MECH433-Energy and the Environment IW 2012Resit Answers(1)

PAPER CODE MECH433 of 11

PAPER CODE EXAMINER : TEL. NO

MECH433 DEPARTMENT :

RESIT 2012 EXAMINATIONS

Master of Science (Engineering): Year 1

Master of Engineering: Year 4

ENERGY AND THE ENVIRONMENT

TIME ALLOWED: Three Hours

INSTRUCTIONS TO CANDIDATES :

Candidates should answer ONE QUESTION from SECTION A and FOUR QUESTIONS from SECTION B. All answers will be marked but only the best FIVE counted. ____________________________________________________________________


SECTION A – answer one question only Question 1 Comment on the changes required in UK energy usage and the solutions presented by MacKay in " Energy Sustainability – without the hot air” for future energy scenarios.

[20 marks] Question 2 Considering the main sections of the 2009 Parsons Brinkerhoff report, ‘Powering the Future – Mapping our (UK) low-carbon path to 2050’, outline the actions required, and the difficulty in achieving these, in order that the 2050 carbon reduction targets may be reached.

[20 marks]

CONTINUED


SECTION B – answer four questions from five

Question 3 Engineers caused climate change and only engineers can solve it’. Discuss.

[20 marks]

Question 4 (a) A wind turbine has blades of 10 m in length and power coefficient of 0.3. Two sites

have been identified and which one will be selected for building the wind turbine will

depend on the profit to be made. On site A, the wind turbine can operate for 200 days

at wind speed of 8 m/s and for the rest of the year at wind speed of 5 m/s. While on

site B, the wind turbine can operate for 150 days at wind speed of 9 m/s and the rest

of the year at wind speed of 4 m/s. If the profit of 1kWh generated is 3 pence,

determine how much profit in pound can be made at each of the two sites in a year?

Note: The mass density of air is 1.2 kg/m3 and a year is taken as having 365 days

[12 marks]

(b) Explain briefly why the performance of the rotor (blades) needs to match that of the

generator and how.

[4 marks]

(c) Write down at least four main reasons why wind energy is expensive.

[4 marks]

CONTINUED


Question 5 A circular tidal lagoon with diameter 4km is built with vertical sides in a sea where the tides are modelled by the 2 major components M2 and S2. These constituents have amplitudes of 3m and 1m respectively. (a) Write down the range for the spring and neap tides.

[4 marks]

(b) Calculate the potential energy yield from the spring and neap tides. [ρ=1028 kg/m3; g = 9.81 m/s2]

[6 marks]

(c) Assuming an overall generation efficiency of 25% calculate an estimate for the annual energy yield, assuming there are 705 tides per year.

[4 marks]

(d) A tidal straight has a flow speed of U0cos(ωt), where U0 = 3.0 m/s and ω = 2 12.42 rads/hour. The tidal straight has the cross section shown in Figure Q5(d) Assuming a Significant Impact Factor of 20% what is the annual energy yield from this straight using the flux method?

noting

, and there are 705 tides per year.

[6 marks]

200 m 1000 m

50 m

200 m

CONTINUED

Figure Q5(f)


Question 6 Figure Q6 shows the current – voltage (I-V) curves of a silicon solar cell of area 100 cm2 in the dark and under illumination (light): (a) Describe the three characteristic properties of the cell indicated at (1), (2)

and (3). [3 marks]

(b) Using the data in the figure estimate the fill factor (FF) of the solar cell.

[3 marks] (c) If the input irradiance is 100 mW cm-2, what is the cells conversion efficiency?

[3 marks] (d) Describe the recombination mechanisms that limit the conversion efficiency.

[5 marks] (e) Sketch a diagram of a dye sensitised solar cell (DSSC) and briefly describe its

principles of operation. [6 marks]

Question 7

a) Explain the term ‘U value’ and describe how it is calculated. Relate your answer to the

flow of heat across a cavity wall construction. Suggest what typical U values might be

for good quality double and triple glazing and well-insulated walls, floors and roofs.

[10 marks]

b) Discuss the concept of ‘thermal mass’ and explain the various ways it can be used modify the environmental performance of a building.

[10 marks] END

Figure Q6


Answers Question 1 Write a review, including some of:- Present UK per capita energy consumption ~ 125kWh/day + ~60kWh/day for imported goods Breakdown: Transport ~ 35%; Heating ~34%; electricity (lighting/applianes/process) ~ 30% Ultimate contributions from potential renewables ~ 170kWh/day but only with ‘country sized’ developments taking up huge and impracticable swathes of the available land. He suggests <20kWh/day would be considered tolerable in today’s society. Forward strategy is to reduce consumption by more efficient transport (electric vehicles) and heating (insulation and heat pumps, with some CHP) and restrained use of electrics (standby wastage etc) This brings consumption down to 68kWh/day (excluding imported embedded energy) with electricity at 48kWh/day. The various plans range from requiring ~ 40% from ‘sustainable’ clean coal and nuclear to over 20% from imported solar energy and the renewables typically (for biofuels/wood especially) threaten agricultural food production. It is argued that the costs in the ballpark of £300billion should e affordable against other national priorities. The final solutions have been based on not markedly reducing living standards and as a result do not demonstrably achieve contraction and convergence CO2 reduction targets.

