generation · May 2016

Question 6 of 6: Solar Energy — central receiver system, photovoltaics and land area

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, May 2016 — 07-Mec-B3 Energy Conversion and Power Generation. Three hours, closed book. The paper has two sections: Section A (calculative) carries Questions 1 to 4 and Section B (descriptive) carries Questions 5 and 6. A candidate answers three questions from Section A and one from Section B, so four questions constitute a complete paper of 60 marks and every question is worth 15 marks. Reference data for particular questions are bound in as pages 9 to 12 (the Van der Kloof waterway cross-section, the Mollier enthalpy-entropy diagram, the Belledune Generating Station heat balance diagram and the coal fired boiler outline), reference formulae and constants as pages 13 to 16, and steam tables from Granet and Bluestein are supplied. All six questions are solved here, because the set is a study resource rather than a three-hour sitting.

Reference texts.

Question 6: Solar Energy — central receiver system, photovoltaics and land area (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — the central receiver system with heliostats. Of the two options offered, the central receiver is described here. The plant has three distinct parts: a field of heliostats, a receiver on a tower, and an entirely conventional steam power block.

Q6(a) Central receiver system with heliostats - collection and power blockSUNabout 1 kW/m2 at noonRECEIVERTOWERheliostat field, westheliostat field, eastSTEAM GENERATORTURBINEGnet electrical outputCONDENSERheat rejected to the condenser cooling circuitmolten salt at about 565 Csalt returned at about 290 Csuperheated steamEach heliostat tracks the sun on two axes and reflects the beam onto the receiver, so a field of several thousand mirrorsconcentrates several hundred suns onto one aperture.The receiver heats molten nitrate salt, which is stored and then boils water in a conventional steam generator - the power block isan ordinary Rankine plant.Heat is received at the receiver, rejected at the condenser, and power is produced at the generator.Overall solar-to-electricity efficiency at design insolation is about 20 %: roughly 55 to 60 % optical times 90 % receiver times 40 %cycle.
Question 6(a) — central receiver system. Heliostats concentrate the beam onto a tower-mounted receiver; molten salt carries the heat to a steam generator, and the power block is an ordinary Rankine plant rejecting heat at the condenser.

Each heliostat is a large, nearly flat mirror — typically 100 to 150 square metres — on a two-axis drive that tracks the sun so that the reflected beam stays fixed on the receiver aperture as the sun moves. Because each mirror must bisect the angle between the sun and the tower, and that angle differs for every mirror in the field, the tracking is computed individually for each unit. A field of several thousand heliostats surrounding a tower 100 to 200 metres high delivers a concentration ratio of several hundred to one at the aperture, and flux densities on the receiver surface of the order of half a megawatt per square metre.

The receiver is an array of tubed panels, either an external cylinder or a cavity behind an aperture, through which molten nitrate salt is pumped. Cold salt is drawn from a tank at about 290 degrees Celsius, heated to about 565 degrees Celsius in a single pass, and returned to a hot tank. That two-tank arrangement is the reason the central receiver is preferred to the trough for large plant: the salt is both the heat transfer fluid and the storage medium, so several hours of full-output generation can be stored cheaply and the plant can follow the evening load peak after sunset. Hot salt is then pumped through a conventional steam generator train — preheater, evaporator and superheater — raising superheated steam at around 540 degrees Celsius and 12 MPa for a reheat Rankine cycle with a surface condenser and either wet or dry cooling.

Following the energy through the flow diagram: heat is received at the tower receiver, power is produced at the turbine-generator in the power block, and heat is rejected at the condenser to the cooling circuit. The overall solar-to-electric efficiency at design insolation is about 20 per cent: roughly 55 to 60 per cent optical efficiency in the field (cosine losses, shading and blocking, mirror reflectance, spillage at the aperture), about 90 per cent receiver thermal efficiency after radiation and convection losses from a very hot surface, and about 40 per cent for the steam cycle. The Carnot bound on a 565 degree receiver against a 40 degree sink is 62.7 per cent, so the cycle term is where the largest irreversible loss sits, exactly as it would in a fossil plant.

Part (b) — the photovoltaic generating system. A photovoltaic plant has no thermodynamic cycle at all: the conversion is direct, from photon to electron, and the system is a ladder of series and parallel connections plus power electronics.

Q6(b) Photovoltaic generating system - cell to gridSUNcell0.5 to 0.6 Vmodule60 cells, about 36 Vstring of modules in series20 modules, about 700 V dcCOMBINERINVERTERdc to ac at 98 % efficiencyTRANSFORMERlow voltage acto the transmission systemarray of tracking or fixed-tilt tablesA silicon cell delivers only about 0.5 V, so cells are wired in series into modules, modules into strings and strings into arrays.Series connection raises voltage, parallel connection raises current; the inverter needs several hundred volts dc to synthesisegrid-quality ac.Absorbed photons create electron-hole pairs; the built-in field of the p-n junction separates them and drives a current through theexternal circuit.Overall system efficiency at design insolation is about 15 %: module about 17 % less inverter, wiring, mismatch and temperaturelosses.
Question 6(b) — photovoltaic system. Cells are built up into modules, strings and arrays to reach a usable dc voltage, which the inverter converts to grid ac.

