(i) Outline, with reference to the energy conversions in the machine, the main features of a conventional horizontal-axis wind generator.

(ii) The mean wind speed on the island is 8.0 ms^–1. Show that the maximum power available from a wind generator of blade length 45 m is approximately 2 MW.
                                                                              Density of air = 1.2 kg m^-3

(iii) The efficiency of the generator is 24%. Deduce the number of these generators that would be required to provide the islanders with enough power to meet their energy requirements.

The graph below shows how the wind speed varies with height above the land and above the sea.

(i) Suggest why, for any given height, the mean wind speed above the sea is greater than the mean wind speed above the land.

(ii) There is a choice of mounting the wind generators either 60m above the land or 60m above the sea.

Calculate the ratio

\[\frac{{{\rm{power available from a land - based generator}}}}{{{\rm{power available from a sea - based generator}}}}\]

at a height of 60m.

Distinguish between photovoltaic cells and solar heating panels.

The diagram shows 12 photovoltaic cells connected in series and in parallel to form a module to provide electrical power.

Each cell in the module has an emf of 0.75V and an internal resistance of 1.8Ω.

(i) Calculate the emf of the module.

(ii) Determine the internal resistance of the module.

(iii) The diagram below shows the module connected to a load resistor of resistance 2.2Ω.

Calculate the power dissipated in the load resistor.

(iv) Discuss the benefits of having cells combined in series and parallel within the module.

The intensity of the Sun’s radiation at the position of the Earth’s orbit (the solar constant) is approximately 1.4×10³Wm^–2.

(i) Explain why the average solar power per square metre arriving at the Earth is 3.5×10² W.

(ii) State why the solar constant is an approximate value.

(iii) Photovoltaic cells are approximately 20% efficient. Estimate the minimum area needed to supply an average power of 850kW over a 24 hour period.

(i) mention of blades/propeller and turbine/generator/dynamo;    

kinetic energy of wind → kinetic energy of turbine;    

(rotational) kinetic energy → electricity/electrical energy;

 Award [1 max] for statement of (unqualified) kinetic energy to electrical energy

(ii) A(=πr²)=6.4×10³(m²);
(P=)1.95 MW;

(iii) 0.24×1.95MW (=0.47 MW/0.48 MW);  
(0.47 MW = 470 kW thus) two generators would meet the maximum demand;
Allow only two generators for the second mark.  Do not accept fractional generators.

(i) sea is smoother (does not interrupt wind flow) / no obstacles on sea /  less friction / less turbulence (vice versa for land) / OWTTE; Allow named obstacles, eg trees/buildings/hills, etc.

(ii) \(\frac{{{v_{land}}}}{{{v_{sea}}}} = \frac{{10}}{{12.4}}\);
\(\frac{{{P_{land}}}}{{{P_{sea}}}} = {\left[ {\frac{{10}}{{12.4}}} \right]^3} = 0.52\);
Award [1 max] for 1.9 due to inverted ratio.

photovoltaic cells generate emf/electricity;  
solar panels generate thermal energy/heat / OWTTE; 

(i) emf=3.0 (V); 

(ii) series combination of resistance=7.2(Ω);
use of parallel resistance formula;  
2.4(Ω);
Award [3] for a bald correct answer

(iii) attempted use of IV, I²R or \(\frac{{{V^{\rm{2}}}}}{R}\);
0.94 (W);
Allow ECF from (d)(i) and (d)(ii).
Must see values substituted to gain first mark as compensation.

(iv) (series) increases the total emf/voltage;
(parallel) increases the current/decreases internal resistance/ensures some power if single cell fails / OWTTE;

(i) the solar radiation is captured by a disc of area πR² where R is the radius of the Earth;
but is distributed (when averaged) over the entire Earth’s surface which has an area four times as large;

or

rays make an angle θ with area of Earth’s half-sphere and so average intensity is proportional to average of cos² θ i.e. \(\frac{1}{2}\);
there is an additional factor of \(\frac{1}{2}\) due to the other half of the sphere;

(ii) variation of solar emission / Earth’s orbit is elliptical/not quite circular;

(iii) input power needed =(5×850(kW)=) 4.25×10⁶ (W);
\(\frac{{4.25 \times {{10}^6}\left( {\rm{W}} \right)}}{{3.5 \times 10^2\left( {{\rm{W}}{{\rm{m}}^{ - 2}}} \right)}} = 1.2 \times {10^4}\left( {{{\rm{m}}^2}} \right)\);
Award [2] for a bald correct answer.

(i) Many did not mention the kinetic energy of the wind (often referring to ‘wind energy’). All types of kinetic energy were referred to as ‘mechanical’ energy by many candidates. The general structure of this type of wind generator was generally well-known.

(ii) This part was generally well answered with those candidates completing the area calculation usually going on to gain both marks.

(iii) Again, this was well answered with nearly all candidates recognising that is not possible to have fractional generators and, therefore, rounding up their answers to 2 from the 1.7 calculation.

(i) Most candidates were able to suggest why the winds above the sea are higher than those above the land for the same height. A minority incorrectly answered this in terms of convection currents and sea versus land temperatures.

(ii) Nearly all candidates were able to correctly read the two values from the graph and a slight majority of these went on to correctly cube the ratio.

Most candidates knew the difference between photovoltaic cells and solar heating panels. A minority believed that both would normally produce electricity.

(i) This was not well known and many candidates simply added the emfs to give a value of 9.0 V rather than the correct 3.0 V.

(ii) Nearly all candidates correctly calculated the resistance of the series portions of the modules but there were frequent errors in combining these to find the total resistance – with the parallel formula often being incorrectly written in shorthand

(iii) Although many candidates recognised how they should use the power formula, very few were able to used the correct resistance and the correct voltage.

(iv) Many candidates knew that a failing cell would still allow current in other parallel branches, but few explained that the series combination increased the emf and the parallel combination increased the current in a module.

(i) A significant minority of candidates insisted that the reduction in the Sun’s intensity was due to radiation reflected from atmosphere. Few went on to do the calculation to support their answer but there were a small number of very good answers to this part.

(ii) Here again, many mentioned radiation reflected by atmosphere rather than variations in solar emissions or the non-circularity of Earth’s orbit.

(iii) This part was generally poorly done. The ‘24-hour period’ confused many candidates and few were able to follow the argument through to a logical conclusion.