[20 marks] Question 2 Actions required include: Transport: reduced road travel (-15%); biofuels (15%); road transport to battery power (~80%), some cars to hydrogen Domestic: improvement in insulation (80-100% of homes); domestic/community biomass (replacing all oil and coal burning homes); domestic CHP (40%); heat pumps (10%); solar water heating (60%); appliance efficiency (-44%); solar pv (30%); wind (10%). Industry: transport fuel (-74%); insulation (-50%); electrical efficiency (~30%); efficiency improvements (-50%); convert oil & coal to gas/biomass and CHP (70% of processes); renewables wind & solar (20/10% of floorspace); industrial CCS (80% of emitters) Commercial: insulation (heat reduction to 25%); cooling % vent (-30%); lighting efficiency (-80%); hot water (-30%); convert oil/coal/electric heating to gas or biomass, CHP or heat pumps (100%); solar hot water (-20%); wind and solar pv (30/10% floorspace) Electricity: coal+CCS (12GW); nuclear (20GW); CCGT(28GW); wind (15GW); tidal(8GW) Scale of challenge is indicated by the penetration figures given above and the investment requirements that these imply.


[20 marks] Question 3 Answer could include: Human use of energy has increased by 2.9% (mean annual) since before the Industrial Revolution; The increase is energy demand is unsustainable; Conventional energy sources (coal, oil, gas, uranium) are incapable of meeting demand (current or future levels) sustainably; The theoretical capacity of renewables (wind, wave, tidal, solar, biomass) may meet current demand, but has external consequences; Energy supply has geopolitical consequences (specifically oil from Middle East presently and solar from North Africa potentially); Developing economies have aspirations for Western-style ‘standard of living’. Should we make room for them, or will they take what is ‘ours’?; Are resources, including energy availability, sufficient for the global population to attain Western-style ‘standards of living’?; The potential for international conflict (A top answer might include reference to internal unrest and suppression); The difference between ‘standard of living’ and ‘quality of life’ (citing Max-Neef); Political agreement, including of target figure for atmospheric CO2 concentrations, citing Montreal Protocol, Kyoto Protocol, Contraction and Convergence; Time available for mitigation; Methods and purpose of adaptation to climate change impacts; Media portrayal of science, public perceptions, activities of contrarians (‘denialists’); Political understanding and will; Non-linear change.

[20 marks]

Question 4

(a) The power extracted is

and the energy extracted is

.

[2 marks]

The total energy on site A

This means profit of about £5009 a year. [5 mark]

This means profit of about £4177 a year. [5 mark] (b) Blades rotate at low speeds because high rotational speeds generate high

centrifugal force and big vibration. However, a generator works best at high

rotational speeds. This mismatch is resolved by gearing up the rotor speeds to

generator speeds. [4 marks]

(c) The main reasons are (any four of the below):

(1) The intermittency of wind, (2) the fluctuation of wind speed and direction, (3) the

low power coefficient of wind turbines, (4) the time and cost of initial consultation

and planning, (5) the maintenance, (6) the storage of electricity, (7) the

connection to the grid. [4 marks]


Question 5 (a) ASpring = M2 + S2 = 3 + 1 = 4m (2) ANeap = M2 - S2 = 3 - 1 = 2m (2) [4 marks] (b) Area of the Lagoon S S = πr2 = π(4/2)2=4π = 12.57 km2 S = 12.57 x 106 m2 (2) Potential Energy Spring Tide ES = 4xρxgxSxASpring

2 ES = 4x1028x9.81x12.57x106x42 ES = 8.1129 x 1012 Ws ES = 2.25 GWh (2) Potential Energy Neap Tide EN = 4xρxgxSxANeap

2 EN = 4x1028x9.81x12.57x106x22

EN = 2.0282 x 1012 Ws EN = 0.5634 GWh (2) [6 marks] (c) Average Tide energy = 0.5*(ES+EN) = 1.4067 GWh Annual Generation Energy = 705*1.4067*0.25 = 247.93 GWh [4 marks]

(d) ) E =

(2)

A = 1000x50+2x0.5x200x50 = 1200x50 = 60000 m2 (2) E = 0.5x1028x60000x2.43x0.2x705x8/(3ω) = 618.9 GWh (2) [6 marks] Question 6 (a) (1) Jsc is the short-circuit current density (estimate 32mA cm-2); (2) Pmax is the maximum power (estimate 30mA cm-2 x 100 cm-2 x 0.42V = 1.26W); and (3) is Voc the open-circuit voltage (estimate 0.55V). [3 marks] (b) The solar cell fill factor (FF) is given by


FF = 30mA cm-2 x 100 cm-2 x 0.42V ~ 0.72 0.55V x 32mA cm-2 x 100 cm-2

[3 marks] (c) The cell conversion efficiency for an input irradiance of 100 mW cm-2 is

η = 0.72 x 0.55V x 32mA cm-2 ~ 13%

100 mW cm-2 [3 marks]

(d) Recombination mechanisms that limit the conversion efficiency. The recombination of

charge carriers (electrons and holes) created by the light / photon absorption are excited to higher energies. Instead of being collected by the junction and the external electrical contacts, these charge carriers can come back together, resulting in light (luminescence), or heat (non-radiative recombination). Photoluminescent emission and non-radiative recombination compete with current extraction and power production. Light is absorbed by the material to produce electrons in the Conduction Band, which can recombine in three distinct ways: radiatively, giving up the excitation energy in the form of an emitted luminescent photon; non-radiatively through traps, recombination centres (RC); or through excitation of CB electrons to higher levels producing a phonon or lattice vibration.