The conversion principle is the photovoltaic effect in a semiconductor p-n junction. A photon whose energy exceeds the band gap of the material — about 1.1 electron volts for crystalline silicon — is absorbed and lifts an electron from the valence band to the conduction band, leaving a hole behind. The built-in electric field of the junction sweeps the electron towards the n-side and the hole towards the p-side before they can recombine, so a potential difference appears across the terminals and a current flows if an external circuit is connected. The open-circuit voltage of a silicon cell is set by the band gap and the junction physics, not by its size: it is only about 0.6 volts, while the current is proportional to the illuminated area. Photons below the band gap pass straight through and contribute nothing, and the excess energy of photons well above it is lost as heat; those two effects together are the fundamental reason a single-junction cell cannot exceed about 33 per cent.

Grid voltages are reached by connection, not by making bigger cells. Sixty to seventy-two cells are wired in series and laminated into a module of about 36 to 45 volts open circuit; fifteen to twenty-five modules are wired in series into a string at 600 to 1500 volts direct current, which is the highest voltage the insulation and the safety standards allow; strings are then paralleled in combiner boxes to build up current. The inverter converts that direct current to alternating current synchronised with the grid, holding the array at its maximum power point as irradiance and cell temperature change, and a step-up transformer takes the output to distribution or transmission voltage. Bypass diodes across each cell group protect a shaded cell from being reverse-biased by the rest of its string.

The overall efficiency of the whole system at design insolation is about 15 per cent: a module efficiency of about 17 per cent for good crystalline silicon, less inverter losses of about 2 per cent, wiring and mismatch losses of a few per cent, soiling, and the temperature coefficient of roughly minus 0.4 per cent per kelvin, which matters because cells in the field run 20 to 30 kelvin above ambient.

Part (c) — comparison, peak insolation and land area. On efficiency the central receiver is ahead, about 20 per cent against about 15 per cent, and it has the decisive advantage of cheap thermal storage: molten salt lets it dispatch power into the evening peak and gives it a capacity factor near 45 per cent against 25 to 30 per cent for a fixed-tilt photovoltaic plant. It also delivers synchronous inertia to the grid through a real rotating machine. Against that, it needs direct beam radiation only, so it is confined to arid high-insolation sites and is nearly useless under diffuse cloud; it needs cooling water or an efficiency penalty for dry cooling; it involves high-temperature salt handling, freeze protection and a full steam plant staff; and it has no economy at small scale. Photovoltaics use diffuse as well as beam radiation, have no moving parts, no water consumption, negligible operating staff, and scale linearly from a rooftop to a hundred megawatts, which is why they dominate deployment in practice, including everywhere in Canada. In a Canadian context the choice is not close: with annual insolation well below the threshold for concentrating solar and a great deal of diffuse radiation, photovoltaics are the only practical solar technology here, and central receiver plant belongs to the sun belts.

The maximum solar radiation intensity at the earth's surface, at sea level with the sun overhead and a clear sky, is about 1000 W/m2 — the standard 1 kW/m2 used as the rating condition for both technologies, reduced from the 1367 W/m2 solar constant by atmospheric absorption and scattering. Sizing a 1000 MW plant at that intensity:

  1. Step 1 — collector area required. Dividing the electrical output by the intensity and the overall efficiency,$$A_{CRS}=\dfrac{P_e}{I\,\eta}=\dfrac{10^{6}\ \text{kW}}{1\ \text{kW/m}^{2}\times 0.20}=5.0\times 10^{6}\ \text{m}^{2}=5.0\ \text{km}^{2}$$$$A_{PV}=\dfrac{10^{6}}{1\times 0.15}=6.7\times 10^{6}\ \text{m}^{2}=6.7\ \text{km}^{2}$$
  2. Step 2 — land area, allowing for spacing. Collectors cannot be laid edge to edge. Heliostats must be spaced so that they do not shade or block one another as the sun moves, and photovoltaic rows must be spaced so that they do not shade the row behind at low winter sun angles; the usable ground cover ratio is about 0.25 for a heliostat field and about 0.40 for fixed-tilt photovoltaic tables:$$L_{CRS}=\dfrac{5.0}{0.25}=\boxed{20\ \text{km}^{2}}\qquad L_{PV}=\dfrac{6.7}{0.40}=\boxed{16.7\ \text{km}^{2}}$$so either plant occupies roughly a square 4 to 4.5 km on a side — about 20 square kilometres for a 1000 MW station. Note that this is the peak rating; because neither plant runs at peak for more than a few hours a day, matching the annual energy of a 1000 MW thermal station would need three to four times that area.
Question 6 — comparison at 1000 MW electrical
QuantityCentral receiverPhotovoltaic
Overall solar-to-electric efficiencyabout 20 %about 15 %
Peak insolation assumed1.0 kW/m21.0 kW/m2
Collector or module area5.0 km26.7 km2
Ground cover ratio0.250.40
Land area required20 km2 (4.5 km square)16.7 km2 (4.1 km square)
Storageseveral hours in molten saltbattery only, at extra cost
Radiation useddirect beam onlybeam and diffuse
Typical capacity factorabout 45 %25 to 30 %
Water usecondenser coolingwashing only
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