[5 marks] (e) Sketch a diagram of a dye sensitised solar cell (DSSC) and briefly describe its

principles of operation.


Dye sensitised solar (or Grätzel) cells consist of a porous layer of titanium dioxide nanoparticles, covered with a molecular dye that absorbs sunlight. The titanium dioxide is immersed under an electrolyte solution, above which is a platinum-based catalyst. Sunlight passes through the transparent electrode into the dye layer where it can excite electrons that then flow into the titanium dioxide. The electrons flow toward the transparent electrode where they are collected for powering a load. After flowing through the external circuit, they are re-introduced into the cell on a metal electrode on the back, flowing into the electrolyte. The electrolyte then transports the electrons back to the dye molecules. Dye-sensitized solar cells separate the two functions provided by silicon in a traditional cell design. Normally the silicon acts as both the source of photoelectrons, as well as providing the electric field to separate the charges and create a current. In the dye-sensitized solar cell, the bulk of the semiconductor is used solely for charge transport, the photoelectrons are provided from a separate photosensitive dye. Charge separation occurs at the surfaces between the dye, semiconductor and electrolyte.

[6 marks]

Question 7 (a) Thermal transmittance indicates the rate at which heat flows through a building element. It depends on the thermal properties of the building materials and the location/exposure of the building. The measure of overall thermal transmittance is called the U-value (unit W/m2K). A high U-value indicates high heat loss and poor energy performance whilst a low U-values indicate low heat loss and good energy performance. Heat flow across building elements involves conduction, radiation and convection. The heat flow meets a thermal resistance at each stage of its flow. The various resistances to heat flow include resistances due to the surface layers of the building component (Rso, Rsi) and resistances from any air cavities (Ra). The resistance of each solid building layer (R1, R2 etc.) is found by dividing the thickness (m) by the thermal conductivity whilst surface and air cavity resistances are found in standard tables

Heat flow through ground floors has two components - an edge loss and a ground loss. Edge losses depend upon inside/outside temperature differences and ground losses are governed by inside/earth temperature differences. The total thermal resistance of the building component is found by adding the individual resistances together

(R) = Rso + R1 + R2 + Ra + Rsi The U value is calculated by taking the reciprocal of the total thermal resistance

U = 1/(R)


For complex combinations of materials in a component each heat flow path needs to be taken into account because some flow paths are weaker and provide less resistance (i.e. mortar in a brick wall). Typical good U values for well-insulated, energy efficient building components would be: Roof: 0.2 W/m2K External wall: 0.3 W/m2K Ground floor: 0.35 W/m2K Double glazing: 2.0 W/m2K Triple glazing 0.8 W/m2K

[10 marks]

(b) Thermal mass refers to a material’s ability to store thermal energy for extended periods of time (i.e. several hours). Thermal mass can be used effectively to absorb daytime heat gains (reducing cooling load) and release the heat during the night (reducing heating load). The use of thermal mass in shelter dates back to the dawn of humans, and until recently was the prevailing strategy for building climate control in hot regions. Traditional types of thermal mass include water, rock, earth, brick, concrete, fibrous cement, and ceramic tile. Modern phase change materials (PCM) store energy while maintaining constant temperatures, using chemical bonds to store and release latent heat The basic properties that indicate the thermal behaviour of materials are: density (p),

specific heat (cm) and conductivity (). The specific heat for most masonry materials is

similar (about 0.2-0.25Wh/kgC). Thus, the total heat storage capacity is a function of the total mass of masonry materials, regardless of its type (concrete, brick, stone, and earth) Material Density (kg/m3) Concrete 600-2200 Stone 1900-2500 Bricks 1500-1900 Earth 1000-1500 (uncompressed) Earth 1700-2200 (compressed) Thermal mass can be used to ‘damp’ oscillations in the internal air temperature or limit peak temperatures when used in conjunction with ‘night’ flushing (ventilation). However, thermal mass can be problematic in buildings that are intermittently occupied and require a fast thermal response. There is also a risk of under-heating on colder mornings and condensation and mould growth. The use of exposed thermal mass is typically employed in buildings (or spaces) likely to experience overheating such as sunspaces, areas of high occupancy period and areas with high equipment loads. Thermal mass is used extensively in passive solar design, which is where buildings try to use the heating benefits of the sun at appropriate times of the day and the year. The main mechanisms for utilising mass in passive solar design are direct gain, thermal storage walls and sunspaces.

[10 marks]

MECH433-Energy and the Environment IW 2012Resit Answers(1)

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